Re: A Property
- --- In firstname.lastname@example.org, "gordon_as_number"
>Finite images (about which I assume you speak) are just countable
> Here is just a comment.
> It is possible to compile a data image as a number,
data, so can be mapped to and from N.
> as inNot my kind of place. Better than arxiv's GM section though; barely.
> That number, when factoredWhy? Data compression _solely_ relies of redundancy.
> in primes, can be stored in the latter to data compress that
> image by a factor of an approximate 10^12.
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 labelinga) What proportion of large numbers are factorable into primes that
> 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.
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
> 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 toWrong, as indicated above.
> 10^15 or so, and each prime in accordance with integers, a
> digit is lost with each prime factor.
> And assuming that thereSo wrong that it beggars belief.
> 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.
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 asAh, so this did belong in GM after all.
> 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.
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
- On 10/20/2012 10:07 AM, email@example.com wrote:
> 2.2. Re: QuestionThank you David.
> 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
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.
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