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Re: A Property

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  • thefatphil
    ... Finite images (about which I assume you speak) are just countable data, so can be mapped to and from N. ... Not my kind of place. Better than arxiv s GM
    Message 1 of 64 , Oct 21, 2005
      --- In primenumbers@yahoogroups.com, "gordon_as_number"
      <gordon_as_number@y...> wrote:
      >
      >
      >
      > Here is just a comment.
      >
      > It is possible to compile a data image as a number,

      Finite images (about which I assume you speak) are just countable
      data, so can be mapped to and from N.

      > as in
      > xxx.arXiv.org/physics/0510148.

      Not my kind of place. Better than arxiv's GM section though; barely.

      > That number, when factored
      > in primes, can be stored in the latter to data compress that
      > image by a factor of an approximate 10^12.

      Why? Data compression _solely_ relies of redundancy.
      Losslessly converting data from one format _never_ introduces
      redundancy. (But in the case of the BWT, may reveal redundancies
      in ways that are easier to exploit.)

      > As the labeling
      > of primes is about a digit or two below the number in accordance
      > with numbers up to 10^17 (the known primes), with density of
      > an approximate x/ln(x) 10^17.

      a) What proportion of large numbers are factorable into primes that
      are <10^17.
      b) You've not thought about how you are going to separate your tokens.
      It takes nearly lg(n)+lglg(n)+2lg(n) bits to transmit an arbitrary n
      in a uniquely decodable way.
      c) What did you think was actually /more redundant/ about the
      expression of a number as its prime factorisation than in any other
      form? Why?

      > The point is,

      That you've reinvented "compression by coincidence", as I like to call
      it, just like hundreds before you who wanted to secrete data into the
      digits of Pi, square roots of non-squares, radix-n expansions of
      ratios of coprime integers, etc., etc.

      > with the prime factorization of a number up to
      > 10^15 or so, and each prime in accordance with integers, a
      > digit is lost with each prime factor.

      Wrong, as indicated above.

      > And assuming that there
      > a maximum ln(N) prime factors, or rather less, most data
      > compression schemes can be achieved with a factor of 10^12,
      > which is actually quite amazing.

      So wrong that it beggars belief.
      Can't you even see that it even contradicts your previous wrong
      statement? You're not just wrong - you're internally consistent as well.

      >(Images could be written as
      > a number with approximately 10^260 digits.)
      >
      > So, depending on how fast you can factor a number with 10^100
      > digits or so, compressions of 10^10 are achievable.

      Ah, so this did belong in GM after all.

      Please learn at least a little about data compression before making
      statements about data compression. Start with the comp.compression
      FAQ. Don't even begin to _think_ about making outlandish claims until
      you have read Cover and Thomas from cover to cover. And then don't
      make them here, as they will _not_ be concerned with prime numbers.


      Compression (at an intellectual level, rather than an "oooh shiny"
      level) was my bag a long time before prime numbers were. Just don't go
      there.
      Phil
    • Kermit Rose
      ... Thank you David. I had been confused by consideration of the following: if x y = z and y x, then if we set t = (y+x)/2 and s = (y-x)/2 then z = t^2 - s^2
      Message 64 of 64 , Oct 20, 2012
        On 10/20/2012 10:07 AM, primenumbers@yahoogroups.com wrote:
        > 2.2. Re: Question
        > Posted by: "djbroadhurst"d.broadhurst@... djbroadhurst
        > Date: Fri Oct 19, 2012 9:02 am ((PDT))
        >
        >
        >
        > --- Inprimenumbers@yahoogroups.com,
        > Kermit Rose<kermit@...> wrote:
        >
        >> >Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
        >> >
        >> >for what values of x would
        >> >x^2 + 5 x + 6 be a perfect square.
        > Set A = 1, B = 0, C = -1, D = 5, E = 0, F = 6.
        >
        > and you will get the obvious answer from Dario:
        >
        > x = -2
        > y = 0
        > and also:
        > x = -3
        > y = 0
        > Calculation time: 0h 0m 0s
        >
        > David

        Thank you David.

        I had been confused by consideration of the following:

        if x y = z and y > x,
        then if we set t = (y+x)/2 and s = (y-x)/2

        then z = t^2 - s^2
        and x = y - 2 t.

        So z = x y = (y - 2t) y = y^2 - 2 t y

        y^2 - 2 t y - z = 0

        Transforming w = y - k yields

        (w + k)^2 - 2 t (w + k) - z = 0

        w^2 + (2 k - 2 t) w + (k^2 - 2 t k - z) = 0

        which has integral solution if and only if

        (2k - 2 t)^2 + (z + 2 t k - k^2) is an integral square

        In all that derivation, I forgot that I still did not know what value t was.

        I confused myself too easily.

        Kermit








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