Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 2 pps

... d(n − 2) + 2lgn = cn lg(n − 2) −2c lg(n − 2) +dn −2d + 2lgn > cn lg(n − 2) −2c lg n + dn − 2d + 2lgn (since −lg n < −lg(n 2) ) = cn lg(n − 2) 2( c − 1) lg n + dn − 2d ≥ cn lg(n /2) 2( c − ... probability 1, 2, 3 4 /27 1, 3, 2 5 /27 2, 1, 3 5 /27 2, 3, 1 5 /27 3, 1, 2 4 /27 3, 2, 1 4 /27 6-6 Lecture Notes for Chapter 6: Heapsort (a) (b) (c) (d) (e) 123 47 2 13...

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... second edition of the book.] 15 -2 Lecture Notes for Chapter 15: Dynamic Programming a 1,1 e 1 enter e 2 a 2, 1 a 1 ,2 a 2, 2 t 1,1 t 2, 1 t 1 ,2 t 2, 2 a 1,3 a 2, 3 t 1,3 t 2, 3 a 1,4 a 2, 4 t 1,4 t 2, 4 a 1,5 a 2, 5 x 1 x 2 exit S 1,1 S 2, 1 S 1 ,2 S 2, 2 S 1,3 S 2, 3 S 1,4 S 2, 4 S 1,5 S 2, 5 2 4 7 8 2 2 9 3 1 5 348 3 13 22 645 6 Automobile ... Programming a...

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... to go to get to 1 or 1/4, starting from 1 /2, rate of increase of  differs. • For α to go from 1/ 2to1 ,num increases from size /2tosize, for a total increase of size /2.  increases from 0 to size. ... sort takes 2m time to merge two sorted lists of size m each. If all the arrays A 0 , A 1 , ,A k 2 are full, then the running time to Þll array A k−1 would be T (n) = 2 ( 2...

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... the preface. • 2 June 20 03. Added the solution to Problem 24 -6. Corrected solutions to Ex- ercise 23 .2- 7 and Problem 26 -4. Affected chapters: Chapters 23 , 24 , and 26 ; index. • 20 May 20 03. Added solutions to ... solutions to Exercises 24 .4-10 and 26 .1-7. Affected chapters: Chapters 24 and 26 ; index. • 2 May 20 03. Added solutions to Exercises 21 .4-4,...

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... (n) ] ≤ 2 n n−1  k=  n /2  ck + an = 2c n  n−1  k=1 k −  n /2  −1  k=1 k  + an = 2c n  (n − 1)n 2 − (  n /2  − 1)  n /2  2  + an ≤ 2c n  (n − 1)n 2 − (n /2 2) (n /2 − 1) 2  + an = 2c n  n 2 − ... 2) (n /2 − 1) 2  + an = 2c n  n 2 − n 2 − n 2 /4 −3n /2 + 2 2  + an = c n  3n 2 4 + n 2 − 2  + an = c  3n 4 + 1 2 − 2 n  + an ≤ 3cn 4...

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... mod (2 p − 1) = ((x a 2 ap + x b 2 bp ) − (y a 2 ap + y b 2 bp )) mod (2 p − 1) = ((x a 2 ap + x b 2 bp ) − (x b 2 ap + x a 2 bp )) mod (2 p − 1) = ((x a − x b )2 ap − (x a − x b )2 bp ) mod (2 p − ... +(u k + 1)u k /2. We have the following sequence of equivalent inequalities: u j < u k + 1 2u j < 2u k + 2 −u k < u k − 2u j + 2 u 2 j − u j + u 2 k −...

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... Thus, n−1  k =2 k lg k <(lg n − 1)  n /2  −1  k =2 k + lg n n−1  k=  n /2  k = lg n n−1  k =2 k −  n /2  −1  k =2 k ≤ 1 2 n(n − 1) lg n − 1 2  n 2 − 1  n 2 ≤ 1 2 n 2 lg n − 1 8 n 2 if n ≥ 2. Now ... substitution P(n) = 2 n n−1  k =2 P(k) + (n) ≤ 2 n n−1  k =2 (ak lg k + b) + (n) = 2a n n−1  k =2 k lg k + 2b n (n − 2) + (n) ≤ 2a n  1 2 n...

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... all the LINK operations take place before any of the 22 -18 Solutions for Chapter 22 : Elementary Graph Algorithms Solution to Exercise 22 .4-5 TOPOLOGICAL-SORT(G) ✄ Initialize in-degree, (V ) time for ... A. To form T  from T : • Remove (x, y). Breaks T into two components. • Add (u,v). Reconnects. Solutions for Chapter 22 : Elementary Graph Algorithms 22 -15 Perform as many B...

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... + E) to compute G  . • O(VE) to run BELLMAN-FORD. • (E) to compute w. • O(V 2 lg V +VE) to run Dijkstra’s algorithm | V | times (using Fibonacci heap). • (V 2 ) to compute D matrix. Total: ... since v 0 has no entering edges.) c corresponds to the constraints x 2 − x 1 ≤ w(v 1 ,v 2 ), x 3 − x 2 ≤ w(v 2 ,v 3 ), . . . x k−1 − x k 2 ≤ w(v k 2 ,v k−1 ), x k − x k−1 ≤...

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... 22 -14 Exercise 22 .2- 5, 22 -14 Exercise 22 .2- 6, 22 -14 Exercise 22 .3-4, 22 -15 Exercise 22 .3-7, 22 -15 Exercise 22 .3-8, 22 -16 Exercise 22 .3-10, 22 -16 Exercise 22 .3-11, 22 -16 Exercise 22 .4-3, 22 -17 Exercise 22 .4-5, ... 21 -6 Exercise 21 .2- 5, 21 -7 Exercise 21 .3-3, 21 -7 Exercise 21 .3-4, 21 -7 Exercise 21 .4-4, 21 -8 Exercise 21 .4-5, 21 -8 Exerci...

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