Booths multiplication algorithm extra bit at LSB

Question

I have a confusion about the booth's multiplication algorithm. suppose we want to multiply two binary numbers A ( 7 bits) and B (7 bits) . A= 00101011 (Multiplicand) B= 00001100 (Multiplier)

initial product =00000000000000

now in the next step, according to the algorithm, we make a product (14 bits) = product + multiplier(on the right half of the product) + we add an extra bit (0) at the LSB position.

for the above case: initial product =0000000000011000

My question is, why we need this extra bit and why it is assigned to 0?

Answer 1

That extra bit may be used to serve the purpose of temporarily changing the comparison of A.i and B.i into the equivalent of A.i and B.i-1. Primarily, the reason for using it is a way to safely invoke the requirement that bit Y_i be compared to bit Y_i-1. By adding it, we prevent the ambiguity of deciding whether bit -1 leaks into other memory.

In the binary system, a value represented by position 0 may represent either the presence or absence of 1 unit. If there is a position -1, it is assumed that its value will always be 0 because the absence of 0 units is exactly the same as the presence of 0 units.