Node +key : int +left : Node +right : Node. root where m is either 0..1, 0..*, or * Node +key : int +left : Node +right : Node 0..

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1 Question 1: è12 pointsè CISC323 Midter Exa Saple Solution March 19, 2003 J. Dingel Consider the following code fragent for ipleenting binary search trees. class BinSearchTree í public Node root;... í class Node í public int key; public Node left; public Node right;... í Below is a partial UML class diagra for the code above. Coplete the diagra with associations and navigability and ultiplicity constraints such that it describes binary search trees adequately. Here's a suary of the correct answers: left BinSearchTree +root : Node root 0..1 Node +key : int +left : Node +right : Node where is either 0..1, 0..*, or * right BinSearchTree +root : Node root 0..1 Node +key : int +left : Node +right : Node 0..2 children æ Places of variability in the above diagras: í The diagras do not contain aggregations or copositions. However, one could put the on the source ends of all three associations. æ Coon istakes: í navigability constraints issing or incorrect í association between Node objects issing í ultiplicity constraints issing or incorrect

2 Question 2: è12 pointsè For your convenience, here is a copy of the code of the previous page. class BinSearchTree í public Node root;... í class Node í public int key; public Node left; public Node right;... í Recall that every node n in a binary search tree satisæes the following invariant: ëthe key of every node in the left subtree of n is less or equal to the key in n and the key of every node in the right subtree of n is greater than the key in n." Draw an activity diagra for searching for a node with a given key k in a binary search tree with root node r. Your search doesn't have to return a list position or any other inforation. It just has to report success èif the key is found in the treeè or failure èif the key is not foundè. set current node to r key not found yes no current node is null? key found yes no k is equal to key of current node? k is less than key of current node? no set current node to right child yes set current node to left child æ Places of variability: í There are any correct ways of expressing the search as an activity diagra. However, the above is one of the ost succint. æ Coon istakes: í Inappropriate assuptions, that is, diagra assues, e.g., that key is always contained in tree, or tree is coplete èevery node has either no or two childrenè í Access to non-existent ènullè objects. E.g., tree ay be epty, that is, r ay be null. í Syntax of diagras wrong

3 Question 3: è8 pointsè Brieæy and inforally, list the ost iportant diæerences between the Visitor and the Iterator design pattern. æ Both the Iterator and the Visitor pattern give the user unifor accesss to the eleents of a collection. The Visitor pattern, however, is ore general than the Iterator pattern because it allows the eleents of the collection to have diæerent types ècollection is heterogenousè. The Iterator pattern requires that all eleents retrieved have the sae type ècollection is hoogeneousè. æ The Visitor pattern requires the addition of an accept ethod to the classes the eleents of the collection are instances of. Coon istakes: æ Many students siply wrote a paragraph about Iterators, then a paragraph about Visitors. That didn't answer the question of what the diæerences were. Full credit was given if the diæerences could be picked out fro corresponding stateents in your paragraphs èfor exaple, if you said in one place that an Iterator was for hoogeneous data structures and in another that a Visitor was for heterogenous data structuresè. If your stateents did not indicate the diæerences, you only got partial arks, even if you ade true stateents about each pattern. æ Another coon istake was saying that Iterators were for arrays or lists and Visitors were for trees. That's not really correct. People write Iterators for data structures such as binary search trees and hash tables, as long as the eleents of these are of the sae type. The real diæerence is that Visitors are for data structures containing ultiple node types, of which parse trees are only one exaple.

4 Question 4: è6 pointsè Brieæy and inforally list three of the biggest diæerences between the Waterfall process and Microsoft's ësync & Save" process. æ The Waterfall process prescribes the existence of a coplete requireents docuent before the design and ipleentation can begin. Mircosoft's planning phase only calls for a ëvision stateent" that is expected to change and evolve. æ A siilar situation exists with respect to testing. The Waterfall process expects the code to be coplete before testing begins. Microsoft's ësync & Save", however, asks for diæerent kinds of testing during the entire developent. æ The developentèipleentation phases in both the Waterfall and the ësync & Save" process ask for the identiæcation of independent subprojects that can be ipleented in parallel. However, the Waterfall process expects all subprojects to be copleted fully before they are integrated. The daily builds of ësync & Save", however, require even incoplete ipleentations of the subprojects to be cobined and integrated. Coon istakes: æ See ærst coon istake for previous question. A paragraph about the Waterfall process followed by a paragraph about ësynch & Save" did not get you full arks unless three distinct diæerences were ebedded in your paragraphs. æ Many people said that there was no possibility of parallelis in the Waterfall process. You got only partial arks for that. What we said in class èand on the slidesè is that there is ore opportunity for parallelis in Microsoft's process. In the Waterfall process, each step ust be ænished before the next starts, so there is no parallelis between steps. However, it is quite usual to have any groups working in parallel within certain steps, especially ipleentation ècodingè. In the Microsoft process, you can also have parallelis between steps - for exaple, soe eleents of analysis, design, coding and testing going on at once. æ Soe people said that the Microsoft process takes less tie than the Waterfall process. That's not always true; it depends a lot on the task and how each process is anaged. In theory, the Microsoft process gets things to arket quicker, but in practice things can get unorganized and slow down because the project hasn't been planned and coordinated carefully enough. So you only got partial arks for that.

5 Question 5: è6 pointsè For each of the following three software quality criteria, write down at least one feature of the Java prograing language that supports that criterion. For each criterion, write one or two sentences explaining how the Java feature supports the criterion. Correct answers include the following. Soe of the answers are deliberately a bit ore verbose than necessary. EncapsulationèInforation Hiding: í Classes in Java èand other OO languagesè allow the easy grouping of data and behaviours that ëbelong together". í Java's visibility odiæers èpublic, protected, and privateè give the prograer æne-grained control over which entities in a class are visible fro the outside. í Interfaces in Java allow ipleentation details to be hidden fro the clients of the code. Only the publically accessible data and behaviours are entioned in the interface. So, if the client only sees the interface èas it should be the caseè, the ipleentation reains hidden. Reuse: í In Java èand other OO languagesè, a subclass autoatically inherits data and behaviours fro its superclass. Code that has already been developed for a class can thus be shared aong and reused for the handling of the objects in a subclass. í The Java Application Prograer Interface èapiè contains a large nuber of pre-deæned packages to do all kinds of things èe.g., inputèoutput, array and vector anipulation, GUI prograingè. All of these packages can be used in diæerent contexts by just iporting the èstandard packages like java.lang are iported autoatically. Modularity: í Java's notion of interface allows data and behaviours be separated fro the details of their ipleentation. í Classes in Java allow code to be broken up into pieces. All data and behaviours that belong together èbecause, for instance, they accoplish a speciæc taskè can be put into a class. Classes deæne their own nae space, that is, two diæerent classes can have an attribute with the sae nae. í Packages allow you to take grouping to the next level. Classes that for a logical unit can be grouped into a package. The Java API, for instance, contains java.ath a package for supporting high-precision atheatical operations. The package java.io contains all classes that deal with input and output. Coon istakes: æ Design patterns are not a Java feature. They can be ipleented in Java, but they are not Java speciæc. æ Confusion about what ëodularity", ëinheritance", or ëpolyorphis" ean

6 Question 6: è8 pointsè Draw a use case diagra for QCard. Your diagra should contain at least two actors and three use cases. Here's one correct answer. QCard Add Course student Drop Course Adinistrator Check Marks Enter Marks æ Places of variability: í Obviously, there are lots of correct answers to this question. æ Coon istakes: í The ain proble was that people did not really understand the proper forat and choices for use case diagras. Soe people picked QCard as an actor. QCard is the whole syste; actors are people or other hardware or software systes who interact with the syste. An actor is eant to be soeone or soething who initiates interactions with the syste, so a database is not a good choice either. The choice of a backup syste was ok, since soe backup systes will initiate backup operations according to a schedule. í Another error wesawwas in the choice of use cases. A use case is soe kind of interaction between a user and the syste. So ëenter arks" or ëlook up arks" are good use cases, but ëarks" is not. Use cases such as ëarks" and labeled the arrows with actions such as ëenter" or ëlook up" do not æt the syntax of use case diagras. í Only huan actors are drawn using a stick ægure. í Relationships èëuses" or ëextends"è between use cases were not required. If you provided soe and used the correctly, that was æne. If you used the incorrectly, you lost a ark or two. If you put lines between use cases without any labels, that is not correct use case syntax.

7 Question 7: è8 pointsè Hans attepted to use a sequence diagra to describe the steps necessary to pay with your debit card èbank card, interac cardè at a grocery store. Here is what he cae up with: Custoer Cashier CardReader Bank enable slidecard askforcard askforinfo inputinfo askforinfo checkinfo no ok? displayerror yes displaysuccess transactioncoplete where the essages have the following eaning: enable: The cashier enables the reader by sending it the aount to be deducted. slidecard: The custoer slides hisèher card. askforcard: The reader propts the custoer to slide hisèher card. askforinfo: The reader propts the custoer for the account to be debited and PIN. inputinfo: The custoer selects the account and inputs hisèher PIN. checkinfo: The reader sends the PIN, the aount, and the account inforation to the bank for veriæcation purposes. displayerror: The reader infors the custoer that the attepted debit was denied. transactioncoplete: The reader infors the cashier that the attepted debit was copleted successfully. displaysuccess: The reader infors the custoer that the attepted debit was copleted successfully. Continued on next page

8 Question 7 ècontinuedè: In the space below, brieæy list at least four diæerent incorrectnesses in Hans's diagra on the previous page. Here're the incorrectnesses: 1. Wrong sender: The enable essage should be sent by the cashier, not the custoer. 2. Wrong order: In the diagra, the custoer slides his card, before he has actually received the essage propting hi to do so. In other words, the order of askforcard and slidecard should be reversed. 3. No branching: ëbranching" is not allowed in sequence diagras. Each sequence diagra describes a single scenario only. In other words, in a given diagra, the essage fro the bank to the custoer represents either success or failure never both, as Hans attepted to do. Another proble related to branching is that according to the diagra, after displaying the error, control ëæows into" the positive case and the reader also indicates a successful copletion of the transaction. 4. No crossing: A essage cannot ëcross" other essages in sequence diagras. Message askforinfo, however, does. The only way to indicate that a sequence of essages repeats is through a loop which is expressed by a rectangular box around the repeated essages together with a possibly inforal terination condition. Answers that did not receive points: æ Not enough detail: How uch detail is enough, depends on the contextèrequireents. There is nothing in the question that indicates that the diagra contains an insuæcient aount of inforation. So, the level of detail does not count as an incorrectness. æ No upper liit on nuber of unsuccessful attepts: Siilar.