12/11/2023 0 Comments Random permutation python![]() You can find the cycles by choosing an element, and following its orbit: what values is that element successively transformed into? Here we see that: Permutations of rank 2 are products of disjoint swaps, and we can see that this is not the case of a. The only permutation of rank 1 is the identity itself. This tells us that a can be written as a product of swaps and length-4 cycles. Since a4 is the identity permutation, this tells us that the rank of a divides 4. Or without the unneeded parentheses: a = np.array() Thus your python code is equivalent to: a = np.array() ![]() Well, it turns out that composition of permutations is associative as well! Using numpy's masks, this means that: a] = (a) You already knew that addition and multiplication were associative: x+(y+z) = (x+y)+z and x(yz) = (xy)z. However, a cool result of algebra is that composition of permutations is associative. This last line looks a little bit complicated. Your code is equivalent to: a = np.array() Let us dissipate some confusion by using a different variable name for every different value. How many times did you apply permutation a? Note that because of the assignment a =, array a changed between the first and the second lines a = a. The rank of a product of disjoint cycles is the least common multiple of the ranks of cycles. The rank of a cycle is the length of the cycle. What is k in general? A well-known theorem of algebra states: every permutation can be written as a product of disjoint cycles. The overall period is the least common multiple of the two periods, which is 6. How about two disjoint cycles? Let's try a cycle of length 3 on the first three elements, simultaneously with swapping the last two elements: a = Īs you can see by carefully examining the intermediary results, there is a period of length 3 on the first three elements, and a period of length 2 on the last two elements. For instance, swapping elements 0 and 1: a = How about a cycle of length 2? A cycle of length 2 is just "swapping two elements". In our previous example, a was a cycle of length 3, so it took three applications of a before we found the identity permutation again. If a is a cycle, then the minimum possible k is the length of the cycle. The minimum value of k is called the rank of a permutation. In algebraic notations: for every permutation a, there exists an integer k such that a^k = id. It is a well-known result of algebra that if you apply the same permutation again and again, you will eventually end up on the identity permutation. This is because b was initialised to, which represent the identity permutation id thus, the permutations that we find by repeatedly applying a to b are:Ī^3 = id Can we always go back where we started? or The rank of a permutation You might see an interesting, but unsurprising fact: The first line is equal to a in other words, the first result of applying a to b is equal to a itself. Of course, I could have used numbers from range(n): a = ![]() Note that I used strings for the elements of b ton avoid confusing them with indices. Since the length of the cycle is 3, if you apply it three times, you will end up where you started: a = In this example, array a represents the permutation (2 0 1), which is a cycle of length 3. When you use an array a obtained by shuffling range(n) as a mask for an array b of same size n, you are applying a permutation, in the mathematical sense, to the elements of b. What is the operation c = b? or Applying a permutation ![]() There is no reason to expect a = a to sort the array. ![]()
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