![]() $$$n - 1$$$-th and $$$n$$$-th elements are in place, this means that we represented our permutation as 3-cycles, so our permutation is even and we've found representation for it. After that there are two possible situations. Let's do cycle $$$n \to j \to i$$$, if $$$j \neq n$$$, and if $$$j = n$$$, do $$$n - 1 \to j \to i$$$, so we won't corrupt first $$$i - 1$$$ elements, and we will place $$$i$$$-th element. If $$$i$$$-th element is not in place, then now in current permutation it is on $$$j$$$-th place, $$$j > i$$$. On $$$i$$$-th iteration we assume, that first $$$i - 1$$$ elements of permutation are in place, and we will try to place $$$i$$$-th element. We will try to build our permutation from identical applying 3-cycles. 3-cycle is even, so composition of 3-cycles must be even, so all that we need to do is proof, that for any even permutation exists such representation. Permutation can be represented as composition of 3-cycles if and only if it is even. There are many ways to count number of inversions in $$$O(n \log n)$$$ time.īut there is a simple way to count parity of permutation, without directly counting number of inversions. Counting inversions, parity of permutation So if $$$f$$$ can be represented as $$$k$$$ swaps, and $$$g$$$ can be represented as $$$m$$$ swaps, then $$$f \circ g$$$ can be represented as $$$k + m$$$ swaps, but parity of $$$f \circ g$$$ is parity of $$$k + m$$$, which is sum of parities of $$$f$$$ and $$$g$$$, QED. Applying swap to permutation change its parity, thus if permutation $$$f$$$ can be represented as $$$k$$$ swaps, then parity of $$$k$$$ is parity of $$$f$$$. Parity of $$$(f \circ g)$$$ is sum of parities of $$$f$$$ and $$$g$$$. This lemma is very important, so it is left to reader as an exercise.įact. Applying swap to permutation will change its parity. Since $$$p$$$ is permutation, indegree and outdegree of each vertex is 1, hence our graph is collection of cycles, that's what we want. Consider directed graph where vertices are numbers from $$$1$$$ to $$$n$$$, and edges are $$$i \to p_i$$$ for all $$$i$$$. ![]() Every permutation is composition of non-intersecting cycles. Inversion is pair $$$(i, j)$$$ such that $$$i p_j$$$.Ĭomposition of permutations defined as follows $$$(f \circ g)_x = f_$$$. ![]() Parity of permutation $$$p$$$ is parity of number of inversions in it. Basic conceptsįirst of all, let's define some basic things, you can skip it if you know it. ![]() The fifteen puzzle is a classic application.This blog is inspired by problem 1591D - Yet Another Sorting Problem, hope you will enjoy it.It is therefore unambiguous to call the elements of S n represented by even-length words "even", and the elements represented by odd-length words "odd". ![]() 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. All relations keep the length of a word the same or change it by two. Since with this definition it is furthermore clear that any transposition of two adjacent elements has signature -1, we do indeed recover the signature as defined earlier.Ī third approach uses the presentation of the group S n in terms of generators τ 1.,τ n-1 and relations Consider the permutation σ of the set (\tau) ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |