       Re: How to do quickest

• To: mathgroup at smc.vnet.net
• Subject: [mg123055] Re: How to do quickest
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Tue, 22 Nov 2011 05:33:31 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201111210929.EAA14830@smc.vnet.net>

```Sorry, I had a couple of typos. Correct is:

Clear[x, y]
kk = m /.
First@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}];
Timing[qq = First@Last@Reap@Do[y = x^3 // Sqrt // Round;
(x^3 - y^2) !=
0 &&
(len =
Length@CoefficientList[MinimalPolynomial[kk][z], z]) < 12 &&
Sow@{x, y, kk, len}, {x, 2, 3000}]]

{13.5446, {{1942, 85580, -3 Sqrt[2/5], 3}, {2878, 154396, -Sqrt,
3}}}

and

Clear[x, y]
kk = m /.
First@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}];
Timing[qq =
First@Last@
Reap@Do[SquareFreeQ@x && (y = x^3 // Sqrt // Round; True) && (
len = Length@CoefficientList[MinimalPolynomial[kk][z], z]) <
12 && Sow@{x, y, kk, len}, {x, 2, 3000}]]

{8.47326, {{1942, 85580, -3 Sqrt[2/5], 3}, {2878, 154396, -Sqrt,
3}}}

Bobby

On Mon, 21 Nov 2011 16:28:08 -0600, DrMajorBob <btreat1 at austin.rr.com>
wrote:

> Here's your code timed with an upper limit of 3000:
>
> Timing[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}][];
>     ll = CoefficientList[MinimalPolynomial[kk][], #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, 3000}]]
>
> {971/1377495072,3 Sqrt[2/5],1942,85580,52488}
>
> {3 Sqrt[2/5],1942,85580}
>
> {1439/117596448,Sqrt,2878,154396,15336}
>
> {Sqrt,2878,154396}
>
> {187.257, Null}
>
> This is better:
>
> Clear[x, y]
> kk = m /.
>     First@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}];
> Timing[qq = First@Last@Reap@Do[y = x^3 // Sqrt // Round;
>        (x^3 - y^2) != 0 &&
>
>         Length@CoefficientList[MinimalPolynomial[kk][z], z] < 12 &&
>         Sow@{x, y, kk, len}, {x, 2, 3000}]]
>
> {14.0493, {{1942, 85580, -3 Sqrt[2/5], 13}, {2878, 154396, -Sqrt,
>     13}}}
>
> And this, even better:
>
> Clear[x, y]
> kk = m /.
>     First@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}];
> Timing[qq =
>    First@Last@
>      Reap@Do[SquareFreeQ@x && (y = x^3 // Sqrt // Round; True) &&
>
>           Length@CoefficientList[MinimalPolynomial[kk][z], z] < 12 &&
>         Sow@{x, y, kk, len}, {x, 2, 3000}]]
>
> {8.39548, {{1942, 85580, -3 Sqrt[2/5], 13}, {2878, 154396, -Sqrt,
>     13}}}
>
> All this is VERY slow nonetheless. Maybe there's another way to
> characterize the problem?
>
> Bobby
>
> On Mon, 21 Nov 2011 03:29:38 -0600, Artur <grafix at csl.pl> wrote:
>
>> 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}][];
>>     ll = CoefficientList[MinimalPolynomial[kk][], #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*)
>>
>
>

--
DrMajorBob at yahoo.com

```

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