[sage-devel] Re: Question to Martin Albrecht

2008-11-12 Thread Michel
Thanks! I will think about your suggestions. I asked you since I know you are a cryptographer. I assume cryptographers know all about reversing hashes! Of course the question was in fact addressed to the whole list. Thanks again, Michel On Nov 12, 4:30 pm, Martin Albrecht <[EMAIL PROTECTED]

[sage-devel] Re: Question to Martin Albrecht

2008-11-12 Thread Martin Albrecht
On Wednesday 12 November 2008, Michel wrote: > Ok that hint was not sufficient. > > What function(s) from the m4ri library would allow me to attack this > problem? > How do I express that the solution of the system is supposed to be > sparse > (which is a non-linear condition)? > As far as I can t

[sage-devel] Re: Question to Martin Albrecht

2008-11-12 Thread Michel
Ok that hint was not sufficient. What function(s) from the m4ri library would allow me to attack this problem? How do I express that the solution of the system is supposed to be sparse (which is a non-linear condition)? As far as I can tell there is no real reference manual for m4ri. Regards, Mi

[sage-devel] Re: Question to Martin Albrecht

2008-11-12 Thread Michael Brickenstein
http://m4ri.sagemath.org/performance.html On 12 Nov., 11:07, Michel <[EMAIL PROTECTED]> wrote: > > If the equations are really linear, then it's trivial. > > Ah, can you tell me more? > > Regards, > Michel --~--~-~--~~~---~--~~ To post to this group, send email to

[sage-devel] Re: Question to Martin Albrecht

2008-11-12 Thread Michel
> If the equations are really linear, then it's trivial. > > Ah, can you tell me more? Regards, Michel --~--~-~--~~~---~--~~ To post to this group, send email to sage-devel@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more

[sage-devel] Re: Question to Martin Albrecht

2008-11-12 Thread Michael Brickenstein
Hi! What do you mean by the word "linear": x+y+z+1 (which is the normally considered as linear and inhomogeneous). or something like deg bound 1 per variable: x*y*z+1 If the equations are really linear, then it's trivial. On 12 Nov., 09:03, Michel <[EMAIL PROTECTED]> wrote: > Hmm this question i

[sage-devel] Re: Question to Martin Albrecht

2008-11-12 Thread Michel
Hmm this question is going to have (too) many solutions: Take any solution with 781-64 arbitrarily assigned zeros. Chances are big that this solution has at most 38 ones. What if we replace 38 by a smaller number. Say 20? Michel On Nov 12, 8:43 am, Michel <[EMAIL PROTECTED]> wrote: > Hi, > >