Re: JSH: One other option

On Sep 11, 9:46 pm, JSH <jst...@xxxxxxxxx> wrote:

On Sep 11, 8:17 pm, marcus_b <marcus_bruck...@xxxxxxxxx> wrote:

On Sep 11, 7:19 pm, JSH <jst...@xxxxxxxxx> wrote:
[...]
It turns out Marcus pointed something out in another thread so I can
give a simpler definitive test of the latest theory which oddly takes
away any factoring of surrogates. Thanks Marcus!!!

Why ... I feel so unworthy.

Kind of odd really, but fascinating to contemplate as by the theory
the following algorithm should be applicable against an RSA sized
number factoring it in a maximum of 32 trials:

Note, still with x^2 = y^2 mod T, and k = 2x mod T.

So then, with x = floor(sqrt(T)), k = 2x,

n_min = floor((4(x+k-1) - 2k^2)/T)

and

n_max = floor(((x+k)^2 - 2k^2)/T)

you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min to
n_max, looking for a
perfect square and the theory says there must be at least one,

Let's see. Let T = 77. Then:

x = floor(sqrt(77)) = 8

k = 2x = 16

nmin = floor(-420/77) = -6

nmax = 0.

So now we have the following table:

n (x + k)^2 - 2k^2 - nT
----- ---------------------------
-6 526
-5 449
-4 372
-3 295
-2 218
-1 141
0 64

Only the last of these is a perfect square, and
it yields y = 8. Since x = 8,

x^2 - y^2 = 0 = 0 mod 77,

but of course this gives no useful factors of T.

So over the specified range of n's, none of
the y's is a useful perfect square.

You're right. It seems though that 8 is too small to work as the
theory is setup with the assumption that abs(x)>abs(y).

Try x=9.

Have fun!

Oh yeah, it seems that with dinky numbers you can find these trivial
exceptions near the edges.

What conditions on T are required so that it's not "dinky"?

I wonder why.

If you find that x=9 does not work, then, hey, maybe I need to re-
check the theory!

But you know it will work.

Hmmmm...interesting though that the limits didn't handle that case
with a dinky number.

Something to wonder about....

Something else to wonder about: JSH claims to be a programmer. The
above could EASILY have been coded in any language (even Java). Why
doesn't JSH take the ten minutes necessary to write the program and
run it, seeing that it doesn't work? I mean, all you have is write
something like

int SF(int T)
{
int x, k, n, y;

x = floor(sqrt(T)); k = 2*x;
for (n = floor((4(x+k-1) - 2*k^2)/T); n <= floor(((x+k)^2 -
2*k^2)/T); n++) {
y = sqrt((x+k)^2 - 2*k^2 - n*T);
if (floor(y)*floor(y) == (x+k)^2-2*k^2 - n*T)
return gcd(x+y, T);
}
return -1;
}

Much easier than, say, a "class viewer".

--- Christopher Heckman

.

Relevant Pages

  • Re: JSH: One other option
    ... away any factoring of surrogates. ... JSH claims to be a programmer. ... int SF ... Because all that Java development was done on the company ...
    (sci.math)
  • Re: JSH: One other option
    ... away any factoring of surrogates. ... JSH claims to be a programmer. ... It's mathematical research people. ...
    (sci.math)
  • Re: I & J
    ... debugger once. ... wouldn't remotely expect any programmer to know what an oscilloscope, DSP, ... skilled and/or smart person by some arbitrary criteria. ... Your statements about factoring code in C ...
    (comp.lang.forth)
  • Re: I & J
    ... they didn't last long in the interview. ... wouldn't remotely expect any programmer to know what an oscilloscope, DSP, ... factoring for C. "Factoring" as applied to C will result in large ... space on a the stack for local variables, etc. I.e., overhead. ...
    (comp.lang.forth)
  • Re: Simple integer factorization algorithm
    ... This does the timing of the factoring algorithm ... final int loBits = 16; ... int misfactors; ... System.out.printf("%n%nAverage timings over %d numbers:%n", ...
    (comp.lang.java.programmer)