Count the no of various Gates required in Carry Generator and look ahead adder

Question

Count the 2 Input And, OR and EX-OR gates required in Carry Generator and Look ahead adder.

Note :- In the question it is asking about 2 input, not multiple input,

for Multiple input AND gates in n bit carry generator we require (n(n+1))/2 and OR Gates n.

Comments

Question is about 2 Input And Gate

Answer 1

If we look at the expression we get an idea

C1 = G0 + P0C0

C2= G1 + G0P1 + P0P1C0

C3= G2 + P2G1 + P0P1P2C0 + G0P1P2

For C1 no. Of and gates= 1

For C2 = 1+2

For C3 = 1+2+3

For Cn = 1+2+3+4....+n

So total  = ∑r(r+1)/2  where r varies from 1 to n.

Or gates for C1 = 1

                For C2 = 2

                For C3 = 3

                For Cn = n

So total = n(n+1)/2

Answer 2

its n+((n+1)n)/2

Comments

And gate = sum of natural no

Or gate= no of bit adder