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I'm trying to generate all integer partitions where the first entry is a fixed number i.e. all young diagrams with first row/column fixed. I'm aware of the function IntegerPartitions[n] and I could generate all partitions of say $n^2$ or less and then only keep the ones with a given first element but I'm wondering if there is a more efficient way to achieve this.

Thanks

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    $\begingroup$ You could subtract the fixed first element from $n$ and then look for integer partitions of what is left. At the end, prepend the fixed first element to all partitions thus found. $\endgroup$
    – Roman
    Commented May 27, 2019 at 15:48
  • $\begingroup$ Would you mind writing a code example? I'm not sure I follow $\endgroup$
    – Samuel Crew
    Commented May 27, 2019 at 16:46

1 Answer 1

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Roman's suggestion combined with using the third argument of IntegerPartitions to restrict the result to partitions that involve only numbers below the first number:

ClearAll[f]
f[n_, k_] := Prepend[#, k] & /@ IntegerPartitions[n - k, All, Range[k]]
f[n_, k_, l_] := Prepend[#, k] & /@ IntegerPartitions[n - k, {l - 1}, Range[k]]

f[6, 3]

{{3, 3}, {3, 2, 1}, {3, 1, 1, 1}}

f[6, 3, 3]

{{3, 2, 1}}

f[10, 4]

{{4, 4, 2}, {4, 4, 1, 1}, {4, 3, 3}, {4, 3, 2, 1}, {4, 3, 1, 1, 1}, {4, 2, 2, 2}, {4, 2, 2, 1, 1}, {4, 2, 1, 1, 1, 1}, {4, 1, 1, 1, 1, 1, 1}}

f[10, 4, 3]

{{4, 4, 2}, {4, 3, 3}}

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  • $\begingroup$ Thanks a lot for your answer. I realised I actually explained the problem a big ambiguously in my post - I was looking for partitions with a fixed first element and a fixed length. I.e inputting 3 should return {3,3,3},{3,3,2},{3,3,1},{3,2,2},{3,2,1},{3,1,1}. But the third argument of InterPartitions is what I was missing to achieve this so thanks! $\endgroup$
    – Samuel Crew
    Commented May 27, 2019 at 16:54
  • $\begingroup$ Actually that still doesn't quite do the job. In the example above I'd want something that generates all partitions of 3^3 or less using only 1,2,3 $\endgroup$
    – Samuel Crew
    Commented May 27, 2019 at 17:02
  • $\begingroup$ @SamuelCrew, please see the updated version. $\endgroup$
    – kglr
    Commented May 27, 2019 at 17:04
  • $\begingroup$ @SamuelCrew, you can use the three-argument form of f as follows: Join @@ (f[#, 3, 3] & /@ Range[3, 9]) gives {{3, 1, 1}, {3, 2, 1}, {3, 3, 1}, {3, 2, 2}, {3, 3, 2}, {3, 3, 3}} $\endgroup$
    – kglr
    Commented May 27, 2019 at 17:09
  • $\begingroup$ Thanks for your help! $\endgroup$
    – Samuel Crew
    Commented May 27, 2019 at 19:37

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