Data Structures and Algorithms Winter term 2012

Transcription

1 Page 0 German University in Cairo November 4, 2012 Faculty of Media Engineering and Technology Prof. Dr. Slim Abdennadher Data Structures and Algorithms Winter term 2012 Midterm Exam Bar Code Instructions: Read carefully before proceeding. 1) Please tick your major Major IET/MET Mechatronics BI 2) Duration of the exam: 2 hours (120 minutes). 3) (Non-programmable) Calculators are allowed. 4) No books or other aids are permitted for this test. 5) This exam booklet contains 14 pages, including this one. Three extra sheets of scratch paper are attached and have to be kept attached. Note that if one or more pages are missing, you will lose their points. Thus, you must check that your exam booklet is complete. 6) Write your solutions in the space provided. If you need more space, write on the back of the sheet containing the problem or on the four extra sheets and make an arrow indicating that. Scratch sheets will not be graded unless an arrow on the problem page indicates that the solution extends to the scratch sheets. 7) When you are told that time is up, stop working on the test. Good Luck! Don t write anything below ;-) Exercise Possible Marks Final Marks

2 Data Structures and Algorithms, Midterm Exam, November 4, 2012 Page 1 Exercise 1 (12+12=24 Marks) A priority queue is a data structure that supports storing a set of objects, each of which has an associated key encapsulated in the object itself. The basic operations on a priority queue are: insert(o) which inserts object o according to its key into the priority queue remove() which returns and removes from the priority queue the entry with the smallest key size() which returns the number of entries in the queue isempty() which returns true if the queue is empty and false otherwise. isfull() which returns true if the queue is full and false otherwise. In the lectures, an implementations of a priority queue using arrays was introduced. a) Implement a priority queue class with the five methods described above using only stacks with the following interface. class Stack { public void push(comparable o); public Comparable pop(); public Comparable top(); public boolean isempty(); public boolean isfull(); public int size(); For each method, determine its time complexity in terms of order of magnitude.

3 Data Structures and Algorithms, Midterm Exam, November 4, 2012 Page 2 class PriorityQueue_stack { StackObj s; private Object Object; public PriorityQueue_stack(int size) { s = new StackObj(size); public void insert(comparable o) { // complexity O(n) if (s.isempty()) s.push(o); else { int count = s.size(); StackObj temp = new StackObj(count); for (int i = 0; i < count; i++) { if (o.compareto(s.top()) > 0) { temp.push(s.pop()); else break; s.push(o); while (!temp.isempty()) s.push(temp.pop()); public Comparable remove() {// complexity O(1) return s.pop(); public boolean isempty() {// complexity O(1) return s.isempty(); public boolean isfull() { return s.isfull(); // complexity O(1) public int size() { return s.size(); // complexity O(1)

4 Data Structures and Algorithms, Midterm Exam, November 4, 2012 Page 3 // Another implementation for the insert(o) and remove() methods: public void insert(comparable o) { // complexity O(1) s.push(o); public Comparable remove() { // complexity O(n) Comparable result = null; if (!s.isempty()) { Comparable min = (Comparable) s.pop(); int count = s.size(); StackObj temp = new StackObj(count); for (int i = 0; i < count; i++) { if (min.compareto(s.top()) > 0) { temp.push(min); min = s.pop(); else temp.push(s.pop()); while (!temp.isempty()) { s.push(temp.pop()); return result; b) Implement a priority queue class with the five methods described above using only queues with the following interface. class Queue { public void enqueue(comparable o); public Comparable dequeue(); public Comparable peek(); public boolean isempty(); public boolean isfull(); public int size(); For each method, determine its time complexity in terms of order of magnitude.

5 Data Structures and Algorithms, Midterm Exam, November 4, 2012 Page 4 class PriorityQueue_Queue { QueueObj q; public PriorityQueue_Queue(int size) { q = new QueueObj(size); public void insert(comparable o) { if (q.isempty()) q.enqueue(o); else { int count = q.size(); QueueObj temp = new QueueObj(count + 1); for (int i = 0; i < count; i++) { if (o.compareto(q.peek()) > 0) { temp.enqueue(q.dequeue()); else break; temp.enqueue(o); while (!q.isempty()) temp.enqueue(q.dequeue()); while (!temp.isempty()) q.enqueue(temp.dequeue()); public Comparable remove() { return q.dequeue(); public boolean isempty() { return q.isempty(); public boolean isfull() { return q.isfull(); public int size() { return q.size();

6 Data Structures and Algorithms, Midterm Exam, November 4, 2012 Page 5 // Another implementation for the insert(o) and remove() methods: public void insert(comparable o) { q.enqueue(o); public Comparable remove() { Comparable result = null; if (!q.isempty()) { Comparable min = (Comparable) q.dequeue(); int count = q.size(); QueueObj temp = new QueueObj(count); for (int i = 0; i < count; i++) { if (min.compareto(q.peek()) > 0) { temp.enqueue(min); min = q.dequeue(); else temp.enqueue(q.dequeue()); while (!temp.isempty()) { q.enqueue(temp.dequeue()); return result;

7 Data Structures and Algorithms, Midterm Exam, November 4, 2012 Page 6 Exercise 2 ( =28 Marks) A graph is a collection of some finite vertices (nodes) and some finite edges which happen to connect these vertices. If you consider a given list of cities as a list of vertices then you can, probably, call the air routes between them as to be the edges for them. Based on this analogy, we can say that a directed graph is a one in which you have one way air routes and that you can not come back through that same route; means, you can fly from Cairo to Munich but can not come back through that same route. A directed graph can be represented in different ways. A simple representation is to use an adjacency matrix. This matrix shows you all the direct linkages between the vertices; meaning the direct edges between the vertices. For example, the following graph can be represented by the following adjacency matrix A B C D E A false false false true false B false false false true false C false true false false false D false false false false true E false false true false false where true means that there is an edge between the vertices false means that there is no edge between the vertices Your task is to implement a class called mygraph with instance variables to store information about vertices and to store information about the adjacencies/edges between the vertices and the following methods: a) Constructor that sets aside as much capacity as specified by the user public mygraph(int capacity) b) Method to calculate the number of vertices public int numberofvertices() c) Method to calculate the number of adjacencies/edges public int numberofedges() d) Method to find the location at which a vertex is stored. It returns -1 if vertex not found public int getindex(string vertex)

8 Data Structures and Algorithms, Midterm Exam, November 4, 2012 Page 7 e) Method to add a new vertex public void addvertex(string newvertex) Note that if the data structure where vertices are stored, is full, then you need to expand it and also to expand the data structure of edges. f) Method to add a new edge public void addedge(string vertex1, String vertex2) g) Method to delete a given vertex public void deletevertex(string vertex) h) Method to delete a given edge public void deleteedge(string vertex1, String vertex2) public class mygraph { String[] names; // 1-d array to store the vertices boolean[][] Edges; // 2-d array to store adjacencies between // vertices, int numvertices; int numedges; // Constructor that sets aside as much capacity as specified by the user public mygraph(int capacity) { names = new String[capacity]; Edges = new boolean[capacity][]; // We use only the portion of the matrix below the main diagonal to // store the edges for (int i = 0; i < Edges.length; i++) { Edges[i] = new boolean[i + 1]; for (int j = 0; j < i; j++) Edges[i][j] = false; public int numberofvertices() { return numvertices; public int numberofedges() { return numedges; // Finds the location at which a vertex is stored in Vertices. // Returns -1 if vertex not found public int getindex(string vertex) { for (int i = 0; i < numvertices; i++) if (vertex.equals(names[i])) return i; return -1;

9 Data Structures and Algorithms, Midterm Exam, November 4, 2012 Page 8 // Resizes the array of vertices. Can make it larger or smaller, depending // on what newsize is. public String[] resize(string[] array, int newsize) { String[] temp = new String[newSize]; int smallersize = newsize; if (array.length < smallersize) smallersize = array.length; for (int i = 0; i < smallersize; i++) temp[i] = array[i]; return temp; // Resizes the array of adjacencies. Can make it larger or smaller, // depending // on what newsize is. public boolean[][] resize(boolean[][] array, int newsize) { boolean[][] temp = new boolean[newsize][]; int smallersize = newsize; if (array.length < smallersize) smallersize = array.length; for (int i = 0; i < newsize; i++) { temp[i] = new boolean[i + 1]; for (int j = 0; j < i + 1; j++) { if (i < smallersize) temp[i][j] = array[i][j]; else temp[i][j] = false; return temp; // Adds a new vertex public void addvertex(string newvertex) { if (getindex(newvertex)!= -1) { System.out.print("addVertex: "); System.out.print(newVertex); System.out.println(" failed -- vertex already exists."); return; // if array of vertices is full, we need to expand it and // also expand Edges if (names.length == numvertices) { names = resize(names, 2 * numvertices + 1); Edges = resize(edges, 2 * numvertices + 1); names[numvertices++] = newvertex; // delete the given vertex public void deletevertex(string vertex) {

10 Data Structures and Algorithms, Midterm Exam, November 4, 2012 Page 9 int i = getindex(vertex); if (i == -1) { System.out.print("deleteVertex: "); System.out.print(vertex); System.out.println(" failed -- it does not exist."); return; // Move the last vertex up to fill the hole made by vertex i names[i] = names[numvertices - 1]; names[numvertices - 1] = null; for (int j = 0; j < i; j++) { // For every edge incident on vertex i, decrement numedges if (Edges[i][j]) numedges--; // Move the first i elements of the last row in the adj. matrix into // the ith row Edges[i][j] = Edges[numVertices - 1][j]; for (int j = i + 1; j < numvertices - 1; j++) { // For every edge incident on vertex i, decrement numedges if (Edges[j][i]) numedges--; // Move the remaining elements of the last row in the adj. matrix // into the ith collumn Edges[j][i] = Edges[numVertices - 1][j]; // Delete the last row in the matrix for (int k = 0; k < numvertices; k++) Edges[numVertices - 1][k] = false; numvertices--; // Adds a new edge public void addedge(string vertex1, String vertex2) { int i = getindex(vertex1); if (i == -1) { System.out.print("addEdge failed: "); System.out.print(vertex1); System.out.println(" does not exist."); return; int j = getindex(vertex2); if (j == -1) { System.out.print("addEdge failed: "); System.out.print(vertex2); System.out.println(" does not exist."); return; Edges[i][j] = true; numedges++;

11 Data Structures and Algorithms, Midterm Exam, November 4, 2012 Page 10 // deletes a given edge public void deleteedge(string vertex1, String vertex2) { int i = getindex(vertex1); if (i == -1) { System.out.print("deleteEdge failed: "); System.out.print(vertex1); System.out.println(" does not exist."); return; int j = getindex(vertex2); if (j == -1) { System.out.print("deleteEdge failed: "); System.out.print(vertex2); System.out.println(" does not exist."); return; if (Edges[i][j]) { Edges[i][j] = false; numedges--; // end of class

12 Data Structures and Algorithms, Midterm Exam, November 4, 2012 Page 11 Exercise 3 Given the following program: ( =18 Marks) public class Mystery { public static void main(string[] args) { int[] array = {10,12,9,71,5,8; mystery(array, 0, array.length-1); public static void mystery (int[] a, int lo, int hi) { int i=lo, j=hi, h; int x=a[(lo+hi)/2]; do { while (a[i]<x) i++; while (a[j]>x) j--; if (i<=j) { h=a[i]; a[i]=a[j]; a[j]=h; i++; j--; printlist(a); while (i<=j); if (lo<j) mystery(a, lo, j); if (i<hi) mystery(a, i, hi); public static void printlist(int[] list) { for(int n = 0; n < list.length; n++) System.out.print(list[n]+" "); System.out.println(); a) Trace your algorithm for the following input: array = {10,12,9,71,5,8 You should execute the method and display the output of the printlist method after each execution of the do-while loop b) what is the functionality of the program for any input array. Sorting the numbers in the array in ascending order.

13 Data Structures and Algorithms, Midterm Exam, November 4, 2012 Page 12 c) What is the invariant of the mystery method after each execution of the do-while loop. After the loop execution the pivot element is in its correct position. The elements before the pivot element are less than it. The elements after the pivot element are greater than it. d) When does the best case behavior of the algorithm occurs. Explain it in in English. The best-case behavior of the sorting algorithm occurs when in each recursion step the partitioning produces two parts of equal length. In order to sort n elements, in this case the running time is in O(nlog(n)). e) When does the worst case behavior of the algorithm occurs. Explain it in in English. The worst case occurs when in each recursion step an unbalanced partitioning is produced, namely that one part consists of only one element and the other part consists of the rest of the elements. The sorting algorithm runs in time O(n 2 ).