By Herbert S. Wilf
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Extra resources for Algorithms and Complexity (Second edition)
Ck denote the connected components of G0 . Each of them has only vertices of even degree because that was true of G and of the walk W that we subtracted from G. Since each of the Ci has fewer edges than G had, there is, by induction, an Eulerian circuit in each of the connected components of G0 . We will thread them all together to make such a circuit for G itself. 6. 6. Begin at the same v and walk along 0 or more edges of W until you arrive, for the first time, at a vertex q of component C1 .
Suppose x0 = y0 and x1 = y1 , where yn+1 = ayn + byn−1 (∀n ≥ 1). If furthermore, xn+1 ≤ axn + bxn−1 (∀n ≥ 1), can we conclude that ∀n : xn ≤ yn ? If not, describe conditions on a and b under which that conclusion would follow. 7. Find the asymptotic behavior in the form xn ∼? 33). 8. 3. 9. 3 we find the phrase “... ” Prove that this phrase is justified, in that the equation shown always has exactly one positive real root. Exactly what special properties of that equation did you use in your proof?
Has exactly two) vertices of odd degree. 4. Not every graph has a Hamiltonian path. 4(b) doesn’t. 42 1. 5. Likewise, not every graph has an Eulerian path. 5(b) doesn’t. Proof. Let G be a connected multigraph in which every vertex has even degree. We will find an Eulerian circuit in G. The proof for Eulerian paths will be similar, and is omitted. The proof is by induction on the number of edges of G, and the result is clearly true if G has just one edge. Hence, suppose the theorem is true for all such multigraphs of fewer than m edges, and let G have m edges.
Algorithms and Complexity (Second edition) by Herbert S. Wilf