You can put all word arrays into one array and use a recursive function like this: function concat(array $array) { $current = array_shift($array); if(count($array) > 0) { $results = array(); $temp = concat($array); foreach($current as $word) { foreach($temp as $value) { $results[] = $word . ‘ ‘ . $value; } } return $results; } else … Read more
google-sheets It is a hard task for native functions. Try a script and use it as a custom function: function getTournament(teams_from_range) { // teams_from_range — 2D Array var teams = []; // convert to list teams_from_range.forEach(function(row) { row.forEach(function(cell) { teams.push(cell); } ); } ); return getTournament_(teams); } function getTournament_(teams) { var start = 0; var … Read more
The simplest way is to use itertools.product: a = [“foo”, “melon”] b = [True, False] c = list(itertools.product(a, b)) >> [(“foo”, True), (“foo”, False), (“melon”, True), (“melon”, False)]
You are looking for the Cartesian Product. In mathematics, a Cartesian product (or product set) is the direct product of two sets. In your case, this would be {1, 2, 3, 4, 5, 6} x {1, 2, 3, 4, 5, 6}. itertools can help you there: import itertools x = [1, 2, 3, 4, 5, … Read more
The limitations described below are because of lambda functions. The first solution can be successfully implemented without lambda: =ARRAYFORMULA(QUERY(BASE(SEQUENCE(PERMUTATIONA(7,7)),7,7),”where not Col1 matches ‘.*((“&JOIN(“)|(“,SEQUENCE(7,1,0)&”.*”&SEQUENCE(7,1,0))&”)).*'”,0)) The trick here is to use regex to find unique elements using query…match. The only problem with this is memory size needed will exceed 10 million for 8 items PERMUTATIONA(8,8). But that … Read more
Returning Combinations Using a numbers table or number-generating CTE, select 0 through 2^n – 1. Using the bit positions containing 1s in these numbers to indicate the presence or absence of the relative members in the combination, and eliminating those that don’t have the correct number of values, you should be able to return a … Read more
There is an important observation (in a comment) that a partition of a set of n elements can be represented as an integer sequence of the form [p1, … pn] where pi is the partition number of element i. For such a sequence to be valid, it must obey the rules that p1 is 1, … Read more
OK, here’s your code (and btw, thanks for posting such an interesting and challenging problem – at least for me… :-)) – using recursion for all possible permutations (by N) given an array of elements) Code : <?php function permutations($arr,$n) { $res = array(); foreach ($arr as $w) { if ($n==1) $res[] = $w; else … Read more
Here is a conceptual answer and a code based on how lex ordering works. (So I guess my answer is like that of “moron”, except that I think that he has too few details and his links have too many.) I wrote a function unchoose(n,S) for you that works assuming that S is an ordered … Read more