#self doubt

Question

A full adder can be implemented with half adder and OR gate.

A 4-bit full adder without any initially carry require?

a) 8 half adder 4 OR gate      b) 7 half adder 4 OR gate

c) 8 half adder 3 OR gate       d) 7 half adder 3 OR gate

Comments

check the comments https://gateoverflow.in/230970/work-book

if you get it, then close your question

Answer 1

Is it D

FA .  FA .    FA .     HA

For 1 full adder using HA 2 Half adder required and 1 OR gate

So 2+2+2+1=7 half adder

And 1+1+1 =3 OR gate

Comments

For initial we use direct half adder because no carry

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yes

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but there is direct formula for (2n-1)H.A +(n-1) OR gates .

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this is simple so no need to remember formula because in next question my change something