poita.org

Pythagorean Polygons - Part 1

Posted: 2012-09-08 - Link

It’s been a while since I solved any coding problems, so I’ve decided to re-solve a Project Euler problem that I looked at years ago: Pythagorean Polygons. Back then I used C++, but this time I’ll use the D programming language :-)

In summary, the problem is to find the number of distinct convex polygons with vertices on integer coordinates, integer length sides, and a perimeter less than or equal to 120. The name of the problem comes from the fact that the edges are the hypotenuses of Pythagorean triples.

The brute force method is quite simple:

  1. Generate all the Pythagorean triples (x, y, d) with d < 120.
  2. Do a depth first search on the integer lattice from (0, 0) to (0, 0), using the triples as your edges, ensuring that you always turn anti-clockwise (or always clockwise).

Generating the Pythagorean triples is easy enough:

enum N = 120;    // Maximum perimeter
enum M = N / 2;  // Maximum edge length

struct V { int x, y, d; }
V[] vs;

foreach (dx; -M..M+1)
    foreach (dy; -M..M+1)
        foreach (dd; 1..M+1)
            if (dx * dx + dy * dy == dd * dd)
                vs ~= V(dx, dy, dd);

It will also help to have them in anti-clockwise order.

double arg(V v)
{
    import std.math;
    return atan2(cast(real)v.y, cast(real)v.x);
}

import std.algorithm;
sort!((a, b) => arg(a) < arg(b))(vs);

Our triples are likely to contain several colinear edges. So we’ll need a way of avoiding those:

bool colinear(V a, V b)
{
    // a and b are colinear if a.x / a.y == b.x / b.y
    // We can avoid the division by multiplying out.
    return a.x * b.y == a.y * b.x;
}

int next(int i)
{
    // Find next, non-colinear edge
    int j = i + 1;
    while (j < vs.length && colinear(vs[i], vs[j]))
        ++j;
    return j;
}

Now all that’s left is to do the depth-first search. I’m just going to use simple recursion to implement it, just to save code.

// Returns number of paths from (x, y) to (0, 0) using less
// than d perimeter length, and only using edges with
// index >= i (to ensure convexity).
long dfs(int x, int y, int d, int i)
{
    if (d > N) return 0; // Perimeter too long
    if (x == 0 && y == 0 && d > 0) return 1;

    // Iterate edges, add them on, and recurse.
    long total = 0;
    foreach (int k; i..cast(int)vs.length)
        if (!(d == vs[k].d && x == -vs[k].x && y == -vs[k].y))
            total += dfs(x + vs[k].x, y + vs[k].y, d + vs[k].d, next(k));
    return total;
}

The if-condition inside the loop is to protect against degenerate polygons with two edges e.g. (0, 0), (60, 0), (0, 0), and the use of next ensures that we don’t use two colinear edges in a row, as this is disallowed in the problem.

Finally, we need to call it with the starting position:

import std.stdio;
writeln(dfs(0, 0, 0, 0));

If we change N to 30, we can test it using the given solutions in the problem.

% dmd -inline -O -release p292
% time ./p292
3655

Great. That matches with the solution and runs in under 2 seconds on my laptop. Unforunately, trying it with N = 60 took just shy of 50 minutes.

The problem with this is that the branching factor of the search is equal to the number of triples (hundreds) and the depth exponent is large enough to make the running time impractical for larger N. We need a way to speed it up.

Let’s look at the signature of dfs.

long dfs(int x, int y, int d, int i)

What’s the domain of each input?

x and y will be from -60 to 60.
d will be between 0 and 120.
i will be between 0 and the number of triples (448).

That gives us a maximum of 121 x 121 x 121 x 448 possible inputs. Since dfs is a pure function, that means we can memoize the results and re-use them for subsequent recursive calls. It’s a large number, but now the running time is polynomial rather than exponential, so it will scale better.

Fortunately, D’s standard library, Phobos, provides a memoize template just for this purpose. With a couple of quick modifications to dfs, we can memoize it.

long dfs(int x, int y, int d, int i)
{
    import std.functional;
    alias memoize!dfs mdfs;

    // ...
            total += mdfs(x + vs[k].x, y + vs[k].y, d + vs[k].d, next(k));
    // ...
}

We just introduce an alias for the memoized version of the function, and change the recursive call to use that version instead. Internally, memoize just maintains a hashtable of input tuples to outputs, so it’s quite fast.

With these changes, the N = 60 version is reduced from 50 minutes to 42 seconds. Unfortunately, N = 120 is still just out of reach. It will likely run in under an hour if you have enough memory, but we’d like it to run a lot faster than that.

In the next post, I’ll show how we can get N = 120 running in sub-second times.

comments powered by Disqus