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Re: [PrimeNumbers] Re: Infallible isprime(p) routine for 0<2^32

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  • Phil Carmody
    ... How many yield square roots of -1 that aren t the same (or additive inverses) from the two SPRP tests?  ... Once you ve got a SPRP, on average each new
    Message 1 of 37 , Mar 11, 2013
      --- On Mon, 3/11/13, WarrenS <warren.wds@...> wrote:
      > If it is your goal to choose X so
      > there are few spsp{2, X},
      > then X=735 does pretty well.
      > There are 57 spsp{2, 735} below 2^32
      > and   1149508 spsp{2, 735} below 2^64. 

      How many yield square roots of -1 that aren't the same (or additive inverses) from the two SPRP tests? 

      > The latter is substantially better than Broadhurst's
      > X=858945; there are 1311216 spsp{2,858945} below 2^64.
      > I make no claim 735 is best, it is merely comparatively
      > good.
      > It would probably be possible to devise an infallible
      > isprime test for
      > <=64-bit integers by performing spsp(2) and spsp(735)
      > tests, and then hashing
      > the 1149508 failures into 8192 bins (each bin representing
      > about 140; need to
      > devise the hash function to get good equidistribution)
      > and then the content of that table entry tells you which
      > base B to use for a final spsp(B)
      > test.  (B chosen to kill all fakes in its hash-table
      > bin.)

      Once you've got a SPRP, on average each new SPRP test removes only a quarter of the fakes. Culling 140/140 with one test seems unlikely. I'm not sure if that quarter includes the square (and higher) root hack.

      > This primality test, in net, would involve three spsp tests
      > and one hash-table lookup in a table with 8192 entries; you
      > could probably compute the table in a few days.

      The $620 test (not $640) can already be performed with a cost of 3 SPRP tests, so this wouldn't be an improvement over what's already done.

      > I am in fact working on an infallible 64-bit isprime test,
      > but not using the method I just outlined,instead based on
      > combining an spsp(2) test with a Lehmer test.
      > I'll report on that later, if I succeed.

      In the kind of science I like, reports of failure are useful too. Knowing what alleyways are blind is useful, but also someone else might have a final insight to patch things up.

      () ASCII ribbon campaign () Hopeless ribbon campaign
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      [stolen with permission from Daniel B. Cristofani]
    • David Cleaver
      ... Apologies, my Pari/GP script-fu is definitely not on par with DJB s. I just wanted to provide a script so that people could plug and chug an r,p pair to
      Message 37 of 37 , Mar 14, 2013
        On 3/13/2013 5:47 AM, Phil Carmody wrote:
        > No I couldn't. That's so overly verbose and redundant it makes me twitch, I can
        > barely bring myself to repeat it!
        > "lift(Mod(p*s, lift(znorder(Mod(2,s))))) == 1" is just
        > "p*s % znorder(Mod(2,s)) == 1"
        > Having 3 exit conditions to the loop is overkill too.

        Apologies, my Pari/GP script-fu is definitely not on par with DJB's. I just
        wanted to provide a script so that people could plug and chug an r,p pair to see
        what psp's would be generated. Also, I didn't know that the % (mod) operator
        still worked in Pari. I thought everything had to be done with Mod(). Thanks
        for that insight.

        > The latter worries me a bit, as it might imply wasted effort. I'm trying to
        > picture how these duplicates arise. Given a n, the maximal prime factor p|s is
        > uniquely defined, and r as order_2(p) is uniquely defined. Therefore n can only
        > appear with pair (r,p)?

        And apologies here too, I mis-remembered a statement from his Category S page
        and mis-spoke by applying it to the Category E psp's.

        On 3/14/2013 11:02 AM, WarrenS wrote:
        > 2. Consulting the Cunningham project pages,
        > http://homes.cerias.purdue.edu/~ssw/cun/index.html
        > every Mersenne-form number 2^r - 1 now is fully factored if
        > r<929. Apparently the first two open cases are r=929 and 947
        > yielding 214 and 217 digit numbers to factor.

        An update here: M929 has been factored, and a group of people have already
        started factoring M947. You can find the factor for M929 here:
        And, you can see who is factoring which Cunningham number here:
        And more importantly, you can find all known*1 factors for all important*2
        numbers in the online factor database here:
        Once there, you can type in 2^929-1, and it will show you all the factors of
        that number and that it is Fully Factored (FF). Currently, you can type in
        2^947-1 and see that it is a CF, composite number, with factors known, but not
        yet fully factored.

        *1 = All known factors that have been stored into the factordb.
        *2 = All numbers that people are interested in and store in the factordb.

        Also, the factordb stores prime numbers too. Below 300 digits it will just
        prove the number prime, and above that it will accept Primo certificates and
        verify them locally.

        -David C.
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