Code to produce edges of a triangle

Question

I am using some code I found that tests two triangles for intersection. The portion of the code I am having trouble with is:

/*tr_tri_intersect3D - C1 is a vertex of triangle A. P1,P2 are the two edges originating from this vertex. D1 is a vertex of triangle B. P1,P2 are the two edges originating from this vertex. Returns zero for disjoint triangles and non-zero for intersection.*/ int tr_tri_intersect3D (double *C1, double *P1, double *P2, double *D1, double *Q1, double *Q2);

What I don't understand is, if I have the three vertices of 2 triangles, what do I need to do to produce the input for this function?

The full source code can be found at:

Triangle Triangle Code

The source code had a test function, it was: for (i = 0; i<10000; i++){ for (j = 0; j<3; j++){ for (k = 0; k<3; k++){ PS[i][j][k] = drand48(); QS[i][j][k] = drand48(); } } for (j = 0; j<2; j++){ for (k = 0; k<3; k++){ EPS[i][j][k] = PS[i][j + 1][k] - PS[i][0][k]; EQS[i][j][k] = QS[i][j + 1][k] - PS[i][0][k]; } } } double sum = 0; int t0 = clock(); int sums[100] = { 0 }; for (j = 0; j<1000; j++) for (i = 0; i<10000; i++){ int res = tr_tri_intersect3D(PS[i][0], EPS[i][0], EPS[i][1], QS[j][0], EQS[j][0], EQS[j][1]); sums[res]++; }

if that help.

As in, how do you call the function? I would assume pass your six verticies that you already have to it. — David
– David, Commented Nov 30, 2014 at 4:53
From the comments in the code it seemed like it was asking me for edges in P1, P2, Q1, and Q2. Is that no different from a vertex? — ThomYorkkke
– ThomYorkkke, Commented Nov 30, 2014 at 4:59
An edge is made by two vertices. You can't pass an edge, as there is no such thing as an edge, just two points which make a line which we call an edge. — David
– David, Commented Nov 30, 2014 at 5:11
@David Yeah, that's what I thought. I just put the test code that was in the file in my post, does that help? — ThomYorkkke
– ThomYorkkke, Commented Nov 30, 2014 at 5:14

byebyebryan · Accepted Answer · 2014-11-30 05:29:14Z

From the sample code it seems like what it mean by an edge is the vector from the originating vertex to the end vertex. In the case of a triangle with three vertices C1, C2, C3, the two input edges would be P1 = C2-C1, P2=C3-C1

Okay, awesome. The algorithm doesn't seem to work right with every combination of triangles, but you were right about how to feed it input.

Andras · Accepted Answer · 2014-11-30 05:40:50Z

Reading the test code, it looks like C and D are two [3][3] array of 9 doubles, which are points 1, 2 and 3 and their 3D x,y,z coordinates.

P1 and P2 are the x,y,z distances of points 2 and 3 from point 1 in triangle C, and Q1 and Q2 are the x,y,z distances of points 2 and 3 of triangle D from point 1. (Edit: byebyebryan said it better, this is the x,y,z vector from point 1 to points 2 and 3). These two are both an array of [3] doubles (basically, bounding box info).

No idea why the Ps and Qs are not computed inside the function, unless the vars also serve to hold return values somehow.

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