Booth's multiplication algorithm

from 

http://en.wikipedia.org/wiki/Booth%27s_multiplication_algorithm 

How it works

Consider a positive multiplier consisting of a block of 1s surrounded by 0s. For example, 00111110. The product is given by :

M	imes \,^{{prime prime }}0;0;1;1;1;1;1;0\,^{{prime prime }}=M	imes (2^{5}+2^{4}+2^{3}+2^{2}+2^{1})=M	imes 62

where M is the multiplicand. The number of operations can be reduced to two by rewriting the same as

M	imes \,^{{prime prime }}0;1;0;0;0;0{mbox{-1}};0\,^{{prime prime }}=M	imes (2^{6}-2^{1})=M	imes 62.

In fact, it can be shown that any sequence of 1's in a binary number can be broken into the difference of two binary numbers:

(ldots 0overbrace {1ldots 1}^{{n}}0ldots )_{{2}}equiv (ldots 1overbrace {0ldots 0}^{{n}}0ldots )_{{2}}-(ldots 0overbrace {0ldots 1}^{{n}}0ldots )_{2}.

Hence, we can actually replace the multiplication by the string of ones in the original number by simpler operations, adding the multiplier, shifting the partial product thus formed by appropriate places, and then finally subtracting the multiplier. It is making use of the fact that we do not have to do anything but shift while we are dealing with 0s in a binary multiplier, and is similar to using the mathematical property that 99 = 100 − 1 while multiplying by 99.

This scheme can be extended to any number of blocks of 1s in a multiplier (including the case of single 1 in a block). Thus,

M	imes \,^{{prime prime }}0;0;1;1;1;0;1;0\,^{{prime prime }}=M	imes (2^{5}+2^{4}+2^{3}+2^{1})=M	imes 58
M	imes \,^{{prime prime }}0;1;0;0{mbox{-1}};1{mbox{-1}};0\,^{{prime prime }}=M	imes (2^{6}-2^{3}+2^{2}-2^{1})=M	imes 58.

Booth's algorithm follows this scheme by performing an addition when it encounters the first digit of a block of ones (0 1) and a subtraction when it encounters the end of the block (1 0). This works for a negative multiplier as well. When the ones in a multiplier are grouped into long blocks, Booth's algorithm performs fewer additions and subtractions than the normal multiplication algorithm.

原文地址:https://www.cnblogs.com/testset/p/3550789.html