From: Matthew Moss Date: 2008-06-13T06:06:31+09:00 Subject: [SUMMARY] Preferable Pairs (#165) ------=_Part_15389_5027399.1213304819512 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 7bit Content-Disposition: inline Apologies for being a little late today... Took a little bit to understand some of what was going on in the solutions for the Preferable Pairs quiz, and I'm not certain I explained it well enough, but... here it is. This "Preferable Pairs" quiz is, as was noted, very similar to the [Stable Roommates][1] problem, though the goal as originally stated is slightly different from the typical presentation of the Stable Roommates problem. Still, the minor goal difference does not change that this is an [NP-complete][2] problem, and as such, large input sets would require some sort of approximation, heuristics or other techniques to keep the run time reasonable. Fortunately, I had no intention of testing the solutions provided on large sets. My test involved a set of 20 people, to be grouped into 10 pairs according to their preferences. Here are the results of my test on the solutions provided. Solution Time Score Andrea Fazzi 0.159 438 Dustin Barker 0.020 589 Eric Ivancich 4.671 311 Steven Hahn -DNF- Eric Mahurin 0.211 311 Matthew Moss 0.022 589 Thomas ML 0.114 311 I neglected to post my own solution, but it's nearly identical to Dustin's both in algorithm, results and performance (but mine is much uglier). Also, apologies to Steven... I tried to wait, really I did, but it just kept going, and going... Andrea, Dustin and myself made straightforward attempts that were fast but not optimal. While the numbers presented above don't show a huge disparity in performance, I suspect that would be a different story if we were attempting to solve larger datasets, of thousands of people as opposed to twenty. For the tested dataset, I believe there might be a few grumbling people, forced to play with someone they didn't like very much, but the tournament would go on. Eric Ivancich provided a genetic algorithm that is notably slower (at this sample size), and isn't guaranteed to get the most optimal answer, but it happened to do so here. It would be interesting to compare its performance against the optimal solution algorithms for larger data sets. Eric Mahurin and Thomas ML provided algorithms that find the optimal solution. There is a fair bit of setup code in both to get the data into a convenient form for the optimization algorithm to work. I'm going to skip over that and jump right into Thomas' `optimize` method. As input to `optimize`, the `pairings` argument looks like this for the sample data: [ [["David", "Helen"], 0], [["David", "Vicki"], 1], [["Helen", "Vicki"], 2], [["Joseph", "Vicki"], 4], [["Helen", "Joseph"], 5], [["David", "Joseph"], 8] ] Basically, the input is an ordered list of all pairs with the corresponding score (as calculated according to the metric as described by the quiz). It is a combined score, and reflects the preferences of both people in the pair. For example, the pair of Helen and Joseph has a score of 5, because Helen scores Joseph as 4, while Joseph scores Helen as 1: so, 4 + 1 == 5. Before we conquer the whole of the algorithm, let's look at a simplified version of `optimize` to get a feel for the general structure. The input here (for argument `pairings`) is similar to the above, but without the scores. def optimize(pairings, names, pos, posmax) bestpairs = nil while pos < posmax pair = pairings[pos] pos += 1 if names & pair == pair names1 = names - pair if names1.size < 2 bestpairs = [pair] bestpairs << names1 unless names1.empty? return bestpairs elsif (rv = optimize(pairings, names1, pos, posmax) bestpairs = rv bestpairs << pair end end end return bestpairs end `bestpairs` starts off empty, but by the end should be an array of pairs chosen as the solution. Each iteration of the loop grabs the next pair from `pairings` and checks to see if it is still usable by set intersection: if names & pair == pair `names` will contain the names of people still available; that is, those that have not already been chosen previously. The `&` operator treats the two arrays as sets and performs set intersections. If the result of intersection is equal to one of the arguments, that argument must be a subset of the other, and in this context, is safe to choose for the solution. When a pair is chosen, those names are removed by Array difference (not quite the same as set difference, but close enough for this case). So it is with: names1 = names - pair That we remove the chosen pair from the list of remaining people. If that is the last possible pair to be made (there are fewer than 2 names remaining), we finish up by setting and returning the `bestpairs` array bestpairs = [pair] bestpairs << names1 unless names1.empty? return bestpairs In the case that there is at least one more pair remaining, we continue onto the recursive part of this solution: elsif (rv = optimize(pairings, names1, pos, posmax) bestpairs = rv bestpairs << pair We know that `bestpairs`, as returned by the recursive call to `optimize`, contains the best pairs from the subset of `names1`, which we got above after removing `pair`. So we make sure to concatenate `pair` onto `bestpairs` before it is returned to the caller. As simplified, this amounts to a greedy algorithm, since the pairs were initially sorted according to score, and the simplified algorithm, at each level, simply takes the first pair possible. Now let's go back to the complete `optimize` method. def optimize(pairings, names, pos, posmax, maxweight=nil) bestpairs = nil maxweight ||= pairings.size ** 2 + 1 while pos < posmax pair, weight1 = pairings[pos] break unless weight1 * (names.size / 2).floor < maxweight pos += 1 if names & pair == pair names1 = names - pair if names1.size < 2 bestpairs = [pair] bestpairs << names1 unless names1.empty? return [weight1, bestpairs] elsif (rv = optimize(pairings, names1, pos, posmax, maxweight - weight1)) maxweight, bestpairs = rv maxweight += weight1 bestpairs << pair end end end return bestpairs && [maxweight, bestpairs] end The algorithm structure is basically the same: recursive, greedily selecting pairs and removing those names from the set of names available... The major difference here is the inclusion of the weights (i.e. scores) and possible rejection of pairs with respect to those weights. As recursive calls to `optimize` are made, the `maxweight` value is updated via this code: maxweight, bestpairs = rv maxweight += weight1 and checked via this code: break unless weight1 * (names.size / 2).floor < maxweight So a pair may be skipped if it's weight (scaled by the number of names) exceeds the current maxweight, determined the the current set of choices. When a possible solution is found, the `maxweight` variable represents the total score for that solution. But the algorithm does not stop immediately; it keeps checking pairs and solutions of pairs, rejecting those solutions and partial solutions whose total score (i.e. `maxweight`) would exceed that previously known `maxweight`. In the end, the solution reported is the collection of pairs with the lowest total score, and is returned via the original invocation of `optimize`. [1]: http://en.wikipedia.org/wiki/Stable_roommates_problem [2]: http://en.wikipedia.org/wiki/Np_complete -- Matthew Moss ------=_Part_15389_5027399.1213304819512--