# Flajolet P., Sedgewick R.'s Analytic combinatorics MAc PDF

By Flajolet P., Sedgewick R.

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Extra resources for Analytic combinatorics MAc

Example text

The solution to the original coin change problem is found to be [z 99 ] 1 = 213. (1 − z)(1 − z 5 )(1 − z 10 )(1 − z 25 ) In the same vein, one proves that {1,2} Pn =⌈ 2n + 3 ⌋ 4 {1,2,3} Pn =⌈ (n + 3)2 ⌋; 12 here ⌈x⌋ ≡ ⌊x + 21 ⌋ denotes the integer closest to the real number x. Such results are typically obtained by the two step process: (i) decompose the rational generating function into simple fractions; (ii) compute the coefficients of each simple fraction and combine them to get the final result [122, p.

8. Lim-sup of classes. Let {A[ j] } be any increasing sequence of combinatorial classes, in the sense that A[ j] ⊂ A[ j+1] , and the notions of size are compatible. If A[∞] = [ j] is a jA 32 I. 5: Formal power series, p. 730). 7. A specification for an r –tuple A = (A(1) , . . , A(r ) ) of classes is a collection of r equations,  (1) (1) (r ) 1 (A , . . , A )   A(2) = (1) (r ) A = 2 (A , . . , A ) (26) · · ·   (r ) (1) (r ) A = r (A , . . , A ) where each i denotes a term built from the A using the constructions of disjoint union, cartesian product, sequence, set, multiset, and cycle, as well as the initial classes E (neutral) and Z (atomic).

2. ADMISSIBLE CONSTRUCTIONS AND SPECIFICATIONS 27 The class E = {ǫ} consisting of the neutral object only, and the class Z consisting of a single “atomic” object (node, letter) of size 1 have OGFs E(z) = 1 Z (z) = z. and Proof. The proof proceeds case by case, building upon what we have just seen regarding unions and products. Combinatorial sum (disjoint union). Let A = B + C. Since the union is disjoint, and the size of an A–element coincides with its size in B or C, one has An = Bn + Cn and A(z) = B(z) + C(z), as discussed earlier.