. In the case of a square of four points (Pll, Pul, Plr, and Prr where u=upper, l=lower/left, r=right), you expect a line between Pll and Pul, you also expect a line between Pll and Plr. But a line between Pll and Prr is also valid.

No, that cannot happen. A square can only arise from a cross pattern of 5 dots. And a diagonal line connecting the opposite corners would pass through the central dot and could not be 'normal' to any divisor between that dot and any other.

BrowserUk:

Sorry, I should've been more clear. I wasn't referring to the vertices, but the input points to the algorithm. So the diagonal line going through two of the points are the bisector for the other two corners. I put this case into the algorithm (just a few minutes ago):

my @points = ( [0,0], [2, 0],  [0, 2], [2, 2] );
[download]

Yields the following lines:

[
  [1, 0, 1, 0, 1], # X=1 for points 0, 1
  [0, 1, 1, 0, 2], # Y=1 for points 0, 2
  [0, 1, 1, 1, 3], # Y=1 for points 1, 3
  [1, 1, 2, 0, 3], # X+Y=2 for points 0, 3
  [1, 0, 1, 2, 3], # X=1 for points 2, 3
]
[download]

That fourth entry is the diagonal I was talking about. It's clearly the bisector for points 0 and 3, but since it intersects at the same point as the first two, it's irrelevant. Looking at the edge list:

[
  [3, 1, 0],   # Edge from (1,1) to (1,1)
  [1, -1, 1],
  [4, -1, 1],
  [2, 0, -1],
  [0, 0, -1]
]
[download]

Other than the first edge, the rest are as you'd expect to see. I'm suspecting some odd comparison (or roundoff) in the C code to be generating that 0-length segment.

Just for completeness, here's the list of vertices:

[
  [1, 1],
  [1, 1]
]
[download]

Looking at the results, I'm struck by a couple items:

• Some lines are identical, except for the point pairs they were generated from.
• Only one of the diagonal lines is generated, which is why I suspect roundoff or an odd conditional.
• It might be worthwhile to add disallow duplicate vertices and/or edges of zero length.

...roboticus

Update: I just tried the cross that you suggested, and the data showed *absolutely no* anomalies....odd!

Update: Fixed HTML (missed closer for UL).

I put this case into the algorithm...

Hm. What do you mean by "put it into the algorithm"?

Four points (o) in a square, gives six bisectors (.), but just four normals(n). (Four pairs of coincident normals.) But they produce just a single vertex (X) ('scuse the crudity):

+-----------------------------------------------------+
|     n                  n                   n        |
|       n                n                 n          |
|         n              n               n            |
|           n            n             n              |
|            o.......................on               |
|            . .n        n         .n.                |
|            .   .n      n       .n  .                |
|            .     .n    n     .n    .                |
|            .       .n  n   .n      .                |
|            .         . n .n        .                |
| nnnnnnnnnnn.nnnnnnnnnn Xnnnnnnnnnnn.nnnnnnnnnnnnnnnn|
|            .         . n .n        .                |
|            .       .n  n   .n      .                |
|            .     .n    n     .n    .                |
|            .   .n      n       .n  .                |
|            . .n        n         .n.                |
|            o.......................o                |
|          n             n             n              |
|        n               n               n            |
|      n                 n                 n          |
+-----------------------------------------------------+
[download]

And you can't form any polygons from a single vertex!

Unless you are using the boundaries of the coordinate space to form the missing edges of polygons?

But if that was the case in Sam's problem, his boundary polygons would just always be a rectangle coincident with the boundaries of the coordinate space.

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