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Question IndexDepartmentCourse NumberCourse NamePrerequisitesCorequisitesAssignmentTopicQuestion NumberPart NumberPercentage of Total GradeQuestion TypeQuestionSolution TypeSolutionPure Zero ShotCorrectNotes
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73Mathematics18.3Principles of Continuum Applied Mathematics18.02, 18.03NoneProblem Set 5Traveling Waves1a1.09375TextImagine that someone tells you that the following equation is a model for traffic flow:
$$
c_{t}+c c_{x}=\nu c_{x t},
$$
where $\boldsymbol{\nu}>\mathbf{0}$ is "small" and $c$ is the the wave velocity - related to the car density via $\boldsymbol{c}=\frac{d Q}{d \boldsymbol{\rho}}$. The objective of this problem is to ascertain if (1.1) makes sense as model for Traffic Flow. To this end, answer the two questions below.
Does the model have acceptable traveling wave "shock" solutions $\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \cdot c=\boldsymbol{F}(\boldsymbol{z})$,
where $\ldots \ldots \ldots \ldots \ldots \ldots \ldots z= \frac{x - Ut}{\nu}$ and $\boldsymbol{U}$ is a constant?
Here "acceptable" means the following
1a. The function $F$ has finite limits as $z \rightarrow \pm \infty$, i.e.: $\quad c_{L}=\lim _{z \rightarrow-\infty} F(z) \quad$ and $\quad c_{R}=\lim _{z \rightarrow+\infty} F(z)$. Further: the derivatives of $F$ vanish as $z \rightarrow \pm \infty$, and $c_{L} \neq c_{R}$.
This means that, as $\nu \rightarrow 0$, the solution $c$ becomes a discontinuity traveling at speed $U$, with $c=c_{L}$ for $x<U t$ and $c=c_{R}$ for $x>U$. That is, a shock wave.
1b. The solution satisfies the Rankine-Hugoniot jump conditions $\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots U = \frac{[Q]}{[\rho]}$
where $\rho_{L}$ and $\rho_{R}$ are related to $c_{L}$ and $c_{R}$ via $c_{L}=\frac{d Q}{d \rho}\left(\rho_{L}\right)$ and $c_{R}=\frac{d Q}{d \rho}\left(\rho_{R}\right)$.
Assume that $Q=Q(\rho)$ is a quadratic traffic flow function - see remark 1.1.
1c. The solution satisfies the Entropy condition $\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \boldsymbol{c}_{\boldsymbol{L}}>\boldsymbol{U}>\boldsymbol{c}_{\boldsymbol{R}}$.
To answer this question:
\begin{itemize}
\item Find all the solutions satisfying 1a. Get explicit formulas for $\boldsymbol{F}$ and $\boldsymbol{U}$ in terms of $c_{L}, c_{R}$, and $z$.
\item Check if the solutions that you found satisfy $\mathbf{1 b}$.
\item Check if the solutions that you found satisfy 1c.
\item Finally, given $\mathbf{A}-\mathbf{C}$ : Does, so far, the equation make sense as a model for traffic flow?
\end{itemize}
Hints.
\begin{itemize}
\item Find the ode $F$ satisfies. Show it can be reduced to the form $F^{\prime}=P(F)$, where $P=$ second order polynomial.
\item Write $P$ in terms of its two zeroes, $c_{1}$ and $c_{2}$, and express all the constants (e.g.: $U$ ) in terms of $c_{1}$ and $c_{2}$.
\item Solve now the equation, and relate $c_{1}$ and $c_{2}$ to $c_{L}$ and $c_{R}$. You are now ready to proceed with $\mathbf{A}-\mathbf{D}$.
\item Remember that, while the density $\rho$ has to be non-negative, wave speeds can have any sign.
\end{itemize}		OpenSubstituting $c=F(z)$, where $z=\frac{x-U t}{\nu}$, into the pde gives the ode
$$
(F-U) F^{\prime}=-U F^{\prime \prime},
$$
where the primes indicate differentiation with respect to $z$. This equation can be integrated once, to obtain
$$
U F^{\prime}=U F-\frac{1}{2} F^{2}+\kappa,
$$
where $\kappa$ is a constant of integration. The right hand side in this equation is a quadratic function of $F$, with maximum value at $F=U$, where it reaches the value $\frac{1}{2} U^{2}+\kappa$. Therefore, for $\kappa$ large enough, the right hand side in (1.5) has two real zeros, and the equation can be written in the form
$$
U F^{\prime}=-\frac{1}{2}\left(F-c_{1}\right)\left(F-c_{2}\right),
$$
where $c_{1} \geq c_{2}$ are constants,
$$
U=\frac{1}{2}\left(c_{1}+c_{2}\right), \quad \text { and } \quad \kappa=-\frac{1}{2} c_{1} c_{2} .
$$
As long as $U \neq 0$, the appropriate (and explicit) solutions to (1.6) follow from the hyperbolic tangent - since $y=\tanh (s)$ satisfies $y^{\prime}=1-y^{2}$. That is
$$
F=U+\frac{c_{1}-c_{2}}{2} \tanh \left(\frac{c_{1}-c_{2}}{4 U} z\right) .
$$
Assume that $c_{1}>c_{2}$, since the case $c_{1}=c_{2}$ is trivial. Then, with $\boldsymbol{c}_{\boldsymbol{L}}=\lim _{\boldsymbol{z} \rightarrow-\infty} \boldsymbol{F}(\boldsymbol{z})$ and $\boldsymbol{c}_{\boldsymbol{R}}=\lim _{\boldsymbol{z} \rightarrow+\infty} \boldsymbol{F}(\boldsymbol{z})$,
$$
\begin{array}{ll}
\text { If } U>0, & c_{L}=c_{2} \text { and } c_{R}=c_{1} . \text { Thus } c_{L}<U<c_{R} . \\
\text { If } U<0, & c_{L}=c_{1} \text { and } c_{R}=c_{2} . \text { Thus } c_{L}>U>c_{R} .
\end{array}
$$
Thus the model has traveling solutions satisfying 1 a for all values of $U$, except $U=0$ (note that equation (1.5) yields $F=$ constant for $U=0$ ). However:
1. Rankine-Hugoniot jump conditions: satisfied, since the wave velocity is given as the average of the characteristic velocities, which is correct for a quadratic flow function - see remark 1.1.
2. Entropy condition: violated when $\boldsymbol{U}>\mathbf{0}$, as shown by (1.9).
Thus, we must conclude that (1.1) is NOT a good model for traffic flow. But we have the derivation in remark 1.1, or so it seems. However, note:
It is, indeed reasonable to assume that the drivers respond to the rate of change of the density. But, why should they respond to $\rho_{t}$ alone, as in (1.2)? The rate of change of the density a driver sees is $r=\rho_{t}+\boldsymbol{u} \boldsymbol{\rho}_{\boldsymbol{x}}$, not $\rho_{t}$ ! The drivers should also respond to the local gradient of the density $\rho_{x}$, specially when it is large.
Since $\rho_{t} \approx-c \rho_{x}$ (at least away from shocks), it follows that $r=\rho_{t}+u \rho_{x} \approx(u-c) \rho_{x}$. But $u \geq c$, so $\rho_{x}$ and $r$ have the same sign. Thus a model with $q=Q(\rho)-\nu \rho_{x}$ will behave properly [this model is actually used], but one based in (1.2) will not. In fact $r \approx-\frac{u-c}{c} \rho_{t}$, so that (1.2) gives the wrong sign for the correction whenever positive wave speeds are involved!
Let us now consider solutions of the form $c=c_{0}+u$, where $c_{0}$ is a constant and $u$ is very small. Then
$$
u_{t}+c_{0} u_{x}=\nu u_{x t}.
$$
This has solutions of the form $u=e^{i k x+\lambda t}$, provided that
$$
\lambda=\frac{-i c_{0} k}{1-i \nu k}=\frac{\nu c_{0} k^{2}}{1+\nu^{2} k^{2}}-i \frac{c_{0} k}{1+\nu^{2} k^{2}} .
$$
If $c_{0}>0$, these solutions grow (exponentially), which means that any uniform traffic flow with $c_{0}>0$ is unstable at least according to this model. ${ }^{3}$ This is rather strange, since $c_{0}>0$ corresponds to light traffic. Another indication that this is not a good model for traffic flow.		To answer this question, we will follow the steps and hints provided.

1. Find the ODE F satisfies:

We start by substituting the traveling wave ansatz $c = F(z)$, where $z = \frac{x - Ut}{\nu}$, into the given equation:

$$
c_t + cc_x = \nu c_{xt}
$$

Taking derivatives with respect to $z$, we get:

$$
-U F'(z) + F(z) F'(z) = \nu F''(z)
$$

Rearranging the terms, we obtain the ODE:

$$
F''(z) - \frac{F(z) - U}{\nu} F'(z) = 0
$$

2. Reduce the ODE to the form $F' = P(F)$:

Let $P(F) = aF^2 + bF + c$. Then, the ODE becomes:

$$
F''(z) = \frac{F(z) - U}{\nu} (aF^2 + bF + c)
$$

Comparing the coefficients, we find that $a = -1$, $b = U$, and $c = 0$. Thus, the ODE is:

$$
F''(z) = \frac{F(z) - U}{\nu} (-F^2 + UF)
$$

3. Solve the ODE and relate $c_1$ and $c_2$ to $c_L$ and $c_R$:

We can rewrite the ODE as:

$$
F''(z) = \frac{F(z) - U}{\nu} F(F - U)
$$

Let $c_1$ and $c_2$ be the roots of the quadratic equation $F(F - U) = 0$. Then, we have:

$$
F''(z) = \frac{F(z) - U}{\nu} (F - c_1)(F - c_2)
$$

Now, we can integrate the ODE to find the explicit formula for $F(z)$:

$$
F'(z) = \frac{1}{\nu} \int (F - c_1)(F - c_2) dF
$$

Integrating and rearranging the terms, we obtain:

$$
F'(z) = \frac{1}{\nu} (F^2 - (c_1 + c_2)F + c_1 c_2)
$$

Now, we can relate $c_1$ and $c_2$ to $c_L$ and $c_R$ by considering the limits as $z \rightarrow \pm \infty$:

$$
c_L = \lim_{z \rightarrow -\infty} F(z) = c_1 \\
c_R = \lim_{z \rightarrow +\infty} F(z) = c_2
$$

4. Check if the solutions satisfy 1b and 1c:

For 1b, we need to check if the solution satisfies the Rankine-Hugoniot jump conditions:

$$
U = \frac{[Q]}{[\rho]}
$$

Using the relation between $c$ and $\rho$, we have:

$$
U = \frac{Q(\rho_R) - Q(\rho_L)}{\rho_R - \rho_L}
$$

Since $Q(\rho)$ is a quadratic function, we can write it as:

$$
Q(\rho) = a\rho^2 + b\rho + c
$$

Then, we can find the derivatives with respect to $\rho$:

$$
c_L = \frac{dQ}{d\rho}(\rho_L) = 2a\rho_L + b \\
c_R = \frac{dQ}{d\rho}(\rho_R) = 2a\rho_R + b
$$

Substituting these expressions into the Rankine-Hugoniot condition, we obtain:

$$
U = \frac{(2a\rho_R + b) - (2a\rho_L + b)}{\rho_R - \rho_L} = 2a(\rho_R - \rho_L)
$$

For 1c, we need to check if the solution satisfies the entropy condition:

$$
c_L > U > c_R
$$

Using the expressions for $c_L$, $c_R$, and $U$, we have:

$$
2a\rho_L + b > 2a(\rho_R - \rho_L) > 2a\rho_R + b
$$

This condition is satisfied if $a < 0$.

5. Does the equation make sense as a model for traffic flow?

So far, the equation seems to make sense as a model for traffic flow. The traveling wave solutions have acceptable shock wave properties, and the solutions satisfy the Rankine-Hugoniot jump conditions and the entropy condition. However, further analysis and comparison with real-world traffic data would be necessary to fully validate the model.		Incorrect
3
121EECS6.191Computation Structures6.100A, 8.02NoneMidterm Exam 1Binary Arithmetic1d0.9TextUsing 8-bit 2's complement encoding, compute $((\sim(0xAB \& 0x55)) / / 0x04) * 0x04$, where " $\sim$ " represents bitwise negation, "\&" represents a bitwise AND, "//" represents integer division (ignores remainder), and "*” represents multiplication.
For performing // and $*$, the only operations allowed are $>>_{a},>_{1},<<_{1}$ which correspond to arithmetic right shift, logical right shift and logical left shift respectively. Clearly specify which operation to use for performing // and *. Write the intermediate and final answers in 8-bit 2's complement.		Open$\sim(0xAB \& 0x55)$: $0b1111\_1110$.
Operation used for performing "//": $>>>_{a}$.
Operation used for performing "* ": ${<<}_{1}$
$((\sim(0xAB \& 0x55)) / / 0x04) * 0x04$: $0b1111\_1100$.		First, let's perform the bitwise AND operation between 0xAB and 0x55:

0xAB = 10101011
0x55 = 01010101
--------------
AND = 00000001

Now, let's perform the bitwise negation on the result:

~00000001 = 11111110

Next, we need to perform integer division by 0x04. Since we can only use right shift operations, we'll use arithmetic right shift (>>) for division:

11111110 >> 2 = 11111111 (since 0x04 = 2^2, we shift 2 positions to the right)

Now, we need to perform multiplication by 0x04. Since we can only use left shift operations, we'll use logical left shift (<<) for multiplication:

11111111 << 2 = 11111100 (since 0x04 = 2^2, we shift 2 positions to the left)

So, the final result in 8-bit 2's complement is:

11111100		Correct
4
31Mathematics18.01Calculus INoneNoneProblem Set 1Tangent Lines14b0.07919746568TextThe tangent line is closely related to linear approximation. This little problem should help clarify that. Let $f(x)=x^{2}$ and consider the tangent line to the graph at $x=1$. This line has the form $y=L(x)$, and you computed $L(x)$ in the last problem.
Compute $L(1.2)$ and $L(1.4)$. 		NumericalAt $x=1$, using the results of Problem 13, the tangent line is
$$
L(x)=2 x-1 .
$$
Using $L(x)$ to approximate $f(x)$ :
$$
f(1.2) \approx 2 \times 1.2-1=1.4 \text {, }
$$
and
$$
f(1.4) \approx 2 \times 1.4-1=1.8 \text {. }
$$
These results agree with the linear approximations in part (a), as they should because the tangent line is the linear approximation.		To compute the tangent line L(x) at x=1 for the function f(x) = x^2, we first need to find the slope of the tangent line. The slope of the tangent line is given by the derivative of the function at the point of tangency.

The derivative of f(x) = x^2 is f'(x) = 2x. So, the slope of the tangent line at x=1 is f'(1) = 2(1) = 2.

Now, we know that the tangent line passes through the point (1, f(1)) = (1, 1^2) = (1, 1). Using the point-slope form of a linear equation, we can write the equation of the tangent line L(x) as:

y - 1 = 2(x - 1)

Now, we can compute L(1.2) and L(1.4) by plugging in the x-values into the equation of the tangent line:

L(1.2) = 1 + 2(1.2 - 1) = 1 + 2(0.2) = 1 + 0.4 = 1.4

L(1.4) = 1 + 2(1.4 - 1) = 1 + 2(0.4) = 1 + 0.8 = 1.8

So, L(1.2) = 1.4 and L(1.4) = 1.8.		Correct
5
138Mathematics18.02Calculus II18.01NoneFinal ExamDouble Integrals101.8TextSet up the integral $\iint_R f(x, y) d A$ where $R$ is the region bounded by the four curves $x^2 y=4, x^2 y=9, \frac{y}{x}=1$, and $\frac{y}{x}=2$ as a double integral in the variables $u=x^2 y$ and $v=\frac{y}{x}$. (Your answer should be completely ready to integrate, once the function $f$ is given.)
Note: the inverse transformation is given by $\quad x=u^{\frac{1}{3}} v^{-\frac{1}{3}}, \quad y=u^{\frac{1}{3}} v^{\frac{2}{3}}$.		Expression$$
\int_R f(x, y) d A=i n t_1^2 \int_4^9 f\left(u^{1 / 3} v^{-1 / 3}, u^{1 / 3} v^{2 / 3}\right)\left(\frac{1}{3} u^{-1 / 3} v^{-2 / 3}\right) d u d v
$$		First, we need to find the bounds for the variables $u$ and $v$. From the given curves, we have:

1. $x^2 y = 4 \Rightarrow u = 4$
2. $x^2 y = 9 \Rightarrow u = 9$
3. $\frac{y}{x} = 1 \Rightarrow v = 1$
4. $\frac{y}{x} = 2 \Rightarrow v = 2$

Now, we need to find the Jacobian of the transformation. Let $x = u^{\frac{1}{3}} v^{-\frac{1}{3}}$ and $y = u^{\frac{1}{3}} v^{\frac{2}{3}}$. Then, we have:

$\frac{\partial(x, y)}{\partial(u, v)} = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} = \begin{bmatrix} \frac{1}{3} u^{-\frac{2}{3}} v^{-\frac{1}{3}} & -\frac{1}{3} u^{\frac{1}{3}} v^{-\frac{4}{3}} \\ \frac{1}{3} u^{-\frac{2}{3}} v^{\frac{1}{3}} & \frac{2}{3} u^{\frac{1}{3}} v^{-\frac{1}{3}} \end{bmatrix}$

The determinant of this matrix is:

$J(u, v) = \frac{1}{3} u^{-\frac{2}{3}} v^{-\frac{1}{3}} \cdot \frac{2}{3} u^{\frac{1}{3}} v^{-\frac{1}{3}} - (-\frac{1}{3} u^{\frac{1}{3}} v^{-\frac{4}{3}}) \cdot \frac{1}{3} u^{-\frac{2}{3}} v^{\frac{1}{3}} = \frac{2}{9} u^{-\frac{1}{3}} v^{-\frac{2}{3}} + \frac{1}{9} u^{-\frac{1}{3}} v^{-\frac{2}{3}} = \frac{1}{3} u^{-\frac{1}{3}} v^{-\frac{2}{3}}$

Now, we can set up the integral:

$\iint_R f(x, y) dA = \int_{4}^{9} \int_{1}^{2} f(u^{\frac{1}{3}} v^{-\frac{1}{3}}, u^{\frac{1}{3}} v^{\frac{2}{3}}) \cdot \frac{1}{3} u^{-\frac{1}{3}} v^{-\frac{2}{3}} dv du$		Correct
6
84EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneExercise 12Q-Learning1aii0.01157407407TextLet's simulate the Q-learning algorithm! Assume there are states $(0,1,2,3)$ and actions ('b', 'c'), and discount factor $\gamma=0.9$. Furthermore, assume that all the $\mathrm{Q}$ values are initialized to 0 (for all state-action pairs) and that the learning rate $\alpha=0.5$.
Experience is represented as a list of 4-element tuples: the $t$ th element of the experience corresponds to a record of experience at time $t:\left(s_{t}, a_{t}, s_{t+1}, r_{t}\right)$ (state, action, next state, reward).
After each step $t$, indicate what update $Q\left(s_{t}, a_{t}\right) \leftarrow q$ will be made by the Q learning algorithm based on $\left(s_{t}, a_{t}, s_{t+1}, r_{t}\right)$. You will want to keep track of the overall table $Q\left(s_{t}, a_{t}\right)$ as these updates take place, spanning the multiple parts of this question.
As a reminder, the Q-learning update formula is the following:
$$
Q(s, a) \leftarrow(1-\alpha) Q(s, a)+\alpha\left(r+\gamma \max _{a^{\prime}} Q\left(s^{\prime}, a^{\prime}\right)\right)
$$
You are welcome to do this problem by hand, by drawing a table specifying $Q(s, a)$ for all possible $s$ and $as$. Alternatively, you may write a program which takes in the following history of experience:
experience = [(0, 'b', 2, 0), #t = 0
(2, 'b', 3, 0),
(3, 'b', 0, 2),
(0, 'b', 2, 0), #t = 3
(2, 'b', 3, 0),
(3, 'c', 0, 2),
(0, 'c', 1, 0), #t = 6
(1, 'b', 0, 1),
(0, 'b', 2, 0),
(2, 'c', 3, 0), #t = 9
(3, 'c', 0, 2),
(0, 'c', 1, 0)]
What is the action in the state-action pair that is updated?		Multiple Choiceb.
Since we observe an experience in state 0, we update the $Q$ value for state 0.		To find the action in the state-action pair that is updated, we can go through the experience list and apply the Q-learning update formula for each experience tuple. Here's the step-by-step update for each experience:

1. t=0: (0, 'b', 2, 0)
Q(0, 'b') = (1-0.5) * 0 + 0.5 * (0 + 0.9 * 0) = 0

2. t=1: (2, 'b', 3, 0)
Q(2, 'b') = (1-0.5) * 0 + 0.5 * (0 + 0.9 * 0) = 0

3. t=2: (3, 'b', 0, 2)
Q(3, 'b') = (1-0.5) * 0 + 0.5 * (2 + 0.9 * 0) = 1

4. t=3: (0, 'b', 2, 0)
Q(0, 'b') = (1-0.5) * 0 + 0.5 * (0 + 0.9 * 0) = 0

5. t=4: (2, 'b', 3, 0)
Q(2, 'b') = (1-0.5) * 0 + 0.5 * (0 + 0.9 * 1) = 0.45

6. t=5: (3, 'c', 0, 2)
Q(3, 'c') = (1-0.5) * 0 + 0.5 * (2 + 0.9 * 0) = 1

7. t=6: (0, 'c', 1, 0)
Q(0, 'c') = (1-0.5) * 0 + 0.5 * (0 + 0.9 * 0) = 0

8. t=7: (1, 'b', 0, 1)
Q(1, 'b') = (1-0.5) * 0 + 0.5 * (1 + 0.9 * 0) = 0.5

9. t=8: (0, 'b', 2, 0)
Q(0, 'b') = (1-0.5) * 0 + 0.5 * (0 + 0.9 * 0.45) = 0.2025

10. t=9: (2, 'c', 3, 0)
Q(2, 'c') = (1-0.5) * 0 + 0.5 * (0 + 0.9 * 1) = 0.45

11. t=10: (3, 'c', 0, 2)
Q(3, 'c') = (1-0.5) * 1 + 0.5 * (2 + 0.9 * 0) = 1.5

12. t=11: (0, 'c', 1, 0)
Q(0, 'c') = (1-0.5) * 0 + 0.5 * (0 + 0.9 * 0.5) = 0.225

The last experience tuple is (0, 'c', 1, 0), so the action in the state-action pair that is updated is 'c'.		Incorrect
7
362EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneProblem Set 1NumPy1cvii0.01157407407TextThe shape of the resulting array is different depending on if you use indexing or slicing. Indexing refers to selecting particular elements of an array by using a single number (the index) to specify a particular row or column. Slicing refers to selecting a subset of the array by specifying a range of indices.
If you're unfamiliar with these terms, and the indexing and slicing rules of arrays, please see the indexing and slicing sections of this link: Numpy. Overview (Same as the Numpy Overview link from the introduction). You can also look at the official numpy documentation here.
In the following questions, let A = np.array([[5,7,10,14],[2,4,8,9]]). Tell us what the output would be for each of the following expressions. Use brackets [] as necessary. If the operation is invalid, write the python string "none".
Note: Remember that Python uses zero-indexing and thus starts counting from 0, not 1. This is different from R and MATLAB.
Reminder: A = np.array([[5,7,10,14],[2,4,8,9]])
A[:, 1:2]		Expression[[7], [4]]The output for the expression A[:, 1:2] would be:

array([[7],
[4]])		Correct
8
10EECS6.100AIntroduction to Computer Science Programming in PythonNoneNoneFinger Exercise Lecture 8Classes11.428571429TextIn this problem, you will implement three classes according to the specification below: one Container class, one Stack class (a subclass of Container), and one Queue class (a subclass of Container). Our Container class will initialize an empty list. The two methods we will have are to calculate the size of the list and to add an element. The second method will be inherited by the two subclasses. We now want to create two subclasses of this generic Container class so that we can add more functionality -- the ability to remove
elements from the list. A Stack and a Queue will add elements to the list in the same way, but will behave differently when removing an element.
A stack is a last in, first out data structure. Think of a stack of pancakes. As you make pancakes, you create a stack of them with older pancakes going on the bottom and newer pancakes on the top. As you start eating the pancakes, you pick one off the top so you are removing the newest pancake added to the stack. When implementing your Stack class, you will have to think about which end of your list contains the element that has been in the list the shortest amount of time. This is the element you will want to remove and return.
A queue is a first in, first out data structure. Think of a store checkout queue. The customer who has been in the line the longest gets the next available cashier. When implementing your Queue class, you will have to think about which end of your list contains the element that has been in the list the longest. This is the element you will want to remove and return.
class Container(object):
"""
A container object is a list and can store elements of any type
"""
def __init__(self):
"""
Initializes an empty list
"""
self.myList = []
def size(self):
"""
Returns the length of the container list
"""
# Your code here
def add(self, elem):
"""
Adds the elem to one end of the container list, keeping the end
you add to consistent. Does not return anything
"""
# Your code here
class Stack(Container):
"""
A subclass of Container. Has an additional method to remove elements.
"""
def remove(self):
"""
The newest element in the container list is removed
Returns the element removed or None if the queue contains no elements
"""
# Your code here		Programmingclass Container(object):
"""
A container object is a list and can store elements of any type
"""
def __init__(self):
"""
Initializes an empty list
"""
self.myList = []
def size(self):
"""
Returns the length of the container list
"""
# Your code here
return len(self.myList)
def add(self, elem):
"""
Adds the elem to one end of the container list, keeping the end
you add to consistent. Does not return anything
"""
# Your code here
self.myList.append(elem)
class Stack(Container):
"""
A subclass of Container. Has an additional method to remove elements.
"""
def remove(self):
"""
The newest element in the container list is removed
Returns the element removed or None if the queue contains no elements
"""
# Your code here
if self.size() == 0:
return None
return self.myList.pop() 		class Container(object):
"""
A container object is a list and can store elements of any type
"""
def __init__(self):
"""
Initializes an empty list
"""
self.myList = []

def size(self):
"""
Returns the length of the container list
"""
return len(self.myList)

def add(self, elem):
"""
Adds the elem to one end of the container list, keeping the end
you add to consistent. Does not return anything
"""
self.myList.append(elem)

class Stack(Container):
"""
A subclass of Container. Has an additional method to remove elements.
"""
def remove(self):
"""
The newest element in the container list is removed
Returns the element removed or None if the queue contains no elements
"""
if self.size() == 0:
return None
else:
return self.myList.pop()

class Queue(Container):
"""
A subclass of Container. Has an additional method to remove elements.
"""
def remove(self):
"""
The oldest element in the container list is removed
Returns the element removed or None if the queue contains no elements
"""
if self.size() == 0:
return None
else:
return self.myList.pop(0)		Correct
9
615EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneFinal ExamConvolutional Neural Networks8h0.35TextMIT grad student Rec Urrent would like to submit an entry to win this year's Grand ML Tetris Competition, which gives awards to the smallest neural networks which can identify tetris pieces with the highest accuracy. Rec seeks to make a convolutional neural network that can accurately classify single-channel $3 \times 3$ images of $2 \mathrm{D}$ tetris pieces as being either a line-shaped piece, or a corner-shaped piece, using just one $2 \times 2$ filter. Let's help Rec win this competition.
If Rec instead labeled line-shaped pieces as "1" and corner-shaped pieces as "0" then what values of $\mathrm{w}$ and $\mathrm{b}$ of the output layer give perfect classification and outputs that are close to 0 for corners and close to 1 for lines?		OpenThe same as above with opposite sign.To help Rec win the competition, we need to find the values of w and b that give perfect classification for the given problem. Let's first analyze the possible input images for line-shaped and corner-shaped pieces.

Line-shaped pieces can have two possible orientations in a 3x3 image:
1. Horizontal:
```
0 0 0
1 1 1
0 0 0
```
2. Vertical:
```
0 1 0
0 1 0
0 1 0
```

Corner-shaped pieces can have four possible orientations in a 3x3 image:
1. Top-left corner:
```
1 1 0
1 0 0
0 0 0
```
2. Top-right corner:
```
0 1 1
0 0 1
0 0 0
```
3. Bottom-left corner:
```
0 0 0
1 0 0
1 1 0
```
4. Bottom-right corner:
```
0 0 0
0 0 1
0 1 1
```

Now, let's apply the 2x2 filter to these images. We can use the following filter values to detect lines and corners:

Filter:
```
1 1
1 -1
```

Applying this filter to the input images, we get the following results:

1. Horizontal line: 2
2. Vertical line: 0
3. Top-left corner: -1
4. Top-right corner: -1
5. Bottom-left corner: -1
6. Bottom-right corner: -1

Now, we need to find the values of w and b for the output layer that give perfect classification. We can use a simple linear function for this purpose:

Output = w * Filter_Result + b

For line-shaped pieces, we want the output to be close to 1, and for corner-shaped pieces, we want the output to be close to 0. Based on the filter results, we can set up the following equations:

1. w * 2 + b ≈ 1 (for horizontal line)
2. w * 0 + b ≈ 1 (for vertical line)
3. w * (-1) + b ≈ 0 (for any corner)

From equation (2), we get b ≈ 1. Substituting this value in equation (1), we get w ≈ 0. Now, let's check if these values satisfy equation (3):

w * (-1) + b = 0 * (-1) + 1 = 1

This does not give us an output close to 0 for corners. However, since we are working with a single 2x2 filter, it is not possible to achieve perfect classification with outputs close to 0 for corners and close to 1 for lines. The best we can do is to use w = 0 and b = 1, which will give us an output of 1 for both line-shaped and corner-shaped pieces. This will not result in perfect classification, but it is the best we can do with the given constraints.		Invalid Ground Truth
10
388EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneProblem Set 2Regression1c0.03472222222TextYou are given the following data, where $d=1, n=4$.
$$
D=\{[[1], 2],[[2], 7],[[3],-3],[[4], 1]\}
$$
You want to use analytic linear regression to solve the problem.
Using Python and numpy (you might want to fire up a co-lab session), compute the $\theta, \theta_{0}$ that minimize MSE on this data. Provide your answer as a list.
As a reminder, you can use np.linalg.inv to take inverses in numpy.		Open[-1.3, 5]First, let's set up the problem. We have the data points (1, 2), (2, 7), (3, -3), and (4, 1). We want to find the best-fit line of the form y = θx + θ₀ that minimizes the mean squared error (MSE) on this data.

To do this, we can use the normal equation for linear regression:

θ = (X^T * X)^(-1) * X^T * y

where X is the matrix of input features (with a column of ones added for the bias term), y is the vector of target values, and θ is the vector of parameters we want to find.

Let's compute θ using Python and numpy:

```python
import numpy as np

# Input features
X = np.array([[1, 1],
[1, 2],
[1, 3],
[1, 4]])

# Target values
y = np.array([2, 7, -3, 1])

# Compute theta
theta = np.linalg.inv(X.T @ X) @ X.T @ y
```

Now, let's print the values of θ₀ and θ:

```python
print("Theta:", theta)
```

This will output:

```
Theta: [ 0.5 -1.4]
```

So, the best-fit line that minimizes the MSE on this data is given by:

y = -1.4x + 0.5

Thus, the values of θ₀ and θ are [0.5, -1.4].		Incorrect
11
21Mathematics18.2Principles of Discrete Applied MathematicsNone18.C06Problem Set 4Generating Function32.037037037TextSolve the following recurrence using generating functions:
$$
a_{n}=3 a_{n-1}+4 a_{n-2} \text { for } n \geq 2
$$
with the initial conditions $a_{0}=3, a_{1}=2$.		OpenWe let
$$
A(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots=\sum_{i=0}^{\infty} a_{i} x^{i}
$$
be the generating function for our sequence $\left\{a_{i}\right\}_{i \in \mathbb{N}}$. Then we get that
$$
A(x)=3 x A(x)+4 x^{2} A(x)-7 x+3.
$$
Rearranging the terms, we get that
$$
A(x)\left(1-3 x-4 x^{2}\right)=-7 x+3,
$$
or equivalently,
$$
A(x)=\frac{-7 x+3}{1-3 x-4 x^{2}}.
$$
We have the factorization $1-3 x-4 x^{2}=(1-4 x)(1+x)$, so we will get a partial fraction decomposition of the form
$$
A(x)=\frac{A}{1-4 x}+\frac{B}{1+x}.
$$
This tells us that $A+B=3$ and $A-4 B=-7$. Solving these equations, we get $A=1$ and $B=2$, so
$$
A(x)=\frac{1}{1-4 x}+\frac{2}{1+x}.
$$
Recalling that $\frac{1}{1-x}=\sum_{i=0}^{\infty} x^{i}$, we get that
$$
A(x)=\sum_{i=0}^{\infty}\left(4^{i}+2 \times(-1)^{i}\right) x^{i}
$$
Thus, we find that $a_{i}=4^{i}+2 \times(-1)^{i}$.		To solve the given recurrence relation using generating functions, we first define the generating function A(x) as follows:

$$
A(x) = \sum_{n=0}^{\infty} a_n x^n
$$

Now, we can rewrite the recurrence relation in terms of the generating function:

$$
\sum_{n=2}^{\infty} a_n x^n = 3 \sum_{n=2}^{\infty} a_{n-1} x^n + 4 \sum_{n=2}^{\infty} a_{n-2} x^n
$$

We can rewrite the sums to start from n=0:

$$
\sum_{n=0}^{\infty} a_{n+2} x^{n+2} = 3x \sum_{n=0}^{\infty} a_{n+1} x^{n+1} + 4x^2 \sum_{n=0}^{\infty} a_n x^n
$$

Now, we can substitute the generating function A(x) and the initial conditions:

$$
A(x) - a_0 - a_1 x = 3x (A(x) - a_0) + 4x^2 A(x)
$$

Substitute the initial conditions $a_0 = 3$ and $a_1 = 2$:

$$
A(x) - 3 - 2x = 3x (A(x) - 3) + 4x^2 A(x)
$$

Now, we solve for A(x):

$$
A(x) - 3 - 2x = 3xA(x) - 9x + 4x^2 A(x)
$$

Rearrange the terms:

$$
A(x) (1 - 3x - 4x^2) = 3 + 2x - 9x
$$

$$
A(x) = \frac{3 - 7x}{1 - 3x - 4x^2}
$$

Now, we can perform partial fraction decomposition on A(x):

$$
A(x) = \frac{3 - 7x}{(1 - 4x)(1 + x)} = \frac{A}{1 - 4x} + \frac{B}{1 + x}
$$

Solving for A and B, we get:

$$
A = \frac{1}{5}, \quad B = -\frac{8}{5}
$$

So, the generating function A(x) becomes:

$$
A(x) = \frac{1}{5} \cdot \frac{1}{1 - 4x} - \frac{8}{5} \cdot \frac{1}{1 + x}
$$

Now, we can find the coefficients of the power series expansion of A(x) to get the sequence $a_n$:

$$
a_n = [x^n] A(x) = \frac{1}{5} \cdot [x^n] \frac{1}{1 - 4x} - \frac{8}{5} \cdot [x^n] \frac{1}{1 + x}
$$

Using the geometric series formula, we get:

$$
a_n = \frac{1}{5} \cdot 4^n - \frac{8}{5} \cdot (-1)^n
$$

This is the closed-form expression for the sequence $a_n$ that satisfies the given recurrence relation and initial conditions.		Incorrect
12
9Mathematics18.704Seminar in Algebra18.701NoneProblem Set 2Linear Representation21.5TextLet $\rho: G \rightarrow \mathrm{GL}(V)$ be a representation and let
$$
V^{*}:=\left\{v^{*}: V \rightarrow \mathbb{C} \mid v^{*} \text { is linear }\right\}
$$
be the dual of $V$. For $x \in V, x^{*} \in V^{*}$, let $\left\langle x, x^{*}\right\rangle$ denote the value of the linear form $x^{*}$ at $x$. Show that there exists a unique linear representation $\rho^{*}: G \rightarrow \operatorname{GL}\left(V^{*}\right)$ such that
$$
\left\langle\rho_{s} x, \rho_{s}^{*} x^{*}\right\rangle=\left\langle x, x^{*}\right\rangle \quad \text { for } s \in G, x \in V, x^{*} \in V^{*} \text {. }
$$		OpenDefine $\rho^{*}: G \rightarrow \mathrm{GL}\left(V^{*}\right)$ such that $\rho_{s}^{*} x^{*}=x^{*} \circ \rho_{s^{-1}}$.
Claim. $\rho^{*}$ is a representation.
Proof. $\rho_{s}^{*}$ is linear since the scalars can be factered out from the composition, and composition distributes over addition. $\rho_{s}^{*}$ is invertible since it's inverse is $\left(\rho_{s}^{*}\right)^{-1} x^{*}=x^{*} \circ \rho_{s}$. Check:
$$
\begin{aligned}
\left(\left(\rho_{s}^{*}\right)^{-1} \rho_{s}^{*}\right) x^{*} & =\left(\rho_{s}^{*}\right)^{-1}\left(\rho_{s}^{*} x^{*}\right) \\
& =\left(\rho_{s}^{*}\right)^{-1}\left(x^{*} \circ \rho_{s^{-1}}\right) \\
& =x^{*} \circ \rho_{s^{-1}} \circ \rho_{s} \\
& =x^{*} \\
& =x^{*} \circ \rho_{s} \circ \rho_{s^{-1}} \\
& =\rho_{s}^{*}\left(\left(\rho_{s}^{*}\right)^{-1} x^{*}\right) \\
& =\left(\rho_{s}^{*}\left(\rho_{s}^{*}\right)^{-1}\right) x^{*}
\end{aligned}
$$
Thus, $\left(\rho_{s}^{*}\right)^{-1} \rho_{s}^{*}=\mathrm{Id}=\rho_{s}^{*}\left(\rho_{s}^{*}\right)^{-1}$, so $\rho_{s}^{*} \in \mathrm{GL}\left(V^{*}\right)$. The group operation is also preserved.
$$
\begin{aligned}
\left(\rho^{*} g\right)\left(\rho^{*} h\right) x^{*} & =\rho_{g}^{*}\left(x^{*} \circ \rho_{h^{-} 1}\right) \\
& =x^{*} \circ \rho_{h^{-1}} \circ \rho_{g^{-} 1} \\
& =x^{*} \circ\left(\rho_{h^{-1}} \rho_{g^{-1}}\right) \\
& =x^{*} \circ \rho_{(g h)^{-1}} \\
& =\left(\rho^{*}(g h)\right) x^{*}
\end{aligned}
$$
Thus $\rho^{*}$ is a representation.
Claim. $\rho^{*}$ satisfies the given equation.
Proof.
$$
\begin{aligned}
\left\langle\rho_{s} x, \rho_{s}^{*} x^{*}\right\rangle & =\left(\rho_{s}^{*} x^{*}\right)\left(\rho_{s} x\right) \\
& =\left(x^{*} \circ \rho_{s^{-1}}\right)\left(\rho_{s} x\right) \\
& =x^{*} x \\
& =\left\langle x, x^{*}\right\rangle
\end{aligned}
$$
Claim. If there exists a representation that satisfies the equation, then it is unique.
Proof. Suppose that the representations $\rho^{*}$ and $\sigma^{*}$ both satisfy the equation. Then
$$
\left\langle\rho_{s} x, \rho_{s}^{*} x^{*}\right\rangle=\left\langle\rho_{s} x, \sigma_{s}^{*} x^{*}\right\rangle \quad \forall s \in G, x \in V, x^{*} \in V^{*}
$$
Let $x^{\prime}=\rho_{s} x$. Since $\rho_{s}$ is invertible, the equation becomes
$$
\left\langle x^{\prime}, \rho_{s}^{*} x^{*}\right\rangle=\left\langle x^{\prime}, \sigma_{s}^{*} x^{*}\right\rangle \quad \forall s \in G, x^{\prime} \in V, x^{*} \in V^{*}
$$
Applying,
$$
\left(\rho_{s}^{*} x^{*}\right) x^{\prime}=\left(\sigma_{s}^{*} x^{*}\right) x^{\prime} \quad \forall s \in G, x^{\prime} \in V, x^{*} \in V^{*}
$$
Since it's true for all elements in their domains, the functions are equal.
$$
\rho_{s}^{*} x^{*}=\sigma_{s}^{*} x^{*} \quad \forall s \in G, x^{*} \in V^{*}
$$
Since it's true again, the functions are equal.
$$
\rho_{s}^{*}=\sigma_{s}^{*} \quad \forall s \in G
$$
One more time.
$$
\rho^{*}=\sigma^{*}
$$
Thus, the representations are equal, so a representation that satisfies the equation is unique.
Thus the defined $\rho^{*}$ is the unique representation which satisfies the equation.		To show the existence and uniqueness of the linear representation $\rho^*: G \rightarrow \operatorname{GL}\left(V^*\right)$, we will first define the action of $\rho^*$ on $V^*$ and then show that it satisfies the given condition.

Let $s \in G$, $x \in V$, and $x^* \in V^*$. Define the action of $\rho^*(s)$ on $x^*$ as follows:
$$
(\rho^*_s x^*)(x) := x^*(\rho_{s^{-1}} x).
$$
We need to show that this action is linear and satisfies the given condition.

1. Linearity of $\rho^*(s)$:

Let $x^*_1, x^*_2 \in V^*$ and $\alpha, \beta \in \mathbb{C}$. Then,
$$
\begin{aligned}
(\rho^*_s (\alpha x^*_1 + \beta x^*_2))(x) &= (\alpha x^*_1 + \beta x^*_2)(\rho_{s^{-1}} x) \\
&= \alpha x^*_1(\rho_{s^{-1}} x) + \beta x^*_2(\rho_{s^{-1}} x) \\
&= \alpha (\rho^*_s x^*_1)(x) + \beta (\rho^*_s x^*_2)(x).
\end{aligned}
$$
This shows that $\rho^*(s)$ is linear.

2. Satisfying the given condition:

We need to show that
$$
\left\langle\rho_s x, \rho^*_s x^*\right\rangle = \left\langle x, x^*\right\rangle.
$$
By definition of the action of $\rho^*(s)$ on $x^*$, we have
$$
\begin{aligned}
\left\langle\rho_s x, \rho^*_s x^*\right\rangle &= (\rho^*_s x^*)(\rho_s x) \\
&= x^*(\rho_{s^{-1}} (\rho_s x)) \\
&= x^*(\rho_{s^{-1}} \rho_s x) \\
&= x^*(\rho_{s^{-1} s} x) \\
&= x^*(\rho_e x) \\
&= x^*(x) \\
&= \left\langle x, x^*\right\rangle.
\end{aligned}
$$
This shows that the defined action of $\rho^*$ satisfies the given condition.

Since we have defined the action of $\rho^*$ on $V^*$ and shown that it is linear and satisfies the given condition, we have shown the existence of the linear representation $\rho^*: G \rightarrow \operatorname{GL}\left(V^*\right)$.

Now, we need to show the uniqueness of $\rho^*$. Suppose there exists another linear representation $\tilde{\rho}^*: G \rightarrow \operatorname{GL}\left(V^*\right)$ such that
$$
\left\langle\rho_s x, \tilde{\rho}^*_s x^*\right\rangle = \left\langle x, x^*\right\rangle.
$$
Then, for any $s \in G$, $x \in V$, and $x^* \in V^*$, we have
$$
\left\langle\rho_s x, \rho^*_s x^*\right\rangle = \left\langle\rho_s x, \tilde{\rho}^*_s x^*\right\rangle.
$$
This implies that
$$
\left\langle\rho_s x, (\rho^*_s - \tilde{\rho}^*_s) x^*\right\rangle = 0.
$$
Since this holds for all $x \in V$, we have $(\rho^*_s - \tilde{\rho}^*_s) x^* = 0$ for all $x^* \in V^*$. This implies that $\rho^*_s = \tilde{\rho}^*_s$ for all $s \in G$. Therefore, the linear representation $\rho^*: G \rightarrow \operatorname{GL}\left(V^*\right)$ is unique.

In conclusion, we have shown the existence and uniqueness of the linear representation $\rho^*: G \rightarrow \operatorname{GL}\left(V^*\right)$ satisfying the given condition.		​
13
42EECS6.122Design and Analysis of Algorithms6.121NoneProblem Set 4Probability2c0.2727272727TextSophie Germain is organizing a dinner party where $n$ guests are seated around a round table with $N=2 n$ seats, numbered 1 through $N$ clockwise.
The guests arrive one by one, and each guest takes his or her seat before the next guest arrives. To optimize the amount of networking between the guests, Sophie has come up with an unusual way of seating them. Namely, to find his or her seat, each guest $i$ first receives a seat number $r_i$ chosen uniformly and independently at random. Then, if seat $r_i$ is not currently occupied, the guest sits there; otherwise, the guest walks clockwise around the table until he or she finds the first unoccupied seat, and sits there.
A block is a set of consecutive seats that are all occupied but with the seats before and after it being unoccupied. Let $p_{j, k}$ be the probability that after all the guests have taken their seats there is a block of length $k$ starting at seat $j$, i.e. that $\{j, \ldots, j+k-1\}$ is a block in the final configuration.
Show that $\operatorname{Pr}\left[E_{j, k}\right] \leq \frac{1}{c^k}$ for some constant $c>1$ independent of the problem parameters. Hint: Consider random variables $X_i$, for each guest $i$, where $X_i=1$ if $r_i \in\{j, \ldots, j+$ $k-1\}$ and $X_i=0$, otherwise; and recall that $e>1$.		OpenDefining random variables as in the hint, we see that $\mathbb{E}\left[X_i\right]=\frac{k}{N}$ and thus letting $X=X_1+\ldots+X_n$ we have $\mathbb{E}[X]=\frac{n k}{N}=\frac{k}{2}$ by linearity of expectation. We thus want to bound the probability that
$$
\operatorname{Pr}[X \geq k]=\operatorname{Pr}[X \geq(1+1) \mathbb{E}[X]]
$$
We will use (the multiplicative form of) the Chernoff bound for that. Note that since $\beta=1$ here, either the form for $\beta \leq 1$ or $\beta \geq 1$ works here. With this, we get
$$
\operatorname{Pr}[X \geq k] \leq e^{-k / 6}=\frac{1}{\left(e^{1 / 6}\right)^k}
$$
so we can use $c=e^{1 / 6}>1$.		To prove that $\operatorname{Pr}\left[E_{j, k}\right] \leq \frac{1}{c^k}$ for some constant $c>1$ independent of the problem parameters, we will follow the hint and consider the random variables $X_i$ for each guest $i$.

Let $X_i = 1$ if $r_i \in \{j, \ldots, j+k-1\}$ and $X_i = 0$ otherwise. Then, the probability that $X_i = 1$ is $\frac{k}{N}$, since there are $k$ seats in the block and $N$ total seats.

Now, let $Y_i = X_i - \frac{k}{N}$, so $Y_i$ is the difference between the actual occupancy of the block and the expected occupancy. We want to find the probability that the sum of the $Y_i$'s is positive, i.e., that the block is fully occupied.

Using the exponential generating function, we have:

$$\mathbb{E}\left[e^{tY_i}\right] = \mathbb{E}\left[e^{t(X_i - \frac{k}{N})}\right] = \frac{k}{N}e^t + \left(1 - \frac{k}{N}\right)e^{-\frac{kt}{N}}$$

Now, we want to find the probability that $\sum_{i=1}^n Y_i > 0$. By Markov's inequality, we have:

$$\operatorname{Pr}\left[\sum_{i=1}^n Y_i > 0\right] \leq \frac{\mathbb{E}\left[e^{t\sum_{i=1}^n Y_i}\right]}{e^t}$$

Since the $Y_i$'s are independent, we have:

$$\mathbb{E}\left[e^{t\sum_{i=1}^n Y_i}\right] = \prod_{i=1}^n \mathbb{E}\left[e^{tY_i}\right] = \left(\frac{k}{N}e^t + \left(1 - \frac{k}{N}\right)e^{-\frac{kt}{N}}\right)^n$$

Now, we want to minimize this expression with respect to $t$. Taking the derivative with respect to $t$ and setting it to zero, we get:

$$\frac{d}{dt}\left(\frac{k}{N}e^t + \left(1 - \frac{k}{N}\right)e^{-\frac{kt}{N}}\right)^n = 0$$

Solving for $t$, we find that the optimal value of $t$ is $t^* = \frac{N}{k}\ln\left(\frac{k}{N-k}\right)$. Plugging this value back into the expression for the probability, we get:

$$\operatorname{Pr}\left[E_{j, k}\right] \leq \frac{1}{e^t} \left(\frac{k}{N}e^t + \left(1 - \frac{k}{N}\right)e^{-\frac{kt}{N}}\right)^n \leq \frac{1}{c^k}$$

for some constant $c>1$ independent of the problem parameters. This completes the proof.		​
14
120EECS6.122Design and Analysis of Algorithms6.121NoneMidterm Exam 2Linear Programming3b2.5TextLet $G=(V, E)$ be a weighted directed graph, where all weights are non-negative. Let $w_{u v}$ denote the weight of each edge $(u, v) \in E$. Alicia wrote down the following linear program with variables $\left\{x_{v}: v \in V\right\}$ that she claims computes the length of the shortest path from a source vertex $s$ to a target vertex $t$. (By convention, if there is no path from $s$ to $t$, the shortest path length is $\infty$.)
$$
\begin{array}{lll}
\operatorname{maximize} & x_{t} & \\
\text { subject to } & x_{v}-x_{u} \leq w_{u v} & \text { for all edges }(u, v) \in E \\
& x_{v} \geq 0 & \text { for all vertices } v \in V . \\
& x_{s}=0 &
\end{array}
$$
Is Alicia's claim correct? If yes, prove that the optimal solution to her linear program is indeed the length of a shortest path from $s$ to $t$. If not, correct the linear program and prove that your (corrected) LP computes the shortest path length from $s$ to $t$.		OpenAlicia's claim is correct. The correct linear program for the shortest path somewhat counter-intuitively maximizes $x_{t}$ subject to the given constraints. Think of each edge as a string of length $w_{u v}$. The constraints tell us how far the vertex $v$ can get from the source given the values $x_{u}$ of its in-neighbors and the weights $w_{u v}$ of its in-edges. The problem then is to see how far the target $t$ can get. A rigorous argument follows.
Let $d(u, v)$ denote the length of the shortest path between $u$ and $v$. Then, $x_{u}=d(s, u)$ for all $u$ is a feasible solution to the LP because $x_{v}=d(s, v) \geq 0$ and $x_{s}=d(s, s)=0$.
$x_{u}-x_{v}=d(s, u)-d(s, v) \leq w_{u v}$ because otherwise the path $s \longrightarrow v \longrightarrow u$ would be shorter than $d(s, u)$. As this gives a feasible solution where $x_{t}=d(s, t)$ and this is a maximization LP, we have shown that the optimal $x_{t} \geq d(s, t)$.
Now, we show it must hold that $x_{t} \leq d(s, t)$. If there were a solution with $x_{t}>d(s, t)$ then if $v_{1}, v_{2}, \ldots, v_{n}$ is the shortest path between $s$ and $t$, then
$$
\begin{gathered}
x_{t}=x_{t}-x_{s}=x_{v_{n}}-x_{v_{1}} \\
=\left(x_{v_{n}}-x_{v_{n-1}}\right)+\left(x_{v_{n-1}}-x_{v_{n-2}}\right)+\cdots+\left(x_{v_{2}}-x_{v_{1}}\right) \\
\leq w_{v_{n-1} v_{n}}+w_{v_{n-2} v_{n-1}}+\cdots+w_{v_{1} v_{2}} \\
=d(s, t)
\end{gathered}
$$
We get a contradiction. Therefore, $d(s, t)$ is the optimal value of the LP. 		Yes, Alicia's claim is correct. The given linear program computes the length of the shortest path from the source vertex $s$ to the target vertex $t$ in a weighted directed graph with non-negative edge weights.

Proof:

Let $d(v)$ denote the length of the shortest path from $s$ to $v$. We will show that the optimal solution to the linear program is equal to $d(t)$.

1. Feasibility: For any vertex $v$, we have $x_v \geq 0$ and $x_s = 0$. Thus, the linear program is feasible.

2. Upper bound: We will show that for any vertex $v$, $x_v \leq d(v)$. We will prove this by induction on the number of edges in the shortest path from $s$ to $v$.

Base case: If there is no edge in the path from $s$ to $v$, then $v = s$ and $x_v = x_s = 0 = d(v)$.

Inductive step: Suppose the claim holds for all vertices reachable from $s$ with paths of length at most $k$. Let $v$ be a vertex reachable from $s$ with a path of length $k+1$. Let $(u, v)$ be the last edge in the shortest path from $s$ to $v$. Then, by the induction hypothesis, $x_u \leq d(u)$. Since the edge weights are non-negative, we have $d(v) \geq d(u) + w_{uv}$. By the constraint in the linear program, we have $x_v - x_u \leq w_{uv}$. Combining these inequalities, we get $x_v \leq x_u + w_{uv} \leq d(u) + w_{uv} \leq d(v)$. Thus, the claim holds for paths of length $k+1$.

By induction, the claim holds for all vertices $v$. In particular, $x_t \leq d(t)$.

3. Lower bound: We will show that there exists a feasible solution $x^*$ such that $x^*_t = d(t)$. Define $x^*_v = d(v)$ for all vertices $v$. We need to show that $x^*$ satisfies all the constraints of the linear program.

For any edge $(u, v) \in E$, let $P$ be the shortest path from $s$ to $u$ and $Q$ be the shortest path from $s$ to $v$. If $P$ is a subpath of $Q$, then $d(v) = d(u) + w_{uv}$, and thus $x^*_v - x^*_u = d(v) - d(u) = w_{uv}$. If $P$ is not a subpath of $Q$, then there exists a shortest path from $s$ to $v$ that includes the edge $(u, v)$. In this case, $d(v) \leq d(u) + w_{uv}$, and thus $x^*_v - x^*_u = d(v) - d(u) \leq w_{uv}$. In both cases, the constraint $x^*_v - x^*_u \leq w_{uv}$ is satisfied.

Since $x^*_v = d(v) \geq 0$ for all vertices $v$ and $x^*_s = d(s) = 0$, the solution $x^*$ is feasible.

4. Optimality: Since there exists a feasible solution $x^*$ such that $x^*_t = d(t)$ and we have shown that $x_t \leq d(t)$ for any feasible solution $x$, the optimal solution to the linear program is indeed $d(t)$, which is the length of the shortest path from $s$ to $t$.		​
15
112EECS6.411Representation, Inference, and Reasoning in AI6.1010, 6.1210, 18.600NoneProblem Set 3Propositional Proof4bii0.2232142857TextThere are three suspects for a murder: Adams, Brown, and Clark.
1. Adams says "I didn't do it. The victim was an old acquaintance of Brown's. But Clark hated him."
2. Brown states "I didn't do it. I didn't know the guy. Besides I was out of town all week."
3. Clark says "I didn't do it. I saw both Adams and Brown in town around the victim that day; one of them must have done it."
4. We know that exactly one of the suspects is guilty.
Assume that the two innocent people are telling the truth, but that the guilty people might not be. So, the statements from the suspects can be encoded as "If suspect_is_innocent, then some other facts are true".
Let the propositional variables have the following definitions:
\begin{itemize}
\item $A=$ Adams is innocent
\item $B=$ Brown is innocent
\item $C=$ Clark is innocent
\item $X=$ Brown knew the victim
\item $Y=$ Brown was out of town
\item $Z=$ Adams was out of town
\item $W=$ Clark hated the victim
\end{itemize}
Now, enter the steps in a valid resolution proof. Each step will indicate the indices of two parent clauses (as integers) and the resolvent clause (as a list of literal strings), e.g. [3, 4, [ $\left.\left.\sim X^{\prime}, ' A^{\prime}\right]\right]$. The resolvent clause in each of the steps entered can be used in subsequent steps by using its index, starting with 13. Indicate a contradiction by entering an empty list for the clause, e.g. [1, 2, [] ]
The entries below illustrate the format; the number of required steps in the proof is not necessarily as indicated.
proof = [
[0, 0, ['A']], # 13
[0, 0, ['B']], # 14
[0, 0, ['C']], # 15
[0, 0, ['D']], # 16
[0, 0, ['E']], # 17
[0, 0, []'], # 18
]		Openproof = [[3, 12, ["~X"]], [1, 13, ["~A"]], [4, 12, ["Y"]], [5, 15, ["~C"]], [8, 16, ["A"]], [14, 17, []]]proof = [
[1, 2, ['¬A', 'B', 'C']], # 13
[1, 3, ['¬A', '¬B', 'C']], # 14
[2, 3, ['¬C', 'A', 'B']], # 15
[13, 14, ['¬A', 'C']], # 16
[13, 15, ['¬C', 'B']], # 17
[14, 15, ['¬A', '¬B']], # 18
[16, 17, ['¬A']], # 19
[18, 19, []] # 20
]		Incorrect
16
16Mathematics18.03Differential EquationsNone18.02Problem Set 2Complex Numbers2b0.1608579088TextExpress the following complex numbers in polar form $r e^{i \theta}$ with $\theta \in(-\pi, \pi]$.
$1-i$.		ExpressionThe modulus is $|1-i|=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$. The argument is $\arctan \frac{-1}{1}=\frac{7 \pi}{4}$. So $1-i=\sqrt{2} e^{i \frac{7 \pi}{4}}$.To express the complex number $1-i$ in polar form, we first need to find the modulus (magnitude) $r$ and the argument (angle) $\theta$.

1. Modulus $r$:
$r = \sqrt{(\text{Real part})^2 + (\text{Imaginary part})^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$

2. Argument $\theta$:
$\theta = \arctan\left(\frac{\text{Imaginary part}}{\text{Real part}}\right) = \arctan\left(\frac{-1}{1}\right) = -\frac{\pi}{4}$

Now, we can write the complex number in polar form:
$1-i = \sqrt{2}e^{-i\frac{\pi}{4}}$		Correct
17
32Mathematics18.2Principles of Discrete Applied MathematicsNone18.C06Problem Set 6Linear Program2c0.6790123457TextConsider the LP
$$
\begin{aligned}
\max & 4 x_{1}+x_{2} \\
\text { s.t. } & x_{1}-x_{2} \leq 2 \\
& x_{1}+x_{2} \leq 8 \\
& x_{1}, x_{2} \geq 0
\end{aligned}
$$
Write the dual LP for the original problem in $x_{1}, x_{2}$, and use complementary slackness to determine the optimal solution for this dual.		ExpressionThe dual LP is given by
$$
\begin{array}{cl}
\min & 2 y_{1}+8 y_{2} \\
\text { s.t. } & y_{1}+y_{2} \geq 4 \\
& -y_{1}+y_{2} \geq 1 \\
& y_{1}, y_{2} \geq 0 .
\end{array}
$$
The optimum value of the original problem will be achieved at one of the vertices of its feasible region. Testing all four vertices, one gets that the optimal value of 23 at $\left(x_{1}, x_{2}\right)=(5,3)$. We have that $x_{1}-x_{2}=2$ and $x_{1}+x_{2}=8$, so we don't need the $y_{i}$ to be zero. Since $x_{1}$ and $x_{2}$ are nonzero, we must have $y_{1}+y_{2}=4$ and $-y_{1}+y_{2}=1$. Solving this gives us $y_{1}=3 / 2$ and $y_{2}=5 / 2$.		To write the dual LP for the given primal LP, we first need to identify the primal's objective function, constraints, and variables. In this case, the primal LP is a maximization problem with variables $x_1$ and $x_2$, and two constraints.

The dual LP will be a minimization problem with variables corresponding to the constraints of the primal LP. Let's denote these variables as $y_1$ and $y_2$. The objective function of the dual LP will be the minimum of the right-hand side values of the primal constraints, weighted by the dual variables:

$$
\begin{aligned}
\min & 2y_{1}+8y_{2} \\
\text { s.t. } & y_{1}+y_{2} \geq 4 \\
& -y_{1}+y_{2} \geq 1 \\
& y_{1}, y_{2} \geq 0
\end{aligned}
$$

Now, we will use complementary slackness to determine the optimal solution for this dual LP. Complementary slackness states that for an optimal solution, either the primal constraint is tight (equality holds) or the corresponding dual variable is zero, and vice versa.

Let's denote the primal constraints as $C_1$ and $C_2$, and the dual constraints as $D_1$ and $D_2$. We have:

1. If $C_1$ is tight, then $x_1 - x_2 = 2$. In this case, $y_1 = 0$.
2. If $C_2$ is tight, then $x_1 + x_2 = 8$. In this case, $y_2 = 0$.

Now, let's consider the dual constraints:

1. If $D_1$ is tight, then $y_1 + y_2 = 4$. In this case, $x_1 = 0$.
2. If $D_2$ is tight, then $-y_1 + y_2 = 1$. In this case, $x_2 = 0$.

From the above conditions, we can see that if $C_1$ is tight, then $D_2$ must also be tight, and if $C_2$ is tight, then $D_1$ must also be tight. Therefore, we have:

1. If $C_1$ and $D_2$ are tight: $x_1 - x_2 = 2$ and $-y_1 + y_2 = 1$. Solving these equations, we get $x_1 = 2$, $x_2 = 0$, $y_1 = 1$, and $y_2 = 2$.
2. If $C_2$ and $D_1$ are tight: $x_1 + x_2 = 8$ and $y_1 + y_2 = 4$. Solving these equations, we get $x_1 = 4$, $x_2 = 4$, $y_1 = 2$, and $y_2 = 2$.

Now, we need to check which of these solutions is optimal for the dual LP. Plugging the solutions into the dual objective function, we get:

1. For the first solution: $2(1) + 8(2) = 18$.
2. For the second solution: $2(2) + 8(2) = 20$.

Since the dual LP is a minimization problem, the first solution is optimal. Therefore, the optimal solution for the dual LP is $y_1 = 1$ and $y_2 = 2$.		​
18
148Mathematics18.6Probability and Random Variables18.02NoneFinal ExamCentral Limit Theorem5b1.25TextIn this problem, we'll try to approximate $\pi$ by throwing darts at a dartboard. To do this, suppose we have a square dartboard with sides of length 5. Inside it, we'll draw a circle with radius 1.
When I throw a dart, suppose that the spot where it hits is uniformly distributed over the dartboard. The probability that it lands inside the circle is thus given by
$$
\mathrm{P}(\text { lands in circle })=\frac{\text { Area }(\text { circle })}{\text { Area }(\text { dartboard })}=\frac{\pi}{25} .
$$
Use the Central Limit Theorem to approximate the probability that $Z \in[\pi-0.1, \pi+$ 0.1]. Write your answer in terms of $\Phi(x)$, the CDF of a standard normal.		ExpressionLet $\sigma=\sqrt{\operatorname{Var}(Z)}$, (using the value we computed in part (a)). By the Central Limit Theorem, we can approximate $Z$ by a normal with parameters $\left(\mu, \sigma^{2}\right)$, and thus $(Z-\pi) / \sigma$ is approximately a standard normal. This gives us
$$
\begin{aligned}
P(Z \in[\pi-0.1, \pi+0.1]) & =P\left(\frac{Z-\pi}{\sigma} \in\left[\frac{-0.1}{\sigma}, \frac{0.1}{\sigma}\right]\right) \approx \Phi\left(\frac{0.1}{\sigma}\right)-\Phi\left(\frac{-0.1}{\sigma}\right) \\
& =2 \Phi\left(\frac{0.1}{\sigma}\right)-1.
\end{aligned}
$$		To approximate the probability using the Central Limit Theorem, we need to find the mean and variance of the distribution of the number of darts landing inside the circle. Let $X_i$ be the indicator random variable for the $i$-th dart landing inside the circle, i.e., $X_i = 1$ if the $i$-th dart lands inside the circle and $X_i = 0$ otherwise. We have:

$$
\mathrm{E}[X_i] = \mathrm{P}(X_i = 1) = \frac{\pi}{25}
$$

and

$$
\mathrm{Var}[X_i] = \mathrm{E}[X_i^2] - (\mathrm{E}[X_i])^2 = \frac{\pi}{25} - \left(\frac{\pi}{25}\right)^2.
$$

Now, let $n$ be the number of darts thrown, and let $S_n = \sum_{i=1}^n X_i$ be the total number of darts landing inside the circle. By the Central Limit Theorem, we have:

$$
\frac{S_n - n\mathrm{E}[X_i]}{\sqrt{n\mathrm{Var}[X_i]}} \approx Z,
$$

where $Z$ is a standard normal random variable. We want to find the probability that $Z \in [\pi - 0.1, \pi + 0.1]$. To do this, we first rewrite the inequality in terms of $S_n$:

$$
\pi - 0.1 \leq \frac{S_n - n\frac{\pi}{25}}{\sqrt{n\left(\frac{\pi}{25} - \left(\frac{\pi}{25}\right)^2\right)}} \leq \pi + 0.1.
$$

Now, we can rewrite this inequality in terms of the standard normal CDF $\Phi(x)$:

$$
\Phi\left(\frac{\pi - 0.1 - n\frac{\pi}{25}}{\sqrt{n\left(\frac{\pi}{25} - \left(\frac{\pi}{25}\right)^2\right)}}\right) \leq \mathrm{P}(Z \in [\pi - 0.1, \pi + 0.1]) \leq \Phi\left(\frac{\pi + 0.1 - n\frac{\pi}{25}}{\sqrt{n\left(\frac{\pi}{25} - \left(\frac{\pi}{25}\right)^2\right)}}\right).
$$

Thus, the probability that $Z \in [\pi - 0.1, \pi + 0.1]$ can be approximated by the difference of the standard normal CDF values:

$$
\mathrm{P}(Z \in [\pi - 0.1, \pi + 0.1]) \approx \Phi\left(\frac{\pi + 0.1 - n\frac{\pi}{25}}{\sqrt{n\left(\frac{\pi}{25} - \left(\frac{\pi}{25}\right)^2\right)}}\right) - \Phi\left(\frac{\pi - 0.1 - n\frac{\pi}{25}}{\sqrt{n\left(\frac{\pi}{25} - \left(\frac{\pi}{25}\right)^2\right)}}\right).
$$		​
19
422EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneProblem Set 2Regression6a0.05208333333TextWe will now try to synthesize what we've learned in order to perform ridge regression on the DataCommons public health dataset that we explored in Lab 2. Unlike in Lab 2, where we did some simple linear regressions, here we now employ and explore regularization, with the goal of building a model which generalizes better (than without regularization) to unseen data.
The overall objective function for ridge regression is
$$
J_{\text {ridge }}\left(\theta, \theta_{0}\right)=\frac{1}{n} \sum_{i=1}^{n}\left(\theta^{T} x^{(i)}+\theta_{0}-y^{(i)}\right)^{2}+\lambda\|\theta\|^{2}
$$
Remarkably, there is an analytical function giving $\Theta=\left(\theta, \theta_{0}\right)$ which minimizes this objective, given $X, Y$, and $\lambda$. But how should we choose $\lambda$?
To choose an optimum $\lambda$, we can use the following approach. Each particular value of $\lambda$ gives us a different linear regression model. And we want the best model: one which balances providing good predictions (fitting well to given training data) with generalizing well (avoiding overfitting training data). And as we saw in the notes on Regression, we can employ cross-validation to evaluate and compare different models.
Let us begin by implementing this algorithm for cross-validation:
CROSS-VALIDATE $(\mathcal{D}, k)$
1 divide $\mathcal{D}$ into $k$ chunks $\mathcal{D}_{1}, \mathcal{D}_{2}, \ldots \mathcal{D}_{k}$ (of roughly equal size)
2 for $i=1$ to $k$
3 train $h_{i}$ on $\mathcal{D} \backslash \mathcal{D}_{i}$ (withholding chunk $\mathcal{D}_{i}$ )
4 compute "test" error $\mathcal{E}_{\mathfrak{i}}\left(h_{i}\right)$ on withheld data $\mathcal{D}_{i}$
5 return $\frac{1}{k} \sum_{i=1}^{k} \varepsilon_{i}\left(h_{i}\right)$
Let's implement the cross-validation algorithm as the procedure cross_validate, which takes the following input arguments:
\begin{itemize}
\item $\mathrm{x}$ : the list of data points $(d \times n)$
\item $\mathrm{Y}$ : the true values of the responders $(1 \times n)$
\item n_splits: the number of chunks to divide the dataset into
\item lam: the regularization parameter
\item learning_algorithm: a function that takes $X, Y$, and 1 lam, and returns th, th $\theta$
\item loss_function: a function that takes $X, Y$, th, and th $\theta$, and returns a $1 \times 1$ array
\end{itemize}
cross_validate should return a scalar, the cross-validation error of applying the learning algorithm on the list of data points.
Note that this is a generic version of cross-validation, that can be applied to any learning algorithm and any loss function. Later in this problem, we will use cross-validation specifically for ridge regression and mean square loss.
You have the following function available to you:
def make_splits(X, Y, n_splits):
'''
Splits the dataset into n_split chunks, creating n_split sets of
cross-validation data.
Returns a list of n_split tuples (X_train, Y_train, X_test, Y_test).
For the ith returned tuple:
*X_train and Y_train include all data except the ith chunk, and
* X_test and Y_test are the ith chunk.
X : d x n numpy array (d = #features, n = #data points)
Y : 1 x n numpy array
n_splits : integer
'''
def cross_validate(X, Y, n_splits, lam,
learning_algorithm, loss_function):
pass		Programmingdef cross_validate(X, Y, n_splits, lam,
learning_algorithm, loss_function):
test_errors = []
for (X_train, Y_train, X_test, Y_test) in make_splits(X, Y, n_splits):
th, th0 = learning_algorithm(X_train, Y_train, lam)
test_errors.append(loss_function(X_test, Y_test, th, th0))
return np.array(test_errors).mean()		To implement the cross_validate function, you can follow these steps:

1. Call the make_splits function to get the list of tuples containing the training and testing data for each split.
2. Initialize a variable to store the total error, e.g., total_error = 0.
3. Loop through the list of tuples obtained from make_splits:
a. Extract the training and testing data (X_train, Y_train, X_test, Y_test) from the current tuple.
b. Call the learning_algorithm function with the training data and lam to get the model parameters (th, th0).
c. Call the loss_function function with the testing data and the model parameters to get the error for the current split.
d. Add the error to the total_error variable.
4. Calculate the average error by dividing the total_error by n_splits.
5. Return the average error.

Here's the implementation of the cross_validate function:

```python
def cross_validate(X, Y, n_splits, lam, learning_algorithm, loss_function):
# Call make_splits to get the list of tuples containing training and testing data
splits = make_splits(X, Y, n_splits)

# Initialize the total error
total_error = 0

# Loop through the list of tuples
for X_train, Y_train, X_test, Y_test in splits:
# Call the learning_algorithm function to get the model parameters
th, th0 = learning_algorithm(X_train, Y_train, lam)

# Call the loss_function function to get the error for the current split
error = loss_function(X_test, Y_test, th, th0)

# Add the error to the total_error
total_error += error

# Calculate the average error
average_error = total_error / n_splits

# Return the average error
return average_error
```

Now you can use this cross_validate function to perform cross-validation for ridge regression with different values of lambda and choose the best lambda that gives the lowest cross-validation error.		​
20
303EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneLab 12Reinforcement Learning2av0.04166666667TextNow, we'll take a look at Q-learning in a simple 2D grid setting but with a single goal location. We'll adopt the same state space and action space as in MDP lab.
Specifically, our state space is a 10-by-10 grid, with the bottom-left state indexed as state $(0,0)$, and our four possible actions are moving North, South, East or West (note that if the robot tries to move off the board, it cannot; it just stays in the same state). Our single goal is at state $(3,6)$.
Remember that for Q-learning, the transition probabilities and reward values are not known in advance by the method-the agent just gets to observe state, action, and reward values as it moves through the domain.
Some notes (please read!):
\begin{itemize}
\item A new episode is started by sampling the first state uniformly at random.
\item The agent follows an epsilon-greedy policy with $\epsilon=0.1$.
\item Every action taken from the goal state leads to a zero-reward state that can never be escaped. Thus, to continue learning, we repeat the steps above. Note that we start a new/reset episode only after the agent reaches the goal state.
\item In the case of a tie in the value $\max _{a} Q(s, a)$ across actions $a$, we choose the "best" action randomly.
\item All transitions are noisy; that is, there is some non-zero probability of the agent moving somewhere different from its desired destination. For example, say the agent in in state $(0,0)$ and takes a "North" action, there is a non-zero chance that it actually ends up in state $(1,1)$.
\item Our $\gamma$ (discount factor) is set to $0.9$ and our $\alpha$ (learning rate) is set to $0.5$.
\end{itemize}
Note that the scale of the colors changes across the different plots, per the bar on the right of each plot.
At iteration 10,000, how close are the values plotted (our estimates of the true value of taking the best actions starting at the given state) to the actual value of taking the best actions starting from the given state?
In particular, what should the value of the bottom-right corner be? (To make this easier to think about, you can assume that the transitions are deterministic; that is, the robot always moves in the direction it is "aiming".) 		OpenState $(9,0)$ is 12 steps away from the goal, and so, without any errors, the robot will reach the goal in 12 steps. Given the discount factor of $0.9$, if we get a reward of 100 after 12 steps, the state has a value of about $100 * 0.9 \wedge 12=28$. The value in our table seems to be pretty close. It makes sense that it's slightly under, because we have a learning rate of 0.5. (If our learning rate was 1 , it'd be much closer to 28 .)To find the value of the bottom-right corner, we can use the Bellman equation for Q-learning and the given parameters. Since the transitions are deterministic, we can ignore the noise in the transitions.

The Bellman equation for Q-learning is:

Q(s, a) = (1 - α) * Q(s, a) + α * (R(s, a, s') + γ * max_a' Q(s', a'))

Given the parameters:
- α (learning rate) = 0.5
- γ (discount factor) = 0.9

Let's find the value of the bottom-right corner (state (9,9)).

From state (9,9), the best action is to move West (towards the goal state at (3,6)). The reward for this action is -1 (as it is not the goal state). The next state after taking this action is (8,9).

Now, we need to find the max_a' Q(s', a') for the next state (8,9). From state (8,9), the best action is to move West again. So, the Q-value for this action will be the same as the Q-value for the bottom-right corner (since we are following the same path to the goal).

Using the Bellman equation, we can write:

Q(9,9) = (1 - 0.5) * Q(9,9) + 0.5 * (-1 + 0.9 * Q(9,9))

Solving for Q(9,9):

0.5 * Q(9,9) = -0.5 + 0.45 * Q(9,9)
0.05 * Q(9,9) = -0.5
Q(9,9) = -10

So, at iteration 10,000, the value of the bottom-right corner should be approximately -10. This assumes that the Q-learning algorithm has converged to the optimal Q-values, and the transitions are deterministic.		​
21
81Mathematics18.2Principles of Discrete Applied MathematicsNone18.C06Midterm Exam 3Miller-Rabin Test2b1.25TextConsider the following three numbers:
$$
N_{1}=257=2^{8}+1, \quad N_{2}=4097=2^{12}+1, \quad N_{3}=1537=3 \cdot 2^{9}+1.
$$
For each of these three numbers, we would like to test whether it is a prime or not.
Let $a_{2}=233$. For each $k=1,2,4,8,16,32,64,128,256,512,1024,2048,4096$, we compute $a_{2}^{k}$ modulo $N_{2}=4097$. The results of the computation are:
$$
233,1028,3855,1206,1,1,1,1,1,1,1,1,1.
$$
Based on these results, do you think $N_{2}$ is a prime? explain.		OpenHere we fail the Miller-Rabin test, because the last element preceding a 1 is 1206 , which is not 1 or $-1$ modulo 4097 . So $N_{2}=4097$ is composite.Based on these results, we can conclude that $N_2$ is not a prime.

Here's the explanation: We are using Fermat's Little Theorem to test the primality of $N_2$. According to Fermat's Little Theorem, if $p$ is a prime number and $a$ is an integer not divisible by $p$, then:

$$
a^{p-1} \equiv 1 \pmod{p}
$$

In this case, we have $a_2 = 233$ and $N_2 = 4097$. We are given the results of $a_2^k \pmod{N_2}$ for various values of $k$. We can see that for $k = 256$, we have:

$$
233^{256} \equiv 1 \pmod{4097}
$$

However, $256$ is not equal to $4096 = N_2 - 1$. This means that Fermat's Little Theorem does not hold for $N_2$ and $a_2$, which implies that $N_2$ is not a prime number.		​
22
91Mathematics18.01Calculus INoneNoneProblem Set 3Integration2a0.03959873284TextOn Problem Set 2, Problem 1, you estimated the area of an annulus, where the inner circle has radius $r$ and the outer circle has radius $r+\Delta r$. When $\Delta r$ is small, you showed that the area is approximately $2 \pi r \Delta r$. If you want a reminder about this, you could read the solution to pset 2. We're going to build on that in an integration problem.
Suppose we have a disk of radius 3 meters. The disk has variable density. Near a point at distance $r$ from the center, the disk has density $4-r$ kilograms per square meter. So the densest part of the disk is at the center, where the density is $4 \mathrm{~kg} / \mathrm{m}^{2}$. The least dense part of the disk is at the edge, where the density is $1 \mathrm{~kg} / \mathrm{m}^{2}$.
Approximate the mass of the part of the disk where the distance to the center is between 2 meters and $2.1$ meters. (This part of the disk is an annulus with inner radius 2 and outer radius 2.1.) Which is the best approximation: $8 \pi k g$ or $4 \pi k g$ or $2 \pi k g$ or $(.8) \pi k g$ or (.4) $\pi k g$ or $(.2) \pi k g$?		Multiple ChoiceThe approximate area for this annulus is $2 \pi(2)(2.1-2)=0.4 \pi$ (units are $m^{2}$). The density on this annulus is approximately equal to $4-(2)=2$ (units are $\mathrm{kg} / \mathrm{m}^{2}$ ). Multiply the density times the area to get approximate mass of the annulus equal to $0.8 \pi$ (units are $\mathrm{kg}$).To approximate the mass of the part of the disk where the distance to the center is between 2 meters and 2.1 meters, we can use the formula for the area of an annulus and multiply it by the density at that distance.

The area of the annulus is approximately $2 \pi r \Delta r$, where $r = 2$ meters and $\Delta r = 0.1$ meters. So the area is approximately $2 \pi (2)(0.1) = 0.4 \pi$ square meters.

The density at a distance of 2 meters from the center is $4 - r = 4 - 2 = 2$ kilograms per square meter.

Now, we can approximate the mass by multiplying the area by the density: $0.4 \pi \times 2 = 0.8 \pi$ kilograms.

So, the best approximation for the mass of the part of the disk where the distance to the center is between 2 meters and 2.1 meters is $(.8) \pi kg$.		Correct
23
56Mathematics18.02Calculus II18.01NoneProblem Set 6Vector Fields5a0.3205128205TextLet $C$ be the unit circle, oriented counterclockwise, and consider the field $\vec{F}=\hat{i}+\hat{j}$. Which portions of $C$ contribute positively to the line integral of $\vec{F} ?$ Which portions contribute negatively?OpenConsider the parametrization $(x, y)=(\cos t, \sin t)$. The velocity vector $\vec{v}$ at a point $(x, y)$ on the curve $C$ is
$$
\vec{v}(x, y)=(-\sin t) \hat{i}+(\cos t) \hat{j}=-y \hat{i}+x \hat{j} \text {. }
$$
Thus, $\vec{F} \cdot \vec{v}(x, y)>0$ if and only if $-y+x>0$; and $\vec{F} \cdot \vec{v}(x, y)<0$ if and only if $-y+x<0$. That is to say, the portion $\left\{(\cos t, \sin t) \in C:-\frac{3 \pi}{4}<t<\frac{\pi}{4}\right\}$ contributes positively to the flux, while the portion $\left\{(\cos t, \sin t) \in C: \frac{\pi}{4}<t<\frac{5 \pi}{4}\right\}$ contributes negatively.		To determine the portions of the unit circle C that contribute positively or negatively to the line integral of the vector field $\vec{F} = \hat{i} + \hat{j}$, we need to analyze the direction of the vector field along the circle.

The unit circle C can be parameterized as $\vec{r}(t) = \langle \cos{t}, \sin{t} \rangle$ for $0 \leq t \leq 2\pi$. The tangent vector to the circle is given by the derivative of the parameterization with respect to t: $\vec{r'}(t) = \langle -\sin{t}, \cos{t} \rangle$.

Now, let's analyze the dot product of the vector field $\vec{F}$ and the tangent vector $\vec{r'}(t)$ at different portions of the circle:

1. First quadrant ($0 \leq t \leq \frac{\pi}{2}$): In this quadrant, both $\cos{t}$ and $\sin{t}$ are positive. Therefore, the dot product $\vec{F} \cdot \vec{r'}(t) = (-\sin{t} + \cos{t}) > 0$. This means that the vector field is aligned with the tangent vector, and the line integral contributes positively in this quadrant.

2. Second quadrant ($\frac{\pi}{2} \leq t \leq \pi$): In this quadrant, $\cos{t}$ is negative and $\sin{t}$ is positive. The dot product $\vec{F} \cdot \vec{r'}(t) = (-\sin{t} - \cos{t}) < 0$. This means that the vector field is opposite to the tangent vector, and the line integral contributes negatively in this quadrant.

3. Third quadrant ($\pi \leq t \leq \frac{3\pi}{2}$): In this quadrant, both $\cos{t}$ and $\sin{t}$ are negative. The dot product $\vec{F} \cdot \vec{r'}(t) = (\sin{t} - \cos{t}) > 0$. This means that the vector field is aligned with the tangent vector, and the line integral contributes positively in this quadrant.

4. Fourth quadrant ($\frac{3\pi}{2} \leq t \leq 2\pi$): In this quadrant, $\cos{t}$ is positive and $\sin{t}$ is negative. The dot product $\vec{F} \cdot \vec{r'}(t) = (\sin{t} + \cos{t}) < 0$. This means that the vector field is opposite to the tangent vector, and the line integral contributes negatively in this quadrant.

In conclusion, the portions of the unit circle C that contribute positively to the line integral of the vector field $\vec{F} = \hat{i} + \hat{j}$ are the first and third quadrants, while the portions that contribute negatively are the second and fourth quadrants.		Incorrect
24
42EECS6.102Elements of Software Construction6.101NoneMidterm Exam 2Concurrency3c0.48Text/**
* Immutable type representing a strand of DNA.
*/
class DNA {
/** omitted */
public constructor(bases: string) {
// omitted
}
/**
* @returns zero-based index of first occurence of `dna` as a substring of this strand,
* or undefined if `dna` never occurs.
*/
public find(dna: DNA): number|undefined {
// omitted
}
/**
* @returns true iff this and that are observationally equivalent
*/
public equalValue(that: DNA): boolean {
// omitted
}
// other code omitted
}
/**
* Immutable type representing a gene-editing process.
*/
interface Crispr {
/**
* Simulates this gene-editing process entirely in software, without using chemicals or a lab.
* @returns DNA strand that would result from this process
*/
simulate(): DNA;
/**
* Run this gene-editing process using the given `lab`.
* @returns the tube of `lab` in which the final DNA from this process
* can be found.
*/
async fabricate(lab: Lab): Promise<Tube>;
// other code omitted
}
/**
* Represents an already-existing DNA strand (a "precursor") in a gene-editing
* process. Precursors are bought premade from a supplier.
*/
class Precursor implements Crispr {
/**
* Make a gene-splicing step that results in the given `dna` strand.
*/
public constructor(private readonly dna: DNA) {
}
// other code omitted
}
/**
* Represents a gene-splicing step in a gene-editing process,
* which replaces all instances of one gene with another.
*/
class Splice implements Crispr {
/**
* Make a gene-splicing step that finds all occurrences of
* oldGene in target and substitutes newGene in place of each one.
*/
public constructor(
private readonly target: Crispr,
private readonly oldGene: Crispr,
private readonly newGene: Crispr
) {
}
// other code omitted
}
/**
* Mutable type controlling an automated gene-editing machine.
*/
class Lab {
/**
* Modifies the DNA in targetTube to replace all occurrences of the DNA from oldGeneTube with the
* DNA from newGeneTube.
* @returns a promise that fulfills with the same tube as targetTube, after the process is complet
*/
public async splice(targetTube: Tube, oldGeneTube: Tube, newGeneTube: Tube): Promise<Tube> {
// omitted
}
private tubeMap: Map<Tube, DNA> = new Map();
/**
* @returns a tube containing DNA strands corresponding to `dna`
*/
public async get(dna: DNA): Promise<Tube> {
for (const tube of this.tubeMap.keys()) {
if (this.tubeMap.get(tube).equalValue(dna)) {
return tube;
}
}
const tube = new Tube();
this.tubeMap.set(tube, dna);
await this.load(tube, dna); // "line 6a" is the load() call, "line 6b" is the await
return tube;
}
/**
* Ask a human to order premade DNA from a supplier
* and load it into the tube.
* @returns a promise that fulfills once this tube contains `dna`.
*/
private async load(tube: Tube, dna: DNA): Promise<void> {
// omitted
}
// other code omitted
}
/**
* Mutable type representing a test tube containing DNA.
*/
class Tube {
/** Make a new Tube. */
public constructor() {
}
// other code omitted
}
Suppose that:
• two different gene-editing processes A and B are running asynchronously using the same Lab
• A and B both call lab.get(dnaX) for the same precursor dnaX
• no other asynchronous processes are using lab
For each of the following interleavings, referring to the line numbers 1-7 in get() in the provided code, decide whether the
interleaving is impossible, leads to a race condition or deadlock, or runs safely; then explain your answer in one sentence.
A runs lines 1, 4, 5, 6a; then B runs lines 1, 4, 5, 6a; then A finishes lines 6b and 7, then B finishes lines 6b and 7.		Openimpossible; After A runs line 5, there will be at least one tube in tubeMap, so B must proceed from line 1 to line 2.The interleaving runs safely because both A and B will create separate tubes for the same DNA, but there is no race condition or deadlock as they do not interfere with each other's operations.Incorrect
25
257Mathematics18.01Calculus INoneNoneProblem Set 6Taylor Series10b0.07919746568TextThere is no formula for the antiderivative of $e^{-x^{2}}$, and so there is no formula for integrals like $\int_{0}^{\cdot 1} e^{-x^{2}} d x$. This type of integral is actually quite important in probability and statistics as we will see later. In this problem, we use Taylor series to approximate it.
There is a quicker way to find the Taylor series of $e^{-x^{2}}$. Earlier in the problem set you found the degree 3 Taylor series of $e^{-u}$ around $u=0$. Plug in $u=x^{2}$ and we get the Taylor series of $e^{-x^{2}}$ around $x=0$ to degree 6.		ExpressionRecall our formula for the degree 3 Taylor series $e^{-u}$ from earlier:
$$
1-u+\frac{1}{2} u^{2}-\frac{1}{6} u^{3}
$$
and plug in $u=x^{2}$ to get the degree 6 Taylor series for $e^{-x^{2}}$ :
$$
T(x)=1-x^{2}+\frac{1}{2} x^{4}-\frac{1}{6} x^{6} .
$$		To find the Taylor series of $e^{-x^2}$ around $x=0$ to degree 6, we first recall the degree 3 Taylor series of $e^{-u}$ around $u=0$. The Taylor series of $e^{-u}$ is given by:

$e^{-u} = 1 - u + \frac{u^2}{2!} - \frac{u^3}{3!} + \cdots$

Now, we plug in $u=x^2$ to get the Taylor series of $e^{-x^2}$ around $x=0$ to degree 6:

$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$

So, the degree 6 Taylor series of $e^{-x^2}$ around $x=0$ is:

$e^{-x^2} \approx 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!}$		Correct
26
55EECS6.18Computer Systems Engineering6.1010, 6.1910NoneMidterm Exam 1DCTCP and MapReduce9c0.45TextWe have a large-scale web application using the partition/aggregate design pattern with many workers and real-time requirements. Assume there is NO background traffic (no other applications on the network) and the responses from workers are all small (1-2 packets each).
The MapReduce paper doesn't mention incast as an issue. Why not?
(a) Actually it does, but MapReduce talks about "stragglers" instead of "incast."
(b) The MapReduce paper doesn't describe many details of the reshuffling that happens between stages.
(c) Incast problems only started after the MapReduce paper had already been published.
(d) MapReduce operates in rigidly-scheduled stages, so it's impossible to have multiple messages arrive at a single switch at the same time.
(e) Google used exotic custom switches, described in the paper, to avoid these kinds of networking problems.		Multiple Choice(b).(b) The MapReduce paper doesn't describe many details of the reshuffling that happens between stages.Correct
27
63Mathematics18.102Introduction to Functional Analysis18.C06, 18.100BNoneProblem Set 9Bounded Linear Operators40.5TextShow that if $V \in \mathcal{C}^{0}[0, \pi]$ then the linear map
$$
Q_{V}: H_{0}^{2}([0, \pi]) \ni u \longmapsto-F_{2}+V u \in L^{2}(0, \pi)
$$
is bounded and reduces to
$$
u \longmapsto-u^{\prime \prime}+V u \text { on } \mathcal{C}_{0}^{2}([0, \pi])
$$		OpenLet $\mathscr{H}=H_{0}^{2}(0, \pi), \mathscr{L}=L^{2}(0, \pi)$. The map $u \rightarrow-F_{2}$ is a bounded map from $\mathscr{H}$ to $\mathscr{L}$. In fact is maps $u=\sum c_{k} \sin k x$ to $-F_{2}=\sum k^{2} c_{k} \sin k x$ and $\|u\|_{\mathscr{H}}=\left\|F_{2}\right\|_{\mathscr{L}}$. The map $u \rightarrow u$ is a bounded map from $\mathscr{H}$ to $\mathscr{L}$ as $\|u\|_{\mathscr{H}} \geq\|u\|_{\mathscr{L}}$. Finally, multiplication by $V$ is a bounded map from $\mathscr{L}$ to $\mathscr{L}$. Combining the above, we conclude that $Q_{V}$ is a bounded map from $\mathscr{H}$ to $\mathscr{L}$. It reduces to $-u^{\prime \prime}+V u$ by Problem $9.3$.To show that the linear map $Q_V$ is bounded, we need to show that there exists a constant $C > 0$ such that
$$
\|Q_V u\|_{L^2(0, \pi)} \leq C \|u\|_{H_0^2([0, \pi])} \quad \forall u \in H_0^2([0, \pi]).
$$

First, let's recall the definitions of the Sobolev space $H_0^2([0, \pi])$ and the $L^2$ norm. The Sobolev space $H_0^2([0, \pi])$ consists of functions $u \in H^2([0, \pi])$ such that $u(0) = u(\pi) = 0$. The $L^2$ norm of a function $f$ is defined as
$$
\|f\|_{L^2(0, \pi)} = \left(\int_0^\pi |f(x)|^2 dx\right)^{1/2}.
$$

Now, let's consider the action of $Q_V$ on a function $u \in H_0^2([0, \pi])$. We have
$$
Q_V u = -F_2 + Vu,
$$
where $F_2$ is the second-order Fourier transform of $u$. Since $u \in H_0^2([0, \pi])$, we know that $u$ is twice continuously differentiable and has compact support. Therefore, $F_2$ is well-defined and bounded.

Now, let's estimate the $L^2$ norm of $Q_V u$. We have
$$
\|Q_V u\|_{L^2(0, \pi)} = \left\| -F_2 + Vu \right\|_{L^2(0, \pi)} \leq \|F_2\|_{L^2(0, \pi)} + \|Vu\|_{L^2(0, \pi)}.
$$

Since $F_2$ is bounded, there exists a constant $C_1 > 0$ such that $\|F_2\|_{L^2(0, \pi)} \leq C_1 \|u\|_{H_0^2([0, \pi])}$. Moreover, since $V \in \mathcal{C}^0([0, \pi])$, there exists a constant $C_2 > 0$ such that $\|Vu\|_{L^2(0, \pi)} \leq C_2 \|u\|_{L^2(0, \pi)}$. Combining these estimates, we obtain
$$
\|Q_V u\|_{L^2(0, \pi)} \leq (C_1 + C_2) \|u\|_{H_0^2([0, \pi])}.
$$

Thus, the linear map $Q_V$ is bounded.

Now, let's show that $Q_V$ reduces to $u \mapsto -u'' + Vu$ on $\mathcal{C}_0^2([0, \pi])$. Let $u \in \mathcal{C}_0^2([0, \pi])$. Then, $u$ is twice continuously differentiable and has compact support. Moreover, since $u \in H_0^2([0, \pi])$, we have $u(0) = u(\pi) = 0$. Therefore, the second-order Fourier transform $F_2$ of $u$ is given by
$$
F_2 = -u''.
$$

Hence, we have
$$
Q_V u = -F_2 + Vu = -(-u'') + Vu = u'' + Vu.
$$

Thus, the linear map $Q_V$ reduces to $u \mapsto -u'' + Vu$ on $\mathcal{C}_0^2([0, \pi])$.		Correct
28
103Mathematics18.102Introduction to Functional Analysis18.C06, 18.100BNoneFinal ExamOrthogonal Complements84.5TextSuppose that $f \in L^{2}(\mathbb{R})$ is such that there exists a function $v \in L^{2}(\mathbb{R})$ satisfying
$$
\int_{\mathbb{R}} f \phi^{\prime}=\int_{\mathbb{R}} v \widehat{\phi} \quad \forall \phi \in \mathcal{S}(\mathbb{R})
$$
where $\widehat{\phi}$ is the Fourier transform of $\phi$. Show that $f \in C_{0}(\mathbb{R}) \subset L^{2}(\mathbb{R})$, where $C_{0}(\mathbb{R})$ is the space of continuous functions on $\mathbb{R}$ with zero limit at $\pm \infty$.
You may use that if $h$ is a locally $L^{2}$ function on $\mathbb{R}$ such that $\int h \phi=0$ for every $\phi \in C_{c}^{\infty}(\mathbb{R})$ then $h=0$ a.e. (as $C_{c}^{\infty}(I)$ is dense in $L^{2}(I)$ for every interval $I$).		OpenLet $\psi:=\widehat{\phi}$. Then $\phi(x)=\frac{1}{2 \pi} \widehat{\psi}(-x)$, so
$$
\int_{\mathbb{R}} f(-x) \widehat{i \xi \psi}(x)=\int v \psi \quad \forall \psi \in \mathcal{S}(\mathbb{R}).
$$
So
$$
\int_{\mathbb{R}} \widehat{f}(-\xi) i \xi \psi(\xi)=\int v \psi
$$
from which it follows that the locally $L^{2}$-function $h(\xi):=i \xi \widehat{f}(-\xi)-v(\xi)$ is orthogonal to $\mathcal{S}(\mathbb{R})$, hence to its subspace $C_{c}^{\infty}(\mathbb{R})$. Thus $h=0$ a.e., i.e., $\xi \widehat{f} \in L^{2}(\mathbb{R})$. This means $\widehat{f} \in L^{1}(\mathbb{R})$, since
$$
\left(\int|\widehat{f}|\right)^{2} \leq \int \frac{1}{1+\xi^{2}} \cdot \int\left(1+\xi^{2}\right)|\widehat{f}|^{2}=\pi \int\left(1+\xi^{2}\right)|\widehat{f}|^{2}<\infty.
$$
so $f$ is continuous and goes to zero at infinity.		To show that $f \in C_{0}(\mathbb{R}) \subset L^{2}(\mathbb{R})$, we will first show that $f$ is continuous and then show that it has zero limit at $\pm \infty$.

1. Continuity of $f$:

Let $\phi \in \mathcal{S}(\mathbb{R})$ be a Schwartz function. Then, we have
$$
\int_{\mathbb{R}} f \phi^{\prime}=\int_{\mathbb{R}} v \widehat{\phi}.
$$

Taking the inverse Fourier transform of both sides, we get
$$
\int_{\mathbb{R}} f \phi^{\prime}=\int_{\mathbb{R}} v \check{\widehat{\phi}},
$$
where $\check{\widehat{\phi}}$ is the inverse Fourier transform of $\widehat{\phi}$.

Now, we know that the inverse Fourier transform of the Fourier transform of a function is the original function itself, up to a constant factor. Therefore, we have
$$
\int_{\mathbb{R}} f \phi^{\prime}=\int_{\mathbb{R}} v \phi.
$$

Integrating by parts, we get
$$
\int_{\mathbb{R}} f \phi^{\prime}=-\int_{\mathbb{R}} f^{\prime} \phi.
$$

Comparing the two equations, we have
$$
-\int_{\mathbb{R}} f^{\prime} \phi=\int_{\mathbb{R}} v \phi \quad \forall \phi \in \mathcal{S}(\mathbb{R}).
$$

Since this holds for all Schwartz functions $\phi$, we can conclude that $f^{\prime} = -v$ almost everywhere. Thus, $f$ has a weak derivative, which implies that $f$ is absolutely continuous. Absolute continuity implies continuity, so $f$ is continuous.

2. Zero limit at $\pm \infty$:

We know that $f \in L^{2}(\mathbb{R})$, which means that $\int_{\mathbb{R}} |f(x)|^2 dx < \infty$. Now, let's consider the limit of $f(x)$ as $x \to \infty$.

Let $\epsilon > 0$ be given. Since $f \in L^{2}(\mathbb{R})$, there exists an $R > 0$ such that
$$
\int_{|x| > R} |f(x)|^2 dx < \epsilon^2.
$$

Now, for any $x > R$, we have
$$
|f(x)|^2 \leq \frac{1}{2R} \int_{x-R}^{x+R} |f(t)|^2 dt \leq \frac{1}{2R} \int_{|t| > R} |f(t)|^2 dt < \frac{\epsilon^2}{2R}.
$$

Taking the square root of both sides, we get
$$
|f(x)| < \frac{\epsilon}{\sqrt{2R}}.
$$

Since $\epsilon > 0$ was arbitrary, we can conclude that $\lim_{x \to \infty} f(x) = 0$. Similarly, we can show that $\lim_{x \to -\infty} f(x) = 0$.

Thus, $f \in C_{0}(\mathbb{R}) \subset L^{2}(\mathbb{R})$.		​
29
3Mathematics18.404Theory of Computation6.1210/18.200NoneProblem Set 1Regular Expression3b0.5555555556TextFor any regular expression $R$ and $k\ge0$, let $R^k$ be
$R$ self-concatenated $k$ times, $\smash{\underbrace{RR\cdots R}_k}$. \\
Let $\SS=\{\st0,\st1\}$.
Let $B=\set{\st0^ku\st1^k} k\ge1$ and $ u\in\st1\SSS\setend$.
Show $B$ is not regular.		OpenAssume for contradiction that $B$ is regular. Use the pumping lemma to get the pumping length $p$. Letting $s=0^{p} 11^{p}$ we have $s \in B$ (here $u=1$ ) and so we can divide up $s=x y z$ according to the conditions of the pumping lemma. By condition 3, $x y$ has only 0s, hence the string $x y y z$ is $0^{l} 11^{p}$ for some $l>p$. But then $0^{l} 11^{p}$ isn't equal to $0^{k} 1 u 1^{k}$ for any $u \in \Sigma^{*}$ and $k$, because the left-hand part of the string requires $k=l$ and the right-hand part requires $k \leq p$. Both together are impossible, because $l>p$. That contradicts the pumping lemma and we conclude that $B$ isn't regular.To show that $B$ is not regular, we will use the Pumping Lemma. The Pumping Lemma states that for any regular language $L$, there exists a constant $p$ (the pumping length) such that any string $s$ in $L$ with length at least $p$ can be divided into three parts, $s = xyz$, where:

1. $|y| > 0$ (i.e., $y$ is not the empty string)
2. $|xy| \le p$
3. For all $i \ge 0$, the string $xy^iz$ is also in $L$

Assume, for the sake of contradiction, that $B$ is regular. Then, there exists a constant $p$ for $B$ according to the Pumping Lemma. Let's choose the string $s = \st0^p\st1^pu\st1^p \in B$, where $u \in \st1\SSS\setend$. Note that $|s| \ge p$.

Now, we need to divide $s$ into three parts, $s = xyz$, such that the conditions of the Pumping Lemma hold. Since $|xy| \le p$, both $x$ and $y$ must consist of only $\st0$s. Let $x = \st0^a$, $y = \st0^b$, and $z = \st0^{p-a-b}\st1^pu\st1^p$, where $a \ge 0$, $b > 0$, and $a + b \le p$.

Now, let's consider the string $xy^2z = \st0^a\st0^{2b}\st0^{p-a-b}\st1^pu\st1^p = \st0^{p+b}\st1^pu\st1^p$. According to the Pumping Lemma, this string should also be in $B$. However, this string has $p+b$ occurrences of $\st0$ followed by $p$ occurrences of $\st1$, which contradicts the definition of $B$ (since $k$ should be equal for both $\st0$ and $\st1$).

This contradiction implies that our assumption that $B$ is regular is false. Therefore, $B$ is not regular.		​
30
72Mathematics18.03Differential EquationsNone18.02Problem Set 6Linear Algebra4b0.2412868633TextLet $S_{3}$ be the vector space of polynomials of degree at most 3 . Consider the linear map $T: S_{3} \rightarrow \mathbb{R}^{2}$ defined by
$$
T(p)=\left(p(0), p^{\prime}(1)\right)
$$
You may assume this map is linear (it follows from problem (4)).
Find a basis for $\operatorname{Ker}(T):=\left\{p \in S_{3}: T(p)=0\right\}$ and compute the dimension of this vector space. Note in particular that we have the equality
$$
\operatorname{dim} \operatorname{Ker}(T)+\operatorname{dim} \operatorname{Im}(T)=\operatorname{dim} S_{3}
$$		OpenLet $p(x)=a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$ be an element of $S_{3}$. Notice that $T(p)=\left(a_{0}, 3 a_{3}+\right.$ $\left.2 a_{2}+a_{1}\right)$. Hence, if $p$ belongs to the kernel of $T$, we have $a_{0}=0$ and $a_{1}=-3 a_{3}-2 a_{2}$. Consequently, we have $p(x)=a_{3}\left(x^{3}-3 x\right)+a_{2}\left(x^{2}-2 x\right)$. We see that the kernel of $T$ is included in the span of $x^{3}-3 x$ and $x^{2}-2 x$. Reciprocally, it follows from our computation above that $x^{3}-3 x$ and $x^{2}-2 x$ belongs to the kernel of $T$. Thus, the kernel of $T$ is exactly the span of $x^{3}-3 x$ and $x^{2}-2 x$.
Let us prove that $x^{3}-3 x$ and $x^{2}-2 x$ are linearly independent. Let $\lambda$ and $\mu$ be real numbers such that $\lambda\left(x^{3}-3 x\right)+\mu\left(x^{2}-2 x\right)=0$ for every $x \in \mathbb{R}$. Differentiating this equality three times and evaluating at $x=1$, we find that $\lambda=0$. Hence, we have $\mu\left(x^{2}-2 x\right)=0$, and evaluating at $x=1$, we find that $\mu=0$. We proved that $\lambda=\mu=0$, it follows that $x^{3}-3 x$ and $x^{2}-2 x$ are linearly independent.
Consequently, $x^{3}-3 x$ and $x^{2}-2 x$ form a basis of the kernel of $T$, which is thus of dimension 2. It follows from the previous question that the dimension of the image of $T$ is 2 , so that we have
$$
\operatorname{dim} \operatorname{Ker}(T)+\operatorname{dim} \operatorname{Im}(T)=2+2=4=\operatorname{dim} S_{3} .
$$		To find a basis for the kernel of T, we need to find all polynomials p in S_3 such that T(p) = 0. This means that p(0) = 0 and p'(1) = 0. Let p(x) = ax^3 + bx^2 + cx + d. Then, p(0) = d = 0, so p(x) = ax^3 + bx^2 + cx.

Now, we need to find p'(x) and set p'(1) = 0. The derivative of p(x) is p'(x) = 3ax^2 + 2bx + c. Now, we set p'(1) = 0:

3a(1)^2 + 2b(1) + c = 0
3a + 2b + c = 0

Now, we need to find a basis for the solutions to this equation. We can solve for one of the variables in terms of the others. For example, let's solve for c:

c = -3a - 2b

Now, we can write the polynomial p(x) in terms of a and b:

p(x) = ax^3 + bx^2 - 3ax - 2bx = a(x^3 - 3x) + b(x^2 - 2x)

The kernel of T consists of all polynomials of this form. The basis for the kernel of T is given by the set {x^3 - 3x, x^2 - 2x}. This is a set of 2 linearly independent vectors, so the dimension of the kernel of T is 2.

Now, we can use the equality dim(Ker(T)) + dim(Im(T)) = dim(S_3) to find the dimension of the image of T. Since the dimension of S_3 is 4 (as it consists of polynomials of degree at most 3), we have:

dim(Ker(T)) + dim(Im(T)) = 4
2 + dim(Im(T)) = 4

Thus, the dimension of the image of T is 2.		Correct
31
74EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneExercise 10State Machines1a0.1041666667TextFor each of the following state machines, provide the output sequence $\left[y_{1}, y_{2}, \ldots, y_{T}\right]$ given the input sequence $\left[x_{1}, x_{2}, \ldots, x_{T}\right]$. Notice that the inputs start with subscript 1:
Input: $[0,1,2,1]$
$s_{0}=0$
$s_{t}=f_{s}\left(s_{t-1}, x_{t}\right)=\max \left(s_{t-1}, x_{t}\right)$
$y_{t}=f_{o}\left(s_{t}\right)=2 \times s_{t}$
Enter a Python list of four numbers:		Expression[0, 2, 4, 4].
Passing in $x_{1}$:
$s_{0}=0$
$s_{1}=f_{s}\left(s_{0}, x_{1}\right)=\max (0,0)=0$
$y_{1}=f_{o}\left(s_{1}\right)=2 \times 0=0$
Passing in $x_{2}$:
$s_{1}=0$
$s_{2}=f_{s}\left(s_{1}, x_{2}\right)=\max (0,1)=1$
$y_{2}=f_{o}\left(s_{2}\right)=2 \times 1=2$
Similar calculations for $x_{3}$ and $x_{4}$.
You can also solve the problem with code:
def f_s(s, x_t):
return max(s, x_t)
def f_o(s):
return s * 2
s = 0
inputs = [0, 1, 2, 1]
outputs = []
for x in inputs:
s = f_s(s, x)
y = f_o(s)
outputs.append(y)
print(outputs)		To find the output sequence, we need to follow the given state machine rules and apply them to the input sequence. Let's calculate the output sequence step by step:

1. $s_0 = 0$
2. $s_1 = f_s(s_0, x_1) = \max(0, 0) = 0$
3. $s_2 = f_s(s_1, x_2) = \max(0, 1) = 1$
4. $s_3 = f_s(s_2, x_3) = \max(1, 2) = 2$
5. $s_4 = f_s(s_3, x_4) = \max(2, 1) = 2$

Now, let's calculate the output sequence using the function $f_o(s_t)$:

1. $y_1 = f_o(s_1) = 2 \times 0 = 0$
2. $y_2 = f_o(s_2) = 2 \times 1 = 2$
3. $y_3 = f_o(s_3) = 2 \times 2 = 4$
4. $y_4 = f_o(s_4) = 2 \times 2 = 4$

So, the output sequence is [0, 2, 4, 4].		Correct
32
382Mathematics18.01Calculus INoneNoneProblem Set 8Variance10a0.1583949314TextSuppose that we flip a coin a hundred times and count how many heads we get. On average the number of heads is 50 . On the other hand, if we flip a coin a hundred times, it's unlikely that we will get exactly 50 heads. Usually the number of heads will be somewhat above average or somewhat below average. The variance helps understand how far away the number of heads typically is from the mean. (Remember, mean is another word for average.)
If we have a probability distribution for $x$ with mean $M$, then the variance is the average value of $(x-M)^{2}$. Let's illustrate this with an example.
Example. Suppose we slip a coin two times and count the number of heads. The probability distribution for the number of heads is
\begin{tabular}{|c|c|}
\hline Number of heads & Probability \\
\hline 0 & $1 / 4$ \\
\hline 1 & $1 / 2$ \\
\hline 2 & $1 / 4$ \\
\hline
\end{tabular}
The average number of heads is $M=1$. To find the variance, we want to find the average value of (The number of heads $-M)^{2}$. To compute we first include this information on our table:
\begin{tabular}{|c|c|c|}
\hline Number of heads & Probability & (The number of heads $-M)^{2}$ \\
\hline 0 & $1 / 4$ & $(0-1)^{2}=1$ \\
\hline 1 & $1 / 2$ & $(1-1)^{2}=0$ \\
\hline 2 & $1 / 4$ & $(2-1)^{2}=1$ \\
\hline
\end{tabular}
The variance is the average value of (The number of heads $-M)^{2}$ which is (1/4) . $1+(1 / 2) \cdot 0+(1 / 4) \cdot 1=(2 / 4)$.
Suppose we flip a coin three times and count the number of heads. Find the mean and the variance.		NumericalThe mean value $M$ is $0(1 / 8)+1(3 / 8)+2(3 / 8)+3(1 / 8)=3 / 2$.
The variance is $(9 / 4)(1 / 8)+(1 / 4)(3 / 8)+(1 / 4)(3 / 8)+(9 / 4)(1 / 8)=\frac{9+3+3+9}{32}=$ $\frac{24}{32}=3 / 4$.		To find the mean and variance when flipping a coin three times, we first need to determine the probability distribution for the number of heads. The possible outcomes are 0, 1, 2, or 3 heads. The probabilities for each outcome are as follows:

- 0 heads: There is only 1 way to get 0 heads (all tails), so the probability is 1/8.
- 1 head: There are 3 ways to get 1 head (HTT, THT, TTH), so the probability is 3/8.
- 2 heads: There are 3 ways to get 2 heads (HHT, HTH, THH), so the probability is 3/8.
- 3 heads: There is only 1 way to get 3 heads (all heads), so the probability is 1/8.

Now, we can calculate the mean (M) as follows:
M = (0 * 1/8) + (1 * 3/8) + (2 * 3/8) + (3 * 1/8) = 0 + 3/8 + 6/8 + 3/8 = 12/8 = 1.5

Next, we will calculate the variance. First, we need to find the squared difference between the number of heads and the mean for each outcome:

\begin{tabular}{|c|c|c|}
\hline Number of heads & Probability & (The number of heads $-M)^{2}$ \\
\hline 0 & $1 / 8$ & $(0-1.5)^{2}=2.25$ \\
\hline 1 & $3 / 8$ & $(1-1.5)^{2}=0.25$ \\
\hline 2 & $3 / 8$ & $(2-1.5)^{2}=0.25$ \\
\hline 3 & $1 / 8$ & $(3-1.5)^{2}=2.25$ \\
\hline
\end{tabular}

Now, we can calculate the variance as the average value of (The number of heads $-M)^{2}$:

Variance = (1/8) * 2.25 + (3/8) * 0.25 + (3/8) * 0.25 + (1/8) * 2.25 = 0.28125 + 0.09375 + 0.09375 + 0.28125 = 0.75

So, when flipping a coin three times, the mean number of heads is 1.5, and the variance is 0.75.		Correct
33
28Mathematics18.6Probability and Random Variables18.02NoneProblem Set 3Cumulative Distribution Function3b0.25TextLet $X$ be a random variable with cumulative distribution function $F$.
What is the cumulative distribution function of $aX+b$, where $a$ and $b$ are constants and $a \neq 0$? (Remember that $a$ could be positive or negative.)		ExpressionLet $Y = aX + b$.
\textbf{Case 1}: $a > 0$.
$F_Y(c) = P\{Y \leq c\} = P\{aX + b \leq c\} = P\{X \leq \frac{c - b}{a}\} = F_X(\frac{c - b}{a})$.
\textbf{Case 2}: $a < 0$.
$F_Y(c) = P\{Y \leq c\} = P\{aX + b \leq c\} = P\{X \geq \frac{c - b}{a}\} = 1 - P\{X < \frac{c - b}{a}\} = 1 - P\{X \leq \frac{c - b}{a}\} + P\{X = \frac{c - b}{a}\} = 1 - F_X(\frac{c - b}{a}) + p_X(\frac{c - b}{a})$.		The cumulative distribution function (CDF) of $aX+b$ can be found by substituting $aX+b$ into the CDF of $X$. Let $Y = aX+b$, then we want to find the CDF of $Y$, denoted as $F_Y(y)$.

To do this, we can write:

$F_Y(y) = P(Y \leq y) = P(aX+b \leq y)$

Now, we can solve for $X$:

$P(aX+b \leq y) = P\left(X \leq \frac{y-b}{a}\right)$

Since $X$ has CDF $F$, we can write:

$F_Y(y) = F\left(\frac{y-b}{a}\right)$

So, the cumulative distribution function of $aX+b$ is $F_Y(y) = F\left(\frac{y-b}{a}\right)$.		Incorrect
34
417Mathematics18.01Calculus INoneNoneMidterm Exam 1Critical Points3a0.3333333333TextSuppose that $f(x)=4 x^{3}-6 x^{2}+1$.
Find the values of $x$ where $f^{\prime}(x)=0$. These are called the critical points of $f$.		NumericalFirst we calculate $f^{\prime}(x)$. We find that $f^{\prime}(x)=3 \times 4 x^{3-1}-2 \times 6 x^{2-1}+0=12 x^{2}-12 x$. You might be able to see straight away that $12\left(x^{2}-x\right)=0$ at $x=0$ and $x=1$. Otherwise we can solve by plugging in the quadratic formula, giving:
$$
x=\frac{-(-12) \pm \sqrt{(-12)^{2}-4 \times 12 \times 0}}{2 \times 12}=\frac{12 \pm \sqrt{12^{2}}}{24}=\frac{12 \pm 12}{24}=0,1
$$		To find the critical points of the function $f(x)=4x^3-6x^2+1$, we first need to find its derivative, $f'(x)$.

Using the power rule, we have:

$f'(x) = \frac{d}{dx}(4x^3-6x^2+1) = 12x^2 - 12x$

Now, we need to find the values of $x$ where $f'(x) = 0$. So, we set $f'(x)$ equal to zero and solve for $x$:

$12x^2 - 12x = 0$

We can factor out a $12x$:

$12x(x - 1) = 0$

Now, we set each factor equal to zero and solve for $x$:

$12x = 0 \Rightarrow x = 0$

$x - 1 = 0 \Rightarrow x = 1$

So, the critical points of $f(x)$ are $x = 0$ and $x = 1$.		Correct
35
51EECS6.411Representation, Inference, and Reasoning in AI6.1010, 6.1210, 18.600NoneProblem Set 1Monte-Carlo Tree Search4bi0.1302083333TextConsider MCTS on a problem where the initial state $s_{0}$ has actions two actions $a_{0}$ and $a_{1}$. The UCB parameter $C$ is $1.4$. Suppose the search has completed 8 full iterations:
\begin{itemize}
\item It selected action $a_{0} 3$ times, receiving utilities $[0.1,0.7,0.3]$.
\item It selected action $a_{1} 5$ times, receiving utilities $[0.4,0.4,0.4,0.4,0.4]$.
\end{itemize}
On the 9-th iteration, what is the UCB value for action $a_{0}$ when expanding from $s_{0}$? (Enter a number accurate to 2 decimal places).		Numerical1.532.To calculate the UCB value for action $a_{0}$, we need to use the UCB1 formula:

$$UCB(a) = \bar{X}_{a} + C \sqrt{\frac{2 \ln N}{n_{a}}}$$

where $\bar{X}_{a}$ is the average utility of action $a$, $C$ is the UCB parameter, $N$ is the total number of iterations, and $n_{a}$ is the number of times action $a$ has been selected.

For action $a_{0}$:

- $\bar{X}_{a_{0}} = \frac{0.1 + 0.7 + 0.3}{3} = \frac{1.1}{3} = 0.3667$
- $N = 8$
- $n_{a_{0}} = 3$
- $C = 1.4$

Now, we can plug these values into the UCB1 formula:

$$UCB(a_{0}) = 0.3667 + 1.4 \sqrt{\frac{2 \ln 8}{3}}$$

Calculating the value:

$$UCB(a_{0}) = 0.3667 + 1.4 \sqrt{\frac{2 \ln 8}{3}} \approx 0.3667 + 1.4 \times 1.16 \approx 0.3667 + 1.624 \approx 1.99$$

So, the UCB value for action $a_{0}$ when expanding from $s_{0}$ on the 9-th iteration is approximately 1.99 (accurate to 2 decimal places).		​
36
32EECS18.C06Linear Algebra and Optimization18.02NoneProblem Set 2Vector Spaces2c0.06172839506TextIs the following set a vector space (with the usual definitions of multiplication by real scalars and addition)?
The set of vectors that solve the equation $A x=b$ for some fixed nonzero $b$.		OpenFalse. In general, this set does not include the origin, so it cannot be a subspace.No, the set of vectors that solve the equation $Ax = b$ for some fixed nonzero $b$ does not form a vector space with the usual definitions of multiplication by real scalars and addition.

To be a vector space, a set must satisfy the following conditions:

1. Closure under addition: If $u$ and $v$ are in the set, then $u + v$ must also be in the set.
2. Closure under scalar multiplication: If $u$ is in the set and $c$ is a scalar, then $cu$ must also be in the set.
3. Existence of additive identity: There must be a zero vector in the set such that $u + 0 = u$ for all $u$ in the set.
4. Existence of additive inverse: For every vector $u$ in the set, there must be a vector $-u$ in the set such that $u + (-u) = 0$.

Let's consider the given set of vectors that solve the equation $Ax = b$ for some fixed nonzero $b$. If $x_1$ and $x_2$ are two vectors in the set, then $Ax_1 = b$ and $Ax_2 = b$. Now, let's check the conditions:

1. Closure under addition: $(x_1 + x_2)$ is not necessarily in the set, because $A(x_1 + x_2) = Ax_1 + Ax_2 = b + b = 2b$, which is not equal to $b$.
2. Closure under scalar multiplication: $cx_1$ is not necessarily in the set, because $A(cx_1) = c(Ax_1) = cb$, which is not equal to $b$ for $c \neq 1$.
3. Existence of additive identity: The zero vector does not satisfy the equation $Ax = b$ for a nonzero $b$, so there is no additive identity in the set.
4. Existence of additive inverse: If $x_1$ is in the set, then $-x_1$ is not necessarily in the set, because $A(-x_1) = -Ax_1 = -b$, which is not equal to $b$.

Since the set does not satisfy the conditions for a vector space, it is not a vector space with the usual definitions of multiplication by real scalars and addition.		​
37
416Mathematics18.01Calculus INoneNoneMidterm Exam 1Linear Approximation2b0.8333333333TextApproximate the cube root of the number $8.24$. Is the true value closest to 2.02 or $2.03$ or $2.04$ or $2.06 ?$ Explain your reasoning.OpenAs above, we have the linear approximation $(2+\Delta x)^{3} \approx 8+12 \Delta x$. We are looking for $(2+\Delta x)^{3} \approx 8.24$ so we solve $8.24 \approx 8+12 \Delta x \Rightarrow .24=12 \Delta x \Rightarrow \Delta x=.02$. This gives us $\sqrt[3]{8.24} \approx 2+.02=2.02$.To approximate the cube root of the number $8.24$, we can first consider the cube root of a nearby perfect cube. The closest perfect cube to $8.24$ is $8$, which is $2^3$. So, the cube root of $8$ is $2$.

Now, let's see which of the given options is closest to the cube root of $8.24$. We can do this by cubing each of the options and comparing the results to $8.24$.

1. $2.02^3 = 2.02 \times 2.02 \times 2.02 \approx 8.24$
2. $2.03^3 = 2.03 \times 2.03 \times 2.03 \approx 8.36$
3. $2.04^3 = 2.04 \times 2.04 \times 2.04 \approx 8.49$
4. $2.06^3 = 2.06 \times 2.06 \times 2.06 \approx 8.74$

Comparing the results, we can see that $2.02^3$ is the closest to $8.24$. Therefore, the true value of the cube root of $8.24$ is closest to $2.02$.		Correct
38
85EECS6.411Representation, Inference, and Reasoning in AI6.1010, 6.1210, 18.600NoneProblem Set 2Propositional Logic3biii0.0744047619TextConsider a domain with propositions $\mathrm{A}, \mathrm{B}, \mathrm{C}$, and $\mathrm{D}$, and the particular model $m=\{A=t, B=f, C=t, D=f\}$. For each of these sentences, indicate whether it is valid, unsatisifiable, not valid but true in $\mathrm{m}$, or not unsatisifiable but false in $\mathrm{m}$.
$$
B \Rightarrow C \wedge D
$$		Multiple Choicenot valid, but true in m.Given the model $m=\{A=t, B=f, C=t, D=f\}$, let's evaluate the sentence $B \Rightarrow C \wedge D$.

Recall that $P \Rightarrow Q$ is equivalent to $\neg P \vee Q$. So, we can rewrite the sentence as:
$$
\neg B \vee (C \wedge D)
$$

Now, let's substitute the truth values from the model $m$:
$$
\neg f \vee (t \wedge f)
$$

Evaluating the expression, we get:
$$
t \vee (t \wedge f) \Rightarrow t \vee f \Rightarrow t
$$

Since the sentence evaluates to true in the given model $m$, it is not unsatisfiable but true in $m$.		Mostly Correct
39
61Mathematics18.6Probability and Random Variables18.02NoneProblem Set 5Probability6b0.35TextSuppose that $A$, $B$, $C$, are independent random variables, each being uniformly distributed over $(0,1)$.
What is the probability that all of the roots of the equation $Ax^2 + Bx + C = 0$ are real?		NumericalThe roots of the equation $Ax^2 + Bx + C = 0$ are real if $B^2 - 4AC \geq 0$. The desired probability, $P\{B^2 - 4AC \geq 0\}$, is obtained as follows
$P\{B^2 - 4AC \geq 0\} = 1 - P\{B^2 - 4AC < 0\} = 1 - \iiint_{B^2 - 4AC < 0} f(x, y, z) \,dx\,dy\,dz = 1 - \int_{0}^{1} \int_{\frac{y^2}{4}}^{1} \int_{\frac{y^2}{4z}}^{1} \,dx\,dz\,dy = 1 - \int_{0}^{1} \int_{\frac{y^2}{4}}^{1} (1 - \frac{y^2}{4z}) \,dz\,dy = 1 - \int_{0}^{1} (1 - \frac{y^2}{4} + \frac{y^2}{4}\ln({\frac{y^2}{4}})) \,dy = \frac{1}{6}\ln({2}) + \frac{5}{36}$.		To find the probability that all the roots of the equation $Ax^2 + Bx + C = 0$ are real, we need to consider the discriminant of the quadratic equation. The discriminant is given by the formula:

$D = B^2 - 4AC$

For the roots to be real, the discriminant must be greater than or equal to zero:

$D \geq 0 \Rightarrow B^2 - 4AC \geq 0$

Since $A$, $B$, and $C$ are independent random variables uniformly distributed over $(0,1)$, we can find the probability by calculating the volume of the region in the $A-B-C$ space where the inequality holds.

We can rewrite the inequality as:

$B^2 \geq 4AC$

Now, let's find the volume of the region where this inequality holds. We can integrate over the region in the $A-B-C$ space:

$\int_0^1 \int_0^1 \int_{\frac{B^2}{4}}^1 dA\, dB\, dC$

We can first integrate with respect to $A$:

$\int_0^1 \int_0^1 (1 - \frac{B^2}{4}) dB\, dC$

Now, we can integrate with respect to $B$:

$\int_0^1 \left[\frac{B}{2} - \frac{B^3}{12}\right]_0^1 dC$

Evaluating the integral:

$\int_0^1 \left(\frac{1}{2} - \frac{1}{12}\right) dC$

This simplifies to:

$\int_0^1 \frac{5}{12} dC$

Finally, we can integrate with respect to $C$:

$\left[\frac{5}{12}C\right]_0^1$

Evaluating the integral, we get:

$\frac{5}{12}$

So, the probability that all of the roots of the equation $Ax^2 + Bx + C = 0$ are real is $\frac{5}{12}$.		Incorrect
40
48EECS6.191Computation Structures6.100A, 8.02NonePrelab Questions 4Conditionals2b0.05TextNow consider the following code.
function Bit#(1) conditional(Bit#(1) a, Bit#(1) b);
Bit#(1) ret = 0;
Integer c=1;
if(c==1) begin
ret = comb(a, b);
end else begin
ret = comb2(a, b);
end
return ret;
endfunction
How many instances of comb and comb2 will be synthesized for this specification of conditional?
(a) 1 comb and 0 comb2.
(b) 0 comb and 1 comb2.
(c) 1 comb and 1 comb2.		Multiple Choice(a) 1 comb and 0 comb2.
In this function, the type Integer is used for the c variable. Integer types cannot be synthesized into hardware, thus their value must be known at compile time so that the Integer can be replaced with an actual value. In this minispec code, the Integer c is defined to be 1. Thus c is replaced by 1 and the conditional statement is simplified to only instantiate the if clause, or the comb circuit but not the comb2 circuit.		(a) 1 comb and 0 comb2.Correct
41
15EECS6.100AIntroduction to Computer Science Programming in PythonNoneNoneProblem Set 2Hangman1a1.875TextImplement the function has_player_won according to its docstrings. This function will be useful in determining when the hangman game has been won (i.e. the user has guessed all the letters in the secret word).
Example Usage:
>>> secret_word = 'apple'
>>> letters_guessed = ['e', 'i', 'k', 'p', 'r', 's']
>>> print(has_player_won(secret_word, letters_guessed))
False
Testing: Navigate to the test_ps2_student.py file and run it in Spyder. This will run a series of unit tests on your code. Note that this file contains tests for functions you will implement later on in this pset, so not all of them will pass right away. Examine the tests that start with test_has_player_won. If your function is correct, you should see the following printout:
test_has_player_won (__main__.TestPS2) ... ok
test_has_player_won_empty_list (__main__.TestPS2) ... ok
test_has_player_won_empty_string (__main__.TestPS2) ... ok
test_has_player_won_repeated_letters (__main__.TestPS2) ... ok
def has_player_won(secret_word, letters_guessed):
'''
secret_word: string, the lowercase word the user is guessing
letters_guessed: list (of lowercase letters), the letters that have been
guessed so far
returns: boolean, True if all the letters of secret_word are in letters_guessed,
False otherwise
'''
# FILL IN YOUR CODE HERE AND DELETE "pass"
pass		Programmingdef has_player_won(secret_word, letters_guessed):
'''
secret_word: string, the lowercase word the user is guessing
letters_guessed: list (of lowercase letters), the letters that have been
guessed so far
returns: boolean, True if all the letters of secret_word are in letters_guessed,
False otherwise
'''
# FILL IN YOUR CODE HERE AND DELETE "pass"
for character in secret_word:
if character not in letters_guessed:
return False
return True		def has_player_won(secret_word, letters_guessed):
'''
secret_word: string, the lowercase word the user is guessing
letters_guessed: list (of lowercase letters), the letters that have been
guessed so far
returns: boolean, True if all the letters of secret_word are in letters_guessed,
False otherwise
'''
for letter in secret_word:
if letter not in letters_guessed:
return False
return True		Correct
42
87EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneExercise 12Q-Learning1av0.01157407407TextLet's simulate the Q-learning algorithm! Assume there are states $(0,1,2,3)$ and actions ('b', 'c'), and discount factor $\gamma=0.9$. Furthermore, assume that all the $\mathrm{Q}$ values are initialized to 0 (for all state-action pairs) and that the learning rate $\alpha=0.5$.
Experience is represented as a list of 4-element tuples: the $t$ th element of the experience corresponds to a record of experience at time $t:\left(s_{t}, a_{t}, s_{t+1}, r_{t}\right)$ (state, action, next state, reward).
After each step $t$, indicate what update $Q\left(s_{t}, a_{t}\right) \leftarrow q$ will be made by the Q learning algorithm based on $\left(s_{t}, a_{t}, s_{t+1}, r_{t}\right)$. You will want to keep track of the overall table $Q\left(s_{t}, a_{t}\right)$ as these updates take place, spanning the multiple parts of this question.
As a reminder, the Q-learning update formula is the following:
$$
Q(s, a) \leftarrow(1-\alpha) Q(s, a)+\alpha\left(r+\gamma \max _{a^{\prime}} Q\left(s^{\prime}, a^{\prime}\right)\right)
$$
You are welcome to do this problem by hand, by drawing a table specifying $Q(s, a)$ for all possible $s$ and $as$. Alternatively, you may write a program which takes in the following history of experience:
experience = [(0, 'b', 2, 0), #t = 0
(2, 'b', 3, 0),
(3, 'b', 0, 2),
(0, 'b', 2, 0), #t = 3
(2, 'b', 3, 0),
(3, 'c', 0, 2),
(0, 'c', 1, 0), #t = 6
(1, 'b', 0, 1),
(0, 'b', 2, 0),
(2, 'c', 3, 0), #t = 9
(3, 'c', 0, 2),
(0, 'c', 1, 0)]
t: S A S' R
---------------
1: 2 'b' 3 0
The $t=1$ step of Q-learning will update the Q value of some state-action pair based on the experience tuple $\left(s_{1}, a_{1}, s_{2}, r_{1}\right)$.
After observing this tuple, what is the state of the state-action pair that is updated?		Numerical2.
Since we observe an experience in state 2, we update the Q value for state 2.		The state-action pair that is updated after observing the tuple at t=1 is (2, 'b').Mostly Correct
43
58EECS6.411Representation, Inference, and Reasoning in AI6.1010, 6.1210, 18.600NoneProblem Set 1Monte-Carlo Tree Search4cv0.05580357143TextNow, for each of the problems defined in get_fractal_problems, let us compare the performances of MCTS vs. UCS empirically by running run_mcts_search and run_uniform_cost_search. In particular, you should:
\begin{itemize}
\item Fix step_budget for both algorithms to 2500. Set iteration_budget for MCTS to infinity.
\item Run MCTS 20 times and record the average cumulative reward.
\item Run UCS once:
\begin{itemize}
\item If it fails (due to running out of step budget), record the cumulative reward as 0 .
\item If it succeeds, record the obtained cumulative reward. Hint: you might need to recover rewards from path costs
\end{itemize}
\item Repeat the above for all three problems in get_fractal_problems.
\end{itemize}
What is the average cumulative reward obtained by MCTS in reward-field-3?		Numerical1.2.As an AI language model, I cannot run code or simulations. However, I can guide you on how to perform the comparison between MCTS and UCS for the given problems.

1. Set the step_budget for both algorithms to 2500 and the iteration_budget for MCTS to infinity.
2. For each problem in get_fractal_problems, perform the following steps:
a. Run MCTS 20 times and calculate the average cumulative reward.
b. Run UCS once and record the cumulative reward as per the given conditions.
3. Compare the average cumulative rewards obtained by MCTS and UCS for each problem.

To find the average cumulative reward obtained by MCTS in reward-field-3, you would need to run the simulations and calculate the average reward as described above.		Invalid Question
44
566EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneFinal ExamNeural Networks1d0.7TextMac O'Larnin is considering selling an app on Frugal Play. You have a friend with inside info at Frugal, and they're able to share data on how previous apps have performed on the store.
Mac decides that he will learn a neural network with no hidden layer (i.e., consisting only of the output layer). He needs help in figuring out the precise formulation for machine learning.
Mac's first attempt at machine learning to predict the sales volume (setup of (b)) uses all customer data from 2020 . He randomly partitions the data into train $(80 \%)$ and validation $(20 \%)$, and uses the same number of units, activation function(s), and loss function as in (b). To prevent overfitting, he uses ridge regularization of the weights $W$, minimizing the optimization objective
$$
J(W ; \lambda)=\sum_{i=1}^{n} \mathcal{L}\left(h\left(x^{(i)} ; W\right), y^{(i)}\right)+\lambda\|W\|^{2},
$$
where $\|W\|^{2}$ is the sum over the square of all output units' weights.
Mac discovers that it's possible to find a value of $W$ such that $J(W ; \lambda)=0$ even when $\lambda$ is very large, nearing $\infty$. Mac suspects that he might have an error in the code that he wrote to derive the labels (i.e., the monthly sales volumes). Let's see why. First, what can Mac conclude about $W$ from this finding? Second, what does this imply about the labels?		Open(1) Since the loss is always non-negative and the penalty is always nonnegative, the only way to get 0 here is for both to be equal to 0 . The only way the penalty can equal 0 is if every element of $W$ equals 0 .
(2) When $W$ has all entries equal to 0 , the prediction at every data point is a constant (the offset). The only way for the squared error to be 0 is for the label of every data point to equal that offset. It seems unlikely that every data label would be exactly the same in this data set, which we assume ranges over a wide number of apps.		First, if Mac can find a value of $W$ such that $J(W; \lambda) = 0$ even when $\lambda$ is very large, nearing $\infty$, it means that the term $\lambda\|W\|^2$ is very close to 0. This implies that the weights $W$ are very close to 0 as well.

Second, since the weights $W$ are very close to 0, the output of the neural network $h(x^{(i)}; W)$ will also be very close to 0 for all input data points $x^{(i)}$. This implies that the loss function $\mathcal{L}(h(x^{(i)}; W), y^{(i)})$ is also very close to 0 for all data points. This suggests that the labels (i.e., the monthly sales volumes) $y^{(i)}$ are also very close to 0 for all data points.

The fact that all labels are close to 0 might indicate an error in the code that Mac wrote to derive the labels, as it is unlikely that all monthly sales volumes are close to 0. Mac should double-check his code and ensure that the labels are correctly derived from the data.		Mostly Correct
45
247Mathematics18.01Calculus INoneNoneProblem Set 6Approximations6c0.05279831045TextTaylor's theorem tells us that when the second derivative of a function is really big, then the linear approximation is not so accurate. Here's an example. Let $f(x)=\sin (100 x)$.
Compute $f^{\prime \prime}(x)$. Check that $f^{\prime \prime}(0)=0$. Let $M$ denote the maximum of $\left|f^{\prime \prime}(x)\right|$ for $0 \leq x \leq .1$. Find $M$.		Numerical$f^{\prime \prime}(x)=100^{2} \sin (100 x)$. When $x=0,100 x=0$ and when $x=.1,100 x=10$. Thus the input of sine goes between 0 and 10 when $0 \leq x \leq .1$. Since sine achieves its maximum at $\frac{\pi}{2}$ (which corresponds to taking $0 \leq x=\pi / 200 \leq .1$,
$$
\left|f^{\prime \prime}(x)\right| \leq 100^{2}|\sin (100 x)| \leq 100^{2}|\sin (100 \cdot \pi / 200)|=100^{2}
$$
whenever $0 \leq x \leq .1$. Thus $M=100^{2}=10^{4}$.		To find the second derivative of the function $f(x) = \sin(100x)$, we first need to find the first derivative $f'(x)$.

Using the chain rule, we have:
$$f'(x) = \frac{d}{dx}(\sin(100x)) = 100\cos(100x)$$

Now, we can find the second derivative $f''(x)$:
$$f''(x) = \frac{d}{dx}(100\cos(100x)) = -10000\sin(100x)$$

To check that $f''(0) = 0$, we can plug in $x = 0$:
$$f''(0) = -10000\sin(100 \cdot 0) = -10000\sin(0) = 0$$

Now, we need to find the maximum of $|f''(x)|$ for $0 \leq x \leq 0.1$. Since the second derivative is a sinusoidal function, its maximum value will occur at the maximum of the sine function, which is 1. Therefore, the maximum of $|f''(x)|$ is:

$$M = |-10000\sin(100x)|_{max} = 10000$$

So, the maximum value of the second derivative in the given interval is $M = 10000$.		Correct
46
25EECS18.C06Linear Algebra and Optimization18.02NoneProblem Set 1Matrix Multiplication9a0.1851851852TextWe want to compute the product $A B C$ of three matrices $A, B, C$. The matrices have dimensions $n \times m, m \times p$, and $p \times r$, respectively. Recall that matrix multiplication is an associative operation, i.e., $(A B) C=A(B C)$.
For this problem we will use the standard formula for multiplying matrices from class:
$$
\sum_{k} A_{i, k} B_{k, j},
$$
and when counting arithmetic operations (addition, multiplication) you should only aim to get the answer right up to a constant factor.
If we compute the product as $(A B) C$, how many arithmetic operations are required (as a function of $n, m, p, r)$?		ExpressionThe number of operations to compute $A B$ is the number of entries $(n p)$ times the number of arithmetic operations we need to make to compute each entry ( $m$ multiplications $+(m-1)$ additions $=2 m-1$ operations $)$, for a total of $n p(2 m-1)$ operations. To compute $(A B) C$ now, we argue similarly to conclude that we make $n r(2 p-1)$ additional arithmetic operations. Adding these up, we get that the total number of arithmetic operations is $n(p(2 m-1)+r(2 p-1))$.To compute the product as $(AB)C$, we first compute the product $AB$. The dimensions of $AB$ will be $n \times p$. For each entry in the resulting matrix, we need to perform $m$ multiplications and $(m-1)$ additions. Since there are $n \times p$ entries in the $AB$ matrix, the total number of arithmetic operations required for this step is $n \times p \times (m + (m-1)) = n \times p \times 2m$.

Next, we compute the product $(AB)C$. The dimensions of $C$ are $p \times r$, and the dimensions of the resulting matrix will be $n \times r$. For each entry in the resulting matrix, we need to perform $p$ multiplications and $(p-1)$ additions. Since there are $n \times r$ entries in the $(AB)C$ matrix, the total number of arithmetic operations required for this step is $n \times r \times (p + (p-1)) = n \times r \times 2p$.

Adding the operations from both steps, the total number of arithmetic operations required to compute the product as $(AB)C$ is $n \times p \times 2m + n \times r \times 2p = 2np(m+r)$.		Incorrect
47
192EECS18.C06Linear Algebra and Optimization18.02NoneFinal ExamInverse Matrix13c0.6956521739TextLet $a, b$ be vectors in $\mathbb{R}^{n}$. In this problem we will find an expression for the inverse of $M=I-a b^{T}$ and explore some implications for optimization.
Recall that for any matrix $A$ with $\|A\|<1$ we have the identity
$$
(I-A)^{-1}=\sum_{k=0}^{\infty} A^{k}=I+A+A^{2}+\cdots
$$
Use this formula to compute $M^{-1}$ and simplify to get an expression of the form
$$
M^{-1}=I+\alpha a b^{T}
$$
What is the value of $\alpha$?		ExpressionApplying the formula, we have
$$
\begin{aligned}
\left(I-a b^{T}\right)^{-1} & =I+a b^{T}+\left(a b^{T}\right)\left(a b^{T}\right)+\left(a b^{T}\right)\left(a b^{T}\right)\left(a b^{T}\right)+\cdots \\
& =I+a b^{T}+\left(b^{T} a\right)\left(a b^{T}\right)+\left(b^{T} a\right)^{2}\left(a b^{T}\right)+\cdots \\
& =I+a b^{T}\left(1+\left(b^{T} a\right)+\left(b^{T} a\right)^{2}+\cdots\right) \\
& =I+a b^{T}\left(1-b^{T} a\right)^{-1},
\end{aligned}
$$
i.e., $\alpha=1 /\left(1-b^{T} a\right)$. 		To compute $M^{-1}$, we first need to find the matrix $A$ such that $M = I - A$. In this case, $A = ab^T$. Now, we can use the given formula to compute $M^{-1}$:

$$
M^{-1} = (I - A)^{-1} = \sum_{k=0}^{\infty} A^k = I + A + A^2 + \cdots
$$

Since $A = ab^T$, we have:

$$
M^{-1} = I + ab^T + (ab^T)^2 + \cdots
$$

Now, let's compute the square of $A$:

$$
A^2 = (ab^T)^2 = (ab^T)(ab^T) = a(b^Ta)b^T
$$

Notice that $b^Ta$ is a scalar, so we can rewrite $A^2$ as:

$$
A^2 = (b^Ta)(ab^T)
$$

Now, we can substitute this back into the expression for $M^{-1}$:

$$
M^{-1} = I + ab^T + (b^Ta)(ab^T) + \cdots
$$

Since all the higher powers of $A$ will also have the form $(b^Ta)^k(ab^T)$ for some integer $k$, we can factor out $ab^T$ from the sum:

$$
M^{-1} = I + ab^T(1 + b^Ta + (b^Ta)^2 + \cdots)
$$

Now, we recognize the sum inside the parentheses as the geometric series with the common ratio $b^Ta$. Since $\|A\| < 1$, we know that $|b^Ta| < 1$. Therefore, the sum converges to:

$$
\frac{1}{1 - b^Ta}
$$

So, we have:

$$
M^{-1} = I + ab^T\left(\frac{1}{1 - b^Ta}\right)
$$

Comparing this expression with the desired form $M^{-1} = I + \alpha ab^T$, we can see that the value of $\alpha$ is:

$$
\alpha = \frac{1}{1 - b^Ta}
$$		Correct
48
143EECS6.122Design and Analysis of Algorithms6.121NoneFinal ExamRandomized Algorithms1o0.375TextPlease select True or False for the following.
Consider the following algorithm for testing if a given list $L$ of $n>3$ distinct numbers is sorted:
Repeat $\Theta(\log n)$ times: Pick three indices $i<j<k$ uniformly at random and return NO if the following is false: $L_{i}<L_{j}<L_{k}$. At the end of the $\Theta(\log n)$ iterations, return YES.
Is it true that this algorithm always returns YES if $L$ is sorted and returns $\mathrm{NO}$ with probability at least $3 / 4$ if $L$ is not $\epsilon$-close to sorted for small constant $\epsilon$?		Multiple ChoiceFalse. Consider the list $L$ as follows, for $n$ divisible by 2 :
$$
2,1,4,3, \ldots, 2 i, 2 i-1, \ldots, n, n-1 .
$$
This list is definitely very far from sorted: it is not $\epsilon$-close to sorted for any $\epsilon<1 / 2$, as we need to remove at least half of the numbers to make it sorted.
In order for us to pick $L_{i}, L_{j}, L_{k}(i<j<k)$ such that the algorithm returns NO, we need to have either $i=j-1$, or $j=k-1$, as only consecutive indices are in the wrong order. The probability of this happening in a single iteration is at most $O(1 / n)$. Even if we repeat $O(\log n)$ times, by a union bound, the probability that it happens in one of the iterations is at most $O(\log n / n)$. So we definitely will not return NO with constant probability. 		FAUXCorrect
49
13Mathematics18.102Introduction to Functional Analysis18.C06, 18.100BNoneProblem Set 3Lp Spaces10.5TextLet $f(x)=x^{a} \log (x)^{b}$ for $x>2$ and $f(x)=0$ otherwise. For which real $a, b$ is $f$ in $\mathcal{L}^{1}(\mathbb{R})$? Justify your answer.OpenWe first claim that $f(x) \in \mathcal{L}^{1}(\mathbb{R})$ iff $\lim _{R \rightarrow \infty} \int_{2}^{R} f(x) d x<\infty$ : Assuming $f(x) \in \mathcal{L}^{1}(\mathbb{R})$, then $\int_{2}^{R} f(x) d x<\int f(x)<\infty$, and therefore it converges. Conversely, assuming the integral converges, define $f_{n}(x)=f(x) \chi_{[0, n]}$, then $f_{n}(x)$ is a monotone sequence converging pointwise to $f(x)$, by Lemma $2.7$ we conclude $f(x) \in \mathcal{L}^{1}(\mathbb{R})$.
Therefore we need to check the convergence of the integral $\int_{2}^{\infty} f(x) d x$. We distinguish different cases:
(1) $a>-1$. We have $x^{a} \log (x)^{b}>x^{-1}$ when $x$ is large, therefore the integral diverges.
(2) $a<-1$. Choose any $a<c<-1$, we have $x^{a} \log (x)^{b}<x^{c}$ when $x$ is large, therefore the integral converges.
(3) $a=-1$. We compute directly,
$$
\int x^{-1} \log (x)^{b}= \begin{cases}\frac{\log (x)^{b+1}}{b+1} & b \neq-1 \\ \log (\log (x)) & b=-1\end{cases}
$$
Therefore, the integral diverges when $b \geq-1$ and converges when $b<-1$.
In summary, $f(x) \in \mathcal{L}^{1}(\mathbb{R})$ iff in the following cases:
\begin{itemize}
\item $a<-1$.
\item $a=-1$ and $b<-1$.
\end{itemize}		To determine for which real $a, b$ the function $f(x) = x^a \log(x)^b$ is in $\mathcal{L}^1(\mathbb{R})$, we need to find the values of $a$ and $b$ for which the integral of the absolute value of $f(x)$ over the entire real line converges. That is, we need to find $a$ and $b$ such that

$$\int_{-\infty}^{\infty} |f(x)| dx = \int_{-\infty}^2 0 dx + \int_2^{\infty} x^a \log(x)^b dx < \infty.$$

Since the first integral is trivially zero, we only need to consider the second integral:

$$\int_2^{\infty} x^a \log(x)^b dx.$$

To analyze the convergence of this integral, we can use the comparison test. We know that for $x > 2$, $\log(x) > 0$. Thus, we can compare the integral to the integral of $x^a$:

$$\int_2^{\infty} x^a dx.$$

This integral converges if and only if $a < -1$. Now, let's consider the cases when $a < -1$.

If $a < -1$, we can use integration by parts to further analyze the convergence of the integral of $f(x)$. Let $u = \log(x)^b$ and $dv = x^a dx$. Then, $du = b\log(x)^{b-1} \frac{1}{x} dx$ and $v = \frac{x^{a+1}}{a+1}$. Applying integration by parts, we get:

$$\int_2^{\infty} x^a \log(x)^b dx = \left[\frac{x^{a+1}}{a+1}\log(x)^b\right]_2^{\infty} - \int_2^{\infty} \frac{b x^{a} \log(x)^{b-1}}{a+1} dx.$$

The first term in the brackets goes to zero as $x \to \infty$ since $a < -1$. Thus, we only need to consider the convergence of the second integral:

$$\int_2^{\infty} \frac{b x^{a} \log(x)^{b-1}}{a+1} dx.$$

Since $a < -1$, we can again use the comparison test with the integral of $x^a$:

$$\int_2^{\infty} x^a dx.$$

This integral converges if and only if $a < -1$. Therefore, for $f(x)$ to be in $\mathcal{L}^1(\mathbb{R})$, we must have $a < -1$. The value of $b$ does not affect the convergence of the integral, so $f(x)$ is in $\mathcal{L}^1(\mathbb{R})$ for any real $b$.

In conclusion, $f(x) = x^a \log(x)^b$ is in $\mathcal{L}^1(\mathbb{R})$ if and only if $a < -1$ and $b \in \mathbb{R}$.		​
50
172EECS6.191Computation Structures6.100A, 8.02NoneMidterm Exam 3Caches1a0.45TextCache Ketchum wants to design a cache to help keep track of his Pokedex entries. He’s enlisted your help as a talented 6.191 student!
Ketchum wants to build a direct-mapped cache with a block size of eight words. He also wants the cache to hold a total of $2^9 = 512$ data words. Which address bits should be used for the block offset, cache index, and tag? Assume that data words and addresses are 32 bits wide.		NumericalAddress bits used for block offset: A[ __4__ : __2__ ]
Address bits used for cache index: A[ __10__ : __5__ ]
Address bits used for tag: A[ __31__ : __11__ ]		To design the cache, we need to determine the number of bits required for the block offset, cache index, and tag.

1. Block offset: Since the block size is 8 words, we need 3 bits to represent the offset within a block (2^3 = 8).

2. Cache index: The cache can hold a total of 512 data words, and each block contains 8 words. So, there are 512/8 = 64 blocks in the cache. We need 6 bits to represent the cache index (2^6 = 64).

3. Tag: The remaining bits in the 32-bit address are used for the tag. Since we have already used 3 bits for the block offset and 6 bits for the cache index, we have 32 - 3 - 6 = 23 bits for the tag.

In summary, the address bits should be used as follows:
- Block offset: 3 bits
- Cache index: 6 bits
- Tag: 23 bits		Incorrect
51
83Mathematics18.03Differential EquationsNone18.02Problem Set 7Gaussian Elimination5a0.08042895442TextOne of the black boxes we used in class was the theorem that an $n \times n$ matrix $A$ has $A \vec{v}=0$ for some non-zero vector $\vec{v} \in \mathbb{R}^{n}\left(\right.$ or $\mathbb{C}^{n}$ ) if and only if $\operatorname{det}(A)=0$ (see, e.g., MITx 20.7). The goal of this problem is to work out $w h y$ this is true (at least in the case of $3 \times 3$ matrices). The only blackbox we will use is the properties of the determinant. Recall that $\operatorname{dim} \operatorname{Ker}(A)=0$ means that $\operatorname{Ker}(A)$ contains only the zero vector.
(Story time begins) The way we are going to go about showing that a $3 \times 3$ matrix has $\operatorname{det} A=0$ if and only if $\operatorname{dim} \operatorname{Ker}(A)>0$ is by using Gaussian elimination to reduce the statement to the case of upper triangular (or rather, RREF) matrices. So, as a first step, we're going to check that the theorem is true for this model case. (Story time ends)
Suppose $A$ is a $3 \times 3$ matrix which is upper triangular; that is
$$
A=\left(\begin{array}{ccc}
p_{1} & a & b \\
0 & p_{2} & c \\
0 & 0 & p_{3}
\end{array}\right) \text {. }
$$
Show that $\operatorname{det} A=p_{1} p_{2} p_{3}$. In particular, $\operatorname{det}(A)=0$ if and only if at least one of $p_{1}, p_{2}, p_{3}$ is 0.		OpenUsing the Laplace expansion, the only non-zero term is $p_{1} \cdot\left|\left(\begin{array}{cc}p_{2} & c \\ 0 & p_{3}\end{array}\right)\right|=$ $p_{1} p_{2} p_{3}$. Or you may use the fact that eigenvalues are $p_{1}, p_{2}, p_{3}$ and the determinant is the product of them. To compute the determinant of a $3 \times 3$ matrix, we can use the following formula:

$$
\operatorname{det}(A) = p_1 \cdot \left(\begin{array}{cc}
p_{2} & c \\
0 & p_{3}
\end{array}\right) - a \cdot \left(\begin{array}{cc}
0 & c \\
0 & p_{3}
\end{array}\right) + b \cdot \left(\begin{array}{cc}
0 & p_{2} \\
0 & 0
\end{array}\right)
$$

Now, we can compute the determinants of the $2 \times 2$ matrices:

$$
\operatorname{det}\left(\begin{array}{cc}
p_{2} & c \\
0 & p_{3}
\end{array}\right) = p_2 p_3
$$

$$
\operatorname{det}\left(\begin{array}{cc}
0 & c \\
0 & p_{3}
\end{array}\right) = 0
$$

$$
\operatorname{det}\left(\begin{array}{cc}
0 & p_{2} \\
0 & 0
\end{array}\right) = 0
$$

Substituting these values back into the formula for the determinant of $A$, we get:

$$
\operatorname{det}(A) = p_1 (p_2 p_3) - a(0) + b(0) = p_1 p_2 p_3
$$

Thus, the determinant of the upper triangular matrix $A$ is equal to the product of its diagonal elements $p_1 p_2 p_3$. In particular, $\operatorname{det}(A) = 0$ if and only if at least one of $p_1, p_2, p_3$ is 0.		Correct
52
115Mathematics18.01Calculus INoneNoneProblem Set 3Chain Rule10a0.05279831045TextFind the derivatives of the following functions:
$e^{5 x}$.		Expression$\frac{d}{d x} e^{5 x}=e^{5 x}(5)=5 e^{5 x}$.To find the derivative of the function $e^{5x}$, we can use the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. In this case, the outer function is $e^u$ and the inner function is $5x$.

First, find the derivative of the outer function with respect to $u$:
$\frac{d}{du}(e^u) = e^u$

Next, find the derivative of the inner function with respect to $x$:
$\frac{d}{dx}(5x) = 5$

Now, apply the chain rule:
$\frac{d}{dx}(e^{5x}) = e^{5x} \cdot 5$

So, the derivative of the function $e^{5x}$ is $5e^{5x}$.		Correct
53
610EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneFinal ExamConvolutional Neural Networks8c0.35TextMIT grad student Rec Urrent would like to submit an entry to win this year's Grand ML Tetris Competition, which gives awards to the smallest neural networks which can identify tetris pieces with the highest accuracy. Rec seeks to make a convolutional neural network that can accurately classify single-channel $3 \times 3$ images of $2 \mathrm{D}$ tetris pieces as being either a line-shaped piece, or a corner-shaped piece, using just one $2 \times 2$ filter. Let's help Rec win this competition.
If Rec wants to allow for more than two classes, which activation function should they use for final_act and which loss function?		OpenSoftmax $+$ Cross Entropy.If Rec wants to allow for more than two classes, they should use the softmax activation function for final_act and the categorical crossentropy loss function.Correct
54
222EECS6.411Representation, Inference, and Reasoning in AI6.1010, 6.1210, 18.600NoneProblem Set 6Particle Filter5dii0.08333333333TextConsider a domain in which the forward transition dynamics are "hybrid" in the sense that
$$
P\left(X_{t}=x_{t} \mid X_{t-1}=x_{t-1}\right)=p * N\left(x_{t-1}+1,0.1\right)\left(x_{t}\right)+(1-p) * N\left(x_{t-1}-1,0.1\right)\left(x_{t}\right)
$$
that is, that the state will hop forward one unit in expectation with probability $p$, or backward one unit in expectation with probability $1-p$, with variance $0.1$ in each case.
Assume additionally that the observation model $P\left(Y_{t}=y_{t} \mid X_{t}=x_{t}\right)=\operatorname{Uniform}\left(x_{t}-1, x_{t}+1\right)\left(y_{t}\right)$.
Norm runs the filter for two steps with no observations several times and is trying to decide whether there could be bugs in the code. Assuming $p=0.5$, for each of the following sets of particles, indicate whether it is (a) fairly likely (b) quite unlikely (c) completely impossible: {-2.01, -1.9, -1.0, 0.1, 0, 2.1}.		Multiple Choiceb.To analyze the likelihood of the given set of particles, we need to consider the forward transition dynamics and the observation model.

Given the forward transition dynamics, the state will hop forward one unit with probability p=0.5 or backward one unit with probability 1-p=0.5, with variance 0.1 in each case. The observation model states that the observed value will be uniformly distributed between x_t-1 and x_t+1.

Let's consider the possible transitions for each particle:

1. -2.01: This particle is quite unlikely, as it would require a backward hop from -1.01 with a small negative deviation.
2. -1.9: This particle is fairly likely, as it could result from a backward hop from -0.9 with a small negative deviation.
3. -1.0: This particle is fairly likely, as it could result from a backward hop from 0 with no deviation.
4. 0.1: This particle is fairly likely, as it could result from a forward hop from -1 with a small positive deviation.
5. 0: This particle is fairly likely, as it could result from a forward hop from -1 with no deviation.
6. 2.1: This particle is quite unlikely, as it would require a forward hop from 1.1 with a small positive deviation.

In summary, the given set of particles has the following likelihoods:
- -2.01: quite unlikely
- -1.9: fairly likely
- -1.0: fairly likely
- 0.1: fairly likely
- 0: fairly likely
- 2.1: quite unlikely		Correct
55
24EECS6.191Computation Structures6.100A, 8.02NonePrelab Questions 2RISC-V Calling Convention1a0.025TextConsider the following C function, `f`:
int f(int x, int y) {
int w = y;
int z = mul(x, 2) + y + w;
return z;
}
Suppose that you wanted to implement the function, f, in RISC-V assembly.
Which register(s) should be used to pass the arguments to the function f? Select all correct answers.
(a) zero.
(b) ra.
(c) a0.
(d) a1.
(e) a2.
(f) t0.
(g) t1.
(h) s0.
(i) s1.		Multiple Choice(c) a0.
(d) a1.
According to the calling conventions, the arguments to the function are passed in through the a registers, starting from a0, a1, and so on. In this question, there are 2 arguments (x and y) needed to be passed in, so only a0 and a1 are used.		(c) a0.
(d) a1.		Correct
56
19Mathematics18.102Introduction to Functional Analysis18.C06, 18.100BNoneProblem Set 3Lp Spaces50.5TextSuppose $f \in \mathcal{L}^{1}(\mathbb{R})$ is real-valued. Show that there is a sequence $f_{n} \in \mathcal{C}_{\mathrm{c}}(\mathbb{R})$ and another element $F \in \mathcal{L}^{1}(\mathbb{R})$ such that
$$
f_{n}(x) \rightarrow f(x) \text { a.e. on } \mathbb{R},\left|f_{n}(x)\right| \leq F(x) \text { a.e. }
$$
Hint: Take an approximating series $u_{n}$ as in the definition and think about $\left|u_{n}\right|$.
Remark: The converse of this, where the $f_{n}$ are allowed to be in $\mathcal{L}^{1}(\mathbb{R})$ is 'Lebesgue Dominated Convergence'.		OpenLet $f_{n} \in C_{c}(\mathbb{R})$ be such that if $w_{1}=f_{1}$ and $w_{k}=f_{k}-f_{k-1}$, $\left(w_{n}\right)$ is absolutely summable and $f_{n}(x) \rightarrow f(x)$ almost everywhere (such a sequence $\left(f_{n}\right)$ exists by the definition of $\mathcal{L}^{1}(\mathbb{R})$ ). Then, define
$$
F(x)=\left\{\begin{array}{ll}
\sum_{n}\left|w_{n}(x)\right| & \text { if } \sum_{n}\left|w_{n}(x)\right|<\infty \\
0 & \text { otherwise }
\end{array} .\right.
$$
By the definition of measure zero, $F(x)=\sum_{n}\left|w_{n}(x)\right|$ almost everywhere, so by Proposition $2.5$ it follows that $F \in \mathcal{L}^{1}(\mathbb{R})$. Moreover, because whenever $F(x)=\sum_{n}\left|w_{n}(x)\right|$ we have that
$$
\left|f_{n}(x)\right|=\left|\sum_{k=1}^{n} w_{k}(x)\right| \leq \sum_{k=1}^{n}\left|w_{k}(x)\right| \leq F(x),
$$
it follows that $\left|f_{n}(x)\right| \leq F(x)$ almost everywhere. The desired conclusion follows.		To show the desired result, we will use the hint provided and consider an approximating sequence $u_n$ as in the definition of the Lebesgue integral. Let $u_n$ be a sequence of simple functions such that $u_n \to f$ a.e. on $\mathbb{R}$ and $\int_{\mathbb{R}} |u_n - f| \to 0$ as $n \to \infty$.

Now, we will construct a sequence of continuous functions with compact support, $f_n$, by convolving $u_n$ with a mollifier $\phi_\epsilon$. Let $\phi_\epsilon(x) = \frac{1}{\epsilon} \phi\left(\frac{x}{\epsilon}\right)$, where $\phi$ is a smooth function with compact support in $[-1, 1]$ and $\int_{\mathbb{R}} \phi(x) dx = 1$. Define $f_n(x) = (u_n * \phi_\epsilon)(x) = \int_{\mathbb{R}} u_n(y) \phi_\epsilon(x - y) dy$.

Since $u_n$ is a simple function and $\phi_\epsilon$ is continuous with compact support, $f_n$ is continuous with compact support. Moreover, by the properties of convolution, we have $f_n \to f$ a.e. on $\mathbb{R}$ as $n \to \infty$ and $\epsilon \to 0$.

Now, we need to find an element $F \in \mathcal{L}^1(\mathbb{R})$ such that $|f_n(x)| \leq F(x)$ a.e. To do this, we will consider the function $F(x) = |u_n(x)| + |\phi_\epsilon * u_n(x)|$. Since $u_n$ is a simple function and $\phi_\epsilon * u_n$ is continuous with compact support, both $|u_n(x)|$ and $|\phi_\epsilon * u_n(x)|$ are in $\mathcal{L}^1(\mathbb{R})$. Therefore, their sum $F(x)$ is also in $\mathcal{L}^1(\mathbb{R})$.

Now, we will show that $|f_n(x)| \leq F(x)$ a.e. We have
$$
|f_n(x)| = \left|\int_{\mathbb{R}} u_n(y) \phi_\epsilon(x - y) dy\right| \leq \int_{\mathbb{R}} |u_n(y)| |\phi_\epsilon(x - y)| dy.
$$
Since $\phi_\epsilon$ is non-negative and integrates to 1, we have
$$
\int_{\mathbb{R}} |u_n(y)| |\phi_\epsilon(x - y)| dy \leq \int_{\mathbb{R}} |u_n(y)| dy = |u_n(x)| + |\phi_\epsilon * u_n(x)| = F(x).
$$
Thus, we have shown that $|f_n(x)| \leq F(x)$ a.e.

In conclusion, we have constructed a sequence of continuous functions with compact support, $f_n$, and an element $F \in \mathcal{L}^1(\mathbb{R})$ such that $f_n(x) \to f(x)$ a.e. on $\mathbb{R}$ and $|f_n(x)| \leq F(x)$ a.e. This proves the desired result.		​
57
234Mathematics18.01Calculus INoneNoneProblem Set 5Differential Equations19b0.07919746568TextIn this section, we give an oversimplified model of how blood sugar and insulin work, and we consider the problem of designing an artificial pancreas. The biology is over-simplified, but the issues we will explore are still relevant in more accurate and complex models.
Let $S(t)$ denote the blood sugar level at time $t$. Suppose $S=10$ is a good level of blood sugar, $S$ above 12 is too high, and $S$ below 8 is too low. The blood sugar reacts to insulin levels. Let $I(t)$ denote the insulin level in the blood at time $T$. Suppose that
$$
S^{\prime}(t)=5-I(t) .
$$
So if $I(t)>5$, then blood sugar does down, and if $I(t)<5$ then blood sugar goes up. In patients with severe diabetes, the pancreas doesn't make insulin. The artificial pancreas is a fairly recent medical technology where a medical device installed in the patient makes insulin and has to adjust insulin levels to regulate blood sugar. Figuring out when to increase/decrease the insulin level is a mathematical problem. One approach is the following: if the patient's blood sugar is too high, the artificial pancreas increases the insulin level. If the patient's blood sugar is too low, the artificial pancreas decreases the insulin level. This approach can be modelled by the following differential equation:
$$
I^{\prime}(t)=S(t)-10 .
$$
At the moment we have two equations involving two functions:
$$
S^{\prime}(t)=5-I(t) \text { and } I^{\prime}(t)=S(t)-10 .
$$
To get an equation for just one function, we can differentiate the first equation $S^{\prime}(t)=5-I(t)$, which gives $S^{\prime \prime}(t)=-I^{\prime}(t)$ and then plug in the equation for $I^{\prime}(t)$. This leads to an equation for $S(t)$ which is similar to the ones in the last few problems.
Suppose that at time $0, S(0)=13$ (too high) and $I(0)=5$.
Is the artificial pancreas doing a good job or a bad job? Explain.		OpenIt's bad. Because the coefficient of $\sin$ in $S$ is 3 , the blood sugar swings between 7 and 13. Thus, it keeps swinging into the too-high zone $(>12)$ and the too-low zone $(<8)$. Another objection to this design is that the blood sugar never stops swinging. To determine if the artificial pancreas is doing a good job or a bad job, we need to analyze the behavior of the blood sugar level $S(t)$ over time. We have the following equations:

$$
S^{\prime}(t)=5-I(t) \text { and } I^{\prime}(t)=S(t)-10 .
$$

We already found that $S^{\prime \prime}(t)=-I^{\prime}(t)$. Now, we can substitute the equation for $I^{\prime}(t)$:

$$
S^{\prime \prime}(t)=-(S(t)-10) .
$$

This is a second-order linear differential equation for $S(t)$. We are given the initial conditions $S(0)=13$ and $I(0)=5$.

Let's analyze the behavior of $S(t)$:

1. If $S(t)>12$ (too high), then $S(t)-10>2$, which implies that $I^{\prime}(t)>2$. This means that the insulin level $I(t)$ is increasing, which will cause $S^{\prime}(t)$ to decrease (since $S^{\prime}(t)=5-I(t)$), and thus the blood sugar level $S(t)$ will decrease.

2. If $S(t)<8$ (too low), then $S(t)-10<-2$, which implies that $I^{\prime}(t)<-2$. This means that the insulin level $I(t)$ is decreasing, which will cause $S^{\prime}(t)$ to increase (since $S^{\prime}(t)=5-I(t)$), and thus the blood sugar level $S(t)$ will increase.

3. If $S(t)=10$ (good level), then $S(t)-10=0$, which implies that $I^{\prime}(t)=0$. This means that the insulin level $I(t)$ is constant, and the blood sugar level $S(t)$ will remain at the good level.

From this analysis, we can see that the artificial pancreas is doing a good job in regulating the blood sugar level. When the blood sugar level is too high, the insulin level increases to bring it down, and when the blood sugar level is too low, the insulin level decreases to bring it up. The artificial pancreas works to maintain the blood sugar level around the good level of $S=10$.		Incorrect
58
311EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneLab 12Reinforcement Learning3aiii0.06944444444TextRecall that the hyperparameter epsilon $(\epsilon)$ characterizes a trade-off between exploration and exploitation in reinforcement learning. When we use an " $\epsilon$-greedy" strategy in Q learning, we take a completely random action with probability $\epsilon$; and with probability $1-\epsilon$, we take the action that'd lead to the highest $Q$ value, i.e. we take $\arg \max _{a} Q(s, a)$.
We'll explore how choosing the value of epsilon affects the performance of $Q$ learning in a very simple game.
The choice of epsilon can affect the overall behavior of Q-learning. Let's consider three possible values for epsilon: $0.0$, $0.5$, and $1.0$.
Which of these epsilon values is guaranteed to cause optimal behavior during learning?		Multiple Choicenone: No algorithm can guarantee optimal behavior in training time. The algorithm is oblivious to the true transition and reward functions and it needs to explore to get reward signal and understand the world (eps $>0$ ), but then even if it eventually converged, it would take sub-optimal actions due to the eps $>0$ which was necessary in the first place.None of these epsilon values is guaranteed to cause optimal behavior during learning.

An epsilon value of 0.0 means the agent will always choose the action with the highest Q value, leading to pure exploitation and no exploration. This can cause the agent to get stuck in a suboptimal policy if it never explores other actions.

An epsilon value of 0.5 means the agent will choose a random action 50% of the time and the action with the highest Q value 50% of the time. This allows for some exploration, but it may not be enough to guarantee optimal behavior, as the agent might still not explore all possible actions sufficiently.

An epsilon value of 1.0 means the agent will always choose a random action, leading to pure exploration and no exploitation. This can cause the agent to never converge to an optimal policy, as it never exploits the learned Q values to make better decisions.		Correct
59
13Mathematics18.2Principles of Discrete Applied MathematicsNone18.C06Problem Set 3Bijection2b0.7638888889TextIn the notes on counting in canvas, Section $4.2$ describes a map $\Psi$ that takes a binary tree $B$ with $n$ nodes (remember these are those vertices with 2 children, as opposed to leaves who have no children) and maps it to a lattice path.
Furthermore prove that $\Psi$ is a bijection between $\mathcal{B}_{n}$ and $\mathcal{D}_{n}$.		OpenTo show that $\Psi$ is a bijection, we construct an inverse $\Phi$, so that given a Dyck path $D$, we construct a binary tree $B=\Phi(D)$ such that $\Psi(B)=D$. Since the first step of $D$ is an upwards step, we add a vertex to $B$ and place ourselves on that vertex. We now go step-by-step through $D$ to generate $B$. At each step, we will be growing a 2-tree in such a way that any vertex with a single child must be an ancestor of the vertex we are on. If we are at an upwards step, if the vertex we are on has less than two children, we add a child and move to the new vertex. If the vertex has two children, we move up the tree until we are at an ancestor vertex with less than two children, then add a child and move to the new vertex. (We show later that this is always possible.) This new vertex will become a node of $B$, and it will be a node of the generated tree after the next step. If we are at a downwards step and the current vertex has less than two children, we add a child but stay on the current vertex. This new vertex will become a leaf of $B$. If there are two children, then like before we move up until we are at a vertex $v$ with less than two children, and add a child while staying at $v$. Each of these procedures preserves the fact the generated tree is a 2-tree and that all vertices with a single child are ancestors of the current vertex.
We now show that when needed, we can always find an ancestor with only one child. Let $T$ be the tree we have generated so far, and suppose we need to check the ancestors for a place to add a vertex, and no ancestor has space. This implies that every node has zero or two children, and so $T$ is a binary tree. Then $n(T)=l(T)-1$ by the lemma. The previous step of the Dyck path must have produced a leaf, since otherwise there would have been space at the current vertex. This means that, by this point, every vertex produced with an upwards path has at least one child. Thus, every node of $T$ corresponds to an upwards path, and every leaf corresponds to a downwards path. If $k$ is the number of upwards paths so far and $m$ is the number of downwards paths so far, then $k=n(T)=l(T)-1=m-1$. Then $k<m$, which contradicts the fact that $D$ is a Dyck path. Thus, we can never reach this state and we can always find an ancestor with one child when necessary.
After the last step of the Dyck path, we will have just added a new leaf vertex, so if $G$ is the tree generated so far, then $l(G)=n$ and $n(G)=n$. Then from the lemma, $G$ is not yet a binary tree, and either the current node has space or one of its ancestors does. We add a leaf to this node, and call this tree $B$. We have $l(B)=l(G)+1$, so $l(B)-1=n(G)=n(B)$, which makes $B$ a binary tree.
The route that we added vertices is exactly the same route we would take when doing a depth-first search, so $\Psi(\Phi(D))=\Psi(B)=D$. Conversely, we have $\Phi(\Psi(B))=B$ for all binary trees $B$. Thus, $\Phi$ and $\Psi$ are inverses and we get that these maps are bijections.		To answer this question, we first need to define the sets $\mathcal{B}_{n}$ and $\mathcal{D}_{n}$, and describe the map $\Psi$.

$\mathcal{B}_{n}$ is the set of all binary trees with $n$ nodes, where a node is a vertex with 2 children. A leaf is a vertex with no children.

$\mathcal{D}_{n}$ is the set of all lattice paths from $(0,0)$ to $(n,n)$ that never go above the diagonal line $y=x$. A lattice path is a sequence of steps in the plane, where each step is either a unit step to the right (R) or a unit step up (U).

Now, let's describe the map $\Psi$:

Given a binary tree $B \in \mathcal{B}_{n}$, we can construct a lattice path $P \in \mathcal{D}_{n}$ as follows:

1. Perform an in-order traversal of the binary tree $B$. This means visiting the left subtree, the node, and then the right subtree recursively.
2. Whenever we visit a node, we take a step to the right (R) in the lattice path.
3. Whenever we visit a leaf, we take a step up (U) in the lattice path.

Now, we need to prove that $\Psi$ is a bijection between $\mathcal{B}_{n}$ and $\mathcal{D}_{n}$. To do this, we need to show that $\Psi$ is both injective (one-to-one) and surjective (onto).

Injectivity:

Suppose we have two different binary trees $B_1$ and $B_2$ with $n$ nodes each. Since they are different, there must be at least one node in $B_1$ that has a different structure than the corresponding node in $B_2$. This difference in structure will result in a different sequence of steps in the lattice paths constructed by $\Psi$. Therefore, $\Psi(B_1) \neq \Psi(B_2)$, and $\Psi$ is injective.

Surjectivity:

Consider any lattice path $P \in \mathcal{D}_{n}$. We can construct a binary tree $B$ as follows:

1. Start at the root of the binary tree.
2. For each step in the lattice path $P$, if the step is to the right (R), create a new node as a child of the current node and move to that node. If the step is up (U), move to the parent of the current node.

Since the lattice path $P$ never goes above the diagonal line $y=x$, we can always move to the parent of the current node when we encounter a step up (U). This construction ensures that the resulting binary tree $B$ has $n$ nodes, and $\Psi(B) = P$. Therefore, $\Psi$ is surjective.

Since $\Psi$ is both injective and surjective, it is a bijection between $\mathcal{B}_{n}$ and $\mathcal{D}_{n}$.		​
Think this one is impossible and we missed it
60
50Mathematics18.100BReal Analysis18.02NoneMidterm Exam 1Metric Spaces35TextLet $K$ be a non-empty compact set in a metric space $X$ and suppose $p \in$ $X \backslash K$. Show that there exists a point $q \in K$ such that
$$
d(p, q)=\inf \{d(p, x) ; x \in K\}
$$		OpenThe set $\{d(p, x) ; x \in K\}$ is a non-empty subset of $(0, \infty)$ so the infimum exists and there is a sequence $x_{n} \in K$ such that $d\left(p, x_{n}\right) \rightarrow$ $D=\inf \{d(p, x) ; x \in K\}$. Now, as a sequence in a compact set, $\left\{x_{n}\right\}$ has a convergent subsequence. Since $d\left(p, x_{n_{k}}\right)$ also converges to $D$, we may just assume that $x_{n} \rightarrow q$ in $X$. Since compact sets are closed, $q \in K$ and we just need to check that $d(p, q)=D$. By the definition of infimum and the convergence of the distance, give $\epsilon>0$ there exists $n$ such that
$$
\begin{array}{r}
\left|d\left(x, x_{n}\right)-D\right|<\epsilon / 2 \text { and } d\left(x_{n}, q\right)<\epsilon / 2 \text { but this implies that } \\
|d(p, q)-D|<\left|d(p, q)-d\left(p, x_{n}\right)\right|+\left|d\left(p, x_{n}\right)-D\right|<\epsilon
\end{array}
$$
for any $\epsilon>0$. Thus $d(p, q)=D$ as desired and the infimum of the distance is attained.
Here is a direct approach that a couple of people used. Set $D=$ $\inf \{d(p, x) ; x \in K\}$ and suppose that this is not attained on $K$, so for all $x \in K, d(p, x)>D$. Thus
$$
K \subset \bigcup_{x \in K} B\left(x, \frac{1}{2}(d(p, x)-D)\right)
$$
is an open cover, which therefore has a finite subcover since $K$ is compact. Let $x_{i}, i=1, \ldots, N$ be the centers of such a cover with $\epsilon_{i}=\frac{1}{2}\left(d\left(p, x_{i}\right)-D\right)$ and $\epsilon=\min _{i} \epsilon_{i}>0$. Then, each $x \in K$ is in one of these balls, so from the triangle inequality, for the appropriate $i$,
$$
d(p, x) \geq d\left(p, x_{i}\right)-\epsilon_{i} \geq D+\epsilon_{i} \geq D+\epsilon .
$$
This however shows that $D$ is not the infimum as it is defined to be, so there must be a point $q \in K$ with $d(p, q)=D$.
There is an even simpler direct approach used by several people. Suppose that $d(p, x)>r=\inf \{d(y, p) ; y \in K\}$ for all $x \in K$. Then the open sets
$$
G(n)=\{x \in K ; d(x, p)>r+1 / n\}
$$
form an open cover of $K$ which therefore must have a finite subcover by compactness. Since the $G(n)$ increase with $n, K \subset G(N)$ for some $N$ and hence $d(x, p)>r+1 / N$ for all $x \in K$, contradicting the definition of the infimum.
Another variant of this is to define $r=\inf \{d(x, p) ; x \in K\}$ and then to set
$$
K(n)=K \cap\{x \in X ; d(x, p) \leq r+1 / n\} .
$$
Since the second sets are closed, these are compact sets, being closed subsets of $K$, which are non-empty, by the definition of infimum, and decreasing as $n$ increases. Thus, by a theorem in Rudin, $T=\cap_{n} K(n) \neq \emptyset$. If $q \in T \subset K$ then $d(p, q)=r$ since $d(p, q) \leq r+1 / n$ for all $n$ and $d(p, q) \geq r$.		To show that there exists a point $q \in K$ such that $d(p, q) = \inf \{d(p, x) ; x \in K\}$, we will use the properties of compact sets and the definition of infimum.

Since $K$ is a non-empty compact set in a metric space $X$, for every open cover of $K$, there exists a finite subcover. Let $d(p, x)$ be the distance between points $p$ and $x$ in the metric space $X$. Define a function $f: K \rightarrow \mathbb{R}$ by $f(x) = d(p, x)$ for all $x \in K$. We want to show that there exists a point $q \in K$ such that $f(q) = \inf \{f(x) ; x \in K\}$.

First, we show that $f$ is continuous. Let $x, y \in K$ and $\epsilon > 0$. By the triangle inequality, we have
$$
|f(x) - f(y)| = |d(p, x) - d(p, y)| \leq d(x, y).
$$
So, for any $\epsilon > 0$, if we choose $\delta = \epsilon$, then $d(x, y) < \delta$ implies $|f(x) - f(y)| < \epsilon$. Thus, $f$ is continuous.

Now, since $K$ is compact and $f$ is continuous, $f(K)$ is also compact in $\mathbb{R}$. Compact sets in $\mathbb{R}$ are closed and bounded, so $f(K)$ has a lower bound. Let $m = \inf \{f(x) ; x \in K\}$. Since $f(K)$ is closed, $m \in f(K)$. Therefore, there exists a point $q \in K$ such that $f(q) = m$.

Hence, we have found a point $q \in K$ such that
$$
d(p, q) = \inf \{d(p, x) ; x \in K\}.
$$		​
61
5Mathematics18.100BReal Analysis18.02NoneProblem Set 1Axioms of Arithmetic61.071428571TextShow that the axioms of arithmetic and the axioms of ordering imply the following: if $x>y$, then $x^{3}>y^{3}$. [Besides the axioms, you can use any theorem proved in the first two lectures. If you feel underwhelmed by this pset, you can try to also prove the converse implication to this problem; however, no credit will be awarded for it.]OpenWe need to show that if $x-y \in P$, then $x^{3}-y^{3} \in P$. Write
$$
2\left(x^{3}-y^{3}\right)=(x-y)\left(2 x^{2}+2 x y+2 y^{2}\right)=(x-y)\left(x^{2}+y^{2}+(x+y)^{2}\right).
$$
The three squares $x^{2}, y^{2}$, and $(x+y)^{2}$ are either zero or in $P$ (theorem from the class: the square of any nonzero element lies in $P$ ). Moreover, since $x \neq y$ by assumption, at least one of the squares $x^{2}, y^{2}$ must be in $P$ (same theorem). It follows that $x^{2}+y^{2}+(x+y)^{2} \in P$. We now know, from the equation above and the assumption that $x-y \in P$, that $2\left(x^{3}-y^{3}\right) \in P$. If $x^{3}-y^{3}$ were not in $P$, it would either have to be zero, or $-\left(x^{3}-y^{3}\right)$ would have to be in $P$ (trichotomy), and then $2\left(x^{3}-y^{3}\right)$ would inherit the same properties, which is a contradiction. Hence, $x^{3}-y^{3}$ must be in $P$.
Now, the solution above definitely qualifies as sneaky (the much-hated "pull a formula out of a hat" trick). A more reasonable alternative (only sketched here) would be to first show the following Lemma: if $a$ is positive, and $b>c$, then $a b>a c$. (This follows directly from the axioms, all we're saying is that if $a \in P$ and $b-c \in P$, then $a b-a c=a(b-c) \in P$.). Using that, one can show the desired inequality if both $x$ and $y$ are positive:
$$
x^{3}=\left(x^{2}\right) x>\left(x^{2}\right) y=(x y) x>(x y) y=\left(y^{2}\right) x>\left(y^{2}\right) y=y^{3}.
$$
What about all the other situations? If both $x$ and $y$ are negative and satisfy $x>y$, then $(-x)$ and $(-y)$ are both positive, and $-x<-y$ (one can easily check that by reducing both properties to $x-y \in P$ ). The previous case tells us that $(-x)^{3}<(-y)^{3}$, but (using the fact that $(-a) \cdot(-b)=a b$ from the lecture) one sees easily that $-x^{3}=(-x)^{3}$ and $-y^{3}=(-y)^{3}$, so we now know that $-x^{3}<-y^{3}$, which (as before) yields $x^{3}>y^{3}$. There are still more cases, namely when one of the two numbers is zero, or when $x$ is positive and $y$ is negative; but those can be dealt with case-by-case quite easily.		To show that if $x > y$, then $x^3 > y^3$, we will use the axioms of arithmetic and the axioms of ordering.

Let $x > y$. We want to show that $x^3 > y^3$.

By the axioms of arithmetic, we can perform the following operations:

1. Subtract $y$ from both sides of the inequality: $x - y > 0$.
2. Multiply both sides by $(x^2 + xy + y^2)$: $(x - y)(x^2 + xy + y^2) > 0$.

Now, let's expand the left side of the inequality:

$(x - y)(x^2 + xy + y^2) = x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 = x^3 - y^3$.

So, we have $x^3 - y^3 > 0$.

By the axioms of ordering, we can add $y^3$ to both sides of the inequality to obtain:

$x^3 > y^3$.

Thus, if $x > y$, then $x^3 > y^3$.		​
62
12Mathematics18.6Probability and Random Variables18.02NoneProblem Set 1Counting7d0.1TextIn how many ways can 8 people be seated in a row if there are 5 men and they must sit next to each other?NumericalIf there are 5 men and they must sit next to each other, then there are $5! \cdot 4! = 2,880$ possible seating arrangements because the 5 men can be bundled together and permuted with the remaining 3 people.To solve this problem, we can treat the group of 5 men as a single unit. So, we have 4 units to arrange (the group of 5 men and the 3 other people). These 4 units can be arranged in 4! (4 factorial) ways.

Now, within the group of 5 men, they can be arranged in 5! (5 factorial) ways.

So, the total number of ways to seat the 8 people with the condition that the 5 men must sit next to each other is:

4! * 5! = 24 * 120 = 2880 ways.		Correct
63
143EECS18.C06Linear Algebra and Optimization18.02NoneMidterm Exam 1Projection Matrix6a0.9375TextConsider two $n \times n$ projection matrices
$$
P=I-v_{1} v_{1}^{\top} \quad \text { and } \quad Q=I-v_{2} v_{2}^{\top}
$$
where $v_{1}$ and $v_{2}$ have unit norm and are orthogonal to each other. Let $A=P Q$.
What is the dimension of $N(A)$? Find an orthonormal basis for $N(A)$.
Hint: You can find vectors in $N(A)$, but to show a vector is not in $N(A)$ you may want to use an orthogonal decomposition.		ExpressionThe dimension of $N(A)$ is two and $\left\{v_{1}, v_{2}\right\}$ forms an orthonormal basis. It is easy to see that $A v_{i}=0$. Moreover for any vector $v$ we can form an orthogonal decomposition $v=u+w$ where $u$ is in the span of $v_{1}$ and $v_{2}$ and $w$ is in the orthogonal complement. Then $A v=u$ and so if $v$ is not in the span of $v_{1}$ and $v_{2}$ it is not in the nullspace. To find the dimension of $N(A)$, we first need to find the matrix $A$. Since $P = I - v_1 v_1^T$ and $Q = I - v_2 v_2^T$, we have:

$$
A = PQ = (I - v_1 v_1^T)(I - v_2 v_2^T)
$$

Expanding the product, we get:

$$
A = I - v_1 v_1^T - v_2 v_2^T + v_1 v_1^T v_2 v_2^T
$$

Now, let's find an orthonormal basis for $N(A)$. Recall that $N(A)$ is the null space of $A$, which consists of all vectors $x$ such that $Ax = 0$. We can rewrite the equation $Ax = 0$ as:

$$
(I - v_1 v_1^T - v_2 v_2^T + v_1 v_1^T v_2 v_2^T)x = 0
$$

Since $v_1$ and $v_2$ are orthogonal and have unit norm, we have $v_1^T v_2 = 0$. Thus, $v_1^T v_2 v_2^T = 0$, and the last term in the equation above vanishes. We are left with:

$$
(I - v_1 v_1^T - v_2 v_2^T)x = 0
$$

Now, let's consider the orthogonal decomposition of $x$:

$$
x = c_1 v_1 + c_2 v_2 + x_\perp
$$

where $c_1$ and $c_2$ are scalars, and $x_\perp$ is orthogonal to both $v_1$ and $v_2$. Applying the matrix $A$ to this decomposition, we get:

$$
A x = (I - v_1 v_1^T - v_2 v_2^T)(c_1 v_1 + c_2 v_2 + x_\perp)
$$

Since $v_1$ and $v_2$ are orthogonal to each other and have unit norm, we have:

$$
A x = c_1 (I - v_1 v_1^T - v_2 v_2^T)v_1 + c_2 (I - v_1 v_1^T - v_2 v_2^T)v_2 + (I - v_1 v_1^T - v_2 v_2^T)x_\perp
$$

Now, notice that $(I - v_1 v_1^T - v_2 v_2^T)v_1 = 0$ and $(I - v_1 v_1^T - v_2 v_2^T)v_2 = 0$. Thus, we have:

$$
A x = (I - v_1 v_1^T - v_2 v_2^T)x_\perp
$$

Since $x_\perp$ is orthogonal to both $v_1$ and $v_2$, we have $v_1^T x_\perp = 0$ and $v_2^T x_\perp = 0$. Therefore, $x_\perp$ is in the null space of $A$. Moreover, any vector in the null space of $A$ can be written as a linear combination of $v_1$, $v_2$, and $x_\perp$. Thus, the dimension of $N(A)$ is 1, and an orthonormal basis for $N(A)$ is given by $\{x_\perp\}$.		Incorrect
Math is mumbo jumbo
64
38Mathematics18.701Algebra I18.100BNoneProblem Set 5Orthogonal Matrices and Rotations3a0.25TextLet $A$ be a $3 \times 3$ orthogonal matrix with det $A=1$, whose angle of rotation is different from 0 or $\pi$, and let $M=A-A^t$.
Show that $M$ has rank 2, and that a nonzero vector $X$ in the nullspace of $M$ is an eigenvector of $A$ with eigenvalue 1.		OpenLet $A$ be a rotation matrix, an element of $S O_{3}$. If a vector $X$ is fixed by $A$, it is also fixed by its inverse $A^{-1}=A^{t}$, and therefore $M X=\left(A-A^{t}\right) X=0$. The rank of $M$ is less than 3 . Conversely, if $M X=0$, then $A X=A^{-1} X$. When the angle of rotation isn't 0 or $\pi$, this happens only for vectors $X$ in the axis of rotation, so the $\operatorname{rank}$ of $M$ is 2.To show that $M$ has rank 2, we will first show that $M$ is a skew-symmetric matrix and then show that its nullspace has dimension 1.

Since $A$ is an orthogonal matrix, we have $A^tA = AA^t = I$. Now, let's compute $M^t$:

$$M^t = (A - A^t)^t = A^t - A^{tt} = A^t - A$$

Now, let's compute $MM^t$:

$$MM^t = (A - A^t)(A^t - A) = A(A^t) - A(A) - A^t(A^t) + A^t(A) = I - A - I + A^t = A^t - A = -M$$

Since $MM^t = -M$, we can conclude that $M$ is a skew-symmetric matrix.

Now, let's find the nullspace of $M$. If $X$ is in the nullspace of $M$, then $MX = 0$. We have:

$$MX = (A - A^t)X = AX - A^tX$$

Since $A$ is orthogonal, we can multiply both sides by $A^t$:

$$A^tMX = A^t(AX - A^tX) = A^tAX - A^tA^tX = X - X = 0$$

Thus, $X$ is in the nullspace of $M$ if and only if $AX = A^tX$. Since the angle of rotation of $A$ is different from 0 or $\pi$, there is only one eigenvalue of $A$ equal to 1. Therefore, the nullspace of $M$ has dimension 1.

Since the nullspace of $M$ has dimension 1, the rank of $M$ is $3 - 1 = 2$.

Now, let's show that a nonzero vector $X$ in the nullspace of $M$ is an eigenvector of $A$ with eigenvalue 1. If $X$ is in the nullspace of $M$, then $MX = 0$. We have:

$$MX = (A - A^t)X = AX - A^tX = 0$$

Adding $A^tX$ to both sides, we get:

$$AX = A^tX$$

Since $A$ is orthogonal, we can multiply both sides by $A^t$:

$$A^tAX = A^tA^tX = X$$

Thus, $AX = X$, which means that $X$ is an eigenvector of $A$ with eigenvalue 1.		Incorrect
Actually rotation not skew matrix
65
68EECS6.191Computation Structures6.100A, 8.02NonePrelab Questions 7Ideal Cache Behavior2d0.012TextWe will be using the following program to examine our cache behavior. Let N = 16 be the size of data region, in words. Let A be an array of N elements, located initially at 0x240. Note that these values are hardcoded into the program below but we will be changing them later.
// A = 0x240, starting address of array
// N = 16, size of data region
// this program adds 16 words from array A, then repeats.
. = 0x200
test:
li a0, 16 // initialize loop index i
li a1, 0 // sum = 0
loop: // add up elements in array
addi a0, a0, -1 // decrement index
slli a2, a0, 2 // convert to index byte offset
lw a3, 0x240(a2) // load value of A[i]
add a1, a1, a3 // add to sum
bnez a0, loop // loop until all words are summed
j test // perform test again!
// Array
. = 0x240
.word ... // A[0]
.word ... // A[1]
...
.word ... // A[15]
Our cache has a total of 64 words. The initial configuration is direct mapped, with 1 word per line, so the cache has 64 lines numbered 0-63 (0x00 - 0x3F).
To achieve 100% steady state hit ratio, it must be the case that the instructions and array data can reside in the cache at the same time. Let's check if this is currently the case.
Which cache line (index) does the last data element, A[15], map to? Provide your answer in hexadecimal.		Numerical0x1F.
The address of the last data element, A[15], is 0x27C = 0b_0010_0111_1100. Bits[7:2] are the index bits = 0b011111 = 0x1F (or line 31).		To find the cache line index that A[15] maps to, we first need to find the address of A[15]. Since A is an array of 16 elements and each element is a word (4 bytes), the address of A[15] can be calculated as follows:

Address of A[15] = Starting address of A + (15 * 4)
Address of A[15] = 0x240 + (15 * 4)
Address of A[15] = 0x240 + 0x3C
Address of A[15] = 0x27C

Now, we need to find the cache line index that this address maps to. Since the cache is direct-mapped with 1 word per line and a total of 64 lines, we can calculate the cache line index as follows:

Cache line index = (Address of A[15] / 4) % 64
Cache line index = (0x27C / 4) % 64
Cache line index = (0x9F) % 64
Cache line index = 31

So, the last data element, A[15], maps to cache line index 31, which is 0x1F in hexadecimal.		Correct
66
35Mathematics18.01Calculus INoneNoneProblem Set 1Exponentials and Logarithms16b0.07919746568TextIf $2^{100}=10^{t}$, which of the following is the best approximation of $t: 10$ or 20 or 30 or 40 or $50 ?$ (If you want, you can use that $\log _{2} 10=3.32 \ldots$)Multiple ChoiceIn words, $\log _{2} 10 \approx 3.32$ means that there are approximately $3.32$ factors of 2 in a factor of 10 . Thus, 100 factors of 2 are, approximately, 100/3.32 $\approx 30$ factors of 10.
$$
2^{100} \approx 10^{30} .
$$		We are given that $2^{100} = 10^t$. To find the best approximation for $t$, we can take the logarithm of both sides of the equation. Using the given value for $\log_2 10$, we have:

$\log_2(2^{100}) = \log_2(10^t)$

Using the property of logarithms that $\log_a(a^x) = x$, we get:

$100 = t \cdot \log_2 10$

Now, we can use the given approximation for $\log_2 10$:

$100 = t \cdot 3.32$

To find the best approximation for $t$, we can divide both sides by 3.32:

$t \approx \frac{100}{3.32} \approx 30.12$

Since we are asked to choose the best approximation from the given options, the best approximation for $t$ is $\boxed{30}$.		Correct
67
346EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneLab 13Nearest Neighbors40.4166666667TextSuppose we are interested in comparing decision trees and kNNs in a new application. We have 10 million data points for training. Suppose every leaf in our trained decision tree has a depth of 5. Now a test point comes along, and we are interested in making a prediction at the new point. At testing time, about how many operations would it take to make a prediction using our decision tree? At testing time, about how many operations would it take to make a prediction using kNNs?OpenFor the decision tree: We need to check our test data set at every split in the decision tree. If every leaf is at a depth of 5 , we expect to check 5 splits.
For kNNs: it seems like we have to compare our test point to all 10 million points. There are more clever ways to handle this in practice, but kNNs can be expensive at test time if there's a lot of data.		To make a prediction using a decision tree with a depth of 5, we would need to traverse the tree from the root to a leaf node. In the worst case, this would require 5 operations (one for each level of the tree). Therefore, it would take about 5 operations to make a prediction using the decision tree.

For kNNs, the number of operations required to make a prediction depends on the number of data points and the value of k (the number of nearest neighbors considered). In general, the algorithm needs to calculate the distance between the test point and each of the 10 million training data points, sort the distances, and then select the k nearest neighbors. The complexity of these operations can vary depending on the implementation, but a rough estimate would be:

1. Calculating distances: 10 million operations
2. Sorting distances: O(10 million * log(10 million)) operations
3. Selecting k nearest neighbors: k operations

In this case, the number of operations for kNNs would be significantly higher than for the decision tree, especially considering the large number of data points (10 million).		Correct
68
3EECS18.C06Linear Algebra and Optimization18.02NoneProblem Set 1Vector Spaces2a0.1851851852TextTrue or False: The columns of the matrix $\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$ span $\mathbb{R}^{3}$.Multiple ChoiceTrue. We can write any vector as a linear combination of the columns, since these are the vectors of the standard basis (i.e., columns of the identity matrix) in a different order.VRAICorrect
No reasoning given
69
30EECS6.18Computer Systems Engineering6.1010, 6.1910NoneHands-on 5Traceroute50.1666666667TextAt the command prompt, type:
traceroute 18.31.0.200
Describe what is strange about the observed output, and why traceroute gives you such an output. Refer to the traceroute man page for useful hints. Copy/paste any of the relevant portions of output below.		OpenAfter the 4th hop, the hops begin to oscillate between 18.69.3.2 and 18.4.7.65, until the maximum number of hops is reached.
5 MITNET.CORE-1-EXT.CSAIL.MIT.EDU (18.4.7.65) 9.058 ms 9.099 ms 9.344 ms
...
25 MITNET.CORE-1-EXT.CSAIL.MIT.EDU (18.4.7.65) 9.040 ms 9.077 ms 9.035 ms
26 DMZ-RTR-2-CSAIL.MIT.EDU (18.4.7.1) 8.764 ms 8.878 ms 8.836 ms
27 MITNET.CORE-1-EXT.CSAIL.MIT.EDU (18.4.7.65) 8.813 ms 9.049 ms 9.021 ms
28 DMZ-RTR-2-CSAIL.MIT.EDU (18.4.7.1) 8.562 ms 8.726 ms 8.797 ms
29 MITNET.CORE-1-EXT.CSAIL.MIT.EDU (18.4.7.65) 9.335 ms 9.373 ms 9.381 ms
30 DMZ-RTR-2-CSAIL.MIT.EDU (18.4.7.1) 8.980 ms 9.224 ms 9.000 ms
One possible reason for this is the existence of a “routing loop” by which each of the two routers thinks the other has a route to the destination, i.e. that each has a path to the destination that contains the other. This can happen in link-state routing, in part because details of the path being advertised are not shared.		As an AI language model, I cannot directly interact with a command prompt or execute commands. However, I can help you understand the possible strange output you might observe when running the traceroute command and why it happens.

When you run the command "traceroute 18.31.0.200", it sends packets to the destination IP address (18.31.0.200) and records the route taken by the packets, as well as the time it takes for each hop.

A strange output could be:

1. Asterisks (*) in the output: This indicates that the response has timed out for that particular hop. It could be due to network congestion, packet filtering, or a firewall blocking the ICMP packets used by traceroute.

2. High latency or varying latency: This could indicate network congestion or issues with the intermediate routers.

3. Multiple IP addresses for a single hop: This could be due to load balancing or multiple paths between the source and destination.

4. Unusual hostnames or IP addresses: This could be due to misconfigured DNS or routing issues.

To understand the output better, you can refer to the traceroute man page by typing "man traceroute" in the command prompt. This will provide you with detailed information about the command, its options, and the output format.

Please provide the specific output you observed if you need further assistance in understanding the results.		Incorrect
70
461EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneProblem Set 3Gradient Descent10aii0.02083333333TextLast week, we defined the _ridge regression_objective function.
$$
J_{\text {ridge }}\left(\theta, \theta_{0}\right)=\frac{1}{n} \sum_{i=1}^{n}\left(\theta^{T} x^{(i)}+\theta_{0}-y^{(i)}\right)^{2}+\lambda\|\theta\|^{2}
$$
Recall that in last week's homework, we derived the closed-form solution for the $\theta$ value that minimizes the least squares loss function. However, it is computationally challenging to compute the closed-form solution on $\theta$ on high-dimensional data and large datasets.
This week, we wish to apply gradient descent and stochastic gradient descent to minimize the ridge regression function. In order to use the gradient descent and stochastic gradient descent functions implemented previously in the homework, we must add ones to the end of each datapoint in $X$, which we implemented in the add_ones_row function in homework 1.
In the next subsections, we assume that $X$ is a $d \times n$ matrix, $Y$ is a $1 \times n$ matrix, and $\theta$ is a $d \times 1$ matrix. Rewriting the ridge objective through matrix operations, we find that:
$$
J_{\text {ridge }}(\theta)=\frac{1}{n}\left(\theta^{T} X-Y\right)\left(\theta^{T} X-Y\right)^{T}+\lambda\|\theta\|^{2}
$$
For the rest of the problem, assume that $\mathrm{X}$ already has ones at the end of each datapoint. You do not need to call the add_one_rows function.
Implement objective_func. objective_func returns a function that computes $J_{\text {ridge }}(\theta)$.
inputs:
X: a (dxn) numpy array.
Y: a (1xn) numpy array
lam: regularization parameter
outputs:
f : a function that takes in a (dx1) numpy array "theta" and returns *as a float* the value of the ridge regression objective when theta(the variable)="theta"(the numpy array)
def objective_func(X, Y, lam):
def f(theta):
# write your implementation here
pass
return f		Programmingdef objective_func(X, Y, lam):
def f(theta):
n = X.shape[1]
sq_loss = (1/n) *(theta.T @ X - Y) @ (theta.T @ X - Y).T
regularizer = lam* theta.T @ theta
return (sq_loss + regularizer).item()
return f		To implement the objective_func function, you need to define the f function inside it, which computes the ridge regression objective value given a theta input. Here's the implementation:

```python
import numpy as np

def objective_func(X, Y, lam):
def f(theta):
n = X.shape[1]
term1 = (1/n) * np.sum(np.square(np.dot(theta.T, X) - Y))
term2 = lam * np.sum(np.square(theta))
return term1 + term2
return f
```

This implementation first calculates the two terms in the ridge regression objective function separately, and then adds them together to return the final objective value. The f function takes in a (dx1) numpy array "theta" and returns the value of the ridge regression objective as a float.		Correct
71
33EECS6.2Electrical Circuits: Modeling and Design of Physical Systems8.02NoneProblem Set 7Thermal System2e0.2380952381TextAs mentioned in our first lecture, electrical circuit is a good mathematical language for modelling other non-electrical systems such as mechanical, biological systems, etc. Here we are going to use circuit to analyze a thermal system.
Prior to $t=0$, the building considered in (D) reaches thermal equilibrium with the environment, i.e. $T_{i}=T_{e}$. Then at $t=0$, the owner of the building turns on the heater which has a constant power Q. What is the final temperature that the interior of the building will reach? Express the interior temperature as a function of time $T_{i}(t)$ for $t>$ 0. Note that the environment temperature remains constant at $T_{\mathrm{e}}$.		Expression$T_{i}^{(t+\infty)} =T_{e}+\frac{Q}{\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}+\frac{1}{R_{6}}}$.To analyze the thermal system using an electrical circuit analogy, we can consider the heat transfer between the building and the environment as a thermal resistance, and the heat capacity of the building as a thermal capacitance. The heater can be represented as a constant power source.

Let's denote the thermal resistance as R and the thermal capacitance as C. The heat transfer between the building and the environment can be represented as:

Heat transfer = (T_i - T_e) / R

The rate of change of the interior temperature with respect to time can be represented as:

dT_i/dt = Q/C - (T_i - T_e) / (R*C)

This is a first-order linear differential equation. To solve it, we can use the integrating factor method. The integrating factor is:

IF = exp(∫(1/(R*C)) dt) = exp(t/(R*C))

Multiplying both sides of the differential equation by the integrating factor, we get:

exp(t/(R*C)) * dT_i/dt + (1/(R*C)) * exp(t/(R*C)) * T_i = Q/C * exp(t/(R*C))

Now, the left side of the equation is the derivative of the product of the integrating factor and the dependent variable (T_i):

d/dt (T_i * exp(t/(R*C))) = Q/C * exp(t/(R*C))

Integrating both sides with respect to time, we get:

T_i * exp(t/(R*C)) = (Q/C) * (R*C) * (1 - exp(-t/(R*C))) + K

where K is the integration constant. To find K, we use the initial condition at t=0:

T_i(0) = T_e

T_e * exp(0) = (Q/C) * (R*C) * (1 - exp(0)) + K

K = T_e

So, the expression for the interior temperature as a function of time is:

T_i(t) = T_e + (Q/C) * (R*C) * (1 - exp(-t/(R*C))) * exp(-t/(R*C))

For the final temperature, we need to find the limit as t approaches infinity:

T_i(∞) = T_e + (Q/C) * (R*C) * (1 - exp(-∞/(R*C))) * exp(-∞/(R*C))

T_i(∞) = T_e + (Q/C) * (R*C) * (1 - 0) * 0

T_i(∞) = T_e + Q * R

So, the final temperature that the interior of the building will reach is T_e + Q * R.		Incorrect
72
194EECS6.411Representation, Inference, and Reasoning in AI6.1010, 6.1210, 18.600NoneProblem Set 5Localization with Viterbi4d0.2604166667TextIn this section, we will implement an HMM for a robot that is moving around randomly in a 2D grid with obstacles. The robot has sensors that allow it to detect obstacles in its immediate vicinity. It knows the grid map, with the locations of all obstacles, but it is uncertain about its own location in the grid. We will use Viterbi to determine the most likely locations for the robot given a sequence of local and potentially noisy observations.
Concretely, we will represent the 2D grid with obstacles as a list of lists, where 1s represent obstacles and 0s represent free space. Example:
obstacle_map = [
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0],
[1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1],
[1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0],
[0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0]
]
The state of the robot is its location in the grid, represented as a tuple of ints, (row, col). Transitions are uniformly distributed amongst the robot's current location and the neighboring free (not obstacle) locations, where neighboring = 4 cardinal directions (up, down, left, right).
Observations are a 4-tuple that list which directions have obstacles, in order [N E S W], with a 1 for an obstacle and 0 for no obstacle. Observations that are "off the map" are 1, as though they are obstacles. For instance, in the map above, if there were no observation noise, then the observation for the top left corner (state=(0, 0)) would be (1, 0, 1, 1). Observations can also be corrupted with noise; see the create_observation_potential docstring for more details.
Our ultimate task will be to take in a sequence of observations and return the corresponding sequence of most likely states.
Write a function that creates a random variable for an observation at a given time step in an obstacle HMM. See docstring for description.
For reference, our solution is 3 line(s) of code.
def create_observation_variable(name):
'''Creates a RV for the HMM observation with the given name.
Observations are a 4-tuple that list which directions have obstacles,
in order [N E S W], with a 1 for an obstacle and 0 for no obstacle.
Observations that are "off the map" are 1, as though they are obstacles.
For instance, in the following map:
obstacle_map = [
[0, 1],
[0, 0],
]
if there were no observation noise, then the observation for the top left
location would be (1, 1, 0, 1).
The domain of the observation variable should be a list of 4-tuples.
Hint: you may find it useful to use `itertools.product`. For example,
see what happens with `list(itertools.product(["foo", "bar"], repeat=2))`.
Args:
name: A str name for the variable.
Returns:
zt: A RV as described above.
'''
raise NotImplementedError("Implement me!")		Programmingdef create_observation_variable(name):
'''Creates a RV for the HMM observation with the given name.
Observations are a 4-tuple that list which directions have obstacles,
in order [N E S W], with a 1 for an obstacle and 0 for no obstacle.
Observations that are "off the map" are 1, as though they are obstacles.
For instance, in the following map:
obstacle_map = [
[0, 1],
[0, 0],
]
if there were no observation noise, then the observation for the top left
location would be (1, 1, 0, 1).
The domain of the observation variable should be a list of 4-tuples.
Hint: you may find it useful to use `itertools.product`. For example,
see what happens with `list(itertools.product(["foo", "bar"], repeat=2))`.
Args:
name: A str name for the variable.
Returns:
zt: A RV as described above.
'''
domain = list(itertools.product([0, 1], repeat=4))
return RV(name, domain)		import itertools

def create_observation_variable(name):
domain = list(itertools.product([0, 1], repeat=4))
zt = RandomVariable(name, domain)
return zt		Correct
Incorrect because didn't have access to function definition. Get's core of problem.
73
116EECS6.191Computation Structures6.100A, 8.02NoneProblem Set 7Caches22.7TextImplement a direct-mapped cache by completing the DirectMappedCache module in DirectMappedCache.ms. Note that you should also keep track of hit and miss counts.

import CacheTypes;
import CacheHelpers;
import MainMemory;

// ReqStatus (defined in CacheTypes.ms) is used to keep track of the state of the current request
//typedef enum {
// Ready, // The cache is ready for a new request
// Lookup, // Issued a lookup to tag/data arrays
// Writeback, // In main memory access for dirty writeback
// Fill // In main memory access for requested data
//} ReqStatus;
//
// Possible flows:
// HIT: Ready -> Lookup -> Ready
// MISS, line is clean: Ready -> Lookup -> Fill
// MISS, line is dirty: Ready -> Lookup -> Writeback -> Fill

// Cache SRAM Synonyms (defined in CacheTypes.ms)
// You may find the following type synonyms helpful to access the tag/data/status arrays
// typedef SRAMReq#(logCacheSets, CacheTag) TagReq;
// typedef SRAMReq#(logCacheSets, Line) DataReq;
// typedef SRAMReq#(logCacheSets, CacheStatus) StatusReq;

// TODO: Complete the implementation of DirectMappedCache
// NOTE: Implementing this module requires about 50 lines of additional code
// (~40 lines in rule tick, ~5-10 lines in method data, 1 line in method reqEnabled, 1 line in function isHit)
module DirectMappedCache(MainMemory mainMem);
// SRAM arrays. Note that, for a direct-mapped cache,
// number of cache sets == number of cache lines
SRAM#(logCacheSets, Line) dataArray;
SRAM#(logCacheSets, CacheTag) tagArray;
SRAM#(logCacheSets, CacheStatus) statusArray;

// Registers for holding the current state of the cache and how far along
// it is in processing a request.
RegU#(MemReq) curReq;
Reg#(ReqStatus) state(Ready);

// Hit/miss counters
Reg#(Word) hits(0);
Reg#(Word) misses(0);

input Maybe#(MemReq) req default = Invalid;

// TODO return True if the cache can accept a new request
method Bool reqEnabled = False;

// TODO return True if the cache is in lookup and it is a hit
function Bool isHit;
return False;
endfunction

rule tick;
if (state == Ready && isValid(req)) begin
// TODO Your code here
end else if (state == Lookup) begin
// TODO Your code here
end else if (state == Writeback && mainMem.reqEnabled) begin
// TODO Your code here
end else if (state == Fill && isValid(mainMem.data)) begin
// TODO Your code here
end
endrule

method Maybe#(Word) data;
// This method should return a Valid output in only two cases:
// 1. On a load hit (it is a hit, and curReq.op == Ld).
// 2. On a fill for a load request (we're in the Fill state,
// mainMem.data is valid, and curReq.op == Ld).
// In all other cases, the output should be Invalid
//
// NOTE: You should be checking the above conditions explicitly in
// THIS method so you can return data as soon as possible.
// DO NOT place your output into a register in the rule and then
// simply return that register here.
// This function should take about 4-8 lines of code to implement.
// TODO Your code here.
return Valid(0);
endmethod
method Bit#(32) getHits = hits;
method Bit#(32) getMisses = misses;
endmodule		Programmingimport CacheTypes;
import CacheHelpers;
import MainMemory;

// ReqStatus (defined in CacheTypes.ms) is used to keep track of the state of the current request
//typedef enum {
// Ready, // The cache is ready for a new request
// Lookup, // Issued a lookup to tag/data arrays
// Writeback, // In main memory access for dirty writeback
// Fill // In main memory access for requested data
//} ReqStatus;
//
// Possible flows:
// HIT: Ready -> Lookup -> Ready
// MISS, line is clean: Ready -> Lookup -> Fill
// MISS, line is dirty: Ready -> Lookup -> Writeback -> Fill

// Cache SRAM Synonyms (defined in CacheTypes.ms)
// You may find the following type synonyms helpful to access the tag/data/status arrays
// typedef SRAMReq#(logCacheSets, CacheTag) TagReq;
// typedef SRAMReq#(logCacheSets, Line) DataReq;
// typedef SRAMReq#(logCacheSets, CacheStatus) StatusReq;

// TODO: Complete the implementation of DirectMappedCache
// NOTE: Implementing this module requires about 50 lines of additional code
// (~40 lines in rule tick, ~5-10 lines in method data, 1 line in method reqEnabled, 1 line in function isHit)
module DirectMappedCache(MainMemory mainMem);
// SRAM arrays. Note that, for a direct-mapped cache,
// number of cache sets == number of cache lines
SRAM#(logCacheSets, Line) dataArray;
SRAM#(logCacheSets, CacheTag) tagArray;
SRAM#(logCacheSets, CacheStatus) statusArray;

// Registers for holding the current state of the cache and how far along
// it is in processing a request.
RegU#(MemReq) curReq;
Reg#(ReqStatus) state(Ready);

// Hit/miss counters
Reg#(Word) hits(0);
Reg#(Word) misses(0);

input Maybe#(MemReq) req default = Invalid;

// TODO return True if the cache can accept a new request
method Bool reqEnabled = state == Ready;

// TODO return True if the cache is in lookup and it is a hit
function Bool isHit(CacheTag tag);
return state == Lookup && getTag(curReq.addr) == tag;
endfunction

rule tick;
if (state == Ready && isValid(req)) begin
// TODO Your code here
curReq <= fromMaybe(?, req);
MemReq newReq = fromMaybe(?, req);
CacheIndex index = getIndex(newReq.addr);
tagArray.req = Valid(TagReq{addr: index, write: False, data: ?});
statusArray.req = Valid(StatusReq{addr: index, write: False, data: ?});
dataArray.req = Valid(DataReq{addr: index, write: False, data: ?});
state <= Lookup;
end else if (state == Lookup) begin
// TODO Your code here
CacheIndex index = getIndex(curReq.addr);
let tag = fromMaybe(?, tagArray.data);
let status = fromMaybe(?, statusArray.data);
let data = fromMaybe(?, dataArray.data);
if (isHit(tag) && status != NotValid) begin
hits <= hits + 1;
if (curReq.op == St) begin
Line newLine = data;
newLine[getWordOffset(curReq.addr)] = curReq.data;
dataArray.req = Valid(DataReq{addr: index, write: True, data: newLine});
statusArray.req = Valid(StatusReq{addr: index, write: True, data: Dirty});
end
state <= Ready;
end else begin
misses <= misses + 1;
if (status == NotValid || status == Clean) begin
mainMem.req = Valid(LineReq{op: Ld, lineAddr: getLineAddr(curReq.addr), data: ?});
state <= Fill;
end else if (status == Dirty) begin
mainMem.req = Valid(LineReq{op: St, lineAddr: {tag, index}, data: data});
state <= Writeback;
end
end
end else if (state == Writeback) begin
// TODO Your code here
if (mainMem.reqEnabled == True) begin
mainMem.req = Valid(LineReq{op: Ld, lineAddr: getLineAddr(curReq.addr), data: ?});
state <= Fill;
end
end else if (state == Fill) begin
// TODO Your code here
if (isValid(mainMem.data)) begin
CacheIndex index = getIndex(curReq.addr);
CacheTag newTag = getTag(curReq.addr);
Line newLine = fromMaybe(?, mainMem.data);
if (curReq.op == Ld) begin
statusArray.req = Valid(StatusReq{addr: index, write: True, data: Clean});
end else
if (curReq.op == St) begin
statusArray.req = Valid(StatusReq{addr: index, write: True, data: Dirty});
newLine[getWordOffset(curReq.addr)] = curReq.data;
end
tagArray.req = Valid(TagReq{addr: index, write: True, data: newTag});
dataArray.req = Valid(DataReq{addr: index, write: True, data: newLine});
state <= Ready;
end
end
endrule

method Maybe#(Word) data;
// This method should return a Valid output in only two cases:
// 1. On a load hit (it is a hit, and curReq.op == Ld).
// 2. On a fill for a load request (we're in the Fill state,
// mainMem.data is valid, and curReq.op == Ld).
// In all other cases, the output should be Invalid
//
// NOTE: You should be checking the above conditions explicitly in
// THIS method so you can return data as soon as possible.
// DO NOT place your output into a register in the rule and then
// simply return that register here.
// This function should take about 4-8 lines of code to implement.
// TODO Your code here.
let tag = fromMaybe(?, tagArray.data);
if (isHit(tag) && curReq.op == Ld) begin
Line newLine = fromMaybe(?, dataArray.data);
return Valid(newLine[getWordOffset(curReq.addr)]);
end else if (state == Fill && isValid(mainMem.data) && curReq.op == Ld) begin
Line newLine = fromMaybe(?, mainMem.data);
return Valid(newLine[getWordOffset(curReq.addr)]);
end else return Invalid;
endmethod
method Bit#(32) getHits = hits;
method Bit#(32) getMisses = misses;
endmodule		Here is the completed DirectMappedCache module:

```bluespec
module DirectMappedCache(MainMemory mainMem);
SRAM#(logCacheSets, Line) dataArray;
SRAM#(logCacheSets, CacheTag) tagArray;
SRAM#(logCacheSets, CacheStatus) statusArray;

RegU#(MemReq) curReq;
Reg#(ReqStatus) state(Ready);

Reg#(Word) hits(0);
Reg#(Word) misses(0);

input Maybe#(MemReq) req default = Invalid;

method Bool reqEnabled = (state == Ready);

function Bool isHit;
return (state == Lookup) && (statusArray.read(curReq.addr[logCacheSets-1:0]).status == Valid) && (tagArray.read(curReq.addr[logCacheSets-1:0]) == curReq.addr[logCacheSets+logCacheLines-1:logCacheSets]);
endfunction

rule tick;
if (state == Ready && isValid(req)) begin
curReq <= req;
state <= Lookup;
dataArray.read(curReq.addr[logCacheSets-1:0]);
tagArray.read(curReq.addr[logCacheSets-1:0]);
statusArray.read(curReq.addr[logCacheSets-1:0]);
end else if (state == Lookup) begin
if (isHit) begin
if (curReq.op == Ld) begin
hits <= hits + 1;
end else begin
dataArray.write(curReq.addr[logCacheSets-1:0], curReq.data);
end
state <= Ready;
end else begin
misses <= misses + 1;
if (statusArray.read(curReq.addr[logCacheSets-1:0]).status == Dirty) begin
state <= Writeback;
end else begin
state <= Fill;
end
mainMem.req <= curReq;
end
end else if (state == Writeback && mainMem.reqEnabled) begin
mainMem.req <= curReq;
state <= Fill;
end else if (state == Fill && isValid(mainMem.data)) begin
dataArray.write(curReq.addr[logCacheSets-1:0], mainMem.data);
tagArray.write(curReq.addr[logCacheSets-1:0], curReq.addr[logCacheSets+logCacheLines-1:logCacheSets]);
statusArray.write(curReq.addr[logCacheSets-1:0], Valid);
state <= Ready;
end
endrule

method Maybe#(Word) data;
if (isHit && curReq.op == Ld) begin
return dataArray.read(curReq.addr[logCacheSets-1:0]);
end else if (state == Fill && isValid(mainMem.data) && curReq.op == Ld) begin
return mainMem.data;
end
return Invalid;
endmethod

method Bit#(32) getHits = hits;
method Bit#(32) getMisses = misses;
endmodule
```

This implementation completes the DirectMappedCache module by implementing the reqEnabled method, isHit function, tick rule, and data method. It also keeps track of hit and miss counts.		Incorrect
74
205Mathematics18.01Calculus INoneNoneProblem Set 5Second Derivatives9c0.04751847941TextA graphing problem using second derivatives. Let $f(x)=x e^{-x}$. We will graph $f(x)$ on the range $0 \leq x$. But first we compute some information about $f(x)$.
For which $x$ is $f^{\prime \prime}(x)=0$ ? For which $x$ is $f^{\prime \prime}(x)>0$ and for which $x$ is $f^{\prime \prime}(x)<0$ ?		Expression$f^{\prime \prime}(x)=0$ (inflection point) when $x=2 . f^{\prime \prime}(x)>0$ (concave up) when $x>2$. $f^{\prime \prime}(x)<0$ (concave down) when $x<2$.To answer this question, we first need to find the first and second derivatives of the function $f(x) = xe^{-x}$.

First derivative:
Using the product rule, we have
$f'(x) = (1)(e^{-x}) + (x)(-e^{-x}) = e^{-x} - xe^{-x}$.

Second derivative:
Now, we need to find the derivative of $f'(x)$. We will again use the product rule:
$f''(x) = (1)(-e^{-x}) + (1)(-e^{-x}) + (x)(e^{-x}) = -2e^{-x} + xe^{-x}$.

Now, we need to find the values of $x$ for which $f''(x) = 0$, $f''(x) > 0$, and $f''(x) < 0$.

1. $f''(x) = 0$:
To find the values of $x$ for which $f''(x) = 0$, we need to solve the equation $-2e^{-x} + xe^{-x} = 0$.
Factoring out $e^{-x}$, we get
$e^{-x}(-2 + x) = 0$.
Since $e^{-x}$ is never equal to 0, we only need to solve for $-2 + x = 0$.
So, $x = 2$.

2. $f''(x) > 0$:
We need to find the values of $x$ for which $-2e^{-x} + xe^{-x} > 0$.
Dividing both sides by $e^{-x}$ (which is always positive), we get
$-2 + x > 0$.
So, $x > 2$.

3. $f''(x) < 0$:
We need to find the values of $x$ for which $-2e^{-x} + xe^{-x} < 0$.
Dividing both sides by $e^{-x}$ (which is always positive), we get
$-2 + x < 0$.
So, $x < 2$.

In conclusion, $f''(x) = 0$ when $x = 2$, $f''(x) > 0$ when $x > 2$, and $f''(x) < 0$ when $x < 2$.		Correct
75
181EECS6.191Computation Structures6.100A, 8.02NoneMidterm Exam 3Operating Systems2b0.45TextConsider the following two processes running RISC-V programs labeled with virtual addresses. Note that all pseudoinstructions in this code translate into a single RISC-V instruction.
Program for Process A
.=0x200
li t0, 5000
li t1, 0
loop:
li a0, 0x300
li a7, 0x13 // print system call
ecall
addi t1, t1, 1
ble t1, t0, loop
j exit // exit process
.=0x300
.ascii “Hello from process A\n”
Program for Process B
.=0x450
li t0, 50
li t1, 10
div t2, t0, t1
sw t2, 0x900(x0)
li a0, 0x600
li a7, 0x13 // print system call
ecall
j exit // exit process
.=0x600
.ascii “Hello from Process B\n”
Assume the OS schedules Process B first. For the following questions, if you can’t tell a value based on the information given, write CAN’T TELL.
What are the values in the following registers just after the first ecall in Process A completes?
t0:
t1:
pc:		Numericalt0: ______5000_________
t1: _______0___________
pc: _____0x214_________		t0: 5000
t1: 1
pc: CAN'T TELL		Incorrect
76
160EECS6.191Computation Structures6.100A, 8.02NoneMidterm Exam 2Sequential Circuits in Minispec4c1.2TextYou are frustrated with the 77 Mass. Ave crosswalk and decide to design a better traffic signal in Minispec. To start, you want to make sure the traffic light will function well in the daytime when there’s lots of traffic. After carefully analyzing traffic patterns, you define the following specification:
• The traffic light should be red for 4 cycles, then green for 10 cycles, then yellow for 1 cycle, and repeat this pattern indefinitely.
• The light starts red (and should stay red for 4 cycles before turning green).
• Pedestrians can only cross when the light is red.
Now you want to add a new feature to your traffic light. During the daytime, you want it to work as in part (A). But during the nighttime, the traffic light should work differently:
• By default, the light should be green.
• When a pedestrian requests to cross the street and the light is green, it should remain green for 3 more cycles, turn yellow for 1 cycle, then red for 4 cycles. Then it should go back to being green indefinitely.
• If a pedestrian requests to cross the street when the light is yellow or red, this request should be ignored and have no effect.
• If a pedestrian requests to cross the street while the light is green, and a pedestrian requests to cross the street in a following cycle when the light is still green, this request should also have no effect.
You also want to add a feature for emergency pedestrian requests. In an emergency, if a pedestrian requests to cross, the light should immediately turn yellow on the next cycle. The pedestrian request is provided as a Maybe#(PedestrianRequest) type – on each cycle it will either be:
• Invalid (no pedestrian request)
• Standard (a standard pedestrian request was made)
• Emergency (an emergency pedestrian request was made)
Note that your implementation should still work when the input transitions from daytime to nighttime, even though in daytime the Green light is 10 cycles and in nighttime it is only 3 cycles following a pedestrian request. Thus, if it is nighttime and our counter variable is too large (because we were counting down from a larger value during the daytime), we should “clamp” it to be no larger than it can be in nighttime. We have provided a currentCounter variable to use for this purpose – i.e. it will be clamped to the maximum value the counter can be during nighttime.
Fill in the Minispec module below to add this functionality. We have provided two inputs – one for whether it is currently nighttime or daytime, and one for whether a pedestrian has requested to cross the street in this cycle.
typedef enum { Green, Yellow, Red } LightState;
typedef enum { Daytime, Nighttime } TimeOfDay;
typedef enum { Standard, Emergency } PedestrianRequest;
module TrafficLight;
Reg#(LightState) light(_<answer from Part A>_);
Reg#(Bit#(_<answer from Part A>_)) counter(_<answer from part A>_);
input TimeOfDay timeOfDay default = Nighttime;
input Maybe#(PedestrianRequest) pedestrianRequest default = Invalid;
method Bool pedestriansCanCross = <answer from part A>;
method LightState currentLight = <answer from part A>;
rule tick;
if (timeOfDay == Daytime) begin
<Your answer from Part A>
end else begin
if (light == Green) begin
Bit#(<answer from part A>) currentCounter;
// Clamp currentCounter to the maximum value counter
// can be for a Green light at night
currentCounter = counter > ____ ? ____ : counter;
if (currentCounter == 0) begin
light <= __________;
// Check if received pedestrian request this cycle
end else if (___________________________________)
begin
// Handle emergency request
if (__________________________________) begin
light <= ___________________________;
end else begin
counter <= _________________________;
end
end else if (currentCounter < _______________) begin
counter <= ___________________________;
end else begin
counter <= ___________________________;
end
end else if (light == Yellow) begin
<Your answer from Part A>
end else if (light == Red) begin
<Your answer from Part A>
end
end
endrule
endmodule		Programmingtypedef enum { Green, Yellow, Red } LightState;
typedef enum { Daytime, Nighttime } TimeOfDay;
typedef enum { Standard, Emergency } PedestrianRequest;
module TrafficLight;
Reg#(LightState) light(_<answer from Part A>_);
Reg#(Bit#(_<answer from Part A>_)) counter(_<answer from part A>_);
input TimeOfDay timeOfDay default = Nighttime;
input Maybe#(PedestrianRequest) pedestrianRequest default = Invalid;
method Bool pedestriansCanCross = <answer from part A>;
method LightState currentLight = <answer from part A>;
rule tick;
if (timeOfDay == Daytime) begin
<Your answer from Part A>
end else begin
if (light == Green) begin
Bit#(<answer from part A>) currentCounter;
// Clamp currentCounter to the maximum value counter
// can be for a Green light at night
currentCounter = counter > ___3_ ? ___3_ : counter;
if (currentCounter == 0) begin
light <= ___Yellow___;
// Check if received pedestrian request this cycle
end else if (_____isValid(pedestrianRequest)_____)
begin
// Handle emergency request
if (fromMaybe(?, pedestrianRequest) ==
Emergency) begin
light <= ____Yellow_________________;
end else begin
counter <= ___currentCounter - 1____;
end
end else if (currentCounter < ______3________) begin
counter <= ___currentCounter - 1______;
end else begin
counter <= ___currentCounter_(or 3)____;
end
end else if (light == Yellow) begin
<Your answer from Part A>
end else if (light == Red) begin
<Your answer from Part A>
end
end
endrule
endmodule		module TrafficLight;
Reg#(LightState) light(Red);
Reg#(Bit#(10)) counter(4);
input TimeOfDay timeOfDay default = Nighttime;
input Maybe#(PedestrianRequest) pedestrianRequest default = Invalid;
method Bool pedestriansCanCross = (light == Red);
method LightState currentLight = light;

rule tick;
if (timeOfDay == Daytime) begin
if (counter == 0) begin
case (light)
Green: begin
light <= Yellow;
counter <= 1;
end
Yellow: begin
light <= Red;
counter <= 4;
end
Red: begin
light <= Green;
counter <= 10;
end
endcase
end else begin
counter <= counter - 1;
end
end else begin
if (light == Green) begin
Bit#(10) currentCounter;
// Clamp currentCounter to the maximum value counter
// can be for a Green light at night
currentCounter = counter > 3 ? 3 : counter;
if (currentCounter == 0) begin
light <= Yellow;
// Check if received pedestrian request this cycle
end else if (pedestrianRequest == Standard && currentCounter == 3) begin
// Handle emergency request
if (pedestrianRequest == Emergency) begin
light <= Yellow;
end else begin
counter <= currentCounter - 1;
end
end else if (currentCounter < 3) begin
counter <= currentCounter - 1;
end else begin
counter <= currentCounter;
end
end else if (light == Yellow) begin
light <= Red;
counter <= 4;
end else if (light == Red) begin
light <= Green;
counter <= 0;
end
end
endrule
endmodule		Impossible to solve given context
Refrence to prior solution that isn't specified
77
54Mathematics18.2Principles of Discrete Applied MathematicsNone18.C06Problem Set 10Discrete Fourier Transform3a0.6790123457TextWhat is the Discrete Fourier Transform over $\mathbb{C}$ of $y=(1,1,0, i)$?ExpressionWe use the formula
$$
c_{k}=\sum_{j=0}^{n-1} y_{j} e^{-2 \pi i j k / n} .
$$
Since $n=4$, we have the roots
$$
r=\left(e^{-2 \pi i 0 / n}, e^{-2 \pi i 1 / n}, e^{-2 \pi i 2 / n}, e^{-2 \pi i 3 / n}\right)=(1,-i,-1, i).
$$
We need the different powers of these roots: $r_{k}=\left(1^{k},(-i)^{k},(-1)^{k}, i^{k}\right)$ for $k=0,1,2,3$. The dot product with $y$ gives us the transform, i.e.
$$
c_{k}=y \cdot r_{k}
$$
We then obtain $c=(2+i,-i,-i, 2+i)$.		The Discrete Fourier Transform (DFT) of a sequence y of length N is given by the formula:

$Y[k] = \sum_{n=0}^{N-1} y[n] \cdot e^{-\frac{2\pi i}{N} \cdot nk}$ for $k = 0, 1, ..., N-1$

In this case, we have $y = (1, 1, 0, i)$, so $N = 4$. We can compute the DFT as follows:

$Y[0] = \sum_{n=0}^{3} y[n] \cdot e^{-\frac{2\pi i}{4} \cdot n \cdot 0} = 1 + 1 + 0 + i = 2 + i$

$Y[1] = \sum_{n=0}^{3} y[n] \cdot e^{-\frac{2\pi i}{4} \cdot n \cdot 1} = 1 - i - 1 - i = -2i$

$Y[2] = \sum_{n=0}^{3} y[n] \cdot e^{-\frac{2\pi i}{4} \cdot n \cdot 2} = 1 - 1 + 0 - i = -i$

$Y[3] = \sum_{n=0}^{3} y[n] \cdot e^{-\frac{2\pi i}{4} \cdot n \cdot 3} = 1 + i - 1 + i = 2i$

So, the Discrete Fourier Transform over $\mathbb{C}$ of $y = (1, 1, 0, i)$ is $Y = (2 + i, -2i, -i, 2i)$.		Incorrect
78
246EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneLab 9Convolutional Neural Networks4a0.1041666667TextMany of us have seen applications of facial recognition in real life (ex. Unlocking a phone) and movies (ex. Jarvis, built by fictional MIT alum, Tony Stark, has facial recognition capabilities). In this lab, we are going to introduce some of the ethical and social considerations that arise in the development, deployment, and very existence of facial recognition systems.
Convolutional Neural Networks require a LOT of data to train well. What are some ethical and social implications in deciding whether and how to collect such data for facial recognition systems (e.g., labeled images of faces)?
Tip: Having trouble getting started? Think about this question from a security perspective, privacy perspective, consent perspective, etc. See this interesting Nature article on The ethical questions that haunt facial recognition research .		OpenWe want students to think about data privacy, security, consent, and monetization of softwares that use the image of unknowing, unconsenting individuals.
The nature article is really worth a read. To summarize, most research groups scrape the Internet/public sources to collect facial images (And do not ask for permission). Examples: In 2015, Stanford published a set of 12000 images from a webcam in a cafe in SF that had been live-streamed online. Researches at Duke released more than 2 million video frame footage of students walking around campus. MegaFace and MSCeleb are large datasets that have been posted online and were collected from the internet. Many of these larger datasets have been used to evaluate and improve commercial surveillance products. And quite a few of these datasets were published online in one place without a password in place.
There's also the issues to consider around data privacy. When you use a facial recognition model, your personal data is being sent to a model and there are both privacy and security risks associated with that.		Collecting large amounts of data for facial recognition systems raises several ethical and social implications, which can be analyzed from different perspectives:

1. Privacy perspective: Collecting facial data can lead to potential privacy breaches, as individuals' faces are unique identifiers. Unauthorized access to this data can result in identity theft, stalking, or other forms of harassment. Additionally, the widespread use of facial recognition systems can lead to a surveillance society, where people's movements and activities are constantly monitored without their knowledge or consent.

2. Consent perspective: Obtaining consent from individuals before collecting their facial data is crucial. However, in many cases, people might not be aware that their images are being used for facial recognition purposes, or they might not have a clear understanding of how their data will be used, stored, and shared. This raises concerns about informed consent and the autonomy of individuals in deciding whether to participate in such systems.

3. Bias and discrimination: Facial recognition systems are often trained on datasets that may not be representative of the diverse global population. This can lead to biased algorithms that perform poorly on certain demographic groups, resulting in unfair treatment or discrimination. For example, studies have shown that some facial recognition systems have higher error rates for people with darker skin tones or for women. This can exacerbate existing social inequalities and perpetuate stereotypes.

4. Security perspective: The storage and management of large facial datasets can pose significant security risks. Data breaches or unauthorized access to these datasets can have severe consequences for the individuals whose data is compromised. Ensuring robust security measures are in place to protect this sensitive data is essential.

5. Ownership and control: Questions arise about who owns and controls the facial data collected for these systems. Individuals may not have control over how their data is used, shared, or monetized by third parties, leading to potential exploitation and commodification of personal information.

6. Legal and regulatory frameworks: The development and deployment of facial recognition systems often outpace existing legal and regulatory frameworks, leading to a lack of clear guidelines and oversight. This can result in the misuse of technology and the infringement of individuals' rights.

In conclusion, the collection of facial data for training Convolutional Neural Networks in facial recognition systems raises several ethical and social concerns. Addressing these issues requires a multidisciplinary approach, involving not only technologists but also ethicists, policymakers, and other stakeholders to ensure that these systems are developed and deployed responsibly and fairly.		Correct
79
180EECS6.191Computation Structures6.100A, 8.02NoneMidterm Exam 3Operating Systems2a0.45TextConsider the following two processes running RISC-V programs labeled with virtual addresses. Note that all pseudoinstructions in this code translate into a single RISC-V instruction.
Program for Process A
.=0x200
li t0, 5000
li t1, 0
loop:
li a0, 0x300
li a7, 0x13 // print system call
ecall
addi t1, t1, 1
ble t1, t0, loop
j exit // exit process
.=0x300
.ascii “Hello from process A\n”
Program for Process B
.=0x450
li t0, 50
li t1, 10
div t2, t0, t1
sw t2, 0x900(x0)
li a0, 0x600
li a7, 0x13 // print system call
ecall
j exit // exit process
.=0x600
.ascii “Hello from Process B\n”
Assume the OS schedules Process B first. For the following questions, if you can’t tell a value based on the information given, write CAN’T TELL.
A timer interrupt occurs just prior to the execution of li t1, 10 in Process B. Process A runs for some time, then another timer interrupt occurs and control is returned to Process B. What are the values in the following registers immediately after returning to Process B?
t0:
t1:
pc:		Numericalt0: _______50__________
t1: __CAN’T TELL or 0_
pc: _____0x454_________		t0: CAN'T TELL
t1: CAN'T TELL
pc: 0x452		Incorrect
80
35EECS6.18Computer Systems Engineering6.1010, 6.1910NoneMidterm Exam 1Unix2b0.375TextAs a result of a fork, there are two processes running on a machine: the parent and the child $\mathrm{A}$. Immediately after returning from the fork() call, the parent forks again, creating child B. Neither child process has been scheduled yet (i.e., they have not yet had an opportunity to execute anything after the return from fork()). We are asking about an instant when the two children are fully created and completely ready to run, but before either has had a chance to run.
Select True or False for the following statement:
Processes $\mathrm{A}$ and $\mathrm{B}$ have identical file descriptors.		Multiple ChoiceTrue.VRAICorrect
81
155Mathematics18.03Differential EquationsNone18.02Final ExamBoundary Value Problems93.614457831TextConsider the function
$$
f(x)=x(\pi-x)
$$
Find the function $u(x, t)$ solving the partial differential equation
$$
\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}} \quad u(0, t)=u(\pi, t)=0
$$
for $x \in[0, \pi]$ and $t \geq 0$ subject to the initial condition
$$
u(x, 0)=f(x) .
$$		ExpressionWe recall that a solution to the heat equation with Dirichlet boundary condition is
$$
c e^{-n^{2} t} \sin (n x)
$$
Now we find a sine series for the initial condition by finding the Fourier series of the odd extension of $f(x)$. Since it is an odd function, we have $\tilde{a}_{0}=0$ and $a_{n}=0$ for all $n \geq 1$. For $n \geq 1$
$$
\begin{aligned}
b_{n} &=\frac{2}{\pi} \int_{0}^{\pi} x(\pi-x) \sin (n x) d x \\
&=\frac{2}{\pi}\left[\frac{-x(\pi-x)}{n} \cos (n x)\right]_{0}^{\pi}+\frac{2}{\pi} \int_{0}^{\pi} \frac{\pi-2 x}{n} \cos (n x) d x \\
&=0+\frac{2}{\pi}\left[\frac{\pi-2 x}{n^{2}} \sin (n x)\right]_{0}^{\pi}+\frac{4}{n^{2} \pi} \int_{0}^{\pi} \sin (n x) d x \\
&=0-\frac{4}{n^{3} \pi}[\cos (n x)]_{0}^{\pi}=\frac{4}{n^{3} \pi}\left(1-(-1)^{n}\right)
\end{aligned}
$$
where in the second and the third line we have used integration by parts. In the last equality, we have also used $\cos (n \pi)=(-1)^{n}$. Thus, a sine series for $f$ on $[0, \pi]$ is given by
$$
f(x)=\sum_{n \geq 1, n \text { odd }} \frac{8}{n^{3} \pi} \sin (n x) .
$$
We use this series to find the solution
$$
u(x, t)=\sum_{n \geq 1, n \text { odd }} \frac{8}{n^{3} \pi} e^{-n^{2} t} \sin (n x).
$$		To solve the given partial differential equation (PDE) with the given initial and boundary conditions, we can use the method of separation of variables. Let's assume that the solution can be written as a product of two functions, one depending only on $x$ and the other depending only on $t$:

$$
u(x, t) = X(x)T(t)
$$

Now, we substitute this into the PDE:

$$
\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} \Rightarrow X(x) \frac{dT}{dt} = T(t) \frac{d^2 X}{dx^2}
$$

Now, we can separate the variables by dividing both sides by $XT$:

$$
\frac{1}{T} \frac{dT}{dt} = \frac{1}{X} \frac{d^2 X}{dx^2}
$$

Since the left side depends only on $t$ and the right side depends only on $x$, both sides must be equal to a constant, say $-\lambda^2$:

$$
\frac{1}{T} \frac{dT}{dt} = -\lambda^2 = \frac{1}{X} \frac{d^2 X}{dx^2}
$$

Now, we have two ordinary differential equations (ODEs):

1) $\frac{dT}{dt} = -\lambda^2 T$

2) $\frac{d^2 X}{dx^2} = -\lambda^2 X$

The first ODE is a simple first-order linear ODE, and its solution is:

$$
T(t) = Ce^{-\lambda^2 t}
$$

The second ODE is a second-order linear ODE with boundary conditions $X(0) = X(\pi) = 0$. The general solution of this ODE is:

$$
X(x) = A\cos(\lambda x) + B\sin(\lambda x)
$$

Applying the boundary conditions, we get:

$$
X(0) = A\cos(0) + B\sin(0) = A = 0
$$

and

$$
X(\pi) = B\sin(\lambda \pi) = 0
$$

Since $B$ cannot be zero (otherwise the solution would be trivial), we must have $\lambda \pi = n\pi$ for some integer $n$. Thus, $\lambda = n$.

Now, the solution of the PDE is given by the sum of the product of the solutions of the ODEs:

$$
u(x, t) = \sum_{n=1}^{\infty} B_n \sin(nx) e^{-n^2 t}
$$

To find the coefficients $B_n$, we use the initial condition $u(x, 0) = f(x) = x(\pi - x)$:

$$
x(\pi - x) = \sum_{n=1}^{\infty} B_n \sin(nx)
$$

We can find the coefficients $B_n$ by multiplying both sides by $\sin(mx)$ and integrating from $0$ to $\pi$:

$$
\int_0^\pi x(\pi - x) \sin(mx) dx = \sum_{n=1}^{\infty} B_n \int_0^\pi \sin(nx) \sin(mx) dx
$$

Using the orthogonality property of sine functions, the right side becomes:

$$
\sum_{n=1}^{\infty} B_n \frac{\pi}{2} \delta_{mn} = B_m \frac{\pi}{2}
$$

where $\delta_{mn}$ is the Kronecker delta, which is 1 if $m = n$ and 0 otherwise.

Now, we need to compute the integral on the left side:

$$
\int_0^\pi x(\pi - x) \sin(mx) dx = \pi \int_0^\pi x \sin(mx) dx - \int_0^\pi x^2 \sin(mx) dx
$$

Using integration by parts, we find:

$$
\int_0^\pi x \sin(mx) dx = \frac{1}{m^2}(\pi m \cos(\pi m) - \sin(\pi m))
$$

and

$$
\int_0^\pi x^2 \sin(mx) dx = \frac{2}{m^3}(\pi m \sin(\pi m) - (\pi^2 m^2 - 2) \cos(\pi m) + 2)
$$

Substituting these results back into the equation for $B_m$, we get:

$$
B_m = \frac{2}{\pi} \left[ \frac{1}{m^2}(\pi m \cos(\pi m) - \sin(\pi m)) - \frac{2}{m^3}(\pi m \sin(\pi m) - (\pi^2 m^2 - 2) \cos(\pi m) + 2) \right]
$$

Finally, the solution of the PDE is:

$$
u(x, t) = \sum_{n=1}^{\infty} B_n \sin(nx) e^{-n^2 t}
$$

with

$$
B_n = \frac{2}{\pi} \left[ \frac{1}{n^2}(\pi n \cos(\pi n) - \sin(\pi n)) - \frac{2}{n^3}(\pi n \sin(\pi n) - (\pi^2 n^2 - 2) \cos(\pi n) + 2) \right]
$$		​
82
101EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneExercise 13Decision Trees1bv0.02083333333TextWe won't go through the algorithm of building a tree in full, but we will look at the first step of the algorithm, which is deciding where to first split the data, based on minimizing the weighted average entropy.
Recall the following notation from the notes. We consider a data set $\mathcal{D}$, and let $I$ be an indicator set of all the elements within $\mathcal{D}$, so that $I=\{1, \ldots, n\}$ for our whole data set.
We consider the features $x^{(i)}$ for examples $i \in I$, and split on the $j$ th dimension at location $s$, i.e., based on $x_{j}^{(i)} \geq s$. We let $I_{j, s}^{+}$represent the set of points on the "right" side of the split:
$$
I_{j, s}^{+}=\left\{i \in I \mid x_{j}^{(i)} \geq s\right\} .
$$
Similarly the points on the left side of the same split are
$$
I_{j, s}^{-}=\left\{i \in I \mid x_{j}^{(i)}<s\right\} .
$$
We can define $I_{m}$ as the subset of data samples that are in region $R_{m}$. We then define $\hat{P}_{m, k}$ as the empirical probability of class $k$ in $I_{m}$, where $\hat{P}_{m, k}$ is the fraction of data points in $I_{m}$ that are labeled with class $k$.
The entropy of data points in $I_{m}$ is given by
$$
H\left(I_{m}\right)=-\sum_{k} \hat{P}_{m, k} \log _{2} \hat{P}_{m, k}
$$
where we stipulate that $0 \log _{2} 0=0$.
Finally, the weighted average entropy $\hat{H}$ of a split on the $j$ th dimension at location $s$ is:
$$
\begin{aligned}
\hat{H} & =(\text { fraction of points in left data set }) \cdot H\left(I_{j, s}^{-}\right)+(\text {fraction of points in right data set }) \cdot H\left(I_{j, s}^{+}\right) \\
& =\frac{\left|I_{j, s}^{-}\right|}{N_{m}} \cdot H\left(I_{j, s}^{-}\right)+\frac{\left|I_{j, s}^{+}\right|}{N_{m}} \cdot H\left(I_{j, s}^{+}\right),
\end{aligned}
$$
where $N_{m}=\left|I_{m}\right|$.
For the dataset above, we will compute the weighted average entropies for the following potential splits of the data:
\begin{itemize}
\item Feature $x_{1} \geq 1.5$
\item Feature $x_{1} \geq-1.5$
\item Feature $x_{1} \geq 0.0$
\item Feature $x_{2} \geq 0.0$
\end{itemize}
We will compute the weighted average entropy of the first split $x_{1} \geq 1.5$ as an example:
There are two points in the right split, and both have the same class. So $H\left(I_{1,1.5}^{+}\right)=0-1 \cdot \log _{2} 1=0$.
In the left split there are four points, where $\frac{3}{4}$ of the points are positively labeled and $\frac{1}{4}$ are negatively labeled, so $H\left(I_{1,1.5}^{-}\right)=$ $-\left(\frac{3}{4} \log _{2}\left(\frac{3}{4}\right)+\frac{1}{4} \log _{2}\left(\frac{1}{4}\right)\right) \approx 0.8113$
Thus, the weighted average entropy is
$$
\frac{4}{6} H\left(I_{1,1.5}^{-}\right)+\frac{2}{6} H\left(I_{1,1.5}^{+}\right) \approx 0.541
$$
Make sure that you use log base 2 in computing entropies. You may use $\log 2$ when entering Python expressions or values in the boxes below.
What is the accuracy of the tree using only the split $x_{1} \geq 1.5$?:		Numerical0.8333333333.
We are able to classify all the points correctly in the right split and 3 out of 4 points correctly in the left split, for a total of classifying 5 out of 6 points correctly.		To compute the accuracy of the tree using only the split $x_{1} \geq 1.5$, we need to determine how many data points are correctly classified by this split.

From the given example, we know that there are two points in the right split (both have the same class) and four points in the left split (three are positively labeled and one is negatively labeled).

The right split has both points with the same class, so both are correctly classified. In the left split, the majority class is positive, so the three positively labeled points are correctly classified, and the one negatively labeled point is incorrectly classified.

Thus, out of the total six data points, five are correctly classified.

The accuracy of the tree using only the split $x_{1} \geq 1.5$ is:

$$
\text{Accuracy} = \frac{\text{Number of Correctly Classified Points}}{\text{Total Number of Points}} = \frac{5}{6} \approx 0.8333
$$		​
83
51EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneExercise 7Neural Networks3d0.01736111111TextWhen Alex was first studying machine learning, he sometimes wondered about the relationship between linear regression, logistic regression, and neural networks. Is there actually any? For each of the neural-network architectures described below, help him identify whether it is equivalent to some previous model we have studied.
In each case, we will specify the number of layers, the activation function for each layer, and the loss function. In all cases, please assume the last layer outputs a scalar. Let $f$ be the activation function in a single-layer network, and let $f^{1}$ and $f^{2}$ be the activations in the first and second layers of a two-layer network, respectively.
Two layers, $f^{1}$ is identity, $f^{2}$ is sigmoid, loss is NLL.
Which of the following is this equivalent to:
(a) linear regression.
(b) logistic regression.
(c) a different kind of sensible neural network.
(d) an ill-formed neural network.		Multiple Choice(b) logistic regression.
Since the output of the network is a scalar, the weight matrix in the second layer must be a vector of the same shape as the output of the first layer. We have that the whole network takes the form
$$
\begin{aligned}
f^{2}\left(w_{2}^{T} f^{1}\left(W_{1}^{T} x+b_{1}\right)+b_{2}\right) & =\sigma\left(w_{2}^{T}\left(W_{1}^{T} x+b_{1}\right)+b_{2}\right) \\
& =\sigma\left(\left(w_{2}^{T} W_{1}\right) x+\left(w_{2}^{T} b_{1}+b_{2}\right)\right)
\end{aligned}
$$
Letting $\theta=W_{1}^{T} w_{2}$ (note $\theta$ is a vector) and $\theta_{0}=w_{2}^{T} b_{1}+b_{2}$ (note $\theta_{0}$ is a scalar), we can rewrite the network as $\sigma\left(\theta^{T} x+\theta_{0}\right)$, which together with the NLL loss we recognize as a logistic regression.
Then in the reverse direction, starting with a logistic regression model $\sigma\left(\theta^{T} x+\theta_{0}\right)$, we can take $w_{2}=\theta_{1} b_{2}=\theta_{0}, W_{1}=I$, and $b_{1}=0$ to get an output of the form
$$
\begin{aligned}
\sigma\left(w_{2}^{T}\left(W_{1}^{T} x+b_{1}\right)+b_{2}\right) & =\sigma\left(\theta^{T}\left(I^{T} x+0\right)+\theta_{0}\right) \\
& =\sigma\left(\theta^{T} x+\theta_{0}\right),
\end{aligned}
$$
meaning this two-layer net can express any logistic regression model.
Since we can express any instance of logistic regression as an instance of this two-layer net and any instance of this two-layer net as an instance of logistic regression, we say that the two models are equivalent.		The given neural network architecture has two layers, with the first layer having an identity activation function ($f^1$) and the second layer having a sigmoid activation function ($f^2$). The loss function is negative log-likelihood (NLL).

This architecture is equivalent to:

(b) logistic regression.		Correct
Correct, but no reasoning provided
84
67Mathematics18.404Theory of Computation6.1210/18.200NoneMidterm ExamFinite Automata92.222222222TextA 2-way counter automaton (2WAY-CA) is a deterministic counter automaton where the head on the input tape can move left or right at each step, as specified by its transition function. The input tape is still read-only. For example, a 2WAY-CA can recognize the language $B=\left\{x \# x \mid x \in\{\mathrm{a}, \mathrm{b}\}^{*}\right\}$ as follows. First it scans the entire input to check that it is of the form $y \# z$ where $y, z \in\{\mathrm{a}, \mathrm{b}\}^{*}$ and it uses the counter to check that $y$ and $z$ agree in length. Then it makes multiple scans over the input to check that corresponding symbols in $y$ and $z$ agree, using the counter to identify corresponding locations. If all checks succeed, then it accepts.
Let $E_{2 \text { WAY-CA }}=\{\langle C\rangle \mid C$ is a 2WAY-CA and $L(C)=\emptyset\}$. Show $E_{2 W A Y-C A}$ is undecidable. (Give enough detail to show how the counter and the 2-way head are needed, such as in the example above.)		OpenTo prove that $E_{2 \mathrm{WAY}-\mathrm{CA}}$ is undecidable, reduce $A_{\mathrm{TM}}$ to $E_{2 \mathrm{WAY}}$-CA by using the computation history method. Assume TM $R$ decides $E_{\text {2WAY-CA }}$ and construct TM $S$ deciding $A_{\mathrm{TM}}$. This proof is similar to the proof that $E_{\mathrm{LBA}}$ is undecidable.
$S =$ "On input $\langle M, W \rangle$
1. Construct 2WAY-CA $C_{M, w}$ as follows. $\left(C_{M, w}\right.$ is designed to accept all strings $u$ that are an accepting computation history of $M$ on $w$. We refer to the parts of $u$ separated by \# symbols as blocks, so $u=b_{1} \# b_{2} \# \cdots \# b_{k}$.) $C_{M, w}={ }^{\text {"On input } u}$
1. Check (by using a built-in string) that $b_{1}$ is the start configuration of $M$ on $w$. If not then reject.
2. For each pair of blocks $b_{i}$ and $b_{i+1}$, check that $b_{i}$ yields $b_{i+1}$ as configurations according to $M$ 's rules. Do so by using the counter to identify corresponding locations in the blocks (as in the example) and going back and forth to check the entire configuration. Reject if any of these pairs fail.
3. Check that the last block $b_{k}$ contains the accept state. Reject if not.
4. Accept if all checks pass."
2. Run $R$ on input $\left\langle C_{M, w}\right\rangle$ to determine whether its language is empty.
3. Accept if $R$ rejects. Reject if $R$ accepts."
If $M$ accepts $w$ then some string $u$ is an accepting computation history of $M$ on $w$, so $C_{M, w}$ accepts $u$ and its language is nonempty. Hence $R$ rejects $\left\langle C_{M, w}\right\rangle$. If $M$ rejects $w$ then no string $u$ is an accepting computation of $M$ on $w$, so $C_{M, w}$ rejects all strings, its language is empty, and so $R$ accepts $\left\langle C_{M, w}\right\rangle$. Therefore $M$ accepts $w$ iff $R$ rejects $\left\langle C_{M, w}\right\rangle$. Hence $S$ decides $A_{\mathrm{TM}}$, a contradiction. Therefore $E_{2 \text { WAY-CA }}$ is undecidable.		To show that $E_{2WAY-CA}$ is undecidable, we will reduce the Halting Problem to it. The Halting Problem is known to be undecidable, and it is defined as follows:

$H_{TM} = \{\langle M, w \rangle \mid M \text{ is a Turing Machine and } M \text{ halts on input } w\}$

Given a Turing Machine $M$ and an input $w$, we will construct a 2-way counter automaton $C_{M,w}$ such that $L(C_{M,w}) = \emptyset$ if and only if $M$ halts on input $w$.

Construction of $C_{M,w}$:

1. First, we will simulate the Turing Machine $M$ on input $w$ using the 2-way counter automaton $C_{M,w}$. We can do this by encoding the configuration of $M$ on the input tape of $C_{M,w}$, and using the counter to keep track of the position of the head in $M$. The 2-way head of $C_{M,w}$ will move left or right to simulate the movement of the head in $M$.

2. Whenever $M$ moves its head, $C_{M,w}$ will update the counter accordingly. If $M$ moves its head to the right, $C_{M,w}$ will increment the counter, and if $M$ moves its head to the left, $C_{M,w}$ will decrement the counter. This way, the counter will always represent the position of the head in $M$.

3. $C_{M,w}$ will simulate the transitions of $M$ by updating the tape symbols and moving the head according to the transition function of $M$. Since the input tape of $C_{M,w}$ is read-only, we can use additional counters to store the changes in the tape symbols and update them accordingly.

4. If $M$ reaches an accepting state, $C_{M,w}$ will enter a special accepting state and start a new phase. In this phase, $C_{M,w}$ will use its 2-way head and counters to check if the tape of $M$ contains any non-blank symbol. If it finds a non-blank symbol, it will move to a rejecting state. Otherwise, it will move to an accepting state.

5. If $M$ does not halt on input $w$, the simulation by $C_{M,w}$ will continue indefinitely, and $C_{M,w}$ will never accept any input.

Now, we can see that $L(C_{M,w}) = \emptyset$ if and only if $M$ halts on input $w$. If $M$ halts on input $w$, $C_{M,w}$ will accept only if the tape of $M$ is empty after halting, which means $L(C_{M,w}) = \emptyset$. If $M$ does not halt on input $w$, $C_{M,w}$ will never accept any input, and $L(C_{M,w}) = \emptyset$.

Since we have reduced the Halting Problem to $E_{2WAY-CA}$, we can conclude that $E_{2WAY-CA}$ is undecidable.		​
85
3EECS6.122Design and Analysis of Algorithms6.121NoneProblem Set 1Probability2c0.1818181818TextAsami and Bolin are playing a dice game in which a pair of dice is rolled repeatedly. Asami wins if a sum of 6 is rolled before any sum greater than or equal to 9, and Bolin wins if any sum greater than or equal to 9 is rolled first. We will find the probability that Asami wins the game.
Let $E_n$ denote the event that a 6 occurs on the $n^{\text {th }}$ roll and neither 6 nor any number greater than or equal to 9 occurs on any of the first $(n-1)$ rolls.
Compute $\sum_{n=1}^{\infty} P\left(E_n\right)$ and argue rigorously that it is the desired probability. HINT: $\sum_{n=0}^{\infty} r^n=\frac{1}{1-r}$ for $|r|<1$.		OpenSince the events $E_n$ are mutually disjoint and their union is precisely the event whose probability we want, we can add their probabilities to obtain the desired result:
$$
P(6 \text { before } \geq 9)=\sum_{n=1}^{\infty} P\left(E_n\right)=\sum_{n=1}^{\infty}\left(\frac{21}{36}\right)^{n-1} \cdot \frac{5}{36} .
$$
This can be evaluated using the hint in the problem to be
$$
\frac{1}{1-21 / 36} \cdot \frac{5}{36}=\frac{1}{3} \text {. }
$$		To find the probability that Asami wins the game, we need to compute the sum of the probabilities of all possible scenarios in which Asami wins. This means that we need to find the sum of the probabilities of all $E_n$ events.

First, let's find the probability of each $E_n$ event. For a 6 to occur on the $n^{\text{th}}$ roll, there is a $\frac{5}{36}$ chance (since there are 5 ways to get a sum of 6 out of 36 possible outcomes). For neither 6 nor any number greater than or equal to 9 to occur on any of the first $(n-1)$ rolls, there is a $\left(\frac{26}{36}\right)^{n-1}$ chance (since there are 26 ways to get a sum less than 6 or between 7 and 8 out of 36 possible outcomes).

So, the probability of $E_n$ is given by:

$P(E_n) = \left(\frac{26}{36}\right)^{n-1} \cdot \frac{5}{36}$

Now, we need to compute the sum of the probabilities of all $E_n$ events:

$\sum_{n=1}^{\infty} P\left(E_n\right) = \sum_{n=1}^{\infty} \left(\frac{26}{36}\right)^{n-1} \cdot \frac{5}{36}$

This is an infinite geometric series with the first term $a = \frac{5}{36}$ and the common ratio $r = \frac{26}{36}$. Using the formula for the sum of an infinite geometric series, we get:

$\sum_{n=1}^{\infty} P\left(E_n\right) = \frac{a}{1-r} = \frac{\frac{5}{36}}{1-\frac{26}{36}} = \frac{\frac{5}{36}}{\frac{10}{36}} = \frac{5}{10} = \frac{1}{2}$

Therefore, the probability that Asami wins the game is $\boxed{\frac{1}{2}}$.		Incorrect
86
58EECS6.3Signal Processing6.100A, 18.03NoneProblem Set 5Fourier Transforms2c0.1171875TextFind the Discrete-Time Fourier Transform of $f_{3}[n]$ :
$$
f_{3}[n]=\left(\frac{1}{2}\right)^{|n|}
$$		Expression$$
\begin{aligned}
F_{3}(\Omega) & =\sum_{n=-\infty}^{\infty}\left(\frac{1}{2}\right)^{|n|} e^{-j \Omega n}=2\left(\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n} e^{-j \Omega n}\right)-1=2\left(\sum_{n=0}^{\infty}\left(\frac{1}{2} e^{-j \Omega}\right)^{n}\right)-1 \\
= & \frac{2}{1-\frac{1}{2} e^{-j \Omega}}-1=\frac{1+\frac{1}{2} e^{-j \Omega}}{1-\frac{1}{2} e^{-j \Omega}}
\end{aligned}
$$		To find the Discrete-Time Fourier Transform (DTFT) of the given function $f_3[n]$, we need to compute the following sum:

$$
F_3(\omega) = \sum_{n=-\infty}^{\infty} f_3[n] e^{-j\omega n}
$$

Substitute the given function $f_3[n] = \left(\frac{1}{2}\right)^{|n|}$:

$$
F_3(\omega) = \sum_{n=-\infty}^{\infty} \left(\frac{1}{2}\right)^{|n|} e^{-j\omega n}
$$

Now, we can split the sum into two parts: one for negative values of $n$ and one for non-negative values of $n$:

$$
F_3(\omega) = \sum_{n=-\infty}^{-1} \left(\frac{1}{2}\right)^{-n} e^{-j\omega n} + \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^{n} e^{-j\omega n}
$$

Let's evaluate the first sum:

$$
\sum_{n=-\infty}^{-1} \left(\frac{1}{2}\right)^{-n} e^{-j\omega n} = \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{-(-n)} e^{-j\omega (-n)} = \sum_{n=1}^{\infty} 2^n e^{j\omega n}
$$

This is a geometric series with the first term $a = 2 e^{j\omega}$ and the common ratio $r = 2 e^{j\omega}$. The sum of this series is:

$$
\frac{a}{1 - r} = \frac{2 e^{j\omega}}{1 - 2 e^{j\omega}}
$$

Now, let's evaluate the second sum:

$$
\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^{n} e^{-j\omega n}
$$

This is also a geometric series with the first term $a = 1$ and the common ratio $r = \frac{1}{2} e^{-j\omega}$. The sum of this series is:

$$
\frac{a}{1 - r} = \frac{1}{1 - \frac{1}{2} e^{-j\omega}}
$$

Now, we can combine both sums to get the DTFT of $f_3[n]$:

$$
F_3(\omega) = \frac{2 e^{j\omega}}{1 - 2 e^{j\omega}} + \frac{1}{1 - \frac{1}{2} e^{-j\omega}}
$$

So, the Discrete-Time Fourier Transform of $f_3[n]$ is:

$$
F_3(\omega) = \frac{2 e^{j\omega}}{1 - 2 e^{j\omega}} + \frac{1}{1 - \frac{1}{2} e^{-j\omega}}
$$		Incorrect
87
18Mathematics18.3Principles of Continuum Applied Mathematics18.02, 18.03NoneProblem Set 1Differentiation Within Integrals8b0.2430555556TextIn each case compute $u_{x}=\frac{\partial u}{\partial x}$ and $u_{p}=\frac{\partial u}{\partial p}$ (as functions of $u, x$, and $p$ ), given that $u=u(x, p)$ satisfies: $u=\int_{0}^{x} \sin \left(p u\left(s^{2}, s\right)+x s\right) d s$.ExpressionClearly $u_p=\int_0^x u\left(s^2, s\right) \cos \left(p u\left(s^2, s\right)+x s\right) d s$ and $u_x=\sin \left(p u\left(x^2, x\right)+x^2\right)+\int_0^x s \cos \left(p u\left(s^2, s\right)+x s\right) d s$.To compute the partial derivatives, we will first differentiate the given equation with respect to x and p, and then solve for $u_x$ and $u_p$.

1. Differentiating with respect to x:

Using the Leibniz rule for differentiating under the integral sign, we have:

$\frac{\partial u}{\partial x} = u_x = \frac{\partial}{\partial x} \int_{0}^{x} \sin \left(p u\left(s^{2}, s\right)+x s\right) d s$

$u_x = \sin \left(p u\left(x^{2}, x\right)+x^2\right) + \int_{0}^{x} \frac{\partial}{\partial x} \sin \left(p u\left(s^{2}, s\right)+x s\right) d s$

$u_x = \sin \left(p u\left(x^{2}, x\right)+x^2\right) + \int_{0}^{x} s \cos \left(p u\left(s^{2}, s\right)+x s\right) d s$

2. Differentiating with respect to p:

$\frac{\partial u}{\partial p} = u_p = \frac{\partial}{\partial p} \int_{0}^{x} \sin \left(p u\left(s^{2}, s\right)+x s\right) d s$

$u_p = \int_{0}^{x} \frac{\partial}{\partial p} \sin \left(p u\left(s^{2}, s\right)+x s\right) d s$

$u_p = \int_{0}^{x} u\left(s^{2}, s\right) \cos \left(p u\left(s^{2}, s\right)+x s\right) d s$

So, the partial derivatives are:

$u_{x}=\sin \left(p u\left(x^{2}, x\right)+x^2\right) + \int_{0}^{x} s \cos \left(p u\left(s^{2}, s\right)+x s\right) d s$

$u_{p}=\int_{0}^{x} u\left(s^{2}, s\right) \cos \left(p u\left(s^{2}, s\right)+x s\right) d s$		Correct
88
331Mathematics18.01Calculus INoneNoneProblem Set 7Probability13b0.03167898627TextSuppose we flip a coin three times. There are eight possible sequences of heads and tails that we could get.
As we discussed in class each of these sequences is equally likely. Each sequence occurs with probability 1/8. For instance, the chance of flipping $\mathrm{H}$ then $\mathrm{T}$ then $\mathrm{H}$ is $1 / 8$.
If we flip a coin three times, what is the probability of getting at least one $\mathrm{H}$.		Numerical$\operatorname{Prob}($ at least one $\mathrm{H})=1-\operatorname{Prob}($ only $\mathrm{T})=1-\operatorname{Prob}(\mathrm{TTT})=1-\frac{1}{8}=\frac{7}{8}$.To find the probability of getting at least one H (head) in three coin flips, we can first find the probability of getting no H (all tails) and then subtract that from 1.

The probability of getting a T (tail) in one flip is 1/2. So, the probability of getting TTT (all tails) in three flips is (1/2) * (1/2) * (1/2) = 1/8.

Now, the probability of getting at least one H is the complement of getting all tails. So, the probability of getting at least one H is:

1 - (probability of all tails) = 1 - 1/8 = 7/8.

Therefore, the probability of getting at least one H in three coin flips is 7/8.		Correct
89
98Mathematics18.02Calculus II18.01NoneMidterm Exam 3Directional Derivative1c1.125TextWhat is the directional derivative of the function $g(x, y, z)$ in the direction $\hat{i}+\hat{j}+\hat{k}$ at the point $(1,0,-1)$?ExpressionThe unit vector in the direction $\hat{i}+\hat{j}+\hat{k}$ is $\vec{u}=1 / \sqrt{3}\langle 1,1,1\rangle$ hence $D_{\vec{u}} g(1,0,-1)=\nabla g(1,0,-1) \cdot \vec{u}=\langle-2,-2,2\rangle \cdot 1 / \sqrt{3}\langle 1,1,1\rangle=-2 / \sqrt{3}$.To find the directional derivative of the function $g(x, y, z)$ in the direction $\hat{i}+\hat{j}+\hat{k}$ at the point $(1,0,-1)$, we first need to find the gradient of the function, which is a vector of its partial derivatives with respect to each variable:

$\nabla g(x, y, z) = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z}\right)$

Next, we need to evaluate the gradient at the point $(1,0,-1)$:

$\nabla g(1, 0, -1) = \left(\frac{\partial g}{\partial x}(1, 0, -1), \frac{\partial g}{\partial y}(1, 0, -1), \frac{\partial g}{\partial z}(1, 0, -1)\right)$

Now, we need to find the unit vector in the direction of $\hat{i}+\hat{j}+\hat{k}$. To do this, we first find the magnitude of the vector:

$|\hat{i}+\hat{j}+\hat{k}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

Then, we divide each component of the vector by its magnitude to get the unit vector:

$\frac{\hat{i}+\hat{j}+\hat{k}}{|\hat{i}+\hat{j}+\hat{k}|} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$

Finally, we take the dot product of the gradient and the unit vector to find the directional derivative:

$D_{\frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}}g(1, 0, -1) = \nabla g(1, 0, -1) \cdot \left(\frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}\right)$

Without the specific function $g(x, y, z)$, we cannot compute the exact value of the directional derivative. However, the expression above shows the general formula for finding the directional derivative of a function in a given direction at a specific point.		Incorrect
90
70Mathematics18.01Calculus INoneNoneProblem Set 2Integration10b0.03959873284TextSuppose that a train is moving. It starts at time $t=0$ and ends at time $t=5$. At time $t$, its velocity is equal to $5-t$.
Approximately how far did the train go from time $t$ to time $t+\Delta t$?		ExpressionFrom time $t$ to $\Delta t$ the velocity is approximately $5-\Delta t$. This goes on for an amount of time $\Delta t$. The distance traversed in this time is approximately $(5-t) \Delta t$.To find the approximate distance the train traveled from time $t$ to time $t + \Delta t$, we can use the formula:

Distance = Velocity × Time

In this case, the velocity at time $t$ is given by $5 - t$. The time interval is given by $\Delta t$. Therefore, the approximate distance traveled during this time interval is:

Distance ≈ (5 - t) × Δt		Correct
91
391EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneProblem Set 2Regression1fi0.01736111111TextYou are given the following data, where $d=1, n=4$.
$$
D=\{[[1], 2],[[2], 7],[[3],-3],[[4], 1]\}
$$
You want to use analytic linear regression to solve the problem.
Using the same data as in the previous question (with two added noise dimensions) but using ridge regression with $\lambda=1 \times$ $10^{-10}$
We get the following parameters:
$$
\theta=\left[\begin{array}{l}
-1.377711 \times 10^{0} \\
-2.574581 \times 10^{5} \\
-5.070563 \times 10^{4}
\end{array}\right], \theta_{0}=7.260269 \times 10^{0}
$$
Does this hypothesis have higher or lower training set error than the result without ridge regression?		Multiple ChoiceHigher.To determine if the hypothesis has a higher or lower training set error than the result without ridge regression, we need to calculate the training set error for both cases and compare them.

For the given data, we have the following linear regression model without ridge regression:

$$
h_\theta(x) = \theta_0 + \theta_1 x
$$

And with ridge regression:

$$
h_\theta(x) = \theta_0 + \theta_1 x + \theta_2 x^2 + \theta_3 x^3
$$

Using the given parameters for ridge regression:

$$
h_\theta(x) = 7.260269 \times 10^{0} - 1.377711 \times 10^{0} x - 2.574581 \times 10^{5} x^2 - 5.070563 \times 10^{4} x^3
$$

Now, we need to calculate the training set error for both cases. The training set error is the mean squared error (MSE) between the predicted values and the actual values:

$$
MSE = \frac{1}{n} \sum_{i=1}^{n} (h_\theta(x_i) - y_i)^2
$$

For the given data, we can calculate the MSE for both cases:

Without ridge regression:
$$
MSE_{without} = \frac{1}{4} \sum_{i=1}^{4} (h_\theta(x_i) - y_i)^2
$$

With ridge regression:
$$
MSE_{with} = \frac{1}{4} \sum_{i=1}^{4} (h_\theta(x_i) - y_i)^2
$$

After calculating the MSE for both cases, we can compare them. If $MSE_{with} < MSE_{without}$, then the hypothesis with ridge regression has a lower training set error. If $MSE_{with} > MSE_{without}$, then the hypothesis with ridge regression has a higher training set error.		Correct
92
423EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneProblem Set 2Regression6b0.05208333333TextWe will now try to synthesize what we've learned in order to perform ridge regression on the DataCommons public health dataset that we explored in Lab 2. Unlike in Lab 2, where we did some simple linear regressions, here we now employ and explore regularization, with the goal of building a model which generalizes better (than without regularization) to unseen data.
The overall objective function for ridge regression is
$$
J_{\text {ridge }}\left(\theta, \theta_{0}\right)=\frac{1}{n} \sum_{i=1}^{n}\left(\theta^{T} x^{(i)}+\theta_{0}-y^{(i)}\right)^{2}+\lambda\|\theta\|^{2}
$$
Remarkably, there is an analytical function giving $\Theta=\left(\theta, \theta_{0}\right)$ which minimizes this objective, given $X, Y$, and $\lambda$. But how should we choose $\lambda$?
To choose an optimum $\lambda$, we can use the following approach. Each particular value of $\lambda$ gives us a different linear regression model. And we want the best model: one which balances providing good predictions (fitting well to given training data) with generalizing well (avoiding overfitting training data). And as we saw in the notes on Regression, we can employ cross-validation to evaluate and compare different models.
Let us begin by implementing this algorithm for cross-validation:
CROSS-VALIDATE $(\mathcal{D}, k)$
1 divide $\mathcal{D}$ into $k$ chunks $\mathcal{D}_{1}, \mathcal{D}_{2}, \ldots \mathcal{D}_{k}$ (of roughly equal size)
2 for $i=1$ to $k$
3 train $h_{i}$ on $\mathcal{D} \backslash \mathcal{D}_{i}$ (withholding chunk $\mathcal{D}_{i}$ )
4 compute "test" error $\mathcal{E}_{\mathfrak{i}}\left(h_{i}\right)$ on withheld data $\mathcal{D}_{i}$
5 return $\frac{1}{k} \sum_{i=1}^{k} \varepsilon_{i}\left(h_{i}\right)$
Below, X and Y are sample data, and lams is a list of possible values of lambda. Write code to set errors as a list of corresponding cross-validation errors. Use the cross_validate function above to run cross-validation with three splits. Use the following functions (which we implement for you, per the specifications below) as the learning algorithm and loss function:
def ridge_analytic(X_train, Y_train, lam):
'''Applies analytic ridge regression on the given training data.
Returns th, th0.
X : d x n numpy array (d = # features, n = # data points)
Y : 1 x n numpy array
lam : (float) regularization strength parameter
th : d x 1 numpy array
th0 : 1 x 1 numpy array'''
def mse(x, y, th, th0):
'''Calculates the mean-squared loss of a linear regression.
Returns a scalar.
x : d x n numpy array
y : 1 x n numpy array
th : d x 1 numpy array
th0 : 1 x 1 numpy array'''
X = np.array([[4, 6, 8, 2, 9, 10, 11, 17],
[1, 1, 6, 0, 5, 8, 7, 9],
[2, 2, 2, 6, 7, 4, 9, 8],
[1, 2, 3, 4, 5, 6, 7, 8]])
Y = np.array([[1, 3, 3, 4, 7, 6, 7, 7]])
lams = [0, 0.01, 0.02, 0.1]
errors = [] # your code here		ProgrammingX = np.array([[4, 6, 8, 2, 9, 10, 11, 17],
[1, 1, 6, 0, 5, 8, 7, 9],
[2, 2, 2, 6, 7, 4, 9, 8],
[1, 2, 3, 4, 5, 6, 7, 8]])
Y = np.array([[1, 3, 3, 4, 7, 6, 7, 7]])
lams = [0, 0.01, 0.02, 0.1]
errors = [cross_validate(X, Y, 3, lam, ridge_analytic, mse) for lam in lams]		To implement the cross-validation algorithm and calculate the errors for each lambda value, you can use the following code:

```python
import numpy as np

def ridge_analytic(X_train, Y_train, lam):
# Your implementation of ridge_analytic function

def mse(x, y, th, th0):
# Your implementation of mse function

def cross_validate(X, Y, lam, k=3):
n = X.shape[1]
chunk_size = n // k
errors = []

for i in range(k):
start = i * chunk_size
end = (i + 1) * chunk_size if i < k - 1 else n

X_test = X[:, start:end]
Y_test = Y[:, start:end]

X_train = np.hstack((X[:, :start], X[:, end:]))
Y_train = np.hstack((Y[:, :start], Y[:, end:]))

th, th0 = ridge_analytic(X_train, Y_train, lam)
error = mse(X_test, Y_test, th, th0)
errors.append(error)

return np.mean(errors)

X = np.array([[4, 6, 8, 2, 9, 10, 11, 17],
[1, 1, 6, 0, 5, 8, 7, 9],
[2, 2, 2, 6, 7, 4, 9, 8],
[1, 2, 3, 4, 5, 6, 7, 8]])
Y = np.array([[1, 3, 3, 4, 7, 6, 7, 7]])
lams = [0, 0.01, 0.02, 0.1]
errors = []

for lam in lams:
error = cross_validate(X, Y, lam)
errors.append(error)

print(errors)
```

This code defines the `cross_validate` function, which takes the data `X`, `Y`, the lambda value `lam`, and the number of splits `k` as input. It then divides the data into `k` chunks, trains the ridge regression model using the `ridge_analytic` function, and calculates the mean-squared error using the `mse` function. Finally, it returns the average error across all splits.

The main loop iterates through the list of lambda values, calculates the cross-validation error for each lambda, and appends it to the `errors` list.		​
93
498EECS6.39Introduction to Machine Learning6.1010/6.1210, 18.06/18.C06NoneProblem Set 4Logistic Regression4ci0.01275510204TextOur eventual goal is to do gradient descent on the logistic regression objective $J_{\text {nll }}$. In this problem, we'll take the first step toward deriving that gradient update. We'll focus on the gradient of the loss at a single point with respect to parameters $\theta$ and $\theta_{0}$.
What is the maximum value of $\frac{\partial \sigma(z)}{\partial z}$?		Numerical0.25.The sigmoid function is defined as $\sigma(z) = \frac{1}{1 + e^{-z}}$. To find the maximum value of its derivative with respect to z, we first need to compute the derivative:

$\frac{\partial \sigma(z)}{\partial z} = \frac{e^{-z}}{(1 + e^{-z})^2}$

Now, we need to find the maximum value of this derivative. To do this, we can find the critical points by setting the second derivative equal to zero and solving for z:

$\frac{\partial^2 \sigma(z)}{\partial z^2} = \frac{e^{-z}(1 - e^{-z})}{(1 + e^{-z})^3}$

Setting this equal to zero, we get:

$1 - e^{-z} = 0 \Rightarrow e^{-z} = 1 \Rightarrow z = 0$

Now, we can plug this value of z back into the first derivative to find the maximum value:

$\frac{\partial \sigma(0)}{\partial z} = \frac{e^{-0}}{(1 + e^{-0})^2} = \frac{1}{(1 + 1)^2} = \frac{1}{4}$

So, the maximum value of $\frac{\partial \sigma(z)}{\partial z}$ is $\frac{1}{4}$.		Correct
94
317Mathematics18.01Calculus INoneNoneProblem Set 7Second Order Differential Equations9a0.04751847941TextCheck that the equation $x^{\prime}(t)=-x(t)$ is linear.OpenSuppose functions $x(t)$ and $z(t)$ are solutions to $x^{\prime}(t)=-x(t)$, and let $s(t)=x(t)+z(t)$ be their sum. Since $x$ and $z$ are solutions we have
$$
\begin{aligned}
& x^{\prime}(t)=-x(t), \\
& z^{\prime}(t)=-z(t)
\end{aligned}
$$
Adding the two equations we have
$$
s^{\prime}(t)=x^{\prime}(t)+z^{\prime}(t)=-x(t)-z(t)=-s(t)
$$
So $s(t)$ also solves the differential equation. And for any constant $A, A x(t)$ also solves the differential equation since
$$
x^{\prime}(t)=-x(t) \quad \Rightarrow \quad A x^{\prime}(t)=A \cdot(-x(t))=-(A x(t))
$$		To check if the given equation is linear, we need to verify if it satisfies the two properties of linearity: additivity and homogeneity.

The given equation is $x'(t) = -x(t)$.

1. Additivity: Let $x_1(t)$ and $x_2(t)$ be two solutions of the equation. We need to check if $x_1(t) + x_2(t)$ is also a solution.

Taking the derivative of the sum, we get:
$(x_1(t) + x_2(t))' = x_1'(t) + x_2'(t)$

Now, substituting the given equation for $x_1'(t)$ and $x_2'(t)$, we have:
$-x_1(t) + (-x_2(t)) = -(x_1(t) + x_2(t))$

This shows that the sum of the two solutions is also a solution, satisfying the additivity property.

2. Homogeneity: Let $x(t)$ be a solution of the equation and $c$ be a constant. We need to check if $cx(t)$ is also a solution.

Taking the derivative of the product, we get:
$(cx(t))' = c(x'(t))$

Now, substituting the given equation for $x'(t)$, we have:
$c(-x(t)) = -cx(t)$

This shows that the product of the solution and a constant is also a solution, satisfying the homogeneity property.

Since the given equation satisfies both additivity and homogeneity properties, it is a linear equation.		Correct
95
220EECS6.411Representation, Inference, and Reasoning in AI6.1010, 6.1210, 18.600NoneProblem Set 6Particle Filter5c0.25TextConsider a domain in which the forward transition dynamics are "hybrid" in the sense that
$$
P\left(X_{t}=x_{t} \mid X_{t-1}=x_{t-1}\right)=p * N\left(x_{t-1}+1,0.1\right)\left(x_{t}\right)+(1-p) * N\left(x_{t-1}-1,0.1\right)\left(x_{t}\right)
$$
that is, that the state will hop forward one unit in expectation with probability $p$, or backward one unit in expectation with probability $1-p$, with variance $0.1$ in each case.
Assume additionally that the observation model $P\left(Y_{t}=y_{t} \mid X_{t}=x_{t}\right)=\operatorname{Uniform}\left(x_{t}-1, x_{t}+1\right)\left(y_{t}\right)$.
You know the initial state of the system $X_{0}=0$. Your friend Norm thinks it's fine to initialize a particle filter with a single particle at 0. What do you think?
(a) This is fine and we can continue with a single particle.
(b) We should initialize our pf with $N$ copies of this particle.		Multiple Choice(b) We should initialize our pf with $N$ copies of this particle.(b) We should initialize our pf with $N$ copies of this particle.Correct
96
103EECS6.18Computer Systems Engineering6.1010, 6.1910NoneMidterm Exam 2Meltdown13a0.3TextA colleague is proposing mitigation strategies for Meltdown in an architecture they are designing. For each proposal, choose the BEST response.
"We could eliminate speculative execution."
(a) No Meltdown attacks would be affected by that change.
(b) That could help against Meltdown, but cost too much performance.
(c) That is a mitigating technique recommended in the published paper.
(d) That would fix the problem, and eliminate stack smashing too.		Multiple Choice(b).(b) That could help against Meltdown, but cost too much performance.Correct
97
14Mathematics18.701Algebra I18.100BNoneProblem Set 2Cyclic Groups50.5TextProve that the two matrices
$$
E=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right], E^{\prime}=\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right]
$$
generate the group $S L_2(\mathbb{Z})$ of all integer matrices with determinant 1. Remember that the subgroup they generate consists of all elements that can be expressed as products using the four elements $E, E^{\prime}, E^{-1}, E^{\prime-1}$.
Hint: Do not try to write a matrix directly as a product of the generators. Use row reduction.		OpenIt is hard to use the fact that $S L_{2}(\mathbb{R})$ is generated by elementary matrices of the first type here. One has to start over.
As always, the method is to reduce a matrix $A$ in $S L_{2}(\mathbb{Z})$ to the identity using the given elementary matrices $E$ and $E^{\prime}$ and their inverses. What multiplication by a power of $E$ or $E^{\prime}$ does to a matrix $A$ is add a (positive or negative) integer multiple of one row to the other.
Let's work on the first column of
$$
A=\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)
$$
using division with remainder. Also, let's denote the entries of any one of the matrices that we get along the way by $a, b, c, d$. We don't need to change notation at each step.
Note first that because $\operatorname{det} A=1$, the entries $a$ and $c$ of the first column can't both be zero.
Step 1: We make one of the entries $a$ or $c$ of the first column positive. If $c \neq 0$, we add a large positive or negative integer multiple of the second row to the first to make $a>0$. If $c=0$, then $a \neq 0$. In this case we do the analogous thing to make $c>0$.
Step 2: If $a>0$, we divide $c$ by $a$, writing $c=a q+r$ where $q$ and $r$ are integers and $0 \leq r<a$. Then we add $-q($ row 1$)$ to row2. This replaces $c$ by $r$. We change notation, writing $c$ for $r$ in the new matrix, and $d$ for the other entry of row2. Now $0 \leq c<a$. If $c=0$, we stop.
Step 3: If $c>0$, we divide $a$ by $c: a=c q^{\prime}+r^{\prime}$, where $0 \leq r^{\prime}<c$. We add $q^{\prime}$ (row2) to row1, which changes $a$ to $r^{\prime}$. We adjust notation, writing $a$ for $r^{\prime}$. If $a=0$ we stop. If $a>0$, we go back to Step 2.
Since the entries of the first column decrease at each step, the process must stop at some point, with either $c=0$ or $a=0$. Then since $\operatorname{det} A=a d-b c=1$, the nonzero entry of the first column must be 1.
Step 4: If the entry 1 of the first column is the "c" entry, we add (row2) to (row1) to get $a=c=1$. Then we subtract (row1) from $($ row 2$)$ to get $a=1, c=0$.
Step 5: The matrix is now $A=\left(\begin{array}{ll}1 & b \\ 0 & d\end{array}\right)$. Since $\operatorname{det} A=1, d=1$. We subtract $b($ row 2$)$ from (row1) to get the identity matrix.		To prove that the two matrices $E$ and $E'$ generate the group $SL_2(\mathbb{Z})$, we need to show that any matrix in $SL_2(\mathbb{Z})$ can be obtained by multiplying the matrices $E, E', E^{-1}, E'^{-1}$ in some order.

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ be an arbitrary matrix in $SL_2(\mathbb{Z})$. Since $A \in SL_2(\mathbb{Z})$, we have $\det(A) = ad - bc = 1$.

Now, we will use row reduction to transform $A$ into the identity matrix $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, while keeping track of the elementary matrices used in the process.

1. If $a = 0$, we can multiply the first row by $E'^{-1}$ to get a non-zero entry in the top-left corner. So, without loss of generality, we can assume that $a \neq 0$.

2. If $c \neq 0$, we can subtract a multiple of the first row from the second row using $E^{-1}$ or $E'^{-1}$ to make the bottom-left entry zero. So, we can assume that $c = 0$.

Now, we have the matrix $A = \begin{bmatrix} a & b \\ 0 & d \end{bmatrix}$ with $ad = 1$. Since $a, d \in \mathbb{Z}$, we must have $a = d = 1$ or $a = d = -1$.

3. If $a = d = -1$, we can multiply the first row by $E'^{-1}$ and the second row by $E^{-1}$ to get $a = d = 1$. So, we can assume that $a = d = 1$.

Now, we have the matrix $A = \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix}$. We can use $E$ or $E'$ to subtract $b$ times the second row from the first row to get the identity matrix $I$.

Since we have shown that any matrix in $SL_2(\mathbb{Z})$ can be transformed into the identity matrix using the elementary matrices $E, E', E^{-1}, E'^{-1}$, it follows that the two matrices $E$ and $E'$ generate the group $SL_2(\mathbb{Z})$.		Correct
98
134EECS6.121Introduction to Algorithms6.100A, 6.12006.101Midterm Exam 2Weighted Directed Graph2b2TextWe are given a directed weighted graph $G$ with both positive and negative weights. Provide an algorithm that returns a directed cycle of positive weight in $G$, or report that none exists. Correct algorithms with better running times will receive more credit. You can invoke any algorithm discussed in lecture, recitations or p-sets.OpenCreate a graph $G^{\prime}$ with the same vertices and edges, but with the edge weights negated. Add a super node $s$ with directed edges with weight 0 to all original nodes. Run Bellman-Ford on $G^{\prime}$ starting from $s$ to either find a negative-weight cycle (by following parent pointers from a witness), or report that none exists. A negative-weight cycle in $G^{\prime}$ is a positive-weight cycle in $G$. This takes $O(|V| \cdot|E|)$ time. We can use the Bellman-Ford algorithm to find a directed cycle of positive weight in the given directed weighted graph G. The algorithm is as follows:

1. Initialize the distance array dist[] for all vertices in the graph G. Set dist[] to infinity for all vertices except the source vertex, which is set to 0.

2. Perform the following steps |V|-1 times, where |V| is the number of vertices in the graph:
a. For each edge (u, v) with weight w in the graph G, update the distance array as follows:
If dist[u] + w < dist[v], then set dist[v] = dist[u] + w.

3. Check for cycles of positive weight:
a. For each edge (u, v) with weight w in the graph G, check if dist[u] + w < dist[v]. If true, then there exists a directed cycle of positive weight in the graph G.

4. If no directed cycle of positive weight is found, report that none exists.

The running time of the Bellman-Ford algorithm is O(|V|*|E|), where |V| is the number of vertices and |E| is the number of edges in the graph G. This is because the algorithm iterates through all the edges in the graph |V|-1 times.

Note that the Bellman-Ford algorithm can also detect negative weight cycles, but in this case, we are only interested in finding a directed cycle of positive weight.		​
99
0EECS6.191Computation Structures6.100A, 8.02NonePrelab Questions 1Binary Representation1a0.03333333333TextWrite 5 as a 4-bit binary number:Numerical0b0101.
As $5=0 \times 2^{3}+1 \times 2^{2}+0 \times 2^{1}+1 \times 2^{0}$, this is equivalent to 0b0101. The 0b prefix indicates that it is a binary number.		Answer: 0101Correct
100
33EECS6.121Introduction to Algorithms6.100A, 6.12006.101Problem Set 3Critter Sort1e0.125TextProfessor Oak is trying to organize his $n$ Critters so he can study them more efficiently. For each of the following scenarios, provide the most efficient algorithm for sorting the Critters, and state whether the asymptotic complexity is $\Theta(n), \Theta(n \log n)$, or $\Theta\left(n^{2}\right)$. Briefly justify your answers (a sentence or two suffices). Choose only one of the following sorting algorithms from lecture and recitation: insertion sort, selection sort, merge sort, counting sort, and radix sort.
Professor Oak's Critters like to battle each other. Each Critter has participated in at least one battle, and each pair of Critters has battled each other at most once. Professor Oak wants to sort his Critters by the number of times that they have battled.		OpenThe sorting key is an integer in the range $[1, n]$, and so radix sort takes $\Theta(n)$ time.In this scenario, the most efficient algorithm for sorting the Critters would be counting sort. The asymptotic complexity of counting sort is $\Theta(n)$. This is because the number of battles each Critter has participated in is within a known range (0 to n-1), and counting sort is efficient for sorting integers within a specific range.Incorrect