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Help with n-Distinct Integer Partitions

Posted 10 years ago
IntegerPartitions displays all the partitions of n
Select[IntegerPartitions[5], Max[Length /@ Split@ #] == 1 &] gives me partitions where each element is at least 1 apart (or distinct
I am trying to find a way to get all the partitions where each element is at least 2 apart, then i want to be able to generalize it to n apart.
for example 4 has two that is two apart because it contains {4} and {1,3}. Any help would be greatly appreciated
POSTED BY: Stephen HIll
2 Replies
Could do this with Reduce, as below.
 f[n_, j_] := Module[
   {vars, x, vals, solns},
   res = Table[
     vars = Table[x[l], {l, k, n, j + 1}];
     vals = Map[First, vars];
     Print[{k, vars}];
     Print[Prepend[Map[0 <= # <= 1 &, vars], vars.vals == n]];
     solns = vars /.
       Solve[Prepend[Map[0 <= # <= 1 &, vars], vars.vals == n], vars,
       Integers];
    solns = Map[vals*# &, solns];
    solns /. 0 :> Sequence[]
    , {k, j}];
  res
  ]
For example:
In[211]:= f[7, 1]
Out[211]= {{7}, {3, 4}, {2, 5}, {1, 6}, {1, 2, 4}}

In[213]:= f[11, 2]
Out[213]= {{11}, {4, 7}, {3, 8}, {2, 9}, {1, 10}, {1, 4, 6}, {1, 3, 7}}
There may also be an approach via generating functions.
POSTED BY: Daniel Lichtblau
Use ListConvolve on the sorted list is helpful: 
In[4]:= Cases[IntegerPartitions[10],x_/;Length[x]>=2 && Min[ListConvolve[{-1,1},x]] >= 2]
Out[4]= {{9,1},{8,2},{7,3},{6,4},{6,3,1}}
ListConvolve with kernel {-1,1} computes the difference of the adjacent elements through the lexicon-reversely-ordered list. If the min value of the new list is larger than 2, I know each pair is at lease 2 apart.  
POSTED BY: Shenghui Yang
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