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#Booth's algorithm can be used to multiple 2 signed binary numbers. It works by shifting and adding bits. This shifting and adding happens for  n(no.bits in multiplicand + no. bits in multiplier). Assuming that adding and shifting takes constant time, we can say that upper bound is O(n).

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BOOTH Algorithm

step 1:i=0 s=0|B    (upper n bit| lower n bit)

do{

step 2:is(s0)Sh =+A (sh=upper bit as in booth takes only in n bit ,after it gots ASR)

step 3:s=s/2;

i++

}while(i<=n)

 

 

for n bit=O(n)

but for 8 bit=O(1)

please correct me if i am wrong

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