mm-1848
===
Given that p=sum of points on a roll of dice, s=number of faces on each
die, and n=number of dice rolled, I know that
k=floor((p-n)/s)
sum(0,k) (-1)^k n!/((n-k)! k!) (p-sk-1)! /((p-sk-n)! (n-1)!)
will give the number of combinations that will yield a particular sum.
If you had an array of such dice, and summed the rows and columns, how
would you calculate the total number of combinations that could yield
that set of sums? For example, if the dice rolled and the sums were:
3 3 3 | 9
3 3 3 | 9
3 3 3 | 9
_____
9 9 9
How would you calculate how many other combinations would also yield
(9,9,9) and (9,9,9)?
I know I could write a loop to just count the combinations, but for the
numbers I am working with it is impractical, so I am looking for a
mathimatical solution if there is one.
-Michael VanDeMar
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
Actually, it was not original. It goes by the name of the
Meissel-Lehmer-Lagarias-Miller-Odlyzko Method, at least according to M.
DELEGLISE AND J. RIVAT's COMPUTING pi(x): THE MEISSEL, LEHMER,
LAGARIAS, MILLER, ODLYZKO METHOD, MATHEMATICS OF COMPUTATION, Volume
65, Number 213, January 1996, Pages 235-245, specifically the bottom
half of page 236.
I have repeatedly asked James Harris to explain why his method is
substantially different from the one presented in that paper. The only
difference is that he uses k instead of the kth prime, and he has never
acknowledged this.
     --- Christopher Heckman
Large number?  It would be a step in the right direction if he could factor
a small number.  15 has been a problem in the past, IIRC.
Alan
-- 
Defendit numerus
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
Uh, he is demonstrating his system by explaining away truths.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
Some posters, mostly from sci.math, are going on and on about
supposedly showing that the SFT doesn't work, but the problem is, you
don't even have to check their math to know they're wrong because of
the two square roots in the SFT.
The first one which is the easy one is
Az = Ax(Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2Ax - 2M^2)
and it's easy as you have
Ax = +/- (k_1 + k_2) + 2j^2
where k_1 k_2 = - j^2 T, so you have that square root no problem.
But notice, k_1 and k_2 are factors of Tj^2, as that's VERY important.
That one is the easy square root.
It's the second one that's problematic:
b_2 f_1 = (-(Az - 2M^2)+/- sqrt((Az - 2M^2)^2 - 4TM^2))/2
as it requires you have factors of TM^2 to evaluate.
But a sci.math poster claims to have evaluated and didn't use factors
of TM^2, so what do you know?
That poster is wrong.
Now that's some easy mathematics, so how can even sci.math posters
claim victory and go on and on as if they don't know very easy algebra?
It's a mystery.
For those of you who are a little lost, imagine you have something
simpler, like
sqrt(x^2 + 4M)
and you wish to evaluate that square root, where x and M are integers.
Well, you can't solve for it without having factors of M, like if
f_1 f_2 = M
you can let
x = f_1 - f_2
and it evaluates as
sqrt(x^2 + 4M) = f_1 + f_2
but you have to have factors of M.
The quadratic
b_2 f_1 = (-(Az - 2M^2)+/- sqrt((Az - 2M^2)^2 - 4TM^2))/2
is more complicated, but you still can't evaluate it without having
factors of M, but a poster named Nora Baron claims to have done so,
so what do you know?
You know that person is wrong.
It's not hard algebra.
Now then, do you think that poster will come out and apologize or even
admit being wrong?
I hope so, as it's nice to hope for the best in people.
After all, mathematics is not an area where it's always easy to just
say something that's not true, as people can at times simply point out
basic algebra that proves you're wrong, which is why mathematics is
such a nice subject, for some people.
Surrogate Factoring Theorem:
Given M, a target natural number to be factored, and j, an integer
b_2 f_1 = (-(Az - 2M^2)+/- sqrt((Az - 2M^2)^2 - 4TM^2))/2
where T = M^2 - j^2, and f_1 is a rational factor of T, and Az is given
by
Az = Ax(Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2Ax - 2M^2)
and Ax is given by
Ax = +/- (k_1 + k_2) + 2j^2
where k_1 k_2 = -Tj^2, and k_1 and k_2 are rationals.
James Harris
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
You shouldn't be calling people liars who do your work for you. One of
them showed that whenever you take square roots, you always end up with
a rational number. You've never shown that. The SF Theorem certainly
doesn't; it just shows that the equations give you a number. And it's
not automatic; sqrt(1/3) is irrational.
A theorem that specifies exactly when sqrt(x) is rational has real
content to it. Infinitely more content than the SF Theorem.
Actually, if you allow k_1 and k_2 to be negative, you can simplify
that to
Ax = k_1 + k_2 + 2 j^2.
That depends on what you mean by evaluate. If you know Az, M, and T,
you can evaluate the right-hand side of the equation by substituting
those values. That is what most of the world means by evaluate: What
number is this equal to?
On the other hand, if you want to know what value of x makes an
equation true, that's called solving the equation.
Look it up in your lying^H^H^H^H^H math book if you don't believe me.
I believe that poster is me.
First of all, you should be THANKING me because I showed that SF wasn't
worthless; I managed to get a non-trivial factor of 15 using SF.
What I did, as a human being, was to pick f_1 = 1, which is certainly a
factor of TM^2. Then I solved for b_2. The other poster (Decker? In any
case, it's someone you should be THANKING as well, because he came up
with an algorithm, something you haven't done) had come up with 8
possible values of b_2. In 2 of those cases, the numerator b_2 had a
non-trivial factor of M in it.
That's 2/8, not 0/8.
Not only did I do YOUR work in finding ONE example where SF works, I
found two, and showed that the rate of success for this example is 25%.
It's not 50%, but it's a start.
Or maybe you WANT SF to FAIL ...
The mystery here is whether you're talking about the ones who said that
SF will never work (which is not a correct description, since they
only took positive square roots), or the ones who made SF work in one
case.
I assume you mean evaluate here, instead of solve for. There is no
equation to solve; there's just an expression which evaluates to be a
number.
Not true. If M = 33 and x = 8, for instance,
sqrt(x^2 + 4M) = sqrt(8^2 + 4 * 33) = sqrt (64 + 132) = sqrt(196) = 14.
There. I never needed to factor 33.
That is true, but you don't HAVE to do it. It just shows that if the
value of x is special, then you can come up with an answer based on the
factors of M.
You don't have to actually FIND f_1 and f_2, though; that is, you don't
need to know what number f_1 is equal to.
It's called algebra, and it lets you work with unknown quantities.
Your terminology is wrong. If you can't communicate, it's not other
people's faults that they misinterpret what you say.
It goes back to algebra; you don't need to actually know the values of
the factors of M. Otherwise, you would be lying when you write an
equation like
p q = M (which you've done in the past).
It also goes back to what she's proving: IF p and q are the factors of
M, THEN (some variable equals some expression involving p and q).
This is a different statement from (some variable equals some
expression involving p and q). Of course, since you haven't taken a
logic course, this is lost on you.
It's the difference between the statements If x = 2, then 2 x = 4 and
2 x = 4. The first one is always true. The second isn't.
Why do you think I haven't written you off as a crank? Same reason.
Then why are you saying all mathematicians are liars? It must be hard
for mathematicians to lie, and the fact that so many can is a miracle
in itself ...
Yes, you can point out where other people have gone wrong, but if they
don't know enough about terminology and methodology, they'll not
comprehend WHY it's wrong and simply repeat what they said before.
Good example.
     --- Christopher Heckman
   posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn
important.
algebra?
integers.
  No, I wouldn't agree with that.  I showed that if
      k_1 k_2 = - j^2 T,
and k_1 and k_2 are both integers, and you take the + square
root in the expression for b_2 f_1, then you get
      b_2 f_1 = E / F,
where E and F are both integers, with
      E = (k_1 + k_2)*j^2 + k_1 (k_1 + 3k_2) + 2 (k_1)^2 k_2/j^2
          - (k_1 + k_2 + 2 j^2)*(k_1 + j^2)
          + 2M^2 (k_1 + k_2 + 2 j^2 - M^2), and
      F = 2 (k_1 + k_2 + 2 j^2 - M^2).
  You can verify this for yourself by letting M = 15,
j = 17, T = M^2 - j^2 = 225 - 289 = -64.  Since
     k_1 k_2 = -T * j^2 = 64 * 289,
I can let k_1 = 289 * 32 = 9248, and k_2 = 2.
  In this case,
      E = (9248 + 2)*289 + 9248*(9248 + 3*2) + 2*9248^2*2/289
          - (9248 + 2 + 578)*(9248 + 289)
          + 2*225*(9248 + 2 + 578 - 225)
        = 2673250 + 85580992 + 1183744 - 93729636 + 4321350
        = 29700
      F = 2 * (9248 + 2 + 578 - 225) = 19206
  Thus, in lowest terms, E/F = 150/97.  This is the same answer that
C. Bond got by applying your theorem directly.  You can check this
by going through your own formulas, as follows:
  First note that Ax = (9248 + 2 + 578) = 9828.
  This yields
    Az = Ax(Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2Ax - 2M^2
       = 9828 (9828 + sqrt(9829 - 578)^2 - 4 * 31 * 256))/19206
       = 9760.45360825.
and this yields
    b_2 f_1 = (-(Az - 2M^2) + sqrt((Az - 2M^2)^2 - 4TM^2))/2
            =  1.54639175 = 150/97.
  The agreement in fact is exact.
  The formula that I gave for the numerator of b_2 f_1, i.e., the
term E above, gives exactly the same answer as your theorem.
  Agreed.
even
  No - the formula that I gave for E in the expression
  b_2 f_1 = E / F
is exactly correct (when the + sign is used in the square roots).
  Note that above, E = 29700, which is exactly divisible by M^2 = 225.
  Note also, that since in lowest terms, b_2 / f_1 = 150/97,
the numerator, 150, is exactly divisible by M = 15 and not by any
additional factors of M.
  So, in contrast to what you said, my formula arrives at the
correct value for b_2 f_1 without any factorization of M.
  Sorry to disappoint.  Do the arithmetic and see for yourself.
out
  Right.  See above.
  Nora B.
given
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
you
of
factors
so,
A single case.
It's not too far-fetched to imagine that you've found an example to
support your position.
Now I could challenge on the use of integers only, except I don't see a
problem with that, and I'll explain below.
What you need to do Nora Baron is step out your calculations.
Start from the beginning, and go through how you solve for b_2 f_1.
If you do so, then I can show you your mistake, or find that I'm wrong.
If you refuse, then I will challenge your use of single cases as single
cases don't mathematically prove, and I have more to explain as to why
I'm sure you have to be wrong.
So?  You can manufacture a formula to match at any number of single
cases.
225.
That shows naivette, which is why I left it in, as in fact, your
formula would indicate that M is being factored trivially.
The equation
b_2 f_1 = (-(Az - 2M^2)+/- sqrt((Az - 2M^2)^2 - 4TM^2))/2
cannot be resolved without a factorization of TM^2, which is basic
algebra, as I noted in the post that started this thread.
If you are correct, then you have proven that the equations I use
always trivially factor, which is why it's a big deal.
However, the SFT connects all points on two hyperbolas, which
contradicts with the possibility that only trivial factors are being
used.
You see, there are mysteries here.
I don't trust you as I've watched you lie a lot, so you don't get the
benefit of the doubt with me, and single cases do not prove in
mathematics.
Show your work Nora Baron, or you can debate as long as you like, and
find that I'll simply point out simple algebra that proves you're
wrong.
For those still lost on that algebra, again, consider
sqrt(x^2 + 4M)
where M is a non-zero integer, not a square, and you wish that to be an
integer.
You MUST use a factorization of M.  There simply is no choice in the
matter as x must be the difference of integer factors of M.
With the more complex square root in
b_2 f_1 = (-(Az - 2M^2)+/- sqrt((Az - 2M^2)^2 - 4TM^2))/2
you MUST use some integer factorization of M, despite the fractions
being used, and if Nora Baron is correct, then the equations I use
rely on a trivial factorization of M, like
M*1 = M
but the problem there is, if the equations are hard-wired to trivially
factor M, then that's what they'll do, so it'd never do anything else
(except in some very special situations).
There are several reasons why that's impossible.
James Harris
   posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn
posters
something
having
a
wrong.
single
why
  OK.  You are at least focussing on the right question.
  The answer is not completely simple.  Bear with me.
  First, I expected that you were going to object that
Rick Decker's answer and my answer do not agree.  Certainly
they do not look the same.  In fact they do agree, as
you can see with the example I gave:  M = 15, j = 17,
T = -64, k_1 k_2 = -T j^2 = 64*189, and k_1 = 9248
and k_2 = 2.  Rick says that
    b_2 f_1 = -k_1 + T or
    b_2 f_1 = T M^2 /(-k1 + T).
  Explicitly, doing the arithmetic,
   b_2 f_1 = -9312 or
   b_2 f_1 = - 14400/(-9312) = 150/97 = 27900/19206.
  The latter result is the same as my result.
  My formula for the numerator of b_2 f_1 is
  b_2 f_1 = E / F, where
  E = (k_1 + k_2)*j^2 + k_1 (k_1 + 3k_2) + 2 (k_1)^2 k_2/j^2
      - (k_1 + k_2 + 2 j^2)*(k_1 + j^2)   and
  F = 2 (k_1 + k_2 + 2 j^2 - M^2).
  So your question is really, how do I prove that this
formula actually works ?
  It follows from an identity.
  There are several identities that one might apply here.
One that *you* believe should work for the computation of
b_2 f_1 is:
     (m - n)^2 + 4*n*m = (m + n)^2.
  Let me call this the +4 identity.
  Another related identity is:
     (m^2 - n^2)^2 + 4*m^2*n^2 = (m^2 + n^2)^2.
  This is the identity that is used to generate
Pythagorean triples.  I will call this the
Pythagorean identity.
  Neither the +4 identity nor the Pythagorean
identity apply to give the expressions that
I gave above for E and F.  There is instead another
identity:
  Let U = c^2 (c + a)(2c + a + b)
          - 2(c^2 - ab)(c^2 + ac + bc + ab).
  Let V = 4 ab(c^2 - ab)(c^2 + ac + bc + ab)^2.
  Let W = (a + b)c^2 + a(a + 3b)c^2 + 2(a^2)bc.
  Then the identity is:
     U^2 + V = W^2.
  This holds for any positive numbers a, b, and c.
  I will call this the UVW identity.
  You can check that this identity is true by
just expanding the left side and the right side and
comparing the terms.
  Now, how do you apply this identity to derive
the expressions I gave for E and F above?
  Let a = k_1, b = k_2, and
      c = M^2 = j^2 - (k_1 k_2)/j^2.
  Then you just crank it out.
  The +4 identity that you thought would work to give
the result of your theorem does not work.  The Pythagorean
identity also does not work.  The UVW identity does work.
  It is probably a new identity.  I don't see it as
particularly interesting by itself - not a publishable
result - but it might fit into a larger pattern of
identities, and it might be of interest if that is the case.
  In my view, this is the only nontrivial result to come
out of your SFT.  And it does not give any support for the
efficacy of your brand of surrogate factoring.
  I didn't manufacture it.  I derived it.  It works for all the
cases I have tried, which is quite a few.  You could have tried
it on multiple examples as well.  It is easy to program.
  No, you're just wrong about this.  The formula I gave above
works, just as Rick Decker's also works, and it does not require
factorization of M.
  No - the +4 identity that you think should apply here does
not apply.
  No - the mysteries are all solved - Rick Decker and I have solved
them differently, but they are solved.  No mysteries are left.
  You didn't have to trust me.  You could have tried it on any number
of other examples.  You could have tried to work through the algebra.
Instead you chose to do nothing and call me a liar.
and
  See above.  Anyone can work out the details.
an
  No, there are other ways to end up with a perfect square inside the
square root that do not fit the x^2 + 4M pattern.  Another identity
than the one you were thinking of applies.
trivially
  Nope, sorry, there are other possibilities than the one you have
thought of.  See above.
  Clearly there are no reasons why it's impossible.  The formulas
that I gave for E and F work.
  Nora B.
<snip, though I hated to delete the accusation that Nora
My hat's off to you, Nora. Cool work.
And, as I noted in another post (and e-mailed to you), your formula
gives a result that's equal to one of mine ( M^2 T / (-k_1 + T), in
particular).
Rick who apparently has too much free time
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
So now you're making up stuff in the factoring arena.
You're just talking about freaking difference of squares and trying to
rename it.
Not very bright Nora Baron as there are too many people who care
about the subject.
There are no new identities beyond
b^2 - a^2 = n
which is the core identity called difference of squares.
Even you aren't so silly as to believe that the cryptology community is
going to accept that some sci.math poster just showed them up by
discovering new fundamental identities.
And yes, readers of sci.crypt, such posts seem to work on sci.math with
a very odd audience.
So the Nora Baron poster is going to re-teach you difference of
squares.
It's freaking DIFFERENCE OF SQUARES.
LOL.  Oh my, laughing hard now, but it's so sad too.
ROFLOL.
So sci.crypt, one of the stars of the sci.math newsgroup is here to
tell you that the difference of squares is really the UVW identity,
which that person say is probably a new identity.
And yes, many times I've faced really bogus statements from the poster
Nora Baron which posters on sci.math would cheer!
I'd say, hey, that's nonsense.  I even gave it a name: voodoo
mathematics.
The sci.math readership does not care.
They're like freaking weird as hell and rather odd people, but
dangerous as a bunch of them got together to shoot down a paper of mine
by gang emailing the editors of that now defunct journal.
So they are journal killers as well.
I left a lot in so readers can see what I have to deal with from
posters who seem to hate algebra.
It turns out I can easily prove them wrong, without going into nearly
so much...chatter.
The difference of squares is an ancient identity, and not some new
thing discovered by Nora Baron no matter how many sci.math'ers might
come along, like Rick Decker to proclaim cool work.
What you see in
b_2 f_1 = (-(Az - 2M^2)+/- sqrt((Az - 2M^2)^2 - 4TM^2))/2
is a fundamantal block which prevents you from resolving the square
root without using some factorization of TM^2.
The triviality argument then boils down to the claim that the SFT gives
a factorization using factors
M and M, M^2 and 1,
and their negatives.
Notice that's an actual mathematical position as to where my work could
fail.
James Harris
Nonsense. b_2 f_1 = -k_i + T or TM^2 / (-k_i + T), where i = 1, 2,
k_1 k_2 = -Tj^2, T = M^2 - j^2.  You don't need to factor TM^2.
Not always, but by any reasonable measure SFT gives nontrivial factors
no more often than just trying random factors.
Rick
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
Hey, your bulb's been dim ever since I started reading your posts.
So Nora Baron has this UVW identity, and the only thing you can do is
to attack her motives? Saying she's trying to take credit for the
difference-of-squares in a modified form? Constructing a case based on
something other than the mathematics?
Did you even BOTHER to check the mathematics? (Oh, that!) Any
fifth-string mathematician would go to the trouble. I don't think you
did, because one simple derivation proves that the math is wrong.
In Maple:
U := c^2* (c + a)*(2*c + a + b) - 2*(c^2 - a*b)*(c^2 + a*c + b*c +
a*b);
V := 4 *a*b*(c^2 - a*b)*(c^2 + a*c + b*c + a*b)^2;
W := (a + b)*c^2 + a*(a + 3*b)*c^2 + 2*(a^2)*b*c;
factor(U^2 + V - W^2);
(I factored it so that it would simplify the expression.) The response
is
c^3*(c-1)*(a+b)*(2*c*a^2+4*a^2*b+c^2*a+6*c*a*b+a*c+c^2*b+b*c)
which isn't even close to zero.
Granted, this takes a bit of work using P&P (pencil and paper), but it
still can be done.
Nora Baron made a mistake, and you goofed. You could have grabbed the
ball and run it in for a touchdown. But you fumbled it.
You responded like a crank when you could have responded like a
mathematician.
You could have done the math and found out that Nora Baron was wrong.
(Granted, she might be right 99% of the time, but if you wanted proof
that she was lying, you couldn't hope for a better opportunity.)
Is it any wonder no one respects you?
Or new variants.
If the readers of sci.math are complete incompentants, then why is it
that no one believes you've solved the factoring problem?
Just like you re-teach hyperbolas.
How you turn one mistake into the plural statements is beyond me.
With the biggest subfield of Voodoo Mathematics being Surrogate
Factoring.
If I didn't care, I wouldn't be writing.
If you think a paper is so easy to kill, get rid of mine: A New Proof
of the Independence Ratio of Triangle-Free Cubic Graphs (with Robin
Thomas), Discrete Mathematics, Vol 233/1-3, pp. 233-237.
How exactly did they kill off a journal? Do you have any evidence of
this?
to the mistake and doesn't sic the CIA and FBI on me, though, because I
have a higher opinion of her than you.
I proved the equality wrong. If you could have, why didn't you? Oh,
yes, you lost your key.
Now you're putting words into her mouth. She never said she had
discovered the difference of squares. She said she discovered an
identity which doesn't look like it could be derived from the
difference of squares, one which she hadn't seen before. That's why she
called it new.
Oh, no! Rick Decker (if that's his real name) will sic the FBI and CIA
on me as well! Oh, whatever shall I do?
Resolving is not the correct terminology. Do you mean evaluating
the square root? Proving that the answer is rational? (A sci.math'er,
maybe Decker or Baron, showed that the square root DOES turn out to be
rational, something that you never bothered to check. If you had ended
up with an equation like
b_2 f_1 = sqrt(5),
there would be no way to connect rational factors, since sqrt(5) is
irrational. And he didn't need to know how to factor M to do it; all he
needed was j and k_1.)
You've never actually said how to turn the SF Theorem into an algorithm
which produces the factorization. Once again, the sci.math people have
done your work for you, evaluating Ax, Az, and (-(Az - 2M^2)+/-
sqrt((Az - 2M^2)^2 - 4TM^2))/2. No one -- not even you -- has figured
out how to take the next step.
The SF Theorem says that once you've evaluated (-(Az - 2M^2)+/-
sqrt((Az - 2M^2)^2 - 4TM^2))/2 (let's call it r for short), then the
rational factors of M can be found from the rational factors of T, by
letting
b_2 = r / f_1.
The problem is that the statement
The rational factors of M can be found from the rational factors of T,
by letting b_2 = r / f_1.
is true for ANY nonzero rational number r, regardless of how we get it.
You may as well choose r = 1 and avoid all the unnecessary
calculations.
Yeah, under Voodoo Mathematics.
     --- Christopher Heckman
P.S. If you think my criticism is harsh, you should remember that I was
one of your supporters for a long time. The last few weeks of
non-mathematical posts of yours has made me doubt that you have
something else up your sleeve.
In other words, you did yourself in.
I should note that although Nora misstated her general
result, her conclusion above is correct.
Be afrain, Prog. Be VERY afraid.
Yup, that was me.
Rick
   posting-account=45mQHQ0AAABhEGyZbLqUdwPik-gRPwwC
is
Yeah!  Who could possibly believe some sci.math poster could make any
significant advance in cryptology?  Like, say, solving the factoring
problem?
   posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn
  As a couple of people noticed here, I had a misprint -
should have been
        W = (a + b)c^3 + a(a + 3b)c^2 + 2(a^2)bc.
  ???
to
  The naming isn't the important part.  The important part
is whether it is correct.  Unlike other posters, you
clearly did not even check it numerically, let alone
algebraically.
is
  I have no idea what they are going to accept.  The identity
I gave above [minus the misprint] is correct and it was the
basis for the formula I derived for b_2 f_1.
with
  Gee, is there a Grandstand out there in sci.crypt also?
I would love to hear what they think about this!
  Oh, it's *freaking*.  Yes, I didn't take into account that
it is *freaking*.  Silly me.
  Yeah, sci.crypt Grandstand.  You gonna let me get away
with that?
poster
mine
  But, I note, you haven't.
  Nor have you even checked things numerically.
  Still terrified to do that?
might
  Really!  I wonder why it keeps working in the numerous
examples where I have tried it.  Would you like to see some
more?
gives
could
  This is a little cryptic [please, no offense sci.crypt - it's
just an expression], but I think you mean that if my formula is
right, then when you combine it with SFT it would say that you can
compute factor of M from just M and M^2.  Which is not what my
formula says at all.  You are confusing what I have shown with
what you think must be true.
  Look, this is really simple.  You may not understand what I
did, you may not like it, you may think it's wrong, you may
roll on the floor laughing at it if you want - fine with me -
but just try checking it numerically with a few dozen examples
and post the results here.  If you do, this argument is going
to go away.
  Nora B.
X-RFC2646:  Original
[Nora Baron]
[also Nora]
FWIW, I ran the corrected identity through Macsyma, and it agrees it's 
correct.  Not news to you, but James claimed that someone reporting the 
result of a symbolic math package is somehow or other less likely to lie to 

him.  LOL -- sometimes I suddenly just can't believe anyone bothers.
Too modest, yes?  I'd say the identity holds for any reals.  Perhaps it's 
that later steps in your derivation don't?
Based on James's deny-ridicule-slander-but-never-ever-do-any-work response, 

BTW, congratulations to you and Rick for cracking the radicals here!  They 
were too messy for me to even consider trying to crack.  Rick's derivation 
is so detailed I can almost believe I might have found it myself (a good 
presentation can trick me that way), but I have no idea how you dreamt up 
your approach, Nora.  That's some impressive prestidigitation. 
   posting-account=EH2x8QsAAABu84CuyjstkC4nRyQ1ZHKW
kudos to HSJ for finding the pythagorean identity;
most people just sort-of learn the pythagorean theorem, period
(and only the 2d version, alas).
--Chair Man George XOR Strep Throat @ W'gate?
http://tarpley.net/bush12.htm
http://larouchepub.com
Oh geez! Those of us who have looked at it agree that
SFT works. The problem is that it is equivalent to several pages
of proof that say that any rational number can be expressed as
the product of two rationals.
Right. Az = ((k_i + j^2)^2 - M^2 j^2) / (k_i - T), for i = 1, 2
No it doesn't. b_2 f_1 = -k_i + T or TM^2 / (-k_i + T), for i = 1, 2
That'd be me. The simplification above is correct. I've posted it once
and I'll be happy to post it again. Perhaps you missed the first one.
Maybe you haven't been keeping up with current events, but, C. Bond has
provided us with a specific numerical example with M = 15, j = 17. If
you check, you'll see that the example is consistent with what I've
written above.
For goodness sake, James, these examples are easy to verify by hand--
you don't even need your lost Mathematica key. Try it and see why the
case M = 15, j = 17, k_1 = 136 is special.  Put up or shut up, laddy
boy.
You claim to have been a programmer, so write a program to verify
your so-called factoring algorithm. I did and it shows that you get
nontrivial factors about as often as you would by picking trial
factors at random.
Rick
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
of
important.
So now you're back to that claim.
The equation
a_1 z + b_1 = f_1
comes from the derivation.
Your claim is equivalent to asserting a_1 = 1.
factors
once
has
the
Well, you may be correct, but before I'm done, you'll explain in
detail, like you're supposed to do, versus making taunting posts and
behaving childishly.
You can taunt as much as you wish Decker, but you're not doing yourself
or anyone else any favors.
You're just showing your true personality, the person you really are
behind the play-acting of civility.
At least I don't play at being civil.  I say I think you're lying scum
who can't be trusted but must be forced to explain out carefully and
objectively, as you won't without pressure like this post.
Sure, some of you may believe that you can act anyway you want in
response to me, as childishly as you want, and refuse to elaborate when
requested, but I'll get it out of you anyway, one way or another.
So put all your cards on the table now.
If not, if you wish to play social games, then this can last for
months.
As make no mistake, I'll let it last that long, drawing you out like I
just did in post after post after post to which you weird people feel
compelled to reply.
James Harris
X-RFC2646:  Original
Qutoing the whole thing in full here, for the benefit of future 
prosecutors.
[JSH]
[Rick Decker]
[JSH]
Rick already explained in detail.  He even politely offered to post it again 

for you, and gave you a ready-made excuse for requesting that he do so:
    I've posted it once and I'll be happy to post it again. Perhaps
    you missed the first one.
Part of me hopes he does (incorporating the sign correction he discovered 
later), because I checked the first half of it at the time, and that much 
was wonderfully easy to follow -- a model for clear exposition of a long, 
messy derivation.
But most of me hopes he ignores you.  You don't deserve more favors from 
Rick, and your reply speaks for itself on that point more damningly than I 
could. 
Oh, what the heck. In response to overwhelming popular demand (N = 1),
here it is for the third time.
his paper)
T = M^2 - j^2                                                [1]
k_1 and k_2 are rationals such that k_1 k_2 = -Tj^2          [2]
Ax = k_1 + k_2 + 2j^2. (James has +/- (k_1 + k_2) + 2j^2,
but that's subsumed by [2].)
Az = Ax(Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2)) / (2(Ax - M^2))  [3]
Then he shows that
(-(Az - 2M^2) +/- sqrt((Az - 2M^2)^2 -4TM^2))) / 2           [4]
will have a rational factor of M, where a rational number q = n/d
is said to contain a rational factor x of M if gcd(n, M) = x.
What follows are two results that simplify these constructions.
RESULT 1. If k_1 is rational, Az will be as well. In particular
    Az = ((k_i + j^2)^2 - j^2 M^2) / (k_i - T)  for i = 1, 2
PROOF.  First, consider the term under the radical in [3]:
    (Ax - 2j^2)^2 + 4Tj^2 = (k_1 + k_2)^2 + 4Tj^2
                          = k_1^2 + 2 k_1 k_2 + k_2^2 - 4k_1 k_2   by [2]
                          = k_1^2 - 2 k_1 k_2 + k_2^2
                          = (k_1 - k_2)^2
Thus we will have
    Az = Ax(k_1 + k_2 + 2j^2 +/- (k_1 - k_2)) / (2(Ax - M^2)
       = Ax(k_i + j^2) / (Ax - M^2), for i = 1 or 2           [5]
Consider now the denominator of [3]:
    Ax - M^2 = k_1 + k_2 + 2j^2 - M^2
             = k_1 + k_2 + j^2 - T      by [1]
Since k_2 = -Tj^2 / k_1 from [2], we have
    Ax - M^2 = (k_1 + j^2) - T (k_1 + j^2) / k_1
             = (k_1 + j^2)(k_1 - T) / k_1
and by symmetry
    Ax - M^2 = (k_2 + j^2)(k_2 - T) / k_2
Combining these two results with [5] yields
    Az = (Ax)(k_i) / (k_i - T)
       = (k_1 + k_2 + 2j^2)(k_i) / (k_i - T)
       = (k_i^2 + k_1 k_2 + 2(j^2)(k_i)) / (k_i - T)
       = (k_i^2 + 2(j^2)k_i + j^4 - j^4 - j^2T) / (k_i - T)
       = ((k_i + j^2)^2 -j^2(j^2 + T)) / (k_i - T)
       = ((k_i + j^2)^2 -j^2 M^2) / (k_i - T)   by [1]
establishing Result 1.
RESULT 2.  Expression [4] is equal to -k_i + T or TM^2 / (-k_i + T)
for i = 1, 2.
PROOF.  Consider first
    Az - 2M^2 = (((k_i + j^2)^2 - j^2 M^2) / (k_i - T)) - 2M^2
              = ((k_i^2 + 2j^2 k_i + j^4 - j^2 M^2 - 2M^2 k_i + 2TM^2)
                  / (k_i - T)
              = (k_i^2 + 2(j^2 - M^2)k_i + j^2(j^2 - M^2) + 2TM^2)
                  / (k_i - T)
              = (k_i^2 - 2T k_i + T^2 - T^2 - j^2 T + 2TM^2)
                  / (k_i - T)
              = ((k_i - T)^2 + T(-T - j^2 + 2(T + j^2))) / (k_i - T)
              = ((k_i - T)^2 + T(-M^2 + 2M^2)) / (k_i - T)
              = ((k_i - T)^2 + TM^2) / (k_i - T)
              = (k_i - T) + ((TM^2) / (k_i - T))
So the term under the radical in [4],
   (Az - 2M^2)^2 - 4TM^2 = ((k_i - T) + ((TM^2) / (k_i - T))))^2 - 4TM^2
              = (k_i - T)^2 + 2TM^2 + ((TM^2) / (k_i - T))^2 - 4TM^2
              = ((k_i - T) - ((TM^2) / (k_i - T)))^2
So [4] is equal to
    (-(k_i - T) - ((TM^2) / (k_i - T))
        +/- ((k_i - T) - ((TM^2) / (k_i - T)))) / 2
which is -k_i + T or T M^2 / (-k_i + T), establishing Result 2.
By the way, for the vanishingly small number of people who still
have any interest in all this, Nora has posted another expression
for [4]. I've demonstrated that her result is equal to one of
my four solutions above and, no, I have no desire to post it,
having already sent it to Nora.
Rick
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
[...]
Uh, he already did.  In detail.  That you choose not to read posts
even when they are referenced to you is no fault of his.
That would be your home turf, right?
How about engaging your brain?  He already explained carefully and
objectively, and not just this one time, but quite a number of times.
And each time you had to admit that he was correct in the end.  So how
does this make him _lying_ scum?  What is lying when you are simply
unable to get the truth at first attempt, because you can't be
bothered reading?
So who is the lying scum here?
He elaborated even before you requested it.
It was already here in the open, Mr. lying scum.
He did already, Mr. lying scum, and you know that.
Sine obviously, Mr. lying scum, you like playing those sort of games
even when unprompted for.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
WRONG!. If you know 'x' and you know 'M', then the term within the
parentheses is a number (in this case, an integer), and if it is positive
it *can* be found directly -- as can the square root of any positive
integer.
--
There are two things you must never attempt to prove: the unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
integers.
positive
Well, x is the difference of factors of M.
That's trivial, as anyone can see if they let
f_1 f_2 = 4M, then with
x = f_1 - f_1
sqrt((f_1 - f_2)^2 + 4M) = sqrt(f_1^2 - 2f_1 f_2 + f_2^2 + 4f_1 f_2)
which works out to
sqrt(x^2 + 4M) = +/-(f_1 + f_2)
and yes, if you know x ahead of time, yes, it works, but x is the
difference of factors, so you have to have M factored, or use trivial
factors like f_1 = M, f_2 = 1.
So, yes, I could easily realize that Nora Baron and Rick Decker were
wrong, without foolishly chasing down the path they were going, as
mathematically their approach was bogus.
If their approach could work, then the factoring problem would never
have been considered difficult, as people would just have solved
sqrt(x^2 + 4M)
and found the difference of squares just by solving it out.
So it was trivial to realize they were wrong, without much effort.
But also, I know that Nora Baron routinely screws up with basic
mathematics, so it was even easier still.
Social crap.  Some of you are in the field of mathematics with no
concept that social crap does not change what's mathematically true.
You learn to bull your way through and think that's what
mathematics is about, so you lie and repeat lies to push a position,
and on the sci.math newsgroup, if you're disagreeing with me, that
usually works.
James Harris
So what? It doesn't matter what relation 'x' has to 'M'. You said if 'x'
and 'M' were integers it was not possible to solve for  sqrt(x^2 + 4M)
without factoring 'M'. That's a preposterous statement.
--
There are two things you must never attempt to prove: the unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
   posting-account=0oMoGgwAAAA45r-W1MV-8N0QRNWHaDRI
So it was trivial to realize they were wrong, without much effort.
But also, I know that Nora Baron routinely screws up with basic
mathematics, so it was even easier still.
You are really funny. You say that people can look at Nora's math and
see how she is wrong and a liar, but just the opposite is true.
People here can follow her math and see that she is right, and that you
apparently can't do rather simple math.
You keep posting that everyone can just look at the math and know who's
right. Thats true, and its not you. Sorry :-(
That's what the And so forth was for.  Well, okay:
              _
1.0010000010010 = 1.4133725644...
              _
10.010000010010 = 1.82758612679...
Again, why are these questions significant?
Could you recheck the second?  I got 1.998..., and it's incorrect to give 
more decimal digits than that, since the number I gave is only bound up to 
V2 ^ -12 (.0156...).
In any case, you should notice I gave two distinct expansions for both 2
and sqrt 2; I could give as many more I cared; I was showing that
expansions are not unique for base sqrt 2; I believe the same happens for 
any other real but noninteger base.
-- 
------------------------
Mark   Jeffrey   Tilford
tilford@ugcs.caltech.edu
Sorry, you're right - I saw the 1s lining up with the first, and
overlooked the decimal point being in a different place.
I didn't say the expansions were unique.  Heck, they're not always unique
for base 10, either:
        _
  1 = 0.9
What I gave was the rule for (arguably) the *canonical* representation
of any number, and I believe *that* is unique for any real number, e.g.
                    _
  2 (base 10) = 100.0 (base sqrt2)            <- canonical
  2 (base 10) = 10.010000010010+ (base sqrt2) <- non-canonical
X-RFC2646:  Original
[Tim Peters, self-servingly snipping his own post so that it looks
 less insulting than it really was]
[Nora Baron]
I have a different working hypothesis for that:  James's sense of morality 
appears defined by what he can get away with, period.  If he can cheat or 
lie, no problem, he's too special to play by external rules, and his 
manifest destiny justifies shortcuts.
So he started his Usenet life reasonably civil, but as time went on he 
learned (sadly, in part because of vicious abuse directed _at_ him too) that 

what he did on Usenet had no visible effect on his real life.  The more he 
learned that he could get away with being an ass, and in oh-so-many 
ways, 
the more he acted like an ass, in oh-so-many ways.
obsessed with it at times.  If posting trash on Usenet had no real-world 
consequences, then how-oh-how could his guinea pigs on sci.math manage to 
get his paper yanked?  There have been lots of posts since then trying to 
reconcile that, see-sawing between insisting Usenet is of no importance, and 

hilarious (to everyone but him) revenge fantasies starring math society 
and increasingly sucking in specific sci.math'ers.
The other thing here is related:  his bizarre offer to compromise on the 
APF paper, agreeing to make an inconsequential change in the conclusion in 
return for sci.math'ers promising not to oppose his paper anymore.  That was 

unmistakable evidence of his morality (ok, so I can't get away with it 
as-is -- what little thing can I change, right or wrong, so I _can_ get away 

with it?) -- and also more thrashing trying to get back to his happy my 
bad Usenet behavior has no bad consequences illusion.
I think there's something to that, but it's not enough:
Yup, he has.  I was more than half serious about that part too.
I agree.  I think he's also less capable of discourse now (point, 
counterpoint, point, counterpoint, ..., he used to be able to keep that up 
for at least a few rounds, at least once a week; now every attempt at 
discussion degenerates at once).  The percentage of pure rant posts 
seems 
to be increasing too; there's not even a pretense of on-topic content in 
most stuff he posts these days.  And I can't remember the last time there 
was evidence that he'd opened a book or read a paper; there used to be. 
Closest it came this year was an announcement that he'd have to study up on 

ideals in order to find the flaws in Dedekind's work, almost instantly 
dropped in favor of mindlessly repetitious promoting of SFT again.
I don't know.  Could just be a natural consequence of increasing 
self-absorption; if James is anything in real life like he his on Usenet 
(and he may well not be at all), I've never met anyone in real life anything 

like him.  Jeez, if he was like that in real life in his Army days, he would 

have been fragged.  I did know a guy so obnoxious that his buddies tossed 
him out of a helicopter in Vietnam; he survived, and when I knew him lived 
in an abandoned oil drum near the U of W (Madison) campus.  We got along 
fine -- but he was a sweetheart compared to James's online persona. 
   posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn
morality
cheat or
  Resembles the Richard Nixon/OJ Simpson/Bill Clinton phenom - I am
special, so I can do things other people are too timid and inhibited
to do and it's OK because I am a boy wonder.
he
too) that
more he
ways,
offer
still
real-world
manage to
trying to
importance, and
society
thing, he bragged about it to his family, he had finally made
good, etc., and then it was snatched away totally against all the
normal rules of publication.  It wasn't so much that he thought the
paper was right - clearly at times some of the logic against it
penetrated for however short a time - it was the rank INJUSTICE
of it.  Journals aren't supposed to do that.  His outrage at the
injustice totally erased any chance after that that he would be
receptive to anyone pointing out flaws.  Hence complete intransigence
regarding his tautologous dead-end constant-terms-are-constant
argument.
  But as far as real-world goes - more actually than most of the
rest of us, he connects what happens on sci.math with the real
world.  Hence his visit to a prof at his old alma mater, Vander-
bilt.  Most of us would regard such a visit as quixotic to say the
least.  Hence his letters to Ken Ribet, Granville, Odlyzko, Wiles
(probably), various others.  Hence too his communications with
Ullrich's chairman and the Attorney General of Oklahoma.  Hence
letters to NSA, letters to Congressmen, probably letters to the FBI
and the President. The real world for him is a continuum.  I at least
think of sci.math as a fairly harmless hobby where most efforts are
ineffectual and most consequences are minor.  He does not.
on the
conclusion in
That was
it
get away
my
  True, a good example.  Some time ago he actually made a bet regarding
a proposed FLT proof - and he lost, and he actually paid up.  That
stung.  He has studiously avoided that ever since.
  Gotten lazier, partly.  And burned so many times.  In early times
he would state lemmas and theorems [he has never understood the
concept of a rigorous definition however].  He got burned on this
so many times that now he will make claims where people cannot
even find a statement of what he is claiming.  This is true even
for the SFT.  A lot of people don't know what the T even is.
We now largely have to guess what he is thinking and try to refute
*that*.  Plus perhaps he thinks that being cryptic and incomprehens-
ible makes him look smarter than the rest of us, who are too dumb
to understand the truths that are obvious to his superior mind.
with
infinite
he
total
that up
seems
in
there
be.
up on
instantly
  The main benefit being that Dedekind can go back to not rotating
in his grave.
  What did he say?  That he was ing terrified to test the
SFT, or some such?  This was assuredly NOT because he feared that
it would destroy the world's economy.  This was entirely because
he feared - KNEW, I think - that it would destroy the James S.
Harris ego.
  Satan would surely be more effective than that.
  Again, I just thank my lucky stars that this guy will never attain
the kind of power that, say, Adolf Hitler had.  His utter lack of an
internal moral clock would make him truly evil.
Usenet
anything
he would
tossed
lived
along
  I think he is at least consistent in real life.  He was in the Army
but I think never in combat.  The Oklahoma episode was real life.  He
seems to have some positive feelings for, and decent relationships,
with his family.  That may be as far as it goes.
  Nora B.
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
Actually, I'd not have thought it to be in him.  The way he forgets
acknowledged truths however trivial at the drop of a hat would make me
think he'd be of the opinion to have won whatever bet in the time it
would take to write out a cheque.
It appears to me that things have been going downhill since then.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
morality
cheat or
Actually, it's the opposite.
Mathematics is very rigid in that what is true is true.
Wishing and hoping don't mean a thing as to whether or not something is
mathematically true, and neither does making people happy.
As for my sense of morality, I'm driven by the need to determine the
truth.
I make lots of mistakes, but I deal with people who refuse to
acknowledge that they make mistakes as well, and they paint me as the
bad guy, when they lie.
So the lying accusations go flying round and round, when there is an
answer, from the mathematics.
he
too) that
more he
ways,
Actually, amazingly enough, sci.math has civilized a bit over the
years, as posters were more, vicious in many ways earlier.
Some of the worst offenders have simply burnt out, and not done any
posting that I've noticed, while a few have hung on.
And there have been some newbies who joined in recently, like C. Bond
and yourself.
And I suspect you will burn out soon enough based on your postings, as
you seem to be dimly understanding how this works.
I always win, as I have from the beginning.  Posters lie about that
reality and go on for a while, but you seem to be a little more
intelligent, and not interested in merely being a lackey to JSH.
I don't like sci.math in many ways, but in quite a few ways, I do own
this newsgroup more than any other person, though I don't try to own
it, and think that Usenet principles should rule.
Many of you have decided to basically become my shadows, but that's
your choice.
James Harris
   posting-account=0oMoGgwAAAA45r-W1MV-8N0QRNWHaDRI
Mathematics is very rigid in that what is true is true.
true
Wishing and hoping don't mean a thing as to whether or not something is
mathematically true, and neither does making people happy.
true
As for my sense of morality, I'm driven by the need to determine the
truth.
Wrong. Your driven by the need to be right, and you rarely are.
I make lots of mistakes, but I deal with people who refuse to
acknowledge that they make mistakes as well, and they paint me as the
bad guy, when they lie.
So the lying accusations go flying round and round, when there is an
answer, from the mathematics.
Blah blah blah.
Broken record.
I always win, as I have from the beginning.  Posters lie about that
reality and go on for a while,
If you mean that your goal is to look like an ignorant ass, then you do
win. In every other way, you are #1, the #1 loser on usenet.
I don't like sci.math in many ways, but in quite a few ways, I do own
this newsgroup more than any other person, though I don't try to own
it, and think that Usenet principles should rule.
Usenet is a place for people to discuss things that interest them. It
is true people can post just about anything they want, but the way most
people use it is to post based on the topic of the NG.
 It wasn't created as a place to demonstrate freedom of speech, it was
created for people to communicate with each other. Just because you can
say something doesn't mean you should. That's what you do when you are
part of society.
 Each newsgroup is its own little community. And usually people try and
get along with their community.
All you do is post a bunch of garbage, and accuse people of lying when
they point out that it is indeed garbage.
Most of what you post James is just a bunch of whining about people
lying and picking on you.
If you would post your ideas, then actually pay attention to what
others say in their replys, you'd be much better off. But all you do is
accuse anyone who disagrees with you, or who demonstrates the flaws and
errors in your math', of lying.
It doesn't even need to be large primes.  He can start with the product of
small primes and show how it works.  If it works there, then larger primes
can be considered.  If (when) it doesn't work with small primes, that will
give him something to look at to try to understand why it doesn't work.
-- 
--Tim Smith
   posting-account=EH2x8QsAAABu84CuyjstkC4nRyQ1ZHKW
also note that two conics intersect in at most 4 points;
you can see that with a cicle & ellipse on the same center.
thus:
b)  the hare-brained analysis of WTC7 assumes
that the substructure of the complex,
which included a large subway station,
was not affected by the two largest buildings,
falling down into it.
 --Chair Man George NAND Strep Throat @ W'gate?
http://tarpley.net/bush12.htm 
http://laroucehpub.com
He may also be at high risk of developing deep-vein thrombosis from spending 

long periods of time sitting immobile in front of his computer.
Hey that would apply to a lot of us!
X-RFC2646:  Original
Hey everyone,
    I need some help.  I'm working on a project modeling a hard disk and I
fear my probability theory is rusty.  The problem is, given that a hard
drive head is at some random location on the disk and must then move
radially for a read at another random location on the disk, what is the
average distance it will travel?  I'm ignoring for now acceleration and 
disk
rotate time.
It gets harder.  Now suppose the disk always has a choice of n random
locations to read, and it always chooses the closest one.  Now how far will
it travel on average?
    I appreciate any help,
    Dan P.
X-Proxy-User: $$rnw$j2sdg_
On Fri, 22 Apr 2005 16:56:48 -0400, Daniel
If you must post the same question to multiple
newsgroups, please cross-post. There's already
an answer in alt.math.recreational.
---
Stan Liou
X-RFC2646:  Original
In the book Speed Mathmatics Simplified on page 157 Edward Stoddard 
states...
the use of both 9's and 11's [digit sum checks] gives an absolute, 
unquestioned proof of accuracy.
I believe that is not true.
------------
The equation 3 * 9,999 results in a correct answer of 29,997. However an 
answer of 2,999,997 also results in both 9's and 11's digitsum checks 
indicating that to also be a correct answer.
3 * 9,999 = 29,997
digitsum(3) * digitsum(0) = digistum(9) {digistum(9) =  digistum(0) }
digitsum(0) =  digistum(0) //// VALID
3 * 9,999 = 2,999,997
digitsum(3) * digitsum(0) = digistum(9) {digistum(9) =  digistum(0) }
digitsum(0) =  digistum(0) //// incorrectly shows as a VALID answer
3 * 9,999 = 29,997
digitsum(3) * digitsum(0) = digistum(0)
digitsum(0) =  digistum(0) //// VALID
3 * 9,999 = 2,999,997
digitsum(3) * digitsum(0) = digistum(0)
digitsum(0) =  digistum(0) //// incorrectly shows as a VALID answer
So, did Stoddard overstate the facts in saying that using both 9's and 11's 

digitsum checks results in an absolute proof of an answers correctness?
Jeff 
   posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK
an
and 11's
correctness?
Sounds like it. Casting out 9's and 11's are checks of CONSISTENCY,
not VALIDITY.
Co9 only checks that you have the right digits, but can't tell you
if you have them in the right order. For example, Co9(1234) = 1 but
so does Co9(1243).
Co11 can only tell you that adjacent pairs of digits are in the
correct order. Although it can spot that 1243 isn't the correct
result, because Co11(1234) = 2 and Co11(1243) = 0, it will tell you
that 3412 is consistent since Co11(3412) = 2 also.
And in your examples, why would you even bother to do the Co9 and
Co11 checks when the false answer has more digits than the number
of digits in the operands?
Of course, one can do the cast out tricks in any standard positional
number system. For an exotic example, see
Slot Machine Arithmetic: Casting Out Cherries
http://members.aol.com/rotanasnem/cherries/cherries.htm
X-RFC2646:  Original
Makes sense.
I see. I hadn't figured that part out.
The way I was thinking of it is that casting out is really dividing and 
finding remainders... so if the wrong answer didn't change the remainder 
then it would show that wrong answer as valid. Since in Co9 you ignore nines 

then adding more nines wouldn't change the result... and since in Co11 two 
of the same number in a row cancel each other out then adding nn 
wouldn't 
change the result. I reasoned that adding 99 met both of those critira.
It was a contrived answer I created just because I didn't believe what I was 

reading in the book. I am sure that doing both Co9 and Co11 will most all 
the time find naturally ocurring errors, but I just didn't believe that it 
gives an absolute, unquestioned proof of accuracy.
And thank you for taking the time to answer my question.. I'm sure you can 
tell I'm not very good with math... but I am trying to improve. So thank you 

again.
Jeff.
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
The form of the surrogate factoring theorem that I typically give is
slanted towards the factoring problem and how I solved it, so I thought
I'd give a more abstract SFT, and consider the question of triviality.
My position is that if the equations are trivial, ok, but there needs
to be proof that answers some basics from algebra.
Abstracted Surrogate Factoring Theorem:
Given non-zero integer A and B, let
f_1 f_2 = A^2 (A^2 - B^2)
then
f_1 = (-(z - 2A^2)+ sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
and
f_2 = (-(z - 2A^2) - sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
and
z = x(x +/- sqrt((x - 2B^2)^2 + 4B^2 (A^2 - B^2)))/(2x - 2A^2)
and x is given by
x = +/- (g_1 - g_2) + 2B^2
where
g_1 g_2 = B^2(A^2 - B^2).
***************************************
Now notice I didn't say much about rings beyond A and B being non-zero
integers as that isn't required, but now let g_1 and g_2 be rationals,
and it follows easily enough that f_1 and f_2 are rationals, as given
before.
The issue of triviality then is the claim that with
f_1 = (-(z - 2A^2)+ sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
and
f_2 = (-(z - 2A^2) - sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
you will tend to get f_1 and f_2 such that you only have trivial
factors of A.
However, you can choose g_1 and g_2 with rationals in such a way that
you use non-trivial factors of B.
Now let A = B, and use g_1 and g_2 equal to f_1 and f_2 from before,
and you get the non-trivial factorization you used before.
That's the simple proof that the SFT cannot only give trivial
factorizations.
Now focus on
f_1 = (-(z - 2A^2)+ sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
and on the square root, and notice that the square root requires the
factorization of A^2.
If there is a full solution for f_1, then it requires the solution for
the factorization of A^2.
So there is no full solution that covers any non-zero integer A^2, as
that would be a solution to the factors of any non-zero integer.
James Harris
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
thought
triviality.
non-zero
rationals,
Well there's understandably confusion at that point, as it should
be--reverse.
That is, if you now let B equal your previous A and A equal what was B,
then you can get back out the g_1 and g_2 that you used before, as f_1
and f_2.
The other way to see it is the second form:
Abstracted Surrogate Factoring Theorem:
Given non-zero integer A and B, let
g_1 g_2 = B^2 (B^2 - A^2)
then
g_1 = (-(z - 2B^2)+ sqrt((z - 2B^2)^2 - 4B^2(B^2 - A^2)))/2
and
g_2 = (-(z - 2B^2) - sqrt((z - 2B^2)^2 - 4B^2(B^2 - A^2)))/2
and
z = x(x +/- sqrt((x - 2A^2)^2 + 4B^2 (B^2 - A^2)))/(2x - 2B^2)
and x is given by
x = +/- (f_1 - f_2) + 2A^2
where
f_1 f_2 = A^2(B^2 - A^2).
So I meant change out A and B to get the way to go back to your
original g_1 and g_2.
James Harris
If you stick your hand in your back pocket, you can hold yourself out at
arm's length, too.
--
There are two things you must never attempt to prove: the unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
was B,
f_1
at
Clearly you're not a mathematician, whether you claim to be one or not.
The reversability in the SFT is not only crucial; it's beautiful.
It shows that the SFT maps over the complex plane as well, connecting
two hyperbolas--for a geometric interpretation--as I've noted.
That means that *every* solution to
f_1 f_2 = M^2 (M^2 - j^2)
is available through the SFT, but more importantly, as the link is to
g_1 g_2 = j^2 (M^2 - j^2)
you have a problem claiming preferential factorization.
If the SFT tends to give trivial factorization for g_1 and g_2 given
non-trivial and trivial factorizations of f_1 and f_2, then you'd have
a contradiction.
Logically, it shows no preference either way over infinity, though I
have wondered if there might be grouping issues depending on which way
g_2 to f_1 and f_2, such that you might find blocks of values that tend
to give mostly trivial or mostly non-trivial factorizations over a
particular range.
Then again, the discussion is over infinity, so necessarily, any
pattern imaginable can occur, so I think maybe I was just being too
James Harris
   posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn
as
out
not.
have
  It might seem so.  At least your claim is provocative and deserves
an explanation.  Here is an analogy that explains why it might not
work the way you expect.
  Let Q+ be the set of positive rationals.  Define a function
      h(r) = r / 37.
  Now: given a positive rational r, is it likely that the
numerator of h(r) is divisible by 37 ?
  Of course there are infinitely many positive rationals, and
it is not possible to define a uniform probability distribution
on the entire set.  So the question seems meaningless.
  However I could restrict the question to rationals where
the numerator and denominator are both less than some very
large number, say, 37^1000.  Assume you choose rationals
by choosing the numerator and denominator independently
from the set of integers between 1 and 37^1000, with
equal probability.  [Choosing numerators and denominators
with this kind of restriction is certainly reasonable - there is
a practical limit to the size of numbers we can represent.]
  The numerator of r has probability 1/37 of being divisible
by 37.  The numerator of h(r) has substantially lower
probability of being divisible by 37.  In fact, the
latter probability is 1/1369.
  Now, of course h(r) has an inverse function, g(r),
defined by:
       g(r) = 37*r.
  Note that g(r) has a very high probability of having
a numerator that is divisible by 35.  It's not quite
36/37 = 97%, but it is pretty close to that.
  So here is a function, h(r), which has the property
that the numerator of h(r) has a *low* probability of
having an nontrivial factor of, say, M = 37*43.  The
inverse function g(r) produces numerators which have
a very *high* probability of having a nontrivial factor
of M = 37*43.
  True, as you note for the SFT transformation
there is a kind of symmetry, but in your case, unlike the
you mean different things when you are talking about (f_1, f_2)
and (g_1, g_2).  For (g_1, g_2), you are talking about nontrivial
factorizations of j^2 (M^2 - j^2), whereas, when you consider
(f_1, f_2), you are talking about nontrivial factorizations of
*different number*:  M^2 (M^2 - j^2).  When you understand that,
you see that the term nontrivial is fooling you - nontrivial
with respect to what?  *Different things!*  The symmetry is verbal,
not arithmetic or mathematical, and it fools you.  It doesn't
lead to the conclusion you that you want and expect.
-------------------------------------------------------------------
  Of course with the SFT or the generalized SFT, you can test
your implied claim directly.  If you were right about the symmetry,
then the SFT would give a nontrivial factorization of M = p*q
(where p and q are large primes) about 50% of the time.  So all you
numerator and denominator independently from a uniform distribution
on {1, 2, ..., M^1000} if you want), and define
       f_2 = j^2*(j^2 - M^2)/f_1,  so that
       f_1 f_2 = j^2 (j^2 - M^2),
and then compute g_1 and g_2 using the SFT, so that
       g_1 g_2 = M^2 (j^2 - M^2).
  The SFT does correctly predict that g_1 and g_2 will be
rational.  All you now have to do is see how often the
numerator of g_1 is divisible by p but not by q, or
divisible by q but not by p. Simple little program to write,
really.  If you are right, you should get a nontrivial
divisor of M about 50% of the time.
  But you don't.
  Nora B.
way
and
tend
   posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn
An obvious misprint.  Where I said:
     Note that g(r) has a very high probability of having
     a numerator that is divisible by 35.
I should have said:
     Note that g(r) has a very high probability of having
     a numerator that is divisible by 37.
N.B.
That's right.
That's right, too. Of course, you should be careful not to draw any
further conclusions after this point.
If A = B, you get f_1 = 0, a trivial factor.
That's actually the simple proof that you get ONLY trivial
factors in that case.
No it doesn't.
    f_1 = (A^2 - B^2)M^2 / (-g_1 + A^2 - B^2)  or
          (A^2 - B^2)M^2 / (g_2 + A^2 - B^2)
or the negative of those two. Notice that I haven't factored A^2.
No. See above.
Respectfully, that's bunk.
Rick
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
Once again, you're making things overly complicated. If you allow g_1
and g_2 to be negative (i.e., if you use both branches of the
hyperbola), then your formula for x will simplify to
x = g_1 - g_2 + 2 B^2.
Not necessarily; it depends on the sign of z - 2A^2. However, one of
f_1 and f_2 will be 0. You can also get this directly from the equation
Note that if A = B, then you also get of of g_1 or g_2 being zero,
since
g_1 g_2 = B^2 (A^2 - B^2) = 0.
That's correct; to get f_1 (and f_2), you only need to choose g_1,
calculate g_2, calculate x, and calculate z, in that order. This is
because g_2 only depends on A, B, and g_1; x only depends on g_1, g_2,
and B; and z only depends on x, A, and B. f_1 and f_2 will only depend
on z, A, and B.
You need to factor B^2(A^2 - B^2) but not A^2.
     --- Christopher Heckman
Demonstrate your solution to the factoring problem by factoring 15 using
your methods, and show your work.
-- 
--Tim Smith
If you've solved the factoring problem and if you've posted the solution
time and time again at several newsgroups, among which sci.crypt, why is
it that nothing happened? I mean, anyone around the world who has access
to the Internet can read your posts. Why don't they use your solution to
the factoring problem to break the RSA code?
Here's the answer: indeed, that has already happened. Right now, banks
all over the world are being victims of massive frauds, at a level never
seen before. Secret services around the world have ceased to work,
because communications aren't secure anymore. And why haven't you seen
all this on TV. Because we, at sci.math, have been able to cover it up!
Say, James, are we good, or are we good? :-)
Jose Carlos Santos
By Jove, right on...
See this?
http://www.sft.ru/
David
Excellent post.
You are good *and* you are good :-)
Dirk Vdm
Wow...you are now claiming that there exists an N such that flipping a 
coin N times *must* produce at least one tail?
-- 
--Tim Smith
For sufficiently small values of (1 - must), yes.
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
Abstracted Surrogate Factoring Theorem:
Given non-zero integer A and B, let
f_1 f_2 = A^2 (A^2 - B^2)
then
f_1 = (-(z - 2A^2)+ sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
and
f_2 = (-(z - 2A^2) - sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
and
z = x(x +/- sqrt((x - 2B^2)^2 + 4B^2 (A^2 - B^2)))/(2x - 2A^2)
and x is given by
x = +/- (g_1 - g_2) + 2B^2
where
g_1 g_2 = B^2(A^2 - B^2).
Now notice that I can reverse:
-g_1 g_2 = B^2 (B^2 - A^2)
then
g_1 = (-(z - 2B^2)+ sqrt((z - 2B^2)^2 - 4B^2(B^2 - A^2)))/2
and
g_2 = (-(z - 2B^2) - sqrt((z - 2B^2)^2 - 4B^2(B^2 - A^2)))/2
and
z = x(x +/- sqrt((x - 2A^2)^2 + 4B^2 (B^2 - A^2)))/(2x - 2B^2)
and x is given by
x = +/- (f_1 - f_2) + 2A^2
where
-f_1 f_2 = A^2(B^2 - A^2).
So you can pick g_1 and g_2, find f_1 and f_2, and then reverse using
f_1 and f_2 to find g_1 and g_2, where there just sign changes.
That proves that you can get non-trivial factorizations using the SFT,
as pick your A and B, and use non-trivial factors one way, then
reverse.
You will get your non-trivial factors back, proving that the SFT can do
non-trivial factorizations.
James Harris
Or, equivalently, f_1 = -A^2 (A^2 - B^2) / (g - (A^2 - B^2))
where g = g_1 or -g_2.
Or, equivalently, f_2 = -(g - (A^2 - B^2)).
I think you made some minor goofs in your derivation, but there's
no question that one can run the process in reverse: starting
with f_1 and f_2, one can arrive at g_1 and g_2.
That's simply not true. This generalization of SFT has exactly the same
problems as did the original: if you allow rational factors g_1 and
g_2, then you can get ANY nonzero rational for, say, f_1.  If, on
the other hand, you restrict your attention to integers, then
there is no reason you'll get useful factors at any better rate
than random choice (and if the number to be factored is the product
of two large primes, then that rate will be VERY small).
Rick
Let K = z-2A^2,
    L = A^2(A^2-B^2)
Then what you are saying is:
    f_1 f_2 = L
    f_1 = -K + sqrt(K^2-4L)/2
    f_2 = -K - sqrt(K^2-4L)/2
However, multiplying f_1 by f_2 gives:
    f_1 f_2 = K^2 - (K^2-4L)/4 = 3K^2/4 + L
This is only equal to L when K = 0.  Thus, you have produced the following
factorization of L:
    L = sqrt(-L) * sqrt(-L).
We are not impressed.
-- 
--Tim Smith
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
No, he isn't. You've misplaced the parentheses.
f_1 = (-K + sqrt(K^2-4L))/2
f_2 = (-K - sqrt(K^2-4L))/2
     --- Christopher Heckman
X-RFC2646:  Original
[JSH]
[Tim Smith]
Sorry, nope, you misread the scope to which /2 applies; i.e.,
    (-(z - 2A^2)+ sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
      ^--------^       ^--------^         ^---------^
                      ^------------------------------^
    ^-------------------------------------------------^
substitutions:
     [so sue me for dropping the clumsy underscores]
     f1 = (-K + sqrt(K^2-4L))/2
     f2 = (-K - sqrt(K^2-4L))/2
Then multiplying f1 by f2 gives (K^2-(K^2-4L))/4 = 4L/4 = L, as claimed.
   posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn
SFT,
do
  Basically what you are saying is that
  So, no matter whether (f1, f2) in the middle is trivial or
nontrivial, it gives rise to (g1, g2) which is nontrivial.
Therefore the SFT *can* give rise to nontrivial
factorizations.  This seems OK to me.  We already knew that
this was true for the original SFT.
  Though it somewhat begs the question, doesn't it?  What you
REALLY want to know is this: If you choose a factorization
(g_1, g_2) of B^2(A^2 - B^2), *what are the chances* that
(f_1, f_2) is a nontrivial factorization?
  For example, let A = 15, B = 16, g_1 = -248, g_2 = 32.
  We let x = g_1 - g_2 + 2*B^2 = 232
  Thus z = x(x + sqrt((x - 2B^2)^2 + 4B^2 (A^2 - B^2)))/(2x - 2A^2)
         = 7424, and
  f_1 = (-(z - 2A^2)+ sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
      = 1.
  Thus f_1 is a TRIVIAL factor of A^2(A^2 - B^2).
  Which proves that this more general SFT can also give rise to
TRIVIAL factorizations.
  Naively, one might think that, because of the symmetry of
this process, the chance that you will get a nontrivial
factorization is about 50%.  What I don't see here is a proof
of that.  'Nontrivial' in the applications of interest means that
when A = M = p*q, where p and q are primes, then the numerator of
f_1 is divisible by one of p or q but not both.  Since when p and
q are large, most integers are not divisible by either p or q,
one might expect that most factorizations are 'trivial', if the
theorem here does not perform any better than the random-gcd
approach.  And there is no hint here that it does.
  It would seem here that Mr. Harris, having totally failed to
prove anything of interest with his original theorem, has now
decided to branch out to a more general but equally worthless
theorem.  
  Nora B.
What would be interesting to me (although I don't know how easy it
would be - certainly difficult for me in particular) is to define
what sorts of additional restrictions could be put on the fs and gs
to increase the likelihood of non-trivial factorizations.
However, considering I can't think of any reason offhand why such
restrictions should change the likelihood in the first place, I
will probably only spend a little while thinking about it.
Matt
   posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn
using
can
  It may be that some form of surrogate factoring produces
candidate factors which have a higher-than-random chance of
factoring.  What Harris is doing is trying to relate one of
his favorite kinds of factoring (factoring of quadratics)
to the general factoring problem.  He defines T as partially
dependent on his factoring target M, then factors T, then
does some algebra and ends up with candidate factors for M.
There are lots of ways to approach this.  Harris's approach
does not depend on any particular insight.  He appears to
think that simply because T is related to M, the factors
of T should be related to the factors of M.  But there is
no rationale.  He just claims it ought to work and then sits
back and waits for us to show it doesn't.  This is even
worse than the usual Harris behavior of the past because this
time he is refusing to even test it (in mortal terror of
what the result might do to the world's economy, supposedly) and
doesn't believe us when we test it either.
  He has this belief that he has special insight and
good.  Never mind that this special insight has been
proven wrong dozens of times.
  Methods which help a little bit do so by narrowing down
searches.  Suppose you want to factor a large integer M
which ends in the digit 1.  You could try dividing M by
every integer up to sqrt(M).  But you can speed things up
substantially by not bothering to test numbers that end in
0, 2, 4, 6, 8, or 5.  In other words confine your searching
to arithmetic progressions of the form 10n + 1, 10n + 3,
10n+7, and 10n + 9.  This cuts the searching to about 40%
base 30, then you could eliminate numbers that end in 0, 2,
3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20,
21, 22, 24, 25, 26, 27, and 28. The only
numbers you need to think about are 1, 7, 11, 13, 17,
19, 23, and 29, i.e., only numbers in 8 arithmetic
progressions, i.e., only about 27% of the numbers below
sqrt(M).  By using, say, the first million primes instead of
2, 3, and 5, you should be able to speed up the search
substantially (though not nearly enough for RSA-cracking
purposes) because the subset of numbers in which you
search is considerably smaller than before.  However here
at least you can see the rationale, i.e., you can see why
the restriction makes sense.
  What Harris was proposing was also a form of restricted
search: try only candidate factors which were related to
factors of his surrogate, T.  The problem was, there wasn't
any explicit rationale for the restriction.  Harris's core
reason for thinking it might work seemed to be, Because
I thought of it, and I am a certified Boy Wonder.
  I would not deny that some form of surrogate factoring
might work, but a lot more insight is going to be needed
to relate the factors of some function of M to those of M itself
- a lot more than Harris has come up with.  It appears to
me that some of the subsets defined by Harris's process -
e.g., numbers based on taking the + sign in front of
all his square root expressions - end up being WORSE for
finding nontrivial factors than randomly choosing numbers less
than sqrt(M).  It's conceivable that numbers based on using
-'s in front of one or more of the square roots could
have a BETTER than random chance of producing nontrivial
factors.  Maybe.  If so, I think it would not be a real
strong effect, and real strong effects are needed to make a
difference in this.
  Nora B.
   posting-account=EH2x8QsAAABu84CuyjstkC4nRyQ1ZHKW
not only should there be some sort of hypothesis,
as to how a particular method should prove to be better
than random factoring, but
how can it be that all approaches to factoring, are not
in some sense surrogatory?
--Chair Man George XOR Strep Throat @ W'gate?
http://tarpley.net/bush12.htm
http://larouchepub.com
Another method that can do non-trivial factorizations of a natural
number M is this one: test whether or not M is a multiple of p, for
each prime number p from 2 to the square root of M.
Care to explain why does your method solve the factoring problem
better than the one I gave?
Jose Carlos Santos
: That proves that you can get non-trivial factorizations using the SFT,
: as pick your A and B, and use non-trivial factors one way, then
: reverse.
Please demonstrate using an example.
Justin
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
SFT,
Why?  I suggest you follow the algebraic proof.  What does a single
example tell you anyway beyond absolute proof?
James Harris
: Why?  I suggest you follow the algebraic proof.  What does a single
: example tell you anyway beyond absolute proof?
It might help me see how g_1 and g_2 maybe be chosen.  As you have pointed 
out and I am not debating, the choice of these is critical.  If you could 
please demonstrate such a choice it would be appreciated.
Justin
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
pointed
could
They're rational.  So it's easy and your request is strange.
Pick any rationals that fit.
James Harris
The absolute lack of a single example tells much more.
I can't work out how to use your algebraic proof as an algorithm to 
factor (for example) 15. I'm obviously missing something here. Why don't 
you humour me and Justin and help us see your method?
Mark Atherton
   posting-account=5oD5og0AAABnw3d8lGyFKtwNb2noovfx
Could someone please explain to me how to find the annihilator for a
function. I have tried to look online for concrete examples but cannot
find anything. I am unsure how to find the annihilator for equations
like e^sqrt(2*x) and 5^x and e^(2*x)*cos(3x)-x sin(3x). But I do think
I get that x^10 has annihilator D^11 and x^10*e^-x has annihilator
(D-1)^11, but I am not sure if this is right could someone please help
me with this topic or direct me to a really good website to help me
(D - 1)e^x = 0
(D - 1/sqr 2x)e^(sqr 2x) = 0
On April 22, 2005, the Dilbert comic strip contained what I think is a
monumental math error, but I'm not sure.
And even if I am right I'm not sure how large that error is.
Please take a look at the part of my web page that starts at
http://barelybad.com/fpdilbert.htm#permutations and let me know whether
I've said anything wrong.  Note that there's a link to the Excel
spreadsheet I used, if that's of any help.
right.
--Johnny
johnnyg aattssiiggnn barelybad.com
http://barelybad.com
X-RFC2646:  Original
Dogbert does not quite disagree, but Dilburt would. The way you get 625 is 
if 
you have 25 for the first fix and 25 for the second - allowing you to repeat 

the first fix. I believe the answer you are looking for is 25!. (Or the 
permutaions of 25 things taken all together.) You have 25 choices for the 
first, 24 for the second, 23 for the third, etc.
Bill 
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
Dogbert didn't say to try them all in pairs; he said ALL COMBINATIONS.
(I posted a thread about this at alt.comics.dilbert.) That means doing
just one (with 25 ways), doing two of them (300 ways to do two things),
doing three of them, etc.
The answer turns out to be 2^25 - 1 = 33,554,431, the -1 being because
doing nothing will not solve the problem.
If the order matters, then the number of combinations of fixes is
P(25,25) + P(25,24), P(25,23) + ... + P(25,1)
     = 25! (1/0! + 1/1! + ... + 1/24!)
     = 42,163,840,398,198,058,854,693,625.
And, as I noted in the other thread, this doesn't include the
possibility of doing one of the fixes more than once.
     --- Christopher Heckman
things),
April 24, 2005
Mr. Heckman,
As you know, we ended up with the same math result assuming the order of
the fixes makes a difference.
Incidentally, don't you agree it is only reasonable to assume it does?
After all, there's a difference between Perform Fix #16, then perform
Fix #22 and Perform Fix #22, then perform Fix #16.
As you might have seen from my spreadsheet, I got only
42,163,840,398,198,100,000,000,000.  May I know how you arrived at all
those extra significant digits?
Also, was your second formula, where you multiply the series by 25!,
supposed to run from 1/0! to 1/24! or to 1/25! ?
Perhaps on the same topic, is it always true that P(N,N) = P(N,N-1) ?
For example, my spreadsheet says P(25,25) = P(25,24) =
15,511,210,043,331,000,000,000,000.
I look forward to your response.
--Johnny
In my own experience I have had several instances where applying a fix has
not solved the problem, but then removing it again *has* solved it.
Illogical perhaps, but that's how M*******t W*****s is.
-- 
Paul Townsend
Pair them off into threes
Interchange the alphabetic letter groups to reply
   posting-account=YNS_mw0AAACdToALz-_WABH4mN-g5uKH
COMBINATIONS.
doing
things),
because
Well, as long as we're speculating on what he meant, maybe you don't
have to restart whenever you finish with one combination. So perhaps
doing the following fixes:
ABCA
counts for doing the combinations ABC and BCA, as well as AB, BC, CA,
A, B, and C. If that's the case, then we only need to try ~25! =
1.551e25 fixes total, which is only 1.533% as many as if we did each
combination individually. Doing it this way, you only have to do ONE
combination of fixes, although that combination is, admittedly,
1.551e25 fixes long.
That misses (among others) the combination of BC *without* A which may well
be the required fix.
-- 
Paul Townsend
Pair them off into threes
Interchange the alphabetic letter groups to reply
Night, your speculation is interesting, and it's one I hadn't thought
of.  May I quote you on that page?
--Johnny
johnnyg aattsssiggn b_a_r_e_l_y_b_a_d_DOT_c_o_m
http://barelybad.com
X-RFC2646:  Original
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
  + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
This post also appears on the alt.math.recreational newsgroup.
Please learn to use cross-posting when posting something to more
than one newsgroup.
Aristotle Polonium
  + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
a
whether
this
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+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Aristotle Polonium,
I have no idea what you're trying to say, but I gather I've broken some
rule of netiquette.
If it is your duty to send this message frequently to ignorant
transgressors, you might want to consider re-wording it so it isn't
quite so rude and so it's a lot more helpful.
--Johnny
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
The surrogate factoring theorem is mathematics at its most beautiful,
as it's compact, highly symmetrical, with one interesting asymmetry,
and it works over infinity, giving us a view into the infinite that
hasn't been there before.
Abstracted Surrogate Factoring Theorem:
Given non-zero integer A and B, let
f_1 f_2 = A^2 (A^2 - B^2)
then
f_1 = (-(z - 2A^2)+ sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
and
f_2 = (-(z - 2A^2) - sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
and
z = x(x +/- sqrt((x - 2B^2)^2 + 4B^2 (A^2 - B^2)))/(2x - 2A^2)
and x is given by
x = +/- (g_1 - g_2) + 2B^2
where
g_1 g_2 = B^2(A^2 - B^2).
*************************
Notice that you start with
f_1 f_2 = A^2 (A^2 - B^2)
and end with
g_1 g_2 = B^2(A^2 - B^2)
so there's a high degree of symmetry, where the only way you can say
there is a direction is by the sign of A^2 - B^2.
What the SFT is, is a connection betwee factorizations that covers the
complex plane.
To bring it down a bit you can use rational f_1 and f_2 or g_2 and g_2.
Even from a pure math standpoint the SFT is beautiful and very
important, and there's just no way to say what it might eventually tell
us about deep properties of numbers.
However, first I have to get past social realities which have to do
with practical matters, and first off is the mystery of how a result I
say is very important can be attacked by posters from math newsgroups.
And hey, what about the factoring problem?
Well, there are a lot of reasons why the SFT should lead to practical
factoring  algorithms that would allow factoring of large integers.
If the world weren't rather strange, by now I should at least be with
some research group talking about my number theory research.
But instead I'm still labeled a crank on the Internet, with people
challenging mathematics that's been solidly proven to be correct, which
can lead to a breakdown of Internet security.
What gives?
Well, a lot depends on some ideas which I've shown to be lacking, but
the difference is so great that some people may have decided that if
that's true, they don't want to deal with reality anymore.
Like, to them, if I'm right, not only is life a total sham and waste,
but that reality is a sick joke and they don't want any part of it, so
they just wander of into their own little world.
Trouble is, other people don't realize it, so they depend on these
people who no longer are following the same rules, and those people
refuse to acknowledge my research.
Now they've been doing it for years with results that aren't as
practical, so to them, it's probably just more of the same with the
SFT.
But for some of you, it's not, as with the SFT, some of my number
theory research is finally fully practical in a way that may force
itself upon you.
As we speak, if the idea has been exploited, and if Internet security
works the way I think it does, then your public keys can be cracked,
and even communications that your browser says are secure, may not be.
People may be casually reading what you do over the Internet, getting
your passwords, getting your credit card information, or anything else
you choose to put out there, without regard to what your browser tells
you about how secure your communication is.
If that is happening now, then it is probably just a matter of time
before it becomes public knowledge no matter what mathematicians do.
I hope it's not happening.  I hope that there's something I'm missing
which means the SFT can't lead to that kind of practical application.
But if I'm not wrong, how long can it last that no one acknowledges it
publicly?
Maybe a few months.
I haven't talked about news items as I can't be sure, but it seems to
me that there are a lot of security related issues in the news lately,
from an attempt at stealing money from a bank, where the crooks had all
the passwords, to hacking of Paris Hilton's cellphone, to the repeated
stories about lost customer data.
But for each there is an explanation given.
I don't know.  Could be an amazing coincidence, you know?
So we wait.  If a couple of months pass and nothing happens, then I
say, breathe a sigh of relief and wander off.
I'll probably be on to other--impractical--math research by then, and
maybe some day I'll re-visit to try and see what I missed and how I got
it so wrong.
James Harris
Yes, there is a way! Here it is: The SFT will never tell us anyting
deep about numbers! See how easy it is?
It's because it's not important at all, crank.
That's funny, I was about to ask that same question.
Name one.
*Then* the world would be rather strange indeed! :-)
You have lots and lots of experience with that.
Not probably, certainly.
Bad news for you: it hasn't!
My guess is that that you're missing something that should be located
between your ears.
That's funny, actually. Me too, if I were able to crack big prime
numbers, the first thing that I would do would be to hack Paris Hilton's
cellphone. The second one, of course, would be to steal a few billion
dollars from the world's largest banks.
What? Me worried?
Don't bother, I'll tell you: the same way you got everyting else wrong.
Jose Carlos Santos
   posting-account=v6xdSg0AAAD63tq-yRAAhvNtO5tjUP7G
Personally I think all of you who reply to JSH are the cranks.  He
posts once and 15 people reply ... every time.
Just get over yourself and realize that JSH does what he does TO START
Basically what I'm trying to suggest is that you act smarter than
JSH... right now he has you all by the balls.  He says something stupid
like
I've discovered the jacobian connection to surrogate complex factoring
rationals and you feel compelled to reply thereby further polluting
usenet and feeding his massively abusive ego.
Can't you figure it out by now?  All he wants is people to reply to
him.  Stop replying and he'll go away.
Tom
You so obviously know so little mathematics -- how can you claim to know 
what's been there before?
James, I've never seen you prove that your mapping is bijective (look it 
up). The important part would be surjectivety (look it up).
Beauty is in the eye of the beholder, so one can't argue with you there, 
but it is very funny when you say that it's very important. You have 
no clue, other than learning a few catchwords, as to what's been done 
and what's important in the field. Your saying that the SFT is important 
does not make it so. It's very funny..., I've seen you try to explain 
certain points, then when challenged or refuted, you come back and say 
that you have proven it without a doubt. I never know where you might 
have proved it. All the posts I've read of yours have wishy-washy 
handwaving in them. If this is the proof you speak of, again, you're way 
out of your league.
You have no idea what's been done and what's important, so you won't be 
extended an offer to join any serious research groups. But that 
shouldn't stop you from doing research on your own, or finding 
like-minded people as yourself to collaborate with -- by saying 
like-minded I'm saying aim lower, not higher. You are less than (much, 
much, much, less than) Euler, Gauss, Dedekind, etc. You are not greater 
or equal to these giants.
James, your stoooopid is showing again! How the hell can public keys be 
cracked. Jeez man, pick up a book!
I don't care enough about you to type everything that's wrong with the 
connections you're trying to make here, but it is damn funny -- keep em 
comming.
Swarmy McDickelson
No.
Even if you are right, no one will *ever* acknowledge it.
This is not just a sci.math thing, you know.
This is a well-orchestrated global conspiracy against you.
How long will it take before you finaly realize that?
Dirk Vdm
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
You have not even been able to factor 15 using it, let alone large
integers.
You have not even been able to factor 15 using it.  That's why you are
called a crank in the light of your claims.
Like proclaiming some breakthrough advance in factoring that is not
even able to factor 15, seemingly?
Prove that it is practical.  Factor some number.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum