Summary The full combinations of 4 to 9 distinctive numbers, minus the combinations which include invalid “jump”s. The Long Version The rule for Android 3×3 password grid: one point for once cannot “jump” over a point The author of the original post used Mathematica to generate all 985824 combinations. Because there is no “jump”, several … Read more
You can make in manually, using the fact that n-bit binary number naturally corresponds to a subset of n-element set. private IEnumerable<int> constructSetFromBits(int i) { for (int n = 0; i != 0; i /= 2, n++) { if ((i & 1) != 0) yield return n; } } List<string> allValues = new List<string>() { … Read more
If you want to keep the key:value in the permutations you can use: import itertools keys, values = zip(*my_dict.items()) permutations_dicts = [dict(zip(keys, v)) for v in itertools.product(*values)] this will provide you a list of dicts with the permutations: print(permutations_dicts) [{‘A’:’D’, ‘B’:’F’, ‘C’:’I’}, {‘A’:’D’, ‘B’:’F’, ‘C’:’J’}, … ] Disclaimer: not exactly what the OP was asking, … Read more
What about: [ (x,y) | (x:rest) <- tails xs , y <- rest ]
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
You can implement this like an odometer, which leads to the following (works for different-sized vectors): Say you have K vectors in an array v: v[0], v[1], … v[K-1] Keep an array of iterators it (size K) into your vectors, starting with it[i] = v[i].begin(). Keep incrementing it[K-1] in a loop. When any iterator hits … Read more
A variation on Neil’s solution which doesn’t need an extension is: create table friendz ( from_id int, to_id int ); create unique index ifriendz on friendz(greatest(from_id,to_id), least(from_id,to_id)); Neil’s solution lets you use an arbitrary number of columns though. We’re both relying on using expressions to build the index which is documented https://www.postgresql.org/docs/current/indexes-expressional.html
This answer missed one aspect: the OP asked for ALL combinations… not just combinations of length “r”. So you’d either have to loop through all lengths “L”: import itertools stuff = [1, 2, 3] for L in range(len(stuff) + 1): for subset in itertools.combinations(stuff, L): print(subset) Or — if you want to get snazzy (or … Read more
itertools can do this: import itertools for comb in itertools.combinations([1, 2, 3, 4], 3): print(comb) Outputs: (1, 2, 3) (1, 2, 4) (1, 3, 4) (2, 3, 4)
What you’re looking for is essentially the power set (minus perhaps the empty set). Guava actually has a method for this: Sets.powerSet(). You can view the source of the Sets class to see how the method is implemented if you want to write it yourself; you might need to modify it to return a List … Read more