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Re: How to do quickest

  • To: mathgroup at smc.vnet.net
  • Subject: [mg116481] Re: How to do quickest
  • From: Artur <grafix at csl.pl>
  • Date: Wed, 16 Feb 2011 06:45:02 -0500 (EST)

Dear David
MY procedure count types of different factorizations given polynomial 
modulo successive primes which don't divided discriminant
(also stored sets of primes of each kind of factorizatios, these 
poddible factorizations are identical as IntegerPartitions
Best wishes
Artur

W dniu 2011-02-16 10:34, David Bailey pisze:
> On 15/02/2011 11:33, Artur wrote:
>> Dear Mathematica Gurus,
>> How to do following procedure quickest?
>> (*start*)
>> pol = x^8 - x - 1; nn = Length[CoefficientList[pol, x]] - 1; If[
>>    IrreduciblePolynomialQ[pol], pp = IntegerPartitions[nn]; aa = {};
>>    Do[AppendTo[aa, {}], {n, 1, Length[pp]}]; Print[aa];
>>    ff = FactorInteger[Discriminant[pol, x]]; bb = {};
>>    Do[AppendTo[bb, ff[[n]][[1]]], {n, 1, Length[ff]}]; n = 1; cn = 0;
>>    While[cn<   nn!, p = Prime[n];
>>     If[MemberQ[bb, p], , cn = cn + 1;
>>      kk = FactorList[pol, Modulus ->   p]; ww = {};
>>      Do[cc = Length[CoefficientList[kk[[m]][[1]], x]];
>>       AppendTo[ww, cc - 1], {m, 2, Length[kk]}]; ww = Reverse[ww];
>>      pos = Position[pp, ww][[1]][[1]]; AppendTo[aa[[pos]], Prime[n]]];
>>     n++]]; Table[Length[aa[[m]]], {m, 1, Length[aa]}]
>> (*end*)
>> Best wishes
>> Artur
>>
> It might be better to describe what you want Mathematica to do, rather
> than leave people to decipher the code you have written!
>
> David Bailey
> http://www.dbaileyconsultancy.co.uk
>
>
>


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