Question

Any Boolean function of n+1 variables can be implemented with ___ multiplexer

• A. \( 2^n \) to 1
• B. \( 2^{2n} \) to 1
• C. \( 2^{n-1} \) to 1
• D. \( 2^{n+1} \) to 1

Hint:

A common method to implement a Boolean function of \(N\) variables is to use a \(2^{N-1} \times 1\) multiplexer.
\(N-1\) variables are used as select inputs.
The remaining variable (or constants 0, 1, or its complement) is connected to the \(2^{N-1}\) data inputs.
In this question, total variables = \(n+1\). So \(N-1 = (n+1)-1 = n\) select lines. MUX size is \(2^n \times 1\).

Answer 1

The Correct Option is A

A multiplexer (MUX) with \(k\) select lines can select one of \(2^k\) data inputs. If we want to implement a Boolean function of \(N\) variables using a multiplexer:
We can use an \(N-1\) to \(2^{N-1}\) decoder and OR gates.
Or, we can use a multiplexer. If we use \(N-1\) variables as select lines for the MUX, then the MUX will have \(2^{N-1}\) data inputs. The remaining 1 variable (or functions of it like 0, 1, itself, or its complement) will be connected to these data inputs.
If we use all \(N\) variables to generate minterms which then feed the data inputs of a \(2^N \times 1\) MUX with some fixed select logic, that's also possible but less direct. The standard way to implement a function of \(N\) variables is to use \(N-1\) variables for the select lines of a \(2^{N-1} \times 1\) MUX. The data inputs are then functions of the Nth variable (0, 1, Nth var, or Nth var complement). The question states a function of \(n+1\) variables. Let \(N = n+1\). We can use \(N-1 = (n+1)-1 = n\) variables as select lines. This requires a multiplexer with \(n\) select lines, which means it is a \(2^n \times 1\) (or \(2^n\) to 1) multiplexer. The \(2^n\) data inputs will be connected to 0, 1, the \((n+1)^{th}\) variable, or its complement, based on the truth table. Example: Function of 3 variables (A,B,C). Here \(N=3\), so \(n+1=3 \implies n=2\). Use A,B as select lines (so \(n=2\) select lines). MUX size is \(2^2 \times 1 = 4 \times 1\). Data inputs \(I_0, I_1, I_2, I_3\) will be functions of C (i.e., 0, 1, C, or \(\bar{C}\)). So, a function of \(N = n+1\) variables can be implemented using a \(2^{(n+1)-1} \times 1 = 2^n \times 1\) multiplexer. This matches option (a). \[ \boxed{2^n \text{ to } 1} \]