GATE | GATE-CS-2003 | Question 11
Consider an array multiplier for multiplying two n bit numbers. If each gate in the circuit has a unit delay, the total delay of the multiplier is
(B) Θ(log n)
Explanation: Number of gates used in ‘n’ bit array multiplier (n * n) = (2n – 1)
Each gate in the circuit has a unit delay.
= 1 * (2n – 1) = O(2n – 1) = O(n)
Suppose 2 numbers A and B.
A = A0 A1 A2 A3 B = B0 B1 B2 B3
For multiplying A and B we need to do,
- Multiply A0 A1 A2 A3 with B1 need 1 AND gate
- Multiply A0 A1 A2 A3 with B2 need 1 AND gate
- Multiply A0 A1 A2 A3 with B3 need 1 AND gate
- Multiply A0 A1 A2 A3 with B4 need 1 AND gate
Also 3 OR gates are required to add terms got by AND gates.
So, total gates required = 4+3 = 7 that is 2N-1.
Time complexity = ϴ(n)
Option (C) is correct.
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