Midterm solutions. n f 3 (n) = 3

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1 Introduction to Computer Science 1, SE361 DGIST April 20, 2016 Professors Min-Soo Kim and Taesup Moon Midterm solutions Midterm solutions The midterm is a 1.5 hour exam (4:30pm 6:00pm). This is a closed book, closed note exam. No calculators or programmable devices are permitted. No cell phones or other communications devices are permitted. Problem 1-1. [15 points] Asymptotic orders of growth For each of the three pairs of functions given below, rank the functions by increasing order of growth. (a) [5 points] Group 1: Solution: 2, 1, 3 (b) [5 points] Group 2: Solution: 1, 3, 2 (c) [5 points] Group 3: Solution: 3,2,1 f 1 (n) = 7 n f 2 (n) = f 3 (n) = ( 5) log 2 n f 1 (n) = n log 2 n f 2 (n) = n log 2 ( ) n n f 3 (n) = 3 f 1 (n) = log 2 n! f 2 (n) = 1000n log 2 n f 3 (n) = n(log 2 n) 3 4 Problem 1-2. [10 points] Recurrence Relation Resolution Find an asymptotic solution of the following functional recurrence. Express your answer using O-notation, and give a brief justification. (a) [5 points] ( n ) T (n) = 9 T + n 3 3 Solution: Using the Master Theorem, we obtain O(n 3 ).

2 2 Midterm solutions (b) [5 points] T (n) = T ( n) + 1 (Note that n 1 log n = 1.) Solution: T (n) = Θ(log log n). To see this, note that n = n 1/2i. So once i becomes log log n we will have n 1/2i = n 1/ log n = 1. Thus, the recursion stops after log log n levels and each level contributes 1, hence T (n) = Θ(log log n). Problem 1-3. [27 points] True/False For each of the following questions, choose either T (True) or F (False). Explain your choice. (Your explanation is worth more than your choice of true or false.) (a) [3 points] (T/F) 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. (b) [3 points] (T/F) If you know the numbers stored in a BST and you know the structure of the tree, you can determine the value stored in each node. Solution: True. You can do an inorder walk of the tree, which would order the nodes from smallest key to largest key. You can then match them with the values. (c) [3 points] (T/F) In max-heaps, the operations insert, max-heapify, find-max, and findmin all take O(log n) time. Solution: False. The minimum can be any of the nodes without children. There are n/2 such nodes, so it would take Θ(n) time to find it in the worst case. (d) [3 points] (T/F) We can sort 7 numbers with 10 comparisons. Solution: False. To sort 7 numbers, the binary tree must have 7! = 5040 leaves. The number of leaves of a complete binary tree of height 10 is 2 10 = This is not enough. (e) [3 points] (T/F) A Θ(n 2 ) algorithm always takes longer to run than a Θ(log n) algorithm. Solution: False. The constant of the Θ(log n) algorithm could be a lot higher than the constant of the Θ(n 2 ) algorithm, so for small n, the Θ(log n) algorithm could take longer to run. (f) [3 points] (T/F) The depths of any two leaves in a max heap differ by at most 1. Solution: True. A heap is derived from an array and new levels to a heap are only

3 Midterm solutions 3 added once the leaf level is already full. As a result, a heaps leaves are only found in the bottom two levels of the heap and thus the maximum difference between any two leaves depths is 1. (g) [3 points] (T/F) The height of any binary search tree with n nodes is O(log n). Solution: False. In the best case, the height of a BST is O(log n) if it is balanced. In the worst case, however, it can be Θ(n). (h) [3 points] (T/F) A tree with n nodes and the property that the heights of the two children of any node differ by at most 2 has O(log n) height. Solution: True. Using the same approach as proving AVL trees have O(log n) height, we say that n h is the minimum number of elements in such a tree of height h. n h 1 + n h 1 + n h 3 (1) n h > 2n h 3 (2) n h > 2 h/3 (3) h < 3 log n h (4) h = O(log n) (5) (i) [3 points] (T/F) Quicksort can always sort n numbers with O(n log n) running time. Solution: False. In worst case, Quicksort can run in Θ(n 2 ).

4 4 Midterm solutions Problem 1-4. [24 points] Short answers (a) [8 points] What order should we insert the elements {1, 2,..., 7} into an empty AVL tree so that we dont have to perform any rotations on it? Solution: 4,2,6,1,3,5,7. (b) [8 points] What is the max-heap resulting from performing on the node storing 6? Solution:

5 Midterm solutions 5 (c) [8 points] What does the following AVL tree look like after we perform Insert 12 and Delete 3? (Draw two resulting trees after each operation.) Solution: After Insert 12, we have Then, after Delete 3, we have

6 6 Midterm solutions Problem 1-5. [24 points] Computing Fibonacci numbers The Fibonacci numbers are defined by the following recurrence: F 0 = 0, F 1 = 1, F i = F i 1 + F i 2 for i 2, yielding the sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,... There is a closed-form formula for F n given by F n = φn (1 φ) n 5, where φ = is the golden ratio. However, this formula is not practical for computing the exact value of F n as it would require increasing precision on 5 and φ as n increases. In this problem, we are interested in obtaining practical algorithms for computing the n-th Fibonacci number F n for any given n. Assume that the cost of adding, subtracting, or multiplying two integers is O(1), independent of the size of the integers we are dealing with. (a) [5 points] From the recurrence definition of the Fibonacci sequence, one can use the following simple recursive algorithm: Give the running time of this algorithm. Express your answer using Θ-notation. Solution: Let T (n) be the time taken to compute F n. We have T (n) = T (n 1) + T (n 2) + Θ(1), and so T (n) = Θ(F n ) = Θ(φ n ). (b) [5 points] Give an algorithm that computes F n in Θ(n) and justify its running time. Solution: To obtain the desired algorithm we develop an iterative (non-recursive) version of the previous algorithm in which we memorize the last two computed Fibonacci numbers, so we are able to compute the next number in constant time. An example code looks as follows: To analyze the complexity of this algorithm it is sufficient to note that we have n iterations of the loop and each iteration can be performed in O(1) time. ( ) ( ) 0 1 (c) [14 points] Consider the matrix A =. Show that for i 1, A 1 1 i Fi 1 F = i F i F i+1. Provide a Θ(log n) algorithm for computing F n. You can give a simple pseudo-code.

7 Midterm solutions 7 (Hint: From the fact that A 2i = (A i ) 2, and A 2i+1 = A A 2i, use divide and conquer to show that A n can be calculated in time Θ(log n)). Solution: First, to show that A i is as given in the problem, we simply use induction. Then, for the algorithm, We first show how to compute A n in Θ(log n) time. Consider the following algorithm: The recurrence describing the running time T (n) of this algorithm is T (n) = T ( n/2 )+ O(1). (Note that we work here with 4-by-4 matrices so multiplying them takes O(1) time.) Using Master Theorem we get T (n) = Θ(log n). Now, to compute Fn we just call C := Matrix power(a, n) and return the entry C[1, 2] of the computed answer. Clearly, this takes Θ(log n) time.