can someone explain this SSE BigNum comparison?

Question

If you look at this answer, the author manages to create a compact comparison algorithm for 2 integer bignums, stored in 2 SSE registers. I am not following it too well :)

What I did so far:

if l = a < b = {a[i] < b[i] ? ~0 : 0} and

e = a == b = {a[i] == b[i] ? ~0 : 0}

then a < b == l[3] v e[3]l[2] v e[3]e[2]l[1] v e[3]e[2]e[1]l[0]

But this does not seem to be what the author is doing. What am I missing? What need is there for a greater than comparison?

He uses greater-than comparison because it's the only integer compare that's available for SSE. — Mysticial, Jan 13 '15 at 19:15
But otherwise he calculates less than according to the same formula I use? — user1095108, Jan 13 '15 at 19:17
@Mysticial Ahhh, I see, nasty trickery and specific to 64-bit bignums. — user1095108, Jan 13 '15 at 19:42

user1095108 · Answer 1 · 2015-01-13T21:36:53.347

0

I've overlooked than the answer was not generic, but limited to 64-bit bignums, composed out of 32-bit elements. If you have 2 64-bit vectors a = {a0, a1}, b = {b0, b1}, then the program calculates:

a < b = ((a1 < b1) | (a0 < b0)) & ~(a1 > b1)

In my question I was aiming at arbitrarily long BigNums, implemented with SSE/AVX registers.

edited Jan 13 '15 at 21:36

answered Jan 13 '15 at 20:16

user1095108

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can someone explain this SSE BigNum comparison?

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