查找数组的所有键唯一排列

Question

I have an array that looks something like this:

[["Sunday", [user1, user2]], ["Sunday", [user1, user4]], ["Monday", [user3, user2]]]

The array essentially has all permutations of a given day with a unique pair of users. I obtained it by running

%w[Su Mo Tu We Th Fr Sa].product(User.all_pairs)

where User.all_pairs is every unique pair of users.

My goal now is to compose this set of nested arrays into schedules, meaning I want to find every permutation of length 7 with unique days. In other words, I want every potential week. I already have every potential day, and I have every potential pair of users, now I just need to compose them.

I have a hunch that the Array.permutation method is what I need, but I'm not sure how I'd use it in this case. Or perhaps I should use Array.product ?

Answer 1

If I understand you correctly, you want all possible weeks where there is one pair of users assigned to each day. You can do it like this:

User.all_pairs.combination(7)

This will give you all possible ways of how you can pick 7 pairs and assign them to the days of the week. But if you are asking for every possible week, then it also matters into which day is which pair assigned, and you also have to take every permutation of those 7 pairs:

User.all_pairs.combination(7).map{|week| week.permutation().to_a}.flatten(1)

Now this will give you all possible weeks, where every week is represented as array containing 7 pairs. For example one of the weeks may look like this:

[(user1, user2), (user1, user3), (user2, user3), (user3, user4), (user1, user4), (user2, user4), (user3, user4)]

However the amount of the weeks will be huge! If you have n users, you will have k = n!/2 pairs, there is p = k! / (7! * (k - 7)!) p = k! / (7! * (k - 7)!) ways of selecting 7 pairs and p * 7! possible weeks. If you have just 5 users, you get 1946482876800 possible weeks! No matter what you are planning to do with it, it won't be possible.

If you are trying to find the best schedule for a week, you can try to make some greedy algorithm.