RE: Yet another factoring puzzle
- View Source--- In email@example.com, <david.broadhurst@...> wrote:>But now, J-L, are you able to explain my comment:
>> What is remarkable about this exercise is that it can be>Can you quantify my suprise?
>> solved so quickly. Heuristically, that was not to be expected.
Well, I would say that for n >= 600, both algebraic factors are more than 415 digits and trial factoring them with pfgw up to 20000000 with -d option you may remove a 20000000-smooth composite factor of say 10 or 15 digits. This leave 2 factors of more than 400 digits. The "probability" for one such factor to be prime is roughly 1/l(10^400), that is about 0.001. So for both factors to be prime the probability is about 0.000001.
Am I right ?
- View SourceJean-Louis Carton wrote:
> So for both factors to be prime the probability is about 0.000001.We seek complete factorization of both of the cofactors
25^k-4*5^k-1 and 25^k+4*5^k-1
for some k > 300, where each has more than
400 decimal digits. We had better avoid
the case k = 0 mod 3, where each cofactor
has an algebraic factorization, which is
here a distinct disadvantage.
Suppose that we sieve out primes to depth d
and hope for what is left to yield a
a pair of PRPs as here, with k = 304:
(25^304-4*5^304-1)/(2^2*11*29*1289*1759*9511*27851) is 3-PRP!
(25^304+4*5^304-1)/(2^2*1439*17390951) is 3-PRP!
The probability for success for a single value of k
coprime to 3 is of order
Setting p = 2*10^7 and summing over /all/ k > 300,
the probability that Exercise 6 has /no/ solution is
exp(-2/3*(exp(Euler)*log(2*10^7)/log(25))^2/301) =~ 83%
In fact it has a solution almost immediately, at k = 304.
- View Source
On Mon, 9/16/13, djbroadhurst wrote:
--- "djbroadhurst" <d.broadhurst@...> wrote:
> > Exercise 6: Find the complete factorization of F(n) for at
> > least one even integer n > 600.
> As far as I can tell, no-one (apart from the setter)
> yet solved Exercise 6, which can be done in less
> than 2 minutes, using OpenPFGW. What is remarkable
> about this exercise is that it can be solved so
> quickly. Heuristically, that was not to be expected.
But is that more or less remarkable than the expectation of
any one of Phil Taylor's darts landing in the region around
where it actually landed? You only chose that target after
the arrow had landed, I'm sure.
How many mathematical diversions have you looked at
via the medium of numerical computation? How many of
them would you expect to be remarkably easier than
expected to solve? Probably a non-zero answer. Don't
be surprised that one particular example was one.
Knowing what he's trying to say, even if he's not getting
it across clearly,
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- View Source
---In firstname.lastname@example.org, <thefatphil@...> wrote:
> You only chose that target after
> the arrow had landed, I'm sure.
It happened thus:
1) I determined to factorize F(n)=((n^2-9)/4)^2-5 for
n <= 300, completely. As later shown in "factordb", I succeeded.
2) Meanwhile I ran OpenPFGW on n in [301,600], hoping for a
quick outlier and found none.
3) I estimated the probability of an easily discoverable
completely factorization for n>600 and found it to be small.
4) Recalling how I had once been caught out before by
a "probably no more" heuristic, I set a lone process running on
n in [601, 10000] so as not to be caught out again by Jens.
5) When I later looked and pfgw.log, it had found a hit at
So yes, Phil, you are quite correct that the puzzle was set
after this finding. However the heuristic that I gave was
made prior to my discovery, else I would not have said that
I was surprised.
The point that you are making (I think) is that I do such
expsriments often and only notice when the result is unexpected.
I don't tell folk about all the boring times when a negative
heuristic is borne out by a null result. That is the selection
David (guilty of not boring folk with what is routine)