By Herbert S. Wilf

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18). The reason for choosing diﬀerentiation is that it will put the missing multipliers 1, 2, 3, . . 18). 18) becomes 1 + 2x + 3x2 + 4x3 + · · · + (n − 1)xn−2 = 1 − nxn−1 + (n − 1)xn . 19) (1 − x)2 Now it’s easy. 19), to obtain, after simplifying the right-hand side: 1 + 2 · 2 + 3 · 4 + 4 · 8 + · · · + N 2N −1 = 1 + (N − 1)2N . 21) If we rewrite the series using summation signs, it becomes: ∞ X j=2 1 . 16). 16), it tells us that ∞ X j=1 1 = log (3/2). 22) 26 1. 21) is equal to log (3/2) − 1/3.

Exactly how many labeled graphs of n vertices and E edges are there? 6. In how many labeled graphs of n vertices do vertices {1, 2, 3} form an independent set? 7. How many cliques does an n-cycle have? 8. True or false: A Hamilton circuit is an induced cycle in a graph. 9. Which graph of n vertices has the largest number of independent sets? How many does it have? 10. Draw all of the connected, unlabeled graphs of 4 vertices. 11. Let G be a bipartite graph that has q connected components. Show that there are exactly 2q ways to properly color the vertices of G in 2 colors.

However, we can make our own splitters, with some extra work, and that is the idea of the Quicksort algorithm. Let’s state a preliminary version of the recursive procedure as follows (look carefully for how the procedure handles the trivial case where n=1). {quicksortprelim} This preliminary version won’t run, though. It looks like a recursive routine. It seems to call itself twice in order to get its job done. But it doesn’t. It calls something that’s just slightly diﬀerent from itself in order to get its job done, and that won’t work.