Modified Baugh-Wooley乘法算法：取反项与额外'1'的技术疑问

Great questions about the Modified Baugh-Wooley (MBW) multiplier—this is a clever design for signed two's complement multiplication, so let's break down your questions clearly:

1. Why use inverted product terms like !(A0*B5) instead of direct products? #

The short answer: this is how MBW handles the negative contributions from sign-bit cross products in two's complement arithmetic.

Let's unpack that. For a 6-bit signed multiplier (like your example with A0-A5 and B0-B5), the highest bits (A5, B5) are sign bits. Any product term involving a sign bit and an opposite operand's lower bit (e.g., A5B0, A0B5) represents a negative value in two's complement.

If we tried to use the direct product (e.g., A0*B5) in our adder tree, we'd have to handle subtraction alongside addition, which complicates the hardware. Instead, MBW uses the inverted product of these negative terms. Here's why:

• In two's complement, a negative value -x can be written as ~x + 1 (bitwise NOT plus 1).
• By using !(A0*B5) (the bitwise NOT of the product), we're representing the "~x" part of the negative value. Later, we'll add the necessary +1 corrections (which ties into your second question).
• This lets us use a uniform carry-ripple or carry-lookahead adder tree for all terms—no separate subtractors needed, which saves hardware and simplifies the design.

Note that the product of the two sign bits (A5*B5) doesn't get inverted, because negative * negative = positive, so that term is a positive value and fits directly into the adder tree.

2. Why are there two extra '1's in the algorithm? #

Those two '1's are the correction terms needed to fix the inverted negative product terms we talked about.

Remember, each inverted product term !(A_i*B5) (for i=0-4) or !(A5*B_j) (for j=0-4) represents ~(A_i*B5) which equals -(A_i*B5) - 1. That means each of these inverted terms is off by -1 compared to the actual negative value we need.

Instead of adding a +1 for each of the 10 inverted terms (which would be inefficient), MBW combines these corrections into two single +1 terms. Here's how it works:

• The 5 inverted terms from A5*B0 to A5*B4 and the 5 from A0*B5 to A4*B5 each introduce a collective offset that can be fixed with one +1 per group.
• These two '1's are placed at specific bit positions in the adder chain to correct the two's complement conversion for both sets of sign-bit cross products. Without these corrections, the final product would be consistently offset, leading to wrong results for signed numbers.

内容的提问来源于stack exchange，提问作者kevin998x