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# 5 Variable K-Map in Digital Electronics

**K-Map** or **Karnaugh Map** is a simplification technique used to minimize a given complex Boolean function. K-Map or Karnaugh Map is a graph or chart which is composed of an arrangement of adjacent cells, where each cell of the K-Map represents a particular combination of variables in either sum or product form. The K-map can be used to simplify Boolean functions involving any number of variables. But, the simplification of a Boolean function using K-map becomes very complex for expressions involving five or more variables. Therefore, in actual practice, the K-map is limited to six variables.

The number of cells in a K-map depends upon the number of variables in the given Boolean function. A K-map will have 2^{n} cells or squares, where n is the number of variables in the Boolean expression. Therefore, for a two variable function, the K-map will have 2^{2} = 4 cells, for a three variable Boolean function, the K-map will have 2^{3} = 8 cells, and for four variable Boolean function, the K-map will have 2^{4} = 16 cells, and so on.

In this tutorial, we will discuss five variable K-Map and will use it to simplify Boolean functions in 5 variables. So let’s start with the introduction of 5 variable K-map.

## Five Variable K-Map

A five variable K-map is used to minimize a 5-variable Boolean expression to its reduced form. The following are the important characteristics of a 5 variable K-map −

A five variable K-map have 32 (2

^{5}) cells or squares, and each cell of the K-map represents either a minterm or a maxterm of the Boolean expression.If the given Boolean function is expressed in SOP (Sum of Products) form, then the minterms of five variables Boolean function are designated as m

_{0},m_{1},m_{2},m_{3}…m_{31}. Where, m_{0}is corresponding to $\mathrm{\lgroup\overline{A}\:\overline{B}\:\overline{C}\:\overline{D}\:\overline{E} \rgroup} $, m_{1}is corresponding to $\mathrm{\lgroup\overline{A}\:\overline{B}\:\overline{C}\:\overline{D}E\rgroup} $,… and m_{31}is corresponding to $\mathrm{\lgroup ABCDE\rgroup}$.On the other hand, if the 5 variable Boolean function is expressed in POS (Product of Sums) form, then the maxterms of the function are designated as M

_{0}, M_{1}, M_{2},… M_{31}. Where, M0 represents $\mathrm{\lgroup A+B+C+D+E\rgroup}$, M_{1}represents $\mathrm{\lgroup A+B+C+D+\overline{E}\rgroup}$,… and M_{31}represents $\mathrm{\lgroup\overline{A}+\overline{B}+\overline{C}+\overline{D}+\overline{E} \rgroup} $.

The 32 cells of the five variable K-map are divided into two blocks of 16 cells each, which are arranged side by side. The left block represents minterms (or maxterms) from m_{0} to m_{15} (or M_{0} to M_{15}. Wheck, thck, th). In the left block, the first variable (let A) is a 0. The right block represents minterms (or maxterms) from m_{16} to m_{31} (or M_{16} to M_{31}), in this block A is a 1.

In the five variable K-map, we can form 2-squares, 4-squares, 8-squares, 16-squares, or 32-squares by involving its two blocks. Also, squares are considered adjacent in these two blocks, when one block superimposes on the top of another.

A five variable SOP K-map is shown in Figure 1.

A five variable POS K-map is represented in Figure 2.

Now, let us discuss some solved examples to understand the application of a 5 variable K map in reducing a given 5-variable Boolean function in either SOP form or POS form.

## Example 1

Reduce the following 5 variable Boolean function in SOP form using the five variable K map.

$$\mathit{f}\mathrm{\lgroup A,B,C,D,E\rgroup=\sum m\lgroup 0,1,2,4,7,8,12,14,15,16,17,18,20,24,28,30,31\rgroup }$$

### Solution

The SOP K-map representation of the given SOP Boolean function is shown in Figure 3.

### Explanation

The minimization of the given 5 variable Boolean function using the five variable K-map (Figure 3) is done as per the following steps −

There are no isolated 1s in the K-map.

The minterm m

_{0}can form an 8-square with m_{4}, m_{8}, m_{12}, m_{16}, m_{20}, m_{24}, and m_{28}. So make it and read it as $\mathrm{\lgroup\overline{D}\:\overline{E} \rgroup} $.The minterms m

_{0}, m_{1}, m_{16}, and m_{17}form a 4-square. Make it and read it as $\mathrm{\lgroup\overline{B}\:\overline{C}\:\overline{D} \rgroup} $.The minterms m

_{0}, m_{2}, m_{16}, and m_{18}form a 4-square. Make it and read it as $\mathrm{\lgroup\overline{B}\:\overline{C}\:\overline{E} \rgroup} $.The minterms m

_{7}and m_{15}form a 2-square. Make it and read it as $\mathrm{\lgroup\overline{A}CDE \rgroup} $.The minterms m

_{14}, m_{15}, m_{30}and m_{31}form a 4-square. Make it and read it as $\mathrm{\lgroup BCD \rgroup} $.Finally, write all the product terms in SOP form.

Hence, the minimal SOP expression of the given 5 variable Boolean function is,

$$\mathit{f}\mathrm{(A,B,C,D,E)=\overline{A}CDE+\overline{B}\:\overline{C}\:\overline{D}+\overline{B}\:\overline{C}\:\overline{E}+BCD+\:\overline{D}\:\overline{E}}$$

## Example 2

Minimize the following 5 variable Boolean function in POS form using the five variable K map.

$$\mathit{f}\mathrm{\lgroup A,B,C,D,E\rgroup=\prod M\lgroup 3,5,6,9,10,11,13,19,21,22,23,25,26,27,29\rgroup}$$

### Solution

The POS K-map representation of the given POS Boolean function is shown in Figure 4.

### Explanation

The minimization of the given 5 variable Boolean function using the five variable K-map (figure-4) is done as per the following steps −

There are no isolated zeros in the K-map.

The maxterms M

_{9}, M_{13}, M_{25}, and M_{29}form a 4 – square. Make it and read it as $\mathrm{\lgroup\overline{B}+D+\overline{E} \rgroup} $.The maxterms M

_{3}, M_{11}, M_{19}, and M_{27}form a 4 – square. Make it and read it as $\mathrm{\lgroup C+\overline{D}+\overline{E} \rgroup} $.The maxterms M

_{5}, M_{13}, M_{21}, and M_{29}form a 4 – square. Make it and read it as $\mathrm{\lgroup \overline{C}+D+\overline{E} \rgroup} $.The maxterms M

_{6}and M_{22}form a 2 – square. Make it and read it as $\mathrm{\lgroup B+\overline{C}+\overline{D}+E \rgroup}$.The maxterms M

_{10}, M_{11}, M_{26}, and M_{27}form 4 – square. Make it and read it as $\mathrm{\lgroup \overline{B}+C+\overline{D} \rgroup}$.The maxterms M

_{22}and M_{23}form 2 – square. Make it and read it as $\mathrm{\lgroup \overline{A}+B+\overline{C}+\overline{D} \rgroup}$.Finally, write all the sum terms in POS form.

Therefore, the minimal POS expression of the given Boolean function in five variables is,

$$\mathit{f}\mathrm{\lgroup A,B,C,D,E\rgroup=\lgroup \overline{B}+D+\overline{E}\rgroup \lgroup C+\overline{D}+\overline{E}\rgroup\lgroup \overline{C}+D+\overline{E}\rgroup\lgroup B+\overline{C}+\overline{D}+E\rgroup\lgroup \overline{B}+C+\overline{D}\rgroup\lgroup \overline{A}+B+\overline{C}+\overline{D}\rgroup}$$

## Tutorial Problems

Try to solve the following tutorial problems to get better command on the utilization of five variable K-map to reduce a Boolean expression.

**Q1.**Reduce the following five variable Boolean expression in SOP form using K-Map.

$$\mathit{f}\mathrm{\lgroup A,B,C,D,E\rgroup=\sum m\lgroup 0,3,4,6,8,10,11,12,15,17,18,22,25,26,27,30,31\rgroup }$$

**Q2.**Reduce the following five variable Boolean expression in POS form using K-map.

$$\mathit{f}\mathrm{\lgroup A,B,C,D,E\rgroup=\prod M\lgroup 0,1,2,4,6,7,9,10,11,13,15,16,18,19,25,26,28,29,31\rgroup }$$

## Conclusion

This is all about the five variable K-map. From the above discussion, we can conclude that a five variable Boolean function can be reduced to the minimal form using the 5 variable K-map. A five variable K-map has 32 squares or cells from 0 to 31. These 32 cells are arranged in two blocks 16 cells each. However, the split form of the 5 variable K-map into two blocks makes the use of it to minimize a Boolean function slightly complex.