# Quine–McCluskey algorithm

The Quine–McCluskey algorithm (or the method of prime implicants) is a method used for minimization of Boolean functions that was developed by Willard V. Quine and extended by Edward J. McCluskey. It is functionally identical to Karnaugh mapping, but the tabular form makes it more efficient for use in computer algorithms, and it also gives a deterministic way to check that the minimal form of a Boolean function has been reached. It is sometimes referred to as the tabulation method.

The method involves two steps:

1. Finding all prime implicants of the function.
2. Use those prime implicants in a prime implicant chart to find the essential prime implicants of the function, as well as other prime implicants that are necessary to cover the function.

## Complexity

Although more practical than Karnaugh mapping when dealing with more than four variables, the Quine–McCluskey algorithm also has a limited range of use since the problem it solves is NP-complete. The running time of the Quine–McCluskey algorithm grows exponentially with the number of variables. For a function of n variables the number of prime implicants can be as large as 3nln(n), e.g. for 32 variables there may be over 534 * 1012 prime implicants. Functions with a large number of variables have to be minimized with potentially non-optimal heuristic methods, of which the Espresso heuristic logic minimizer was the de facto standard in 1995.[needs update]

Step two of the algorithm amounts to solving the set cover problem;NP-hard instances of this problem may occur in this algorithm step.

## Example

### Input

In this example, the input is a Boolean function in four variables, $f:\{0,1\}^{4}\to \{0,1\}$  which evaluates to $1$  on the values $4,8,10,11,12$  and $15$ , evaluates to an unknown value on $9$  and $14$ , and to $0$  everywhere else (where these integers are interpreted in their binary form for input to $f$  for succinctness of notation). The inputs that evaluate to $1$  are called 'minterms'. We encode all of this information by writing

$f(A,B,C,D)=\sum m(4,8,10,11,12,15)+d(9,14).\,$

This expression says that the output function f will be 1 for the minterms $4,8,10,11,12$  and $15$  (denoted by the 'm' term) and that we don't care about the output for $9$  and $14$  combinations (denoted by the 'd' term).

### Step 1: finding prime implicants

First, we write the function as a table (where 'x' stands for don't care):

A B C D f
m0 0 0 0 0 0
m1 0 0 0 1 0
m2 0 0 1 0 0
m3 0 0 1 1 0
m4 0 1 0 0 1
m5 0 1 0 1 0
m6 0 1 1 0 0
m7 0 1 1 1 0
m8 1 0 0 0 1
m9 1 0 0 1 x
m10 1 0 1 0 1
m11 1 0 1 1 1
m12 1 1 0 0 1
m13 1 1 0 1 0
m14 1 1 1 0 x
m15 1 1 1 1 1

One can easily form the canonical sum of products expression from this table, simply by summing the minterms (leaving out don't-care terms) where the function evaluates to one:

fA,B,C,D = A'BC'D' + AB'C'D' + AB'CD' + AB'CD + ABC'D' + ABCD.

which is not minimal. So to optimize, all minterms that evaluate to one are first placed in a minterm table. Don't-care terms are also added into this table, so they can be combined with minterms:

Number
of 1s
Minterm Binary
Representation
1 m4 0100
m8 1000
2 m9 1001
m10 1010
m12 1100
3 m11 1011
m14 1110
4 m15 1111

At this point, one can start combining minterms with other minterms. If two terms vary by only a single digit changing, that digit can be replaced with a dash indicating that the digit doesn't matter. Terms that can't be combined any more are marked with an asterisk (*). When going from Size 2 to Size 4, treat '-' as a third bit value. For instance, -110 and -100 can be combined, as well as -110 and -11-, but -110 and 011- cannot. (Trick: Match up the '-' first.)

Number
of 1s
Minterm 0-Cube Size 2 Implicants Size 4 Implicants
1 m4 0100 m(4,12) -100* m(8,9,10,11) 10--*
m8 1000 m(8,9) 100- m(8,10,12,14) 1--0*
m(8,10) 10-0
m(8,12) 1-00
2 m9 1001 m(9,11) 10-1 m(10,11,14,15) 1-1-*
m10 1010 m(10,11) 101-
m(10,14) 1-10
m12 1100 m(12,14) 11-0
3 m11 1011 m(11,15) 1-11
m14 1110 m(14,15) 111-
4 m15 1111

Note: In this example, none of the terms in the size 4 implicants table can be combined any further. Be aware that this processing should be continued otherwise (size 8 etc.).

### Step 2: prime implicant chart

None of the terms can be combined any further than this, so at this point we construct an essential prime implicant table. Along the side goes the prime implicants that have just been generated, and along the top go the minterms specified earlier. The don't care terms are not placed on top—they are omitted from this section because they are not necessary inputs.

4 8 10 11 12 15 A B C D
m(4,12)*     1 0 0
m(8,9,10,11)       1 0
m(8,10,12,14)       1 0
m(10,11,14,15)*       1 1

To find the essential prime implicants, we run along the top row. We have to look for columns with only 1 "✓". If a column has only 1 "✓", this means that the minterm can only be covered by 1 prime implicant. This prime implicant is essential.

For example: in the first column, with minterm 4, there is only 1 "✓". This means that m(4,12) is essential. So we place a star next to it. Minterm 15 also has only 1 "✓", so m(10,11,14,15) is also essential. Now all columns with 1 "✓" are covered.

The second prime implicant can be 'covered' by the third and fourth, and the third prime implicant can be 'covered' by the second and first, and neither is thus essential. If a prime implicant is essential then, as would be expected, it is necessary to include it in the minimized boolean equation. In some cases, the essential prime implicants do not cover all minterms, in which case additional procedures for chart reduction can be employed. The simplest "additional procedure" is trial and error, but a more systematic way is Petrick's method. In the current example, the essential prime implicants do not handle all of the minterms, so, in this case, one can combine the essential implicants with one of the two non-essential ones to yield one equation:

fA,B,C,D = BC'D' + AB' + AC

or

fA,B,C,D = BC'D' + AD' + AC

Both of those final equations are functionally equivalent to the original, verbose equation:

fA,B,C,D = A'BC'D' + AB'C'D' + AB'C'D + AB'CD' + AB'CD + ABC'D' + ABCD' + ABCD.