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Computer Organization. CSC 405 Quine-McKluskey Minimization.

Computer Organization

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Computer Organization

CSC 405

Quine-McKluskey Minimization

The practical use of algebraic manipulation, Venn diagrams or Karnaugh maps for minimization of Boolean expressions is limited to problems with a few variables. The Quine-McKluskey procedure for minimization can be implemented as a computer algorithm and is applicable to expressions with any number of variables.

0 00000 no 1's

1 00001 one 1's

4 00100

16 10000

3 00011 two 1's

6 00110

18 10010

24 11000

11 01011 three 1's

14 01110

28 11100

15 01111 four 1's

27 11011

31 11111 five 1's

Example: Given the boolean expression

F(p,q,r,s,t)=

m(0,1,3, 4,6,11,14,15,16,18,24,27,28,31)

minimize using the Quine-McKluskey procedure.

Step 1: Separate the minterms into groups based on the number of 1's in their binary representations.

Step 2: Compare neighboring groups and replace pairs of terms with only one bit different with a common term containing a dash at the location of the unmatched bit. Be sure to mark the minterms that are used to create common terms so that they can be removed from the list.

Step 3: Repeat the search until no new adjacent terms are found. Terms containing dashes are adjacent only if all dash positions match and all but one other position contain the same values.

Step 4: Remove duplicates and list all terms surviving terms. Indicate which of the original minterms are covered by the surviving terms called implicants.

x 00000 0000-

x 00001 00-00

x 00100 -0000

x 10000 000-1

x 00011 001-0

x 00110 100-0

x 10010 1-000

x 11000 0-011

x 01011 0-110

x 01110 11-00

x 11100 01-11 x -1-11

x 01111 -1011 x -1-11

x 110110111-

x 11111 -1111 x

11-11 x

0 1 3 4 6 11 14 15 16 18 24 27 28 31

0000- x x

00-00 x x

-0000 x x

000-1 x x

001-0 x x

100-0 x x

1-000 x x

0-011 x x

0-110 x x

11-00 x x

-1-11 x x x x

I.

II.

III.

IV.

V.

VI.

VII.

VIII.

IX.

X.

XI.

More than one minterm is covered by each implicant and some minterms are covered by more than on implicant. We need to find the smallest number of implicants that cover all minterms. These will be called the prime implicants.

minterms

implicants

Step 5: Notice that prime implicants VI, IX, X and XI are the only ones to cover minterms 18, 14, 28 and 31 respectively. These implicants must be included in our selected list. They are called essential prime implicants.

Step 6: Next we must select the minimum number of remaining implicants so as to cover all minterms. This is not an easy problem, in general. Can you find a better (i.e. smaller) set of prime implicants that cover all the minterms?

Step 7: Finally we use the selected set of prime implicants to generate the simplified Boolean expression.

(00-00) (000-1) (100-0) (0-110) (11-00) (-1-11)

F(p,q,r,s,t) = p'q's't' + p'q'r't + pq'r't + p'rst' + pqs't' + qst