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OK, let's finish up this year and this series. We have an algorithm that can compute what cells in the zero octant are in view to a viewer at the origin when given a function that determines whether a given cell is opaque or transparent. It marks the visible points by calling an action with the visible cells. We would like that to work in any octant, and for the viewer at any point, not just the origin.

We can solve the "viewer at any point" problem by imposing a coordinate transformation on the "what is opaque?" function and the "cell is visible" action. Suppose the algorithm wishes to know whether the cell (3, 1) is opaque. And suppose the viewer is not at the origin, but rather at (5, 6). The algorithm is actually asking whether the cell at (3 + 5, 6 + 1) is opaque. Similarly, if that cell is determined to be visible then the transformed cell is the one that is visible from (5, 6). We can easily transform one delegate into another:

private static Func<int, int, T> TranslateOrigin<T>(Func<int, int, T> f, int x, int y)
{
return (a, b) => f(a + x, b + y);
}

private static Action<int, int> TranslateOrigin(Action<int, int> f, int x, int y)
{
return (a, b) => f(a + x, b + y);
}

Similarly we can perform an octant transformation; if you want to map a point in octant one into a point in octant zero, just swap its (x,y) coordinates! Every octant has a simple transformation that reflects it into octant zero:

private static Func<int, int, T> TranslateOctant<T>(Func<int, int, T> f, int octant)
{
switch (octant)
{
default: return f;
case 1: return (x, y) => f(y, x);
case 2: return (x, y) => f(-y, x);
case 3: return (x, y) => f(-x, y);
case 4: return (x, y) => f(-x, -y);
case 5: return (x, y) => f(-y, -x);
case 6: return (x, y) => f(y, -x);
case 7: return (x, y) => f(x, -y);
}
}

(And similarly for actions.)

Now that we have these simple transformation functions we can finally implement the code that calls our octant-zero-field-of-view algorithm:

int x, int y, int radius,
Func<int, int, bool> isOpaque,
Action<int, int> setFoV)
{
Func<int, int, bool> opaque = TranslateOrigin(isOpaque, x, y);
Action<int, int> fov = TranslateOrigin(setFoV, x, y);

for (int octant = 0; octant < 8; ++octant)
{
ComputeFieldOfViewInOctantZero(
TranslateOctant(opaque, octant),
TranslateOctant(fov, octant),
}
}

Pretty slick, eh?

One minor downside of this algorithm is that it computes the visibility of the points along the axes and major diagonals twice; however, the number of such cells grows worst case linearly with radius. The algorithm as a whole is worst case quadradic in the radius, so the extra linear cost is probably irrelevant.

I've posted the project that builds the Silverlight control from the first episode here, if you want to get a complete working example.

Happy New Year everyone and we'll see you in 2012 for more Fabulous Adventures in Coding!

• "Suppose the algorithm wishes to know whether the cell (3, 1) is opaque. And suppose the viewer is not at the origin, but rather at (5, 6). The algorithm is actually asking whether the cell at (3 + 5, 6 + 1)"

I think there is a typo there. I guess the algorithm should be asking if cell at (3 - 5, 1 - 6) is visible.

Very nice series by the way!

• Freakin' awesome man. Thanks a lot for doing this. This is definitely going into some of my future games :)

• No need for the TranslateOctant() switch.  Create an 8 entry array of the value to multiply X and Y by.  entry 1 is (1,1), entry 2 is (1,-1), ...

Much faster than the switch.

• @Tom: ...except that doesn't account for the cases whee X & Y are swapped.  You could create an array of 2x2 matritices representing two linear equations (X & Y) each with two inputs (X & Y).  That might still be faster than the switch, but short branches that stay within the L1 cache are very fast, so a single branch might be faster than 4 array lookups. Only profiling would tell if it's worht the trouble.

As to which is easier to understand, I think that depends on your background.  People familiar with graphics would no doubt say that the matrix multiply method is clearer, while other would likely chose the switch as clearer.

• @Federico: At first, I thought the same. However notice that he only does the correction the other way around. The algorithm always asks for opacity (or sets visibility) in its local coordinates. The isOpaque and setFoV methods then translate them to global ones.

• Wow, great articles! Very good.. Thank you for the contribution!

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