Explain xkcd: It's 'cause you're dumb.
Title text: It's like the traveling salesman problem, but the endpoints are different and you can't ask your friends for help because they're sitting three seats down.
Cueball is upset at the way he and his friends have sat down at the movie theater. Part of the problem is that two people who are apparently in a relationship do not sit together. He therefore tries to use a social graph to calculate the best way for him and his seven friends to sit in a row, while taking into account all of the social connections among them. In mathematics, this type of problem is called combinatorial optimisation. The most popular example, the "Travelling Salesman Problem", is referenced in the title text, as well as in comics 287 and 399.
The title text shows that another part of the problem is that Cueball's two friends who could have helped him calculate a solution are each sitting three seats away from him, and so he cannot ask them for help.
- At the movies, I get frustrated when we file into our row haphazardly, ignoring the computationally difficulty problem of seating people together for maximum enjoyment.
- [Map of relationships between 8 people.]
- Single line: friends.
- Double line: in a relationship.
- Arrow: one-way crush.
- Dashed line: acquaintances
- [The eight friends sitting in a row in a dark cinema. Cueball and one other are between two lovers.]
- Cueball: Guys! This is not socially optimal!
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I think that "two friends ... each sitting three seats away" takes "they" a little too literally. Daddy (talk) 06:17, 28 April 2013 (UTC)
Look at it like this. Placing them _friends_ apart guarantees lower background noise so you can actually watch the movie. 18.104.22.168 (talk) (please sign your comments with ~~~~)
Truthfully, I feel one-way crush should be second priority 05:09, 14 November 2015 (UTC) Lonely...
I know I'm horrifically late on this one, but surely swapping 4 and 7 is the solution which minimizes strangers in "Do not keep 8 away from 6".
There is currently no path from 8 to 4. Swapping 4 and 7 downgrades the link to 5 to an Acquaintance from Friendship, but offers a viable path between all members. Unless we're intentionally forcing 8 to talk to 6? 22.214.171.124 08:20, 15 December 2015 (UTC)
The solutions are making assumptions about the weighting of friends, relationships, and acquaintances. For example, maybe someone in a relationship would achieve maximum enjoyment if the SO is seated next to a stranger so all focus is on them. -- Flewk (talk) (please sign your comments with ~~~~)
If we take the "keep 8 away" solution and put 8 in middle of 6 and 3, wouldn't the result be strictly superior to the "do not keep 8 away" one? It'd have the same number of adjacent relationships, but 6 would be seated next to a friend instead of an aquaintance, which is better. 126.96.36.199 16:38, 23 May 2016 (UTC)
8 doesn't know 1 or 5, so "keep 8 away" solution is bad. He would be alone then. 188.8.131.52 15:44, 24 December 2016 (UTC)
Maybe it's best to separate the couples so that no one has to be the third wheel. I propose [1,3,4,5,7,8,6,2]. Now, it might be unfair to #1, since his only connection is his girlfriend, so [3,4,5,7,8,6,2,1] works as well. 184.108.40.206 19:52, 15 August 2017 (UTC)