A round robing pairing generator where you are control the round in which certain players get the bye or meet certain other players. If a large number of rounds need to be fixed in this manner and they involve players other than the Bye, there is no guarantee that a pairing exists.

round_robin.py

import sys
import random
from itertools import permutations

class RoundRobin(object):
    '''
    Usage:
        r = RoundRobin(['Bye','Nigel','Wellignton','Andrew','Sherwin','Chollapat','David','Craig'])
        r.fix_pairing('Nigel',5, 'Bye')
        r.fix_pairing( 'Craig',0,'Bye')
        r.brute_force()
        
        Note: if you have an odd number of players you need to add a player called 'Bye' as in the
        example above.
    '''
        

    def __init__(self,players):
        self.players = players
        self.num_players = len(self.players)
        self.num_rounds = self.num_players

        self.games = [[None for x in range(self.num_rounds)] for x in range(self.num_players)]
        self.fixed_rounds = [ [] for x in range(self.num_rounds)]
        self.matches = 0

    def print_table(self, dr = False):
        '''
        Prints the generated pairing table. If this is a double round robin
        tournament set dr = True
        '''
        sys.stdout.write("{:<10}".format(''))
        for rnd in range(1,self.num_players):
            if dr:
                sys.stdout.write("{:<10}".format('{0},{1}'.format(rnd*2-1,rnd*2)))
            else :
                sys.stdout.write("{:<10}".format(rnd))

        print()

        for i in range(self.num_players):
            sys.stdout.write("{:<10}".format(self.players[i]))
            
            for j in self.games[i]:
                if j is not None:
                    if j == -1:
                        p = 'XX'
                    else :
                        p = self.players[j]

                    sys.stdout.write("{:<10}".format(p))
                else :
                    sys.stdout.write("{:<10}".format('None'))

            print()


    def berger_table(self):
        '''
        Produces a Berger table. Standard pairing with no fixing
        '''
        rotator = [i for i in range(self.num_players)]
        mode = len(rotator)//2
        
        for rnd in range(self.num_rounds):
            first = rotator[0:mode]
            second = rotator[mode:]
            second.reverse()

            #print first, second
            for idx in range(mode):
                player = first[idx]
                opponent = second[idx]
                self.games[player][rnd] = opponent
                self.games[opponent][rnd] = player

            popped = rotator.pop()
            rotator.insert(1,popped)
        
    
    def berger_table_with_fixes(self, permutation):
        '''
        Brute force a solution to the pairing problem.
        A solution might not always exists. The possibilities that needs to be
        examined grows as N! so for large number it might never complete
        '''
        self.players = permutation
        rotator = [i for i in range(self.num_players)]

        mode = len(rotator)/2
        
        for rnd in range(self.num_rounds):
            first = rotator[0:mode]
            second = rotator[mode:]
            second.reverse()

            #print first, second
            for idx in range(mode):
                player = first[idx]
                opponent = second[idx]
                self.games[player][rnd] = opponent
                self.games[opponent][rnd] = player

            if self.fixed_rounds[rnd]:
                for p,o in self.fixed_rounds[rnd]:
                    player_idx = self.players.index(p)
                    opponent_idx = self.players.index(o)

                    if self.games[player_idx][rnd] != opponent_idx :
                        return False

            popped = rotator.pop()
            rotator.insert(1,popped)

        self.matches +=1

        return True

    def fix_pairing(self, player, rnd, opponent):
        ''' Marks player has having to play opponent in round n '''
        self.fixed_rounds[ rnd ].append(  (player, opponent) )

        
    def brute_force(self):
        '''
        Brute force method to find a pairing that meets all player requirements.
        '''
        i = 0
        for perm in permutations(r.players):
            i +=1
            if self.berger_table_with_fixes( list(perm)):
                self.print_table() 
                break
            if i == 100000000:
                break

Congratulations, this is just what I was looking for.

An usual case in chess championships is a round-robin tournament with several players from the same club. These players must be paired as soon as possible to avoid cheating (i.e. a player who has no chances of winning plays against a player with chances in the last round, and both are of the same club).

This is solved with a simple Berger table where the players are assigned, but it's not a good solution, often having pairings in the last rounds.

I have tested your algorithm manually entering the first rounds of a real use case (8 players from three clubs, with 4 players in one club and 2 players in the others). It works nice, but soon the fixed pairings are impossible and some players of the same club are paired in the last rounds.

An algorithm which would receive some players of several clubs and would find the best permutation for this use case would be highly useful in chess world! Just finding the Berger table where all players of the same clubs are paired before.

If you manage to find the solution, please share it!
Thanks and best regards.