Recognizing even or odd permutation easily?

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Discussion Overview

The discussion revolves around identifying whether a given permutation of a set of numbers is even or odd. Participants explore various methods to determine the parity of permutations without resorting to exhaustive counting of swaps.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant suggests counting the number of swaps needed to return to the identity permutation, noting that odd permutations require an odd number of swaps.
  • Another participant provides an example with a specific permutation, proposing a method of applying swaps to determine its parity.
  • A different approach is introduced, where participants count inversions in the permutation, stating that an odd number of inversions indicates an odd permutation.
  • One participant discusses using the cycle structure of the permutation, explaining how the parity of cycles contributes to the overall parity of the permutation.

Areas of Agreement / Disagreement

Participants present multiple methods for determining the parity of permutations, and while they share their approaches, there is no consensus on a single best method. The discussion remains open with various techniques being explored.

Contextual Notes

Some methods rely on understanding specific concepts such as cycle structures and inversions, which may not be universally familiar to all participants.

vulpe
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Hello folks, I'm not even sure if this is the place to put this but with some luck it might be.

I just have a general question about permutations. I understand the concept of even and odd permutations of a set of numbers, what I am hoping for is an easy way to figure out if (given a scrambled arrangement of the original set) the new set is even or odd? Obviously other than the trivial solution of going through it permutation by permutation and counting.

Thanks for any input!
 
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The most basic odd permutations are those that swap 2 numbers. So count how many of those permutations must be applied to reach the identity. For example, 321 is odd (swap 3 and 1) but 312 is even (two swaps are required). It's a simple as that.
 
I understand what odd and even permutations are, my question makes more sense if I use an example perhaps. Imagine that instead of 1,2,3 you have a set that's 1,2,3,4,5,6 . Now tell me, is 5,2,1,4,6,3 even or odd without taking forever haha :P
 
Hi, vulpe,
I'd attempt the swaps that put each peg in place. In your example,
5,2,1,4,6,3
1,2,5,4,6,3 (1,5)
1,2,3,4,6,5 (3,5)
1,2,3,4,5,6 (5,6)
hence this one is odd. At worst, you'd do as many swaps as elements are (minus 1, actually).
 
Go through all 2-element subsets of the numbers in the permutation and see for how many of these the greater number comes before the smaller in the permutation, call it an inversion if this is the case. If the number of inversions is odd, the permutaion is odd, otherwise even.

For example, in the permutation given by (2 6 1 4 3 5), (that is: 1 is mapped to 2, 2 to 6, 3 to 1, etc.) there are 6 inversions: (2,1), (6,1), (6,4), (6,3), (6,5), (4,3). 6 is en even number, so the permutation is even.
 
Another approach using the cycle structure.

In your example: the permutation taking 1, 2, 3, 4, 5, 6 to 5, 2, 1, 4, 6, 3 has cycle structure (1 5 6 3) (2) (4). (If you don't know how to write down the cycle structure of a permutation, learn!) Now, a 4-cycle is odd and a 1-cycle is even, so we have odd × even × even = odd, so the permutation is odd.
 

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