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I'm missing something: > The downside to radix-8 multiplication is that multiplying by a number from 0 to 7 is much more complicated than multiplying by 0 or 1

I'm missing something: > The downside to radix-8 multiplication is that multiplying by a number from 0 to 7 is much more complicated than multiplying by 0 or 1, which is almost trivial. Fortunately, there are some shortcuts. Note that multiplying by 2 is the same as shifting the number to the left by 1 bit position, which is very easy in hardware—you wire each bit one position to the left. Similarly, to multiply by 4, shift the multiplicand two bit positions to the left. > Multiplying by 7 seems inconvenient, but there is a trick, known as Booth's multiplication algorithm. Instead of multiplying by 7, you add 8 times the number and subtract the number, ending up with 7 times the number. You might think this requires two steps, but the trick is to multiply by one more in the digit to the left, so you get the factor of 8 without an additional step. > Thus, you can get the ×7 by subtracting. Similarly, for a ×6 term, you can subtract a ×2 multiple and add ×8 in the next digit. > Thus, the only difficult multiple is ×3. (What about ×5? If you can compute ×3, you can subtract that from ×8 to get ×5.) If we can easily compute ×2 for the purposes of exploiting 6x = 8x - 2x, and we can easily compute ×4 for the purposes of exploiting 4x = 4x... why is it more difficult than that to compute 3x as the sum of 2x and 1x, or the difference of 4x and 1x? Also, if we can easily compute ×6 by any means, why not compute ×3 as a right shift of that value? That is admittedly an extra step, but the extra step is a shift.

kens 2y ago
For the 64-bit multiplication, you're adding 22 terms, one for each base-8 digit. (Think of grade-school multiplication.) Each term needs to be trivial to compute, so you can do a shift and/or a negation to get the term, but you can't do another addition. The point is that ×3 is precomputed once, so you can then simply feed it into any of the 22 terms as needed. You can't do ×2 and ×1 into a term to get ×3 because every term would need another adder. In other words, you want one circuit to compute ×3 rather than 22 circuits. For your ×6 question, this value is computed by putting negative ×2 into the term and conceptually adding one more to the next digit to get ×8. You can't right-shift this ×8 value since it's part of a completely different term. Hopefully this makes sense. There are a lot of digits and sums flying around.
- thaumasiotes 2y ago
  How do you do a negation without doing an addition? Can you do better than -x = ~x + 1?
  - kens 2y ago
    I think that the +1 is accomplished simply by setting carry-in on the adder for that term, so you don't need a separate adder. (Disclaimer: I haven't looked at that part of the circuit yet.) Another interesting question is how sign extension is implemented for a negative term. The problem is that you're adding 64-bit shifted terms to create a 128-bit result. When negating a 64-bit term, all the unused bits to the left need to be set to 1. But the chip doesn't do this. (Among other things, the adders would need to be much wider than 64 bits.) So it must be using a trick for the sign extension.
  - sifar 2y ago
    If you are negating a single term you can't. If you are adding n terms using an adder tree, the 1 feeds to the carry in of the next level ( it is simply a concatenation since the carry term is shifted left by 1). Only thw final carry propagate adder will have the add with 1.

kens 2y ago

For the 64-bit multiplication, you're adding 22 terms, one for each base-8 digit. (Think of grade-school multiplication.) Each term needs to be trivial to compute, so you can do a shift and/or a negation to get the term, but you can't do another addition. The point is that ×3 is precomputed once, so you can then simply feed it into any of the 22 terms as needed. You can't do ×2 and ×1 into a term to get ×3 because every term would need another adder. In other words, you want one circuit to compute ×3 rather than 22 circuits. For your ×6 question, this value is computed by putting negative ×2 into the term and conceptually adding one more to the next digit to get ×8. You can't right-shift this ×8 value since it's part of a completely different term. Hopefully this makes sense. There are a lot of digits and sums flying around.

thaumasiotes 2y ago
How do you do a negation without doing an addition? Can you do better than -x = ~x + 1?
- kens 2y ago
  I think that the +1 is accomplished simply by setting carry-in on the adder for that term, so you don't need a separate adder. (Disclaimer: I haven't looked at that part of the circuit yet.) Another interesting question is how sign extension is implemented for a negative term. The problem is that you're adding 64-bit shifted terms to create a 128-bit result. When negating a 64-bit term, all the unused bits to the left need to be set to 1. But the chip doesn't do this. (Among other things, the adders would need to be much wider than 64 bits.) So it must be using a trick for the sign extension.
- sifar 2y ago
  If you are negating a single term you can't. If you are adding n terms using an adder tree, the 1 feeds to the carry in of the next level ( it is simply a concatenation since the carry term is shifted left by 1). Only thw final carry propagate adder will have the add with 1.