Re: Solve - takes very long time

• To: mathgroup at smc.vnet.net
• Subject: [mg121860] Re: Solve - takes very long time
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Wed, 5 Oct 2011 04:01:17 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <664167423.1722926.1317752736583.JavaMail.root@jaguar8.sfu.ca>

```Or even better (9 times faster):

interpret[s_List] :=
Flatten@{FromDigits /@ Transpose@Take[#, 3],
FromDigits@Flatten@Take[#, -3]} &[
Flatten@Position[s, #, {1}, 2] & /@ {100, 10, 1, -100, -10, -1}]
Timing[
dual = Permutations@{100, 10, 1, 100, 10, 1, -100, -10, -1};
interpret /@ dual[[Flatten at Position[dual.Range@9, 0]]] // Sort]

{0.010079, {{124, 659, 783}, {125, 739, 864}, {127, 359, 486}, {127,
368, 495}, {128, 439, 567}, {134, 658, 792}, {142, 596, 738}, {142,
695, 837}, {143, 586, 729}, {152, 487, 639}, {152, 784,
936}, {162, 387, 549}, {162, 783, 945}, {173, 286, 459}, {173, 295,
468}, {182, 394, 576}, {182, 493, 675}, {214, 569, 783}, {214,
659, 873}, {215, 478, 693}, {215, 748, 963}, {216, 378, 594}, {216,
738, 954}, {218, 349, 567}, {218, 439, 657}, {234, 657,
891}, {235, 746, 981}, {241, 596, 837}, {243, 576, 819}, {243, 675,
918}, {251, 397, 648}, {271, 593, 864}, {271, 683, 954}, {281,
394, 675}, {314, 658, 972}, {317, 529, 846}, {317, 628, 945}, {324,
567, 891}, {324, 657, 981}, {341, 586, 927}, {342, 576,
918}, {352, 467, 819}}}

Bobby

On Tue, 04 Oct 2011 15:52:09 -0500, DrMajorBob <btreat1 at austin.rr.com>
wrote:

> I missed the fact that you'd already explained this, but the same idea
> yields THIS solution:
>
> interpret[s_List] :=
>   Flatten@{FromDigits /@ Transpose@Take[#, 3],
>       FromDigits@Flatten@Take[#, -3]} &[
>    Flatten@Position[s, #] & /@ {100, 10, 1, -100, -10, -1}]
> nine = Range@9;
> interpret /@
>    Select[Permutations@{100, 10, 1, 100, 10,
>       1, -100, -10, -1}, #.nine == 0 &] // Timing
>
> {0.096459, {{127, 359, 486}, {127, 368, 495}, {128, 439, 567}, {125,
>     739, 864}, {124, 659, 783}, {182, 394, 576}, {162, 387, 549}, {182,
>      493, 675}, {162, 783, 945}, {142, 596, 738}, {142, 695,
>     837}, {152, 487, 639}, {152, 784, 936}, {173, 295, 468}, {173, 286,
>      459}, {143, 586, 729}, {134, 658, 792}, {218, 349, 567}, {216,
>     378, 594}, {218, 439, 657}, {216, 738, 954}, {214, 569, 783}, {214,
>      659, 873}, {215, 478, 693}, {215, 748, 963}, {317, 529,
>     846}, {317, 628, 945}, {314, 658, 972}, {281, 394, 675}, {251, 397,
>      648}, {271, 593, 864}, {271, 683, 954}, {241, 596, 837}, {341,
>     586, 927}, {243, 576, 819}, {243, 675, 918}, {342, 576, 918}, {352,
>      467, 819}, {234, 657, 891}, {235, 746, 981}, {324, 567,
>     891}, {324, 657, 981}}}
>
> That uses far less memory (1/8 as many permutations), and it's also
> faster:
>
> FromDigits /@ Partition[#, 3] & /@
>    Select[Permutations@
>      Range@9, #[[1]] < #[[4]] && #[[2]] < #[[5]] && #[[3]] < #[[6]] && \
> #.{100, 10, 1, 100, 10, 1, -100, -10, -1} == 0 &] // Timing
>
> {2.02554, {{124, 659, 783}, {125, 739, 864}, {127, 359, 486}, {127,
>     368, 495}, {128, 439, 567}, {134, 658, 792}, {142, 596, 738}, {142,
>      695, 837}, {143, 586, 729}, {152, 487, 639}, {152, 784,
>     936}, {162, 387, 549}, {162, 783, 945}, {173, 286, 459}, {173, 295,
>      468}, {182, 394, 576}, {182, 493, 675}, {214, 569, 783}, {214,
>     659, 873}, {215, 478, 693}, {215, 748, 963}, {216, 378, 594}, {216,
>      738, 954}, {218, 349, 567}, {218, 439, 657}, {234, 657,
>     891}, {235, 746, 981}, {241, 596, 837}, {243, 576, 819}, {243, 675,
>      918}, {251, 397, 648}, {271, 593, 864}, {271, 683, 954}, {281,
>     394, 675}, {314, 658, 972}, {317, 529, 846}, {317, 628, 945}, {324,
>      567, 891}, {324, 657, 981}, {341, 586, 927}, {342, 576,
>     918}, {352, 467, 819}}}
>
> Timing[Length[
>    solns = FromDigits /@ Partition[#, 3] & /@
>      Select[Permutations@
>        Range@9, #[[1]] < #[[4]] && #.{100, 10, 1, 100, 10,
>            1, -100, -10, -1} == 0 &]]]
>
> {1.56286, 168}
>
> Surely "interpret" could be simpler, but I haven't thought of a way, as
> yet... and it doesn't need to be fast.
>
> Bobby
>
> On Tue, 04 Oct 2011 13:25:36 -0500, Ray Koopman <koopman at sfu.ca> wrote:
>
>> The basic condition can be written as
>>
>> 100*(x2 + y2) + 10*(x1 + y1) + (x0 + y0) = 100*z2 + 10*z1 + z0,
>>
>> in which form it is clear that we can always swap corresponding xi and
>> yi, and that solutions therefore come is sets of 8. Requiring xi < yi
>> for all i is just a way of picking a "canonical" member of each set.
>>
>> ----- DrMajorBob <btreat1 at austin.rr.com> wrote:
>>> The conditions #[[2]] < #[[5]] and #[[3]] < #[[6]] do not belong,
>>> however.
>>>
>>> Bobby
>>>
>>> On Tue, 04 Oct 2011 00:30:53 -0500, Ray Koopman <koopman at sfu.ca> wrote:
>>>
>>>> On Oct 3, 1:26 am, Fredob <fredrik.dob... at gmail.com> wrote:
>>>>> 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
>>>>
>>>> For each of the 42 solutions found by the brute force search given
>>>> below there are seven other solutions that may be obtained by
>>>> interchanging x0,y0 and/or x1,y1 and/or x2,y2.
>>>>
>>>> FromDigits/@Partition[#,3]& /@ Select[Permutations@Range@9,
>>>>   #[[1]] < #[[4]] && #[[2]] < #[[5]] && #[[3]] < #[[6]] &&
>>>>   #.{100,10,1,100,10,1,-100,-10,-1} == 0 &]
>>>>
>>>> {{124,659,783}, {125,739,864}, {127,359,486},
>>>>  {127,368,495}, {128,439,567}, {134,658,792},
>>>>  {142,596,738}, {142,695,837}, {143,586,729},
>>>>  {152,487,639}, {152,784,936}, {162,387,549},
>>>>  {162,783,945}, {173,286,459}, {173,295,468},
>>>>  {182,394,576}, {182,493,675}, {214,569,783},
>>>>  {214,659,873}, {215,478,693}, {215,748,963},
>>>>  {216,378,594}, {216,738,954}, {218,349,567},
>>>>  {218,439,657}, {234,657,891}, {235,746,981},
>>>>  {241,596,837}, {243,576,819}, {243,675,918},
>>>>  {251,397,648}, {271,593,864}, {271,683,954},
>>>>  {281,394,675}, {314,658,972}, {317,529,846},
>>>>  {317,628,945}, {324,567,891}, {324,657,981},
>>>>  {341,586,927}, {342,576,918}, {352,467,819}}
>
>

--
DrMajorBob at yahoo.com

```

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