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CS2110 Final Exam SOLUTION
10 December 2018, 9:00AM–9:30PM
0 2 3 4 5 6 7 Total Question Name Short Answer Recursion^ Hashing^
Object Oriented
Data Structures Graphs Max 0 34 7 9 19 18 13 100 Score
The exam is closed book and closed notes. Do not begin until instructed.
You have 150 minutes. Good luck!
Write your name and Cornell NetID, legibly, at the top of every page! There are 6 questions on 10 numbered pages, front and back. Check that you have all the pages. When you hand in your exam, make sure your pages are still stapled together. If not, please use our stapler to reattach all your pages!
We have scrap paper available. If you do a lot of crossing out and rewriting, you might want to write code on scrap paper first and then copy it to the exam so that we can make sense of what you handed in.
Write your answers in the space provided. Ambiguous answers will be considered incorrect. You should be able to fit your answers easily into the space provided.
In some places, we have abbreviated or condensed code to reduce the number of pages that must be printed for the exam. In others, code has been obfuscated to make the problem more difficult. This does not mean that it’s good style.
Academic Integrity Statement: I pledge that I have neither given nor received any unauthorized aid on this exam. I will not talk about the exam with anyone in this course who has not yet taken the final.
(signature)
- Name (0 points)
Write your name and NetID, legibly, at the top of every page of this exam.
- Short Answer (34 points)
(a) True / False (10 points) Circle T or F in the table below.
(a) T F The try-catch block can be used to handle an Exception but not an Error. False. (b) T F A for-loop with a control variable that ranges over the numbers 1..n takes expected time O(n). False. It depends on what the body of the loop does. (c) T F Suppose class C implements interface I1. Then a subclass of C cannot implement I1. False. (d) T F To create and start a new thread, (1) have a class C that implements interface Runnable, (2) in C implement method run(), and (3) create an instance M of class C. False. You also have to call M.start(). (e) T F A graph is bipartite if it is 2-colorable. True. (f) T F There can be two shortest paths between two nodes of a graph. True. Suppose all weights are 1. Let edges be (A, B), (B, D), (A, C), and (C, D). The 2 paths (A, B, D) and (A, C, D) both have distance 2. (g) T F If a graph is drawn so that 2 edges cross, it is non-planar. False. (h) T F Execution of System.out.println(′a′^ + 1); prints the character ′b′. False. it prints the integer 98. (i) T F To resize the array used in implementing hashing using linear probing, double its size and copy the old array into the beginning of the new one. False. Every element in the array has to be rehashed. (j) T F A recursive method must return a value. False.
(b) Proof that f (n) is O(g(n)) (3 points) Prove that 8n^3 + 2n^4 is O(10 + n^4 ).
8 n^3 + 2n^4 ≤ <Assume 1 ≤ n, so n^3 ≤ n^4 > 8 n^4 + 2n^4 = 10 n^4 ≤ 10 n^4 + 100 = 10(n^4 + 10) So choose c = 10 and N = 1
(c) Complexity (2 points) State the time complexities:
(i) Worst-case insertion in the beginning of a linked list of size n: O(1) (ii) Worst-case insertion in an ArrayList of size n: O(n)
(iii) Worst case insertion time in a heap of size n: O(log n) (iv) Worst-case search for a value in a HashSet of size n: O(n)
(f ) Binary search (5 points) To the right are the precondition, postcondition, and invari- ant for a binary search, looking for v in b[0..n]. We assume with stating it anywhere that array b is sorted (in ascending order). Below, complete the algorithm.
h= -1; k= n+1; while ( h+1 < k ) { int e= (h+k)/2; if (b[e] < v) h= e; else k= e; }
Pre: b
0 n ?
Post: b
0 k n < v ≥^ v
Inv: b
0 h k n < v? ≥^ v
(g) New-expression (3 points) Write the steps in evaluating the new-expression C(5, 3).
- Create (draw) an object of class C. 2. Execute the constructor call C(5, 3). 3. Yield as value of the new-expression the name of (pointer to) the newly created object.
- Recursion (7 points)
Consider the following class Person. We give only the fields in the class and the specification of method wasBornIn. Note carefully the part in the class invariant about the relation between the child’s and par- ent’s birth years.
Complete the body of method wasBornIn. A portion of the grade will depend on using the class in- variant to cut the search short where possible.
public class Person { int yrBorn; // The year this person was born. // It is greater than the years its known parents were born Person par1; // one parent, null if unknown Person par2; // another parent, null if unknown
/** Return true iff someone in the tree with this as root was born in year y. */ public boolean wasBornIn(int y) { if (y == yrBorn) return true; if (y < yrBorn) return false; if (par1 != null && par1.wasBornIn(y)) return true; return par2 != null && par2.wasBornIn(y); } }
- Hashing, Iterator, and Iterable (9 points)
Below is part of class HshSet for hashing using linear probing. It is iterable. The constructor of inner class SetIterator is complete. Yes, its body is empty.
(a) (7 points) Complete method iterator() and methods hasNext() and next() of inner class SetIterator. Read carefully the class invariant of SetIterator and the comment in method hasNext().
(b) (2 points) Do any changes have to be made in the methods you wrote if we decide to use quadratic probing instead of linear probing? No. All array elements must still be checked and enumerated if they are not null.
public class HshSet implements Iterable { private E[] b; // Each b[i] is either null or a non-null value of type E; // in the latter case, that value is in the set.
/** = an Iterator over this hashed set. */ public Iterator iterator() { return new SetIterator(); }
/** An iterator over the elements of the hashed set. */ private class SetIterator implements Iterator { // 0 <= k <= b.length. Elements in b[0..k-1] have been enumerated // If k < b.length, b[k] may be null or an element of the set private int k;
/** Constructor: an iterator over the set. */ public SetIterator() {}
@Override public boolean hasNext() { // Note: When hasNext() returns true, b[k] IS next element to enumerate while (k < b.length && b[k] == null) k= k+1; return k < b.length; }
@Override public E next() { if (!hasNext()) throw new NoSuchElementException(); k= k+1; return b[k-1]; } } }
(b) 3 points Which methods does class SUNY need to implement in order to compile, if any? If none, explain why in one sentence. None, since it is abstract.
(c) 4 points Above, complete the constructor Cornell(String pName) according to its specification.
(d) 3 points The IRS has imposed an extremely large tax penalty for all private donations exceeding MAX PRIV DON in PrivateUniv. Hoping to circumvent this penalty and take advantage of its affluent alumni base, Cornell would like to assign a larger value to MAX PRIV DON in its second constructor to allow for larger donations. Is this possible? If not, explain why in one sentence. No, final fields cannot be modified. Note: final in interface fields is optional.
(e) 3 points The NYS Education Dept. decided that each university should be in charge of its own admission process and created a method admitStudents() in class SUNY. How can it ensure that all SUNY universities implement admitStudents()? Make admitStudents() abstract.
(f ) 3 points Cornell has mandated that its students pass a rudimentary swim test before graduat- ing and have modified method public void graduateStudents() to public void graduateStudents( boolean swimPassed). Is this code valid Java code? If not, explain in a sentence. No, Cornell must declare method graduateStudents() with no parameters, since it is declared that way in class Univ.
- Data Structures (18 points)
(a) Queues (5 points)
To the right below are the fields of class ArrayQueue, which implements a queue of bounded size —the size of array b. Notice how queue elements wrap around in the array: if a queue element is in the last array element b[b.length-1], the next one is in b[0]. We discussed this implementation in lecture. Below, implement methods peek and put.
/** Return first element of the queue.
- Precondition: queue is not empty. */ public int peek() { return b[h]; }
/** Add e to the queue..
- Precondition. queue is not full */ public void put(int e) { b[(h+n) % b.length]= e; n= n+1; }
public class ArrayQueue { // The n elements of the queue are in // b[h], b[h+1], ... b[h+n-1] (all // indexes mod b.length) // 0 <= h < b.length private int[] b; private int n; private int h; }
(b) Printing a B-Tree (9 points)
The Binary Search Tree is difficult to keep balanced, In the 1970’s other data structures were developed to overcome this difficulty. Hopcroft of Cornell CS came up with a 2-3 tree. But the data structure that received the most attention was Bayer and McCreight’s B-tree. B-trees are new to you. This question explores some aspects of B-trees, using B-trees containing sets of integers, called keys.
To the right is an example of a leaf of a B-tree: a leaf contains an ascending sequence of keys, in this case, the integers 3, 5, 7, 8, 9.
Underneath the leaf we show an internal node of a B-tree. An internal node consists of a sequence of at least two children, each a BTree node, and between each pair of children is a key.
Here are the important properties of all internal nodes (non-leaves):
node1 node2 node
- The keys (in this case, integers) are in ascending order.
- The keys in the subtree before a key are less than the key. For example, all keys in subtree node are less than 22 and all keys in subtree node2 are less than 80.
- The keys in the subtree after a key are greater than the key. For example, all keys in subtree node are greater than 22 and all keys in subtree node3 are greater than 80.
Below is part of class BTree, showing the fields that contain the keys and the children, both as ArrayLists. Also shown is method print. Write the body of method print. Assume that method printArray(ArrayList b) prints the values in b, and use the standard println statement to print a single integer.
public class BTreeNode { private ArrayList keys; // These are the integers private ArrayList children; // null only if this is a leaf
/** Print the values of this Btree in ascending order. */ public void print() { if (children == null) { printArray(keys); return; }
children.get(0).print(); for (int k= 0; k < keys.size(); k= k+1) { System.out.println(keys.get(k)); children.get(k+1).print(); } }
- Graphs (13 points)
(a) (2 points)
If all edge weights on a graph are 1, to what algorithm is Dijkstra’s shortest-path algorithm equivalent? BFS —Breadth-first search.
(b) (3 points)
Will Dijkstra’s shortest-path algorithm work if the edge weights are negative? If no, explain why briefly. No. Here’s 1 reason. The theorem we proved about choosing a node from F with minimum d-value cannot be proved. Another reason. If there is a cycle, a shortest path doesn’t exist because going around the cycle again reduces the ”distance”.
(c) (2 points)
Kruskal’s algorithm for constructing a spanning tree starts with all nodes and no edges. It repeatedly adds a minimum-weight edge that does not form a cycle until no more edges can be added. What does Prim’s algorithm do? Same thing, except that the added edge must be connected to all the other previously added edges.
(d) (3 points)
You have a directed graph of all airplane flights in the U.S. Thus, there is an edge from city C1 to city C2 if there is a flight from C1 to C2. Of the several graph algorithms we have discussed, which one is best suited to find a flight from Ithaca to Kalamazoo with as few stops as possible? BFS —Breadth-first search. Use it to find all cities 1 stop from Ithaca, then all cities 2 stops from Ithaca, etc.
(e) (3 points)
How many edges does a complete undirected graph of n nodes have? Describe or draw the smallest non-planar complete graph. A complete graph has as many edges as possible: n(n − 1)/2 edges. Every graph of fewer than 5 nodes is planar. The complete graph K 5 has 5 nodes and 10 edges. It is not planar. We don’t draw it here.
