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Should you ever need to do such a thing, here is a quick guide to the process.

Suppose that you have a grid of vertices that looks something like this:
```*     *
*     *
*     *
*     *
*     *
*     *
*     *
*     *
```
...when all the dots are connected, you’ll have a grid of triangular faces. What you need is a way to determine which three vertices are used by an arbitrary triangular face.

To do this, the vertices first need a coordinate system so they can be identified:
```  0  1  2  3
0*     0*
0*     0*
1*     1*
1(*)    1*
2*     2*
2*     2*
3*     3*
3*     3*
```
A vertex can now be specified with vertex location vX, vY, where vX is the column of the vertex, and vY is how far down the column the vertex is. For instance, the vertex in parentheses is (1, 1).

Now, the faces need a coordinate system:
```   0  1   2
*     *
0    *     *
1 *     *
2    *     *
3 *     * X
4    *     *
5 *     *
*     *
```
So now faces can be specified with a coordinate pair (fX, fY), where fX is the column and fY is the row of the face. The face with the X in it is located at (2, 3).

This is all pretty easy, and probably didn’t hurt your brain. Now, here’s the good bit -- how do you determine what the coordinates of the vertices are that belong to face (fX, fY)?

If fX and fY are ODD:
```   * fX, (fY/2)
* fX + 1, (fY/2) + 1
* fX, (fY/2) + 1
```
If fX and fY are EVEN:
```   *fX, (fY/2)
* fX + 1, (fY/2)
*fX, (fY/2) + 1
```
If fX is ODD and fY is EVEN:
```      *fX + 1, (fY/2)
* fX, (fY/2)
*fX + 1, (fY/2) + 1
```
If fX is EVEN and fY is ODD:
```      * fX + 1, (fY/2)
* fX, (fY/2) + 1
* fX + 1, (fY/2) + 1
```

A couple of hints to perhaps make this intelligible:

- Division shown is considered to be integer division -- that is, you drop any funny numbers after the decimal place. So (3 / 2) = 1. Make sure you drop it, rather than rounding.

- The little formulations are in the spots representing where the vertex is in the face. The *’s are the vertices, not multiplication or anything like that.

Anyway, the chance you’ll ever need to do this is pretty slim (and even if you do, it’s not too hard to figure out), but you never know...

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