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Unformatted text preview: Introduction to Algorithms October 13, 2010 Massachusetts Institute of Technology 6.006 Fall 2010 Professors Konstantinos Daskalakis and Patrick Jaillet Quiz 1 Solutions Quiz 1 Solutions Problem 1. We hope you mastered this question. Your name is that thing you include at the top of your problem set when you submit. Found yourself tonguetied? This question is bound to show up on future quizes, so feel free to put it on your crib sheet for free points. Not that two points is enough to really dominate the quiz, but its a start. Problem 2. Asymptotics & Recurrences [20 points] (3 parts) (a) [10 points] Rank the following functions by increasing order of growth. That is, find any arrangement g 1 ,g 2 ,g 3 ,g 4 ,g 5 ,g 6 ,g 7 ,g 8 of the functions satisfying g 1 = O ( g 2 ) , g 2 = O ( g 3 ) , g 3 = O ( g 4 ) , g 4 = O ( g 5 ) , g 5 = O ( g 6 ) , g 6 = O ( g 7 ) , g 7 = O ( g 8 ) . f 1 ( n ) = n log 2 n f 2 ( n ) = n + n log 4 n f 3 ( n ) = n 4 f 4 ( n ) = n n/ 3 f 5 ( n ) = n n 2 f 6 ( n ) = 2 log 2 n f 7 ( n ) = n log n f 8 ( n ) = n 3 n 2 Solution: f 2 ,f 1 ,f 5 ,f 3 ,f 8 ,f 7 ,f 6 ,f 4 (b) [5 points] Find an asymptotic solution of the following functional recurrence. Express your answer using notation, and give a brief justification. T ( n ) = 16 T ( n/ 4) + n 2 log 3 n Solution: Using Master Theorem, we compare n 2 log 3 n with n log 4 16 = n 2 . This is case 2 of the generalized version of the theorem as treated in class, so we increment the log k n for ( n 2 log 4 n ) . (c) [5 points] Find an asymptotic solution of the following recurrence. Express your answer using notation, and give a brief justification. (Note that n 1 log n = 1 .) T ( n ) = T ( n ) + 1 Solution: T ( n ) = (log log n ) . To see this, note that q ... n  {z } i times = n 1 / 2 i . So, once i becomes log log n we will have n 1 / 2 i = n 1 / log n = 1 . Thus the recursion stops after log log n levels and each level contributes 1 , hence T ( n ) = (log log n ) . 6.006 Quiz 1 Solutions Name 2 Problem 3. True/False [18 points] (9 parts) Circle (T)rue or (F)alse. You dont need to justify your choice. (a) T F [2 points] Inserting into an AVL tree can take o (log n ) time. Solution: False. To answer this question, we need to know the length of the shortest possible path from the root to a leaf node in an AVL tree with n elements. In the best possible case, for each node we pass, the heights of its two children differ by 1, and we move to the child with the lower height. The childs height is then 2 less than the current nodes height. So in the best case, each time we move to a new node, the height decreases by 2. The number of times we do this to get to height 0 is then the height of the root divided by 2. The height of the root is (log n ) , so it takes 1 / 2 (log n ) = (log n ) time to insert into an AVL tree, in the best case. Therefore it cannot take o (log n ) time....
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 Fall '08
 ErikDemaine
 Algorithms

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