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By Herbert S. Wilf

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The answer is known, but the derivation involves Burnside’s lemma about the action of a group on a set, and some fairly delicate counting arguments. We will state the approximate answer to this question, which is easy to write out, rather than the exact answer, which is not. If gn is the number of unlabeled graphs of n vertices then n gn ∼ 2( 2 ) /n!. 6 1. Show that a tree is a bipartite graph. 2. Find the chromatic number of the n-cycle. 3. Describe how you would find out, on a computer, if a given graph G is bipartite.

The first step is to choose one of the subarray elements (the element itself, not the position of the element) to be the splitter, and the second step is to make it happen. The choice of the splitter element in the Quicksort algorithm is done very simply: at random. We just choose, using our favorite random number generator, one of the entries of the given subarray, let’s call it T , and declare it to be the splitter. To repeat the parenthetical comment above, T is the value of the array entry that was chosen, not its position in the array.

That involves some work, but it is rather routine, and its cost is linear in the number of edges of G, say c|E(G)|. Finally we call chrompoly on the graph G/{e}. Let F (V, E) denote the maximum cost of calling chrompoly on any graph of at most V vertices and at most E edges. 5) together with F (V, 0) = 0. If we put, successively, E = 1, 2, 3, we find that F (V, 1) ≤ c, F (V, 2) ≤ 4c, and F (V, 3) ≤ 11c. 6) then we will have such a solution. 5), we find that E f(E) = 2E j2−j j=0 ∼ 2E+1. 3. To summarize the developments so far, then, we have found out that the chromatic polynomial of a graph can be computed recursively by an algorithm whose cost is O(2E ) for graphs of E edges.

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Algorithms and Complexity (Internet edition, 1994) by Herbert S. Wilf

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