Re: How to do quickest
- To: mathgroup at smc.vnet.net
- Subject: [mg123080] Re: How to do quickest
- From: Artur <grafix at csl.pl>
- Date: Tue, 22 Nov 2011 07:23:40 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <febdfdb5-c1cc-472c-afd6-e1768e8723e1@wrimail02.wolfram.com> <201111221035.FAA28035@smc.vnet.net>
- Reply-to: grafix at csl.pl
P.S.
Generally almost all polynomials on m are of 12 degreen. Only known
exceptions are order 2 (only 23 such cases for x from 1 to 1248412) ,
but extremaly intersting will be to find cases of degree 3,4 and 6
(degree have to divided 12).
best wishes
Artur
W dniu 2011-11-22 11:35, Artur pisze:
> Dear Daniel and Rest,
> Thank Your very much for improove my procedure!
> m have simplest form but can be + or - n is little bit more complicated
> but have only one sign
> Generally because I don't find another cases as m is root of quadratic
> polynopmial or 12 degree polynomial when non zero coefficient occured in
> even powers of variable (that mean is square root of 6 degree polynomial
> root) but for me aren't interesting this last cases we can uses
> IntegerQ[m^2]
>
> Best wishes
> Artur
>
>
> W dniu 2011-11-21 23:18, Daniel Lichtblau pisze:
>> ----- Original Message -----
>>> From: "Artur"<grafix at csl.pl>
>>> To: mathgroup at smc.vnet.net
>>> Sent: Monday, November 21, 2011 3:29:38 AM
>>> Subject: How to do quickest
>>>
>>> Dear Mathematica Gurus,
>>> How to do quickest following procedure (which is very slowly):
>>>
>>> qq = {}; Do[y = Round[Sqrt[x^3]];
>>> If[(x^3 - y^2) != 0,
>>> kk = m /. Solve[{4 m^2 + 6 m n + n^2 ==
>>> x, (19 m^2 + 9 m n + n^2) Sqrt[m^2 + n^2] == y}, {m,
>>> n}][[1]];
>>> ll = CoefficientList[MinimalPolynomial[kk][[1]], #1];
>>> lll = Length[ll];
>>> If[lll< 12, Print[{x/(x^3 - y^2)^2, kk, x, y, x^3 - y^2}];
>>> If[Length[ll] == 3, Print[{kk, x, y}]]]], {x, 2, 1000000}];
>>> qq
>>>
>>>
>>> (*Best wishes Artur*)
>> This is somewhat faster,
>>
>> rootpoly =
>> m /. First[
>> Solve[{4 m^2 + 6 m n + n^2 -
>> xx, (19 m^2 + 9 m n + n^2)^2 (m^2 + n^2) - yy^2} == 0, {m, n}]]
>>
>> -Sqrt[Root[
>> xx^6 - 2*xx^3*yy^2 + yy^4 + (270*xx^3 - 270*yy^2)*#1^3 -
>> 2916*xx*#1^5 +
>> 3645*#1^6& , 1]]
>>
>> I only had patience to go to 10^3.
>>
>> nn = 3;
>> Clear[x, y, kk, lll]
>> Timing[Do[
>> If[! IntegerQ[Sqrt[x]],
>> y = Round[Sqrt[x^3]];
>> kk = rootpoly /. {xx -> x, yy -> y};
>> lll = Exponent[MinimalPolynomial[kk][[1]], #1];
>> If[lll< 12, Print[{lll, x/(x^3 - y^2)^2, kk, x, y, x^3 - y^2}]];
>> ], {x, 2, 10^nn}];]
>>
>> Faster still would be to figure out a priori conditions for which rootpoly is of degree less than 12. Then just test {x,y} pairs to see which ones qualify. Other possibilities for this might include working with
>>
>> Resultant[
>> 4 m^2 + 6 m n + n^2 - x, (19 m^2 + 9 m n + n^2)^2 (m^2 + n^2) -
>> y^2, n]
>>
>> Out[9]= 3645*m^12 - 2916*m^10*x + 270*m^6*x^3 + x^6 - 270*m^6*y^2 -
>> 2*x^3*y^2 + y^4
>>
>> Specifically one wants to know for which {x,y} pairs in the loop this will factor.
>>
>> Daniel Lichtblau
>> Wolfram Research
>>
>>
>
- References:
- Re: How to do quickest
- From: Artur <grafix@csl.pl>
- Re: How to do quickest