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Re: [PrimeNumbers] QS multipliers

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  • Phil Carmody
    ... Squaresome (i.e. not squarefree) multipliers don t need tobe tried, as a^2.b ... It seems most people use the Knuth multiplier. A quick root around here
    Message 1 of 4 , Feb 3, 2003
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      --- D�cio Luiz Gazzoni Filho <decio@...> wrote:
      > -----BEGIN PGP SIGNED MESSAGE-----
      > Hash: SHA1
      >
      > Hello,
      >
      > I figured out about multiplying the number n to be factored in QS by a small
      > number m to have mn be a quadratic residues modulo many small primes. I was
      > trying to code something that would find out this multiplier automatically.
      > In order to do that, I test quadratic residuosity of mn (for many small m)
      > modulo primes up to a small bound, say 100. I came up with a performance
      > metric, which is: set a = 0, and if mn is a quadratic residue modulo p, add
      > 1/p to a. Then look for the highest value of a in all multipliers tested. The
      > reasoning here is that smaller numbers contribute the most. But using that
      > metric, here's what I obtained for RSA-129 (for m <= 30):

      Squaresome (i.e. not squarefree) multipliers don't need tobe tried, as a^2.b
      will give you the same result as b, as the following show:

      > 6 1.557914324605730750503938962
      > 24 1.557914324605730750503938962

      > 5 1.497243896414790462459092847
      > 20 1.497243896414790462459092847

      > 8 1.190858309479667878600839507
      > 18 1.190858309479667878600839507
      > 2 1.190858309479667878600839507

      > 3 1.129039847090011466380876713
      > 27 1.129039847090011466380876713
      > 12 1.129039847090011466380876713

      > 7 1.076872592433713518032436873
      > 28 1.076872592433713518032436873

      > 4 0.9818042063484476677777650724
      > 16 0.9818042063484476677777650724
      > 25 0.9818042063484476677777650724


      > And here are the quadratic residues modulo 21, 6 and 5:
      >
      > prime n 21n
      > Quadratic residues: 12 19
      > Quadratic non-residues: 13 6
      >
      > prime n 6n
      > Quadratic residues: 12 15
      > Quadratic non-residues: 13 10
      >
      > prime n 5n
      > Quadratic residues: 12 15
      > Quadratic non-residues: 13 10
      >
      > But actually, the multiplier chosen by the team that factored RSA-129 was 5.
      > So I assume my performance metric is wrong. Is there a better way to measure
      > the effect of multipliers on QS?

      It seems most people use the 'Knuth' multiplier.
      A quick root around here indicates that the weight is 2*log(p)/(p-1) or
      log(p)/p probably depending on whether there's one or two roots, not 1/p.
      That's more or less what I'd expect, because larger primes do a proportionally
      (~log) better job of shrinking the cofactor. The (p-1) I can't explain off the
      top of my head though.

      > Also, 21 is clearly better than 5, but 21 adds about 2 bits to the number
      > being factored compared to 5. In these cases, how do I measure whether the
      > additional 2 bits are worth the increase in quadratic residues?

      Numbers that are ~n^(1/2) will have 1 more bit. The largest factor of
      such numbers will have .6 more of a bit. Therefore they will be 52% bigger.
      If you were to extend your factor base then you'd give yourself a much
      larger LA step, so you've got to take the hit primarily in sieving.

      I just did some fag-packetty calculations, and I came to the conclusion that
      the increase in sieving is fractional. I'd like to see someone else come up
      with the same figure though.

      Phil


      =====
      Polish inventions to the world cuisine include the Hamburger, (originally
      named Hzczambrzurszyngerschandwicz), which is in its traditional form a
      meat coupon in between two bread coupons.
      http://softavenue.fi/u/henry.w/poland/ parody of http://jpzr.com/finland/

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    • Paul Leyland
      In the immortal phrase: I think you ll find it in Knuth. In particular, Section 4.5.4 (p 383 in the 2nd edition) and Ex. 28 in that section. The analysis
      Message 2 of 4 , Feb 3, 2003
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        In the immortal phrase: "I think you'll find it in Knuth." In particular, Section 4.5.4 (p 383 in the 2nd edition) and Ex. 28 in that section.

        The analysis there is described in terms of optimizing CFRAC but the problem and its solution are exactly the same for QS and its variants.


        Paul

        > -----Original Message-----
        > From: Phil Carmody [mailto:thefatphil@...]
        > Sent: 03 February 2003 10:59
        > To: primenumbers
        > Subject: Re: [PrimeNumbers] QS multipliers
        >
        >
        > --- Décio Luiz Gazzoni Filho <decio@...> wrote:
        > > -----BEGIN PGP SIGNED MESSAGE-----
        > > Hash: SHA1
        > >
        > > Hello,
        > >
        > > I figured out about multiplying the number n to be factored
        > in QS by a small
        > > number m to have mn be a quadratic residues modulo many
        > small primes. I was
        > > trying to code something that would find out this
        > multiplier automatically.
        > > In order to do that, I test quadratic residuosity of mn
        > (for many small m)
        > > modulo primes up to a small bound, say 100. I came up with
        > a performance
        > > metric, which is: set a = 0, and if mn is a quadratic
        > residue modulo p, add
        > > 1/p to a. Then look for the highest value of a in all
        > multipliers tested. The
        > > reasoning here is that smaller numbers contribute the most.
        > But using that
        > > metric, here's what I obtained for RSA-129 (for m <= 30):
        >
        > Squaresome (i.e. not squarefree) multipliers don't need tobe
        > tried, as a^2.b
        > will give you the same result as b, as the following show:
        >
        > > 6 1.557914324605730750503938962
        > > 24 1.557914324605730750503938962
        >
        > > 5 1.497243896414790462459092847
        > > 20 1.497243896414790462459092847
        >
        > > 8 1.190858309479667878600839507
        > > 18 1.190858309479667878600839507
        > > 2 1.190858309479667878600839507
        >
        > > 3 1.129039847090011466380876713
        > > 27 1.129039847090011466380876713
        > > 12 1.129039847090011466380876713
        >
        > > 7 1.076872592433713518032436873
        > > 28 1.076872592433713518032436873
        >
        > > 4 0.9818042063484476677777650724
        > > 16 0.9818042063484476677777650724
        > > 25 0.9818042063484476677777650724
        >
        >
        > > And here are the quadratic residues modulo 21, 6 and 5:
        > >
        > > prime n 21n
        > > Quadratic residues: 12 19
        > > Quadratic non-residues: 13 6
        > >
        > > prime n 6n
        > > Quadratic residues: 12 15
        > > Quadratic non-residues: 13 10
        > >
        > > prime n 5n
        > > Quadratic residues: 12 15
        > > Quadratic non-residues: 13 10
        > >
        > > But actually, the multiplier chosen by the team that
        > factored RSA-129 was 5.
        > > So I assume my performance metric is wrong. Is there a
        > better way to measure
        > > the effect of multipliers on QS?
        >
        > It seems most people use the 'Knuth' multiplier.
        > A quick root around here indicates that the weight is
        > 2*log(p)/(p-1) or
        > log(p)/p probably depending on whether there's one or two
        > roots, not 1/p.
        > That's more or less what I'd expect, because larger primes do
        > a proportionally
        > (~log) better job of shrinking the cofactor. The (p-1) I
        > can't explain off the
        > top of my head though.
        >
        > > Also, 21 is clearly better than 5, but 21 adds about 2 bits
        > to the number
        > > being factored compared to 5. In these cases, how do I
        > measure whether the
        > > additional 2 bits are worth the increase in quadratic residues?
        >
        > Numbers that are ~n^(1/2) will have 1 more bit. The largest factor of
        > such numbers will have .6 more of a bit. Therefore they will
        > be 52% bigger.
        > If you were to extend your factor base then you'd give yourself a much
        > larger LA step, so you've got to take the hit primarily in sieving.
        >
        > I just did some fag-packetty calculations, and I came to the
        > conclusion that
        > the increase in sieving is fractional. I'd like to see
        > someone else come up
        > with the same figure though.
        >
        > Phil
        >
        >
        > =====
        > Polish inventions to the world cuisine include the Hamburger,
        > (originally
        > named Hzczambrzurszyngerschandwicz), which is in its
        > traditional form a
        > meat coupon in between two bread coupons.
        > http://softavenue.fi/u/henry.w/poland/ parody of
        http://jpzr.com/finland/

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      • Phil Carmody
        ... Yuppers. I guessed it might be, even though I was aware of the absence of QS. Edition 3 has CFRAC on 397-398, but the explanation of the Schroeppel
        Message 3 of 4 , Feb 3, 2003
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          --- Paul Leyland <pleyland@...> wrote:
          > In the immortal phrase: "I think you'll find it in Knuth." In particular, Section 4.5.4 (p 383
          > in the 2nd edition) and Ex. 28 in that section.
          >
          > The analysis there is described in terms of optimizing CFRAC but the problem and its solution
          > are exactly the same for QS and its variants.

          Yuppers. I guessed it might be, even though I was aware of the absence of QS.

          Edition 3 has CFRAC on 397-398, but the explanation of the Schroeppel
          multiplier on 400. However, the magical function is left as an exercise
          (Exercise 28).

          Fortunately it's both instructive and easy to write a simply GP script that
          will print out counts of divisors for each residue for each small prime.
          I've done this for my Fermat-with-sieves-alike, and it was a _real_
          eye-opener. (I even found a way to visualise themin 2D, and some very
          interesting (but obvious in retrospect) patterns emerged. I don't remember
          the results, but they were trivial to reproduce so I'm not worried about
          that. (And who needs GP - I did those in Perl!). I'd hope that the (p-1)
          in the formula will drop out immediately from actually counting them!

          Phil


          =====
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