Re: JSH: What will you do?
- From: Matt Gutting <tchrmatt@xxxxxxxxx>
- Date: Fri, 22 Apr 2005 01:27:40 -0400
Tim Peters wrote:
[C. Bond, tries his best to factor 15, but gets stuck at the end] [...]
The product be_2 if_1, for this example (and my choice for j and k_1), is what James method leads to. There are eight possible rational numbers representing this product. I *still* don't see how these number yield the integer factors of 15. Do they? By what method or process? Is SF incomplete?
This was already adequately explained, but in the face of JSH's relentless hysterical propaganda perhaps it might help to finish your example. So I will.
Please be patient. SFT is mind-numbingly ... well, "dumb" is the only accurate word that comes to mind ... and in more than one way. In my opinion, that makes the few posters who look at it seriously too often strive to give it more benefit-of-the-doubt than it actually deserves. If you stick to what's _there_, it barely merits a one-sentence critique.
First, James doesn't have an algorithm this time, just "a theorem". That's the real reason you got stuck. It's an important reason too, as will soon become clear.
His Google Groups "SF Algorithm" post (which does not actually give an algorithm) ends with this:
b_2 f_1 = (-(Az - 2M^2)+/- sqrt((Az - 2M^2)^2 - 4TM^2))/2
and if you have done it right the square root will be rational, so that's a way to check your implementation.
Here b_2 is a rational factor of your target M, so you need to check the gcd of its numerator with M.
That's where you got stuck. It's the same place James got stuck, because he has no idea what to do at this point either (although he won't admit this, and either lies about what he thinks will happen next, or is incurably deluded by hope).
What you need to do after you compute b_2 f_1 is divide by f_1, giving b_2. So what's f_1? He doesn't say in that post. You have to go to the "Factoring theorem" post, where it is explained:
f_1 is a rational factor of T
Jaw-dropping idiocy #1: That's an insane way of saying "f_1 is a non-zero rational". T has nothing to do with it (except in an accidental way having to do with irrelevant algebra in the middle of the theorem -- that will become clear below too).
But which non-zero rational? James has no advice to give here -- it's _some_ non-zero rational. You have to pick it, and hope you get lucky. It wasn't enough that you already picked j and k_1, you also need to pick f_1. Here I'll pick f_1 for you that happen to work, completing your example calculations successfully:
b_2 f_1 = (-(Az_1 -2M^2) + sqrt((Az_1 - 2M^2)^2 -4TM^2))/2 = (-(Az_1 -2M^2) - sqrt((Az_1 - 2M^2)^2 -4TM^2))/2 = (-(Az_2 -2M^2) + sqrt((Az_2 - 2M^2)^2 -4TM^2))/2 = (-(Az_2 -2M^2) - sqrt((Az_2 - 2M^2)^2 -4TM^2))/2 = (-(Az_3 -2M^2) + sqrt((Az_3 - 2M^2)^2 -4TM^2))/2 = (-(Az_3 -2M^2) - sqrt((Az_3 - 2M^2)^2 -4TM^2))/2 = (-(Az_4 -2M^2) + sqrt((Az_4 - 2M^2)^2 -4TM^2))/2 = (-(Az_4 -2M^2) - sqrt((Az_4 - 2M^2)^2 -4TM^2))/2
These reduce to:
b_2 f_1 = 150/97 = -9312 = 2400/11 = -66 = 7200/31 = -62 = 9184 = -450/287
b_2 f_1 f_1 b_2 = (b_2 f_1)/f_1 gcd(num(b_2), 15) -------- ----------- ------------------- ----------------- 150/97 (150/97)/3 3 3 -9312 (-9312)/3 3 3 2400/11 (2400/11)/3 3 3 -66 (-66)/3 3 3 7200/31 (7200/31)/3 3 3 62 (62)/3 3 3 9184 (9184)/3 3 3 -450/287 (-450/287)/3 3 3
Get the picture? You succeeded 8 of 8 times! Feel cheated? Well, unless you've thought it all the way through, you don't yet feel cheated _enough_ <0.9 wink>.
Here is a true thing, which I'll state without proof (I omit it just because a proof is probably obvious to most):
Given any non-zero integer M and any non-zero rational K, then for every positive integer divisor i of M (regardless of whether i is prime or composite, and including i=1 and i=|M|), there are an infinite number of rationals f_1 such that gcd(numerator(K/f_1), M) = i.
Note that his holds for _any_ non-zero rational K. That leads to jaw-dropping idiocy #2: all the algebra you endured to compute b_2 f_1 products was _almost_ entirely irrelevant. The only relevant thing is that the equations produce _some_ non-zero rational. You didn't need to compute 8, or 4, or 2 of 'em -- every one of them "works". And you didn't need to use JSH's bizarre equations to come up with them either. There's nothing special about them; you may as well always use b_2 f_1 = K = 1, or K = -13/31, or K = floor(e^google), ... and skip all the obfuscating algebra.
It just doesn't matter: for any fixed non-zero rational K, the function
b_2(f_1) = K/f_1
is a bijection from Q-{0} onto itself, and if M is composite then an infinite number of f_1 exist with 1 < gcd(numerator(b_2(f_1)), M) < M.
And that's jaw-dropping idiocy #3: While it's true that an infinite number of f_1 exist that yield a non-trivial factor of a composite M, SFT gives no method, nor even a shadow of a hint of a method, to find even one such f_1, short of (say) systematically picking f_1 so that the sequence of resulting b_2's has numerators divisible by the primes in 2 .. sqrt(M), one at a time.
JSH's "50% claim" is a jaw-dropping idiocy too, but that claim isn't actually made in SFT.
That's what's actually _there_ in SFT: a pile of complicated but ultimately irrelevant algebra, leading to a true conclusion that's fairly characterized as all of trivial, uninteresting, and useless.
James will howl like a banshee about it, but that's the plain truth of this.
What you seem to be saying is that James has developed a long, complicated way of choosing a nearly random trial divisor of M.
Which would explain why it works about as well.
Matt .
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