Sign in to follow this  

can the number of triangles in a convex model be odd

This topic is 4813 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Recommended Posts

pyramid has an even number of triangles *if it is convex*

if it is three sided, and it must be convex, then the bottom also is a triangle, and so it actually has 4 triangles because the bottom is also a triangle


and then, if it is four sided, then of course there are at least four triangles, but the bottom forms a quad, which then forms two more triangles, for a total of six, which is again, even.

Share this post


Link to post
Share on other sites
Good question.
Number of triangles in piramid is even.[edit: wow. first time beaten by OP ! [grin]]
And if you split _one_ triangle into two without touching other triangles it's cheating, it's no longer an closed convex polyhedron.

Share this post


Link to post
Share on other sites
Proven . Number of edges =(number of triangles * 3) / 2
So number of triangles must be even(otherwise we have non-integer number of edges), no matter if it's convex or not, it only need to be closed.

Share this post


Link to post
Share on other sites
Guest Anonymous Poster
from what I can tell, it does not depend if the model is convex or not, but if it is manifold or not

if and only if every edge has exactly two faces touching it, then the mesh is manifold.

if I recall correctly:
Faces + Verticies = Edges + 2

for a triangular pyramid:
4 + 4 = 6 + 2
a 4 sided pyramid with two triangles on the bottom
6 + 5 = 9 + 2
a cube:
6 + 8 = 12 + 2

somehow I feel that you could derive from this something like what you are asking, but having not taken anything like a topology course I don't think I could explain exactly how

in practcice, two triangles are needed minimum for a manfold mesh. If you subdivide the middle of one, you will essentially end up with that triangular pyramid [and you added an even number of faces, 2 to the 2 original faces]. If you subsdivide an edge you add an even number of faces. I guess you could argue that all mesh modeling tools [bridge, extrude...] as long as they result in a mainfold mesh would add an even number of faces.

but that is a rather mystical proof, there is no guarntee that you can generate all shapes with those tools [as in, somethimes you may need to delete faces and recreate them], but it also isn't elegant in requiring my to prove something for each of those tools.

[this has been enough ranting for a while.... back to bed with me]

Share this post


Link to post
Share on other sites
yes, that "manifold " probably is exactly what i meant by closed, that is, "if every edge is shared by exactly 2 triangles, 3*number of triangles must be divisible by 2, therefore, number of triangles must be divisible by 2". ...

BTW,it would be alot harder to prove if someone said that number of edges is multiply of 3... because of misleading "triangle have 3 edges" [grin]

Share this post


Link to post
Share on other sites
F+V = E+2 isn't valid if the polyhedra has holes in it.
Of course, it wouldn't be convex then.

V: the number of vertices
E: the number of edges
F: the number of faces
G: the number of holes that penetrate the solid, usually referred to as genus in topology
S: the number of shells. A shell is an internal void of a solid. A shell is bounded by a 2-manifold surface, which can have its own genus value. Note that the solid itself is counted as a shell. Therefore, the value for S is at least 1.
L: the number of loops, all outer and inner loops of faces are counted.
Then, the Euler-Poincaré formula is the following:

V - E + F - (L - F) - 2(S - G) = 0

Info from this site:
http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/model/euler.html

Share this post


Link to post
Share on other sites
I think by holes they mean holes like in torus...
anyway, if we have cube with one triangle removed, we have one hole and odd number of triangles. If we have 2 triangles removed, we have even number of triangles and either have 2 holes or we have one hole.

Iff number of boundary edges is even, number of triangles is even.

Share this post


Link to post
Share on other sites
Yes, holes through the polyhedron like in a torus.
I was just clarifying the AP's "If I recall correctly" formula.
Euler's rule for graphs (Nodes + Regions = Arcs + 2) only applies to graphs drawn on a plane as well, graphs drawn on toruses [edit: that pass through the hole] break that rule also.
Topology is an interesting topic, I found it very hard though. Maybe I shouldn't have skived all my lectures all those years ago.

Share this post


Link to post
Share on other sites
Quote:
Original post by Dmytry
Proven . Number of edges =(number of triangles * 3) / 2
So number of triangles must be even(otherwise we have non-integer number of edges), no matter if it's convex or not, it only need to be closed.


That's actually why I've posted...I came to the conclusion on my own (although probably everybody here on this forum already knew this) that to get the number of edges you multiply the number of triangles by 1.5, or 3/2, but only if the model is convex, and the number of triangles must always be even

EDIT:
and, like I said, I only care about convex models (so I can do things like run the brush algorithm on a transformed model for collision detection, stencil shadows, etc)

Share this post


Link to post
Share on other sites

This topic is 4813 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this