can the number of triangles in a convex model be odd

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Im just wondering, can the number of triangles in a convex model ever be odd

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Of course.
Think pyramids. Or you could just split any triangle into 2 triangles.

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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.

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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.

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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.

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Nice, Dmytry!
rating++ for an elegant proof

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Oops I was thinking of planes rather than triangles.
Although a single triangle is of course convex. Presumably we're not considering it closed unless it has a back side also.

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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]

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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]

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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

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