q1-f2009-sol

# q1-f2009-sol - Introduction to Algorithms October 14, 2009...

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Unformatted text preview: Introduction to Algorithms October 14, 2009 Massachusetts Institute of Technology 6.006 Spring 2009 Professors Srini Devadas and Constantinos (Costis) Daskalakis Quiz 1 Solutions Quiz 1 Solutions Problem 1. Asymptotic orders of growth [9 points] (3 parts) For each of the three pairs of functions given below, rank the functions by increasing order of growth; that is, find any arrangement g 1 ,g 2 ,g 3 of the functions satisfying g 1 = O ( g 2 ) , g 2 = O ( g 3 ) . lg n represents logarithm of n to the base 2. (a) f 1 ( n ) = 8 n, f 2 ( n ) = 25 1000 , f 3 ( n ) = ( 3) lg n Solution: f 2 ( n ) ,f 1 ( n ) ,f 3 ( n ) (b) f 1 ( n ) = 1 100 , f 2 ( n ) = 1 n , f 3 ( n ) = lg n n Solution: f 2 ( n ) ,f 3 ( n ) ,f 1 ( n ) (c) f 1 ( n ) = 2 lg 3 n , f 2 ( n ) = n lg n , f 3 ( n ) = lg n ! Solution: f 3 ( n ) ,f 2 ( n ) ,f 1 ( n ) 6.006 Quiz 1 Solutions Name 2 Problem 2. Balanced Augmented BST [20 points] (2 parts) In this question, we use balanced binary search trees for keeping a directory of MIT students. We assume that the names of students have bounded constant length so that we can compare two different names in O (1) time. Let n denotes the number of students. (a) Say we have a binary search tree with students last names as the keys with lexico- graphic dictionary ordering. Let k be the number of students whose last name starts with a given prefix x . For example, if the tree contains 5 names { ABC, ABD, ADA, ADB, ADC } there are k=2 names, ABC and ABD respectively, starting with the prefix x=AB. How can we output a list of all those students in O ( k + log n ) time? Solution: Consider the following procedure for traversing the tree. traverse ( tree , prefix ): if tree is empty then return name := the last name at the root of tree if prefix is a prefix of name then traverse (left subtree of tree , prefix ) output( name ) traverse (right subtree of tree , prefix ) elseif prefix &lt; name then traverse (left subtree of tree , prefix ) else traverse (right subtree of tree , prefix ) It suffices to run traverse (the entire balanced BST, x ) to find all last names with prefix x . The algorithm has the following properties: It finds all names with prefix x , because it never excludes a subtree that could contain a name with prefix x . It only outputs names with x as a prefix. The running time is O ( k +log n ) , because at each depth it visits at most two nodes for which x is not a prefix. Therefore, it visits at most 2 O (log n ) + k nodes. Another correct solution: Another correct solution is to find the lexicographically first last name with prefix x , and then to continue visiting consecutive nodes in lexi- cographic order as in the in-order tree traversal, until we encounter a name for which x is not a prefix. This procedure visits exactly the same nodes as the first solution, and also runs in O ( k + log n ) time. 6.006 Quiz 1 Solutions Name 3 Common errors: It is not true that the nodes with prefix x must constitute a connected component...
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## q1-f2009-sol - Introduction to Algorithms October 14, 2009...

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