![]() ![]() The good thing is that the algorithm works even when you have a sequence with repeated elements: with, say, "232254421", it would find the tail as "54421", swap the "2" and "4" (so "232454221"), reverse the rest, giving "232412245", which is the next permutation. The number of permutations on a set of elements is given by ( factorial Uspensky 1937, p. It might seem that it can take O(n) time per permutation, but if you think about it more carefully, you can prove that it takes O(n!) time for all permutations in total, so only O(1) - constant time - per permutation. We have directly imported the permutations object from the itertools module in the above code. A permutation, also called an 'arrangement number' or 'order,' is a rearrangement of the elements of an ordered list into a one-to-one correspondence with itself. ![]() The C++ code does precisely this (look at the source in /usr/include/c++/4.0.0/bits/stl_algo.h on your system, or see this article) it should be simple to translate it to your language: template īool next_permutation(BidirectionalIterator first, Once you do this, you can avoid using a sorting algorithm for (3.), because the tail was, and is still (think about this), sorted in decreasing order, so it only needs to be reversed. You can do (2.) by just swapping the "4" with the '2", so you'll have "34521". You can do (1.) efficiently by starting at the end and going backwards as long as the previous element is not smaller than the current element. Change the number just before the tail (the "2") to the smallest number bigger than it in the tail (the 4).Find the longest "tail" that is ordered in decreasing order.The algorithm is to implement precisely that line of reasoning: Now, once you've decided that the permutation will start as "34", the rest of the numbers should be in increasing order, so the answer is "34125".So you'll have to change the "2" to something bigger - in fact, to the smallest number bigger than it in the "541" part, namely 4.there is no way to keep the "32" fixed and find a later permutation in the "541" part, because that permutation is already the last one for 5,4, and 1 - it is sorted in decreasing order.And your thoughts were probably something this: In "32541", ![]() If you think about it, you'll see that it is "34125". The idea is this: suppose you are given a sequence, say "32541". The algorithm actually finds the next permutation - the lexicographically next one. The C++ standard template library (STL) even has a function called next_permutation. Yes, there is a "next permutation" algorithm, and it's quite simple too. Indices, indices = indices, indicesĪnd another, based on itertools.product: def permutations(iterable, r=None):įor indices in product(range(n), repeat=r): If len(elements) AB AC AD BA BC BD CA CB CD DA DB DC If for some reason you want to implement it yourself or are just curious to know how it works, here's one nice approach, taken from : def all_perms(elements): There's a function in the standard-library for this: itertools.permutations. ![]()
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