#PYTHON PERMUTATION HOW TO#
#PYTHON PERMUTATION CODE#

List valuesLeft = new List(_sortedValues) įor (int index = 0 index < size index++) / Calc the index, relative to _sortedValues, of the permutation received

Throw new ArgumentException('sortIndex should be less than factorial(the lenght of items)') Public void GetValuesForIndex(long sortIndex) fruits 'Apple', 'Banana', 'Orange', 'Peach', 'Avocado' def permutation (menu, lst): if len (lst) 0: here prints of each permutation print (' '.join ( fruits i for i in menu)) return the first try which gives the permutations as ever expanding list for i in range (len (lst)): fruit lst i menu.append (fruit) re. / Value is not used as inpu, only as output. / Sort Index is 0 based and should be less than MaxIndex. / Return the permutation relative to the index received, according to Public PermutationOuelletLexico3(T sortedValues) Public readonly long MaxIndex // long to support 20! or less

#PYTHON PERMUTATION CODE#

C# code that should give you the result you expect: using System 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): 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', '1','2','3', '4','5']įor x in map(''.join, itertools.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. This code seems to be a copy of itertools.permutations, so I will use that here: from itertools import permutations allpermutations for index in range (1, nCols+1): allpermutations + permutations (range (nCols), index) print (allpermutations) Note that the second parameter r specifies the length of the permutations to generate. I tried to use map, but am not sure if this works because my p圜harm doesn't result anything.

#PYTHON PERMUTATION HOW TO#

I am not sure how to access each individual permutation at a time without going through another list, since I am not able to store it in a variable, because itertools.permutations returns a permutations object. I don't necessarily want to go through another list of permuted strings and try and access all of them. Basically what I want to do is after I get a permutation, I want to hash it using hashlib and check it against a certain value, which I am getting from a password.txt file before moving on to the next permutation. I want to find all the possible permutations of them, and I am doing so with itertools.permutations. I am using Python, and I have a list of strings (around 30).

Python permutation