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Re: Simple n-tuple problem - with no simple solution
*To*: mathgroup at smc.vnet.net
*Subject*: [mg115805] Re: Simple n-tuple problem - with no simple solution
*From*: DrMajorBob <btreat1 at austin.rr.com>
*Date*: Fri, 21 Jan 2011 04:31:52 -0500 (EST)
This may do the trick.
Clear[m]
n = 10;
addends = Rationalize@Range[0, 1, .05];
multipliers = Array[m, Length@addends];
sum = Total@multipliers;
nonNegative = And @@ Thread[multipliers >= 0];
soln = Sort[
multipliers /.
Solve[{multipliers.addends == 1, nonNegative, sum == n},
multipliers, Integers]];
Length@soln
soln.addends // Union
Total /@ soln // Union
530
{1}
{10}
When 0 is one of the addends, you can also write:
Clear[m]
n = 10;
addends = DeleteCases[Rationalize@Range[0, 1, .05], 0];
multipliers = Array[m, Length@addends];
sum = Total@multipliers;
nonNegative = And @@ Thread[multipliers >= 0];
soln = Sort[
multipliers /.
Solve[{multipliers.addends == 1, nonNegative, sum <= n},
multipliers, Integers]];
Length@soln
soln.addends // Union
Total /@ soln // Union
530
{1}
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
You may also want to look at IntegerPartitions and FrobeniusSolve.
Bobby
On Thu, 20 Jan 2011 05:33:21 -0600, Don <donabc at comcast.net> wrote:
> Problem: Given an n-tuple (n >= 1). with each element able to take
> on the values in
> Range[0, 1.0, .05] , produce all the n-tuples that sum to 1.0.
>
> The most direct way to solve this problem is to generaate all possible
> n-tuples and Select out all those that sum to 1.0.
>
> For example, when n = 2 :
>
> n = 2;
> Select[Tuples[Table[Range[0, 1.0, .05], {n}]], Total[#] == 1 &]
>
> The problem with this solution is that the number of n-tuples that are
> generated before the Select is used grows exponentially fast as a
> function
> of n - causing the system to run out of memory (RAM) very quickly.
>
> Is there a more memory efficient way to solve this problem that
> doesn't
> use so much memory but still is not too slow in terms of processor
> time?
>
> Thank you.
>
--
DrMajorBob at yahoo.com
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