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Re: Solve - takes very long time

  • To: mathgroup at smc.vnet.net
  • Subject: [mg121830] Re: Solve - takes very long time
  • From: Peter Pein <petsie at dordos.net>
  • Date: Tue, 4 Oct 2011 01:32:09 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <j6bris$8m4$1@smc.vnet.net>

Am 03.10.2011 10:26, schrieb Fredob:
> Hi,
>
> I tried the following on Mathematica 8 and it doesn't seem to stop
> running (waited 40 minutes on a 2.6 Ghz processor w 6 GB of primary
> memory).
>
> Solve[
>   {100*Subscript[x, 2] + 10*Subscript[x, 1] + Subscript[x, 0] +
>      100*Subscript[y, 2] + 10*Subscript[y, 1] + Subscript[y, 0] ==
>     100*Subscript[z, 2] + 10*Subscript[z, 1] + Subscript[z, 0],
>     Subscript[x, 0]>  0, Subscript[y, 0]>  0, Subscript[z, 0]>  0,
>    Subscript[x, 1]>  0, Subscript[y, 1]>  0, Subscript[z, 1]>  0,
>    Subscript[x, 2]>  0, Subscript[y, 2]>  0, Subscript[z, 2]>  0,
>    Subscript[x, 0]<= 9, Subscript[y, 0]<= 9, Subscript[z, 0]<= 9,
>    Subscript[x, 1]<= 9, Subscript[y, 1]<= 9, Subscript[z, 1]<= 9,
>    Subscript[x, 2]<= 9, Subscript[y, 2]<= 9, Subscript[z, 2]<= 9,
>    Subscript[x, 0] != Subscript[y, 0] != Subscript[z, 0] != Subscript[
>     x, 1] != Subscript[y, 1] != Subscript[z, 1] != Subscript[x, 2] !=
>     Subscript[y, 2] != Subscript[z, 2]},
>   {Subscript[x, 2], Subscript[y, 2], Subscript[z, 2], Subscript[x, 1],
>    Subscript[y, 1], Subscript[z, 1], Subscript[x, 0], Subscript[y, 0],
>    Subscript[z, 0] },
>   Integers]
>
> The problem was a homework for my daugther where you are supposed to
> use all digits to build - but only once - 2 three digit numbers and
> addition.
>
>

Hi,

so you want to pick some of the permutations of {1,2,...,9}?

Mathematica can do this very fast.

In[1]:= 
Length[allsoln=Pick[#,#.{100,10,1,100,10,1,-100,-10,-1},0]&[Permutations[Range[9]]]]//AbsoluteTiming
Out[1]= {0.032373,336}

Shall your daughter give all solutions (then a computer printout should 
be accepted ;-))

Just to print three of the 336 solutions (first, middle, last sol.):

In[2]:= MatrixForm[Partition[#,3]]&/@allsoln[[{1,168,-1}]]
Out[2]=
  {(1	2	4
    6	5	9
    7	8	3
   ),
   (3	4	2
    5	7	6
    9	1	8
   ),
   (7	8	4
    1	5	2
    9	3	6
)}

Peter



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