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MathGroup Archive 2005

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Re: Partition Function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57634] Re: Partition Function
  • From: "Ray Koopman" <koopman at sfu.ca>
  • Date: Thu, 2 Jun 2005 05:17:44 -0400 (EDT)
  • References: <d7k3qf$ooq$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Shug Boabby wrote:
> is there any way to get mathematica to return the actual partitions
> of an integer? the only functions related to this, (e.g.
> http://mathworld.wolfram.com/PartitionFunctionP.html) only return
> the *number* of partitions, not the actual partititons (including
> permutations) themselves.
>
> e.g. for 4, i'd like a function to return
> {{1,1,1,1}, {2,1,1}, {1,2,1}, {1,1,2}, {3,1}, {2,2}, {1,3}, {4}}
>
> with the ordering of the partitions not being important

Here is the code I use to get ordered (i.e., unpermuted) partitions:

In[1]:=
P[0,m_] := {Table[0,{m}]};
P[n_,m_]/;n<m := With[{z=Table[0,{m-n}]},Join[#,z]&/@P[n,n]];
P[n_,1] := {{n}};
P[n_,2] := Table[{n-i,i},{i,0,n/2}];
P[n_,m_] := ToExpression[ "Block[" <>
ToString[Table[SequenceForm["n",k],{k,m-2}]] <>
",Flatten[Table[" <>
ToString[Table[If[k==m,SequenceForm["n",m-2,"-i",m-1],
SequenceForm["i",k]],{k,m,1,-1}]] <> "," <>
StringTake[ToString[Table[Which[
k==1,StringForm["{i1,0,``/``}",n,m],
k==2,StringForm["{i2,i1,(n1=``-i1)/``}",n,m-1],
True,StringForm["{i`3`,i`2`,(n`2`=n`1`-i`2`)/`4`}",
k-2,k-1,k,m+1-k]], {k,m-1}]], {2,-2}] <> "]," <>
ToString[m-2] <> "]]" ]

In[6]:= P[4,4]
Out[6]= {{4,0,0,0},{3,1,0,0},{2,2,0,0},{2,1,1,0},{1,1,1,1}}

This will get what you want:

In[7]:=
P[n_] := Flatten[Map[Permutations,Select[#,Positive]&/@P[n,n]],1];

In[8]:= P[4]
Out[8]= {{4},{3,1},{1,3},{2,2},{2,1,1},{1,2,1},{1,1,2},{1,1,1,1}}


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