INTRODUCTION TO ALGORITHMS 3rd phần 2 pdf

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... 41 17 30 24 23 26 35 46 7 21 18 52 38 39 41 17 30 24 23 26 35 46 7 21 18 52 38 39 41 17 30 24 23 26 35 46 7 21 18 52 38 39 41 17 30 24 23 26 35 46 7 21 18 52 38 39 41 17 30 24 23 26 35 46 7 21 18 52 38 39 ... 14,15 u 2 0 1 A proto-vEB (2) 1 1 u 2 0 1 A proto-vEB (2) 0 0 u 2 0 1 A proto-vEB (2) 0 1 u 2 0 1 A proto-vEB (2) 0 0 u 2 0 1 A proto-vEB (2) 0...

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... ∞ 7–1 2 2 (a) xtsryz 25 16 34 7–1 2 2 (c) xtsryz 25 16 34 7–1 2 2 (e) xtsryz 25 16 34 7–1 2 2 (g) xtsryz 2 5 16 34 7–1 2 2 (b) xtsryz 25 16 34 7–1 2 2 (d) xtsryz 25 16 34 7–1 2 2 (f) xtsryz ∞ 0 ∞ 26 ∞ 0 26 54 ∞ 0 26 53 ∞ 0 26 53 ∞ 0 2 664 ∞ ∞0 ... 659 0 ∞∞ ∞∞ 0 ∞ ∞ 1 2 10 5 (c) 10 5 0 8 5 14 7 0 8 5 13 7 0 8 5 9 7 0 5 9 7 8 64 32 9 7 s tx yz 1 2 10 5 (f) 64 32 9...

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... Elementary Graph Algorithms 589 22 .1 Representations of graphs 589 22 .2 Breadth-first search 594 22 .3 Depth-first search 603 22 .4 Topological sort 6 12 22. 5 Strongly connected components 615 23 Minimum ... worst-case linear time 22 0 III Data Structures Introduction 22 9 10 Elementary Data Structures 23 2 10.1 Stacks and queues 23 2 10 .2 Linked lists 23 6 10.3 Implementin...

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... the number of collisions. When m D n 2 , the expected number of collisions is E ŒX  D n 2 !  1 n 2 D n 2  n 2  1 n 2 <1 =2: (This analysis is similar to the analysis of the birthday paradox ... language, or the set of file names on a CD-ROM. We 29 0 Chapter 12 Binary Search Trees 2 4 3 13 7 6 17 20 18 15 9 Figure 12. 2 Queries on a binary search tree. To search...

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... its pixels. 15.3 Elements of dynamic programming 385 1 4 1 1 2 4 1 2 3 4 1 3 4 4 2 2 3 4 2 3 4 4 1 1 2 2 3 3 4 4 1 1 2 3 1 2 3 3 3 3 4 4 2 2 3 3 2 2 3 3 1 1 2 2 Figure 15.7 The recursion tree for the computation ... contribution k 1 1 0.15 0.30 k 2 0 0.10 0.10 k 3 2 0.05 0.15 k 4 1 0.10 0 .20 k 5 2 0 .20 0.60 d 0 2 0.05 0.15 d 1 2 0.10 0.30 d 2 3 0.05 0 .20 d...

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... p 2 / C 1 and is not provided as an input. P-M ERGE.T; p 1 ;r 1 ;p 2 ;r 2 ;A;p 3 / 1 n 1 D r 1  p 1 C 1 2 n 2 D r 2  p 2 C 1 3 if n 1 <n 2 // ensure that n 1  n 2 4 exchange p 1 with p 2 5 ... r 2 6 exchange n 1 with n 2 7 if n 1 == 0 // both empty? 8 return 9 else q 1 D b .p 1 C r 1 / =2 c 10 q 2 D BINARY-SEARCH.T Œq 1 ; T; p 2 ;r 2 / 11 q 3 D p 3 C .q 1...

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... power of 2. 31.3 Modular arithmetic 941 0 123 45 0 1 2 3 4 5 0 123 45 0 123 45 0 123 45 0 123 45 0 123 45 0 123 45 (a) 1 2 4 7 8 11 13 14 1 2 4 7 8 11 13 14 1 2 4 7 8 11 13 14 2 4 8 14 1 7 11 13 4811 321 4711 7141341 12 ... for s D 1 to lg n 2 for k D 0 to n 1 by 2 s 3 combine the two 2 s1 -element DFTs in AŒk : : k C 2 s1  1 and AŒk C 2 s1 ::kC 2 s  1 int...

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... L 1 to L 2 in polynomial time, and A 2 is a polynomial-time algorithm that decides L 2 . Algorithm A 1 decides whether x 2 L 1 by using F to transform any input x into f.x/ and then using A 2 to ... a powerful tool for proving that various lan- guages belong to P. Lemma 34.3 If L 1 ;L 2 Â f 0; 1 g  are languages such that L 1 Ä P L 2 ,thenL 2 2 P implies L 1 2 P. Pr...

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... be partitioned into two sets V 1 and V 2 such that .u; / 2 E implies either u 2 V 1 and  2 V 2 or u 2 V 2 and  2 V 1 . That is, all edges go between the two sets V 1 and V 2 . An acyclic, undirected ... that 0 Ä k Ä n =2 to prove inequality (C.6), and use equation (C.3) to extend it to all integers k such that 0 Ä k Ä n. C.1-13 ? Use Stirling’s approximation to p...

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