I hope this is the right section for this.

Here's what I have, if this isn't sufficient info let me know. I have a basic line struct, made up of two Vector2 coordinates, Start and End.

Here is a visual representation of lines making triangles and then making a rectangle.

http://imageupper.co...354333708244635

What I would like to do is remove the lines that are interior to the outer part of the shape. Now obviously I have all the data making up the lines and it would be trivial to do it manually, but I need a formula that can handle this for me for the sake of time consumption it would take.

To be clear, what I'm considering exterior and interior; if the outward most points were N,S,E,W, I want the lines connecting N to E & W, S to E & W. The two interior triangles making the smaller square, all of that can go.

And yes the end result would be eight Line structs, with two touching lines between N to E and etc.

The best approach I can think of is to use a flood-fill algorithm to determine which lines "touch" exterior space, and then remove everything else. But it really depends on what you consider exterior space. For instance, let's say that there was a small hole in the side of one of the exterior triangles. Would the interior smaller triangles then be considered exterior?

One way to identify the exterior shell of your set of points (so we're assuming there are not any 'holes' that should be considered exterior to the volume).

a) Identify a point on the exterior of the line set.
b) Walk around the exterior of the line set till you get back to start.

This identifies all the lines required to define the exterior; and then ones that you can remove are all those not belonging to the exterior.

For a) you can choose the lexicographically minimal point (a, b) < (c, d) iff a < c || (a == c &amp;amp;&amp;amp; b < d)
For b) you need to be able to; from any given point of the exterior identify the next point of the exterior. This comes in two cases; the initial case where you have 2 exterior lines to choose and you have to pick one of them (for this, abuse that your start point is lexicographically minimal), and in the other case you can make use of the point of the exterior you have just came from to order the outgoing edges.

Presume that we choose, as a result of the first case of b) the point that is clockwise around the exterior (We'll come back to this after).

If you label the previous point 'p' and the current point 'c'; then you can order the outgoing edges based on their relative rotation from the vector (c - p) http://www.zimagez.com/zimage/screenshot-021212-223758.php here, we want to pick vertex v0.

We can do this by computing the perp-dot products of (v - c) with (c - p) normalised by the length of (v - c); and choosing the minimal (or maximal depending on coordinate system); since the perp-dot of two vectors u,v is |u||v|sin(theta). We can also avoid any division if sorting of the outgoing edges uses a comparison function instead of a key.

Back to the first case of b); choosing the second vertex of exterior we can use the same technique, and use the vector (1, 0) instead of (c - p) above. This is valid given that our first vertex was lexicographically minimal since the clockwise direction of the next vertex can in the worst case be in the direction (1, 0) and there cannot exist any vertex of the set that would give us a 'negative' perp-dot since that point would need to have a lower y-coordinate and would have been chosen as the lexicographically minimal one.

^^ this is very similar to gift wrapping algorithm for computing a convex hull and is exactly equivalent if we join every vertex to every other one using this algorithm.

Paradigm shifter, that only works if the lines always form triangles..

eg: http://www.zimagez.com/zimage/screenshot-021212-225045.php

here, your solution would choose to remove the crossed-out red line. My algorithm would give us the green polygon.

Your step doesn't really need a triangulation. Eitherway given purely a set of lines you need to determine what the triangles are in the first place; and that same step can be used to simply determine the sub-polygons which would then lead to the same rule of an internal line has more than one sub-polygon using it and the triangulation step is pointless.

Eitherway, I don't see how you could determine those internal sub-polygons any faster than just traversing the exterior directly.

For fun, I implemented my solution in Haxe and checked it worked with a somewhat degenerate case as well of having a dangling vertex connected outside the main exterior set

http://pastebin.com/uyu2ZehA

Further examples shown with results in red (additional number being number of times in total that vertex appears in output)

http://www.zimagez.com/zimage/screenshot-031212-021000.php