Problem: Given a number $n,$ we want to find out the subsets of $\{1,2,\ldots,n\}$ that add up to the given number $n.$
Example: If $n=6,$ then the output is: $\{1,5\}, \{2,4\}, \{1,2,3\}.$
Can anyone suggest a program for this?
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You mean subsets of {1,2,...,N} right? Numbers don't have subsets. – anon Mar 27 '12 at 13:45
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What about $[ 3,3 ]$? Are repetitions allowed? – draks ... Mar 27 '12 at 13:46
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So you don't want to include $[1,1,4]$ or $[3,3]$ then? In that case what you want is called "partitions of an integer into distinct parts". – MJD Mar 27 '12 at 13:47
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see here: Partitions – draks ... Mar 27 '12 at 13:48
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Did I edit too much? – Mar 27 '12 at 23:38
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Ordered partitions of an integer... – draks ... Apr 05 '12 at 06:51
2 Answers
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First, if $N=6$, then the output should also include $\{{6\}}$.
Second, it's not clear whether you want to write a program that can do this, or whether you want to be directed to an already-existing program that can do it. If it's the former, it shouldn't be too hard. The largest number used has to be at least $(1/2)(\sqrt{8N+1}-1)$; let's call this quantity (after rounding up to an integer) $r$. So for each value of $n$ from $r$ to $N$, you compute, recursively, the partitions of $N-n$ into distinct parts less than $n$, and then stick $n$ into each of them.
E.g., for $N=11$, we get $r=5$, so first we compute the useful partitions of $N-r=6$, which are $\{{2,4\}}$ and $\{{1,2,3\}}$, and that gives us $\{{2,4,5\}}$ and $\{{1,2,3,5\}}$; then let $n=6$, $N-n=5$, partition 5 as $\{{5\}},\{{1,4\}},\{{2,3\}}$, so we have $\{{5,6\}},\{{1,4,6\}},\{{2,3,6\}}$; and so on.
Gerry Myerson
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Thanks Gerry I am getting lost in recursion somewhere The integer partition problem gives u sets where elements are repeated also but my problem requires only distinct elements.
Thanks, Anshul
– Anshul Mar 28 '12 at 15:00 -
The word "useful" was intended to take care of that: in the example, when we compute partitions of 6, we discard ${{1,5}}$ as not being useful because it would lead to ${{1,5,5}}$. – Gerry Myerson Mar 29 '12 at 06:01
