Consider the Karnaugh map given below, where X represents “don't care” and blank represents 0. Assume for all inputs $(a, b, c, d)$, the respective complements $(\overline{a},\overline{b},\overline{c},\overline{d})$ are also available. The above logic is implemented using 2-input NOR gates only. The minimum number of gates required is _________.

From the given Karnaugh map with column variables $ba$ and row variables $dc$:

1. Identify the 1-cells and don't-care (X) cells:

   • 1-cells are at $(ba, dc) \in \{(00, 01), (00, 11), (10, 11)\}$.

   • X-cells are at $(ba, dc) \in \{(01, 00), (11, 00), (10, 01), (01, 10), (11, 10)\}$.

2. Group the cells to find the minimal SOP expression:

   • We can form a 4-cell group (quad) using columns $ba = 00, 10$ and rows $dc = 01, 11$:

     • $(00, 01) = 1$

     • $(00, 11) = 1$

     • $(10, 01) = \text{X}$

     • $(10, 11) = 1$

   • In this group, $a = 0$ (common to columns $00$ and $10$) and $c = 1$ (common to rows $01$ and $11$).

   • This covers all three 1s in the K-map, giving the minimal Boolean expression:

     $$F = \overline{a}c$$

3. Implement using 2-input NOR gates:

   • Using the NOR logic function $\text{NOR}(x, y) = \overline{x + y}$ and De Morgan's laws:

     $$\overline{a + \overline{c}} = \overline{a} \cdot \overline{\overline{c}} = \overline{a}c$$

   • Since all complemented literals ($\{a, b, c, d, \overline{a}, \overline{b}, \overline{c}, \overline{d}\}$) are freely available, $F$ can be generated directly as:

     $$F = \text{NOR}(a, \overline{c})$$

Hence, the minimum number of 2-input NOR gates required is 1.