Yes, but not a polygon (or polyhedron).
Consider a quadrilateral with one diagonal. The end points of the diagonal are at odd vertices while the other two are even.
Break up the odd shape into even shapes and add the areas.
It is traversable if there is an even number of edges at each vertex, or at every vertex except two. In the latter case the traverse must start at one of the "odd vertices" and finish at the other.
All you do is add up all the sides to find the perimeter even though it is a odd shape
If it is a polygon with an even number (>2) of vertices, join any two pairs of opposite vertices. These lines will meet in the centre. If it is a polygon with an odd number (>1) of vertices, join any two vertices to the midpoints of the opposite sides. These lines will meet in the centre.
pyramid
No. Not can it have an odd number of vertices.
In order for a network to be transversable, it either needs to have all of the vertices even, or just 2 odd vertices
Break up the odd shape into even shapes and add the areas.
No it is not possible to get a even number by adding odd and even number. 1+2=3-odd number.3+4=7-odd number.
It is traversable if there is an even number of edges at each vertex, or at every vertex except two. In the latter case the traverse must start at one of the "odd vertices" and finish at the other.
No. even and even or even and odd give even only
All you do is add up all the sides to find the perimeter even though it is a odd shape
No.
no
3 if n is odd 2 if n is even where n is the number of vertices.
I'm gonna guess no, because numbers are even and odd by the last number, so... No.
No, it cannot.