Eliminating diagonals?

Question

I have written you before. I have made an image showing my problem to
make things easier to explain. Here you can view it:

http://img152.imageshack.us/img152/8866/new1ja8.png

As you can see, I need to eliminate these diagonal edges, and I have
made a few attempts that worked to a degree.

I tried iterating on each triangle and getting the edge that is faced
by the largest angle. But this doesn't always work as shown in the
image. So I have tried getting the faces of an edge, then from there I
collected all edges of them, and get the largest one (mostly the
diagonal). This doesn't always work too.

Is there a way to separate these edges from the others clearly?

Thank you!

Answer 1

If the picture is accurate, and all sets of lines are parallel, then a = b. So instead of making the criteria "a > c and b > c" you could just use "a = b"....?

Answer 2

Jim: I don't think that he meant to imply that the lines are are parallel and equidistant.

If you calculate the distance from a point to its surrounding points then those distances fall into two catagories with the diagonal distance being significantly larger than the others. This works for boundary as well as interior point.

I assume that your figure is represented by a collection of (x,y) coordinates?

An angle approach might work. Take dot products of all vectors representing line segments and calculate the angle. It will be 0 or 90 for most and around 45 for the diagonals.

Answer 3

So a+b+c=180 and c-a>0 and c-b>0; now add LHSides and RHSides, thence 3c>180 and c>60;
Exclude 2 longest criss-cross diags as their halves are clearly opposite to c<60; but rhombus is OK!

oh sorry, rhombus is also bad!

Answer 4

From your diagram it appears to me that the slopes of all diagonals tha have a positive slope have a slope = +1. All diagonals that have a negative slope have a slope of -1.

Perhaps I'm not understanding the problem, but I would just eliminate all diagonals that have a slope = +/- 1.