[Python] Find ternary relationships with no duplicates for game scheduling algorithm

I'm working on assigning groups in Python using rules.

You could think about it like building a season for a three player based game.

Each player needs to face the other players on the list exactly once.

So suppose I have this list:


players = ['a','b','c','d','e','f','g','h','i','j','k']



Here's the expected output if we start with 'a':

game schedule for a:

game 1 = a, b, c

game 2 = a, d, e

game 3 = a, f, g

game 4 = a, h, i

game 5 = a, j, k

I'll need a schedule like that above for each player.

What I've got right now gets me 45 combinations for player 'a'... trying to figure out how to whittle it down:


players = ['a','b','c','d','e','f','g','h','i','j','k']

from itertools import combinations

num = 3

all_games = list(combinations(players, num))

a_s = []

for i in all_games:
if 'a' in i:
a_s.append(i)



I'm just not sure how to build in the condition I want.

I suppose I could (in pseudo code):


create a list without the player I want:

player = a
opponents = ['b','c','d','e','f','g','h','i','j','k']

extract/get every 2 elements from the list

add player a to the extracted opponent pairs



But I don't know if that's close to the best way to solve this.

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