Minimum NAND Gates for (A+B)CD Expression

The given Boolean expression is \(((A + B)C) D\). Our goal is to implement this expression using the minimum possible number of 2-input NAND gates.

First, let's look at the structure of the expression. It's an AND operation of \((A+B)\), \(C\), and \(D\). We can also expand this expression using the distributive property:

\(((A + B)C) D = (AC + BC) D = ACD + BCD\)

The expression \(ACD + BCD\) is in Sum of Products (SOP) form. We can implement this SOP form using only NAND gates.

A common technique to implement SOP expressions like \(P+Q\) using NAND gates is \(\overline{\overline{P+Q}} = \overline{\overline{P} \cdot \overline{Q}} = \text{NAND}(\overline{P}, \overline{Q})\). So, if we can generate the inverted terms \(\overline{ACD}\) and \(\overline{BCD}\), we can get the final result using one more NAND gate.

Let's implement the terms \(\overline{ACD}\) and \(\overline{BCD}\) and then combine them:

• The term \(ACD\) is an AND operation. \(\overline{ACD} = \text{NAND}(A, \text{NAND}(C,D))\) is incorrect. \(\overline{XYZ} = \overline{X} + \overline{Y} + \overline{Z}\).
• However, we can implement \(ACD\) and \(BCD\) by noticing they share the term \(CD\).

Implementing (A+B)CD with 5 NAND Gates

We can implement the expression \(ACD + BCD\) using the following 5 NAND gates:

1. Gate 1: Take inputs \(C\) and \(D\). The output is \(\text{NAND}(C, D) = \overline{CD}\).
2. Gate 2: Take the output of Gate 1 (\(\overline{CD}\)) and feed it into both inputs of a NAND gate. The output is \(\text{NAND}(\overline{CD}, \overline{CD}) = \overline{\overline{CD} \cdot \overline{CD}} = \overline{\overline{CD}} = CD\). Now we have the term \(CD\).
3. Gate 3: Take inputs \(A\) and the output of Gate 2 (\(CD\)). The output is \(\text{NAND}(A, CD) = \overline{A \cdot CD} = \overline{ACD}\). This is the inverted first term of the SOP form.
4. Gate 4: Take inputs \(B\) and the output of Gate 2 (\(CD\)). The output is \(\text{NAND}(B, CD) = \overline{B \cdot CD} = \overline{BCD}\). This is the inverted second term of the SOP form.
5. Gate 5: Take the outputs of Gate 3 (\(\overline{ACD}\)) and Gate 4 (\(\overline{BCD}\)). The output is \(\text{NAND}(\overline{ACD}, \overline{BCD})\).

Let's evaluate the output of the final gate (Gate 5) using De Morgan's theorem:

\(\text{Output of Gate 5} = \text{NAND}(\overline{ACD}, \overline{BCD}) = \overline{\overline{ACD} \cdot \overline{BCD}}\)

Applying De Morgan's theorem (\(\overline{X \cdot Y} = \overline{X} + \overline{Y}\)):

\(\overline{\overline{ACD} \cdot \overline{BCD}} = \overline{\overline{ACD}} + \overline{\overline{BCD}} = ACD + BCD\)

As we established earlier, \(ACD + BCD\) is equivalent to the original expression \(((A + B)C) D\).

Therefore, the expression \(((A + B)C) D\) can be implemented using 5 NAND gates as described above.

This circuit structure provides the minimum number of NAND gates required.