[curves] The Pareto frontiers of sleeveless primes
Ben Harris
mail at bharr.is
Tue Oct 28 19:01:08 PDT 2014
I just quickly played around with this too. Looking at 96 potential
Crandalls (2^n - c, 80 <= n <= 1024, c < 32, no prime for smaller c),
looking at those that aren't dominated for size, 32-bit length, 28-bit
limbs or 58-bit limbs gets 60 potentials. Ignoring the 28-bit kills 9 more.
prime mod 4 28 32 58
2^95 - 15 1 4 3 2
2^96 - 17 3 4 3 2
2^109 - 31 1 4 4 2
2^110 - 21 3 4 4 2
2^116 - 3 1 5 4 2
2^122 - 3 1 5 4 3
2^127 - 1 3 5 4 3
2^137 - 13 3 5 5 3
2^140 - 27 1 5 5 3
2^152 - 17 3 6 5 3
2^158 - 15 1 6 5 3
2^166 - 5 3 6 6 3
2^174 - 3 1 7 6 3
2^189 - 25 3 7 6 4
2^191 - 19 1 7 6 4
2^196 - 15 1 7 7 4
2^206 - 5 3 8 7 4
2^221 - 3 1 8 7 4
2^226 - 5 3 9 8 4
2^230 - 27 1 9 8 4
2^235 - 15 1 9 8 5
2^251 - 9 3 9 8 5
2^255 - 19 1 10 8 5
2^266 - 3 1 10 9 5
2^285 - 9 3 11 9 5
2^291 - 19 1 11 10 6
2^321 - 9 3 12 11 6
2^336 - 3 1 12 11 6
2^338 - 15 1 13 11 6
2^369 - 25 3 14 12 7
2^383 - 31 1 14 12 7
2^389 - 21 3 14 13 7
2^401 - 31 1 15 13 7
2^414 - 17 3 15 13 8
2^444 - 17 3 16 14 8
2^452 - 3 1 17 15 8
2^468 - 17 3 17 15 9
2^489 - 21 3 18 16 9
2^495 - 31 1 18 16 9
2^521 - 1 3 19 17 9
2^529 - 31 1 19 17 10
2^537 - 9 3 20 17 10
2^546 - 11 1 20 18 10
2^550 - 5 3 20 18 10
2^563 - 9 3 21 18 10
2^583 - 27 1 21 19 11
2^607 - 1 3 22 19 11
2^610 - 27 1 22 20 11
2^620 - 15 1 23 20 11
2^664 - 17 3 24 21 12
2^694 - 3 1 25 22 12
2^699 - 9 3 25 22 13
2^715 - 7 1 26 23 13
2^717 - 25 3 26 23 13
2^729 - 9 3 27 23 13
2^810 - 5 3 29 26 14
2^848 - 17 3 31 27 15
2^850 - 3 1 31 27 15
2^869 - 21 3 32 28 15
2^923 - 31 1 33 29 16
On 27 October 2014 17:17, Michael Hamburg <mike at shiftleft.org> wrote:
>
> On Oct 26, 2014, at 11:57 PM, Mike Hamburg <mike at shiftleft.org> wrote:
>
>
> Right. In my try, I had calculated it by multiplication not requiring
> internal carry propagation, which depends on c as well as nail length.
> This can be computed by expanding the prime into polynomial P in the radix,
> and comparing the largest coefficient of ((x^limbs - 1) / (x-1))^2 mod P to
> 2^(2*wordsize - 2*radix - extra). Here extra is some small amount (0.1) to
> account for not having reduced perfectly the first time; + 1 if the
> polynomial is signed;
>
>
> +1 if the polynomial is signed isn’t quite right actually. It should be
> something more like, always treat the non-leading coefficients of the
> polynomial as negative, so that when computing the reduction they always
> add to each other rather than canceling.
>
> — Mike
>
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