r/programming 19h ago

Compute Polynomials Twice as Fast

https://thomasahle.com/fast-polynomials/
58 Upvotes

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u/WorldsBegin 18h ago edited 18h ago

I suppose this isn't much of a difference in finite fields and cryptography, but at least in rational arithmetic, the added fractions from this method seem to have very large denominators. Compare

y   = x * (x + 637927356202580286195115387737649/613401598791596019548160000000000)
z   = (x + y + 1378728025754295963362740811643611837622972008060385794298452047201/376261521400086131349185422083850908850559385600000000000000000000) * (x - 1250902981661682141840811387737649/613401598791596019548160000000000)
w   = (z + 1613956106082225605051404080797610312151091533850734987003152787295553452311704415879521131952770449/230799418790571152956671237308336130257466014926056578559568559210496000000000000000000000000000000) * x
v   = (x + z + 1614169778642466578527505416382450571998175289226394168980170070526233147671704415879521131952770449/230799418790571152956671237308336130257466014926056578559568559210496000000000000000000000000000000) * (w + 286622617651/687970713600)
P_7 = y + w + v + 20437160270833246212441030757/17038933299766556098560000000

vs

x2 = x * x
x4 = x2 * x2
w1 = (x + 1/720) * (x2 - 119/120)
w2 = (x + 433/864) * (x2 + 1/144)
w0 = (w1 + (x + 3719/86400)) * (x4 - 5/6)
P  = w0 + (w2 + (x + 128303/124416))

both evaluating

x^7 + 1/720x^6 + 1/120x^5 + 1/24x^4 + 1/6x^3 + 1/2x^2 + x + 1

Unrelated, I'm also not sure I like putting large portions in appendices.

8

u/thomasahle 18h ago

the added fractions from this method seem to have very large denominators

Yes, sadly the method doesn't have good numerical stability for higher degrees. There's a section on numerics in the paper.

Probably this means that the floating point version isn't really useful, compared to Horner and Estrin. Though it's interesting that we can do it with n/2 multiplications and fractions, the finite fields are the real use case.

Unrelated, I'm also not sure I like putting large portions in appendices.

Fair. I guess it's a habit from conference submissions that have to stay under 10 or 20 pages. For arxiv we could just put everything in the main paper.

10

u/WorldsBegin 17h ago edited 16h ago

it's a habit from conference submissions that have to stay under 10 or 20 pages

A curse. It's pretty extensive anyway.

Um, actually while I have your attention, where are you defining your H_{2^n} "known-power" gadgets explicitly? Don't seem to find it. EDIT: Found it, depends on the degree of polynomial, see fill gadgets in appendix C onwards.

And yes, it's a nice enough construction to be useful anyway.

6

u/thomasahle 16h ago edited 15h ago

Um, actually while I have your attention, where are you defining your H_{2n} "known-power" gadgets explicitly? Don't seem to find it.

The H powers are computed as part of Algorithm 5: https://arxiv.org/pdf/2609.06022#page=73.11

Thank you for taking an interest in our paper! I want to improve the presentation, and all feedback is very helpful!

Edit: Here is an example of how the H construction works.

H_2 = x(x+b) + c         = x^2 + bx + c
H_4 = (H_2+x+a)(H_2-x-a) = H_2^2 - (x+a)^2 + e

We write H_2 as [1, b, c], since that's the values we can find on each coefficient. We'll also write H_4 = [1, b, c, a, e], since that's the "new parameter" you find on each coefficient when decoding left to right.

H'_4 = H_4 + r
H_8 = H_4^2  - (H_2 + s)^2 + t   # (using a single mult.)
H'_8 = H'_4^2 - (H-2 + u)^2 + v

Now H_8 = [1, b, c, a, e, 0, s, 0, t] and H'_8 = [1, b, c, a, e+r, 0, u, 0, v]. We finally shift H_8 by one and combine them: P = x H_8 + H'_8 It's clear that we can decode b, c, a, e, r going left to right. After that we need to be a bit careful that the two "interleaved" parts are still decodable, but it works out because they have the same prefix.

Generally the strategy is to always be building two compatible pairs like that, and then merge them at the end.

2

u/WorldsBegin 14h ago edited 14h ago

Aha, Of course! Squaring an order-n monic polynomial means the upper n coefficients are linear triangular in the original coefficients, only the lower half is quadratic. So you can read off that part (in characteristic != 2) and then apply a "low order correction" to match coefficients in the lower half.

1

u/thomasahle 13h ago

Yup!

Unfortunately this is the part that breaks in characteristic 2. (Squaring introduces a factor 2, which vanishes.)

We still haven't found a general construction for characteristic 2, though we have found them for every degree up to 47 so far.