Minimization of Boolean Functions

Minimization of Boolean Functions

What is Minimization?

• In mathematics, expressions are simplified for a number of reasons, for instance simpler expression are easier to understand and easier to write down, they are also less prone to error in interpretation but, most importantly, simplified expressions are usually more efficient and effective when implemented in practice.

• A Boolean expression is composed of variables and terms. The simplification of Boolean expressions can lead to more effective computer programs, algorithms and circuits.

What is Minimization?

• Minimisation can be achieved by a number of methods, three well known methods are:1. Algebraic Manipulation of Boolean

Expressions

2. Tabular Method of Minimization

3. Karnaugh Maps

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Algebraic Manipulation of Boolean Expressions

• We can now start doing some simplifications

x’y’ + xyz + x’y= x’(y’ + y) + xyz [ Distributive: x’y’ + x’y = x’(y’ + y) ]= x’1 + xyz [ complement: x’ + x = 1 ]= x’ + xyz [ identity: x’1 = x’ ]= (x’ + x)(x’ + yz) [ Distributive ]= 1 (x’ + yz) [ complement: x’ + x = 1 ]= x’ + yz [ identity]

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Algebraic Manipulation of Boolean Expressions

• Here are two different but equivalentcircuits.

• In general the one with fewer gates is “better”:

– It costs less to build

– It requires less power

– But we had to do some work to find the second form

Tabular Method of Minimization

• The tabular method which is also known as the Quine-McCluskey method is particularly useful when minimising functions having a large number of variables, e.g. The six-variable functions. Computer programs have been developed employing this algorithm.

• The method reduces a function in standard sum of products form to a set of prime implicants from which as many variables are eliminated as possible.

• These prime implicants are then examined to see if some are redundant.

Note: A prime implicant is a product term which cannot be further simplified by combination with other terms.


• We will show how the Quine–McCluskey method can be used to find a minimal expansion

equivalent to:

• We will represent the minterms in this expansion by bit strings. The first bit will be 1 if x occurs and 0 if x occurs. The second bit will be 1 if y occurs and 0 if y occurs. The third bit will be 1 if z occurs and 0 if z occurs.

• We then group these terms according to the number of 1s in the corresponding bit strings. This information is shown in Table 1.



• Step 1:– Minterms that can be combined are those that differ in

exactly one literal. Hence, two terms that can be combined differ by exactly one in the number of 1s in the bit strings that represent them.

– When two minterms are combined into a product, this product contains two literals. A product in two literals is represented using a dash to denote the variable that does not occur.

– For instance, the minterms xyz and x yz, represented by bit strings 101 and 001, can be combined into yz, represented by the string _01.


• Step 1 (continued):– All pairs of minterms that can be combined and the

product formed from these combinations are shown in Table 2.

Note: (1,2) means that term 1 and term 2 are combined.(4,5) means that term 4 and term 5 are combined, etc.


• Step 2:– Next, all pairs of products of two literals that can be

combined are combined into one literal. Two such products can be combined if they contain literals for the same two variables, and literals for only one of the two variables differ.

– In terms of the strings representing the products, these strings must have a dash in the same position and must differ in exactly one of the other two slots.

– We can combine the products yz and yz, represented by the strings _11 and _01, into z, represented by the string _ _1.


• Step 2 (continued):– We show all the combinations of terms that can be

formed in this way in Table 3.

Note: (1,2,3,4) means that terms (1,2) and (3,4) are combined OR

terms (1,3) and (2,4) are combined.


• Step 3:– In Table 3 we also indicate which terms have been used

to form products with fewer literals; these terms will not be needed in a minimal expansion.

– The next step is to identify a minimal set of products needed to represent the Boolean function.

– We begin with all those products that were not used to construct products with fewer literals.


• Step 3:– Next, we form Table 4, which has a row for each

candidate product formed by combining original terms, and a column for each original term; and we put an X in a position if the original term in the sum-of-products expansion was used to form this candidate product.

Note: z covers xyz, xyz, xyz and xyz (i.e. there is an “X”).xy covers xyz and xyz.


• Step 3 (continued):– In this case, we say that the candidate product covers

the original minterm. We need to include at least one product that covers each of the original minterms.

– Consequently, whenever there is only one X in a column in the table, the product corresponding to the row this X is in must be used.

– From Table 4 we see that both z and x y are needed.

– Hence, the final answer is z + x y.

Karnaugh Maps

• So far we can see that applying Boolean algebra can be awkward in order to simplify expressions.

• Apart from being laborious (and requiring the remembering all the laws) the method can lead to solutions which, though they appear minimal, are not.

• The Karnaugh map provides a simple and straight-forward method of minimising boolean expressions.

• With the Karnaugh map Boolean expressions having up to four and even six variables can be simplified.

Karnaugh Maps

• So what is a Karnaugh map?– A Karnaugh map (K-map) provides a pictorial method of

grouping together expressions with common factors and therefore eliminating unwanted variables. The Karnaugh map can also be described as a special arrangement of a truth table.

– The diagram below illustrates the correspondence between the Karnaugh map and the truth table for the general case of a two variable problem.

Same As:

Karnaugh Maps

• 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. y is along the top and x is down the left hand side.

• The diagram below explains this:

Karnaugh Maps (K-maps)

• Example 1:

We can identify minterms that can be combined from the K-map. Whenever there are 1s in two adjacent cells in the K-map, the minterms represented by these cells can be combined into a product involving just one of the variables.


• Example 1 (continued):– For instance, and are represented by adjacent

cells and can be combined into , because

– Moreover, if 1s are in all four cells, the four minterms can be combined into one term, namely, the Boolean expression 1 that involves none of the variables.

– We circle blocks of cells in the K-map that represent minterms that can be combined and then find the corresponding sum of products.

– The goal is to identify the largest possible blocks, and to cover all the 1s with the fewest blocks using the largest blocks first and always using the largest possible blocks.


• Example 1 (continued):– Simplify the sum-of-products expansions.

Note: Part (b) cannot be simplified, that is why the solution is the same expression we started with.


• Example 2:

Note: A K-map in three variables is a rectangle divided into eight cells. The cells represent the eight possible minterms in three variables. Two cells are said to be adjacent if the minterms that they represent differ in exactly one literal.


• Example 2 (continued):– The K-maps for these sum-of-products expansions are

shown below:


• Example 2 (continued):– The grouping of blocks shows that minimal expansions

into Boolean sums of Boolean products are:

Note: In part (d) note that the prime implicants xz and x y are essential prime implicants, but the prime implicant yz is not essential, because the cells it covers are covered by the other two prime implicants.

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Minimization of Boolean Functions