![]() ![]() So, you may be better off for both small and large n if you don't insist on it being theoretically O(n).Ī bit too late. In practice, for small n, O(n^2) or O(n log n) methods to rank or unrank a permutation, and also methods requiring O(2^n) or O(n!) memory to store some precomputed values, may be faster than an O(n) method involving integer division, which is a relatively slow operation on modern processors.įor n large enough so that the n! does not fit into a machine word, the "O(n) if order-n! integer operations are O(1)" argument stops working. This is, strictly speaking, not true: for large n, the number n! contains on the order of O(n log n) binary digits, and that division's cost is proportional to the number of digits. Now, if we consider division of our number by an integer from 1 to n to be O(1) operation, transforming the number to factorial system is O(n) such divisions. Thus we will get a unique sequence to feed the above pseudocode with. ![]() , and there are n possibilities from 0 to n-1 for the first digit. What we want to do now is to write an integer from 0 to n!-1 in factorial number system: the last digit is always 0, the previous one is 0 or 1. (2) There are n! permutations of order n. , the last one can be any integer from 0 to n-1. So, we substitute random integers by non-random ones: the first one is 0, the second one 0 or 1. This transform is injective: if we give it a different sequence of random integers, it is guaranteed to produce a different permutation. ![]() (1) Consider Fisher-Yates shuffle method to generate a random permutation (pseudocode below): Note that although there is an in-built method, understanding the logic behind it and implementing on our own is a good way to practice.įeel free to leave any sort of feedback, suggestions, doubts below.If you don't care about which permutations get which indices, an O(n) solution becomes possible if we consider that arithmetic operations with arbitrary integers are O(1).įor example, see the paper " Ranking and unranking permutations in linear time" by Wendy Myrvold and Frank Ruskey. Given below is the output for the same array We import the specific function “permutations” from the itertools library, call the function and print the set of values returned by the function The post simply shows the way to use it!Ĭonsider the following program from itertools import permutations Yes, python does have an in-built library function to generate all possible permutations of a given set of elements. Method 2 – In-Built Method – All permutations While calling the function, we obviously have to pass the array and indexes as 0 and length-1. (Refer to this)īelow is an output printing all permutation for an array. Note here that we use list(arr) to make sure we are doing a deep copy and not a shallow copy. We append all the permutation results into an array final. Notice that we keep passing smaller parts of the same array to the same function by modifying the index values and hence generate all possible permutations. At the end of that recursion, we will have all possible permutations Implementation in PythonĬonsider the following program, final = list() Hence, this process continues until we reach the last element and try it as the first element, in the first recursion depth. Now here, 3 is tried as the first element. In this recursion, this has to be the one and only first element, hence a permutation is printed and control goes back by one recursion depth. Then we call the array with the remaining element i.e. This recursion will take 2 as the first element. We take 1 as first element, then for the remaining part of the array, we call the same function. The idea is to take up every element in the array and place it at the beginning and for every such case, recursively do the same for a smaller instance of the same array. ![]() Method 1: generate all possible permutations in Python Prerequisites: Basics of loops and conditionals in Python. Hence if there is a repetition of elements in the array, the same permutation may occur twice. We consider numeric elements in an array here and do not consider repetition of the same elements. This post deals with methods to generate all possible permutations in Python, of a given set of elements. ![]()
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