0

Wrong solution

I have tried to make a schedule for player a, b, c, d, e with some criteria:

  • Two people are playing on each team.
  • Each pair should play against each pair possible (each pair is playing against 3 different pairs).
  • There should be equal time between breaks, e.g. if someone takes a break, the player should play 4 matches before the person is taking a break.
  • After every 5 rounds, each unique pair should only have played once together, e.g. player a and player b should not play two games together in the first 5 games. (This is not fulfilled in my example since player b and d are already playing together in match 6 and 8).

To explain the last criteria: Overview of each unique pair. Each of these pairs should only have played 1 time after 5 rounds, 2 times after 10 rounds, 3 times after 15 rounds.

I think it should be possible to make or is this an impossible problem? Have been using the last 5 hours trying to solve this but maybe there is a smart way?

Maurice Schneider
  • 9
  • Maybe I am misunderstanding the problem, but if the first 5 games are fine, then how about using permutation? After replacing b and c, change the order of games so that breaks for players are fair. – MathDrifter Jan 11 '24 at 01:00
  • ab, cd, e. ac, be, d. ae, bd, c. ad, ce, b. bc, de, a. – Gerry Myerson Jan 11 '24 at 01:32
  • @MathDrifter Yes, I have the all the combinations. But arranging the order where the breaks are even and that a pair is not playing two times in 5 rounds is the problem. – Maurice Schneider Jan 11 '24 at 11:26
  • @GerryMyerson Thanks, that works for the first 5 rounds but it's the last 10 rounds which is difficult. – Maurice Schneider Jan 11 '24 at 11:26
  • @MauriceSchneider So the 2nd & 4th criteria are about a streak of 15 rounds, right? Each combination should be done once in a streak. I saw it is impossible. Not answering because I did brute-force. – MathDrifter Jan 11 '24 at 11:58
  • @MathDrifter Yes, I thought 15 since after 15 rounds, each pair has played against all possible combinations. Oh, so it is actually impossible? Then I understand why I couldn't find a solution. – Maurice Schneider Jan 11 '24 at 13:46

0 Answers0