Re: Simple n-tuple problem - with no simple solution

*To*: mathgroup at smc.vnet.net*Subject*: [mg115869] Re: Simple n-tuple problem - with no simple solution*From*: "Mr. Wizard" <gleam at flashmail.com>*Date*: Sun, 23 Jan 2011 05:37:00 -0500 (EST)

DrMajorBob wrote: >I didn't think rearrangements of a solution mattered, so my approach was >different... and I found that Solve is pretty darn fast! > >My approach also allows Range[0,1,.05] to be replaced with almost any list >of reals, though I didn't experiment with that. > >Here are some timings. > >n = 25; >Length@t1[addends, n] // Timing > >{2.57619, 627} > >I'd say Solve (which must use Integer Linear Programming for this) is the >way to go. Before replies started appearing on the list, I sent a private message to Don with my recommendation. Leonid Shifrin gave a similar solution, and you also suggested a look at IntegerPartitions. In my testing it is far and away faster. Compared to the original method: f1 = Select[Tuples[Table[Range[0, 1.0, .05], {#}]], Total[#] == 1 &] &; f2 = Union @@ Permutations /@ # &@N@IntegerPartitions[1, {#}, Range[0, 1, 1/20]] &; Timing[ r1 = f1[5]; ] Timing[ r2 = f2[5]; ] r1 === r2 Out[1]= {8.734, Null} Out[2]= {0.016, Null} Out[3]= True If the permutations are not required: f3 = N@IntegerPartitions[1, {#}, Range[0, 1, 1/#2]] &; Timing[Length@f3[25, 20]] Timing[Length@f3[17, 50]] Out[4]= {0., 627} Out[5]= {0.704, 161144} Range[...] can be replaced with a different list, as in your method. Paul