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Therefore the number of n labeled graphs of n vertices is 2( 2 ) = 2n(n−1)/2 . If we were to ask the corresponding question for unlabeled graphs we would find it to be very hard. 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!.

Each of these two new graphs has s − 1 edges. , will not cost any additional labor. Then the work that we do on G(s, t) will depend only on s, and will be twice as much as the work we do on G(s−1, ·). Therefore G(s, t) will cost at least 2s operations, and our complexity estimate wasn’t a mirage, there really are graphs that make the algorithm do an amount 2|E(G)| of work. ). Let’s look at the algorithm chrompoly in another way. For a graph G we can define a number γ(G) = |V (G)| + |E(G)|, which is rather an odd kind of thing to define, but it has a nice property with respect to this algorithm, namely that whatever G we begin with, we will find that γ(G − {e}) = γ(G) − 1; γ(G/{e}) ≤ γ(G) − 2.

The overall structure of a recursive routine will always be something like this: 31 procedure calculate(list of variables); if {trivialcase} then do {trivialthing} else do {call calculate(smaller values of the variables)}; {maybe do a few more things} end. In this chapter we’re going to work out a number of examples of recursive algorithms, of varying sophistication. We will see how the recursive structure helps us to analyze the running time, or complexity, of the algorithms. We will also find that there is a bit of art involved in choosing the list of variables that a recursive procedure operates on.

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