ifactor problem with new update

[Peter Van Roy]

With the new update, I get the following (occurs with various inputs, the following is just one example):

ifactor(2ˆ101+5)

"Quadratic sieve failure, perhaps number too large Error: Bad Argument Value"

On the TI-nspire, factor(2ˆ101+5) works just fine (and the factorization it gives is correct, as a multiplication proves).

Peter

PS: Is there an official bug submission site for the Prime? I hate to create a new thread whenever I find a bug.

[Francois Lanciault]

This is strange because ifactor(2^128+1) works, even if the number to factorize is bigger.

F.

[toml_12953]

(05-30-2014 08:05 PM)Peter Van Roy Wrote:  With the new update, I get the following (occurs with various inputs, the following is just one example):

ifactor(2ˆ101+5)

"Quadratic sieve failure, perhaps number too large Error: Bad Argument Value"

ifactor(2^102+5) works fine. I wonder what the bug is caused by.

[Helge Gabert]

This might not be a bug. Depends on how the quadratic sieve factoring algorithm was implemented in the Prime, what the cutoff is for the running time, what primes are selected to be in the factor base, etc. Not all integers of comparable length can be expected to be factored in the same amount of time.

[DMaier]

By the way:

ifactor(2^101+5)=47*1721*253719757*123536856746739823 (this fails)

ifactor(2^102+5)=3*85060127*19870659260881492319689

ifactor(2^128+1)=59649589127497217*5704689200685129054721



It doesn't fail with either:

ifactor(47*1721*253719757)

or

ifactor(47*1721*123536856746739823)

but the following does fail:

ifactor(253719757*123536856746739823)

[parisse]

I'm going to increase the number of tries for Pollard-rho.

[rprosperi]

(05-31-2014 06:52 AM)parisse Wrote:  I'm going to increase the number of tries for Pollard-rho.

Oh, of course! The old Pollard-rho count wasn't set right. I should have thought of that....

You amaze me Bernard. In a good way.