![]() ![]() For instance, looking at $ be the cycle decomposition of $\sigma. ![]() Now, consider the smallest possible permutations - the ones that interchange two elements, but otherwise leave everything as-is. Define the identity permutation (that is, the one that doesn't move any elements) as an even permutation, since applying it twice will produce itself. "Odd" and "Even" are defined in terms of how they interact. Stack Exchange network consists of 182 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. If you add an odd and an even number, you'll get another odd number. If you add two odd numbers, you'll get an even number. If you add two even numbers, you'll only ever get another even number. Simply put, it's like adding odd and even numbers. ![]() Why? Well, note that we can represent the identity permutation by the product, say, of $(12)(12) = (12)(12)(3)\cdots(n) = (1)(2)(3)\cdots (n)$, so it is indeed an even permutation. One final note: the identity permutation (i.e., the "do nothing" permutation): the permutation which can be represented as the product of one-cycles sending $1 \mapsto 1,\ 2\mapsto 2,\ \cdots, n\mapsto n\ $ is always considered to be an EVEN permutation. If you have a permutation that is the product of disjoint cycles: say three cycles, corresponding to lengths $n_1, n_2, n_3$, then the number of transpositions representing this permutation can be computed by the parity of $(n_1 - 1)+(n_2 - 1) + (n_3 - 1)$ or simply the parity (oddness/evenness) of $n_1+n_2+n_3 - 1$ So a cycle with a length that is even (has an even number of elements) is ODD, and a cycle with a length that is odd (has an odd number of elements) is EVEN. You'll see that the number of transpositions in a product corresponding to a permutation that is a cycle of length $n$ can be expressed as the product of $n - 1$ transpositions. One can always resort to following the pattern: The meaning of EVEN PERMUTATION is a permutation that is produced by the successive application of an even number of interchanges of pairs of elements. If you know cycle notation, knowing the parity (oddness/evenness) can be found fairly easily. There are many ways to write a permutation as the product of transpositions, and they can vary in length, but those products will have either an odd or an even number of factors, never both. an even number of transpositions $\iff$ even permutation.Now, notice that once you write any permutation allowed, it is written as a product of an even number of 2-cycles (you always move the 'empty' tile, it starts in the corner and it has to be still there at the end of your moves). an odd number of transpositions $\iff$ odd permutation to rearrange the puzzle, you have to perform a permutation of the 15 tiles.One thing to note: This still works even if $\sigma$ is not written in terms of disjoint cycles.Every permutation can be expressed as the product of one and only one of the following: So we reverse the list of cycles and then write each one backwards - thus the inverse is just the whole thing written backwards. To find the inverse of a permutation just write it backwards. ![]()
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