TITLE: Contiguous edges of a cube (and other regular solids)
QUESTION [5 upvotes]: I had the desire to create a cube out of a single piece of string, where each edge is represented only once. Through experimentation it appears that this is impossible, and the closest you can get is to create a cube missing 3 of the 12 edges.
A tetrahedron leaves 1 left over. An octahedron leaves none. Why is the limit 3 for a cube, and 1 and 0 for those? Is there a general rule that predicts this? What about an icosahedron, or a tesseract?

REPLY [4 votes]: As already mentioned, in order to have a Eulerlian path you need to have at most one pair of odd degree vertices (nodes in the corresponding graph). You have 4 such pairs (8 vertices) in a cube, so you need to delete at least 3 edges, just as you said in your question.
For tetrahedron you have 4 (2 pairs of) odd degree vertices, hence you need to delete (miss) 1 edge. For octahedron you don't have odd degree vertices, so you don't need to miss anything. Dodecahedron has 20 odd degree vertices (10 pairs) -- need to miss 9 edges. Icosahedron has 12 odd degree vertices (6 pairs) -- need to miss 5 edges. In tesseract all 16 vertices have degree 4, so no odd degrees, and we don't miss anything.