6 Variable K-Map in Digital Electronics

Read this article to learn how you can use the K-Map (Karnaugh Map) to reduce a Boolean function in six variables. Let's start with a brief introduction to Karnaugh Map (K-Map).

Karnaugh Map (K-Map)

The Karnaugh Map or K-Map is a graphical method of reducing a Boolean function to its minimal form. The K-Map can be defined as a chart or a graph that is composed of an arrangement of adjacent squares or cells, where each cell represents a particular combination of variables of the Boolean expression either in sum or product form. A typical 2 Variable K-Map is represented in Figure-1.

In actual practice, we generally use a K-map upto 6 variables. However, K-map can be used for any number of variables, but for variables 5 or more, it becomes tedious.

Now, let us discuss the six variable K-map and its application to reduce a Boolean function to the minimal form.

Six Variable K-Map

A six variable K-Map is used to reduce a Boolean expression of six variables (say A, B, C, D, E, F). It has 64 (2⁶) adjacent cells or squares. Where, each cell represents a combination of variables of the function.

For a 6-variable Boolean function in SOP form, the possible combinations of the input variables are as follows −

$$\bar{A}\bar{B}\bar{C}\bar{D}\bar{E}\bar{F},\bar{A}\bar{B}\bar{C}\bar{D}\bar{E}F,\bar{A}\bar{B}\bar{C}\bar{D}E\bar{F},\cdot \cdot \cdot ABCDEF$$

The minterm designations of these combinations of variables are m₀, m₁, m₂ … m₆₃ respectively.

Similarly, for a 6-variables Boolean function in POS form, the possible combinations of input variables are as follows −

$$\left ( A\,+\,B\,+\,C\,+\,D\,+\,E\,+\,F \right ),\left ( A\,+\,B\,+\,C\,+\,D\,+\,E+\bar{F} \right ),\left ( A\,+\,B\,+\,C\,+\,D+\bar{E}+F \right ),\cdot \cdot \cdot \cdot \left ( \bar{A}+\bar{B}+\bar{C}+\bar{D}+\bar{E}+\bar{F} \right )$$

The maxterm designations of these combinations of variables are M₀, M₁, M₂, …M₆₃ respectively.

As already mentioned, the six variable K-map has 64 cells which are divided into 4-blocks of 16 squares each. Each cell on the K-map represents a minterm or a maxterm. In the case of 6 variable K-map, the values of variables A and B remain the same for all minterms (or maxterms) in each block of 16 squares.

The 64 cells of a six variable K-map are divided into 4 blocks as follows −

• Block 1 − This is the top left block. This block represents minterms from m₀ to m₁₅, (or maxterms from M₀ to M₁₅). In this block, the variable A is a 0 and the variable B is also a 0.

• Block 2 − This is the top right block. This block represents minterms from m₁₆ to m₃₁, (or maxterms from M₁₆ to M₃₁). In this block, the variable A is a 0 and the variable B is a 1.

• Block 3 − This is the bottom left block. This block represents minterms from m₃₂ to m₄₇, (or maxterms from M₃₂ to M₄₇). In this block, the variable A is a 1 and the variable B is a 0.

• Block 4 − This is the bottom right block. This block represents minterms from m₄₈ to m₆₃, (or maxterms from M₄₈ to M₆₃). In this block, the variable A is a 1 and the variable B is also a 1.

While simplification, the 6-variable K-map may contain 2-squares, 4-squares, 8-squares, 16- squares, 32-squares, or a 64-square by involving all the four blocks of the map.

In a six variable K-Map, when a block is superimposed on the top of another block, which is either above or below or beside of the first block, and the squares coincide with one another. Then, the squares are considered adjacent in the two blocks. It is also important to note that the diagonal elements such as m₁₀ and m₅₈, m₁₅ and m₆₃, m₁₈ and m₃₄, m₂₉ and m₄₅ are not adjacent to each other.

A six variable SOP K-map is represented in Figure-2.

A six variable POS K-map is represented in Figure-3.

Now, let us understand the utilization of the six variable K-map for minimization of a six variable Boolean function with the help of solved numerical examples.

Example 1

Minimize the following 6 variable Boolean function using K-map.

$$f=\sum m( 1, 3, 4, 5, 6, 9, 11, 12, 14, 15, 17, 19, 20, 21, 22, 23,25, 27, 28, 30, 33, 35, 36, 38, 41, 43, 44, 46, 49, 51, 52, 54, 57, 59, 60, 62)$$

Solution

The 6-variable SOP K-map representation of the given Boolean function is shown in Figure-4.

The reduction of this function is done as per the following steps −

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

• The minterm m₁ forms a 16-square with minterms m₃, m₉, m₁₁, m₁₇, m₁₉, m₂₅, m₂₇, m₃₃, m₃₅, m₄₁, m₄₃, m₄₉, m₅₁, m₅₇, and m₅₉. Make it and read it as $\bar{D}F$.

• The minterm m₄ forms a 16-square with minterms m₆, m₁₂, m₁₄, m₂₀, m₂₂, m₂₈, m₃₀, m₃₆, m₃₈, m₄₄, m₄₆, m₅₂, m₅₄, m₆₀, and m₆₂. Make it and read it as $\bar{F}D$.

• The minterm m₅ can form a 4-square with minterms m₄, m₂₀, and m₂₁, or with m₁, m₁₇, and m₂₁. We will make it with minterms m₄, m₂₀, and m₂₁, and read it as $\bar{A}\bar{C}D\bar{E}$.

• The minterm m₂₁ forms a 4-square with minterms m₁₇, m₁₉, and m₂₃. Make it and read it as $\bar{A}B\bar{C}F$.

• The minterm m₁₅ can make a 2-square with the minterm m₁₁ or m₁₄. We will make it with m₁₄, and read it as $\bar{A}\bar{B}CDE$.

• Write all the product terms in SOP form.

Therefore, the minimal SOP expression is,

$$f\left ( A,B,C,D,E,F \right )=\bar{D}F+D\bar{F}+\bar{A}\bar{C}D\bar{E}+\bar{A}B\bar{C}F+\bar{A}\bar{B}CDE$$

Example 2

Minimize the following six variable Boolean function using K-map.

$$f=\Pi M(0, 2, 7, 8, 10, 13, 16, 18, 24, 26, 29, 31,32, 34, 37, 39, 40, 42, 45, 47, 48, 50, 53, 55, 56, 58, 61, 63)$$

Solution

The 6-variable POS K-map representation of the given Boolean function is shown in Figure-5.

The reduction of this function is done as per the following steps −

• There are no isolated 0s in the K-map.

• The maxterm M₀ forms a 16-square with maxterms M₂, M₈, M₁₀, M₁₆, M₁₈, M₂₄, M₂₆, M₃₂, M₃₄, M₄₀, M₄₂, M₄₈, M₅₀, M₅₆, M₅₈. Make it and read it as $\left ( D+F \right )$.

• The maxterm M₃₇ forms an 8-square with maxterms M₃₉, M₄₅, M₄₇, M₅₃, M₅₅, M₆₁, and M₆₃. Make it and read it as $\left ( \bar{A}+\bar{D}+\bar{F} \right )$.

• The maxterm M₁₃ makes a 4-square with maxterms M₂₉, M₄₅, and M₆₁. Make it and read it as $\left( \bar{C}+\bar{D}+E+\bar{F} \right )$.

• The maxterm M₃₁ makes a 4-square with M₂₉, M₆₁, and M₆₃. Make it and read it as $\left( \bar{B}+\bar{C}+\bar{D}+\bar{F} \right )$.

• The maxterm M₇ forms a 2-square with the maxterm M₃₉. Make it and read it as $\left( B+C+E+\bar{D}+\bar{E}+\bar{F} \right )$.

• Write all the sum terms in POS form.

Thus, the minimal POS expression for the given Boolean function is,

$$f\left ( A,B,C,D,E,F \right )=\left ( D+F \right )\left ( \bar{A}+\bar{D}+\bar{F} \right )\left( \bar{C}+\bar{D}+E+\bar{F} \right )\left( \bar{B}+\bar{C}+\bar{D}+\bar{F} \right )\left( B+C+E+\bar{D}+\bar{E}+\bar{F} \right )$$

This is all about six variable K-map and its application in minimization of Boolean functions. Try solving the following tutorial problems to excel in the concept of six variable K-map and its application to reduce Boolean expressions.

Q. 1 − Minimize the following six variable Boolean function in SOP form using K-map.

$$f\left ( A,B,C,D,E,F \right )=\sum m(1, 3, 5, 7, 9, 10, 11, 12, 14, 16, 17, 19, 20, 22, 24, 25, 26, 31, 32,34, 36, 37, 39, 40, 42, 45, 49, 50, 53, 54, 57, 60, 61, 63) $$

Q. 2 − Minimize the following six variable Boolean function in POS form using K-map.

$$f\left ( A,B,C,D,E,F \right )=\Pi M( 0, 1, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 17, 20, 21, 22, 25, 26, 27,29, 30, 32, 33, 35, 36, 37, 39, 40, 41, 42, 45, 47, 48, 50, 52, 53, 55, 56, 58, 59, 60, 62 )$$