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Karnaugh map can be said to be a special arrangement of truth table. It provides excellent method for eliminating unwanted variables. This method is a pictorial method of grouping together expressions with common factors hence, eliminating unwanted variables.

For a general case of a two variable problem, the below diagram shows the correspondence between Truth Table and Karnaugh map:

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.

Karnaugh map is used to simplify the boolean expressions. To understand more, see the example solved below.

Simplify the boolean expression for the truth table given below using Karnaugh map:

The Boolean expression is:

m = a'bc + ab'c + abc' + abc.

Now we do the minimization by replicating abc term and combining it with other terms:

m = a'bc + abc + ab'c + abc + abc' + abc

= (a' + a)bc + a(b' + b)c + ab(c' + c)

= bc + ac + ab

To use a Karnaugh map we draw the following map which has a position (square) corresponding to each of the 8 possible combinations of the 3 Boolean variables. The upper left position corresponds to the 000 row of the truth table, the lower right position corresponds to 110. Each square has two coordinates, the vertical coordinate corresponds to the value of variable a and the horizontal corresponds to the values of b and c.

The 1s are in the same places as they were in the original truth table. The 1 in the first row is at position 011 (a = 0, b = 1, c = 1). The vertical coordinate, variable a, has the value 0. The horizontal coordinates, the variables b and c, have the values 1 and 1.

The minimization is done by drawing circles around sets of adjacent 1s. Adjacency is horizontal, vertical, or both. The circles must always contain 2n 1s where n is an integer.

We have circled two 1s. The fact that the circle spans the two possible values of a (0 and 1) means that the a term is eliminated from the Boolean expression corresponding to this circle. The bracketing lines shown above correspond to the positions on the map for which the given variable has the value 1. The bracket delimits the set of squares for which the variable has the value 1. We see that the two circled 1s are at the intersection of sets b and c, this means that the Boolean expression for this set is bc.

Now we have drawn circles around all the 1s. The left bottom circle is the term ac. Note that the circle spans the two possible values of b, thus eliminating the b term. Another way to think of it is that the set of squares in the circle contains the same squares as the set a intersected with the set c. The other circle (lower right) corresponds to the term ab. Thus the expression reduces to: bc + ac + ab.

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