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# Pythagorean Polygons - Part 2

Posted: 2012-09-14 - Link

In my last post I showed how to solve the Project Euler problem, Pythagorean Polygons for N = 60 in under a minute. In this post I’ll show how to solve it for the target N = 120, and in under a second.

Our current algorithm uses a brute force depth first search, memoized using the generic Phobos memoize template. One common way to speed up searches is to prune branches that have no possibility of reaching the goal. In this problem, we can prune a search when the distance to the goal is greater than the remaining perimeter:

This alone speed up the N = 60 solution to under 2 seconds, and gives us the solution to N = 120 in just over 2 minutes.

We can do better.

In the N = 120 case, the memoization cache uses up roughly 200MB of memory. This is much larger than my L2 cache, and since hash maps distribute elements randomly, we’ll be spending a lot of time going out to memory checking the memoization cache. Reducing memory usage and improving lookup patterns should help performance significantly.

One observation we can make is that dfs(..., i) always calls dfs(..., next(i)). If we change the algorithm from a top-down memoization algorithm to a bottom-up dynamic programming algorithm, then we could do the depth-first search backwards, in layers of different values of i. We start by computing dfs(..., vs.length-1) for all x, y, and d. That gives us everything we need to compute all of dfs(..., prev(vs.length-1)).

prev is just the opposite function to next – it returns the index j, previous to i such that vs[i] is not colinear with vs[j].

Since we aren’t going to be caching all of dfs anymore, we’ll get rid of the memoization and instead just use a big array as our cache. We’ll also need two of them: one for the previously computed layer of i, and one for the next layer that is being computed.

Here, dp[r] stores the previously computed layer. dp[r][M+x][M+y][d] represents the number of paths of length d from (0, 0) to (x, y) using the only the triples from the previous layers. Using dp[r] we compute dp[w] by using the next layer down from what dp[r] used. To do this, we start by copying all of dp[r] into dp[w] and then simply iterate over all x, y, and d, as well as all colinear edges in the current layer.

The algorithm looks something like this:

Once all this has run for each layer, dp[r] will contain the final path counts for each (x, y, d) tuple. All that’s left to do is count up the number of paths that reach back to (0, 0) (dp[r][M][M][...]).

With this improvement, the N = 120 case now runs in just over 1 second, which is easily fast enough for us to be satisfied, but there’s much more that can be done.

Like many geometrical problems, Pythagorean Polygons has a symmetry to it that can be exploited for speed. Notice that we generate the Pythagorean triplets redundantly from all quadrants. If (3, 4) is a triplet, then so is (-3, 4), (3, -4), (-3, 4), (4, 3), (-4, 3), (4, -3), and (-4, -3). What this also means is that if there is a path from (0, 0) to (x, y, d) then there are also paths to (-x, y, d), (x, -y, d) etc. but we compute all of these redundantly!

Taking full advantage of the symmetry is tricky, but there’s one simple way that can speed up our algorithm up significantly. Notice that the first half of triplets travel upwards, and the second half travel downwards (due to the sort). This means our algorithm calculates all paths down to some point (x, y, d) using the second half of the triplets, then calculates all paths from there back to (0, 0) using the second half of the triplets. But we know from the symmetry that if there is a path to (x, y, d) using the first half then there is a path back to (0, 0) using the second half. In fact, there are the same number of paths, so the total unique paths is equal to the product of all combinations where the total distance is less than N.

To implement this, we only need to make one simple change, and add a combining step. The change is to modify the main loop to only use the first half of the triplets, i.e.

becomes

Next, instead of just summing up the dp[r][M][M][1..N] entries, we need to sum up the path products. To do this, we iterate over all (x, y, d) tuples and then multiply the cache value with those for (x, y, d2) where d2 <= N - d.

That last line needs some explanation. Previously our algorithm accounted for paths with only two edges, but we don’t do that anymore, so they will be included in the totals. What that line of code does is remove those paths from the total. That value is vs.length / 2 because every edge (dx, dy) has a corresponding edge (-dx, -dy) that will be included, so the total of these two-edge paths is equal to half the triplets.

With this improvement, the N = 120 case gives the solution in 0.36 seconds, which is more than fast enough for my liking. The other axes of symmetry can be exploited further, but I’ll leave that as an exercise for the reader :-)

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