Hey everyone,

Last night I was trying to solve challenge #15 from Project Euler:

Starting in the top left corner of a 2×2 grid, there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a 20×20 grid?

I figured this shouldn't be so hard, so I wrote a basic recursive function:

const int gridSize = 20;

// call with progress(0, 0)
static int progress(int x, int y)
{
    int i = 0;

    if (x < gridSize)
        i += progress(x + 1, y);
    if (y < gridSize)
        i += progress(x, y + 1);

    if (x == gridSize && y == gridSize)
        return 1;

    return i;
}

I verified that it worked for a smaller grids such as 2×2 or 3×3, and then set it to run for a 20×20 grid. Imagine my surprise when, 5 hours later, the program was still happily crunching the numbers, and only about 80% done (based on examining its current position/route in the grid).

Clearly I'm going about this the wrong way. How would you solve this problem? I'm thinking it should be solved using an equation rather than a method like mine, but that's unfortunately not a strong side of mine.

Update:

I now have a working version. Basically it caches results obtained before when a n×m block still remains to be traversed. Here is the code along with some comments:

// the size of our grid
static int gridSize = 20;

// the amount of paths available for a "NxM" block, e.g. "2x2" => 4
static Dictionary<string, long> pathsByBlock = new Dictionary<string, long>();

// calculate the surface of the block to the finish line
static long calcsurface(long x, long y)
{
    return (gridSize - x) * (gridSize - y);
}

// call using progress (0, 0)
static long progress(long x, long y)
{
    // first calculate the surface of the block remaining
    long surface = calcsurface(x, y);
    long i = 0;

    // zero surface means only 1 path remains
    // (we either go only right, or only down)
    if (surface == 0)
        return 1;

    // create a textual representation of the remaining
    // block, for use in the dictionary
    string block = (gridSize - x) + "x" + (gridSize - y);

    // if a same block has not been processed before
    if (!pathsByBlock.ContainsKey(block))
    {
        // calculate it in the right direction
        if (x < gridSize)
            i += progress(x + 1, y);
        // and in the down direction
        if (y < gridSize)
            i += progress(x, y + 1);

        // and cache the result!
        pathsByBlock[block] = i;
    }

    // self-explanatory :)
    return pathsByBlock[block];
}

Calling it 20 times, for grids with size 1×1 through 20×20 produces the following output:

There are 2 paths in a 1 sized grid
0,0110006 seconds

There are 6 paths in a 2 sized grid
0,0030002 seconds

There are 20 paths in a 3 sized grid
0 seconds

There are 70 paths in a 4 sized grid
0 seconds

There are 252 paths in a 5 sized grid
0 seconds

There are 924 paths in a 6 sized grid
0 seconds

There are 3432 paths in a 7 sized grid
0 seconds

There are 12870 paths in a 8 sized grid
0,001 seconds

There are 48620 paths in a 9 sized grid
0,0010001 seconds

There are 184756 paths in a 10 sized grid
0,001 seconds

There are 705432 paths in a 11 sized grid
0 seconds

There are 2704156 paths in a 12 sized grid
0 seconds

There are 10400600 paths in a 13 sized grid
0,001 seconds

There are 40116600 paths in a 14 sized grid
0 seconds

There are 155117520 paths in a 15 sized grid
0 seconds

There are 601080390 paths in a 16 sized grid
0,0010001 seconds

There are 2333606220 paths in a 17 sized grid
0,001 seconds

There are 9075135300 paths in a 18 sized grid
0,001 seconds

There are 35345263800 paths in a 19 sized grid
0,001 seconds

There are 137846528820 paths in a 20 sized grid
0,0010001 seconds

0,0390022 seconds in total

I'm accepting danben's answer, because his helped me find this solution the most. But upvotes also to Tim Goodman and Agos :)

Bonus update:

After reading Eric Lippert's answer, I took another look and rewrote it somewhat. The basic idea is still the same but the caching part has been taken out and put in a separate function, like in Eric's example. The result is some much more elegant looking code.

// the size of our grid
const int gridSize = 20;

// magic.
static Func<A1, A2, R> Memoize<A1, A2, R>(this Func<A1, A2, R> f)
{
    // Return a function which is f with caching.
    var dictionary = new Dictionary<string, R>();
    return (A1 a1, A2 a2) =>
    {
        R r;
        string key = a1 + "x" + a2;
        if (!dictionary.TryGetValue(key, out r))
        {
            // not in cache yet
            r = f(a1, a2);
            dictionary.Add(key, r);
        }
        return r;
    };
}

// calculate the surface of the block to the finish line
static long calcsurface(long x, long y)
{
    return (gridSize - x) * (gridSize - y);
}

// call using progress (0, 0)
static Func<long, long, long> progress = ((Func<long, long, long>)((long x, long y) =>
{
    // first calculate the surface of the block remaining
    long surface = calcsurface(x, y);
    long i = 0;

    // zero surface means only 1 path remains
    // (we either go only right, or only down)
    if (surface == 0)
        return 1;

    // calculate it in the right direction
    if (x < gridSize)
        i += progress(x + 1, y);
    // and in the down direction
    if (y < gridSize)
        i += progress(x, y + 1);

    // self-explanatory :)
    return i;
})).Memoize();

By the way, I couldn't think of a better way to use the two arguments as a key for the dictionary. I googled around a bit, and it seems this is a common solution. Oh well.