[Date Index]
[Thread Index]
[Author Index]
Re: Solve - takes very long time
*To*: mathgroup at smc.vnet.net
*Subject*: [mg121870] Re: Solve - takes very long time
*From*: Ray Koopman <koopman at sfu.ca>
*Date*: Wed, 5 Oct 2011 04:03:06 -0400 (EDT)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
I was wondering if something like that might be possible, but it
didn't jump out at me. If I had figured it out, I might have used
Pick[dual, dual.Range@9, 0]
instead of
dual[[Flatten at Position[dual.Range@9, 0]]].
It's a little easier to read, and on my system it's just as fast.
----- DrMajorBob <btreat1 at austin.rr.com> wrote:
> 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
>>>>>> addition.
>>>>>
>>>>> 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}}
Prev by Date:
**Re: Solve - takes very long time**
Next by Date:
**Re: Solve - takes very long time**
Previous by thread:
**Re: Solve - takes very long time**
Next by thread:
**Re: Solve - takes very long time**
| |