U N I V E R S I T Y O F W E L L I N G T O N EXAMINATIONS 2018 TRIMESTER 2 COMP 103 PRACTICE EXAM

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1 T E W H A R E W Ā N A N G A O T E Ū P O K O O T E I K A A M Ā U I VUW VICTORIA U N I V E R S I T Y O F W E L L I N G T O N EXAMINATIONS 2018 TRIMESTER 2 COMP 103 PRACTICE EXAM Time Allowed: TWO HOURS ******** WITH SOLUTIONS ********** CLOSED BOOK Permitted materials: Silent n-programmable calculators or silent programmable calculators with their memories cleared are permitted in this examination. Instructions: October 24, 2018 Questions: Printed foreign language English dictionaries are permitted. No other material is permitted. Attempt ALL Questions. The real examination is out of 120 marks; BUT, The marks on this practice exam DON T ADD UP RIGHT!. Brief Documentation is at the end of the examination script Collection Types [16] Lists, Maps, and Sorting [31] Complexity, Big-O costs [20] Simulation with Collections [23] Traversing General Trees [20] Traversing Graphs [10] COMP 103 Page 1 of 16

2 Question 1. Collection Types [16 marks] (a) [4 marks] What is the key difference between Stacks and Queues? Stacks are LIFO, add and remove from same end Queues are FIFO, add and remove from opposite ends (b) [4 marks] Suppose you are writing a program to keep track of all the words that an ESL student has already learned and the words that they still get wrong. What collection type(s) would you use? Justify your choice. Set(s) because need to add and access words easily, but there is order required (c) [4 marks] Suppose you are writing a program to manage the keystroke signals from a keyboard. What Collection type should you use, and why? Queue - the signals need to be processed in the order they were received - FIFO (Question 1 continued on next page) COMP 103 Page 2 of 16

3 (Question 1 continued) (d) [4 marks] A Set has the same operations as a Collection. but it behaves differently. How is a Set different from a Collection? When would you use a Collection that isn t a Set? Set doesn t allow duplicates. When we need to allow duplicates. COMP 103 Page 3 of 16

4 Question 2. Concordances [31 marks] A Concordance is an index of words in a document. It should have an entry for each distinct word, and for each word, it should identify all the positions in the document where the word occurs. Suppose you are writing a program that creates concordances. Assume that the program can read a document into a List of words: private List <String> allwords; (a) [3 marks] Assume the positions of the words should be the index of the word in allwords (ie, the number of the word in the document). Typically, a concordance should be printed out in alphabetical order of the words. Define an appropriate Map type for the concordance field and initialise it. private Map <String, List <Integer>> concordance = new TreeMap<String, List<Integer>>();. (b) [11 marks] Complete the following makeconcordance method which should process the words in allwords to produce a concordance in the concordance field. public void makeconcordance(){ for ( int pos = 0; pos<allwords.size (); pos++){ String word = allwords.get(pos); if (!concordance.containskey(word)){ concordance.put(word, new ArrayList<Integer>()); concordance.get(word).add(pos); (Question 2 continued on next page) COMP 103 Page 4 of 16

5 (Question 2 continued) (c) [7 marks] Complete the following printalphabetically method to print out the concordance, in alphabetic order of the words, listing each word on a new line, followed by the positions of the word. E.g. afraid: [10, 49, 84, 143, 243, 589, 745] all: [53, 205, 301, 372, 632, 746] : Note: you can print all the values in a List or Collection with UI.print(myData). public void printalphabetically (){ for ( String word : concordance.keyset()){ UI. println (word+": "+concordance.get(word)); (Question 2 continued on next page) COMP 103 Page 5 of 16

6 (Question 2 continued) (d) [10 marks] [Hard and rather nasty!] It is sometimes useful to print a concordance in order of the frequency of the words. Complete the following printbyfrequency method to print out all the words in the concordance, in order of frequency. Each word should be on a new line, followed by the frequency. E.g. the: 89 of: 74 and: 68 : Hint: Copy the words and their frequencies to a different collection and sort them before printing them out. public void printbyfrequency (){ List <Map.Entry<String, Collection<Integer>>> list = new ArrayList<Map.Entry<String, Collection<Integer>>>(concordance.entrySet()); Collections. sort ( list, (Map.Entry<String, Collection<Integer>> me1, Map.Entry<String, Collection<Integer>> me2) > { if (me1.getvalue(). size ()>me2.getvalue().size ()) {return 1; if (me1.getvalue(). size ()<me2.getvalue().size ()) {return 1; return 0; ); for (Map.Entry<String, Collection<Integer>> me : list){ UI. println (me.getkey()+": "+me.getvalue().size()); COMP 103 Page 6 of 16

7 Question 3. Complexity: Big-O costs [20 marks] (a) [12 marks] What are the Big-O costs of the fragments of code below. Assume the size of the list is n. for ( int k = 0; k < list. size (); k++) { list.add( list.remove(0)); // cost= O( n ) times= // Total Cost = O( nˆ2 ) n for ( int k = 0; k < list. size (); k++) { for ( int j = 0; j < k; j++) { [or n(n 1)/2] if ( list.get(k). equals( list.get(j ))){ // cost= O( 1 ) times= nˆ2, UI. println ( list.get(k )); // cost= O( 1 ) times= 0.. nˆ2 // Total Cost = O( nˆ2 ) Map<String, Integer> mydata = new TreeMap<String, Integer>(); for ( String word : list ){ if (mydata.containskey(word)){ // cost = O( log(n) ) times= n mydata.put(word, mydata.get(word) 2); // cost = O( log(n) ) times= 0.. n 1 else { mydata.put(word, 1); // cost = O( log(n) ) times= 1.. n int max = 0; for ( String word : mydata.keyset()){ if (mydata.get(word) > max){ // cost = O( log(n) ) times= 1.. n max = mydata.get(word); // cost = O( log(n) ) times= 1.. n UI. println (max); // Total Cost = O( n log(n) ) (Question 3 continued on next page) COMP 103 Page 7 of 16

8 (Question 3 continued) (b) [4 marks] A program uses an ArrayList to store all the words in a document, and then searches for duplicate words by comparing each word in the list with every other word. When the List has 1000 words, the program takes 50 milliseconds to check for duplicates. If the List had 10,000 words, how long would you expect the program to take to check for duplicates? Explain why. 5000mS = 5 seconds. The algorithm is O(n 2 ). If you multiply the size of the list by 10, you will multiply the cost by 10 2 = 100 Therefore cost is 50 x 100 = 5000 (c) [4 marks] A better program uses a HashSet to check for duplicates by stepping through the list of words, checking the word is already in the HashSet, and then adding it to the HashSet if it is t present. When the document has 1,000 words, the program takes 10 milliseconds to check for duplicates If the document had 10,000 words, how long would you expect the program to take to check for duplicates? Explain why. 100 ms. The algorithm is O(n) - check costs O(1), do it n times If you multiply the size of the list by 10, you will multiply the cost by 10) Therefore cost is 10 x 10 = 100 COMP 103 Page 8 of 16

9 Question 4. Simulation with Collections [23 marks] Suppose you are writing a program to simulate a minibus service at an airport where minibuses take groups of passengers to lots of different suburbs. There is a separate queue for passengers for each destination Each time tick, a passenger may arrive at the minibus stop and joins the queue for their destination. If there is queue for their destination, then the passenger starts a new queue. Each time tick, if a minibus is ready, it is allocated to the destination with the longest queue, loads as many passengers as it can from the queue, and departs. If a queue is empty, it is removed. You are to write the passengerarrive and loadbus methods. The Passenger class has the following constructor and methods: Passenger class : public Passenger(); // make a new passenger (with a destination ) public String getdestination (); // removes 1 tick from remaining screening time The Bus class has the following constructor and methods: Bus class: public Bus(); // make a new bus public void loadpassenger(passenger p) // load a passenger onto the bus public boolean hasspace() // if space on bus for ather passenger public void depart() // make bus depart The MiniBusSimulation class has the following field, constructor and run method. public class MiniBusSimulation{ private Map<String, Queue<Passenger>> station; // map of queues for each destination public MiniBusSimulation(){ station = new HashMap<String, Queue<Passenger>>(); public void run (){ int time = 0; while (true){ time++; if (Math.random()<0.2) { passengerarrive(new Passenger()); if (Math.random()<0.01) { loadbus(new Bus()); // subquestion (a) // subquestion (b) Hint: sketch a diagram of the content of station. (Question 4 continued on next page) COMP 103 Page 9 of 16

10 (Question 4 continued) (a) [11 marks] Complete the following passengerarrive method which should add the passenger to the queue for their destination, creating the queue if there isn t one already. public void passengerarrive (Passenger p){ String dest = p.getdestination (); if ( station. get(dest)==null){ station. put(dest, new ArrayDeque<Passenger>()); station. get(dest ). offer (p); (b) [12 marks] Complete the following loadbus method which should choose the largest queue load passengers from the queue (up to the CAPACITY of the bus) remove the queue if it is empty make the bus depart public void loadbus(bus b){ Queue<Passenger> maxq = new ArrayDeque<Passenger>(); for ( String dest : station. keyset()){ if ( station. get(dest ). size ()> maxq.size()){ maxq = station.get(dest); while (!maxq.isempty() && b.hasspace()){ b.loadpassenger(maxq.poll()); b.depart (); COMP 103 Page 10 of 16

11 Question 5. Traversing Trees [20 marks] This question uses GTNodes in a program to record a tree of the evolutionary history of plant species. Each GTNode contains a Plant object, which contains information about a particular plant species. The children of a de are the plants that were direct descendents of the plant in the de. class GTNode<E> implements Iterable<GTNode<E>>{ public GTNode(E item); // constructor public E getitem(); // return item in the de public void addchild(gtnode<e>); // add a new child public void removechild(gtnode<e>); // remove a child public Iterator <GTNode<E>> iterator(); // enables the for (E child : de ){... class Plant public Plant( String name, boolean flowers ); // constructor public String getname(); public boolean hasflowers (); public String getleafshape(); (a) [9 marks] Complete the following hasplant, method which will return true if any plant in the subtree starting at de has the given name. To do this, it will need to traverse the tree under de. public boolean hasplant(gtnode<plant> de, String name){ if (de.getitem(). getname().equals(name)){ return true; for (GTNode<Plant> child : de){ if (hasplant( child, name)){ return true; return false ; (Question 5 continued on next page) COMP 103 Page 11 of 16

12 (Question 5 continued) (b) [11 marks] It is uncommon for a plant without flowers to have several descendent plants with flowers. In some cases, this indicates there is a missing plant in the evolutionary history. Complete the following fixtree method which should find any plant in the tree which does t have flowers, has at least one child without flowers, and at least two children with flowers. For each such plant, it should insert a new unkwn plant (name:?? ) with flowers as a child of the de and as parent of the children with flowers. For example: AA AA BB CC FF BB CC FF DD EE KK GG DD?? GG HH LL EE KK HH LL (Each de in the figure contains the name of a plant and whether it has flowers.) public void fixtree (GTNode<Plant> tree){ if (! tree.getitem (). hasflowers()){ int numflowerchildren = 0; int numnoflowerchildren = 0; for (GTNode<Plant> child : tree){ if ( child. getitem (). hasflowers()) {numflowerchildren++; else {numnoflowerchildren++; if (numflowerchildren>=2 && numnoflowerchildren>=1){ GTNode<Plant> unkwn = new GTNode<Plant>(new Plant("??", true)); for (GTNode<Plant> child : tree){ if ( child. getitem (). hasflowers()) {unkwn.addchild(child); for (GTNode<Plant> child : unkwn){tree.removechild(child); tree. addchild(unkwn); for (GTNode<Plant> child : tree){ fixtree ( child ); COMP 103 Page 12 of 16

13 Question 6. Traversing Graphs [10 marks] You are writing a program to analyse social networks. Your program stores information about all the people in the network in a Collection of Person objects. private Collection <Person> socialnetwork; Each Person object contains a Collection of the person s friends. class Person public String getname(); public Collection <Person> getfriends(); (a) [5 marks] Complete the following isconnected method which should return true if two Persons are connected via a sequence of friend relations in the network. Hints: Use a visited set. You will need an additional recursive method. public boolean isconnected(person p1, Person p2){ Set<Person> visited = new HashSet<Person>(); return isconnected(p1, p2, visited ); public boolean isconnected(person p1, Person p2, Set<Person> visited){ if ( visited. contains(p1)) { return false ; if (p1. equals(p2)) { return true; visited. add(p1); for (Person friend : p1. getfriends ()){ if (isconnected( friend, p2, visited ) ) { return true; return false ; (Question 6 continued on next page) COMP 103 Page 13 of 16

14 (Question 6 continued) (b) [5 marks] A more interesting question is whether two people are connected over at most n links, where n is a small integer like 4 or 5. Complete a new version of isconnected that returns true if two Persons are connected over at most n links: public boolean isconnected(person p1, Person p2, int n){ return isconnected(p1, p2, n, visited ); public boolean isconnected(person p1, Person p2, int n, Set<Person> visited){ if (n < 0) { return false ; if ( visited. contains(p1)) { return false ; if (p1. equals(p2)) { return true; visited. add(p1); for (Person friend : p1. getfriends ()){ if (isconnected( friend, p2, n 1, visited ) ) { return true; visited. remove(p1); return false ; * * * * * * * * * * * * * * * COMP 103 Page 14 of 16

15 Documentation for COMP 103 Exam Brief, simplified specifications of some relevant Java collection types and classes. Note: E stands for the type of the item in the collection. interface Collection <E> public boolean isempty() // cost : O(1) for standard collection classes public int size () // cost : O(1) for standard collection classes public void clear () public boolean add(e item) public boolean contains(object item) public boolean remove(object element) interface List <E> extends Collection<E> // Implementations: ArrayList public boolean isempty() public int size () public void clear () public E get( int index) public E set( int index, E element) public boolean contains(object item) public void add(int index, E element) public E remove(int index) public boolean remove(object element) // cost : O(1) // cost : O(1) // cost : O(n) // cost : O(n) ( unless index close to end.) // cost : O(n) ( unless index close to end.) // cost : O(n) interface Set extends Collection<E> // Implementations: HashSet, TreeSet public boolean isempty() public int size () public void clear () public boolean add(e item) // cost : O(1) for HashSet, // O(log(n)) for TreeSet public boolean contains(object item) // cost : O(1) for HashSet, // O(log(n)) for TreeSet public boolean remove(object element) // cost : O(1) for HashSet, // O(log(n)) for TreeSet interface Map<K, V> // Implementations: HashMap, TreeMap public V get(k key) // cost : O(1) for HashMap, O(log(n)) for TreeMap public V put(k key, V value) // cost : O(1) for HashMap, O(log(n)) for TreeMap public V remove(k key) // cost : O(1) for HashMap, O(log(n)) for TreeMap public boolean containskey(k key) // cost : O(1) for HashMap, O(log(n)) for TreeMap public Set<K> keyset() // cost : O(1) public Collection <V> values() // cost : O(1) public Set<Map.Entry<K,V>> entryset() // returns set of key value pairs // ( get returns null if the key is t present ) // (put and remove return the old value, if any)

16 interface Queue<E> extends Collection<E> // Implementations: ArrayDeque, LinkedList, PriorityQueue public boolean isempty() public int size () public void clear () public E peek () // cost : O(1) for ArrayDeque, LinkedList, // O(1) for PriorityQueue public E poll () // cost : O(1) for ArrayDeque, LinkedList, // O(log(n)) for PriorityQueue public boolean offer (E element) // cost : O(1) for ArrayDeque, LinkedList, // O(log(n)) for PriorityQueue // (peek and poll return null if the queue is empty) class Stack<E> implements Collection<E> public boolean isempty() public int size () public void clear () public E peek () // cost : O(1) public E pop () // cost : O(1) public E push (E element) // cost : O(1) // (peek and pop return null if the queue is empty) class Collections public void sort ( List <E> list); // cost = O(n log(n)) in general // O(n) almost sorted public void sort ( List <E> list, (E e1, E e2) >{..);// cost = O(n log(n)) in general // O(n) almost sorted public void swap(list<e> list, int i, int j ); // cost = O(1) public void reverse ( List <E> list); // cost = O(n) public void shuffle ( List <E> list); // cost = O(n) interface Comparable<E> // Items can be compared for sorting or a priority queue. public int compareto(e other); // Comparable objects must have a compareto method: // returns ve if this comes before other ; // +ve if this comes after other, // 0 if this and other are the same interface Iterable <E> // can use a foreach loop on these items public Iterator <E> iterator() ; // returns an Iterator object Integer and Double constants: Integer. MAX VALUE; Integer.MIN VALUE; Double.MAX VALUE; Double.NaN; Double.POSITIVE INFINITY; Double.NEGATIVE INFINITY;