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

*To*: mathgroup at smc.vnet.net*Subject*: [mg115883] Re: Simple n-tuple problem - with no simple solution*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Mon, 24 Jan 2011 05:22:27 -0500 (EST)

Brilliant! Bobby On Sun, 23 Jan 2011 16:32:03 -0600, Dana DeLouis <dana.del at gmail.com> wrote: > On Jan 22, 3:23 am, DrMajorBob <btre... at austin.rr.com> wrote: >> Here are some timings. > >> n = 5; addends = Rationalize@Range[0, 1.0, 0.05]; >> Length@t1[addends, n] // Timing >> >> {1.44372, 192} >> {23.6318, 192} > >> n = 7; >> Length@t1[addends, n] // Timing >> {1.85135, 364} > >> n = 9; >> Length@t1[addends, n] // Timing >> {2.04216, 488} > >> The Compositions method rendered Mathematica unresponsive at n = 10, but >> here's n = 25 for my approach: >> 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. > > > Hi. Just some thoughts. With the smallest unit 1, the most n can be > is 20. > > If we assume that solution sizes smaller than n have implied 0's, then... > > IntegerPartitions[20,5]//Length //Timing > {0.000085,192} > > IntegerPartitions[20,20]//Length //Timing > {0.000244,627} > > If we want to look at each exact n, then the function uses {n} > > Table[Length[IntegerPartitions[20,{j}]],{j,20}] > {1,10,33,64,84,90,82,70,54,42,30,22,15,11,7,5,3,2,1,1} > > Accumulate[%] > = > {1,11,44,108,192,282,364,434,488,530,560,582,597,608,615,620,623,625,626,627} > > By adding them up, we see that for n=5, we get the 192. For n=7, 364, > etc. > > If the question was "Count all possible combinations" as in the last > total above, then one additional way... > > Timing[SeriesCoefficient[ > 1/Times @@ (1 - n^#1 & ) /@ > Range[20], {n, 0, 20}]] > > {0.0003, 627} > > = = = = = = = = = = > HTH : >) > Dana DeLouis > > -- DrMajorBob at yahoo.com