Problem 31 - Discrete Mathematics with Applications

Question

Find the DNFs of the boolean functions $$\begin{array}{|ccc||c|} \hline \boldsymbol{x} & \boldsymbol{y} & \boldsymbol{z} & \boldsymbol{f}(\boldsymbol{x}, \boldsymbol{y}, \boldsymbol{z}) \\ \hline 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

The Disjunctive Normal Form (DNF) of the given boolean function f(x, y, z) is: $$f(x, y, z) = (x \cdot \overline{y} \cdot \overline{z}) \oplus (x \cdot y \cdot z)$$

Step 1: Identify the minterms (rows with output 1)

Analyze the provided Truth Table to find the rows in which the output is True (1). In this exercise, we have the following minterms identified by the input values (x, y, z): (1, 0, 0) and (1, 1, 1).

Step 2: Create conjunctions (AND operations) for each minterm

For each minterm, create a conjunction (AND operation) using the corresponding input values of x, y, and z. If the input value is 0, use the negation of the variable, and if the input value is 1, use the variable itself. For the minterm (1, 0, 0), the conjunction is: \(x \cdot \overline{y} \cdot \overline{z}\) For the minterm (1, 1, 1), the conjunction is: \(x \cdot y \cdot z\)

Step 3: Combine the conjunctions using disjunctions (OR operations)

Join the conjunctions created in Step 2 with the disjunction (OR) operation to form the Disjunctive Normal Form (DNF) of the boolean function. The DNF is as follows: $$f(x, y, z) = (x \cdot \overline{y} \cdot \overline{z}) \oplus (x \cdot y \cdot z)$$

Key Concepts

Boolean Functions

Boolean functions are the foundation of digital logic design and computer science. They provide a systematic way of mapping binary inputs to a single binary output. A Boolean function can be represented using a truth table, which lists all possible combinations of its input variables and the resultant output for each combination. This function is defined using logical operations like AND, OR, and NOT.

For example, in the exercise, the Boolean function \( f(x, y, z) \) has three input variables \( x, y, z \) and one output. The function is evaluated for every possible input combination, and the corresponding output is noted. Boolean functions are immensely useful in designing circuits and algorithms because of their ability to precisely define logical relationships between inputs and outputs.

Minterms in Truth Tables

Minterms play a crucial role when analyzing truth tables in the context of Boolean functions. A minterm is a product (AND operation) of all input variables in a function, where each variable is either in its true form or complemented (negated). Each minterm corresponds to a particular row in the truth table where the function's output is 1 (true).

In the given exercise, we observe the truth table and identify rows where the function outputs a 1. For these rows, we construct the corresponding minterms. Since we are dealing with binary values, a minterm for an input combination where the variable is 0 will include the negation of that variable, and if it's 1, we use the variable itself. This creates a unique minterm for every row where the function's output is true.

Conjunctions in Boolean Algebra

The core of constructing minterms lies in the use of conjunctions, which are AND operations in Boolean algebra. A conjunction connects two or more binary variables such that the output is true only if all of the individual variables are true. This operation is denoted by a dot \( \cdot \) or simply by placing variables side by side.

When solving the exercise, we translate each identified minterm into a conjunction. For instance, the minterm (1, 0, 0) becomes \( x \cdot \overline{y} \cdot \overline{z} \) because the function's output is true when \( x \) is 1 and both \( y \) and \( z \) are 0, hence their negations are involved. These conjunctions accurately reflect the conditions under which the Boolean function yields a true value. Additionally, these conjunctions can be combined using disjunctions (OR operations), as seen in the Disjunctive Normal Form, to create a complete expression for the Boolean function.