By Herbert S. Wilf

**Read or Download Algorithms and Complexity (Internet edition, 1994) PDF**

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**Extra info for Algorithms and Complexity (Internet edition, 1994)**

**Sample text**

In fact, the theorem applies also to multigraphs, which are graphs except that they are allowed to have several different edges joining the same pair of vertices. 1. A (multi-)graph has an Eulerian circuit (resp. path) if and only if it is connected and has no (resp. has exactly two) vertices of odd degree. 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.

It is extremely difficult computationally to decide if a given graph has a Hamilton path or circuit. We will see in Chapter 5 that this question is typical of a breed of problems that are the main subject of that chapter, and are perhaps the most (in-)famous unsolved problems in theoretical computer science. 1) it is easy to decide if a graph has an Eulerian path or circuit. Next we’d like to discuss graph coloring, surely one of the prettier parts of graph theory. Suppose that there are K colors available to us, and that we are presented with a graph G.

Of course the mere fact that our proved time estimate is O(2E ) doesn’t necessarily mean that the algorithm can be that slow, because maybe our complexity analysis wasn’t as sharp as it might have been. However, consider the graph G(s, t) that consists of s disjoint edges and t isolated vertices, for a total of 2s + t vertices altogether. If we choose an edge of G(s, t) and delete it, we get G(s − 1, t + 2), whereas the graph G/{e} is G(s − 1, t + 1). Each of these two new graphs has s − 1 edges.