/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Let \(\left[B ; V_{1} \wedge_{+}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(\left[B ; V_{1} \wedge_{+}-\right]\) be a Boolean algebra of order \(4,\) and let \(f\) be a Boolean function of two variables on \(B\). (a) How many elements are there in the domain of \(\mathrm{f?}\) (b) How many different Boolean functions are there of two, variables? Three variables? (c) Determine the minterm normal form of \(f\left(x_{1}, x_{2}\right)=x_{1} \vee x_{2}\). (d) If \(B=\\{0, a, b, 1\\},\) define a function from \(B^{2}\) into \(B\) that is not a Boolean function.

Short Answer

Expert verified
(a) 16 elements; (b) 16 for 2 variables, 256 for 3 variables; (c) \( x_{1}\overline{x_{2}} \vee \overline{x_{1}}x_{2} \vee x_{1}x_{2} \); (d) Define using non-standard mappings.

Step by step solution

01

Determine the Domain Size for Two Variables

In a Boolean algebra of order 4, the set \( B \) contains 4 elements. The domain of a Boolean function with two variables \( f(x_{1}, x_{2}) \) consists of all possible pairs \( (x_{1}, x_{2}) \) where \( x_{1}, x_{2} \in B \). Hence, there are \( 4 \times 4 = 16 \) possible input pairs.
02

Calculate Number of Boolean Functions for Two Variables

A Boolean function has a truth value for each possible outcome. For two variables taken from algebra of order 2 (e.g., \( \{0,1\} \)), there are \( 4 \) possible input pairs: (00, 01, 10, 11). Each pair maps to either 0 or 1. Hence, the total number of different Boolean functions is \( 2^4 = 16 \). In the case of two variables from algebra of order 4, it also follows that a function can take any of the 4 values for each possible input pair, hence \( 4^{16} \).
03

Calculate Number of Boolean Functions for Three Variables

For three variables in order 2, there are \( 2^3 = 8 \) possible input combinations. As before, each output can be either 0 or 1, so the number of functions is \( 2^8 = 256 \).
04

Determine the Minterm Normal Form

The function \( f(x_{1}, x_{2}) = x_{1} \vee x_{2} \) can be expressed in the minterm (or disjunctive normal) form by considering each minterm where the function's value is 1. These minterms correspond to the cases: (1,0), (0,1), (1,1). Thus, the minterm expression is \( m_{1} \vee m_{2} \vee m_{3} = x_{1}\overline{x_{2}} \vee \overline{x_{1}}x_{2} \vee x_{1}x_{2} \).
05

Define a Non-Boolean Function from \(B^2\) Into \(B\)

Since the set \( B = \{0, a, b, 1\} \), a valid function \( g: B^2 \to B \) must map every pair in \( B^2 \) to an element in \( B \). A non-Boolean function may break Boolean rules like commutativity, associativity, or have undefined behavior for some inputs. For instance, selecting values arbitrarily for combinations such as \( g(a,b) = 1 \), \( g(a,a) = a + b \) (not a single element of \( B \)).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boolean function
A Boolean function is a mathematical construct used within the realm of Boolean algebra. It takes binary values of variables, such as 0 and 1, as inputs and produces a binary output. These functions underpin digital circuit design, computer logic, and programming.
For instance, a function with two variables, say \(f(x_1, x_2)\), will have outcomes based on combinations of these variables. If the set \(B\) has four elements as in the exercise, the domain for \(f\) consists of all possible pairs from these elements. This means, if \(B = \{0, a, b, 1\}\), \(f\) can take 16 different input pairs, making it expansive in terms of possibilities.
The number of unique Boolean functions, indeed grows as the number of variables increases. With two binary variables, each giving a binary output, there are \(2^4 = 16\) potential functions, while with three, the possibilities expand to \(2^8 = 256\). The complexity of creating comprehensive computing systems builds from these exponential possibilities.
Minterm normal form
Every Boolean function can be expressed in a format known as the **Minterm Normal Form** (MNF), which is a systematic way to represent Boolean expressions. It is formed by OR-ing (disjoining) the minterms. A minterm is essentially a unique AND-ed (conjoined) sequence of variables where the function evaluates to 1.
Consider the function \(f(x_1, x_2) = x_1 \vee x_2\). In MNF, this function is expressed by focusing on combinations where the function gives the result 1:
  • \((1,0)\): The minterm is \(x_1\overline{x_2}\).
  • \((0,1)\): The minterm is \(\overline{x_1}x_2\).
  • \((1,1)\): The minterm is \(x_1x_2\).
Therefore, the minterm expression becomes \(x_1\overline{x_2} \vee \overline{x_1}x_2 \vee x_1x_2\). This systematic approach not only organizes the Boolean function neatly but also serves as the foundation for designing complex logical circuits.
Domain of a function
In the context of Boolean algebra, understanding the "domain of a function" is central to determining how many inputs a Boolean function can take. The domain refers to all the possible input combinations that the function might receive.
For instance, if a Boolean function \(f\) is defined over a domain \(B\) with four elements \(\{0, a, b, 1\}\), and \(f\) uses two variables, the domain becomes \(B^2\). This denotes all pairs of inputs from \(B\) multiplied by itself, resulting in \(16\) possible input combinations like \((0,0), (0,a), (0,b), \ldots, (1,1)\).
Understanding the domain is vital for determining how many different Boolean functions can exist. Larger domains mean more input combinations, which expands the variety of function outputs. This consideration is essential for tasks ranging from logical circuit design to algorithmic problem-solving.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the Boolean expression \(f\left(x_{1}, x_{2}, x_{3}\right)=\left(\overline{x_{3}} \wedge x_{2}\right) \vee\left(\overline{x_{1}} \wedge x_{3}\right) \vee\) \(\left(x_{2} \wedge x_{3}\right)\) on \(\left[B_{2}^{3} ; V, \wedge,-\right]\) (a) Simplify this expression using basic Boolean algebra laws. (b) Write this expression in minterm normal form. (c) Write out the table for the given function defined by \(f\) and compare it to the tables of the functions in parts a and \(\mathrm{b}\). (d) How many possible different functions in three variables on \(\left[B_{2} ; V_{1} \wedge_{1}-\right]\) are there?

Corollary 13.4.7 implies that there do not exist Boolean algebras of orders \(3,5,6,7,9,\) etc. (orders different from \(\left.2^{n}\right)\). Without this corollary, directly show that we cannot have a Boolean algebra of order 3 . Hint. Assume that \([B ; \vee, \wedge,-]\) is a Boolean algebra of order 3 where \(B=\\{0, x, 1\\}\) and show that this cannot happen by investigating the possibilities for its operation tables.

Formulate a definition for isomorphic Boolean algebras.

Prove an element of a Boolean algebra is an atom if and only if it covers the zero element.

(a) There are many different, yet isomorphic, Boolean algebras with two elements. Describe one such Boolean algebra that is derived from a power set, \(\mathcal{P}(A),\) under \(\subseteq .\) Describe a second that is described from \(D_{n},\) for some \(n \in P,\) under "divides." (b) Since the elements of a two-element Boolean algebra must be the greatest and least elements, 1 and \(0,\) the tables for the operations on \\{0,1\\} are determined by the Boolean algebra laws. Write out the operation tables for \([\\{0,1\\} ; \vee, \wedge,-]\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.