Question

How many different Boolean functions \({\bf{F(x,y,z)}}\) are there such that \({\bf{F(\bar x,y,z) = F(x,\bar y,z) = F(x,y,\bar z)}}\) for all values of the Boolean variables \({\bf{x,y}}\), and \({\bf{z}}\)\({\bf{?}}\)

Short answer

There are \(4\) different Boolean functions of\({\bf{ }}F\left( {x,y,z} \right)\).

Step-by-step solution

Definition

The complement of an element: \(\bar 0 = 1\) and \(\bar 1 = 0\)

The Boolean sum \( + \) or \(O{\bf{ }}R\) is \(1\) if either term is \(1\) .

The Boolean product or \(A{\bf{ }}N{\bf{ }}D\) is \(1\) if both terms are \(1\) .

The dual of a Boolean expression interchanges the Boolean sum and the Boolean product, and interchanges \(0\) and \(1\) .

Product rule If one event can occur in \(m\) ways \(A{\bf{ }}N{\bf{ }}D\) a second event can occur in \(n\) ways, then the number of ways that the two events can occur in sequence is then \(m \times n\)

Finding different Boolean functions

Given: \(F\left( {\bar x,y,z} \right) = F\left( {x,\bar y,z} \right) = F\left( {x,y,\bar z} \right)\)

There are three variables \(x,y,z\) and each variable can take on the value of \(0\)or \(1\) .

\(\begin{array}{l}\begin{array}{*{20}{l}}{\left( {x,{\bf{ }}y,{\bf{ }}z} \right) = \left( {0,0,0} \right){\bf{ }}}\\{\left( {x,{\bf{ }}y,{\bf{ }}z} \right) = \left( {1,0,0} \right){\bf{ }}}\\{\left( {x,{\bf{ }}y,{\bf{ }}z} \right) = \left( {0,1,0} \right){\bf{ }}}\end{array}\\\left( {x,{\bf{ }}y,{\bf{ }}z} \right) = \left( {1,1,0} \right)\end{array}\)

Substituting the values

\(\begin{array}{l}F\left( {\bar x,y,z} \right) = F\left( {1,0,0} \right)\\F\left( {x,\bar y,z} \right) = F\left( {0,1,0} \right)\\F\left( {x,y,\bar z} \right) = F\left( {0,0,1} \right)\\F\left( {1,0,0} \right) = F\left( {0,1,0} \right) = F\left( {0,0,1} \right)\end{array}\)

\(\begin{array}{l}F\left( {\bar x,y,z} \right) = F\left( {0,0,0} \right)\\F\left( {x,\bar y,z} \right) = F\left( {1,1,0} \right)\\F\left( {x,y,\bar z} \right) = F\left( {1,0,1} \right)\\F\left( {0,0,0} \right) = F\left( {1,1,0} \right) = F\left( {1,0,1} \right)\end{array}\)

\(\begin{array}{l}F\left( {\bar x,y,z} \right) = F\left( {1,1,0} \right)\\F\left( {x,\bar y,z} \right) = F\left( {0,0,0} \right)\\F\left( {x,y,\bar z} \right) = F\left( {0,1,1} \right)\\F\left( {0,0,0} \right) = F\left( {1,1,0} \right) = F\left( {1,0,1} \right) = F\left( {0,1,1} \right)\end{array}\)

\(\begin{array}{l}F\left( {\bar x,y,z} \right) = F\left( {0,1,0} \right)\\F\left( {x,\bar y,z} \right) = F\left( {1,0,0} \right)\\F\left( {x,y,\bar z} \right) = F\left( {1,1,1} \right)\\F\left( {1,0,0} \right) = F\left( {0,1,0} \right) = F(0,0,1) = F(1,1,1)\end{array}\)

Notes that in total, you need to assign \(2\) values \(\left( {F\left( {0,0,0} \right),{\bf{ }}F\left( {1,1,1} \right)} \right)\) to uniquely determine \(F\) and the other \(6\) values will follow from \(F\left( {0,0,0} \right) = F\left( {1,1,0} \right) = F\left( {1,0,1} \right) = F\left( {0,1,1} \right)\) and \(F\left( {1,0,0} \right) = F\left( {0,1,0} \right) = F\left( {0,0,1} \right) = F\left( {1,1,1} \right)\). Since you can assign a \(0\) or \(1\) to each of the \(2\) values, there are \(2\) options for \(F\left( {0,0,0} \right)\) and \(2\) options for \(F\left( {1,1,1} \right)\).

By the product rule:\(2 \times 2 = {2^2} = 4\).

Therefore, in total, there are then \(4\) different Boolean functions\({\bf{ }}F\left( {x,y,z} \right)\) such that \(F\left( {\bar x,y,z} \right) = F\left( {x,\bar y,z} \right) = F\left( {x,y,\bar z} \right)\)