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Home/ Questions/Q 1928842
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T07:01:59+00:00 2026-05-17T07:01:59+00:00

I’m trying to write a program to automate a ticket draft. We have a

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I’m trying to write a program to automate a ticket draft.

We have a certain number of season ticket passes and want to split up the tickets among a group of people. There are X number of games, Y number of season passes, and Z number of people. Each of Z people has ranked the X games.

My code basically goes through the draft order and back picking out the tickets from their ranking if available, otherwise, picking the next highest ranking. For the most part it works. The problem is, there’s a point where most of the tickets are taken and the remaining tickets left are ones you already have so you just don’t pick them. People therefore have different numbers of tickets. Is there a good way to get around this?

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  1. Editorial Team
    Editorial Team
    2026-05-17T07:01:59+00:00Added an answer on May 17, 2026 at 7:01 am

    If there were no preferences involved, this would be a straight min-cut max flow problem. http://en.wikipedia.org/wiki/Maximum_flow_problem, as follows:

    Create a source vertex A. From A, create Z vertices, one for each person. The capacity can be infinite (or very, very large). Create a sink B, and create X vertices, one for each game, linked to B; the capacity should be Y (you have Y tickets per game). From each person, link to each game they’ve ranked, with capacity 1.

    If you look at the wiki link above, there are about 10 algorithms to solve this basic problem. Find one you understand and can implement yourself, because you’ll need to modify it slightly. I’m not familiar with all of them, but the ones I know about have a step ‘pick an edge’ or ‘pick a path.’ You should modify the ‘how you pick an edge’ logic to take the priority ordering of the games into account. I’m not sure exactly what the ordering should be (you’ll probably need to experiment), but if you say the lowest ranked game is 1, the next is 2, up to X, then a score like ‘ranking of the edge – number of games the person is already signed up for’ might work.

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