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Title: Expecto Inversio! Post by TenaliRaman on Jul 1st, 2004, 8:47pm In a permutation a1..an of n distinct integers, an inversion is a pair(ai,aj) such tht i<j and ai>aj. If all permutations are equally likely , what is the expected no. of inversions in a randomly chosen permutation of 1..n? |
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Title: Re: Expecto Inversio! Post by pedronunezmd on Jul 2nd, 2004, 7:28am Sounds like another way of asking "[hide]How many combinations of the n integers taken 2 at a time are there?[/hide]" |
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Title: Re: Expecto Inversio! Post by Barukh on Jul 2nd, 2004, 7:46am What is [hide]E(n) - E(n-1)[/hide]? |
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Title: Re: Expecto Inversio! Post by Grimbal on Jul 2nd, 2004, 12:08pm I would just [hide]compare the expected number of inversions and the expected number of non-inversions. (a non-inversion being a pair(ai,aj) such tht i<j and ai<aj). They should be equal and the sum is the number of pairs[/hide]. |
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Title: Re: Expecto Inversio! Post by towr on Jul 2nd, 2004, 1:57pm ::[hide]And of course the non-inversion change to inversions and vice versa if you take the mirror image of the permutation, which is itself another permutation. So on average there must be half of each..[/hide]:: |
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Title: Re: Expecto Inversio! Post by TenaliRaman on Jul 2nd, 2004, 2:22pm gee i knew this was too easy even for easy :-/ |
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Title: Re: Expecto Inversio! Post by towr on Jul 2nd, 2004, 2:39pm I wouldn't say it's too easy for easy. We're just too smart for the easy section, really :P |
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Title: Re: Expecto Inversio! Post by Grimbal on Jul 2nd, 2004, 2:57pm Expecto Inversio! sounds like a Harry Potter spell. |
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Title: Re: Expecto Inversio! Post by Barukh on Jul 3rd, 2004, 4:31am I like that, Grimbal and towr. Here's another approach: [hide]E(n) - E(n-1) = (n-1)/2[/hide]. |
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Title: Re: Expecto Inversio! Post by TenaliRaman on Jul 3rd, 2004, 9:31am Grimbal, yes i am a harry potter fun obviously excited abt the 6th book. ;D Barukh, Nice approach! It never hit me though i think it should have been obvious! |
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