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- and (en)
- Consider the elements that are sandwiched by the two elements of a transposition. Each one lies completely above, completely below, or in between the two transposition elements.
An element that is either completely above or completely below contributes nothing to the inversion count when the transposition is applied. Elements in-between contribute .
As the transposition itself supplies inversion, and all others supply 0 inversions, a transposition changes the parity of the number of inversions. (en)
- , we have
:
Now for a given permutation σ of the numbers {1, ..., n}, we define
:
Since the polynomial has the same factors as except for their signs, it follows that sgn is either +1 or −1. Furthermore, if σ and τ are two permutations, we see that
:
It can be shown that any transposition of two elements has signature −1, and thus we do indeed recover the signature as defined earlier. (en)
- is called an inversion. We want to show that the count of inversions has the same parity as the count of 2-element swaps. To do that, we can show that every swap changes the parity of the count of inversions, no matter which two elements are being swapped and what permutation has already been applied.
Suppose we want to swap the ith and the jth element. Clearly, inversions formed by i or j with an element outside of (en)
- A third approach uses the presentation of the group Sn in terms of generators τ1, ..., τn−1 and relations
* for all i
* for all i < n − 1
* if
[Here the generator represents the transposition (en)
- .] All relations keep the length of a word the same or change it by two. Starting with an even-length word will thus always result in an even-length word after using the relations, and similarly for odd-length words. It is therefore unambiguous to call the elements of Sn represented by even-length words "even", and the elements represented by odd-length words "odd". (en)
- Recall that a pair x, y such that (en)
- inversions are formed. The count of inversions i gained is thus (en)
- Let σ be a permutation on a ranked domain S. Every permutation can be produced by a sequence of transpositions . Let the following be one such decomposition
:σ = T1 T2 ... T'k
We want to show that the parity of k is equal to the parity of the number of inversions of σ.
Every transposition can be written as a product of an odd number of transpositions of adjacent elements, e.g.
: = .
Generally, we can write the transposition on the set {1,...,i,...,i+d,...} as the composition of 2d−1 adjacent transpositions by recursion on d:
* The base case d=1 is trivial.
* In the recursive case, first rewrite as . Then recursively rewrite as adjacent transpositions.
If we decompose in this way each of the transpositions T1 ... T'k above, we get the new decomposition:
:σ = A1 A2 ... Am
where all of the A1...Am are adjacent. Also, the parity of m is the same as that of k.
This is a fact: for all permutation τ and adjacent transposition a, aτ either has one less or one more inversion than τ. In other words, the parity of the number of inversions of a permutation is switched when composed with an adjacent transposition.
Therefore, the parity of the number of inversions of σ is precisely the parity of m, which is also the parity of k. This is what we set out to prove.
We can thus define the parity of σ to be that of its number of constituent transpositions in any decomposition. And this must agree with the parity of the number of inversions under any ordering, as seen above. Therefore, the definitions are indeed well-defined and equivalent. (en)
- [i, j] (en)
- elements within the interval (en)
- n − 2vi (en)
- n − vi (en)
- n = 3 (en)
- n = j − i − 1 (en)
- will not be affected.
For the (en)
- x < y (en)
- , assume v'i of them form inversions with i and v'j of them form inversions with j. If i and j are swapped, those vi inversions with i are gone, but (en)
- , which has the same parity as n.
Similarly, the count of inversions j gained also has the same parity as n. Therefore, the count of inversions gained by both combined has the same parity as 2n or 0. Now if we count the inversions gained by swapping the ith and the jth element, we can see that this swap changes the parity of the count of inversions, since we also add 1 to the number of inversions gained for the pair .
Note that initially when no swap is applied, the count of inversions is 0. Now we obtain equivalence of the two definitions of parity of a permutation. (en)
- An alternative proof uses the Vandermonde polynomial
:
So for instance in the case (en)
- σ > σ (en)
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