Question 21

From the Karnaugh map:

• Row variables are $dc$ in order: 00, 01, 11, 10.
• Column variables are $ba$ in order: 00, 01, 11, 10.
• The cells containing 1 are:
  • $(dc = 01, ba = 00)$
  • $(dc = 11, ba = 00)$
  • $(dc = 11, ba = 10)$
• The cells containing X (don't care) are:
  • $(dc = 00, ba = 01)$
  • $(dc = 00, ba = 11)$
  • $(dc = 01, ba = 10)$
  • $(dc = 10, ba = 01)$
  • $(dc = 10, ba = 11)$

Consider grouping the 1s:
Notice that cells $(dc=01, ba=00)=1$, $(dc=01, ba=10)=X$, $(dc=11, ba=00)=1$, and $(dc=11, ba=10)=1$ form a quad of 4 adjacent cells across rows $dc \in \{01, 11\}$ and columns $ba \in \{00, 10\}$.

In this quad:

• For rows $dc \in \{01, 11\}$, variable $c$ is constant at 1.
• For columns $ba \in \{00, 10\}$, variable $a$ is constant at 0 (i.e., $\overline{a}$).
  Hence, this quad corresponds to the product term:

This single term covers all three 1s in the K-map.

We are given that all inputs $(a, b, c, d)$ and their complements $(\overline{a}, \overline{b}, \overline{c}, \overline{d})$ are available. We need to implement $F = c \overline{a}$ using 2-input NOR gates only.
Recall that a 2-input NOR gate computes:

Setting $X = \overline{c}$ and $Y = a$:

Because $\overline{c}$ and $a$ are directly available, exactly 1 two-input NOR gate is sufficient to produce $F$.