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Chapter 15: Problem 8

Let \(F_{n}=\left\\{f: B^{n} \rightarrow B\right\\}\) be the Boolean algebra of all Boolean functions on \(n\) Boolean variables. How many atoms does \(F_{n}\) have?

Short Answer

Expert verified
The Boolean Algebra \(F_{n}\) on \(n\) variables has \(2^n\) atoms.

Step by step solution

01

Understanding Atoms in Boolean Algebra

An atom in Boolean algebra is the smallest non-empty subset of a Boolean algebra. These are the elements that cannot be expressed as the union of other elements in the algebra. For a Boolean function with \(n\) variables, atoms are represented by 'minterms'. A 'minterm' is a product (AND) of \(n\) literals (A literal is a variable or its complement). Therefore, for a Boolean function of \(n\) variables, atoms are the \(n\)-variable minterms.
02

Counting Atoms in Boolean Algebra

There are two possibilities for each literal in a minterm; it can be the variable or its complement. So for \(n\) variables, there are \(2^n\) possible minterms. Therefore, the number of atoms in the Boolean algebra of all Boolean functions on \(n\) variables, i.e., \(F_{n}\) is \(2^n\).

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Most popular questions from this chapter

If \(g: B^{7} \rightarrow B\) is a Boolean function of the Boolean variables \(x_{1}, x_{2}, \ldots, x_{7}\), how many 1 's are needed in the Karnaugh map of \(g\) in order to represent the product term (a) \(x_{1}\); (b) \(x_{1} x_{2} ;\) (c) \(x_{1} \bar{x}_{2} x_{3} ;\) (d) \(x_{1} x_{3} x_{3} x_{7}\) ?

Find a minimal-sum-of-products representation for a) \(f(w, x, y, z)=\sum m(1,3,5,7,9)+d(10,11,12,13,14,15)\) b) \(f(w, x, y, z)=\sum m(0,5,6,8,13,14)+d(4,9,11)\) c) \(f(v, w, x, y, z)=\sum m(0,2,3,4,5,6,12,19,20,24,28)+d(1,13,16,29,31)\)

Let \(g: B^{4} \rightarrow B\) be defined by \(g(w, x, y, z)=(w z+x y z)(x+\bar{x} \bar{y} z)\). a) Find the d.n.f. and c.n.f. for \(g\). b) Write \(g\) as a sum of minterms and as a product of maxterms (utilizing binary labels).

Simplify the following Boolean expressions. a) \(x y+(x+y) \bar{z}+y\) b) \(x+y+\overline{(\bar{x}+y+z)}\) c) \(y z+w x+z+[w z(x y+w z)]\) d) \(x_{1}+\bar{x}_{1} x_{2}+\bar{x}_{1} \bar{x}_{2} x_{3}+\bar{x}_{1} \bar{x}_{2} \bar{x}_{3} x_{4}+\cdots\)

For (a) \(n=60\), and (b) \(n=120\), explain why the positive divisors of \(n\) do not yield a Booiean algebra. (Here \(x+y=\operatorname{lcm}(x, y), x y=\operatorname{gcd}(x, y), \bar{x}=n / x, 1\) is the zero element, and \(n\) is the one element.)
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