A variation of IntegerPartition?

Question

IntegerPartitions[n, {3, 10}, Prime ~Array~ 10]

In Mathematica this will give a list of all the ways to get n as the sum of from three to ten of the first ten prime numbers, allowing duplicates as needed.

How can I efficiently find the sums that equal n, allowing each element to only be used once?

Using the first ten primes is only a toy example. I seek a solution that is valid for arbitrary arguments. In actual cases, generating all possible sums, even using polynomial coefficients, takes too much memory.

I forgot to include that I am using Mathematica 7.

Answer 1

The following will build a binary tree, and then analyze it and extract the results:

Clear[intParts];
intParts[num_, elems_List] /; Total[elems] < num := p[];
intParts[num_, {fst_, rest___}] /; 
   fst < num := {p[fst, intParts[num - fst, {rest}]], intParts[num, {rest}]};
intParts[num_, {fst_, rest___}] /; fst > num := intParts[num, {rest}];
intParts[num_, {num_, rest___}] := {pf[num], intParts[num, {rest}]};


Clear[nextPosition];
nextPosition = 
  Compile[{{pos, _Integer, 1}},
     Module[{ctr = 0, len = Length[pos]},
       While[ctr < len && pos[[len - ctr]] == 1, ++ctr];
       While[ctr < len && pos[[len - ctr]] == 2, ++ctr];
       Append[Drop[pos, -ctr], 1]], CompilationTarget -> "C"];

Clear[getPartitionsFromTree, getPartitions];
getPartitionsFromTree[tree_] :=
  Map[Extract[tree, #[[;; -3]] &@FixedPointList[nextPosition, #]] &, 
     Position[tree, _pf, Infinity]] /. pf[x_] :> x;
getPartitions[num_, elems_List] := 
    getPartitionsFromTree@intParts[num, Reverse@Sort[elems]];

For example,

In[14]:= getPartitions[200,Prime~Array~150]//Short//Timing

Out[14]= {0.5,{{3,197},{7,193},{2,5,193},<<4655>>,{3,7,11,13,17,19,23,29,37,41},      
       {2,3,5,11,13,17,19,23,29,37,41}}}

This is not insanely fast, and perhaps the algorithm could be optimized further, but at least the number of partitions does not grow as fast as for IntegerPartitions.

Edit:

It is interesting that simple memoization speeds the solution up about twice on the example I used before:

Clear[intParts];
intParts[num_, elems_List] /; Total[elems] < num := p[];
intParts[num_, seq : {fst_, rest___}] /; fst < num := 
    intParts[num, seq] = {p[fst, intParts[num - fst, {rest}]], 
          intParts[num, {rest}]};
intParts[num_, seq : {fst_, rest___}] /; fst > num := 
    intParts[num, seq] = intParts[num, {rest}];
intParts[num_, seq : {num_, rest___}] := 
    intParts[num, seq] = {pf[num], intParts[num, {rest}]};

Now,

In[118]:= getPartitions[200, Prime~Array~150] // Length // Timing

Out[118]= {0.219, 4660}

Answer 2

Can use Solve over Integers, with multipliers constrained between 0 and 1. I'll show for a specific example (first 10 primes, add to 100) but it is easy to make a general procedure for this.

primeset = Prime[Range[10]];
mults = Array[x, Length[primeset]];
constraints01 = Map[0 <= # <= 1 &, mults];
target = 100;

Timing[res = mults /. 
  Solve[Flatten[{mults.primeset == target, constraints01}],
    mults, Integers];
  Map[Pick[primeset, #, 1] &, res]
 ]

Out[178]= {0.004, {{7, 11, 13, 17, 23, 29}, {5, 11, 13, 19, 23, 29}, {5, 7, 17, 19, 23, 29}, {2, 5, 11, 13, 17, 23, 29}, {2, 3, 11, 13, 19, 23, 29}, {2, 3, 7, 17, 19, 23, 29}, {2, 3, 5, 7, 11, 13, 17, 19, 23}}}

---edit--- To do this in version 7 one would use Reduce instead of Solve. I'll bundle this in one function.

knapsack[target_, items_] := Module[
  {newset, x, mults, res},
  newset = Select[items, # <= target &];
  mults = Array[x, Length[newset]];
  res = mults /.
    {ToRules[Reduce[
       Flatten[{mults.newset == target, Map[0 <= # <= 1 &, mults]}],
       mults, Integers]]};
  Map[Pick[newset, #, 1] &, res]]

Here is Leonid Shifrin's example:

Timing[Length[knapsack[200, Prime[Range[150]]]]]

Out[128]= {1.80373, 4660}

Not as fast as the tree code, but still (I think) reasonable behavior. At least, not obviously unreasonable.

---end edit---

Daniel Lichtblau Wolfram Research