Dartmouth College Computer Science 10, Fall 2015 Midterm Exam

Transcription

1 Dartmouth College Computer Science 10, Fall 2015 Midterm Exam pm, Monday, October 19, Dartmouth Hall Professor Prasad Jayanti Print your name: Print your section leader name: If you need more space to answer a question, use the other side of the paper, which I have left blank. If you need still more space, you may use additional sheets of paper and attach them to the exam. Make sure that we know where to look for your answer! The exam has 12 pages. Read each question carefully and make sure that you answer everything asked for. Write legibly so that we can read your solutions. Please do not write anything in red. You have until 9.00pm to complete this exam. No extra time is allowed. Question Value Score Total 90

2 CS 10 Fall 2015, Midterm 2 Question 1 14 points (a) (4 points) What is the distinction between an ADT and its implementation? Why is this distinction important? Answer: ADT only specifies the set of methods with which a user can manipulate the object, and what the methods do (from the user s perspective). Implementation defines the code (for the methods) that ensures that the methods behave as specified in the ADT. The distinction makes it possible for the user of the object to write their program without any knowledge of how the object is implemented. Consequently, if the object is implemented differently tomorrow, there will be no need to change the user program (since it only depends on the ADT, and not the implementation, and the ADT has not changed). (b) (10 points) Consider the following pieces of code (you can assume that each class is in a separate file). public class A { public void print() { System.out.println("A"); public class B extends A { public void print() { super.print(); System.out.println("B"); For each of the following fragments of code (one below and four more on the next page), state whether the code fragment is legal. If the code fragment is legal, state (i) whether it illustrates polymorphism, (ii) whether it illustrates dynamic binding, and (iii) what, if anything, is printed. Briefly justify (no more than a phrase or a sentence, where a justification is appropriate). 1. A a = new A(); B b = new B(); a.print(); b.print(); 2. Answer: Legal. Does not illustrate polymorphism. Does not illustrate dynamic binding. Prints: A A B (one letter per line). A a = new B(); B b = new B(); a.print(); b.print();

3 CS 10 Fall 2015, Midterm Answer: Legal. Illustrates polymorphism (a is declared to be a reference to class A object, but references class B object). Illustrate dynamic binding (a.print() executes the print method of class B, even though a is declared to be of class A). Prints: A B A B (one letter per line). A a = new A(); B b = new A(); a.print(); b.print(); Answer: Ilegal (because the subclass variable b cannot be assigned a reference to a superclass object, as in Line 2). A a = new A(); B b = new B(); b = a; b.print(); Answer: Ilegal (because the subclass variable b cannot be assigned a reference to a superclass object, as in Line 3). public void fun(a x) { System.out.println("funA"); public void fun(b x) { System.out.println("funB"); // assume that the following code fragment is in the main method A a = new B(); fun(a); Answer: Legal. Static type checking is used to figure out which method to invoke: since a is declared to be a reference to class A, the first fun method is executed. So, funa gets printed. There is polymorphism because a is declared as reference to class A and assigned a reference to class B. Question 2 20 points (a) (10 points) Consider an ADT whose state is a collection of integers. The ADT supports two operations, specified as follows:

4 CS 10 Fall 2015, Midterm 4 /* Adds e to the collection of integers. */ public void insert(int e) /* If the collection has at least two distinct integers, returns the second largest integer in the collection; else returns -1. For example, if 8, 4, 10, 6, 10, 2, 10 were previously inserted, a call to secondlargest() returns 8. As another example, if 8, 8, 8 are the only integers previously inserted, a call to secondlargest() returns -1 (since there are no two distinct integers in the collection). */ public int secondlargest() Define a class that efficiently implements this ADT. State the running times of your methods in asymptotic notation. (It is not enough for the implementation to be correct: very little credit is awarded unless the implementation is as efficient as possible in terms of both running time and space.) public class MyClass { private int n; // 0, 1, or 2, based on // number of distinct integers inserted so far private int max1; // largest if n > 0 private int max2; // second largest if n > 1 public MyClass() { n = 0; public void insert(int e) { if (n == 0) { max1 = e; n = 1; else if (n == 1) { if (e > max1) { max2 = max1; max1 = e; n = 2; else if (e < max1) { max2 = e; n = 2; else if (e > max1) { max2 = max1; max1 = e;

5 CS 10 Fall 2015, Midterm 5 else if ((e < max1) && (e > max2)) max2 = e; public int secondlargest() { if (n < 2) return -1; else return max2; Each of the above three methods runs in O(1) time. (b) (10 points) Consider the following interface for a stack of integers. public interface MyStackADT { /* adds e to the top of the stack. */ public void push(int e); /* legal only if stack is nonempty; removes and returns top item of stack */ public int pop(); /* returns true if the stack is empty; else returns false */ public boolean isempty(); Write a class that implements the above interface, representing the stack using an array of integers, and expanding the array whenever necessary. State the worst-case running times of your methods in asymptotic notation. public class MyStack implements MyStackADT { private int[] s; private int n; // number of elements in stack public MyStack() { n = 0; s = new int[10]; // Any positive integer works, not just 10 public boolean isempty() { return (n==0);

6 CS 10 Fall 2015, Midterm 6 public int pop() { n = n-1; return s[n]; public void push(int e) { int[] new_s; if (s.length == n) { new_s = new int[2*s.length]; for (int i=0; i < n; i++) new_s[i] = s[i]; s = new_s; s[n] = e; n++; The constructor, isempty, and pop run in O(1) time. The worst-case time of push is O(n). Question 3 20 points Consider the following ADT that maintains a sequence of integers and supports the following three methods. public interface SeqADT { /* inserts e at the front of the sequence */ public void addfirst(int e); /* Removes alternate items from sequence, starting from first. For example, if sequence is a1, a2, a3, a4, a5 before method is called, the sequence would be a2, a4 after the method completes. Similarly, if sequence is a1, a2, a3, a4 before method is called, the sequence would be a2, a4 after the method completes. As a final example, if sequence has only one item before method is called, the sequence would be empty after the method completes. */ public void removealternate(); /* prints the items in the sequence in order, from first to last */ public void print(); (a) (10 points) Write a class S that implements the above interface using a singly linked list with sentinel. Define an

7 CS 10 Fall 2015, Midterm 7 inner class E that has two public instance variables: data (of type int) and next (of type E). Make sure that your class S has only one instance variable sentinel that references the sentinel node. Show the code for only the constructor, addfirst, and removealternate methods (no need to write the code for the print() method). public class S implements SeqADT { private Element sentinel; private class Element { public int data; public Element next; public Element(int d) { data = d; next = null; public S() { sentinel = new Element(0); public void addfirst(int e) { Element x = new Element(e); x.next = sentinel.next; sentinel.next = x; public void removealternate() { Element prev = sentinel; // references node before the node to remove while (prev.next!= null) { prev.next = prev.next.next; prev = prev.next; (b) (10 points) Redo the problem with the sequence represented using a singly linked list without a sentinel. (Again, there is no need to write the code for the print() method.) I require that your class S has only one instance variable. public class S implements SeqADT {

8 CS 10 Fall 2015, Midterm 8 private Element head; // references node containing first item of sequence private class Element { public int data; public Element next; public Element(int d) { data = d; next = null; public S() { head = null; public void addfirst(int e) { Element x = new Element(e); if (head == null) head = x; else { x.next = head; head = x; public void removealternate() { if (head!= null) { head = head.next; // remove first element if (head!= null) { prev = head; // references node before the node to remove while (prev.next!= null) { prev.next = prev.next.next; prev = prev.next; Question 4 20 points In the two problems below, which ask you to write methods for the BinaryTree class, you must assume that there are only three instance variables in the class: data, left, and right. Specifically, there is no instance variable parent. When writing the methods for the following two problems, you are not allowed to introduce any extra instance variables.

9 CS 10 Fall 2015, Midterm 9 To help you recall the instance variables of the BinaryTree class, I am showing below the beginning lines of that class: public class BinaryTree<E> { private E data; private BinaryTree<E> left, right;... (a) (10 points) For the BinaryTree class, write a recursive method public int numtwochildrennodes() that returns the number of nodes in the tree this that have both a left child and a right child. public int numtwochildrennodes() { if ((left==null)&&(right==null)) return 0; else if (left==null) return right.numtwochildrennodes(); else if (right==null) return left.numtwochildrennodes(); else return 1+left.numTwoChildrenNodes()+right.numTwoChildrenNodes(); (b) (10 points) For the BinaryTree class, write a method public int numdeep&highnodes(int d, int h) that returns the number of nodes in the tree this whose depth and height in this are at least d and h, respectively. You might need to define an inner class and a helper method. I require you to use recursion. Recall that the depth of a node x in a tree T is the distance from T s root to x, and the height of x in T is the distance from x to its furthest descendant leaf. private Element { public int num; public int ht; // inner class to help methods return two integers

10 CS 10 Fall 2015, Midterm 10 public Element(int i, j) { num = i; ht = j; // Henceforth, let T denote the tree "this" // returns number of nodes of T whose depth is at least d_limit // and height is at least h_limit public int numdeep&highnodes(int d_limit, int h_limit) { Element x = helper(d_limit, h_limit, 0); return x.num; // When the method is called, the parameters have following values: // cur_d = depth of "this" in T, and // d_limit and h_limit are as described earlier // When the method completes, it returns an element x where // x.num is set to number of nodes in "this" whose depth in T is // at least d_limit and height in T is at least h_limit, and // x.ht is set of height of "this" private Element helper(int d_limit, int h_limit, int cur_h) { Element e1, e2; if (left == null) e1 = new Element(0, -1); else e1 = left.helper(d_limit, h_limit, cur_h+1); if (right == null) e2 = new Element(0, -1); else e2 = right.helper(d_limit, h_limit, cur_h+1); int h = 1 + Math.max(e1.ht, e2.ht); if ((cur_d >= d_limit) && (h >= h_limit)) return new Element(1+e1.num+e2.num, h); else return new Element(e1.num + e2.num, h); Question 5 16 points Write an applet where a dot moves continuously along the periphery of a square. When the mouse is clicked, the dot stops; when the mouse is clicked again, the movement resumes; if the mouse is clicked once more, the dot stops again. Thus, each mouse click pauses or restarts the dot s motion; in the beginning the dot is in motion. Here is some useful information that you already know, but I am repeating to help you recall. The constructor for the Timer class takes two parameters an integer and a reference to an

11 CS 10 Fall 2015, Midterm 11 ActionListener object. For instance, if r is a variable that holds a reference to an ActionListener object, the call new Timer(100, r) creates a timer object with two guarantees: (i) once the timer is turned on, it goes off every 100 milliseconds, and (ii) every time the timer goes off, r.actionperformed() gets called. The Timer class supports start(), stop(), isrunning() methods. You must call repaint() after you make changes to the representation of what appears in the Applet window. I have supplied the beginning lines of the code. Some of these lines might be incomplete, so complete them if necessary. And, of course, add many more lines to complete the applet class. The meaning of the constants in my code are as follows: X_TOP_LEFT, Y_TOP_LEFT are the x and y coordinates of the top left corner of the square. SIDE is the length (in pixels) of each side of the square. DT denotes how often the drawing should be updated (when the point is in motion) to create the illusion of continuous movement; its unit is milliseconds. Specifically, when the point is in motion, it should be redrawn at the appropriate location every DT milliseconds. SPEED denotes the distance (in pixels) that the point travels per second. DOT_RADIUS is the radius of the dot (in pixels). public class MyApplet extends JApplet implements ActionListener, MouseListener { private final int X_TOP_LEFT = 100; private final int Y_TOP_LEFT = 100; private final int SIDE = 200; private final int DT = 50; private final int SPEED = 300; private final int DOT_RADIUS = 5; private int len; // distance traveled modulo square s perimeter private Timer timer; // draws the dot on page with center at (x,y) private void drawdot(graphics page, int x, int y) { page.filloval(x-dot_radius, y-dot_radius, 2*DOT_RADIUS, 2*DOT_RADIUS); public void init() { setsize(500, 500); // set size of applet window setvisible(true); addmouselistener(this); (getcontentpane()).add(new MyPanel());

12 CS 10 Fall 2015, Midterm 12 len = 0; timer = new Timer(DT, this); timer.start(); repaint(); public void actionperformed(actionevent e) { len = (len + SPEED * DT/1000) % (4*SIDE); repaint(); public void mouseclicked(mouseevent e) { if (timer.isrunning()) timer.stop(); else timer.start(); public void mousepressed(mouseevent e) { public void mousereleased(mouseevent e) { public void mouseentered(mouseevent e) { public void mouseexited(mouseevent e) { private class MyPanel extends JPanel { public void paintcomponent(graphics page) { super.paintcomponent(page); int segment = len/side; // which of the four sides is the point at int x = X_TOP_LEFT; int y = Y_TOP_LEFT; if (segment == 0) x = x + len % SIDE; else if (segment == 1) { x = X_TOP_LEFT + SIDE; y = y + len % SIDE; else if (segment == 2) { x = x + SIDE - len % SIDE; y = Y_TOP_LEFT + SIDE; else y = y + SIDE - len % SIDE; drawdot(page, x, y);