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Home/ Questions/Q 108703
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Asked: May 11, 20262026-05-11T01:57:12+00:00 2026-05-11T01:57:12+00:00

(This is no homework and no work issue. It’s just my personal interest/occupation and

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(This is no homework and no work issue. It’s just my personal interest/occupation and completly fictional. But I am interested in a good algorithm or data structure.)

Let’s assume, that I would run a dating site. And my special feature would be that the singles were matched by movie taste. (Why not?)

In that case I would need a way to store the movie ratings for each user. (So far no problem.) And I would need a data structure to find the best fitting user. The distance between two taste patterns would be the average distance between all ratings that both users made.

Example

movies   A B C D E F G H I J K L M ... user Xm  9 5   1   1   5 user Ym      4 6 1         8 user Zf  9   6 4           7 

Distance(X,Z) = avg( abs(9-9) + abs(1-4) ) = 1.5

Distance(Y,Z) = avg( abs(4-6) + abs(6-4) + abs(8-7) ) = 1.666

So Mr. X fits slightly better to Mrs. Z, than Mr. Y does.

I like soulution that …

  • … don’t need many operations on the database
  • … don’t need to handle a lot of data
  • … run fast
  • … deliver the best matching
  • Ok, maybe I would consider good approximations too.

Try to keep in mind that this should also work with thousands of possible movies, users that rate only about 20-50 movies, and thousands of users.

(Because this is a mental puzzle and not a real problem, work-arrounds are not really helping.)

What would be your search algorithm or data structure?

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  1. 2026-05-11T01:57:13+00:00Added an answer on May 11, 2026 at 1:57 am

    Sounds a lot like the Netflix Prize challenge, more specifically the first half of the most popular approach. The possible implementations of what you are trying to do are numerous and varied. None of them are exceptionally efficient, and the L1 metric is not a particularly good option for reliable correlations.

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