By Paul Ricoeur

Textual content IN FRENCH.

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It is therefore incumbent upon us to clarify the notion of homeomorphism of surfaces. It is clear that homeomorphic surfaces have identical presentations. For, if / : S —» Sf is a homeomorphism of surfaces, and if Π = {Pi, Pi,.. ,Pn } is a presentation of S, then Π' = {/(Pi),/(P2), · · · >/(^5i)} is an identical presentation of 5', assuming that the arcs of Ξ(Π') inherit their labels from Π. It therefore makes sense to define two surfaces to be homeomorphic provided they have identical presentations.

One difference is that a has two distinct arcs joining the same two vertices, which is not the case for b. 23 are clearly not homeomorphic. 24. 23 4 A Two nonhomeomorphic graphs. 24 Two homeomorphic graphs. verifying that similarly labeled pairs of vertices are either both adjacent or both nonadjacent in the two graphs. Unfortunately, even when they exist, such corresponding labelings are not easily found, and all that can be done here is to describe one method that works in principle but is not practical.

Prove that a connected graph is traversable if and only if it has either two or no nodes of odd degree. 10. 7 can be drawn without lifting the pencil and without retracing any lines? 11. Prove that if G is a graph with p > 3 nodes such that deg v > p/2 for each node v, then G is Hamiltonian. 12. Which of the following graphs are Hamiltonian? (a) #7 (b) #8 (c) K3,4 (d) #4,4 (e) #3,5 (i) ^2,2,2 (0 *4,5 (g) *i,2,2 00 *1,2,3 (j) ^1,2,3,4 (k) #2,2,2,2 0) #3,3,3,3 13. For which values of n\, «2 is Kn] 5„2 Hamiltonian?