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I think it will be 7 And gates and 7 or gates.

Please Correct me if I'm wrong
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Carry Generator Gi=Ai.Bi and  Carry Propogater Pi=Ai xor Bi 

lets take an example with 4 bits to make it clear

G0=A0.B0  and  P0= A0 xor B0

G1=A1.B1  and  P1= A1 xor B1

G2=A2.B2  and  P2= A2 xor B2

G3=A3.B3  and  P3= A3 xor B3

we know that   C0 is initial carry

                      C1=C0.P0 + G0  ------(a)

                      C2=C1.P1 + G1  -------(b)

                      C3=C2.P2 + G2   -------(c)

                      C4=C3.P3 + G3 ---------(d)

substitute (a) in (b)

C2=C0.P0.P1+G0.P1+G1 ----- (e)

and (e) in (c), and so on till you get the C4 i.e.,

 

C4=C0.P0.P1.P2.P3+G0.P1.P2.P3 + G1.P2.P3 + G2.P3 + G3

 

if you clearly observe C4= (4 AND gates) + (3 AND gates) + (2 AND gates) + ( 1 AND gate) + ( 0 GATES)

 

it is 4+3+2+1=10. In general for n-bit carry look-ahead adder, n(n+1)/2 AND gates are needed.

We can easily say from the C4 that  only n no.of OR gates are required.

so Answer is 7*8/2=28 AND gates and 7 OR gates.

 

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