Lecture 5: Karnaugh Map Minimisation
Using algebraic techniques to minimise Boolean expressions has two disadvantages. Firstly,
the techniques are not systematic and therefore do not lend themselves to computer aided
design (CAD). Secondly, insofar as you cannot tell if the expression you arrive at is indeed
the minimum, the technique is uncertain. The Karnaugh map represents an alternative
means of representing and simplifying a Boolean function. When applied systematically, the
technique will yield a minimum expression.
Learning Outcomes:
On completing this lecture you will be able to:
 Represent two-, three-, and four-variable Boolean functions on a Karnaugh map;
 Minimise such functions in sum-of-product form;
 Distinguish between prime implicants and essential prime implicants;
 Carry out minimisation taking into account “don’t care” states.
5.1
Karnaugh Map Representation
Consider a three-variable Boolean function F(A,B,C). Corresponding to the truth table
representation (8 rows, one for each input combination or minterm), the Karnaugh map
representation comprises a rectangle divided into eight cells, one cell for each input
combination or minterm.
B=0
B=1
A=0
m0
m1
m3
m2
A=1
m4
m5
m7
m6
C=0
C=1
C=0
The base rectangle is cut in two in three different ways, one way for each of the three
variables; one side of each cut-line corresponds to the variable having the value 0, the other
side corresponds to the variable having the value 1. Note that the resulting lay-out of the
minterms does not follow a strictly linear sequence  cell m2 does not sit beside m1.
Consider the truth table representation of a Boolean function F:
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
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F
f0
f1
f2
f3
f4
f5
f6
f7
If F is specified by a set of fi values in the truth table, then the Karnaugh map representation
is generated by inserting the fi values into the corresponding cells.
B
A
f0
f1
f3
f2
f4
f5
f7
f6
C
Normally, we only write in the 1’s; an empty cell is understood to contain a 0.
Example 5.1:
Let
F   1,2,3,5,7
B
A
1
1
1
1
1
C
As shown, the Karnaugh map has 1’s in cells m1, m2, m3, m5, and m7.
5.2
Adjacencies and Simplification
Consider a Karnaugh Map with 1’s in adjacent cells m2 and m3 as shown:
B
1
1
A
C
F  m 2  m3
 ABC   ABC
 AB C   C 
 AB
ie two adjacent 1’s, comprising six literals, simplify to a single product term of two literals.
Alternatively viewed, the variable corresponding to the cut-line crossed by the “adjacency
loop” drops out altogether leaving a common two-literal term.
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Note that cells at the left and right edges are adjacent:
B
1
1
A
C
F  m0  m 2
 AB C   ABC 
 AC 
Consider a Karnaugh map with four adjacent 1’s:
B
A
1
1
1
1
C
F  m1  m3  m5  m7
 AB C  ABC  AB C  ABC
 AC B   B   AC B   B 
 AC  AC
C
ie four adjacent 1’s, comprising twelve literals, simplify to a single-literal term. The essence
of Karnaugh map simplification then is to search for the largest possible groups of adjacent
1’s. The number of adjacent 1’s must always be an integral power of two.
If a three-variable Boolean function is represented on an 8-cell Karnaugh map, a four
variable function must have 16 cells, corresponding to the 16 rows in the truth table.
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C
m0
m1
m3
m2
m4
m5
m7
m6
B
m12
m13
m15
m14
m8
m9
m11
m10
A
D
A
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
B
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
C
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
D
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
Minterm
m0
m1
m2
m3
m4
m5
m6
m7
m8
m9
m10
m11
m12
m13
m14
m15
The four-variable map has the following properties, as illustrated below:
Any single, isolated cell with a 1 yields a four-literal term.
Two adjacent 1’s yield a three-literal term.
Four adjacent 1’s yield a two-literal term, and eight adjacent 1’s yield a single-literal term.
C
A'B'C'D
1
1
B'CD'
B
1
1
ABD
A
1
D
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C
1
CD'
1
B
AC'
1
1
1
1
1
1
A
D
5.3
Karnaugh Map Minimisation
As stated earlier, the essence of Boolean simplification by means of the Karnaugh map is the
search for clusters of adjacent 1’s of the largest possible size. We illustrate with an example.
Example 5.2:
Minimise the Boolean function
F  A, B, C    1,2,3,7
The first step is to represent the function on a Karnaugh map.
B
1
1
A
1
1
C
We then go through the 1’s looking for unique ways of putting each into a cluster. Thus in
the example under consideration, the 1 in cell m3 is not particularly helpful because it
clusters in three possible ways. However, the 1 in cell m 1 forms a two-cell cluster in only one
way  corresponding to A’C. Similarly, the 1 in m2 forms a unique cluster, A’B while the 1 in
m7 forms BC.
The result then is that
F  AC  AB  BC
The following example illustrates the process for a four-variable function.
Example5.3: Minimise the Boolean function
F  A, B, C, D   3,5,7,11,12,13,15
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C
1
1
1
B
1
1
1
A
1
D
As shown, the Karnaugh map forms into two clusters of four cells and one cluster of two
cells resulting in
F  BD  CD  ABC
5.4
Prime Implicants and Essential Prime Implicants
Consider the following four-variable minimisation problem:
Example 5.4:
Minimise the Boolean function
F  A, B, C, D   2,3,5,7,10,11,13,15
C
1
1
1
1
BD
B
1
1
A
1
1
B'C
D
CD
Note first that when we carry out the looping of 1’s we end up with the following clusters:
BD , B C , and
CD
Formally these clusters are referred to as prime implicants of the specified function.
However, when it comes to representing the function we note that each 1 in the Karnaugh
map only has to be included in one cluster. For this particular example then, the function
minimises to
F  BD  BC
We say that the essential prime implicants of the function are
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BD
and
B C
A prime implicant then is any possible cluster of 8-, 4-, 2-, or single cells that can be formed
from the Karnaugh Map. An essential prime implicant is a prime implicant which has to be
included in the final Boolean expression. Here is another example.
Example 5.4:
Minimise the Boolean function
F  A, B, C, D   0,2,3,5,7,8,10,11,13,15
C
1
1
1
1
1
B
1
1
A
1
1
1
D
Examining the adjacencies yields the following set of prime implicants
PI  BD, BD, BC, CD
For the minimised Boolean expression, we see that we have to include B D  and BD . We
then have a choice as to whether we include BC or CD . The set of essential prime
implicants is given by
EPI  BD, BD
but the the required Boolean specification would be given by
F  BD  BD  either BC or CD as illustrated


C
1
1
1
C
1
1
1
1
1
1
1
B
1
1
1
1
A
A
1
1
1
1
D
5.5
B
1
1
D
Don’t Care States
In some applications, the Boolean function is undefined for some input combinations; they
simply do not occur due to particular physical constraints that apply. A common example
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would be the number pad associated with a calculator or mobile phone handset. These
essentially are systems which yield a four-bit binary output when the corresponding key is
depressed. It is physically impossible for binary combinations 1010 through to 1111 to
occur.
7
8
9
4
5
6
X2
1
2
3
X1
X3
X0
0
If now we have inputs to a logic circuit coming from such a keypad, we know that, for the
relevant four-variable Boolean function, the six input combinations 1010 through to 1111
are not going to occur and consequently it does not matter what values are assigned to the
output for these input combinations. Stated formally, we don’t care whether the output for
these input combinations is defined to be 1 or 0. We represent such a state by inserting an x
in the relevant cell of the Karnaugh map.
From the point of view of minimisation, the idea is that we can assign values to these don’t
care states to optimise the minimisation. This is illustrated as follows.
Example 5.5:
Minimise the Boolean function
don’t care set
F  A, B, C, D   2,3,5,7 subject to the
d   10,11,12,13,14,15
C
1
1
1
1
B
x
x
x
x
x
x
A
D
Clearly in this case we optimise the minimisation by assigning
m10  1 m11  1 m13  1 m15  1
and
m12  0 m14  0
The resulting Boolean equation is
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F  BD  BC
5.6
Conclusion
In this lecture we have covered Boolean representation and minimisation by means of the
Karnaugh map. It is clearly a very use method for pencil-and-paper working mode for up to
four variables, essentially reducing the problem of algebraic manipulation to one of “visual
algebra.” Although not covered here, the approach can be extended to five and even six
variables. More importantly, however, the idea of systematically searching a defined space
for adjacent 1’s would suggest taking a computer-oriented approach to the task of Boolean
minimisation, ie computer-aided design.
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