ECE 242 Fall 14 Exam II Profs. Gao and Zink

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1 ECE 242 Fall 14 Exam II Profs. Gao and Zink Name: ID Number: Maximum Achieved Question 1 20 Question 2 17 Question 3 20 Question 4 20 Question 5 23 Total 100 Instructions: Put your name and student number on each sheet of paper! The exam is closed book, closed notes. No electronic devices (including calculators) are allowed. You have 120 minutes to complete the exam. Be a smart exam taker - if you get stuck on one problem go on to another problem. Also, don't waste your time giving irrelevant (or not requested) details. The total number of points for each question is given above. There are 100 points total. The exam is 120 minutes long. Show all your work. Partial credit is possible for an answer, but only if you show the intermediate steps in obtaining the answer. If you make a mistake, it will also help the grader show you where you made a mistake. Write legibly. When writing code, please indent appropriately and give your variables meaningful names. Good luck. 1

2 Question 1 Recursion (20 points): Answer the following questions about the recursive definition of multiplication, m*n=n+(m-1)*n and 1*n=n, and the following implementation in Java. You may assume that m>=1 for all initial calls to this method. public static int mult(int m, int n) { if (m==1) { return n; return n+mult(m-1,n); a) For the computation of mult(5,7), list all calls to the method mult() with the values of m and n. (5 points) mult(5,7) mult(4,7) mult(3,7) mult(2,7) mult(1,7) b) How many times is mult() called for the computation of mult(5000,7)? (2 points) 5000 times c) How many times is mult() called for the computation of mult(7,5000)? (2 points) 7 times d) The recursive definition of computing the factorial of n is n! = n * (n-1)! with 1! = 1. Develop a Java function that recursively calculates the factorial of n. (6 points) static private int factorial(int n) { 2

3 static private int factorial(int n) { // System.out.println("method call: factorial("+n+")"); if (n==1) { // check for termination condition // System.out.println("method factorial("+n+") returning 1"); return 1; else { // break problem into smaller problem int fac = n*factorial(n-1); // System.out.println("method factorial("+n+") returning "+fac); return fac; e) The definition for recursively calculating a Fibonacci number F(n) is F(n) = F(n-1) + F(n- 2) with F(1) = 1 and F(0) = 0. A function that implements this recursive calculation in Java is shown below. Illustrate the call sequence of this function for the case of F(4). (5 points) static private int fibonacci(int n) { // System.out.println("method call: fibonacci("+n+")"); if (n==0) { // check one termination condition // System.out.println("method fibonacci("+n+") returning 0"); return 0; else if (n==1) { // check other termination condition // System.out.println("method fibonacci("+n+") returning 1"); return 1; else { // break problem into smaller pieces int fib = fibonacci(n-1)+fibonacci(n-2); // System.out.println("method fibonacci("+n+") returning "+fib); return fib; F(4) 3

4 F(4) F(3) F(2) F(2) F(1) F(1) F(0) F(1) F(0) 4

5 Question 2 Mergesort (17 points): a) Given the following array as input, illustrate how the Mergesort algorithm performs. To illustrate the Mergesort s behavior, start with the dividing of the array until the end condition of the recursive function is met and then show how the merge is performed. (6 points) b) Given the function below, that divides the original array in two arrays of half size, develop the Java code that combines two sorted arrays. Use the result from problem 4a) to develop that code. (6 points) int middle = values.length/2; // divide array into two arrays of half size int[] left = new int[middle]; for (int i=0; i<middle; i++) { left[i] = values[i]; int[] right = new int[values.length-middle]; for (int i=0; i<values.length-middle; i++) { right[i] = values[middle+i]; sort(left); //recursively call sorting function on each smaller array sort(right); int l=0, r=0; // combine sorted arrays for (int i=0; i<values.length; i++) { if (r>=right.length (l<left.length && left[l]<right[r])) { values[i]=left[l]; l++; else { values[i]=right[r]; r++; 5

6 c) Answer the following three questions: What is the complexity of dividing and merging an array of length n? What is the complexity of dividing arrays in Mergsort? What is the overall complexity of Megesort? Give your answers in big O notation. (5 points) Dividing and merging arrays: O(n) time (two passes through array) (2 points) Dividing arrays: One division: O(n) + 2*O(n/2), Second division: O(n) + 2*O(n/2) + 4*O(n/4) O(log n) (2 points) Overall complexity: O(n log n) (1 point) 6

7 Question 3 Quicksort (20 points): a) Given the following array [10, 5, 3, 9, 22, 24, 28, 27,?] and assuming that Quicksort will be used to sort this array in ascending order, select a value for the last element of the array (indicated by? ) such that the partitioning performed by Quicksort is most balanced. Explain why this makes Quicksort perform efficiently. (4 points) any value from 11 to 21 is okay. b) Show the results of the first two rounds of the Quicksort algorithms based on the number you have chosen. (4 points) c) The Java code shown below represents the implementation of a function that partitions an array around a pivot element. Briefly explain what the purpose of the two while loops in lines 8 and 9 is. (4 points) 1: private static int partition(int[] values, int leftbound, int rightbound) 2: { 3: int pivot = values[rightbound]; 4: int left = leftbound-1; 5: int right = rightbound; // rightmost is pivot 6: 7: while (true) { 8: while (values[++left]<pivot) { ; 9: while (right>leftbound && values[--right]>pivot) { ; 10: 11: if (left >= right) { 12: // cross-over indicates end of partitioning process 13: break; 14: else { // found misplaced items; swap 15: swap(values, left, right); 16: 17: 18: swap(values, left, rightbound); 19: // move pivot from right position to "middle 7

8 20: return left; // return index of pivot 21: Line 8: move left index to the right (in the array) until a value is reached that is >= the pivot element. Line 9: move right index to the left (in the array) until a value is reached that is larger than the pivot element d) Given the partition function shown in problem 3c), complete the Jave code shown below for the QuickSort function (4 points). private static void quicksort(int[] values, int left, int right) { printarray(values); if (right-left <= 0) { else { private static void quicksort(int[] values, int left, int right) { printarray(values); if (right-left <= 0) { return; else { int pivotindex = partition(values, left, right); quicksort(values, left, pivotindex-1); quicksort(values, pivotindex+1, right); e) Quicksort and Mergesort have the same complexity. Briefly explain why, despite similar complexity, Quicksort is the more popular sorting algorithm? (4 points) Quicksort does not require any additional arrays, which results in less memory use. This important in cases in which large numbers of elements have to be sorted. 8

9 Question 4 Binary Tree (20 points): a) If a perfect (complete) binary tree has n leaves and all levels are fully populated, how many nodes does the tree have in terms of n? (2 points) 2n-1 b) Given the following BST. Draw the binary search tree after each of the following operations. (Operations are cumulative.) Assume the remove operations on nodes with two children take a replacement node from the right child s subtree. (6 points) Remove 10. Add 33. Remove

10 c) Write the Java code that finds the element with smallest value in BST. (8 points) private Node findmin(node n) { if (n==null) { return null; else { if (n.left==null) { return n; else { return findmin(n.left); d) We have seen that in some cases BST can be very inefficient (e.g., remember the resulting BST when the values 1, 2, 3, 4, 5, 6, 7 are in that exact sequence). Red-black trees address this issue by creating a more balanced tree. For this problem, show how the sequence 1, 2, 3, 4 is added to a red-black tree. (Make sure you show any rotation and recoloring step! Also indicate red nodes by not filling them out and black nodes by filling them out) (4 points) 10

11 Left recolo

12 Question 5 Graphs (23 points): a) For this problem the Java classes Vertex and Graph are shown below. Based on the given code write a function that adds a vertex to a graph. (4 points) public class Vertex { public String name; public int graphindex; //index of adj. matrix position of node in graph public boolean visited; public Vertex (String s) { name = s; graphindex = -1; // invalid position by default visited = false; public class Graph { private int maxvertices; private Vertex[] vertices; // array of nodes private int[][] edges; // adjacency matrix int activevertices; public Graph(int maxsize) { maxvertices = maxsize; vertices = new Vertex[maxVertices]; edges = new int[maxvertices][maxvertices]; // allocating adjacency matrix activevertices = 0; public void addvertex(vertex v) { if (activevertices >= maxvertices) { System.out.println("Graph full"); return; vertices[activevertices] = v; // add vertex to list of vertices v.graphindex = activevertices; // record index of vertex in graph activevertices++; // increment vertex count b) For this problem familiarize yourself with the graph shown below first. Assume that (by using the Java code shown in problem 5a)) the vertices shown in the graph have already been added. For this problem write a Java function that adds the edges (i.e., populates the adjacency matrix) shown in the figure to the graph AND also list the function calls that actually add these edges. (10 points) 12

13 public void addedge(vertex v1, Vertex v2) { public void addedge(vertex v1, Vertex v2) { edges[v1.graphindex][v2.graphindex] = 1; edges[v2.graphindex][v1.graphindex] = 1; addedge(1,2); addedge(1,3); addedge(2,3); addedge(2,4); addedge(2,5); addedge(2,6); addedge(3,5); addedge(4,6); c) Given the following adjacency matrix, draw the resulting graph for this matrix. (6 points) ALB BDL LGA JFK EWR HPN PVD BOS MHT BTV SWF ALB BDL LGA JFK EWR HPN PVD BOS MHT BTV SWF

14 BTV MH T ALB SWF BDL BOS LGA JFK HPN PVD d) Perform a DFS graph traversal, starting from BDL. In your answer list all the vertices in the correct sequential order (the order they are traversed by the DFS algorithm) (3 points) EW R DFS from BDL: BDL EWR MHT PVD JFK ALB LGA BTV BOS HPN SWF 14

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