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Adding short description: "Algorithm that multiplies two signed binary numbers in two's complement notation" (Shortdesc helper) |
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Booth's algorithm examines adjacent pairs of [[bit]]s of the 'N'-bit multiplier ''Y'' in signed [[two's complement]] representation, including an implicit bit below the [[least significant bit]], ''y''<sub>−1</sub> = 0. For each bit ''y''<sub>''i''</sub>, for ''i'' running from 0 to ''N'' − 1, the bits ''y''<sub>''i''</sub> and ''y''<sub>''i''−1</sub> are considered. Where these two bits are equal, the product accumulator ''P'' is left unchanged. Where ''y''<sub>''i''</sub> = 0 and ''y''<sub>''i''−1</sub> = 1, the multiplicand times 2<sup>''i''</sup> is added to ''P''; and where ''y''<sub>i</sub> = 1 and ''y''<sub>i−1</sub> = 0, the multiplicand times 2<sup>''i''</sup> is subtracted from ''P''. The final value of ''P'' is the signed product.
The representations of the multiplicand and product are not specified; typically, these are both also in two's complement representation, like the multiplier, but any number system that supports addition and subtraction will work as well. As stated here, the order of the steps is not determined. Typically, it proceeds from [[Least significant bit|LSB]] to [[Most significant bit|MSB]], starting at ''i'' = 0; the multiplication by 2<sup>''i''</sup> is then typically replaced by incremental shifting of the ''P'' accumulator to the right between steps; low bits can be shifted out, and subsequent additions and subtractions can then be done just on the highest ''N'' bits of ''P''.<ref name="Chen_1992"/> There are many variations and optimizations on these details
The algorithm is often described as converting strings of 1s in the multiplier to a high-order +1 and a low-order −1 at the ends of the string. When a string runs through the MSB, there is no high-order +1, and the net effect is interpretation as a negative of the appropriate value.
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