Recognizing even or odd permutation easily?

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SUMMARY

This discussion focuses on identifying whether a given permutation of a set of numbers is even or odd. The key methods discussed include counting the number of swaps needed to reach the identity permutation and calculating inversions within the permutation. For example, the permutation (5, 2, 1, 4, 6, 3) is determined to be odd through a series of swaps. Additionally, the cycle structure method is highlighted, where the permutation's cycle structure (1 5 6 3)(2)(4) indicates an odd permutation due to the presence of a 4-cycle.

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  • Understanding of permutations and their properties
  • Familiarity with the concepts of even and odd permutations
  • Knowledge of inversions in permutations
  • Ability to represent permutations using cycle notation
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  • Learn how to determine the cycle structure of a permutation
  • Explore algorithms for efficiently counting swaps in permutations
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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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