CDEFGHIJKL
1
NameLocationAre you ok with your solution being shared?Your thoughts, progress, or solution. Links to solutions are welcome and encouraged (github or colab notebooks are free and great for sharing solutions!)
2
IZUMIHARA RyomaJapanYesI may have misunderstood the problem, but my solution can be found at the following link:
https://drive.google.com/file/d/1Evarz2FDtYRAmTQI2dMJSiC3k_pAz1sW/view?usp=sharing
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Peter JiMadison, WIYes10
S = 1 + AS^2 + BS^3

Explanation:
A nested inside A: A(A,*), 2 ways
B nested inside A: A(B,*), 2 ways
B nested inside B: B(B,*,*), 3 ways
A nested inside B: B(A,*,*), 3 ways
For a total of 10 ways.

Let S be the total cost of all possible signal structures. Then we have the cost of A(S1, S2) = A*cost(S1)*cost(S2), which is equal to AS^2. Similarly, cost of B(S1, S2, S3) = BS^3. We have the recursion S = 1 + AS^2 + BS^3. We can check by substituting successive costs into the equation:
S_0 = 1 (cost of pulse)
S_1 = 1+A+B (cost of above in addition to A(*,*) plus B(*,*,*)
S_2 = 1 + A(1+A+B)^2 + B(1+A+B)^3
= 1+A+B+5AB+2A^2+3B^2
We see that the total number of ways to have 2 signal structures is 5+2+3 = 10, as found before.
4
Lise Andreasen Valby København Danmark Yes1st question: 10.

https://docs.google.com/document/d/1Xl9D0eQG5JPZ7PxA_RN7cdkAlP4gARI4mRH551_g2X4/edit?usp=drivesdk
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