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10762752
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0
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2012-05-26T00:33:31.880
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2013-12-10T20:15:46.017
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2013-12-10T20:15:46.017
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1085128
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1085128
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2161406
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2
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I have an evenly distributed partition generator. Where n := the integer to be partitioned, r:= the number of slices: The algorithm is a patched version of the naive method of simply inserting partings at random. The problem with this method, as it appeared to me when I looked at its output, was that scenarios where partings are placed in the same spot are less likely to occur. There is only one way to get {1,1,1}, while there are 3! ways of getting {2,4,9}, any of {4,2,9},{2,4,9},{9,4,2}... will lead to the same partition placement when sorted. This has been amended by providing additional explicit opportunities for repeats. For each parting insertion, there's a chance that the position of the parting wont be random, but will be selected as a repeat of a formerly selected value. This balances the uneven probability distribution of the naive method right out. I have proved by exhaustion that each partitioning is perfectly equally likely for r = 3, n = 2. I cbf proving it for higher values but healfhearted ventures to do so found only promising signs. I also tested it on random input, finding that it is at least roughly even for every values I tried[but probably perfectly even]. here it is in C++11: [the output format is different to what you're expecting, it's the positions of the partings rather than the size of the space between them. The conversion is easy, though] <pre><code>#include <vector> #include <algorithm> #include <random> #include <cassert> template <typename Parting, typename Seed> vector<Parting> partitionGen(unsigned nparts, unsigned bandw, Seed seed){//nparts is the number of parts, that is, one greater than the number of dividers listed in the output vector. Bandw is the integer being partitioned. assert(nparts > 0); vector<Parting> out(nparts-1); srand(seed); unsigned genRange = bandw; for(auto i=out.begin(); i<out.end(); ++i, ++genRange){ unsigned gen = rand()%genRange; *i = ((gen<bandw)? gen: *(i-(gen-bandw+1))); } sort(out.begin(), out.end(), less<Parting>()); return out; } </code></pre> I don't like the fact that I have to sort it though. If Vlody's version has an even distribution, it appears that it'd be better.
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1. POHow do I generate a uniform random integer partition?
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1. USmako
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