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Comment by stromgo | Hacker News Reader

How do you correctly round, say, 6172293634027511 * 7059478094414279 / 2^53 = 4837593847918366.50000000000000011102 without essentially computing the full 106-bit product?

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The idea is that you can use guard digits instead of full products and get correct rounding (as if to infinite precision). For IEEE-754 math you need 3 guard digits, typically named Guard, Round, and Sticky. The trick at the hardware level is that Sticky represents the OR of a bunch of bits. Instead of keeping the full 106 bit partial product at each step you only need to keep the 53+3 bit product. As you shift+add the partial products you use special rules to keep track of the Sticky bit and you will still get the right answer.

I'm aware of the 53+3 bit algorithm for addition and subtraction, but I don't see how this could possibly work for multiplication. Could you please point me to a paper that explains how to compute a correctly rounded 53-bit product without computing the full 106-bit product?

I can't find a handy reference, but here is a 5x5 multiplication:

     1XXXX
   * 1XXXX
 _________
     XXXXX
    XXXXX
   XXXXX
  XXXXX
 1XXXX

Using a sticky bit means computing the logical-or of all the bits that are sufficiently far to the right in all these partial results, instead of computing them individually and summing them.

Like for addition, the idea is that the result you obtain is different from the exact result, but rounds the same. The sticky bit only serves to round (midpoint + a small quantity) correctly up.

Let's say that the full product in your example is

  1XXXXYZZZZ

To round it we need:

(1) the X bits for our answer

(2) the Y bit for rounding

(3) the logical-or of the Z bits for rounding

Ok so there might be a trick to compute (3) without computing ZZZZ, but we still want to know how to compute 1XXXXY without computing ZZZZ.

Perhaps a better way to describe it is that during the additions, the accumulator only ever needs to be shifted right in order to be aligned with the next number to add, never left. So the accumulator can be 53+3-bit wide and carry at each step the 53+3-bit representation of the partial sum already computed, and this is enough to produce the correctly rounded result after all numbers have been added.

This explanation assumes that the N partial products are added sequentially. I don't know what techniques are used to achieve the 4- or 5-cycle latencies of modern desktop processors. Implementations with more parallelism and/or that are part of an FMA unit have it harder, but I still expect that they don't naively compute a full width partial result before rounding it (in the case of the FMA, it would be impractical as the partial result could be thousands of bits wide)

Looks like you're actually computing the full product. The only optimization you're doing is forgetting the low order bits as soon as you're done with them. The author is asking to keep them, which sounds reasonable. You initially claimed they were "never computed in the first place", and we're asking you to back up that claim if you still think it's true.

I see what you mean now. I agree that each of the low-order bits of the 106-bit product is available at some point. Not necessarily simultaneously, but yes, it seems like little additional effort for processor designers to keep these somewhere and assemble them in another register or something.

For some reason (maximum depth reached?) there is no “reply” link at the bottom of the post of yours at which I understood, so I am replying here.

IEEE Floating point rounds numbers in binary, not in decimal. So it is trivial by simply having a 54-bit register to hold the 53-bit mantissa.

So the answer to your question is... Floating Point numbers "don't round correctly". They're actually quite complicated, and require careful study.

My example has nothing to do with decimal rounding. It's an example of a product (6172293634027511 * 7059478094414279) whose result is extremely close to the fence between two representable numbers (4837593847918366.5 * 2^53). If the 106-bit product happened to be 2 units smaller, then the rounded result would change.

As stated before, Floating Point numbers "don't round correctly" according to IEEE Floating Point arithmetic.

There are very simple rules on how to round. This makes calculation of errors a bit complicated. http://en.wikipedia.org/wiki/IEEE_floating_point#Rounding_ru...

At best, Intel at one point provided 80-bit rounding (ie: 68-bit mantissa). That is, if you use the x87 floating-point coprocessor. But these technically do not match IEEE specifications for rounding.

Er, what? Floating point numbers do "round correctly" (in binary) -- that's pretty much the whole point of the IEEE standard.

In particular, to compute a product of two 53-bit floating-point numbers with correct rounding (as the standard mandates), it is certainly not sufficient in general to compute just a 54-bit approximation and round that.

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