Let’s look at some permutations: 1 2 3 4 1 2 4 3 1 3 2 4 1 3 4 2 1 4 2 3 1 4 3 2 2 1 3 4 … How do we go from one permutation to the next? Firstly, let’s look at things a little differently. We can view … Read more
After some research I was able to implement the “early refusal” algorithm as described e.g. in this paper. It goes like this: import random def random_derangement(n): while True: v = [i for i in range(n)] for j in range(n – 1, -1, -1): p = random.randint(0, j) if v[p] == j: break else: v[j], v[p] … Read more
This method will generate all integers with exactly N ‘1’ bits. From https://graphics.stanford.edu/~seander/bithacks.html#NextBitPermutation Compute the lexicographically next bit permutation Suppose we have a pattern of N bits set to 1 in an integer and we want the next permutation of N 1 bits in a lexicographical sense. For example, if N is 3 and the … Read more
This might be what you’re looking for. private static bool NextPermutation(int[] numList) { /* Knuths 1. Find the largest index j such that a[j] < a[j + 1]. If no such index exists, the permutation is the last permutation. 2. Find the largest index l such that a[j] < a[l]. Since j + 1 is … Read more
The idiomatic way to do this with Pandas is to use the .sample method of your data frame to sample all rows without replacement: df.sample(frac=1) The frac keyword argument specifies the fraction of rows to return in the random sample, so frac=1 means to return all rows (in random order). Note: If you wish to … Read more
To describe a permutation of n elements, you see that for the position that the first element ends up at, you have n possibilities, so you can describe this with a number between 0 and n-1. For the position that the next element ends up at, you have n-1 remaining possibilities, so you can describe … Read more
You can use a back tracking based approach to systematically generate all the permutations: // function to generate and print all N! permutations of $str. (N = strlen($str)). function permute($str,$i,$n) { if ($i == $n) print “$str\n”; else { for ($j = $i; $j < $n; $j++) { swap($str,$i,$j); permute($str, $i+1, $n); swap($str,$i,$j); // backtrack. … Read more
The itertools module has a useful method called permutations(). The documentation says: itertools.permutations(iterable[, r]) Return successive r length permutations of elements in the iterable. If r is not specified or is None, then r defaults to the length of the iterable and all possible full-length permutations are generated. Permutations are emitted in lexicographic sort order. … Read more
Little late, but like to add a slightly more elegant version here. Can be any array… function permutator(inputArr) { var results = []; function permute(arr, memo) { var cur, memo = memo || []; for (var i = 0; i < arr.length; i++) { cur = arr.splice(i, 1); if (arr.length === 0) { results.push(memo.concat(cur)); } … Read more
As stated by RickyBobby, when considering the lexicographical order of permutations, you should use the factorial decomposition at your advantage. From a practical point of view, this is how I see it: Perform a sort of Euclidian division, except you do it with factorial numbers, starting with (n-1)!, (n-2)!, and so on. Keep the quotients … Read more