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CS 20 Lecture 14 Karnaugh Maps

Learn how to simplify Boolean expressions using Karnaugh Maps to minimize terms and literals. Applications include logic rule reduction and hardware design optimization. Examples and explanations provided.

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CS 20 Lecture 14 Karnaugh Maps

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  1. CS 20 Lecture 14Karnaugh Maps Professor CK Cheng CSE Dept. UC San Diego

  2. Karnaugh Maps • Introduction • The Maps • Boolean Optimization

  3. Introduction Definitions: Literal: xi or xi’ Product Term: x2x1’x0 Sum Term: x2 + x1’ + x0 Minterm of n variables: A product of n literals in which every variable appears exactly once. f(a,b,c,d): ab’cd’, a’bc’d’ Maxterm of n variables: A sum of n literals in which every variable appears exactly once. f(a,b,c,d): (a’+b+c+d), (a’+b’+c+d)

  4. Introduction Input: Boolean expression of n binary variables Goal: Simplification of the expression. E.g. we want to minimize # terms and # literals. Applications: Logic: rule reduction Hardware Design: cost and performance optimization. Cost (wires, gates): # literals, product terms, sum terms Performance: speed, reliability

  5. Introduction An example of 2-variable function f(A,B)=A+B

  6. Function can be represented by sum of minterms: f(A,B) = A’B+AB’+AB This is not minimal however! We want to minimize the number of literals and terms. We factor out common terms – A’B+AB’+AB= A’B+AB’+AB+AB =(A’+A)B+A(B’+B)=B+A Hence, we have f(A,B) = A+B

  7. K-Map: Truth Table in 2 Dimensions A = 0 A = 1 AB’ 0 2 0 1 1 1 B = 0 B = 1 1 3 A’B AB f(A,B) = A + B

  8. Another Example f(A,B)=B f(A,B)=A’B+AB=(A’+A)B=B

  9. On the K-map: A = 0 A= 1 0 2 0 0 1 1 B= 0 B = 1 1 3 AB A’B f(A,B)=B

  10. Using Maxterms f(A,B)=(A+B)(A’+B)=(AA’)+B=0+B=B

  11. The Maps: Representation of k-Variable Functions (1,1,0) (1,1,1) • Boolean Expression • Truth Table • Cube • K Map • Binary Decision Diagram A (0,1,0) (0,1,1) B (1,0,1) C (0,0,0) (0,0,1) A cube of 3 variables: (A,B,C)

  12. Representation of k-Variable Func. (0,1,1,0) (0,1,1,1) (1,1,1,0) (1,1,1,1) • Boolean Expression • Truth Table • Cube • K Map • Binary Decision Diagram B (0,0,1,1) (0,0,1,0) (1,0,1,0) (1,0,1,1) C (0,1,0,1) (1,1,0,1) D (0,0,0,0) (0,0,0,1) (1,0,0,0) (1,0,0,1) A A cube of 4 variables: (A,B,C,D)

  13. Three-Variable K-Map Id a b c f (a,b,c) 0 0 0 0 f(0,0,0) 1 0 0 1 f(0,0,1) 2 0 1 0 f(0,1,0) 3 0 1 1 f(0,1,1) 4 1 0 0 f(1,0,0) 5 1 0 1 f(1,0,1) 6 1 1 0 f(1,1,0) 7 1 1 1 f(1,1,1)

  14. Three-Variable K-Map Id a b c f (a,b,c) 0 0 0 0 1 1 0 0 1 0 2 0 1 0 1 3 0 1 1 0 4 1 0 0 1 5 1 0 1 0 6 1 1 0 1 7 1 1 1 0

  15. Corresponding K-map b = 1 Gray code (0,0) (0,1) (1,1) (1,0) 0 2 6 4 c = 0 1 1 1 1 1 3 7 5 c = 1 0 0 0 0 a = 1 f(a,b,c) = c’

  16. Karnaugh Maps (K-Maps) • Boolean expressions can be minimized by combining terms • K-maps minimize equations graphically

  17. K-map • Find rectangles to cover 1’s in adjacent entries. • Rectangles can overlap but should not include 0’s. • Use the rectangle that corresponds to a product term. y(A,B)=A’B’C’+A’B’C= A’B’(C’+C)=A’B’

  18. Another 3-Input example Id a b c f (a,b,c) 0 0 0 0 0 1 0 0 1 0 2 0 1 0 1 3 0 1 1 0 4 1 0 0 1 5 1 0 1 1 6 1 1 0 - 7 1 1 1 1 Don’t Care Entry: “-” means the entry is not relevant either at input or output. In other words, we are free to assign either 0 or 1 to reduce the Boolean expression.

  19. Corresponding K-map b = 1 (0,0) (0,1) (1,1) (1,0) 0 2 6 4 c = 0 0 1 - 1 1 3 7 5 c = 1 0 0 1 1 a = 1 f(a,b,c) = a + bc’

  20. Yet another example Id a b c f (a,b,c,d) 0 0 0 0 1 1 0 0 1 1 2 0 1 0 - 3 0 1 1 0 4 1 0 0 1 5 1 0 1 1 6 1 1 0 0 7 1 1 1 0

  21. Corresponding K-map b = 1 (0,0) (0,1) (1,1) (1,0) 0 2 6 4 c = 0 1 - 0 1 1 3 7 5 c = 1 1 0 0 1 a = 1 f(a,b,c) = b’

  22. Quiz • Given Boolean function, • f(a,b,c)= a’b’+a’c’+bc’+ab+b’c • Write the truth table • Use Karnaugh map to derive the function in a minimal expression of sum of product form.

  23. 4-input K-map

  24. 4-input K-map

  25. 4-input K-map

  26. K-maps with Don’t Cares

  27. K-maps with Don’t Cares

  28. K-maps with Don’t Cares

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