## Q. 3.19

Use a Karnaugh map to minimize the following POS expression:

$F = (A + B + C)(A + B +\bar{C})(A + \bar{B} + C)(A + \bar{B} + \bar{C})(\bar{A} + \bar{B} + C)$

## Verified Solution

The given POS expression has a domain of three variables A, B, and C. You can map this POS expression on a 3-variable Karnaugh map.

Since all sum terms are standard sum terms which can be mapped directly in the corresponding cells.

$A + B + C\rightarrow 000$;

$A + B + \bar{C}\rightarrow 001$;

$A + \bar{B} + C\rightarrow 010$;

$A + \bar{B} + \bar{C}\rightarrow 011$;

$\bar{A} + \bar{B} + C\rightarrow 110$;

The mapping result is shown in Figure 3.6.13.

Totally, there are five cells placed 0s, and the others are placed 1s. 1s can be left off.

Next step is to group 0s. One group of four cells is formed since the cells are adjacent. Another group of two cells is formed to pick up the remaining one 0.

The final step is to determine the minimum sum terms for each group and multiply them together. Thus the resulting minimum POS expression is

$F(A,B,C)=A(\bar{B}+C)$