By Herbert S. Wilf

ISBN-10: 1568811780

ISBN-13: 9781568811789

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**Extra resources for Algorithms and Complexity**

**Example text**

A graph of n vertices has ¢ a maximum of n2 edges. To construct a graph, we would decide which ¡ ¢ of these possible edges would be used. We can make each of these n2 decisions independently, and for every way of deciding where to put the edges, we would get a diﬀerent graph. Therefore, the number of labeled n 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.

The first one ‘n ≥ 0’ tells us for what values of n the recurrence formula is valid, and the second one ‘x0 = 1’ gives the starting value. If one of these is missing, the solution may not be uniquely determined. 25) needs two starting values in order to “get going,” but it is missing both of those starting values and the range of n. 25) (which is a second-order recurrence) does not uniquely determine the sequence. The situation is rather similar to what happens in the theory of ordinary diﬀerential equations.

The first step is to define a new unknown function as follows. 29) define a new unknown sequence y1 , y2 , . . 28), getting b1 b2 · · · bn+1 yn+1 = bn+1 b1 b2 · · · bn yn + cn+1 . We notice that the coeﬃcients of yn+1 and of yn are the same, and so we divide both sides by that coeﬃcient. 4. Recurrence Relations 29 where we have written dn+1 = cn+1 /(b1 · · · bn+1 ). Notice that the ds are known. We haven’t yet solved the recurrence relation. 30). 30) is quite simple, because it says that each y is obtained from its predecessor by adding the next one of the ds.

### Algorithms and Complexity by Herbert S. Wilf

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