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Quinne-McCluskey's Method:

Step 1. Group Minterms according to the number of ones they contain. (For example,the Minterm 1101, corresponding to square 13, contains 3 ones.)

Step 2. Recall that ABCD + ABCD = ABD(C+ C) = ABD

Using this knowledge, compare all pairs of terms in adjacent groups, looking for a changein one and only one position. Create new groups from the composite terms obtained, if any and replace the position in which any 2 terms disagree with a dash (-). Looking atthe first entry (2,3) we can see that minterm 2 and minterm 3 agree in three of the four

possible places. There is only one disagreement, however -- the rightmost digit. Wetherefore place a dash in this position. Notice that in this step we only compare members

of adjacent groups; obviously, members of non-adjacent groups differ in more than two places by definition. Every time a member of a group is combined with any other member of a different group, they both are eliminated from the list of implicants to beconsidered later. At the termination of step 3 only the prime implicants (not contained inany larger size implicants) remain. Therefore, the purpose of Quinne-McCluskey'smethod is to generate the list of prime implicants for a given function.

Group #

1 0010 (2)

1000 (8)

2 0011 (3)

0101 (5)

1010 (10)

1100 (12)

3 0111 (7)

1101 (13)

4 1111 (15)

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Step 3. Now, repeat the above step comparing all terms in the new 1st group against allterms in the new 2nd group, etc. Continue this until no more reduction can be obtained.In this step, we again compare adjacent groups. The result from this step is that allmembers of all groups contain two dashes.

from 2nd with 3rd new groups, we get these terms,

- 1 - 1 (5,7,13,15)- 1 - 1 (5,13,7,15)

The following results are obtained;

(2,3) (2,10) (8,10) (8,12) (3,7)

Group 1 Against 2

(First New Group)

001- (2,3)

-010 (2,10)

10-1 (8,10)

1-00 (8,12)

Group 2 Against 3

(Second New Group)

#2 in not reducible

0-11 (3,7)

01-1 (5,7)

-101 (5,13)

110- (12,13)

Group 3 Against 4

(Third New Group)-111 (7,15)

11-1 (13,15)

Group #

1 0010 (2)

1000 (8)

2 0011 (3)

0101 (5)

1010 (10)

1100 (12)

3 0111 (7)

1101 (13)

4 1111 (15)

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(12,13) (5,7,13,15) these are the PRIME IMPLICANTS.

Note: We only need to exhaustively compare all terms of any 2 consecutive groups. (Why?)

Now generate a "Cover Table".Minterms 2 3 5 7 8 10 12 13 15P.Is2,3 X X2,10 X X8,10 X X8,12 X X3,7 X X12,13 X X5,7,13,15 X X X X

Now to find a "Minimum Cover Table",use the following procedure:1. Eliminate all identical rows except for one of them.

2. If any COLUMN contains only one X, then the associated ROW (P.I.) is ESSENTIALand must be in our final selection. Eliminate this row and the columns it covers.

3. If any ROW dominates any other ROW, then the dominated row is eliminated. (weshould not see that in this example as there are only PIs in the column at the left)

4. If any COLUMN dominates any other COLUMN, then the dominating column iseliminated!! (this is trying to minimize the redundant terms).

After step 2 above, our table looks as follows (here we have no repetition of rows):

Minterms 2 3 8 10 12P.Is2,3 X X2,10 X X8,10 X X8,12 X X3,7 X12 X

Visual inspection could yield the following covers;(2,3) (8,10) (12,13) or (2,3) (2,10) (8,12)

The minimal PI's for cover are then added to the essential PI's to form the complete cover.