## Invitation to Computer Science 8th Edition

Each 1 -CE circuit contains two NOT gates, two AND gates, and one OR gate. This will require $(2 \times 1)+(2 \times 3)+(1 \times 3)=2+6+3=11$ transistors. The 32 -bit compare-for-equality circuit of Figure 4.28 has the following components: 32 1-CE circuits $=32 \times 11=352$ transistors 31 AND gates $=31 \times 3=93$ transistors Total $=352+93=445$ transistors
Each 1 -CE circuit contains two NOT gates, two AND gates, and one OR gate. This will require $(2 \times 1)+(2 \times 3)+(1 \times 3)=2+6+3=11$ transistors. The 32 -bit compare-for-equality circuit of Figure 4.28 has the following components: 32 1-CE circuits $=32 \times 11=352$ transistors 31 AND gates $=31 \times 3=93$ transistors Total $=352+93=445$ transistors