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My thoughts on the MVP changes


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#1 Renaldoo

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Posted 26 September 2011 - 01:06 PM

posterity


Edited by Renaldoo, 24 October 2018 - 09:34 PM.

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#2 cRoc

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Posted 26 September 2011 - 01:17 PM

what does an anime question have anything to do with mvp?? trying to troll? need to try harder.....
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#3 Azyrk

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Posted 26 September 2011 - 01:39 PM

You forgot to carry the one. :o
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#4 Fureedo

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Posted 26 September 2011 - 01:41 PM

The total number of permutations if you were allowed to repeat would be 14^14

Since you're (I guessing) not allowed repetitions, the result is actually 14!

You start with 14 episodes, then 13, then 12, then 11, etc.

so 14x13x12x11x10...

Which is the definition of 14!


Overcomplicated troll is overcomplicated.

ohh, wait, I see what you did there. note to self: read past paragraph 2~3 next time

Edited by Fureedo, 26 September 2011 - 01:47 PM.

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#5 Renaldoo

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Posted 26 September 2011 - 01:48 PM

You must not have read the whole post-- the question isn't the total number of permutations, it's the shortest string of elements that contain all possible permutations.

For a 3-element set (0, 1, and 2), the number permutations are:

012
021
102
120
210
201

However, rather than exploring 6 permutations with 3 episodes per-permutation, and thus 18 episodes, you'd run it in a string:

012
-120
--201
----102
-----021
------210

that's 6 permutations in 012010210, or 9 episodes. Take note that this is a palindrome, as are most results of cycle-grouping. Proving palindrome efficiency is a whole different can of worms though.

Edited by Renaldoo, 26 September 2011 - 01:49 PM.

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#6 Trixdee

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Posted 26 September 2011 - 01:53 PM

I am sorry but I do not understand how this had anything to do with MVPs. :S Explain?
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#7 Mefistofeles

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Posted 26 September 2011 - 02:03 PM

I am sorry but I do not understand how this had anything to do with MVPs. :S Explain?

troll
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#8 Trixdee

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Posted 26 September 2011 - 02:08 PM

No, it's not a troll. Honest. O_o
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#9 HayrohsLegacy

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Posted 26 September 2011 - 02:09 PM

Pointless thread... /sigh
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#10 BoingBoing

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Posted 26 September 2011 - 02:44 PM

Being decimal, you have not assumed the cases of permutations containing 13 and 14 being both 1,3|1,4 and 13|14, no?
Either that or you have to work in quinquadecimal, no?
If the episode from 1 to 14 are to be treated like alphabets, the answer would be n(n-2) +3, NOT what you have got there, no?
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#11 Renaldoo

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Posted 07 November 2011 - 01:11 AM

Any thoughts, Renouille?
:|
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