1. CS 140 Lecture 6 Professor CK Cheng UC San Diego

2. Part I. Combinational Logic –Implementation K-map Quine-McCluskey

3. Quine-McCluskey Method Given two sets of F R D find min sum of products. 1)Exploit the adjacency to find prime implicants. 2)Find essential primes among primes via prime implicant chart.

4. Example Id a b c d f (a,b,c,d) 0 0 0 0 0 1 1 0 0 0 1 1 2 0 0 1 0 1 3 0 0 1 1 0 4 0 1 0 0 0 5 0 1 0 1 0 6 0 1 1 0 0 7 0 1 1 1 0 8 1 0 0 0 1 9 1 0 0 1 - 10 1 0 1 0 - 11 1 0 1 1 0 12 1 1 0 0 0 13 1 1 0 1 0 14 1 1 1 0 1 15 1 1 1 1 0 Given f(a,b,c,d) F =  m(0,1,2,8,14) D =  m(9,10)

5. Corresponding 4-variable K-map f (a, b, c, d) = b’c’ + b’d’ + acd’ b c a d 1 0 0 1 1 0 0 - 0 0 0 0 1 0 1 - 0 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10

6. Quine-McCluskey Approach 1)Draw truth table that only include F and D. Order by The number of ones. Divide into groups. Id 0 1 2 8 9 10 14 a0001111a0001111 b0000001b0000001 c0010011c0010011 d0100100d0100100 f1111--1f1111--1 I II III IV Each group has an equal number of ones in the input combination. Each minterm can only be adjacent to the minterms in the next group.

7. Continue again. Pair up rows from adjacent regions if they differ by exactly one bit. Put a dash where they differ. (0,1) (0,2) (0,8) (1,9) (2,10) (8,9) (8,10) (10,14) a00---111a00---111 b0000000-b0000000- c0-0010-1c0-0010-1 d-0010-00d-0010-00 I&II II&III III&IV

8. We stop when we can no longer combine rows. Make sure that we cover the entire onset. (0,1,8,9) (0,2,8,10) (10,14) a--1a--1 b00-b00- c0-1c0-1 d-00d-00 Primes  m(0,1,8,9)  m(0,2,8,10)  m(10,14) Draw the prime implicant chart: primes versus the minterms in on-set. Circle essential primes.  m(0,1,8,9)  m(0,2,8,10)  m(10,14) m0 X m1 X m2 X m8 X m14 X f(a,b,c,d) =  m(0,1,8,9) +  m(0,2,8,10) +  m(10,14) f (a,b,c,d) = b’c’ + b’d’ + acd’

9. Another example Id a b c d f (a,b,c,d) 0 0 0 0 0 1 1 0 0 0 1 0 2 0 0 1 0 1 3 0 0 1 1 0 4 0 1 0 0 1 5 0 1 0 1 0 6 0 1 1 0 0 7 0 1 1 1 1 8 1 0 0 0 1 9 1 0 0 1 - 10 1 0 1 0 0 11 1 0 1 1 0 12 1 1 0 0 - 13 1 1 0 1 0 14 1 1 1 0 0 15 1 1 1 1 1 Given f(a,b,c,d) w/ F =  m(0,2,4,7,8,15) D =  m(9,12)

10. Id 0 2 4 8 9 12 7 15 a00011101a00011101 b00100111b00100111 c01000011c01000011 d00001011d00001011 I II III IV V (0,2) (0,4) (0,8) (4,12) (8,9) (8,12) (7,15) a00--11-a00--11- b0-010-1b0-010-1 c-000001c-000001 d0000-01d0000-01 (0,4,8,12) - - 0 0  m(0,2)  m(7,15)  m(0,4,8,12) m0 X m2 X m4 X m7 X m8 X m15 X f(a,b,c,d) =  m(0,2) +  m(7,15) +  m(0,4,8,12) f (a,b,c,d) = a’b’d + bcd + c’d’