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Re: Bill Krys' sieve - more Q&A

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  • Bill Krys
    Andrey, ... IT WON T BE THAT KIND OF SIEVE BECAUSE THAT SIEVE MUST START AT ONE AND CONTINUOUSLY REFERENCE ONE, BUT MINE CAN START ANYWHERE. ALSO I PLAN ON
    Message 1 of 3 , Dec 29, 2000
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      Andrey,


      --- Andrey Kulsha <Andrey_601@...> wrote:
      > Hello Bill!
      >
      > > Audrey,
      >
      > Well, I'm not offended :-)
      >
      > > the sieve as it stands cannot avoid division,
      > unless i
      > > figure out something else.
      >
      > I think it will be Eratosthen's sieve :-)

      IT WON'T BE THAT KIND OF SIEVE BECAUSE THAT SIEVE MUST
      START AT ONE AND CONTINUOUSLY REFERENCE ONE, BUT MINE
      CAN START ANYWHERE. ALSO I PLAN ON HAVING AN ALGORITHM
      THAT DOESN'T REQUIRE DIVISIONS TO SIMPLIFY THE
      SIEVING. AFTER THAT I PLAN TO HAVE A REVERSE ALGORITHM
      TO FIND THE PRIMES DIRECTLY.
      >
      > > So the sieve can be started anywhere on the
      > fragment,
      > > but I thought the square of a prime would be as
      > good a
      > > place as any.
      > >
      > > Some definitions:
      > >
      > > N# = a number to be factored
      > > R# = the rank of a number to be factored
      > > F# = the factor of a composite
      > > ArmA = the left half of the fragment
      > > ArmB = the right half of a fragment
      >
      > Well, if M is a center of the fragment, then
      > R=|N-M|.

      YES, THAT'S RIGHT.
      >
      > > For example, on the frag 2^6 to 2^7, we start at
      > 121
      > > on Arm B. Its factors are F11 and F11. its rank is
      > > R25. look for all ranks that occupies a gcd=1
      > position
      >
      > All N of a kind 6k+-1, and only them, occupies a
      > gcd=1
      > position. Hence, theirs (and only theirs) ranks have
      > this
      > form too.

      TRUE.
      >
      > > on arm A and which, when added to R25, will be
      > > divisible by F11.
      >
      > Let this rank will be Rx, and number of this rank
      > will be
      > Nx=M-Rx.
      > M+R25 is divisible by 11. R25+Rx must be divisible
      > by R11.

      IT MUST BE DIVISIBLE BY F11, NOT R11.

      > Hence, (M+R25)-(R25+Rx)=Nx will be too.

      RIGHT.
      >
      > > That rank is R19 and the number is N77.
      >
      > All that you are doing there, - a trial sieving the
      > numbers
      > 6k+-1 divisible by 11 from the ArmA. There are
      > around M/33
      > such numbers, so if you take M, for example, 3*2^50,
      > you
      > will have more than 10^14 such numbers!

      RIGHT. I BOO-BOO'ED HERE. I WILL ENDEAVOR TO HAVE THE
      SIEVE SIEVE FOR COMPOSITES DIRECTLY WITHOUT SIEVING
      FOR RANKS, AND EVENTUALLY, AS I'VE MENTIONED ABOVE, NO
      SIEVING. BUT EVEN AS IT STANDS, WE DON'T NEED TO
      REFERENCE "1" ALL THE TIME AND CAN SIEVE ANY FRAGMENT
      NEEDED, UNLIKE ERATOSTHENE'S SIEVE.
      >
      > > N77/F11=F7, so
      > > now look for a rank on arm B where gcd=1 which,
      > when
      > > added to R19 is divisible by F7 or F11.
      >
      > And now you have sieved:
      >
      > a) all the numbers 6k+-1 divisible by 11 from the
      > ArmB (and
      > now you have sieved them from both fragments);
      >
      > b) all the numbers 6k+-1 divisible by 7 from the
      > ArmB (at
      > the next step you will sieve such numbers from
      > another
      > fragment too).
      >
      > > There are none for F11 so stop this thread. But
      > there
      > > is one for F7. N119 at R23: (N119/F7=F17). So look
      > for
      > > a number on arm A that occupies gcd=1, whose rank
      > when
      > > added to R23 is divisible by F17 or F7.
      > >
      > > There are 2 new numbers: N85 at R11,
      > (R23+R11)/F17=2
      > > and N91 at R5 (R23+R5/F7=4). N91 has no number or
      > rank
      > > that fits the criteria on arm B, so stop this
      > thread.
      > >
      > > N85/F17=5, so continue as above. There are 2
      > numbers
      > > on arm B derived from N85: N115 and N125 at ranks
      > R19
      > > and R29 respectively.
      > >
      > > There are no ranks on arm B that when added to R29
      > = a
      > > number divisible by 25, so stop this thread. The
      > other
      > > number, 115 when divided by 5=23. Look for a
      > number on
      > > arm B at gcd=1 whose rank when added to 19 is
      > > divisible by 5. 65 and 95 fulfil this criteria at
      > > ranks 31 and 1 respectively. 95's rank (rank=1)
      > has no
      > > number whose rank will fit the criteria, so stop
      > this
      > > thread. Continue with 65. There should be no
      > > composites left at gcd's=1.
      >
      > Well, now you have sieved all the numbers 6k+-1
      > divisible by
      > 11,7 and 5 from both fragments. Hence, all left
      > numbers
      > 6k+-1 must be prime.
      >
      > But Eratosthen's sieve is much faster, because:
      >
      > a) generating a list of primes less or equal to
      > sqrt(N) is
      > much faster than irregular receiving them when
      > dividing
      > ranks (lots of them appears for large N, about
      > 3*2^50);

      TRUE, BUT EVEN IF I SIEVE IN AN ERATOSTHENIAN WAY, I
      CAN SIEVE WITHOUT REFERENCING "1" AND THEREFORE CAN
      SIEVE OTHER FRAGMENTS.
      >
      > b) we can work with any fragments, not with two
      > given;

      NO, YOU MUST ALWAYS START AT 1 AND SIEVE UP TO THE
      DESIRED POINT.
      >
      > c) there are no squares or any another powers of
      > primes in
      > Eratosthen's sieve.

      THE SQUARES OR POWERS IS JUST A STARTING POINT BECAUSE
      THEY ARE EASY TO FIND.
      >
      > > I am working on finding a pattern to the factors
      > > without relying on division. Some interesting
      > teasers
      > > appear, but I need more numbers to look at.
      >
      > I've just showed you a method.
      >

      ARE YOU REFERRING TO ERATOSTHENE'S? BECAUSE I DON'T
      SEE ANY OTHER METHOD YOU'VE SHOWN.

      > > Could you
      > > generate, in an excel-readable file, fragments
      > that
      > > include the gcd's and the greatest factor (not the
      > > gcd), for those numbers that are at gcd=1, but are
      > > composite.
      >
      > There will be 1/3 of all composites, namely of form
      > 6k+-1,
      > i.e. non-divisible by 2 or 3.

      BUT ANDREY, THE COMPOSITES NOT DIVISIBLE BY 2 OR 3
      ALSO SHOW A SYMMETRY. I REALLY THINK THIS IS YOUR AND
      OTHER CRITICS' MAIN AREA OF CONFUSION. I AM TRYING TO
      FIND A PATTERN FOR THIS SYMMETRY WITHOUT SIEVING FOR
      THEM AND THERE ARE PATTERNS, I JUST NEED TO SEE MORE
      NUMBERS.

      >
      > > I've included the excel files for reference.
      >
      > I'm surprized that you missed that your "symmetry"
      > exists
      > only for divisibility by 2 and 3. All of exceptions
      > from
      > your symmetry are divisible by prime greater than 3,
      > and you
      > are "restoring" this symmetry by trial sieving of
      > these
      > numbers.

      KIND OF RIGHT. THE FACT THAT THERE IS SYMMETRY OF
      THESE OTHER COMPOSITES IS A PATTERN WORTHY OF
      INVESTIGATION, NOT DEBUNKING BEFORE IT'S UNDERSTOOD
      WHAT YOU'RE DEBUNKING.
      >
      > > Reimann
      >
      > Well, I'm not offended on "Audrey", but please don't
      > distort
      > a surname of great scientist!

      WHAT DOES THIS MEAN? I COULD GO ON AND ON ABOUT
      UNQUESTIONED AUTHORITY IS DANGEROUS, ETC, ETC. I'M NOT
      CLAIMING TO BE IN HIS LEAGUE OR ATTEMPTING TO BELITTLE
      HIM OR HIS LEGACY.
      >
      > > will wait.
      >
      > I think he never will wait. :-)
      >
      I DON'T KNOW WHAT THIS MEANS. BUT ALL I UNDERSTAND
      ABOUT REIMANN IS THAT THERE IS A LINE WHICH ON EITHER
      SIDE HAS AN EQUAL NUMBER OF PRIMES. IF THIS
      UNDERSTANDING IS WRONG THEN I'M SORRY. HOWEVER, IF
      EACH FRAGMENT CAN BE SHOWN TO HAVE AN EQUAL NUMBER OF
      PRIMES AND POWERS OF PRIMES (PERHAPS ONLY SQUARES OF
      PRIMES), THAT GOES A LONG WAY TO PROVING PRIMES ALONE
      ARE BALANCED IF YOU SHOW POWERS OF PRIMES ARE
      BALANCED.

      PLEASE ATTACK WHEN YOU ARE READY.

      BILL

      =====
      Refining my sieve into a simple, non-stochastic algorithm

      Bill Krys
      Email: billkrys@...
      Toronto, Canada (currently: Beijing, China)

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    • Jack Brennen
      ... What he means is that you are misspelling the name -- it s spelled: Riemann
      Message 2 of 3 , Dec 29, 2000
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        Bill Krys wrote:

        > > > Reimann
        > >
        > > Well, I'm not offended on "Audrey", but please don't
        > > distort
        > > a surname of great scientist!
        >
        > WHAT DOES THIS MEAN? I COULD GO ON AND ON ABOUT
        > UNQUESTIONED AUTHORITY IS DANGEROUS, ETC, ETC. I'M NOT
        > CLAIMING TO BE IN HIS LEAGUE OR ATTEMPTING TO BELITTLE
        > HIM OR HIS LEGACY.

        What he means is that you are misspelling the name -- it's spelled:

        Riemann
      • Bill Krys
        Jack and Andrey, Thanks for the correction. I guess I ve been in my little mirror world so long that ie and ei are the same to me. Just joking. I m sorry
        Message 3 of 3 , Dec 29, 2000
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          Jack and Andrey,

          Thanks for the correction.

          I guess I've been in my little "mirror" world so long
          that "ie" and "ei" are the same to me. Just joking.
          I'm sorry to have been so dense.

          Bill


          --- Jack Brennen <jack@...> wrote:
          > Bill Krys wrote:
          >
          > > > > Reimann
          > > >
          > > > Well, I'm not offended on "Audrey", but please
          > don't
          > > > distort
          > > > a surname of great scientist!
          > >
          > > WHAT DOES THIS MEAN? I COULD GO ON AND ON ABOUT
          > > UNQUESTIONED AUTHORITY IS DANGEROUS, ETC, ETC. I'M
          > NOT
          > > CLAIMING TO BE IN HIS LEAGUE OR ATTEMPTING TO
          > BELITTLE
          > > HIM OR HIS LEGACY.
          >
          > What he means is that you are misspelling the name
          > -- it's spelled:
          >
          > Riemann
          >


          =====
          Refining my sieve into a simple, non-stochastic algorithm

          Bill Krys
          Email: billkrys@...
          Toronto, Canada (currently: Beijing, China)

          __________________________________________________
          Do You Yahoo!?
          Yahoo! Photos - Share your holiday photos online!
          http://photos.yahoo.com/
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