Computer Organization CSC 405 QuineMc Kluskey Minimization The
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Computer Organization CSC 405 Quine-Mc. Kluskey 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-Mc. Kluskey procedure for minimization can be implemented as a computer algorithm and is applicable to expressions with any number of variables. 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-Mc. Kluskey procedure. Step 1: Separate the minterms into groups based on the number of 1's in their binary representations. 0 1 4 16 3 6 18 24 11 14 28 15 27 31 00000 no 1's 00001 one 1's 00100 10000 00011 two 1's 00110 10010 11000 01011 three 1's 011100 01111 four 1's 11011 11111 five 1's
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 x 00001 x 00100 x 10000 x 00011 x 00110 x 10010 x 11000 x 01011 x 01110 x 11100 x 01111 x 11011 x 11111 000000 -00 -0000 000 -1 001 -0 100 -0 1 -000 0 -011 0 -110 11 -00 01 -11 x -1011 x 0111 -1111 x 11 -11 x -1 -11
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 0 1 x 3 4 6 11 14 15 16 18 24 27 28 31 I. 0000 - x II. 00 -00 x III. -0000 x IV. 000 -1 V. 001 -0 VI. 100 -0 x VII. 1 -000 x VIII. 0 -011 IX. 0 -110 X. 11 -00 XI. -1 -11 x x x x x
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 + p'rst' + pqs't' + qst
