�Cardamom Planters’ Association College, BODINAYAKANUR�

K-MAP (KARNAUGH MAP)

Prepared By

V. Mani Mala MCA.,M.Phil.

Guest Lecturer,

Department of CS &IT.

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K-MAP (KARNAUGH MAP)

• In Many Digital Circuits And Practical Problems, We Need To Find Expressions With Minimum Variables.

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• We Can Minimize Boolean Expressions Of 3, 4 Variables Very Easily Using K-map Without Using Any Boolean Algebra Theorems.

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• K-map Can Take Two Forms:

Sum Of Product (SOP)

Product Of Sum (POS)

������� SOP -2 variable

• SOP form is way to simplify and write Boolean expressions using AND to combine inputs K-map for 2 variables.

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• In the 2 variable k-map, four squares are constructed. Each square contains one term of expression with two variables.

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SOP-3 variable

K-map SOP form for 3 variables �Z= ΣA,B,C(1,3,6,7)

SOP-4 variable

����������������K-map SOP form for 4variables�F(A,B,C,D)=Σ(0,2,5,7,8,10,13,15)�

���POS -2 variable�

• POS form is a way to simplify and write Boolean expressions using OR to combine terms inside parentheses and then AND to combine those groups.

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K-map of 2 variables

• In the 2 variable k-map, four squares are constructed. Each square contains one term of expression with two variables.

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POS -3 variable

K-map 3 variable POS form�F(A,B,C)=Σ(0,3,6,7)

K-map 3 variable POS form

• From red group we find terms A & B   
• Taking complement of these two  A&B
• From brown group we find terms  B   C 
• Taking complement of these two terms  B’  C’ 
• Now sum up them  (B’+C’) 
• Taking complement of these two  A B C 
• Now sum up them  (A + B + C) 
• We will take product of these three terms : 
• (A + B’) (B’ + C’) (A + B + C)

POS -4 variable�

K-map 3 variable POS form�F(A,B,C)=Σ(3,5,7,8,10,11,12,13)

K-map 3 variable POS form

• From green group we find terms  C’  D  B 
• Taking their complement and summing them  (C+D’+B’) 
• From red group we find terms  C  D  A’ 
• Taking their complement and summing them  (C’+D’+A) 
• From blue  group we find term A  C’  D’ 
• Taking their complement and summing them  (A’+C+D) 
• From brown  group we find terms A  B’  C 
• Taking their complement and summing them  (A’+B+C’) 
• Finally we express these as product -

(C+D’+B’).(C’+D’+A).(A’+C+D).(A’+B+C’)