In C There’s a pretty straightforward description of an algorithm (plus implementation) at geeksforgeeks: Given a string, print all permutations of it in sorted order. For example, if the input string is “ABC”, then output should be “ABC, ACB, BAC, BCA, CAB, CBA”. We have discussed a program to print all permutations in this post, … Read more
java.util.Collections.shuffle(List); javadoc link for Collections.shuffle List<Integer> list = new ArrayList<Integer>(); list.add(1); list.add(2); list.add(3); java.util.Collections.shuffle(list); It’s worth noting that there are lots of algorithms you can use. Here is how it is implemented in the Sun JDK: public static void shuffle(List<?> list, Random rnd) { int size = list.size(); if (size < SHUFFLE_THRESHOLD || list instanceof … Read more
I suggest you use List::Permutor: use List::Permutor; my $permutor = List::Permutor->new( 0, 1, 2); while ( my @permutation = $permutor->next() ) { print “@permutation\n”; }
Given the R data.frame: > df1 a b c 1 1 1 0 2 1 0 0 3 0 1 0 4 0 0 0 Shuffle row-wise: > df2 <- df1[sample(nrow(df1)),] > df2 a b c 3 0 1 0 4 0 0 0 2 1 0 0 1 1 1 0 By default sample() … Read more
Here is a sample solution in Scala, which I will explain in detail: /** example: index:=15, list:=(1, 2, 3, 4) */ def permutationIndex (index: Int, list: List [Int]) : List [Int] = if (list.isEmpty) list else { val len = list.size // len = 4 val max = fac (len) // max = 24 val … Read more
Note: this answer is for 1-based rankings, as specified implicitly by example. Here’s some Python that works at least for the two examples provided. The key fact is that suffixperms * ctr[y] // ctr[x] is the number of permutations whose first letter is y of the length-(i + 1) suffix of perm. from collections import … Read more
In C++ you can use std::next_permutation to go through permutations one by one. You need to sort the characters alphabetically before calling std::next_permutation for the first time: cin>>anagrama; int len = strlen(anagrama); sort(anagrama, anagrama+len); do { cout << anagrama << endl; } while (next_permutation(anagrama, anagrama+len)); Here is a demo on ideone. If you must implement … Read more
Here is a non-recursive algorithm in C++ from the Wikipedia entry for unordered generation of permutations. For the string s of length n, for any k from 0 to n! – 1 inclusive, the following modifies s to provide a unique permutation (that is, different from those generated for any other k value on that … Read more
You want to do recursion, so you first have to find out how the recursion would work. In this case it is the following: permutation [a,b,c,…] = [a + permutation[b,c,…], b + permutation[a,c,..], …] And as a final condition: permutation [a] = [a] So the recursion splits up the list in sublists with one element … Read more
If allPossibleItems contains two different elements, x and y, then you successively write x and y to the list until it reaches DESIRED_SIZE. Is that what you really want? If you pick DESIRED_SIZE sufficiently large, you will have too many recursive calls on the stack, hence the SO exception. What I’d do (if original has … Read more