Introduction

The Karnaugh map provides a simple and straight-forward method of minimising
boolean expressions. With the Karnaugh map Boolean expressions having up to four
and even six variables can be simplified. *So what is a Karnaugh map?*

A Karnaugh map
provides a pictorial method of grouping together expressions with common factors
and therefore eliminating unwanted variables. The Karnaugh map can also be
described as a special arrangement of a truth
table.

The diagram below illustrates the correspondence between the Karnaugh map and
the truth table for the general case of a two variable problem.

The values
inside the squares are copied from the output column of the truth table,
therefore there is one square in the map for every row in the truth table.
Around the edge of the Karnaugh map are the values of the two input variable. A
is along the top and B is down the left hand side. The diagram below explains
this:

The values around the edge of the map can be thought of as coordinates. So
as an example, the square on the top right hand corner of the map in the above
diagram has coordinates A=1 and B=0. This square corresponds to the row in the
truth table where A=1 and B=0 and F=1. Note that the value in the F column
represents a particular function to which the Karnaugh map corresponds.

Consider the following map. The function plotted is: Z = f(A,B) = A + AB

- Note that values of the input variables form the rows and columns. That is the logic values of the variables A and B (with one denoting true form and zero denoting false form) form the head of the rows and columns respectively.
- Bear in mind that the above map is a one dimensional type which can be used to simplify an expression in two variables.
- There is a two-dimensional map that can be used for up to four variables, and a three-dimensional map for up to six variables.

Using algebraic simplification,

Variable B becomes redundant due to Boolean Theorem T9a.

Referring to the map above, the two adjacent 1's are grouped together. Through inspection it can be seen that variable B has its true and false form within the group. This eliminates variable B leaving only variable A which only has its true form. The minimised answer therefore is Z = A.

**Example 2:**

Consider the expression Z = f(A,B) = + A + B plotted on the Karnaugh map:

Pairs of
1's are *grouped* as shown above, and the
simplified answer is obtained by using the following steps:

Note that
two groups can be formed for the example given above, bearing in mind that the
largest rectangular clusters that can be made consist of two 1s. Notice that a 1
can belong to more than one group.

The first group labelled I, consists of
two 1s which correspond to A = 0, B = 0 and A = 1, B = 0. Put in another way,
all squares in this example that correspond to the area of the map where B = 0
contains 1s, independent of the value of A. So when B = 0 the output is 1. The
expression of the output will contain the term

For group labelled II corresponds to the area of the map where A = 0. The
group can therefore be defined as . This implies that when A = 0 the output is 1. The output is
therefore 1 whenever B = 0 and A = 0

Hence the simplified answer is Z = +

Verify this *algebraically *in your
notebooks.

Z = f(A,B,C) = + B + AB + AC

Z = f(A,B,C) = B + B + BC + A

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