Hi Ben,

Gotcha. No my code won't guarantee transitive points as written, but you can make it guarantee it.

In SetPointSame(i,j) you'd look through all pairs already set to see if i or j already had a match. So if i1==i2 and i1==i3 then it will set i3==i2.

I'd be very surprised if you can ever really guarantee bitwise compatibility with floats after any operations. Even Roland's code cannot do so*. Indeed, if you ever compile with Intel's compilers with warnings on it will highlight every bitwise float match as a potential error.

Tim

* Although as I note in my comment below that it can be made to do so.

Gotcha. No my code won't guarantee transitive points as written, but you can make it guarantee it.

Ben Supnik Wrote:For example, if 3 vertices A, B, and C are near each other so that the distance from A to B is < TOL and the distance from B to C is < TOL but the distance from A to C is > TOL, are all three points the same? If not, do we say that A == B and B == C but A != C? :-)

My smoothing algorithm assumes transitivity of equality for point locations, so a fuzzy match has to be transitive too. As an experiment, I implemented a simple and stupid grid-snap, which does result in transitive behavior - in the case above, if A, B, and C all snap together, then A == B == C. If the snap separates B and C but keeps A and B together then A == B, A != C, B != C and things are still sane.

In SetPointSame(i,j) you'd look through all pairs already set to see if i or j already had a match. So if i1==i2 and i1==i3 then it will set i3==i2.

Quote:As an aside, OpenGL (maybe D3D too??) has some pretty specific rules about invariance, water tight meshes, and cracks; if the part library has bit-wise exact transform data and point locations then there will be no mesh cracks, but if that isn't true then I think there could be rendering artifacts..that's why I thought there might be some hope of having bitwise comparable data.* :-) The case that hosed me was a line and triangle not having the same bitwise definitions, which isn't surprising; no one needs watertight rendering between a tri and a line.

* For X-Plane we basically require bit-wise exact vertices to get manifold rendering in our model files, because the graphics cards require it...but the output is coming from 3-d modeling programs that more or less do this for free. Thus we ensure the same mathematical function is applied on the same input bits, so while we don't know what the floating point output is, we know it's the same every time.

I'd be very surprised if you can ever really guarantee bitwise compatibility with floats after any operations. Even Roland's code cannot do so*. Indeed, if you ever compile with Intel's compilers with warnings on it will highlight every bitwise float match as a potential error.

Tim

* Although as I note in my comment below that it can be made to do so.