Re: LatticeReduce

• To: mathgroup at smc.vnet.net
• Subject: [mg93580] Re: LatticeReduce
• From: Artur <grafix at csl.pl>
• Date: Sun, 16 Nov 2008 07:02:56 -0500 (EST)
• References: <200811151103.GAA16591@smc.vnet.net> <491F2366.3040401@csl.pl>

```P.S.
And for the case:
{a1, a2, a3} = {1, 3^4,
5^4}; a = {{1, 0, 0, -a1}, {0, 1, 0, -a2}, {0, 0, 1, -a3}}; b =
LatticeReduce[a]

Out1:*{{1, 0, 0, -1}, {12, -8, 1, 11}, {-6, -23, 3, -6}}

Mathematica should be return 2 solutions:

23 - 8*3^4 + 5^4=0
12 + 23*3^4 - 3*5^4=0

**After good LatticeReduce should begin from : {{23,-8,1,0}, {12,
23,-3,0}, {c, c1,c2, c3}}

What is wrong in LatticeReduce or what I'm doing wrong ?

Artur
*

Artur pisze:
> Dear Mathematica Gurus,
> I want to find such inetegrs x,y,z that
> x + y*2^15 + z*3^8 = 0
> I'm reading in manual that I can use LatticeReduce
>
> {a0, a1, a2} = {1,
>  2^15, -3^8}; a = {{1, 0, 0, -a0}, {0, 1, 0, -a1}, {0, 0,
>   1, -a2}}; b = LatticeReduce[a]
>
> Out1:{{1, 0, 0, -1}, {18, 1, 5, 19}, {117, -171, -854, 117}}
>
> That mean that computer don't find such solution (solution is finded
> when last number in one of rows should be 0)
>
> If I run again:
> {a0, a1, a2} = {37,
>  2^15, -3^8}; a = {{1, 0, 0, -a0}, {0, 1, 0, -a1}, {0, 0,
>   1, -a2}}; b = LatticeReduce[a]
>
> Out2:{{1, 1, 5, 0}, {1, 0, 0, -37}, {170, -7, -34, 12}}
>
> Now solution is finded by Mathematica OK!
> k = Transpose[{{37, 2^15, -3^8, anything}}]; b[[1]].k
>
> Out3: {0}
>
> Is OK!
>
> Is another method finding coefficients x, y, z as LatticeReduce and
> why Mathematica don't reduced
> {a0, a1, a2} = {1,  2^15, -3^8}; a = {{1, 0, 0, -a0}, {0, 1, 0, -a1},
> {0, 0, 1, -a2}}; b = LatticeReduce[a]
>
>
> Best wishes
> Artur
>

```

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