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Given an arbitrary amount of ordered segments, with arbitrary lengths is there a way to determine if they can be formed into a simple polygon?

And if so, is it possible to work out the angles needed to do so?


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What are "ordered segments"? Are they oriented segments, or is there a sequential order given: the 1st segment, the 2nd segment, etc.? – Joseph O'Rourke Jan 5 '13 at 13:08
The order is given to me in an array. I can't rearrange them. Hope that's more clear :s Thanks. – Jason Jan 5 '13 at 18:49
up vote 2 down vote accepted

As @Mark Bennet points out, there are lots of possible configurations (for $n>3$); here's one:

Chain the segments end-to-end with their vertices on the circumference of a circle whose diameter is at least as long as the longest segment. Simply adjust the size of the circle until the segment chain makes a single closed loop.

The worst case is the limit of the configuration in which the largest segment "separates" the other segments from the center, with the center moving off to Infinity. In the limit, the segment sub-chains on either side of the largest segment are "folded flat" onto that largest segment as the circle becomes a line; if the ends of the chain don't meet in that circumstance, then it wasn't meant to be.

With this cyclic construction, you can determine the angles between segments by joining all the endpoints to the center and analyzing the various isosceles triangles.

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If I read the question right, the order of the (say $n>2$) segments is given, but the orientation is not. If the longest segment is shorter than the sum of the others then you can make a non-degenerate simple polygon (in a plane).

Since the order is cyclical, make the longest segment the first. Then the next $n-2$ can be made into a straight line (piece 2) with the final piece as a third side - and you can construct a triangle. This configuration, incidentally, minimises the angle between the first two pieces, and maximises the angle between the last piece and the first one. You can put a bound on the angles, therefore, but unless you have only three pieces you cannot determine the angles.

It is clearly possible to purturb such a polygon into a non-planar configuration and there are non-convex possibilities as well ($n>3$ in both cases).

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This doesn't quite do it actually, because the remaining segments could be too long to make a triangle in the way suggested. However it is straightforward to show that there is a vertex which separates the "shorter" sides into left and right portions in such a way that the difference in the lengths of the two portions is less than or equal to the longest side - and that builds a triangle, except in the case of equality, where you can make a parallelogram or rectangle. – Mark Bennet Jan 6 '13 at 9:09

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