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Code Golf

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  • \$\begingroup\$ Please either clarify what the ambiguity is that remains or remove the hold on my question. I believe I have corrected everything you asked, and aside from that I don't believe it violates the rules in any way. \$\endgroup\$ Commented Jan 23, 2017 at 23:07
  • \$\begingroup\$ Can you give an example with n=9, m=7? (If you can then there's a further ambiguity, which is what the distribution of the random cycles should be: I presume it's selecting a random set of valid derangements such that each set is selected with equal probability assuming a perfect random number generator, but that's not entirely clear; however, the task being actually possible is a much bigger problem). \$\endgroup\$ Commented Jan 23, 2017 at 23:17
  • \$\begingroup\$ An answer for n=9, m=7 is {{5, 3, 4, 2, 0, 7, 8, 6, 1}, {2, 0, 8, 6, 5, 3, 1, 4, 7}, {1, 4, 6, 7, 8, 2, 3, 0, 5}, {8, 6, 0, 4, 7, 1, 2, 5, 3}, {3, 7, 5, 1, 2, 0, 4, 8, 6}, {4, 5, 7, 8, 3, 6, 0, 1, 2}, {6, 8, 3, 5, 1, 4, 7, 2, 0}} \$\endgroup\$ Commented Jan 24, 2017 at 1:57
  • \$\begingroup\$ I cleared up ambiguity and stated that all the different ways of doing it with m cycles of length n should have the same probability of occurring \$\endgroup\$ Commented Jan 24, 2017 at 2:05
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    \$\begingroup\$ "The fastest algorithm (in time complexity) wins." There's two parameters here, m and n, so it's not clear how to compare tradeoffs between these. If you mean for m to be linear in n, that would resolve it. \$\endgroup\$ Commented Jan 24, 2017 at 5:11