# Machine Learning (Theory)

## 9/9/2006

### How to solve an NP hard problem in quadratic time

This title is a lie, but it is a special lie which has a bit of truth.

If n players each play each other, you have a tournament. How do you order the players from weakest to strongest?

The standard first attempt is “find the ordering which agrees with the tournament on as many player pairs as possible”. This is called the “minimum feedback arcset” problem in the CS theory literature and it is a well known NP-hard problem. A basic guarantee holds for the solution to this problem: if there is some “true” intrinsic ordering, and the outcome of the tournament disagrees k times (due to noise for instance), then the output ordering will disagree with the original ordering on at most 2k edges (and no solution can be better).

One standard approach to tractably solving an NP-hard problem is to find another algorithm with an approximation guarantee. For example, Don Coppersmith, Lisa Fleischer and Atri Rudra proved that ordering players according to the number of wins is a 5-approximation to the NP-hard problem.

An even better approach is to realize that the NP hard problem may not be the real problem. The real problem may be finding a good approximation to the “true” intrinsic ordering given noisy tournament information.

In a learning setting, the simplest form of ranking problem is “bipartite ranking” where every element has a value of either 0 or 1 and we want to order 0s before 1s. A common way to measure the performance of bipartite ranking is according to “area under the ROC curve” (AUC) = 1 – the fraction of out-of-order pairs. Nina, Alina, Greg and I proved that if we learn a comparison function which errs on k dissimilar pairs, then ordering according to the number of wins yields an order within 4k edge reversals of the original ordering. As a reduction statement(*), this shows that an error rate of e for a learned pairwise binary classifier produces an ordering with an expected AUC of 1 – 4e. The same inequality even holds for a (stronger) regret transform. If r = e – emin is the regret of the binary pairwise classifier, then the AUC regret is bounded by 4r. (Here emin is the error rate of the best possible classifier which predicts knowing the most likely outcome.) The regret result extends to more general measures of ordering than simply AUC.

We were unable to find any examples where ordering according to the degree produced more than a 2r AUC regret. Nikhil Bansal, Don, and Greg have worked out a tightened proof which gives exactly this upper bound. At the end of the day, we have an algorithm with satisfies precisely the same guarantee as the NP hard solution.

There are two lessons here. The learning lesson is that a good pairwise comparator implies the ability to rank well according to AUC. The general research lesson is that an NP hard problem for an approximate solution is not an intrinsic obstacle. Sometimes there exist simple tractable algorithms which satisfy the same guarantees as the NP hard solution.

(*) To prove the reduction, you must make sure that your form pairwise examples in the right way. Your source of pairwise ordering examples must be uniform over the dissimilar pairs containing one example with label 1 and one example with label 0.

Sorry, the comment form is closed at this time.

Sorry, the comment form is closed at this time.