Simplifications Using Boolean Algebra

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What is a Karnaugh Map (K-Map)?

K-Map is a graphical tool used for simplifying Boolean expressions represented in their standard form to obtain their minimal form.

The K-Map is basically a graph or chart that composed of an arrangement of adjacent cells or squares, where each cell represents a particular combination of variables of the function either in sum or product form.

The number of cells in the K-Map depends upon number of variables in the Boolean function, i.e., K-map has 2ⁿ adjacent cells, where n is the number of variables in the Boolean expression. Therefore, the number of cells in a 2 variable K-map are 4 (2²), in 3 variable KMap, the number of cells are 8 (2³), in 4 variable K-map, the number of cells are 16 (2⁴), and so on.

However, we can use the K Map for any number of variables. But, for simplifying Boolean functions in variables more than 5, it becomes tedious.

Now, let us discuss the procedure of simplifying a Boolean expression using K-Map.

Steps to Simplify a Boolean Expression using K Map

The following steps are involving in simplification of a given Boolean expression by using K-Map −

Step 1 − Select a K-Map as per the number of variables in the given Boolean function.

Step 2 − Identify the minterms (in SOP form) or maxterms (in POS form).

Step 3 − For SOP (Sum of Products) Form, put 1s in cells of the K-Map with respect to the minterms of given function. Read the K-Map as follows −

• Read the K-map for 1s which are not adjacent to any other 1. These 1s are the isolated minterms, thus they are to be read as they are, because they cannot be combine in groups.
• Read the K-map for 1s which are adjacent only one other 1. Combine such minterms in 2-squares.
• Read the K-map for quads (4-squares), octets (8-squares), and so on of adjacent 1s even if they have some 1s which are already combined in other groups. The only thing to remember that they must geometrically form a rectangle or a square.
• Read the K-map for any 1s that have not been grouped yet and group them into bigger squares or rectangles if possible.
• Finally, obtain product terms of all the groups, and then sum up them to form the minimal SOP expression.

For POS (Product of Sums) Form, put 0s in cells of the K-Map with respect to the maxterms of given function. Read the K-Map as follows −

• Read the K-map for 0s which are not adjacent to any other 0. These 0s are the isolated maxterms, thus they are to be read as they are, because they cannot be combine in groups.
• Read the K-map for 0s which are adjacent only one other 0. Combine such maxterms in 2-squares.
• Read the K-map for quads (4-squares), octets (8-squares), and so on of adjacent 0s even if they have some 0s which are already combined in other groups. The only thing to remember that they must geometrically form a rectangle or a square.
• Read the K-map for any 0s that have not been combined yet and combine them into bigger squares or rectangles if possible.
• Finally, obtain sum terms of all the groups, and then product them to form the minimal POS expression.

Let us understand this procedure of simplifying Boolean expression using K-map with the help of some solved examples.

Example 1

Simplify the following 3-variable Boolean function in SOP form using K-Map.

$$\mathrm{F(P, \: Q, \:R) \: = \: \sum m ( 0, \: 1, \: 3, \: 5, \: 7)}$$

Solution

The K-Map representation of the given Boolean function is shown in Figure-1.

The simplification of this K-map is done as per the following steps −

• There are no isolated 1s.
• The minterm m₁ forms a 4-square with minterms m₃, m₅, and m₇. Make it and read it as R.
• The minterm m₀ forms a 2-square with the minterm m₁. Make it and read it as $\mathrm{\bar{P} \: \bar{Q}}$
• Write all the product terms in SOP form.

Thus, the simplified SOP expression is,

$$\mathrm{F \: = \: R \: + \: \bar{P}\bar{Q}}$$

Example 2

Simplify the following 3-variable Boolean function in POS form using K-Map.

$$\mathrm{F(A, \: B, \: C) \: = \: \Pi \: M(1, \: 2, \: 4, \: 6)}$$

Solution

The POS K-map representation of the given Boolean function is shown in Figure-2.

The simplification of this POS K-map is done as per the following steps −

• The maxterm M₁ has no adjacency. Thus, keep it as it is and read it as $\mathrm{(A \: + \: B \: + \: \bar{C})}$.
• The maxterm M₂ has only one adjacency M₆. Hence, expand the maxterm M₂ into a 2-square with the maxterm M₆ and read the 2-square as $\mathrm{(\bar{B} \: + \: C)}$.
• The maxterm M₄ also has only one adjacency M₆. Hence, expand the maxterm M₄ into a 2-square with the maxterm M₆ and read the 2-square as $\mathrm{(\bar{A} \: + \: C)}$.
• Write all the sum terms in POS form.

Therefore, the simplified POS expression is,

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

Example 3

Simplify the following 4-variable Boolean function in SOP form to obtain the minimal SOP expression.

$$\mathrm{F(A, \: B, \: C, \:D) \: = \: \sum \: m( 0,\: 1, \:3, \: 5, \: 7, \: 6, \: 10, \: 13, \: 14, \: 15)}$$

Solution

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

The simplification of this SOP K-map is done as per the following steps −

• There are no isolated 1s.
• The minterm m₁ has three adjacencies m₃, m₅, and m₇. So expand m₁ into a 4-square with minterms m₃, m₅, and m₇, and read the 4-square as $\mathrm{\bar{A}D}$.
• The minterm m5 has three adjacencies m₇, m₁₃, and m₁₅. Expand m₅ into a 4-square with minterms m₇, m₁₃, and m₁₅, and read the 4-square as BD.
• The minterm m6 also has three adjacencies m₇, m₁₄, and m₁₅. Expand m6 into a 4- square with minterms m₇, m₁₄, and m₁₅, and read the 4-square as BC.
• The minterm m₁₀ has only one adjacency m₁₄. Expand m₁₀ into a 2-square with minterm m₁₄ and read the 2-square as $\mathrm{AC\bar{D}}$.
• The minterm m₀ also has only one adjacency m₁. Expand m₀ into a 2-square with minterm m₁ and read the 2-square as $\mathrm{\bar{A}\bar{B}\bar{C}}$.
• Write all the product terms in SOP form.

Hence, the simplified SOP expression is,

$$\mathrm{F \: = \: \bar{A}D \: + \: BD \: + \: BC \: + \: AC\bar{D} \: + \: \bar{A}\bar{B}\bar{C}}$$

Conclusion

In this way, we can simplify a given Boolean expression using K-map to obtain the minimal expression. Try solving the following tutorial problems for better understanding.

Q. 1 − Simplify the following 3-variable Boolean function in SOP form to obtain its minimal expression.

$$\mathrm{F(A, \: B, \: C) \: = \: \sum \: m(1, \: 2, \: 4, \: 5, \: 7)}$$

Q. 2 − Simplify the following 4-variable Boolean function in SOP form to obtain the minimal Boolean expression.

$$\mathrm{F(A, \: B, \: C, \: D) \: = \: \sum \: m(0, \: 1, \: 2, \: 4, \: 5, \: 7, \: 8, \: 9, \: 10, \: 12, \: 14, \: 15)}$$