GATE CSE 1999 | Question: 2.16

Question

The number of full and half-adders required to add $16$-bit numbers is

• A. $8$ half-adders, $8$ full-adders

• B. $1$ half-adder, $15$ full-adders

• C. $16$ half-adders, $0$ full-adders

• D. $4$ half-adders, $12$ full-adders

Comments

@Mohnish option is c is incorrect because HA doesn’t take input carry which we need in this case else the answer of addition come out to be wrong. 

So either use 16 FA or use 1HA+15FA.

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Before getting to the solution the question is about implementing the 16 bit number using both full and half adders as combinations that is in the same implementation we will use both full and half adders

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is this equation  generalization or it can work only for few variations?

Answer 1

• For 2-bit Addition, we need 1 Half Adder.
• For 3-bit Addition, we need 1 Full Adder.

So, Total 15-Full Adders and 1-Half Adder are needed .

Correct Ans: (B).

Answer 2

To realize this

We have to realize first the intitution.

Two n bit numbers are given and to add them we need “n Full Adders”

Now In this “n” Full adders The First adder (left Most Full adder) have the cin = 0

which is a loss of a wire and moreover it is Unneseccary. therefore omiting that with a Half adder doesn’t impact .

There fore 1 Half adder and (n-1) Full adders are needed.