Build an algorithm to predict friendships, then actually use it to meet people

Build an algorithm to predict friendships, then actually use it to meet people(joingrouper.com)

60 points by waxman 12 years ago | 37 comments

avalaunch 12 years ago |

This is a cool challenge but the prize is definitely lacking. I think anyone capable of writing an algorithm of the caliber you're looking for isn't likely to participate. I could be wrong but I think you're going to have to pony up some serious cash to get developers taking this seriously. Or you could go the more standard route and just hire someone to do the job.

chegra 12 years ago | |

I would do it cause I'm curious.

Edit: Now looking at the dataset, I wouldn't be able to use the model I developed personally.

benhamner 12 years ago |

Ping me (b@kaggle.com) if you're interested in running this competition more formally on https://kaggle.com.

We've run hundreds of machine learning competitions & offer a real-time leaderboard to encourage competitive participation, a very active community of data scientists, and many other features that simplify running this type of challenge.

mariusz79 12 years ago |

So basically they are asking people to build them an algorithm that will be a critical part of their business, in exchange for a free service that will be based on this algorithm. Right...

nottombrown 12 years ago | |

Sorry if this was unclear. You own any code that you write for this competition.

The prize is that we'll use your algorithm to validate any matches that you go on. If that doesn't seem worthwhile to you, feel free to pass on this contest.

Aloisius 12 years ago | | |

Do you allow closed source entries? Rewriting an algorithm implemented in someone else's code to avoid copyright infringement is trivial not to mention inevitable given the performance requirement differences between a contest and a production site.

TrevorJ 12 years ago | | |

So you won't use the winning solution in your product?

angersock 12 years ago | |

Yep, basically.

EDIT: ah fuck it might hack on this dataset anyways--if i get a beer and a date out of it ill help fiddle while rome burns

mjmahone17 12 years ago |

This is interesting, but given your parameters (predict the most friendships), all you're technically asking for is recall. I'll write an algorithm that has 100% recall: predict that all people become friends with each other.

If this is really a competition (and not just "Here, have fun with our dataset!"), you need to define the rules a little bit more clearly. How are you weighing recall vs. precision? Or are you just looking at % correct labels, where the only two labels possible are "FRIENDS" and "NOT FRIENDS"?

nottombrown 12 years ago | |

Sorry this was unclear. We meant "correctly predict the most friendships"

You get 1 point for each friendship that you correctly predict did or did not occur. In the test data set ~50% of pairs became friends, so predicting "everyone became friends" would get 250 points, whereas a perfect algorithm would get 500 points.

I'm updating the README now to make our scoring system more clear.

glimcat 12 years ago | |

They're also looking for whether people become friends on Facebook.

The dominant factor here is going to be the rate at which the participants send and accept connection requests on Facebook. Some people send them to everyone they meet, some people never use Facebook.

KPI overfitting, yay!

(The best second-order effect is probably a multi-feature similarity measure between the participants and the person's current Facebook Friends, including graph distance to current Friends. In case anyone is taking a run at this.)

thebiglebrewski 12 years ago |

This would be a little more fun if there was a cash prize. No offense meant, groupers look cool, but you'd probably get some more participation that way.

easy_rider 12 years ago | |

But then they could just hire a M.S. in Computer Science?

nottombrown 12 years ago |

Hey HN, Grouper founder here. Let me know if you have any questions about the contest.

maxk42 12 years ago |

I'll be the first to say it: Your data is either incorrect, arbitrary, or we're missing some information here.

Why does everyone have "7.5" - 8 siblings and 7.5 - 8 "weekly workouts" and 7.5 - 8 platinum albums?

maxk42 12 years ago | |

Specifically, you should explain all the columns, including:

- Is that the person's height in inches?

- What does the asterisk in certain column-names indicate?

- Why do the pets, platinum_albums, weekly_workouts, number_of_siblings and pokemon_collected values seem to fall in the range of 7 - 8?

Also, this dataset is far too small. There is a single male-male relationship and that's not going to provide any significant data if we're looking at genders at all.

I would also argue that it's not the best set of metrics to use to determine whether people will become friends. Age and facebook_friends_count might give you some hints, but I seriously doubt that shoe size has as big an impact on the potential for friendship as, say, common interests, shared culture, income class, or other socioeconomic factors.

nottombrown 12 years ago | |

The headers with asterisks are intentionally mislabeled. Updated this to be more clear in the README.

JFoss117 12 years ago | | |

You write in the README that the mislabeled columns are "from our internal ratings". Can you give any more definite sense of what this means? What kind of things are these ratings based off of? What are they designed to reflect? How are they computed (roughly)?

chegra 12 years ago |

Mutual Information for the fields:

I(f_facebook_friends_count,members_became_friends) = 0.117320113379

I(m_facebook_friends_count,members_became_friends) = 0.113972809724

I(m_facebook_photos_count,members_became_friends) = 0.0449092782303

I(f_facebook_photos_count,members_became_friends) = 0.0426531483254

I(m_shoe_size,members_became_friends) = 0.00276175766018

I(m_height,members_became_friends) = 0.00255043390135

I(f_shoe_size,members_became_friends) = 0.00233148724025

I(m_age,members_became_friends) = 0.00198005768283

I(f_height,members_became_friends) = 0.0013606978915

I(m_weekly_workouts,members_became_friends) = 0.00123271513215

I(f_age,members_became_friends) = 0.00122660347743

I(m_platinum_albums,members_became_friends) = 0.00111710129455

I(f_number_of_pets,members_became_friends) = 0.00108593667378

I(f_pokemon_collected,members_became_friends) = 0.000880040104571

I(m_number_of_siblings,members_became_friends) = 0.000830295252089

I(f_platinum_albums,members_became_friends) = 0.000820683185117

I(m_number_of_pets,members_became_friends) = 0.000768855827053

I(m_pokemon_collected,members_became_friends) = 0.000720822383999

I(f_weekly_workouts,members_became_friends) = 0.000620666529567

I(f_number_of_siblings,members_became_friends) = 0.00019278884716

I(f_gender,members_became_friends) = 0.000124279429698

I(m_gender,members_became_friends) = 0.000124279429698

icebraining 12 years ago |

This reminds of http://robrhinehart.com/?p=1005

That fact that the women are depicted as just three pairs of legs doesn't help, though.

joshfraser 12 years ago |

Ok, let's make this more interesting. I'll pay $50 to the first person to de-anonymize their training set.

chbg 12 years ago |

members_became_friends = 1/(1+ exp(-1297.88087 * f_shoe_size + m_shoe_size * m_facebook_friends_count - 11761.6138))