INTRODUCTION TO ALGORITHMS 3rd phần 6 pdf

... ∞ 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 ∞ 0 265 4 ∞ 0 265 3 ∞ 0 265 3 ∞ 0 2 66 4 ∞ ... the inequalities around cycle c gives us 64 8 Chapter 24 Single-Source Shortest Paths (a) (b) (c) 0 6 6 7212 4 3 5 3 s tx yz 39 51...

Ngày tải lên: 13/08/2014, 18:20

... wish to hire. 158 Chapter 6 Heapsort 1 23 4 567 8910 1 23 4 567 8910 1 23 4 567 8910 1 23 4 567 8910 1 23 4 567 8910 1 23 4 567 8910 4 13 2910 14 8 7 (a) 16 41 23 16 9 10 14 8 7 4 13 2910 14 8 7 (b) 16 4 13 14 ... similar to our analysis of the upper bound: 6. 2 Maintaining the heap property 155 16 410 14 7 9 281 (a) 16 14 10 4793 281 (b) 16 14 10 8793 241 (c) 3 1 3 4...

Ngày tải lên: 13/08/2014, 18:20

... 23 26 35 46 7 17 30 2423 26 35 46 7 21 18 52 38 39 41 (a) 3 (b) (f) (g) 21 18 52 38 39 41 (h) 17 30 2423 26 35 46 7 21 18 52 38 39 41 17 30 2423 26 35 46 7 21 18 52 38 39 41 17 30 2423 26 35 46 7 ... 3 clusteru 16 summary proto-vEB( 16) 0 1 cluster u 4 summary proto-vEB(4) 0 1 A proto-vEB(2) 1 1 0 1 cluster u 4 summary proto-vEB(4) 0 1 cluster u 4 summary proto-vEB(4) 0...

Ngày tải lên: 13/08/2014, 18:20

... Maximum bipartite matching 732 ? 26. 4 Push-relabel algorithms 7 36 ? 26. 5 The relabel -to- front algorithm 748 VII Selected Topics Introduction 769 27 Multithreaded Algorithms 772 27.1 The basics of ... multiplication 68 6 25.2 The Floyd-Warshall algorithm 69 3 25.3 Johnson’s algorithm for sparse graphs 700 26 Maximum Flow 708 26. 1 Flow networks 709 26. 2 The Ford-Fulkerson m...

Ngày tải lên: 13/08/2014, 18:20

... key k not already stored in the table is equally likely to hash to any of the m slots. The expected time to search unsuccessfully for a key k is the expected time to search to the end of list ... interest are pointers to other nodes, and they vary according to the type of tree. Binary trees Figure 10.9 shows how we use the attributes p, left,andright to store pointers to the...

Ngày tải lên: 13/08/2014, 18:20

... lines 1–2 and of lines 6 7 each take at least unit time, as 420 Chapter 16 Greedy Algorithms 01234 567 891011121314 time 235 3 06 457 539 65 9 761 0 8811 9812 10214 11 12 16 114 k s k f k a 1 a 2 a 1 a 3 a 1 a 4 a 1 a 4 a 5 a 1 a 4 a 6 a 1 a 4 a 7 a 1 a 4 a 8 a 1 a 4 a 8 a 9 a 1 a 4 a 8 a 10 a 1 a 4 a 8 a 11 a 1 a 4 a 8 a 11 0–0 a 1 a 0 a 0 RECURSIVE-ACTIVITY-SELECTOR(s, ... for a s...

Ngày tải lên: 13/08/2014, 18:20

... defined by  120 344 563  x D  3 7 8  ; where A D  120 344 563  ; b D  3 7 8  ; and we wish to solve for the unknown x. The LUP decomposition is L D  100 0:2 1 0 0 :6 0:5 1  ; U D  56 3 00:80 :6 002:5  ; P ... comes into the picture, executing instructions 4 6. Processor 2 reads x from memory into register r 2 ; increments it, producing the value 1 in r 2 ; and then store...

Ngày tải lên: 13/08/2014, 18:20

... circuit. (a 0 ,a 1 ,a 2 ,a 3 ,a 4 ,a 5 ,a 6 ,a 7 ) (a 0 ,a 2 ,a 4 ,a 6 ) (a 0 ,a 4 )(a 2 ,a 6 ) (a 0 )(a 4 )(a 2 )(a 6 ) (a 1 ,a 3 ,a 5 ,a 7 ) (a 1 ,a 5 ) (a 1 )(a 5 ) (a 3 ,a 7 ) (a 3 )(a 7 ) Figure 30.4 The tree of input vectors to ... 2;3;:::;n. a. Is the sum of two Toeplitz matrices necessarily Toeplitz? What about the prod- uct? b. Describe how to represent a Toeplitz matrix...

Ngày tải lên: 13/08/2014, 18:20

...  is a tautology if it evaluates to 1 for every assignment of 1 and 0 to the input variables. Define TAUTOLOGY as the language of boolean formulas that are tautologies. Show that TAUTOLOGY 2 co-NP. 34.2-9 Prove ... at least as hard as another, to within a polynomial-time factor. That is, if L 1 Ä P L 2 ,thenL 1 is not more than a polynomial factor harder than L 2 ,whichis 1 066 Chapter 3...

Ngày tải lên: 13/08/2014, 18:20

... integers k such 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 prove that 2n n ! D 2 2n p n .1 ... likely: Pr f s g D 1= 36. Define the random variable X to be the maximum of the two values showing on the dice. We have Pr f X D 3 g D 5= 36, sinceX assigns a value of 3...

Ngày tải lên: 13/08/2014, 18:20