: Fundamental Data Structures and Algorithms. June 06, 2011 SOLUTIONS

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1 Midterm Exam Summer 2011 Page 1 of : Fundamental Data Structures and Algorithms June 06, 2011 SOLUIONS

2 Midterm Exam Summer 2011 Page 2 of 10 (10) 1. rue/false questions. Clearly mark each of the following assertions as true () or false (F). Hash tables are a suitable data structure for managing an ordering of data. Deletion from an AVL tree takes O(log n) time, because we do log n rotations. Building a heap of n elements has Θ(n) running time. In a splay tree, immediately after inserting a new element X,a find(x) operation takes a constant time. In a max-heap the smallest element resides always at the leaves assuming all elements are distinct. A tree is a balanced search tree. Amortized constant time for hashtable operations is still guaranteed if our rehash policy increases the size of the array by 5% Deletion of the minimal item from a binomial heap takes Θ(log log n) time. Any function which is Ω(log n) is also Ω(log log n). Priority queue can be implemented using AVL trees F F (8) 2. Match Data Structures with Algorithms. Write the letter corresponding to your answer in the blank to the left of the name of the data structure. In some cases, more than one answer might be argued, but you must put only one letter per algorithm. If you pick the best matches, you will use all the letters. E splay tree A. best use for external storing F dynamic array B. find the minimum in constant time B binomial heap C. quadratic worst case construction cost A tree D. optimal search (bits examined) H treap E. amortized performance guarantee D trie F. amortized constant-time insert C binary search tree G. optimal search (comparisons) G AVL tree H. randomized performance guarantee

3 Midterm Exam Summer 2011 Page 3 of rees. Consider the following tree. 5, 10, , 8 15, 17 30, 40 (4) (a) Draw the modified tree resulting in inserting 18, using the top-down approach , 8 15, 17, 18 30, 40 (4) (b) Now insert 14 into the tree from the previous step , , 8 14, , 40

4 Midterm Exam Summer 2011 Page 4 of 10 (4) 4. Splay rees. In the following tree splay on node 50. Show the resulting tree

5 Midterm Exam Summer 2011 Page 5 of Compression: (3) (a) Circle all the valid Huffman codes: Code 1: A - 0, B - 100, C - 101, D - 11 Code 2: A - 010, B - 11, C - 011, D - 10, E - 00 Code 3: A - 011, B - 010, C - 00, D - 01, E - 11 Solution: Code 1 and 2 are valid (4) (b) Fill in the blanks below so that the dictionary, the input and the output are consistent with the LZW algorithm. In other words, fill in the blanks so that running the LZW algorithm on the input would produce the dictionary and the output. Place an X in any blank which should not be filled in. (4) (c) You have a file with a repeated n times, and no other characters. Consider the LZW compression algorithm and approximate the size of the encoded file for large n. Assume that we do not run out of codewords in the process of compressing this file. Solution: he size of the encoded file is kb, where k is the number of codewords in the dictionary, and b is the codeword size. kb = O( 2n)O(log 2 2n)

6 Midterm Exam Summer 2011 Page 6 of he k Largest Elements. Recall that binary max-heaps are suitable data structure for finding the largest element in Θ(log n), where n is the input size. By repeatedly performing a deletemax k < n times we can find the k largest elements in n+k Θ(log n) time. However, there is a more advanced approach (based on a binary heap) that does this slightly faster, in n + k Θ(log k) time. (5) (a) Describe your algorithm in plain English (or in some reasonably equivalent pseudocode) Solution: Build a max-heap of n elements. Let us call it H n. hen traverse the heap H n using BFS with a priority queue. Let us call it PQ. Run deletemax on PQ and then add its children from H n into PQ. Repeat this step k times. (5) (b) Show the running time of your algorithm. Solution: Build a max-heap H n takes Θ(n). Run deletemax on PQ takes Θ(log k). Repeating this step k times gives Θ(k log k) time. hus, the total complexity is Θ(n + k log k).

7 Midterm Exam Summer 2011 Page 7 of Multiway Merge. Recall that merge sort is based on the fact that two already sorted lists can be merged into a single sorted list in linear time. By repeatedly performing a binary merge we can also handle k > 2 lists, ending up with Θ(nk) runtime complexity. However, there is a more advanced approach that does not involve binary merge. his new approach is a heap-based algorithm with running time Θ(n log k). he algorithm merges k sorted lists of integers into a single sorted list. Here n is the total number of elements in all k lists. he algorithm uses one binary min-heap, and therefore it requires only Θ(k) extra space. Answer the following questions. (5) (a) What integers do you store in that heap? Be precise. Clearly, you cannot store integers from all k lists. Solution: we store first elements from each sorted list. (5) (b) Once you perform a deletemin operation, which integer will go to the heap? Solution: we take a next element from that list where the minimum was from

8 Midterm Exam Summer 2011 Page 8 of Hui Han s table. Given a set of strings of decimal digits. Each string has at most length n. Find the number of possible strings H(n, k) such that the sum of its digit is k. (4) (a) Write initial conditions for H(1, k). Solution: H(1, k) = 1, ifk 9, elseh(1, k) = 0 (5) (b) Write a recurrence formula for H(n, k). Solution: H(n, k) = 9 j=0 H[n 1, k j] (4) (c) What is the runtime of your algorithm in terms of n and k.? Solution: O(n k 2 )

9 Midterm Exam Summer 2011 Page 9 of AVL Checker. Fred Hacker s boss, Bob the AVL ree, has asked Fred to write an algorithm to encode AVL tree (and subtree) balance factors into a string. Here s what he came up with: /* his function populates the heights and balance factors of the given AVL ree. */ public int populateavl(avlree tree) { if (tree == null) return -1; /* Populate data for this sub-tree */ tree.height = 1 + max(populateavl(tree.left), populateavl(tree.right)); tree.balancefactor = populateavl(tree.left) - populateavl(tree.right); } return tree.height; (7) (a) Write a recurrence (n) for the runtime of populateavl(). (n) = Solution: (n) = 4(n/2) + Θ(1) (7) (b) Solve the above recurrence (n) using a tree method. Solution: (n) = Θ(n 2 )

10 Midterm Exam Summer 2011 Page 10 of he Ghosts of 211s Past : In the fall of 2018, you return to campus and decide to check out the Gates-Hillman building. However, as you enter the building, you find that there are (n + 1) 2 previous 211 As standing in your way. As some of these As have completed their studies at GHC and some haven t, each one has a different monetary demand. hey have arranged themselves on a grid, as shown below n 0 fee (0,0) fee (0,1) fee (0,2) fee (0,3) fee (0,n) 1 fee (1,0) fee (1,1) fee (1,2) fee (1,3) fee (1,n) 2 fee (2,0) fee (2,1) fee (2,2) fee (2,3) fee (2,n) 3 fee (3,0) fee (3,1) fee (3,2) fee (3,3) fee (3,n) n fee (n,0) fee (n,1) fee (n,2) fee (n,3) fee (n,n) You start at square (0, 0). At any point, you may go right one square, down one square, or diagonally down and right one square (assuming there is a square there you can land on). However, you must pay the A at the destination square (i, j) according to: 2 fee (i,j), if you go diagonally fee (i,j), otherwise. he final roadblock is that you can only enter the building through cell (n, n). You d like to get into it for the least possible cost, and you are stumped as to what to do. hen you think back to what you learned in and realize that you can solve this using dynamic programming. (8) (a) Write a formula for the minimum cost to get to cell (i, j). (You do not need to consider the corner cases,) Cost(i, j) = Solution: Cost(i, j) = min(cost(i 1, j) + fee (i,j), Cost(i, j 1) + fee (i,j), Cost(i 1, j 1) + 2fee (i,j) ) (4) (b) What is the running time of this algorithm? (Give the IGHES bound.) Solution: Θ(n 2 ).