Re: Solve - takes very long time

*To*: mathgroup at smc.vnet.net*Subject*: [mg121858] Re: Solve - takes very long time*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Wed, 5 Oct 2011 04:00:56 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <664167423.1722926.1317752736583.JavaMail.root@jaguar8.sfu.ca>*Reply-to*: drmajorbob at yahoo.com

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}} -- DrMajorBob at yahoo.com