Algorithm for generation of Voronoi Diagrams
You may use whatever algorithm you like to generate your Voronoi Diagrams, as long as it is yours (no using somebody's Voronoi generating package) and runs in at worst O(n^2) time. The algorithm below is the simplest algorithm we could come up with, and it runs in Theta(n^2) (for the truly curious, this bound holds in part because it can be proven that a Voronoi diagram has at most O(n) edges). There are several algorithms which run in O(n log n) time. If you implement one of these you will definitely exceed our expectations (they are all quite tricky!). Information on these can be found on the web (e.g., in an on-line chapter on Voronoi diagrams or from the Voronoi Page), or you can ask us. If you do choose to use an algorithm other than the one outlined below, make sure you carefully document how it works in comments and in your README.
We shall refer to the points that define the Voronoi Diagram as sites. The sites are the inputs to the algorithm, but are not part of the output. Each site will have a corresponding polygon in the Voronoi Diagram, which we shall call a cell. Cells will be made up of edges, which meet at vertices.
Note: for the numerical calculations we assume that we are dealing with real numbers. When you implement this you should use floating point numbers, but be aware that floating point numbers are not the same thing as real numbers. This probably won't cause any problems but sometimes does in computational geometry problems such as this one.
You may use whatever data structures you deem appropriate to implement the four sets below. For each set, we have listed what operations you need to be able to perform.
You will also need data types for the following:
Input: int width, int height, set of sites S. Initialize E and C to be empty.  Add three or four "points at infinity" to C, to bound the diagram. For example:
The result of this is a set of edges describing the Voronoi diagram. Remember that your output needs to describe the vertices (points), edges (walls), and cells (rooms). Moreover, adjacent rooms should share their common wall, not use two separate instantiations of it. You will need to massage this output into that form. But, barring that...
Yay! We’re done! Wasn’t that fun? Keep in mind that you are free to modify this algorithm however you see fit. But just make sure to document your choices!
Testing Spatial Relationships
In the above algorithm we need to be able to determine which side of a line a given point is on. We can do this using cross products of the vectors describing those points. I will give a very brief explanation here. For more detail refer to a text on Computational Geometry. A vector is essentially like a point; it has an x-coordinate and a y-coordinate. The cross product of two vectors V1 and V2 (with corresponding coordinates (X1, Y1) and (X2, Y2)) is given by V1 x V2 = X1*Y2 - X2*Y1. Note that this is equal to the negative of V2 x V1. If the vectors are collinear (ie they point in the same, or opposite, direction) the cross product will be zero. Now back to the original problem. Suppose we are given a line defined by points A and B, and a point P which we want to test. To do this we calculate the cross product (P - A) x (B - A). If P would be to our right as we walk along the vector from A to B, this value will be positive. If P would be to our left it will be negative. If P is on the line AB then it will be zero.
Check out Yahoo's Computational Geometry page for tips, tricks, and data structures useful in computational geometry.
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