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Why we do right shift in booth algorithm?

I know the working of booths algorithm.

Suppose we have multiplicand M = 01011

and multiplier Q = 01110

We can write Q as (2^4 - 2^1).

So multiplication reduces to 2^4(M) + 2(-M)

Now booths algorithm rules are:-

If Q = 0 and Q(-1)=0 then do arithmetic right shift.

If Q = 1 and Q(-1)=0 then do A-M and arithmetic right shift.

If Q = 0 and Q(-1)=1 then do A+M and arithmetic right shift.

If Q = 1 and Q(-1)=1 then do arithmetic right shift.

Here A is initialized to 00000 and Q(-1) is initialized to 0.

If we see the algorithm then in every step we do right shifting. But as per the calculation shown above which is 2^4(M) + 2(-M) we multiply by 16 and 2 which requires left shift.

So how is booths algorithm working with right shift ?

in CO and Architecture by Junior (853 points) | 112 views

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