student book answers - barringtons€¦ · web viewa wireless network is a broadcast network and...

Student book answers1.1............................................................................................................................................3

End of topic questions.........................................................................................................3Exam practice questions......................................................................................................6

1.2............................................................................................................................................7End of topic questions.........................................................................................................7Exam practice questions......................................................................................................9

1.3............................................................................................................................................9End of topic questions.........................................................................................................9

1.4..........................................................................................................................................14End of topic questions.......................................................................................................14Exam practice questions....................................................................................................16




2.2..........................................................................................................................................36End of topic questions.......................................................................................................36




2.6..........................................................................................................................................98End of topic questions.......................................................................................................98Exam practice questions..................................................................................................106

2.7........................................................................................................................................109End of topic questions.....................................................................................................109Exam practice questions..................................................................................................114



4.2........................................................................................................................................121End of topic questions.....................................................................................................121

AQA Computing A2 Teacher Notes © Nelson Thornes Ltd 2009 1

Exam practice questions..................................................................................................1235.1........................................................................................................................................124

End of topic questions.....................................................................................................124Exam practice questions..................................................................................................125









7.3........................................................................................................................................163Exam practice questions..................................................................................................163

7.4........................................................................................................................................169Exam practice questions..................................................................................................169



1.1End of topic questions1 Information hiding means hiding design details behind a standard interface.2 An interface is a device or combination of devices such as a keyboard/VDU situated between the

user and the internal operation of a machine and through which the user interacts with the machine.3 Possible answers are:

a) By hiding the complexities of the machine behind a well-designed interface users are able to learn how to use a machine with the minimum of training.

b) The operator of a machine does not need to know how the operation of the machine is achieved only how to control it.

c) The internal components of the machine may be replaced by similarly functioning components but designed in a different way, e.g. using a different technology, without the experience of the user changing or the user needing to be re-trained.

4 In order to make the manufacturing, and maintenance of a car manageable, a car is divided into parts with particular interfaces hiding design decisions. Some car parts or components are listed below: Engine Gearbox Transmission (transfers power from engine to wheels) Lights PedalsThis division into separate components or modules allows the car manufacturer to offer different options. For example, a particular model may be offered with different capacities of engine, 1.8 litres, 2.5 litres, 3 litres. All three engines fit the same transmission, the same engine mounts, and the same controls. This is possible because all three engines provide the same interface to the other parts/modules that they connect to. The same applies for maintenance. An engine may be replaced relatively easily because it is designed to fit the engine mounts, transmission system and so on.

5 A well-designed computer program has a solution in which the source code has been decomposed into modules using the principle of information hiding. When a programmer is working on one module he or she does not need to know the detail contained in other modules, only what other modules can do for the programmer. Also changes are much easier to make with this approach because the changes typically are local to a module rather than global changes across all modules.

4 a) Different objects can have identical interfacesb) Sharing a common interface among many different objects means that users of the objects do not

need to be re-trained when changing from using one object to using anothera) To use a module or object a user needs to have no knowledge of the module’s or object’s

internal designb) The internal design can be kept secretc) Allows flexibility, one module may be replaced by another module with an identical interface

but a different internal design or manner in which the module’s function is carried outd) Changes are made easier because changes are local to modules

5 Abstraction means omitting unnecessary details or, as a skill, it means drawing out the essence of a problem, solving it and then seeing what other problems can be solved using the same techniques.


6 Any suitable example, e.g. sketching a map to give someone directions. You might base your sketch on an ordnance survey map. You pick and add four or five major labelled landmarks to your sketch and link these by lines representing the route. You have left out all the unnecessary detail on the ordnance survey map in order to convey the essentials of the route in your sketch.

7

F ← GetF

C ← ConvertFtoC(F)

Display(C)

8 There are several ways of classifying plants. One way to classify plants is according to the way a plant absorbs water. Plants are either vascular or non-vascular. Vascular plants have tube-like structures that transport water from the roots to the stem to the leaves. Non-vascular plants absorb water only through their surfaces.

9 Each person can shake hands with 0 to n − 1 other people. This creates n − 1 possible holes. It is not possible for both the '0' and the 'n − 1' hole to be empty because when one person shakes hands with everybody, it's not possible to have another person who shakes hands with nobody; likewise, when


Parallel veins Branching network of veins

Non-

vascular

Angiosperms

Vascular

Monocots

Seeds on cones

Seeds in flowers

Write('Input F: ')

Readln(FTemp)

GetF ← FTemp

ConvertFtoC ← (F - 32)* 5/9 Writeln('Celsius = ', C)

F ← GetF

ConvertTemperature

Display(C)C ← ConvertFtoC(F)

Parallel veins

Branching network of

veins

one person shakes hands with no one there cannot be a person who shakes hands with everybody. This leaves n people to be placed in at most n − 1 non-empty holes, guaranteeing duplication.

10 Assume that in a box there are 10 black socks and 12 blue socks and you need to get one pair of socks of the same colour. Supposing you can take socks out of the box only once and only without looking, how many socks do you have to pull out together? Use the Pigeonhole principle.The correct answer is three. To have at least one pair of the same colour (m = 2 holes, one per colour), using one pigeonhole per colour, you need only three socks (n = 3 objects). In this example, if the first and second sock drawn are not of the same colour, the very next sock drawn would complete at least one same-colour pair. (m =.

11


ColumbiaSuriname

Argentina

Peru

French

Guyana

Bolivia

VenezuelaGuyana

Ecuador

Brazil

Paraguay

Chile

Uruguay

12(a)

(b) 3

Exam practice questionsQuestion Number Answer Marks1 Information hiding means hiding design details behind a

standard interface.In software development, a programmer can use a library module without needing an intimate knowledge of the module’s design. Information hiding usually means that a software module is self-contained, otherwise its design would have to be revealed. This allows the internal design of a module to be changed whilst isolating users from these changes, i.e. the software that uses the module should still work and probably better than before because of the internal changes to the module. Also, changes are much easier to make with this approach because the changes typically are local to a module rather than global changes across all modules. If a whole software project is divided into modules using the principle of information hiding then it is possible for different programmers to work on different modules at the same time so that the project can be completed more quickly. Information hiding also creates designs that can be kept secret more easily. From a user’s point of view, information hiding means that when the internal design of a module changes the user doesn’t have to be retrained. Outwardly the module should still appear the same.The principle of information hiding allows the division of any piece of equipment, software or hardware, into separate parts. For example, a car is a complex piece of equipment. In order to make the design, manufacturing, and

3


A

ED

C

B

maintenance of a car manageable, a car is divided into parts with particular interfaces hiding design decisions. This division into separate components or modules allows the car manufacturer to offer different options. For example, a particular model may be offered with different capacities of engine, 1.8 litres, 2.5 litres, 3 litres. All three engines fit the same transmission, the same engine mounts, and the same controls. This is possible because all three engines provide the same interface to the other parts/modules that they connect to.

Question Number Answer Marks

2a Abstraction by generalisation proceeds from a detailed picture of specific instances to a less detailed picture of classes to which the instances belong or are examples of. This is an example of classification-type generalisation. Another type of generalisation is of the decomposition-type. In this type, the abstraction proceeds in the direction of more and more detail as in top-down program development.

1

2b Abstraction by representation is applied to problem solving. Details are removed until it becomes possible to represent the problem in a way that is possible to solve.

1

5 Question Number

6 Answer 7 Marks

3a 5

3b 2 tables 2

1.2End of topic questions1 Algorithms are considered to be procedural solutions to problems and problem solving is a human activity. Designing algorithms is a problem solving activity that can be useful regardless of whether a computer is involved. Computer programs would not exist without algorithms. Therefore, a study of algorithms is considered to be a cornerstone of computer science. There may be more than one algorithm for solving a particular problem and algorithms for solving a particular problem may do so with dramatically different speeds and also with different memory requirements.


B

C D

E

A

2 a) Twice as longb) Eight times as longc) 2n times as long

3 Using a standard unit of time measurement, such as a second, to estimate the running time of a program implementing the algorithm, would not be sensible. The estimate would be dependent on the speed of the computer, the quality of the program implementing the algorithm and the compiler used in generating the machine code.4 Compute the number of times the basic operation is executed on inputs of size n. The basic

operation is the operation in the algorithm contributing the most to the total running time.5 a) The computational complexity of an algorithm measures how economical the algorithm is with

time and space.b) Computational complexity of a problem is taken to be the worst-case complexity of the most

efficient algorithm, which solves the problem.

6 a) Time-complexity of an algorithm indicates how fast an algorithm runs.b) Space-efficiency of an algorithm indicates how much memory an algorithm needs.

7 Input size is number of elements in the array, i.e. n. The innermost loop contains a single operation, the comparison of two elements. Consider this as the algorithm’s basic operation. In the worst-case scenario, the input is an array for which the number of element comparisons is the largest amongst all arrays of size n. There are two cases for this worst case scenario:a) arrays with no equal elementsb) arrays in which the last two elements are the only pairs of equal elements.For such inputs, one comparison is made for each repetition of the innermost loop, i.e. for each value of the loop’s variable j between limits i + 1 and n. This is repeated for each value of the outer loop, i.e. for each value of the loop’s variable i between its limits 1 and n - 1. Consider the case when n = 6.

Element no.

No. of comparisons

Element in first column compared with elements

1 5 2, 3, 4, 5, 62 4 3, 4, 5, 63 3 4, 5, 64 2 5, 65 1 6

In general, the number of comparisons becomes 1 + 2 + 3 + 4 + 5 + … + nThis has the sum

b) For large values of n this becomes as (n - 1)n = n2 - n and for large n, n2 dominates.

8 Asymptotic behaviour of a function is the behaviour of the function for very large values of n.

9 Given a function g(n) then big O of g, represents the class of functions that grow no faster than the function g.

10f = O(log2 n)f = O(n)f = O(nlog2 n )f = O(n2)f = O(n3)


n-1)n2

f = O(2n)f = O(n!)

11 (a)Exponential growth exhibits growth rates that vary according to the form kn, e.g. 2n where k = 2 and n = 1, 2, 3, etc.(b) Polynomial growth exhibits growth rates that vary according to the form nk, e.g. n3 where k = 3 and n = 1, 2, 3, 4, etc.(c) An exponential time algorithm is one whose execution time grows exponentially with input size.(d) A polynomial time algorithm is one whose execution time grows as a polynomial of input size.(e) A linear time algorithm is a polynomial time algorithm that executes in O(n) time.

Exam practice questionsQuestion 1 – Specimen Paper COMP 3 Q1

Question number Answer Marks

1a 1. A2. C3. B

1

1b Start at first item; and examine each succeeding item in turn;Until item is found; or the end of the list reached;A algorithmmax 3O(n) as up to n items searched;All items may be searched;max 1

4

1c For each succeeding pair of items;if they are out of sequence they are swapped;the process is repeated;up to n-1 times; or until no more swaps are made;max 3All n items are compared up to n-1 times;1

4

1.3End of topic questions1 A finite state machine (FSM) consists of a set of input symbols (input symbol alphabet) and if it

produces output, a set of output symbols (output symbol alphabet), a finite set of states and a transition function that maps a state-symbol pair (current state and input symbol) to a state (next state) and possibly generates an output (output symbol) depending on the type of FSM.

12 000100


13 00000000, S014

15 Mealy machine: an FSM that determines its outputs from the present state and from the inputs.Moore machine: an FSM that determines its outputs from the present state only.

16

17 = 0 = 1 = 1 = 0c1

18 One-bit binary adder19 An FSM without output is known as a finite state automaton (FSA) or finite automaton, plural finite

state automata. FSA are restricted to decision problems, i.e. problems that are solved by answering YES or NO. Therefore, FSA do not write anything at all whilst processing input whereas an FSM with output does.


20 No Yes No Yes No


21 An FSA is deterministic if every state has exactly one transition or edge leaving it for each of the symbols in the alphabet. As there is exactly one edge for each symbol, an input word will cause the exact same behaviour every time a deterministic finite automaton is run on the word. A finite automaton can be non-deterministic if it does not have exactly one edge for each symbol of the alphabet. A state can have no edges for a symbol or it can have multiple edges for each symbol of the alphabet.

22 On reaching the end of the input if the FSA is in an accepting state then the FSA outputs a YES response.

23 aa|b(a|b)*

1.4End of topic questions1 11111


2 a)

b)

3 Adds 1 to the number on the tape4 A Turing machine (TM) is a finite state machine that controls one or more tapes, where at least one

tape is of unbounded length (i.e. infinitely long).24 Principle of universality: a universal machine is a machine capable of simulating any other

machine.25 A universal Turing machine, U, is an interpreter that reads the description <M> of any arbitrary

Turing machine M and faithfully executes operations on data D precisely as M does. For single-tape Turing machines, imagine that <M> is written at the beginning of the tape, followed by D.

26 Coffee maker27 A consequence of a universal Turing machine is that programs and data are really the same thing.

A program is just a sequence of symbols that looks like any other piece of input but when fed to a universal machine this input wakes up and begins to compute. A computer downloads Java applets, e-mail viruses as data but can then proceed to execute them as programs.

28 A microprocessor is a universal machine. An embedded computer with a microprocessor at its core can be programmed to perform as a microcontroller by choosing the right set of program instructions. The program or programs are changed when a different functionality is required.

29 A task is computable if and only if it can be computed by a Turing machine.


Exam practice questionsQuestion 1– Specimen Paper COMP3 Q6Question Number

Answer Marks

1a (i)1 mark for correctly positioned arrow, ignore data on tape

(ii) 1

1

1

1b 2 marks for correct tape symbols, 1 mark for correct position of arrow 3

1c 3 + 1 or the successor to 3 1

1.5End of topic questions1

a) The answer is not decidableb) Undecidable or noncomputable

21024104857610737418241099511627776112589990684262411529215046068470001.329227995784916 × 10+36

3a)

24 = 16 210 = 1024 18446744073709552000 1.33 × 1050

b) Intractable4

a) Tractable means that a problem has a reasonable (polynomial) time solution as the size of the input increases.b) Intractable means that no reasonable (polynomial) time solution has yet been found.


c) first option = £10.24 and second option = £11 first option = £10485.76 and second option = £21 first option = £45035996273704.96 and second option = £53

30 15 days31 NP-type problem32 No

Exam practice questionsQuestion number Answer Marks

1a Let’s call the liars group A and the truth tellers group B. If the native belongs to B, then since this native tells the truth he/she contradicts themself by saying that they did not say they were a truth teller when asked if they were a truth teller. Therefore the native belongs to the liars group.

1

1b The problem posed in part (a) belongs to the family of logic problems known as decision problems.

1

1c First is from the liars group. If the first is from the truth teller’s group then the statement would be a contradiction. If the first is from the liar’s group then he has lied and so both are not from the liar’s group. Therefore, the second must be from the truth teller’s group.

1

The second from the truth teller’s group. 1

1d 1. No, there is only one haircutter and he cuts the hair of everyone on the island who does not cut their own hair. If the haircutter cut their hair then this would contradict the statement in the previous sentence. Therefore, the haircutter does not cut his own hair. So, the haircutter’s hair must be cut by someone else. However, everyone who cuts another’s hair is called a haircutter but the island has only one haircutter. Therefore, the question cannot be answered.

1

2. This type of problem is undecidable. 1

1e 16 1

210 1

264 1

Intractable 1

The planks would take up a large amount of space and it would take a long time to place all the planks in the squares on the sand.

2



2a Intractable means that no reasonable (polynomial) time solution has yet been found. Processing will take an unreasonable amount of time even for quite small data sets.

2

2b For a class of intractable problems it is possible to guess a solution and check the solution in polynomial time, i.e. a reasonable amount of time. School timetabling is one of these problems. They are known as non-deterministic polynomial type problems or NP-type problems.

2


3 The halting problem is the problem of determining without executing a program whether it will halt on a given input. It relates to answering the question of whether or not it is possible to write a program that can, in general, test any other program supplied with its input and determine, without executing the program under test, if it will halt on this input. The answer to this question is no.

2


a) A valid string consists of one or more as or one or more cs followed by two bs[ll]b) aaccbb, cacccbb

2 (a|c)a*c*bb3

a) (0|1)*b) 1+c) 10*d) (110|011)(0|1)*e) (0|1)+0f) 11(0|1)*11

4 x*y*z*5 beean6 bead7 101018 01296-433006 or 01793 2345899 www.([_a-zA-Z\d\-]+(\.[_a-zA-Z\d\-]+)+)10 li(c|s)en(c|s)e


11 Roman Numerals

12 Identifier


Letter

Digit

V

I

X

I

I

V

I

13 UserID

14a) xy+b) xy+2/c) xy+x4-/d) xy↑e) xy+2↑f) xy+xy-/

15a) x * yb) (x + y) * 6c) (x + y) / (x - y)d) (2 * x + 3 * y) /8e) ((x +y) ↑ 2) * (6 - x)


Digit HDigit

L

D

R

Pa

Ph

A..Z A..Z

Exam practice questionsQuestion 1 – Specimen Paper COMP3 CPT4 Q7 Question number

Answer Marks

1a 4

1b 5 6 2 ;*; +;No Brackets;Easy to Compute;

2

1

Question 2 - KB


2a Yes 1

2b No 1

2c Yes 1

2.1End of topic questionsQ1 This is to produce a table as Table 01 in the book.

Q2 & Q3 are just using the code in the book and running it.

Q4 & Q5 Library unit

Unit Unit2;InterfaceType TClock = Class Private Hours: Integer; Minutes: Integer; Public Function GetHours: Integer; Function GetMinutes: Integer;


Procedure SetTime(h, m : Integer); Procedure IncrementTime ; Virtual; End;Type TAlarmClock = Class (TClock) Private AlarmHours: Integer; AlarmMinutes: Integer; Public Function GetAlarmHours: Integer; Function GetAlarmMinutes: Integer; Procedure SetAlarmTime(h, m : Integer); End;Type TWatch = Class (TClock) Private DayOfWeek: Integer; DayNumber: Integer; Public Function GetDayNumber: Integer; Function GetDayName: String; Procedure SetDayOfWeek(d : Integer); Procedure SetDayNumber(d : Integer); Procedure IncrementTime; Override; End;

ImplementationUses SysUtils;Function TClock.GetHours: Integer; Begin GetHours := Hours; End;

Function TClock.GetMinutes: Integer; Begin GetMinutes := Minutes; End;

Procedure TClock.SetTime (h, m : Integer); Begin Hours := h; Minutes := m; End;

Procedure TClock.IncrementTime;Begin Minutes := Minutes + 1; Sleep(600); // wait one minute If Minutes = 60 Then Begin Hours := Hours + 1; Minutes := 0; If Hours = 24 Then Hours := 0; End;End;

Procedure TWatch.IncrementTime;Begin Inherited IncrementTime;


If Hours = 24 Then Begin Hours := 0; DayNumber := DayNumber + 1; If DayNumber > 31 Then Daynumber := 1; DayOfWeek := DayOfWeek + 1; If DayOfWeek > 7 Then DayOfWeek := 1; End;End;

Function TWatch.GetDayNumber: Integer; Begin GetDayNumber := DayNumber; End;

Function TWatch.GetDayName: String; Begin Case DayOfWeek Of 1: GetDayName := 'Sun'; 2: GetDayName := 'Mon'; 3: GetDayName := 'Tue'; 4: GetDayName := 'Wed'; 5: GetDayName := 'Thu'; 6: GetDayName := 'Fri'; 7: GetDayName := 'Sat'; End; End;

Procedure TWatch.SetDayOfWeek (d : Integer); Begin DayOfWeek := d; End;

Procedure TWatch.SetDayNumber(d: Integer); Begin DayNumber := d; End;

Function TAlarmClock.GetAlarmHours: Integer; Begin GetAlarmHours := AlarmHours; End;

Function TAlarmClock.GetAlarmMinutes: Integer; Begin GetAlarmMinutes := AlarmMinutes; End;

Procedure TAlarmClock.SetAlarmTime (h, m : Integer); Begin AlarmHours := h; AlarmMinutes := m; End;End.


Q4)

Program AlarmClockExample;{$APPTYPE CONSOLE}Uses SysUtils, Unit2 in 'Unit2.pas';

Var AlarmClock: TAlarmClock;

Begin AlarmClock := TAlarmClock.Create; AlarmClock.SetTime(15,10); AlarmClock.SetAlarmTime(15,15); Repeat AlarmClock.IncrementTime; writeln(AlarmClock.GetHours,':',AlarmClock.GetMinutes); Until (AlarmClock.GetAlarmHours = AlarmClock.GetHours) And (AlarmClock.GetAlarmMinutes = AlarmClock.GetMinutes); Writeln('Wake up');; AlarmClock.Free; ReadLn;End.

Q5)

Program WatchExample;{$APPTYPE CONSOLE}uses SysUtils, Unit2 in 'Unit2.pas';

Var Watch: TWatch;

Begin Watch := TWatch.Create; Watch.SetTime(23,50); // give a start time Watch.SetDayNumber(15); // give a start date Watch.SetDayOfWeek(7); // give a day of the week Repeat Watch.IncrementTime; // display the time and date write(Watch.GetHours:2,':',Watch.GetMinutes:2); WriteLn(' ',Watch.GetDayName,' ', Watch.GetDayNumber); Until False; Watch.Free; ReadLn;End.


Q6 & Q7Class definition code:

Unit ClassDefs;InterfaceUses Classes, StdCtrls;Type TClock = Class (TLabel)

Private Hours: Integer; Minutes: Integer; Seconds: Integer; // only required for Exercise 2 Function IntToStrF (i: Integer): String; Public Constructor Create (AOwner:TComponent); Override; Procedure IncrementTime; Procedure DisplayTime; End;

ImplementationUses SysUtils, StrUtils;

Constructor TClock.Create (AOwner: TComponent);Begin Inherited Create(AOwner); Top := 20; Left := 20; Hours := 0; Minutes := 0; Seconds := 0; // only required for Exercise 2 Caption := '00:00:00';End;

Procedure TClock.IncrementTime;Begin Seconds := Seconds + 1; // only required for Exercise 2 If Seconds = 60 // only required for Exercise 2 Then Begin Minutes := Minutes + 1; Seconds := 0; // only required for Exercise 2 If Minutes = 60 Then Begin Hours := Hours + 1; Minutes := 0; If Hours = 24 Then Hours := 0; End; End;End;

Function TClock.IntToStrF (i: Integer): String;Var StringI: String[2];Begin StringI := IntToStr(i); If Length(StringI)=1 Then IntToStrF := '0' + StringI


Else IntToStrF := StringI;End;

Procedure TClock.DisplayTime;Var StrMinutes, StrHours, StrSeconds : String[2];Begin StrMinutes := IntToStrF(Minutes);StrHours := IntToStrF(Hours); StrSeconds := IntToStrF(Seconds); // only required for Exercise 2 Caption := StrHours + ':' + StrMinutes + ':' + StrSeconds;End;

End.

Code for the form:

Unit ClockForm;

Interface

Uses Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms, Dialogs, ClassDefs, ExtCtrls;

Type TForm1 = Class(TForm) tmrClockTick: TTimer; Procedure FormCreate(Sender: TObject); Procedure tmrClockTickTimer(Sender: TObject); Private { Private declarations } Public Clock : TClock; End;

Var Form1: TForm1;

Implementation

{$R *.dfm}

Procedure TForm1.FormCreate(Sender: TObject);Begin Clock := TClock.Create(Self); Clock.Parent := Self; tmrClockTick.Enabled := True; tmrClockTick.Interval := 1000; // tick every secondEnd;

Procedure TForm1.tmrClockTickTimer(Sender: TObject);Begin Clock.IncrementTime; Clock.DisplayTime;End;

End.


Q8)

Unit ClassDefs;InterfaceUses Classes, StdCtrls;Type TClock = Class (TLabel)

Private Hours: Integer; Minutes: Integer; Function IntToStrF (i: Integer): String; Public Constructor Create (AOwner:TComponent); Override; Procedure IncrementTime; Procedure DisplayTime; Function GetHours: Integer; Function GetMinutes: Integer; Procedure SetTime(h, m : Integer); End;Type TAlarmClock = Class (TClock) Private AlarmHours: Integer; AlarmMinutes: Integer; Public Function GetAlarmHours: Integer; Function GetAlarmMinutes: Integer; Procedure SetAlarmTime(h, m : Integer); Procedure Ring; End;

ImplementationUses SysUtils, StrUtils;

Constructor TClock.Create (AOwner: TComponent);Begin Inherited Create(AOwner); Top := 20; Left := 20; SetTime(0,0);End;

Procedure TClock.SetTime(h: Integer; m: Integer);Begin Hours := h; Minutes := m;End;

Function TClock.GetHours: Integer;Begin GetHours := Hours;End;

Function TClock.GetMinutes: Integer;Begin GetMinutes := Minutes;End;


Procedure TClock.IncrementTime;Begin Minutes := Minutes + 1; If Minutes = 60 Then Begin Hours := Hours + 1; Minutes := 0; If Hours = 24 Then Hours := 0; End;End;

Function TClock.IntToStrF (i: Integer): String;Var StringI: String[2];Begin StringI := IntToStr(i); If Length(StringI)=1 Then IntToStrF := '0' + StringI Else IntToStrF := StringI;End;

Procedure TClock.DisplayTime;Var StrMinutes, StrHours : String[2];Begin StrMinutes := IntToStrF(Minutes); StrHours := IntToStrF(Hours); Caption := StrHours + ':' + StrMinutes;End;

Function TAlarmClock.GetAlarmHours: Integer; Begin GetAlarmHours := AlarmHours; End;

Function TAlarmClock.GetAlarmMinutes: Integer; Begin GetAlarmMinutes := AlarmMinutes; End;

Procedure TAlarmClock.SetAlarmTime (h, m : Integer); Begin AlarmHours := h; AlarmMinutes := m; End;

Procedure TAlarmClock.Ring;Begin If (AlarmHours=Hours) And (AlarmMinutes=Minutes) Then Caption := 'Wake Up' // this can be elaborated on!!!End;

End.

Unit ClockForm;

Interface


Uses Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms, Dialogs, ClassDefs, ExtCtrls, StdCtrls;

Type TForm1 = Class(TForm) tmrClockTick: TTimer; btnSetAlarm: TButton; procedure FormCreate(Sender: TObject); procedure tmrClockTickTimer(Sender: TObject); procedure btnSetAlarmClick(Sender: TObject); Private { Private declarations } Public AlarmClock : TAlarmClock; End;

Var Form1: TForm1;

Implementation

{$R *.dfm}

Procedure TForm1.btnSetAlarmClick(Sender: TObject);Begin AlarmClock.SetTime(15,10); // this can made interactive by copying AlarmClock.SetAlarmTime(15,15); // values from textboxes tmrClockTick.Enabled := True;End;

Procedure TForm1.FormCreate(Sender: TObject);Begin AlarmClock := TAlarmClock.Create(Self); AlarmClock.Parent := Self; tmrClockTick.Enabled := False; tmrClockTick.Interval := 600; // tick every minuteEnd;

Procedure TForm1.tmrClockTickTimer(Sender: TObject);Begin AlarmClock.IncrementTime; AlarmClock.DisplayTime; AlarmClock.Ring;End;

End.


9 Inheritance diagram of the shapes classes

Exam practice questionsQuestion 1 – June 2008 CPT4 Q6

Question number

Answer Marks

1a 2

1b Insert a SetColour procedure; A Function into the Public section;R make Colour Public 2


TShape

TRectangle TEllipse

1c Yacht = Class/sub-class (Boat)(Public) Procedure SetBoatDetails (Override) Function GetMasts Function GetEnginePrivate Masts: Interger Engine: Boolean EndA Procedure SetMasts and Procedure SetEngine/ProcedureAddnewYacht/SetYachtsDetails instead of Procedure SetBoatDetailsMast and Engine must be privateP1 if extra functions/variables are includedR any diagramsI any parameters to methods

1

111

11

Question 2 – January 2008 CPT4 Q4

Question number

Answer Marks

2a

1 mark for correct boxes1 mark for correct lines1 mark for correct line endings

3

2b

1 mark for Loan=Class + Public + Private + End1 mark for CreateLoan + DeleteLoan + GetLoanDetails1 mark for Person + BookLoaned1 mark for DateOfLoan + ReturnDate

4


A any reasonable names for operations and data items

2c Add a new data item ShortLoan; of type Boolean; A loanlength; integer;A loantype; string;Modify the code for the operations.

Max 2

Question Number

Answer Marks

3a mouse click// mouse movement// keyboard operation// any interrupt; 1

3b event-driven programs service an event and wait for another;non event-driven programs run to completion/ are sequential;

2

3c contains its own data/fields/variables/properties;contains its ownoperations/methods/functions/procedures/behaviours/code;responds to messages; A Based on a Class definition

Max 2

3d frame/form/window/button/check box/radio button/menu/text box;A any sensible widgetR Plurals

1



Question number

Answer Marks

4a

1 mark for all three classes in appropriate single enclosures.1 mark for correct independent arrows in correct directions.

2

4b (Insert) a SetColour Procedure; A Functioninto the Public section;R make Colour Public

2

4c Van = Class/ subclass (Vehicle), i.e. Clearly identify Van as a (sub) class of vehicle (1 mark)

(Public) Procedure SetVehicleDetails (Override) condone if not included Function GetCapacity Function GetTailLift (penalise extra functions/procedures once.) Private Capacity : Integer/real/fixed/float TailLift : Boolean (penalise once if not private and once if extra variables listed.)End

Max 6

1 mark1 mark1 mark

1 mark1 mark

A Procedure SetCapacity and Procedure SetTailLift/ Procedure AddNewVan instead of Procedure SetVehicleDetails

OR

Public class/subclass Van extends/inherits Vehicle (1 mark){ public void SetVehicleDetails public int GetCapacity public boolean/int GetTailLift private int Capacity private boolean/int TailLift}

A public void SetCapacity and public void SetTailLift// public void AddNewVan instead of public void SetVehicleDetailsR any diagramsI any parameters to methods

1 mark

1 mark1 mark1 mark1 mark1 mark

6 marks(10)


Question number

Answer Marks

5a a class has properties/fields/attributes/characteristics and methods/procedures/functionsof the parent class it is derived from// a subclass/derived class inherits all the properties/fields/attributes/characteristics andmethods/procedures/functions from a super-class/base-class/parent class;

1 mark


5b StockItem (=) Class // Class (=) StockItem; (A Object instead of Class) Public (procedure DisplayDetails) (virtual)(virtual;abstract) (procedure) SetLoan (virtual)(virtual;abstract) Private; A protected Title: String OnLoan: Boolean DateAcquired: String/DateTime/ End Book = Class (StockItem) // Class Book extends/derives from StockItem // Book Sub-class: StockItem; A without keyword Class

1 mark for keywords Class and StockItem1 mark for keyword Public and correct methodsA text instead of string1 mark for correct data fields Date & data typesdon’t allow the other fields

Private Author: String ISBN: String Public (Procedure) DisplayDetails (override)EndCD = Class (StockItem) // Class CD extends/derives from StockItem // CD Subclass: StockItem; Private Artist: String PlayingTime: Integer/Real/Time/DateTime Public (Procedure) DisplayDetails (override)End

No marks for a diagrammatic answer. I method parameters

Java version:Public Class StockItem{Private String title;Private boolean onLoan;Private String dateAquired;

1 mark1 mark

1 mark1 mark

1 mark

1 mark

1 mark


If candidate declared ‘getters’ and ‘setters’ for the base class fields then don’t have to have DisplayDetails as a base class method.

Public void displayDetails ();Public void setLoan ();}Public Class Book extends StockItem{Private string author;Private string isbn;Public void displayDetails ();}Public Class CD extends StockItem{Private string artist;Private integer playingTime;Public void displayDetails();}

Total (max 7)

2.2End of topic questions

2 Product := 1; For i := 1 To n Do Product := i * Product; Factorial := Product;

3Call number n Output Comment

Call number 1 5

4

Call number 2 4

3

Call number 3 3

2

Call number 4 2

1 1 Procedure call completed

Return to previous call

Continue call 3 2 2 Procedure call completed






4Call number n Output Comment

Call number 1 5

4 4

Call number 2 4

3 3

Call number 3 3

2 2

Call number 4 2

1 1 Procedure call completed


Continue call 3 2 Procedure call completed





5Function Fibonacci(n: Integer): Integer;Begin If n=0 Then Fibonacci := 0 Else If n=1 Then Fibonacci := 1 Else Fibonacci := Fibonacci(n-1) + Fibonacci(n-2);End;

6Function Power(a, n: Integer): Integer;Begin If n=0 Then Power := 1 Else Power := a * Power(a, n-1);End;

7 2n-18 A solution using iteration is: on odd-numbered moves, move the smallest disk clockwise. On even-numbered moves, make the single other move which is possible. 9 Requires student to test the code in the book using 3 disks and 3 pegs as per images.


http://www.nist.gov/dads/HTML/iteration.html

2.3End of topic questionsExercise 1

Program Project1;

{$APPTYPE CONSOLE}

Uses SysUtils;

Const MaxListSize=30;Var ExitChosen: Boolean = False; MenuItem: Char; NameList: Array[1..MaxListSize] Of String; NumberOfFriendsStored: Integer = 0;

Function ListEmpty: Boolean;Begin ListEmpty := NumberOfFriendsStored=0;End;

Function ListFull: Boolean;Begin ListFull := NumberOfFriendsStored=MaxListSize;End;

Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add a name to the list'); WriteLn('2) Show number of names in the list'); WriteLn('3) Show list of names'); WriteLn('4) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;

Procedure AddNameToList;Var NewName: String; Finish: Boolean;


Begin Finish := False; If ListFull Then WriteLn('Can''t add any more names') Else Begin Repeat Write('Enter Name to be added to list (to finish type

XXX): '); ReadLn(NewName); If NewName = 'XXX' Then Finish := True Else Begin NumberOfFriendsStored := NumberOfFriendsStored + 1; NameList[NumberOfFriendsStored] := NewName; End; Until Finish Or ListFull; If ListFull Then WriteLn('Can''t add any more names'); End;End;

Procedure PrintListOfNames;Var i: Integer;Begin If Not ListEmpty Then For i := 1 To NumberOfFriendsStored Do WriteLn(NameList[i]);End;

Procedure OutputNumberOfNamesInList;Begin WriteLn('There are ', NumberOfFriendsStored, ' Names in the

list');End;

Begin AddNametoList; Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddNameToList; '2': OutputNumberOfNamesInList;


'3': PrintListOfNames; '4': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;End.

Exercise 2

Program Project1;

{$APPTYPE CONSOLE}

Uses SysUtils;

Const MaxListSize=30;Var ExitChosen: Boolean = False; MenuItem: Char; NameList: Array[1..MaxListSize] Of String; NumberOfFriendsStored: Integer = 0;



Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add a name to the list'); WriteLn('2) Show number of names in the list'); WriteLn('3) Show list of names'); WriteLn('4) Find a name'); WriteLn('5) Delete a name'); WriteLn('6) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);


End;

Procedure AddNameToList;Var NewName: String; Finish: Boolean;Begin Finish := False; If ListFull Then WriteLn('Can''t add any more names') Else Begin Repeat Write('Enter Name to be added to list (to finish type

XXX): '); ReadLn(NewName); If NewName = 'XXX' Then Finish := True Else Begin NumberOfFriendsStored := NumberOfFriendsStored + 1; NameList[NumberOfFriendsStored] := NewName; End; Until Finish Or ListFull; If ListFull Then WriteLn('Can''t add any more names'); End;End;

Procedure PrintListOfNames;Var i: Integer;Begin If Not ListEmpty Then For i := 1 To NumberOfFriendsStored Do WriteLn(NameList[i]);End;


list');End;

Procedure SearchFor(N: String; Var Pos: Integer);Var Found, NotInList: Boolean;Begin


Found := False; NotInList := False; Pos := 1; Repeat If NameList[Pos] = N Then Found := True Else Pos := Pos + 1; If Pos > NumberOfFriendsStored Then NotInList := True; Until Found Or NotInList; If NotInList Then Pos := 0;End;

Procedure Delete(Pos: Integer);Begin While Pos < NumberOfFriendsStored Do Begin NameList[Pos] := NameList[Pos+1]; Pos := Pos + 1; End; NameList[Pos] := ''; // clear space NumberOfFriendsStored := NumberOfFriendsStored - 1;End;

Procedure FindName;Var Name: String; Position: Integer;Begin Write('Please enter the name you wish to find: '); ReadLn(Name); SearchFor(Name, Position); If Position=0 Then WriteLn('Sorry, the name is not in the list') Else WriteLn(Name, ' is name number ', Position, ' in the

list');End;

Procedure DeleteName;Var Name: String; Position: Integer;Begin Write('Please enter the name you wish to delete: '); ReadLn(Name); SearchFor(Name, Position); If Position=0 Then WriteLn('Sorry, the name is not in the list')


Else Delete(Position);End;

Begin AddNametoList; Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddNameToList; '2': OutputNumberOfNamesInList; '3': PrintListOfNames; '4': FindName; '5': DeleteName; '6': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;End.


Exercise 3Program Project1;

{$APPTYPE CONSOLE}

Uses SysUtils;

Const MaxListSize=30;

Type TFriend = Record Name: String; Replied: Boolean; ComingToParty: Boolean; End;

Var ExitChosen: Boolean = False; MenuItem: Char; NameList: Array[1..MaxListSize] Of TFriend; NumberOfFriendsStored: Integer = 0;



Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add a name to the list'); WriteLn('2) Show number of names in the list'); WriteLn('3) Print list of names'); WriteLn('4) Find a name'); WriteLn('5) Delete a name'); WriteLn('6) Print list of all those definitely coming to the

party'); WriteLn('7) Print list of all those who have not replied yet'); WriteLn('8) Exit Program');


WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;

Procedure AddNameToList;Var NewName: String; Finish: Boolean;Begin Finish := False; If ListFull Then WriteLn('Can''t add any more names') Else Begin Repeat Write('Enter Name to be added to list (to finish type

XXX): '); ReadLn(NewName); If NewName = 'XXX' Then Finish := True Else Begin NumberOfFriendsStored := NumberOfFriendsStored + 1; NameList[NumberOfFriendsStored].Name := NewName; NameList[NumberOfFriendsStored].Replied := False; NameList[NumberOfFriendsStored].ComingToParty :=

False; End; Until Finish Or ListFull; If ListFull Then WriteLn('Can''t add any more names'); End;End;

Procedure PrintListOfNames;Var i: Integer;Begin If Not ListEmpty Then For i := 1 To NumberOfFriendsStored Do WriteLn(NameList[i].Name);End;


list');


End;

Procedure SearchFor(N: String; Var Pos: Integer);Var Found, NotInList: Boolean;Begin Found := False; NotInList := False; Pos := 1; Repeat If NameList[Pos].Name = N Then Found := True Else Pos := Pos + 1; If Pos > NumberOfFriendsStored Then NotInList := True; Until Found Or NotInList; If NotInList Then Pos := 0;End;

Procedure Delete(Pos: Integer);Begin While Pos < NumberOfFriendsStored Do Begin NameList[Pos] := NameList[Pos+1]; Pos := Pos + 1; End; NameList[Pos].Name := ''; // clear space NumberOfFriendsStored := NumberOfFriendsStored - 1;End;

Procedure FindName;Var Name: String; Position: Integer; Answer: Char;Begin Write('Please enter the name you wish to find: '); ReadLn(Name); SearchFor(Name, Position); If Position=0 Then WriteLn('Sorry, the name is not in the list') Else Begin Repeat Write('Has ', Name, ' replied to the invite? (y/n) '); ReadLn(Answer); Until ((Answer = 'y') Or (Answer = 'n')); If Answer='y'


Then Begin NameList[Position].Replied := True; Repeat Write('Is ', Name, ' coming to the party? (y/n) '); ReadLn(Answer); Until ((Answer = 'y') Or (Answer = 'n')); If Answer='y' Then NameList[Position].ComingToParty := True; End; End;End;

Procedure DeleteName;Var Name: String; Position: Integer;Begin Write('Please enter the name you wish to delete: '); ReadLn(Name); SearchFor(Name, Position); If Position=0 Then WriteLn('Sorry, the name is not in the list') Else Delete(Position);End;

Procedure AttendanceList;Var i: Integer;Begin WriteLn; WriteLn('The following are definitely coming to the party:'); WriteLn('================================================='); WriteLn; For i := 1 To NumberOfFriendsStored Do If NameList[i].ComingToParty Then WriteLn(NameList[i].Name);End;

Procedure NoReplyList;Var i: Integer;Begin WriteLn; WriteLn('The following have not replied yet:'); WriteLn('==================================='); WriteLn; For i := 1 To NumberOfFriendsStored Do


If Not NameList[i].Replied Then WriteLn(NameList[i].Name);End;

Begin AddNametoList; Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddNameToList; '2': OutputNumberOfNamesInList; '3': PrintListOfNames; '4': FindName; '5': DeleteName; '6': AttendanceList; '7': NoReplyList; '8': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;End.


Exercise 4

Program Project1;

{$APPTYPE CONSOLE}

Uses SysUtils;

Const MaxListSize = 6;

Type TNode = Record Data: String; Pointer: Integer; End;

Var Node: Array[1..MaxListSize] Of TNode; Start: Integer = 0; // list empty NextFree: Integer; MenuItem: Char; ExitChosen: Boolean = False;

Procedure InitialiseList;Var i: Integer;Begin NextFree := 1; For i := 1 To MaxListSize - 1 Do Node[i].Pointer := i+ 1; Node[MaxListSize].Pointer := 0;End;

Function ListFull: Boolean;Begin ListFull := NextFree=0;End;

Function ListEmpty: Boolean;Begin ListEmpty := Start=0;End;

Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn;


WriteLn('Menu'); WriteLn; WriteLn('1) Add a name to the list'); WriteLn('2) Show list of names'); WriteLn('3) Delete a name'); WriteLn('4) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;

Procedure AddNodeToFrontOfList(Name: String);Var Temp: Integer;Begin Node[NextFree].Data := Name; Temp := NextFree; NextFree := Node[NextFree].Pointer; Node[Temp].Pointer := Start; Start := Temp;End;

Procedure AddNodeIntoList(Name: String; ThisNode: Integer);// the new node is to be inserted after ThisNodeVar Temp: Integer;Begin Node[NextFree].Data := Name; Temp := NextFree; NextFree := Node[NextFree].Pointer; Node[Temp].Pointer := Node[ThisNode].Pointer; Node[ThisNode].Pointer := Temp;End;

Procedure DeleteHeadOfList;Var Temp: Integer;Begin Temp := Start; Start := Node[Start].Pointer; Node[Temp].Pointer := NextFree; NextFree := Temp;End;

Procedure DeleteNodeWithinList(PreviousNode, ThisNode: Integer);// ThisNode is the node to be deleted)Var Temp: Integer;Begin Temp := NextFree;


NextFree := Node[PreviousNode].Pointer; Node[PreviousNode].Pointer := Node[ThisNode].Pointer; Node[ThisNode].Pointer := Temp;End;

Procedure OutputListElements;Var ThisNode: Integer;Begin If ListEmpty Then WriteLn('List is empty') Else Begin ThisNode := Start; Repeat WriteLn(Node[ThisNode].Data); ThisNode := Node[ThisNode].Pointer; Until ThisNode = 0; End;End;

Function InsertionPoint(N: String): Integer;Var ThisNode, NextNode: Integer;Begin ThisNode := 0; // start at the beginning of the list NextNode := Start; While NextNode > 0 // not end of list Do Begin ThisNode := NextNode; NextNode := Node[NextNode].Pointer; // follow the

pointer End; InsertionPoint := ThisNode;End;

Procedure SearchFor(N: String; Var This, Previous: Integer);Begin This:= Start; Previous:= 0; While (This > 0) And (Node[This].Data <> N) Do Begin Previous := This; This := Node[This].Pointer; End; If Node[This].Data <> N


Then This := 0; // name not foundEnd;

Procedure AddNameToList;Var Finish: Boolean; NewName: String; Position: Integer;Begin Finish := False; If ListFull Then WriteLn('Can''t add any more names') Else Begin Repeat Write('Enter Name to be added to list (to finish type

XXX): '); ReadLn(NewName); If NewName = 'XXX' Then Finish := True Else Begin If ListEmpty Then AddNodeToFrontOfList(NewName) Else Begin Position := InsertionPoint(NewName); If Position = 0 Then AddNodeToFrontOfList(NewName) Else AddNodeIntoList(NewName, Position); End; End; Until Finish Or ListFull; If ListFull Then WriteLn('Can''t add any more names'); End;End;

Procedure DeleteName;Var Name: String; ThisNode, PreviousNode: Integer;Begin If ListEmpty Then WriteLn('There are no names in the list') Else Begin Write('Please enter the name you wish to delete: '); ReadLn(Name); SearchFor(Name, ThisNode, PreviousNode);


If ThisNode=0 Then WriteLn('Sorry, the name is not in the list')

Else If ThisNode = Start Then DeleteHeadOFList Else DeleteNodeWithinList(PreviousNode, ThisNode); End;End;

Procedure ShowArray;// this is just to check the linked list is built up correctlyVar i: Integer;Begin WriteLn('Start: ', Start); WriteLn('NextFree: ',NextFree); For i := 1 To MaxListSize Do WriteLn('[',i,']', Node[i].Data ,'->',Node[i].Pointer);End;

Begin InitialiseList; Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddNameToList; '2': OutputListElements; '3': DeleteName; '4': ExitChosen := True; '6': ShowArray; // for testing purposes Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;End.


Exercise 5Program Project2; {$APPTYPE CONSOLE}Uses SysUtils;

Type TNodePtr = ^TNode; TNode = Record Data : String; Ptr : TNodePtr; End;

Var Start : TNodePtr; MenuItem: Char; ExitChosen: Boolean = False;

Procedure InitList (Var List: TNodePtr);Begin List := Nil;End;

Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add a name to the list'); WriteLn('2) Show list of names'); WriteLn('3) Delete a name'); WriteLn('4) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;

Procedure AddItem (Var List: TNodePtr; NewItem: String);Var NewNode, ThisNode, NextNode: TNodePtr;Begin New (NewNode); NewNode^.Data := NewItem; NewNode^.Ptr := Nil; If List = Nil Then List := NewNode


Else Begin NextNode := List; While NextNode <> Nil Do Begin ThisNode := NextNode; NextNode := NextNode^.Ptr; End; ThisNode^.Ptr := NewNode; End;End;

Procedure DeleteItem (Var List: TNodePtr; WantedData: String);Var ThisNode, PreviousNode: TNodePtr;Begin If List <> Nil Then Begin ThisNode := List; If List^.Data = WantedData // is it the first node ? Then List := List^.Ptr Else Begin While ThisNode^.Data <> WantedData Do Begin PreviousNode := ThisNode; ThisNode := ThisNode^.Ptr End; PreviousNode^.Ptr := ThisNode^.Ptr; End; Dispose (ThisNode); End Else WriteLn('List empty');End;

Procedure PrintList (List : TNodePtr);Var ThisNode : TNodePtr;Begin If List <> Nil Then Begin ThisNode := List; Repeat WriteLn (ThisNode^.Data);


ThisNode := ThisNode^.Ptr; Until ThisNode = Nil; End;End;

Function GetName: String;Var Name: String;Begin Write('Enter a name: '); ReadLn(Name); GetName := Name;End;

Begin InitList(Start); Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddItem(Start,GetName); '2': PrintList(Start); '3': DeleteItem(Start,GetName); '4': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;

End.


Exercise 6

Program Project2; {$APPTYPE CONSOLE}Uses SysUtils;




Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add a name to the list'); WriteLn('2) Show list of names'); WriteLn('3) Delete a name'); WriteLn('4) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;

Procedure AddItem (Var List: TNodePtr; NewItem: String);Var NewNode, ThisNode, NextNode: TNodePtr;Begin New (NewNode); NewNode^.Data := NewItem; NewNode^.Ptr := Nil; If List = Nil


Then List := NewNode Else Begin NextNode := List; While NextNode <> Nil Do Begin ThisNode := NextNode; NextNode := NextNode^.Ptr; End; ThisNode^.Ptr := NewNode; End;End;

Procedure DeleteItem (Var List: TNodePtr; WantedData: String);Var ThisNode, PreviousNode: TNodePtr; Found: Boolean;Begin If List = Nil Then WriteLn('List empty') Else Begin ThisNode := List; If List^.Data = WantedData // is it the first node ? Then Begin List := List^.Ptr; // Then adjust front pointer Dispose (ThisNode); End Else Begin Found:= False; While (ThisNode^.Ptr <> Nil) And Not Found Do Begin PreviousNode := ThisNode; ThisNode := ThisNode^.Ptr; Found := ThisNode^.Data = WantedData; End; If Found Then Begin PreviousNode^.Ptr := ThisNode^.Ptr; Dispose (ThisNode); End Else WriteLn('Name not in list');


End; End;End;

Procedure PrintList (List : TNodePtr);Var ThisNode : TNodePtr;Begin If List <> Nil Then Begin ThisNode := List; Repeat WriteLn (ThisNode^.Data); ThisNode := ThisNode^.Ptr; Until ThisNode = Nil; End;End;


Begin InitList(Start); Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddItem(Start,GetName); '2': PrintList(Start); '3': DeleteItem(Start,GetName); '4': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;End.


Exercises 7, 8Program Project2; {$APPTYPE CONSOLE}Uses SysUtils;




Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add a name to the list'); WriteLn('2) Show list of names'); WriteLn('3) Find a name'); WriteLn('4) Delete a name'); WriteLn('5) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;

Procedure AddItem (Var List: TNodePtr; NewItem: String);Var NewNode, ThisNode, NextNode: TNodePtr;Begin New (NewNode); NewNode^.Data := NewItem; NewNode^.Ptr := Nil; If List = Nil


Then List := NewNode Else Begin NextNode := List; While NextNode <> Nil Do Begin ThisNode := NextNode; NextNode := NextNode^.Ptr; End; ThisNode^.Ptr := NewNode; End;End;

Procedure SearchFor(N:String; List: TNodePtr; Var This, Previous: TNodePtr);

Var Found: Boolean;Begin This := List; Previous := Nil; Found:= False; While (This^.Ptr <> Nil) And Not Found Do Begin Previous := This; This := This^.Ptr; Found := This^.Data = N; End; If Not Found Then This := Nil;End;

Procedure DeleteItem (Var List: TNodePtr; WantedData: String);Var ThisNode, PreviousNode: TNodePtr; Found: Boolean;Begin If List = Nil Then WriteLn('List empty') Else Begin ThisNode := List; If List^.Data = WantedData // is it the first node ? Then Begin List := List^.Ptr; // Then adjust front pointer Dispose (ThisNode);


End Else Begin SearchFor(WantedData, List, ThisNode, PreviousNode); If ThisNode = Nil Then WriteLn('Name not in list') Else Begin PreviousNode^.Ptr := ThisNode^.Ptr; Dispose (ThisNode); End; End; End;End;

Procedure PrintList (List : TNodePtr);Var ThisNode : TNodePtr;Begin If List = Nil Then WriteLn('List empty') Else Begin ThisNode := List; Repeat WriteLn (ThisNode^.Data); ThisNode := ThisNode^.Ptr; Until ThisNode = Nil; End;End;


Procedure FindName(List: TNodePtr; Name: String);Var ThisNode, PreviousNode: TNodePtr;Begin If List = Nil Then WriteLn('No names in list') Else Begin


SearchFor(Name, List, ThisNode, PreviousNode); If ThisNode = Nil Then WriteLn('Sorry, the name does not exist in the

list'); End;End;

Begin InitList(Start); Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddItem(Start,GetName); '2': PrintList(Start); '3': FindName(Start, GetName); '4': DeleteItem(Start,GetName); '5': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;

End.


Exercise 9Program Project2; {$APPTYPE CONSOLE}Uses SysUtils;




Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add a name to the list'); WriteLn('2) Show list of names'); WriteLn('3) Find a name'); WriteLn('4) Delete a name'); WriteLn('5) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;

Function InsertionPoint(List: TNodePtr; NewItem: String): TNodePtr;Var This, Next: TNodePtr;Begin This := Nil; Next := List; While (Next <> Nil) And (Next^.Data < NewItem) Do Begin


This := Next; Next := Next^.Ptr; End; InsertionPoint := ThisEnd;

Procedure AddItem (Var List: TNodePtr; NewItem: String);Var NewNode, ThisNode, NextNode: TNodePtr;Begin New (NewNode); NewNode^.Data := NewItem; NewNode^.Ptr := Nil; If List = Nil Then List := NewNode Else Begin ThisNode := InsertionPoint(List, NewItem); If ThisNode = Nil Then Begin NewNode^.Ptr := List; List := NewNode; End Else Begin NewNode^.Ptr := ThisNode^.Ptr; ThisNode^.Ptr := NewNode; End; End;End;

Procedure SearchFor(N:String; List: TNodePtr; Var This, Previous: TNodePtr);

Var Found: Boolean;Begin This := List; Previous := Nil; Found:= False; While (This^.Ptr <> Nil) And Not Found Do Begin Previous := This; This := This^.Ptr; Found := This^.Data = N; End; If Not Found


Then This := Nil;End;

Procedure DeleteItem (Var List: TNodePtr; WantedData: String);Var ThisNode, PreviousNode: TNodePtr; Found: Boolean;Begin If List = Nil Then WriteLn('List empty') Else Begin ThisNode := List; If List^.Data = WantedData // is it the first node ? Then Begin List := List^.Ptr; // Then adjust front pointer Dispose (ThisNode); End Else Begin SearchFor(WantedData, List, ThisNode, PreviousNode); If ThisNode = Nil Then WriteLn('Name not in list') Else Begin PreviousNode^.Ptr := ThisNode^.Ptr; Dispose (ThisNode); End; End; End;End;

Procedure PrintList (List : TNodePtr);Var ThisNode : TNodePtr;Begin If List = Nil Then WriteLn('List empty') Else Begin ThisNode := List; Repeat WriteLn (ThisNode^.Data); ThisNode := ThisNode^.Ptr; Until ThisNode = Nil; End;


End;


Procedure FindName(List: TNodePtr; Name: String);Var ThisNode, PreviousNode: TNodePtr;Begin If List = Nil Then WriteLn('No names in list') Else Begin SearchFor(Name, List, ThisNode, PreviousNode); If ThisNode = Nil Then WriteLn('Sorry, the name does not exist in the

list'); End;End;

Begin InitList(Start); Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddItem(Start,GetName); '2': PrintList(Start); '3': FindName(Start, GetName); '4': DeleteItem(Start,GetName); '5': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;End.


Exam practice questionsQuestion 1 – January 2003 CPT4 Q1 Question number

Answer Marks

1 a) i) Data must be given in correct order1 pheasant 22 teal 33 widgeon 5

START 4 partridge 15 woodpigeon 0

End Pointer can be blank1 for correct START and END pointers;1 for correctly numbered nodes and correct pointers (need all birds)

1

ii) 1 pheasant 72 teal 33 widgeon 54 partridge 15 woodpigeon 0

START 6 grouse 47 snipe 2

//correctly amended diagram1 for grouse and snipe physically at end;1 for correct pointers (if not as ms then clear and logical);

1Max 2

b) The amount of memory taken up can vary;//The size/length of the structure/linked list can vary;At run time;

11

c) A heap/stack/a pool of available locations;A pointer holds the address of the allocated block/next available location; 2


2.4End of topic questions1a)

1b)


2Stack

Index DataTopOfStackPointer 3 6

5

4

3 Cup

2 Saucer

1 Plate

3Program StackExampleUsingLinearList;{$APPTYPE CONSOLE}Uses SysUtils;Const MaxStackSize=6;Var ExitChosen: Boolean; MenuItem: Char; Stack: Array[1..MaxStackSize] Of String; TopOfStackPointer: Integer = 0;Function StackEmpty: Boolean;Begin StackEmpty := TopOfStackPointer=0;End;Function StackFull: Boolean;Begin StackFull := TopOfStackPointer=MaxStackSize;End;Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add an item to the stack'); WriteLn('2) Remove an item from the stack'); WriteLn('3) Show contents of stack');

WriteLn('4) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;Procedure AddItemToStack;Begin If StackFull Then WriteLn('Can''t add to a full stack') Else Begin TopOfStackPointer := TopOfStackPointer + 1; Write('Enter item to be added to stack: '); ReadLn(Stack[TopOfStackPointer]); End;End;Procedure RemoveItemFromStack;Begin If StackEmpty Then WriteLn('Can''t take anything from an empty stack') Else Begin WriteLn('Item popped off stack: ', Stack[TopOfStackPointer]); TopOfStackPointer := TopOfStackPointer - 1; End;End;Procedure ShowContentsOfStack;Var Pointer: Integer;Begin Pointer := MaxStackSize; // output stack with bottom of stack

last While TopOfStackPointer < Pointer Do Begin WriteLn(Pointer); Pointer := Pointer − 1; End; While Pointer > 0 Do Begin WriteLn(Pointer,' ',Stack[Pointer]); Pointer := Pointer − 1;

End;

End;Begin ExitChosen := False; Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddItemToStack; '2': RemoveItemFromStack; '3': ShowContentsOfStack; '4': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;End.

4Program LinkedListStackExample;{$APPTYPE CONSOLE}Uses SysUtils;Const MaxStackSize=6;Type TNode = Record Data: String; Pointer: Integer; End;Var ExitChosen: Boolean; MenuItem: Char; Stack: Array[1..MaxStackSize] Of TNode; FreePtr: Integer; TopOfStackPointer: Integer = 0;Procedure InitialiseStack;Var i: Integer;Begin FreePtr := 1; For i := 1 To MaxStackSize − 1 Do Stack[i].Pointer := i + 1;

Stack[MaxStackSize].Pointer := 0;End;Function StackEmpty: Boolean;Begin StackEmpty := TopOfStackPointer=0;End;Function StackFull: Boolean;Begin StackFull := FreePtr=0;End;Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add an item to the stack'); WriteLn('2) Remove an item from the stack'); WriteLn('3) Show contents of stack'); WriteLn('4) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;Procedure AddItemToStack;Var Temp: Integer;Begin If StackFull Then WriteLn('Can''t add to a full stack') Else Begin Temp := TopOfStackPointer; TopOfStackPointer := FreePtr; FreePtr := Stack[FreePtr].Pointer; Write('Enter item to be added to stack: '); ReadLn(Stack[TopOfStackPointer].Data); Stack[TopOfStackPointer].Pointer := Temp; End;End;Procedure RemoveItemFromStack;Var Temp: Integer;Begin

If StackEmpty Then WriteLn('Can''t take anything from an empty stack') Else Begin Temp := TopOfStackPointer; WriteLn('Item popped: ', Stack[TopOfStackPointer].Data); TopOfStackPointer := Stack[TopOfStackPointer].Pointer; Stack[Temp].Pointer := FreePtr; FreePtr := Temp; End;End;Procedure ShowContentsOfStack;Var ThisNode: Integer;Begin If StackEmpty Then WriteLn('empty stack') Else Begin ThisNode := TopOfStackPointer; Repeat WriteLn(Stack[ThisNode].Data); ThisNode := Stack[ThisNode].Pointer; Until ThisNode=0; // end of linked list reached End;End;Begin ExitChosen := False; InitialiseStack; Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddItemToStack; '2': RemoveItemFromStack; '3': ShowContentsOfStack; '4': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;End.

5

Program StackExampleUsingHeap; {$APPTYPE CONSOLE}Uses SysUtils;Type TNodePtr = ^TNode; TNode = Record Data : String; Ptr : TNodePtr; End;Var TopOfStack : TNodePtr; ExitChosen: Boolean; MenuItem: Char;Procedure InitStack (Var Stack: TNodePtr);Begin Stack := Nil;End;Procedure AddItemToStack (Var Stack: TNodePtr);Var NewNode: TNodePtr; NewItem: String;Begin Write('Enter item to be added to stack: '); ReadLn(NewItem); New (NewNode); NewNode^.Data := NewItem; NewNode^.Ptr := Nil; If Stack = Nil Then Stack := NewNode Else Begin NewNode^.Ptr := Stack; Stack := NewNode; End;End;Procedure RemoveItemFromStack (Var Stack: TNodePtr);Var ThisNode: TNodePtr;Begin If Stack <> Nil Then Begin WriteLn('Popped Data: ', Stack.Data);

ThisNode := Stack; Stack := Stack^.Ptr; Dispose (ThisNode); End Else WriteLn('Stack Empty');End;Procedure ShowContentsOfStack (Stack : TNodePtr);Var ThisNode : TNodePtr;Begin If Stack <> Nil Then Begin ThisNode := Stack; Repeat WriteLn (ThisNode^.Data); ThisNode := ThisNode^.Ptr; Until ThisNode = Nil; End Else WriteLn('Stack Empty');End;Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add an item to the stack'); WriteLn('2) Remove an item from the stack'); WriteLn('3) Show contents of stack'); WriteLn('4) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;Begin InitStack (TopOfStack); ExitChosen := False; Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddItemToStack(TopOfStack); '2': RemoveItemFromStack(TopOfStack);

'3': ShowContentsOfStack(TopOfStack); '4': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen; Readln;End.

6Program StackExampleOOP; {$APPTYPE CONSOLE}Uses SysUtils, StackClass in 'StackClass.pas';Var MenuItem: Char; Stack : TStack; ExitChosen: Boolean; NewItem: String;Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add an item to the stack'); WriteLn('2) Remove an item from the stack'); WriteLn('3) Show contents of stack'); WriteLn('4) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;Begin

Stack := TStack.Create; Repeat

GetMenuChoice(MenuItem); Case MenuItem Of '1': Begin Write('Enter item to be added to stack: ');

ReadLn(NewItem); Stack.Push(NewItem); End; '2': WriteLn('Item popped: ',Stack.Pop); '3': Stack.PrintStack; '4': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;End.

7a)Queue

Index Data FrontPointer 1 1 Jones

2 Smith

3 Peters

4 FranklinBackPointer 5 5 Taylor

6b)

Queue

Index Data FrontPointer 3 1

2

3 Peters


6

c)

Queue

Index Data FrontPointer 3 1 Y

2 Z

3 Peters


6 X

8Program LinearQueue;{$APPTYPE CONSOLE}uses SysUtils;Const MaxQueueSize=6;Var ExitChosen: Boolean; MenuItem: Char; Queue: Array[1..MaxQueueSize] Of String; BackPointer: Integer = 0;Function QueueEmpty: Boolean;Begin QueueEmpty := BackPointer=0;End;Function QueueFull: Boolean;Begin QueueFull := BackPointer=MaxQueueSize;End;Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add an item to the queue'); WriteLn('2) Remove an item from the queue'); WriteLn('3) Show contents of queue'); WriteLn('4) Exit Program');

WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;Procedure AddItemToQueue;Begin If QueueFull Then WriteLn('Can''t add to a full queue') Else Begin BackPointer := BackPointer + 1; Write('Enter item to be added to queue: '); ReadLn(Queue[BackPointer]); End;End;Procedure ShuffleQueue(Ptr: Integer);Var i: Integer;Begin For i := 2 To Ptr Do Queue[i-1] := Queue[i]; Queue[Ptr]:=''; // clear the spaceEnd;Procedure RemoveItemFromQueue;Begin If QueueEmpty Then WriteLn('Can''t take anything from an empty queue') Else Begin WriteLn('Item taken from queue: ', Queue[1]); If BackPointer > 1 Then ShuffleQueue(BackPointer); BackPointer := BackPointer - 1; End;End;Procedure ShowContentsOfQueue;Var Pointer: Integer;Begin For Pointer := 1 To MaxQueueSize Do WriteLn(Pointer, Queue[Pointer]);End;Begin

ExitChosen := False; Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddItemToQueue; '2': RemoveItemFromQueue; '3': ShowContentsOfQueue; '4': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;End.

9Program CircularQueueExample;{$APPTYPE CONSOLE}uses SysUtils;Const MaxQueueSize=6;Var ExitChosen: Boolean; MenuItem: Char; Queue: Array[1..MaxQueueSize] Of String; BackPointer: Integer = 0; FrontPointer: Integer = 0; NumberOfItemsInQueue: Integer = 0;Function QueueEmpty: Boolean;Begin QueueEmpty := NumberOfItemsInQueue=0;End;Function QueueFull: Boolean;Begin QueueFull := NumberOfItemsInQueue=MaxQueueSize;End;Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add an item to the queue'); WriteLn('2) Remove an item from the queue'); WriteLn('3) Show contents of queue');

WriteLn('4) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;Procedure IncrementPointer(Var Ptr: Integer);Begin Ptr := Ptr MOD MaxQueueSize + 1;End;Procedure AddItemToQueue;Begin If QueueFull Then WriteLn('Can''t add to a full queue') Else Begin If QueueEmpty Then FrontPointer := 1; IncrementPointer(BackPointer); Write('Enter item to be added to queue: '); ReadLn(Queue[BackPointer]); NumberOfItemsInQueue := NumberOfItemsInQueue + 1; End;End;Procedure RemoveItemFromQueue;Begin If QueueEmpty Then WriteLn('Can''t take anything from an empty queue') Else Begin WriteLn('Item taken from queue: ', Queue[FrontPointer]); Queue[FrontPointer] := ''; // clear space NumberOfItemsInQueue := NumberOfItemsInQueue - 1; If NumberOfItemsInQueue = 0 Then Begin FrontPointer := 0; BackPointer := 0; End Else IncrementPointer(FrontPointer); End;End;

Procedure ShowContentsOfQueue;Var Pointer: Integer;Begin For Pointer := 1 To MaxQueueSize Do Begin Write(Pointer, Queue[Pointer]); If Pointer = FrontPointer Then Write('<-- Front'); If Pointer = BackPointer Then Write('<-- Back'); WriteLn; End;End;Begin ExitChosen := False; Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddItemToQueue; '2': RemoveItemFromQueue; '3': ShowContentsOfQueue; '4': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen;End.

10Program QueueExampleUsingHeap; {$APPTYPE CONSOLE}Uses SysUtils;Type TNodePtr = ^TNode; TNode = Record Data : String; Ptr : TNodePtr; End;Var FrontPointer : TNodePtr;

BackPointer : TNodePtr; ExitChosen: Boolean; MenuItem: Char;Procedure InitQueue;Begin FrontPointer := Nil; BackPointer := Nil;End;Procedure AddItemToQueue;Var NewNode: TNodePtr; NewItem: String;Begin Write('Enter item to be added to Queue: '); ReadLn(NewItem); New (NewNode); NewNode^.Data := NewItem; NewNode^.Ptr := Nil; If BackPointer = Nil Then Begin FrontPointer := NewNode; BackPointer := NewNode; End Else Begin BackPointer^.Ptr := NewNode; BackPointer := NewNode; End;End;Procedure RemoveItemFromQueue;Var ThisNode: TNodePtr;Begin If FrontPointer <> Nil Then Begin WriteLn('Popped Data: ', FrontPointer.Data); ThisNode := FrontPointer; Frontpointer := FrontPointer^.Ptr; Dispose (ThisNode); IF FrontPointer = Nil Then BackPointer := Nil; End Else WriteLn('Queue Empty');

End;Procedure ShowContentsOfQueue;Var ThisNode : TNodePtr;Begin If FrontPointer <> Nil Then Begin ThisNode := FrontPointer; Repeat WriteLn (ThisNode^.Data); ThisNode := ThisNode^.Ptr; Until ThisNode = Nil; End Else WriteLn('Queue Empty');End;Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add an item to the Queue'); WriteLn('2) Remove an item from the Queue'); WriteLn('3) Show contents of Queue'); WriteLn('4) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;Begin InitQueue; ExitChosen := False; Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': AddItemToQueue; '2': RemoveItemFromQueue; '3': ShowContentsOfQueue; '4': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End; Until ExitChosen; Readln;

End.

11Program OOPQueueExample; {$APPTYPE CONSOLE}Uses SysUtils, QueueClass In 'QueueClass.pas';Var Queue: TQueue; MenuItem: Char; ExitChosen: Boolean; NewItem: String;Procedure GetMenuChoice(Var Choice: Char);Begin WriteLn; WriteLn('Menu'); WriteLn; WriteLn('1) Add an item to the queue'); WriteLn('2) Remove an item from the queue'); WriteLn('3) Show contents of queue'); WriteLn('4) Exit Program'); WriteLn; Write('Enter your choice number: '); ReadLn(Choice);End;Begin Queue := TQueue.Create; Repeat GetMenuChoice(MenuItem); Case MenuItem Of '1': Begin Write('Enter item to be added to queue: '); ReadLn(NewItem); Queue.Add(NewItem); End; '2': WriteLn('Item popped: ',Queue.Remove); '3': Queue.PrintQueue; '4': ExitChosen := True; Else WriteLn('Invalid menu choice. Try again'); End;

Until ExitChosen;End.

Exam practice questionsQuestion 1 – May 2007 CPT1 Q5 Question number

Answer Marks

1a Last (item) in, is the first (item) out/first (item) in is the last (item) out;R. LIFO / FILO

1

1b i)

All items in the correct locations

ii)

Correct three items // ft from an incorrect (i) including 605 as the first location used;A. .R. and .Y. entries indicated in some way as . deleted.

iii)

Correct list of five items // ft from an incorrect (i) + a correct ft (ii) including 605 asthe first location used;

1

1

1

1c i) Queue ; A. First In . First Out FIFO / LILOii) Items are removed/popped from the stack (one at a time) (and items are then added to the queue);iii) Items leave the queue on a .first in-first out. basis; A. from the front

11

1

of the queueiv) .Y., .R., .E., .V., .A. on the queue; .Y., .R., .E., .V., .A. on the final stack; A. using 701 for the first queue location

2

Question 2 – January 2007 CPT1 Q4Question number

Answer Marks

2a Next item to be added is at position/location/address (Tail + 1);Position/location/address Tail is the last item in the queue; R. .points to the end of the queue.

Max 1

2b Cat // item at position Head; 1

2c

Snake + Eel + Shark at positions 3,4,5;Tail points to 5;Head points to 2;I. Dog and Cat crossed through

111

2d Tail will eventually reach position 99 (A. 100);Head will eventually reach 99 (A. 100);Memory/queue will become full;Space is not re-useable

Max 2

Question 3 – June 2005 CPT1 Q7 Question Number

Answer Marks

3a First In First Out; or by description

Last In First Out; or by description 2

3b 2

3c Reverse the contents of a queue/list;

Push all contents of queue/list onto stack then pop them off into a new queue/list;

Procedure/function calls;

2

Local variables;

Parameters;

Return address;

Volatile environment; A register contents

State 1 Describe 1

3d List of elements inserted into tree;

to allow rapid/fast searching of the data;

to output sorted/ordered data;

2


Answer Marks

4 Last In First Out; 14b) i) 1

4b) ii) 1

4b) iii) 1

4b) iv) 1

4c) To reverse elements/pass parameters/store volatile environment;A store return address

1

Question 5 – June 2008 CPT4 Q2

Question number

Answer Marks

5a) i) Empty entries waste memory // Maximum size // fixed size; 15a) ii) Memory used by pointers//takes more time to add/delete notes//indirect

access takes more time; R programming difficulties1

5b) Place next item in first location/location 0/location 1//Implement a circular array/queue//allow wraparound;

1

5c) IsFull/IsQueueFull; 1

Question 6 – January 2007 CPT4 Q6 Question number

Answer Marks

6a 3

6b AddItem//Add; 16c i) Full/FullQueue;

ii) No memory used for pointers; I Faster R Easier to programiii) Size is limited by array size; memory wasted when not full;

11

2


Answer Marks

7a) i) Empty entries waste space // Maximum/fixed/static size A stack may overflow

ii) Space used by pointers // more complex to program

1

17b) i) The size of the stack/amount of data is known/limited/predictable

Memory saved since no pointers (if not given in a (ii)) R easier to program

ii) The size of the stack is unknown// The stack is volatile/number of items fluctuates widely;

1

1

Question 8– June 2002 CPT4 Q5Question number

Answer Marks

8a)4

8b)2

8c)In a linear queue data is static, so queue ‘moves’ through storage/In a FIFO structure storage locations are only used once;In a circular queue, the locations will be re-used;Thus a circular queue has a more efficient use of memory;

1 mark for each of 2 points

111

2.5End of topic questions1 The number of actors that the actor at the vertex has starred with in movies.

2

3 A graph is a diagram consisting of circles called vertices, joined by lines called edges or arcs. Each edge joins exactly two vertices.

4

5a)

b) Three cages: one for A, one for D and one for B, C and E6a)

Vertex 1 2 3 4 5

1 0 1 1 0 1

2 1 0 0 1 1

3 1 0 0 1 1

4 0 1 1 0 1

5 1 1 1 1 0

b)Vertex Adjacent vertices

1 2, 3, 5

2 1, 4, 5

3 1, 4, 5

4 2, 3, 5

5 1, 2, 3, 4

7a)Vertex 1 2 3 4 5

1 0 0 1 0 0

2 1 0 0 0 0

3 0 0 0 0 1

4 0 1 1 0 0

5 1 1 0 1 0

b)Vertex Adjacent vertices

1 3

2 1

3 5

4 2, 3

5 1, 2, 4

8a)Vertex 1 2 3 4 5

1 ∞ ∞ 1 ∞ ∞2 20 ∞ ∞ ∞ ∞3 ∞ ∞ ∞ ∞ 81

4 ∞ 30 -3 ∞ ∞5 5 8 ∞ 54 ∞

b)Vertex Adjacent

Vertices

1 3 3 0

2 1 1 20

3 5 5 81

4 2, 3 2 30 3 −3

5 1, 2, 4 1 5 2 8 4 54

9a) When many vertex pairs are connected by edges, then the adjacency matrix is best because it doesn't waste much space, and it indicates whether an edge exists with one access (rather than following a list). b) When the graph is sparse, that is, not many of its vertex pairs have edges between them, the adjacency list is best.

10

11 Imagine a graph of four vertices connected by four edges to create a cycle as follows:

If one edge is removed then the graph becomes a tree – a tree is a connected graph with no cycles. Generalising, a tree of n vertices must have n−1 edges.

12 A tree is a connected undirected graph with no cycles. A rooted tree is a tree in which one vertex has been designated as the root and every edge is directed away from the root.13 Vertices visited in the following order EGFDABC

Exam practice questionsQuestion 1 – Specimen Paper COMP 3 Q3Question number

Answer Marks

1a

or

Max 3

1b i) Max size only limited by memory;Only uses memory it requires;ii) No pointers required;

max 1

1

1c i) Adjacency matrix;

ii) Faster to insert/delete; fixed size; 12

2.6End of topic questions1 Dry-run of iterative algorithm searching for Davidson

ItemSought First Last Midpoint List[Midpoint] ItemFound SearchFailed

Davidson 1 15 8 King False False

1 7 4 Farmer

1 3 2 Bond

3 3 3 Clark True

2 Dry-run of recursive algorithm searching for DavidsonCall

ItemSought

First

Last

Midpoint

List[Midpoint]

ItemFound

SearchFailed

1 Davidson 1 15 8 King False False

2 Davidson 1 7 4 Farmer False False

3 Davidson 1 3 2 Bond False False

4 Davidson 3 3 3 Clark False TrueFour calls are made before the search fails.

3a)Program Project1;{$APPTYPE CONSOLE}Uses SysUtils;Const Max = 13;Var NameToFind: String;Var NumberOfIterations: Integer = 0;Var NumberOfProcedureCalls: Integer = 0;Var SortedList: Array[0..Max] Of String = ('Alan', 'Ben',

'Chris', 'David', 'Eve', 'Fred', 'Greg', 'Heidi', 'Ida', 'Jude', 'Karl', 'Les', 'Mike', 'Noel');

// could read a text file of many names into an array to test instead

Procedure BinarySearch1 (Var List: Array Of String; First, Last: Integer; ItemSought: String);

Var ItemFound, SearchFailed: Boolean; MidPoint: Integer;

Begin ItemFound := False; SearchFailed := False; Repeat NumberOfIterations := NumberOfIterations + 1; // for stats

only Midpoint := (First + Last) DIV 2; // integer part of division If List[Midpoint] = ItemSought Then ItemFound := True Else If First >= Last Then SearchFailed := True Else If List[Midpoint] > ItemSought Then Last := Midpoint − 1 Else First := Midpoint + 1 Until ItemFound Or SearchFailed; If SearchFailed Then WriteLn('No such name') Else WriteLn('Name found at position ',MidPoint);End;c) The iterative program uses less memory as it does not have the procedure call overheads.

Each procedure call saves return address, parameters and local variables on the stack (in main memory).

4First Last CurrentPointer CurrentValue Pointer List

[1] [2] [3] [4] [5] [6]

1 8 2 72 1 87 72 28 45 59 36

10

87 87 28 45 59 36

72 87 28 45 59 36

3 28 2 72 87 28 45 59 36

2 72 87 87 45 59 36

10

72 72 87 45 59 36

28 72 87 45 59 36

4 45 3 28 72 87 45 59 36

3 28 72 87 87 59 36

2 28 72 72 87 59 36

1

28 45 72 87 59 36

5 59 4 28 45 72 87 59 36

4 28 45 72 87 87 36

32

28 45 72 72 87 36

28 45 59 72 87 36

6 36 5 28 45 59 72 87 36

5 28 45 59 72 87 87

4 28 45 59 72 72 87

3 28 45 59 59 72 87

21

28 45 45 59 72 87

28 36 45 59 72 87

5a) & b) Answer to insertion sort:Program Project1;{$APPTYPE CONSOLE}Uses SysUtils;Const Max = 7;Type TList = Array[1..Max] Of Integer; Var UnsortedList: TList ;Procedure OutputList(List:TList; First, Last: Integer);Var i: Integer;Begin Write('('); For i := First To Last−1 Do Write(List[i], ' ,'); WriteLn(List[Last],')');End;Procedure Sort (List: TList; First, Last: Integer);Var CurrentPointer, Pointer: Integer; CurrentValue: Integer;Begin For CurrentPointer := First + 1 To Last Do Begin CurrentValue := List[CurrentPointer];

Pointer := CurrentPointer − 1; Write(First:6, Last:6, CurrentPointer:6, CurrentValue:6,

Pointer:6); OutputList(List, First, Last); While (List[Pointer] > CurrentValue) AND (Pointer > 0) Do Begin List[Pointer+1] := List[Pointer]; Pointer := Pointer − 1; Write(First:6, Last:6, CurrentPointer:6, CurrentValue:6,

Pointer:6); OutputList(List, First, Last); End; List[Pointer+1] := CurrentValue; Write(First:6, Last:6, CurrentPointer:6, CurrentValue:6,

Pointer:6); OutputList(List, First, Last); End;End;Begin UnsortedList[1]:=56; UnsortedList[2]:=23; UnsortedList[3]:=67; UnsortedList[4]:=12; UnsortedList[5]:=45; UnsortedList[6]:=99; UnsortedList[7]:=17; OutputList(UnsortedList,1,Max); Sort(UnsortedList,1,Max); ReadLn;End.

6 Bubble sort programProgram Project1;{$APPTYPE CONSOLE}Uses SysUtils;Const Max = 7;Type TList = Array[1..Max] Of Integer;Var UnsortedList: TList;Procedure OutputList(List:TList; NumberOfItems: Integer);Var i: Integer;

Begin Write('('); For i := 1 To NumberOfItems-1 Do Write(List[i], ' ,'); WriteLn(List[NumberOfItems],')');End; Procedure BubbleSort(List: TList; NumberOfItems: Integer); Var NoMoreSwaps: Boolean; Temp, Element: Integer; Begin Repeat NoMoreSwaps := True; For Element := 1 To NumberOfItems − 1 Do If List[Element] > List[Element+1] Then Begin NoMoreSwaps := False; Temp := List[Element]; List[Element] := List[Element + 1]; List[Element + 1] := Temp; OutputList(List,NumberOfItems); End; Until NoMoreSwaps; End;Begin UnsortedList[1]:=56; UnsortedList[2]:=23; UnsortedList[3]:=67; UnsortedList[4]:=12; UnsortedList[5]:=45; UnsortedList[6]:=99; UnsortedList[7]:=17; OutputList(UnsortedList,Max); BubbleSort(UnsortedList,Max); ReadLn;End.

The bubble sort program does 11 swaps of elements using the given test data. Each swap consists of 3 assignments. This means that 33 assignments are required.

The insertion sort program uses 22 assignments for the same test data.

i) using the list: 12, 23, 45, 56, 67, 99, 17The insertion sort uses 17 assignments.

The bubble sort uses 15 assignments.

ii) using the list: 99, 67, 56, 45, 23, 17, 12The insertion sort uses 66 assignments.

The bubble sort uses 66 assignments.

8Program Project1;{$APPTYPE CONSOLE}Uses SysUtils;Const Max = 7;Type TList = Array[0..Max] Of Integer; Var UnsortedList: TList ;Procedure OutputList(List:TList; First, Last: Integer);Var i: Integer;Begin Write('('); For i := First To Last-1 Do Write(List[i], ' ,'); WriteLn(List[Last],')');End;Procedure Swap(Var A, B : Integer);Var Temp: Integer;Begin Temp := A; A := B; B := Temp;End;Procedure QuickSort (Var List: TList; First, Last:Integer);Var PivotValue, LeftPointer, RightPointer, Pivot: Integer;Begin If First < Last {if there are elements in the list} Then Begin PivotValue := List[First]; LeftPointer := First + 1; RightPointer := Last; Pivot := 0; While LeftPointer <= RightPointer Do Begin While (List[LeftPointer] < PivotValue) And (LeftPointer

<= RightPointer)

Do Begin LeftPointer := LeftPointer + 1; End; While (List[RightPointer] > PivotValue) And (LeftPointer

<= RightPointer) Do Begin RightPointer := RightPointer − 1; End; If LeftPointer < RightPointer Then Swap (List[LeftPointer], List[RightPointer]); End; Pivot := RightPointer; Swap (List[First], List[Pivot]); QuickSort (List, First, Pivot − 1); QuickSort (List, Pivot + 1, Last); End;End;Begin UnsortedList[1]:=56; UnsortedList[2]:=23; UnsortedList[3]:=67; UnsortedList[4]:=12; UnsortedList[5]:=45; UnsortedList[6]:=99; UnsortedList[7]:=17; OutputList(UnsortedList,1,7); QuickSort(UnsortedList,1,7); OutputList(UnsortedList,1,7); ReadLn;End.

Quick sort questions1

List Comment

First Last Pivot value

Left pointer

Right pointer

Pivot [1] [2] [3] [4] [5] [6] [7]

1 7 56 2 7 56 23 67 12 45 99 17 Swap left & right

3 7 56 23 17 12 45 99 67 Swap right & pivot

6 5 5 45 23 17 12 56 99 67

1 4 45 2 4 45 23 17 12 No value larger than pivot value

5 4 4 12 23 17 45

1 3 12 2 3 12 23 17 No value smaller than pivot value

2 1 1 12 23 17

1 0 Last < First return from call

2 3 23 3 3 23 17 No value larger than pivot value

4 3 3 17 23

2 2 Last = First Return from call



6 7 99 7 7 99 67 No value larger than pivot value

8 7 7 67 99

6 6 Last = First return from call


Exam practice questionsQuestion 1 – Jan 2007 CPT4 Q7 Question number

Answer Marks

1a)

1 mark for first row (12, 1, 9)2 marks for second row (1 mark for each 5)2 marks for third row (3 and 3)2 marks for fourth row (1 mark for Lower = 4, 1 mark for upper = 4)1 mark for correct return value

8

1b) Find the position of 12/ a number in the array// search for 12/ a number in the array;

1

Question 2 – June 2006 CPT4 Q5 Question number

Answer Marks

2a) 271;

The required item might be the 271st one/last one/ not be present//Every item accessed;

11

2b) 9;

Each comparison halves the number of items to be accessed//271 lies between 28 and 29

11

2c)

1 mark for Count11 mark for Count21 mark for Temp(ii) (Bubble) sort the items into ascending order;(ii) Reduce the number of tests each pass// stop when no swaps occur during a pass//Add a flag NoSwaps to indicate when no swaps occur// change loop control to Repeat until no swaps// sort variable sized array;

511

Question 3 – June 2004 CPT4 Q8Question number

Answer Marks

3a) A procedure/routine which calls itself//is defined in terms of itself; R re-entrant;A function instead of procedure R programiteration Talked Out (no mark)

1

3b) i)

Accept True in row 3Marks in each row all three/two parts correctAccept empty cell to mean: same as in previous row.Stop marking when logic goes wrong

7

3b) ii) Binary search;;Search;R any other type of search

2Question 4 – January 2003 CPT4 Q3Question number

Answer Marks

4a) i) 8 14a) ii) Each time a comparison is made in a binary search the number of items

to be searched/list is halved;//137 lies between 27 and 28

Could have (ii) even if (i) incorrect

11

4b) i) 137 14b) ii) In a linear search of 137 items, the required item might be the

137th one;need a termination - must explain why 137 is the maximum

4


Answer Marks

5a) 1, 17, 9, 21, 15, 23;(2 if all right, 1 if 4 of 6)If > misinterpreted, follow through for 1 mark

2

5b) A bubble sort 15c) To detect when all the numbers have been sorted

Efficiency (to stop procedure repeating unnecessarily);R to detect when numbers have switched

1


1a entities: lorry, queue, loading bayAttributes of lorry: waiting, being unloadedAttributes of queue: lorries in queue, queue emptyAttributes of loading bay: loading bay free, loading bay in use

b states: no lorry waiting, no lorry being unloadedno lorry waiting, lorry being unloadedlorries waiting in queue, lorry being unloadedpossible events: a lorry arrives, a lorry leaves the loading baypossible activities: unloading a lorry, inter-arrival time of lorries

c

d possible variables: arrival times of lorries, loading times of lorries, number of loading bays

2

Master clock (hours)

Lorries arriving

Lorries being unloaded

Lorries in queue

Lorry hours in queue

Loading bay status

05:00 - - 0 0 Free06:00 Lorry1 Lorry1 0 0 In use

07:00 Lorry1 0 0 In use08:00 Lorry2 Lorry1 1 1 In use09:00 Lorry2 0 1 In use10:00 Lorry3 Lorry2 1 2 In use11:00 Lorry2 1 3 In use12:00 Lorry4 Lorry3 1 4 In use13:00 Lorry3 1 5 In use14:00 Lorry5 Lorry3 2 7 In use15:00 Lorry4 1 8 In use16:00 Lorry4 1 9 In use17:00 Lorry4 1 10 In use18:00 Lorry5 0 10 In use19:00 Lorry5 0 10 In use20:00 Lorry5 0 10 In use

3Master clock (hours)

Lorries arriving

Lorries being unloaded in Bay 1

Lorries being unloaded in Bay 2

Lorries in queue

Loading bay hours free

Loading Bay 1 status

Loading bay 2 status

05:00 - - 0 2 Free Free06:00 Lorry1 Lorry1 0 3 In use Free07:00 Lorry1 0 4 In use Free08:00 Lorry2 Lorry1 Lorry2 0 4 In use In use09:00 Lorry2 0 5 Free In use10:00 Lorry3 Lorry3 Lorry2 0 5 In use In use11:00 Lorry3 0 6 In use Free12:00 Lorry4 Lorry3 Lorry4 0 6 In use In use13:00 Lorry4 0 7 Free In use14:00 Lorry5 Lorry5 Lorry4 0 7 In use In use15:00 Lorry5 0 8 In use Free16:00 Lorry5 0 9 In use Free

4a entities: passenger, queue, check-in deskAttributes for passenger: waiting, checking inAttributes for queue: passengers in queue, queue emptyAttributes for check-in desk: in use, free

b states: no passenger waiting, check-in desk freeno passenger waiting, passenger checking inpassengers waiting, passenger checking inevents: passenger arrives, passenger finishes checking inactivities: checking in a passenger, inter-arrival time of passengers

cMaster clock (minutes)

Passenger arriving

Passenger checking in

Passengers in queue (25 at start)

Passenger minutes in queue (from check-in desk opening)

Check-in desk status

1 Passenger1 24 24 In use2 Passenger1 24 48 In use3 Passenger2 23 71 In use4 Passenger2 23 94 In use5 Passenger26 Passenger3 23 117 In use6 Passenger3 23 140 In use7 Passenger4 22 162 In use8 Passenger4 22 184 In use9 Passenger5 21 205 In use

10 Passenger27 Passenger5 22 227 In use11 Passenger6 21 248 In use12 Passenger6 21 269 In use13 Passenger7 20 289 In use14 Passenger7 20 309 In use15 Passenger28 Passenger8 20 329 In use16 Passenger8 20 349 In use17 Passenger9 19 368 In use18 Passenger9 19 387 In use19 Passenger10 18 405 In use20 Passenger29 Passenger10 19 424 In use21 Passenger11 18 442 In use22 Passenger11 18 460 In use23 Passenger12 17 477 In use24 Passenger12 17 494 In use25 Passenger30 Passenger13 17 511 In use26 Passenger13 17 528 In use27 Passenger14 16 544 In use28 Passenger14 16 560 In use29 Passenger15 15 575 In use30 Passenger31 Passenger15 16 591 In use31 Passenger16 15 606 In use32 Passenger16 15 621 In use33 Passenger32 Passenger17 15 636 In use34 Passenger17 15 651 In use35 Passenger18 14 665 In use36 Passenger33 Passenger18 15 680 In use37 Passenger19 14 694 In use38 Passenger19 14 708 In use39 Passenger34 Passenger20 14 722 In use40 Passenger20 14 736 In use41 Passenger21 13 749 In use42 Passenger35 Passenger21 14 763 In use43 Passenger22 13 776 In use44 Passenger22 13 789 In use45 Passenger36 Passenger23 13 802 In use46 Passenger23 13 815 In use47 Passenger24 12 827 In use48 Passenger37 Passenger24 13 840 In use49 Passenger25 12 852 In use50 Passenger25 12 864 In use51 Passenger38 Passenger26 12 876 In use52 Passenger26 12 888 In use

53 Passenger27 11 899 In use54 Passenger39 Passenger27 12 911 In use55 Passenger28 11 922 In use56 Passenger28 11 933 In use57 Passenger40 Passenger29 11 944 In use58 Passenger29 11 955 In use59 Passenger30 10 965 In use60 Passenger41 Passenger30 11 976 In use61 Passenger31 10 986 In use62 Passenger31 10 996 In use63 Passenger32 9 1005 In use64 Passenger32 9 1014 In use65 Passenger42 Passenger33 9 1023 In use66 Passenger33 9 1032 In use67 Passenger34 8 1040 In use68 Passenger34 8 1048 In use69 Passenger35 7 1055 In use70 Passenger43 Passenger35 8 1063 In use71 Passenger36 7 1070 In use72 Passenger36 7 1077 In use73 Passenger37 6 1083 In use74 Passenger37 6 1089 In use75 Passenger44 Passenger38 6 1095 In use76 Passenger38 6 1101 In use77 Passenger39 5 1106 In use78 Passenger39 5 1111 In use79 Passenger40 4 1115 In use80 Passenger45 Passenger40 5 1120 In use81 Passenger41 4 1124 In use82 Passenger41 4 1128 In use83 Passenger42 3 1131 In use84 Passenger42 3 1134 In use85 Passenger46 Passenger43 3 1137 In use86 Passenger43 3 1140 In use87 Passenger44 2 1142 In use88 Passenger44 2 1144 In use89 Passenger45 1 1145 In use90 Passenger47 Passenger45 2 1147 In use

6 passengers are still waiting to check in half an hour before take off.If the check-in process takes longer, there would be even more passengers not checked in.Queues would be even longer if more passengers arrive during the peak half-hour.

A second check-in desk would then become necessary.

Exam practice questionsQuestion number

Answer Marks

1a A computer model is used to manipulate variables and observe results. 11b We may consider simulation because experimenting with the real

system might cause unknown irremovable effects, or because we can experiment with a model more cheaply and at a time of our choosing.

1

1c Possible examples include queuing for service, traffic flow, population growth.

1

Question number

Answer Marks

2a The entities are: ship, queue, berth.The attributes for ship are: waiting, being unloaded.The attributes for berth are: berth free, berth in use.The attributes for queue are: ships in queue, queue empty.

2b Possible states are:no ship waiting in queue, berth freeno ship waiting in queue, ship being unloaded in berthships waiting in queue, ship being unloaded in berth.

Possible events are:a ship arrivesa ship leaves the berth.

Possible activities are:the unloading of a shipthe inter-arrival time of ships.

2c

Ships Arriving

Ship

Being Unloaded

Ships

Queuing

Berth Free

2d

Master clock (hours)

Ship arriving

Ship being unloaded

Ships in queue

Ship hours in queue

Berth status

1 - - 0 - Free2 Ship1 Ship1 0 - In use3 Ship1 0 0 In use4 Ship2 Ship1 1 1 In use5 Ship2 0 1 In use6 Ship3 Ship2 1 2 In use7 Ship2 1 3 In use8 Ship4 Ship3 1 4 In use9 Ship3 1 5 In use10 Ship5 Ship3 2 7 In use

2e Possible variables: ship arrival times, unloading times, number of berths, maximum space for ships queuing.


Qu. Mantissa Exponent m x 2e Denary

1 a 0.1011 10010 0001102=610 0.101110012 x 26 = 101110.012 46.25

b 1.0101 01000 0001002=410 -0.1010112 x 24 = -01010.112 -10.75

c 0.1100 00000 1111112=-110 0.112 x 2-1 = 0.0112 0.375

d 1.0100 00000 1111012=-310 -0.112 x 2-3 = -0.000112 -0.09375

2 a 1.10100 01000 101002=2010 1.10100012 x 25 = 110100.012 52.25

b 1.01100 00000 011112=1510 1.0112 x 20 = 1.0112 1.375

c 1.01010 00000 011102=1410 1.01012 x 2-1 = 0.101012 0.65625

d -1.10001 11010 101102=2210 -1.1000111012 x 27 = -11000111.012 -199.25

e -1.01100 00000 011012=1310 -1.0112 x 2-2 = -0.010112 -0.34375

f 0.00000 00000 111112 special case infinity +∞

g 0.00000 00000 000002 special case zero 0

h -0.00000 00000 111112 special case infinity -∞

3 a Mantissa Exponent

5.5 0.10112 x 23 0.1011 00000 000011

0.375 0.112 x 2-1 0.1100 00000 111111

-9.75 1.0110012 x 24 1.0110 01000 000100

-0.1875 1.012 x 2-2 1.0100 00000 111110

3 b Sign Exponent Mantissa

5.5 1.0112 x 22 0 10001 01 1000 0000

0.375 1.12 x 2-2 0 01101 10 0000 0000

-9.75 -1.001112 x 23 1 10010 00 1110 0000

-0.1875 -1.12 x 2-3 1 01100 10 0000 0000

4 Precision means the maximum number of significant digits we can represent. Precision is important with floating point numbers because there are an infinite number of values between any two given real numbers. These have to be mapped to a finite number of representations (bit patterns). Therefore some values will be approximated.

5 36.1 is between 1.0010000011 x 25 and 1.0010000100 x 25

1.0010000011 x 25 = 36.09375 1.0010000100 x 25 = 36.125

Rounding to the nearest value, means 36.1 is represented by 1.0010000011 x 25 This introduces an absolute error or 0.00625 or a relative error of 0.000173

6 1.0110000000 x 212 x 27 = 1.0110000000 x 219

The largest possible value in the exponent field in minifloat format is 111102 = 30. As the exponent is stored in excess-15 mode, the largest exponent possible is 15. Therefore the result of this calculation will cause the exponent field to contain 11111 and set the mantissa to 0 (this means +∞).

7 1.0110000000 x 2-9 x 27 = 1.0110000000 x 2-16

The smallest possible value in the exponent field in minifloat format is 000002 = 0. As the exponent is stored in excess-15 mode, the smallest exponent possible is -15. Therefore the result of this calculation will cause the exponent field to contain 0 and set the mantissa to 0 (this means 0).

8 1.10100 000002 x 212 = 1101000000000.02

+ 1.01000 000002 x 2-7 = + 000000000000.0000001012

1101000000000.0000001012 = 1.1010000000000000001012 x 212

However, there are only 11 bits for the mantissa, so the result is 1.10100000002 x 212

Note that the effect of adding a very small number to a large number does not change the result.

Exam practice questionsQuestion 1 – January 2008 CPT4 Q1Question Number

Answer Marks

1a BF2 1

1b -;1038 2(1 mark for sign, 1 mark for value)

1c +;191;.125; A 1/8 3

1d -;2;.03125; A 1/32

(If incorrect part marks as follows1 mark for complemented mantissa 010000011 mark for moving binary point 2 places)

3

1e To maximise precision in a given number of bits;A to maximise accuracy in a given number of bitsTo minimise rounding errors;To allow a wider range of values to be stored;

Max 2

Question 2 – June 2005 CPT4 Q2Question Number

Answer Marks

2a 140 1/4 ;; one mark for correct integer part,140.25;; one mark for correct fractional part

2

2b i) -14.5;;; give 2 marks for 14.5 partial marks for workings if result incorrect: 1 mark for negative number; 1 mark for x24 (accept 16 instead of 24); A showing that binary point moves 4 places right; ii)leftmost 2 digits/bits are different; a significant bit is stored after the binary point; bit after point different from bit before point; (negative number) starts with 10. (positive number starts with 01)..; A the first bit after the sign bit is a .0.; A The second bit is a .0.;

3

max 1

A an answer that clearly implies a .0. follows the .1. iii) to maximise accuracy/precision for a given number of bits

// to minimise rounding errors;A more accurate/precise for a given number of bits;a given number can only be expressed in one way in a given number of bits// a given number can only be expressed in one way in a given format;to simplify arithmetic/logical operations; I range

1


Answer Marks

3a) BE4; must be capital letters 1

b) 190.25 / 190 1/4 ;; one mark for correct integer part, one mark for correct fractional part one mark for correct working (e.g. correct place values)

3

c) -1052;; 1 mark for workings if result incorrect 1 mark for sign, 1 mark for 1052

2

d) i) -8.25 / -81/4;;; partial marks for workings if result incorrect 1 mark for sign, 1 mark for moving binary point 4 places or showing 24

3

ii) starts with 1 0 the first 2 binary digits are different; a significant bit is stored after the (implied) binary point; bit after (implied) binary point different from bit before binary point; A all leading 1.s have been removed // there are no leading 1.s; R there are no leading zeros

1

Question 4 – June 2004 CPT4 Q4(b)Question number

Answer Marks

4a) i)

(if the answer is wrong give:moving e places to right / exponent processing 2e or equivalent: 1 mark

4

correctly identifying negative number: 1 markfollow through if binary representation wrong

ii) To maximise precision in a given number of bits // to minimise rounding errors

// to have just one representation of the decimal number // to simplify arithmetic operations: 1 mark

A to maximise accuracy in a given number of bits;

9


Answer Marks

5a) i) positive 1 ii) < 2-2 15b) Correct answer 194.5 or 194 ½

working

If wrong answer, method marks as follows: exponent 28 [clearly defined] 1 mark application of shift / *28 from correct start point 1 mark correct interpretation of bits 1 mark

(Basically here, if it is a little inaccurate, give 2 marks, if quite inaccurate but slightly correct, give 1 mark.)

21

c) i) Processing fixed point numbers is quicker than floating point / less processing is required;More accurate / greater precision;

1

ii) Where the possible range of numbers to be stored is limited / small;Where number is of a set format / processing integers / Working with currency;Where maximum precision is required

1

4.1End of topic questions1 System programs and application programs.2 Hide the complexities of the hardware from the user.

Manage the hardware resources in order to provide for an orderly and controlled allocation of the processors, memories and I/O devices among the various programs competing for them, manage the storage of data.

3 Processors a) Storage

b) Input/output devices

c) Data4 Processor scheduling a) Memory management b) I/O management c) File management5 Virtual machine: this is the apparent machine that the operating system presents to the user,

achieved by hiding the complexities of the hardware behind layers of operating system software.

6 Application program interface: this is a layer of software that allows application programs to call on the services of the operating system.

7 A standard application program interface (API) allows a software developer to write an application on one computer and have a high degree of confidence that it will run on another computer of the same type, even if the specifications of the latter are different. The program should continue to run when hardware upgrades and updates occur because the operating system, not the application, manages the hardware and the distribution of its resources.

8 Command Line and Graphical user interface.Some embedded systems require no operating system because they function as very simple controllers. Others such as mobile phones do require an operating system. However, a mobile phone operating system has to manage a different set of resources. The resources include the keypad, the screen, the address book, the phone dialler, the battery and the network connection. Operating systems for general purpose computers have to manage resources such as printers, hard drives, DVD R/W drives, high resolution flat screen visual display units, mouse and standard QWERTY keyboard.

Exam practice questionsQuestion 1 – January 2003 CPT2 Q5 Question number Answer Marks

1a) OS hides complexities of hardware from the user; 1

1b) Any three @ 1 eachProcessor(s)/cpu(s);Memory/IAS/Main memory;Disk (space)/backing store; A Hard disk/drive //Floppy disk (drive)//Secondary storageI/O devices//peripherals; R examplesFile space; A files R dataR programs

Max 3

4.2End of topic questions1a) Interactive operating system: An operating system in which the user and the computer are in

direct two-way communication.1b) Real time operating system: inputs are processed in a timely manner so that the output can

affect the source of the inputs.1c) Network operating system: in this system, a layer of software, added to the operating system

of a computer connected to the network, intercepts commands that reference resources elsewhere on network, e.g. File server and then redirects the request to the remote resource in a manner completely transparent to the user.

2 Both the airline reservation system and the system to land an aircraft:have to support application programs which are non-sequential in nature, i.e. programs which do not have a START- PROCESS - END structure;have to deal with a number of events which happen in parallel and at unpredictable moments in time;have to carry out processing and produce a response within a specified interval of time.In the case of the airline reservation system Airline reservation system - up to a 1000 messages per second can arrive from any one of 11000-12000 terminals, situated all over the world. The response time must be less than 3 seconds.In the case of the aircraft landing system - up to 1000 signals per second can arrive from sensors attached to the system being controlled. The response time must be less than one thousandth of a second.

a) C: - local hard drive; D: - DVD R/W drive; E: - flash memory drive; N: - network drive b) Network operating system

c) Washing machine; microwave oven; central heating controller d) Cable TV set-top box; broadband router; computer network firewall3 The mobile phone operating system will be different from that of a desktop computer because

the kind of resources available to a mobile phone is different. In particular, the energy, the physical context and the mobility of the user, and the limited amount of some resources such as memory or CPU. An operating system for a mobile phone needs to manage a limited amount of energy from the battery, a limited amount of memory and a processor of limited capability. It needs to manage a network connection in a fairly autonomous way that continually sends a message packet to base stations so that mobile phone location can be identified. It needs to manage a keyboard that supports predictive text, a physically small display, an address book, a phone dialler and a battery of limited power. It needs to meet real time constraints because, for example, multimedia applications need to run. It needs to take into account the instability of mobile phone networks (bandwidth, connection, etc).

4 The O.S. for a smartphone needs to be more powerful than that of an ordinary mobile phone because the smartphone offers advanced capabilities with PC-functionality beyond a typical mobile phone. The operating system must support advanced features like e-mail and Internet capabilities and/or a full keyboard. The OS must have the capability to allow a user to install new applications on the smartphone.

5 The operating system must take on the tasks of the BIOS in a PC and has to be designed to run on processors with low clock frequency and a main memory of limited capacity. The operating system must use various techniques to save energy and must cater for short reaction times. The operating system must support character input via a stylus and touch sensitive screen and should support high-resolution colour rendition with resolutions of approximately 640x480 pixels.

6 An embedded computer system is a dedicated computer system with a limited or non-existent user interface and designed to operate completely or largely autonomously from within other machinery, e.g. a motorcar. An embedded computer system in a motorcar could be set up to manage the engine of the car carrying out a range of tasks one of which is controlling the ignition of the gases in the cylinders.

7 The embedded computer O.S. must manage multiple tasks that need to meet specific time constraints. In the simplest system, a low-level piece of code switches between tasks or threads based on a timer (connected to an interrupt). Operating systems for embedded systems are designed to work with the constraints of limited memory size and limited processor performance. In portable embedded system the operating system must also take account of limited battery life.The O.S. on which embedded systems rely has had to become more sophisticated as the hardware of embedded systems has become more complex with every generation, and more features have been added. The role of an O.S. is to manage the resources and to hide the complexities of the hardware. In the case of embedded computer systems, it required to hide the complexities from the application programs so making it possible to create new applications without consuming an excessive amount of development time. With more tasks to manage and subject to strict timing constraints a greater reliance on getting this support from the O.S. has become necessary. This in turn requires greater sophistication in the O.S.

Exam practice questionsQuestion 1 - KBQuestion Number Answer Marks

1 An embedded computer system’s operating system has a limited number of resources to manage unlike a desktop operating system, most notably a limited or non-existent user-interface and does not have to support the execution of a large number of different applications. On the contrary the embedded system’s operating system has to support the execution of a limited number of applications.

2

Question 2 - KBQuestion Number Answer Marks

2 A server operating system is an operating system optimised to provide one or more specialised services to networked clients such as file storage, domain control, running applications. They are optimised for the service that they carry out. This leads to a near optimum performance that would not be achieved if a server were carrying out a large amount of general-purpose processing.

2


3 1 - It has one set of tasks to perform; 2 - Very straightforward input to expect (a numbered keypad and a few pre-set buttons); 3 - Simple, never-changing hardware to control.

3


4 Any three from:1) They have to support application programs which are

non-sequential in nature, i.e. programs which do not have a START- PROCESS - END structure;

2) They have to deal with a number of events which happen in parallel and at unpredictable moments in time;

3) They have to carry out processing and produce a response within a specified interval of time;

4) Some systems are safety-critical meaning they must be fail-safe and guarantee a response within a specified

3

time interval.


2

3

Exam practice questionsPlease see the Teacher Notes for Section 5.3: SQL for all of the answer to the Examination-style questions.

5.2End of topic questions 1 Course (CourseCode, CourseName)

Student (StudentID, StudentName, StudentAddress, StudentDateOfBirth)CourseSet (CourseCode, SetNo, TeacherInitials)Enrolment (StudentID, CourseCode, SetNo)

2 Borrower (BorrowerID, BorrowerName, BorrowerAddress)BookTitle (ISBN, Title, Author)BookCopy (AccessionNumber, ISBN)Loan (BorrowerID, AccessionNumber, LoanDate)

Exam practice questionsPlease see the Teacher Notes for Section 5.3: SQL for all of the answer to the Examination-style questions.

5.3End of topic questions 1a)CREATE DATABASE CourseEnrolmentCREATE TABLE Course( CourseCode VARCHAR(5), CourseName VARCHAR(20), PRIMARY KEY (CourseCode));CREATE TABLE Student( StudentID VARCHAR(10), StudentName VARCHAR(25), StudentAddress VARCHAR(50), StudentDateOfBirth Date, PRIMARY KEY (StudentID) );CREATE TABLE CourseSet( CourseCode VARCHAR(5),

SetNo INT, TeacherInitials VARCHAR(3), PRIMARY KEY (CourseCode, SetNo) FOREIGN KEY (CourseCode) REFERENCES Course (CourseCode));CREATE TABLE Enrolment( StudentID VARCHAR(10), CourseCode VARCHAR(5), SetNo INT, PRIMARY KEY (StudentID, CourseCode), FOREIGN KEY (StudentID) REFERENCES Student (StudentID), FOREIGN KEY (CourseCode, SetNo) REFERENCES CourseSet (CourseCode, SetNo) );

1 b) SELECT * FROM Course

1 c) SELECT CourseCode, CourseName, StudentID, StudentName FROM Enrolment, Student, Course WHERE Enrolment.CourseCode = Course.CourseCode AND Enrolment.StudentID = Student.StudentID ORDER BY CourseCode2a)CREATE DATABASE BookLoanSystemCREATE TABLE Borrower( BorrowerID VARCHAR(5), BorrowerName VARCHAR(20), BorrowerAddress VARCHAR(50), PRIMARY KEY (BorrowerID));CREATE TABLE BookCopy( AccessionNumber INT, ISBN VARCHAR(13), PRIMARY KEY (AccessionNumber), Foreign Key (ISBN) REFERENCES BookTitle(ISBN));CREATE TABLE BookTitle( ISBN VARCHAR(13), Title VARCHAR(30), Author VARCHAR(25), PRIMARY KEY (ISBN));CREATE TABLE Loan( BorrowerID VARCHAR(5), AccessionNumber INT, LoanDate Date, PRIMARY KEY (AccessionNumber, LoanDate), Foreign Key (BorrowerID) REFERENCES Borrower(BorrowerID),

Foreign Key (AccessionNumber) REFERENCES BookCopy(AccessionNumber));

2 b)SELECT BorrowerID, BorrowerNameFROM Borrower, LoanWHERE Borrower.BorrowerID = Loan.BorrowerID

2 c)SELECT BorrowerID, BorrowerName COUNT(Loan.AccessionNumber) AS BookCountFROM Borrower, LoanWHERE Borrower.BorrowerID = Loan.BorrowerIDGroup By BorrowerID

Exam practice questionsQuestion 1 – June 2007 CPT5 Q5Question number

Answer Marks

1a i) Recipe table; A Figure 2; ii) Why: contains multiple values in Ingredients field/attribute/column;// data in Ingredients column not atomic // repeating groups.

11

1b i) fully normalised:every attribute is dependent on the key, the whole key and nothing but the key;OR (tables contain no repeating groups of attributes,) no partial dependencies;no non-key dependencies; A rely on instead of depend on;OR if (and only if) every determinant in the relation is a candidate key.ii) Why: to aid consistency of data // to avoid potential data inconsistency problems;// to eliminate data inconsistency // to minimise data duplication;// to eliminate data redundancy; A reduce instead of eliminate;R saving space.

2

1

1c i) Recipe (RecipeID, Dish, PrepTime, CookTime, NoOfServings, CookInstructions); ii) FoodItem (FoodItemID, FoodItemName, PackSize, Price); iii) RecipeIngredient(FoodItemID, RecipeID, Quantity).

1

14

(1 mark for each correct field, 1 mark for correct primary key. Take off 1 mark for every extra field included.)

1d SELECT FoodItemName, Quantity, PackSize, PriceFROM FoodItem, RecipeId = RecipeIngredient, RecipeWHERE (Recipe.RecipeId = RecipeIngreident.RecipeId)AND (RecipedIngredient.FoodItemId = FoodItem.FoodItemId)AND (Recipe.Dish = “Feat Salad”)ORDER BY FoodItemName

field names F/T P1 for fieldname.tablename P1 tbl prefixA ORDER BY FoodItemNameA Dish instead of Recipe.DishA “feta salad” instead of “Feta Salad” A #feta salad# instead of “Feta Salad”

111111

Max 5


Answer Marks

2a Copyright, Designs and Patents Act (1998); if other laws included T.O. 1

2b Boxes for correct entities: SoftwareLicence SoftwareInstallation (1 mark)Correct degree of relationship: 1 to many (1 mark)Suitable name for relationship: (1 mark)

3

2c Any sensible field length accepted except for SoftwareID, ComputerID, StaffID i) SoftwareID VARCHAR(10) PRIMARY KEY (NOT NULL)// SoftwareID VARCHAR(10) PRIMARY KEY(SoftwareID);

SoftwareName VARCHAR(30)Version VARCHAR(10)Supplier VARCHAR(20)DatePurchased DATEExpiryDate/DateValidTo DATENoOfLicences INT

ii )SoftwareID VARCHAR(10)ComputerID VARCHAR(6)DateInstalled DATEStaffID VARCHAR(3)PRIMARY KEY (SoftwareID, ComputerID)FOREIGN KEY (SoftwareID) REFERENCES Software Licence(SoftwareID)

I NOT NULL

(1 mark for any 3 attributes correctP1 if extra symbols used. Ignore spaces and case in attribute names.)

(A char/string/text/alphanumericInstead of VARCHARA Date/Time instead of DateA Integer instead of INTBOD any attributes which are clearly more than 1 word.)

(1 mark for any 2 attributes correct.)

(If not DDL, give 1 mark if composite key identified.)

2dSELECT ComputerID, SoftwareName, Version;FROM SoftwareLicence, SoftwareInstallation;WHERE SoftwareLicence.SoftwareID=SoftwareInstallation.SoftwareID;ORDER BY ComputerID; A ASC or DESC

A LEFT JOIN

4

Question 3 – June 2006 CPT 5 Q5Question number

Answer Marks

3a i)

R one to many relationshipii)Entity-Relationship Diagram; A E-R diagram; A E-R D R E-A-R diagram

2

1

Table names prefixed with tbl, P1 If table name and attribute transposed, P1

P1 for other spurious punctuation inc. semicolons

Accept (instead of FROM WHERE):FROM SoftwareLicence INNER JOINSoftwareInstallation ONSoftwareLicence.SoftwareID = SoftwareInstallation.SoftwareID

Extra attributes: T.O.

3b 6

3c 6

F/T with attribute namesP1 for tbl prefixP1 if table name after attribute nameI extra punctuation


Answer Marks

4a I Minor spelling

i)

ii)

iii)

1

1

1

4b Penalise table name. field name in reverse order once

R Quotes and additional constructs

I Table names unless in wrong order or wrongly expressed

i)Select FirstName, Surname

From Student

ii) Select Student.FirstName, Student.Surname

MarkAwarded.Mark;

From Student, MarkAwareded;

Where MarkAwarded.LifecyclePhaseID = 1;

And Student.StudentID = MarkAwarded.StudentID ;

Order By Student Surname;

1

5


Answer Marks

5a (NB Take note of labelling inside boxes because candidate’s positioning of labels may be opposite to that shown below.)

Alternative symbols:

I table name unless incorrect

Order By Student.Surname AscA AscendingAsc/Ascending must be in correct positionA OrderBy

i)

ii)

iii)

iv)

1

1

1

1

5b (R. Tbl in front of table name – penalise once.)

i) Select Surname From Swimmer Where SwimmerNo = 6;

1

Accept minor mis-spelling or spaces between parts of entity name.A. plural namesI. Box outlines

Select Swimmer.Surname is OKI. Brackets surrounding attributesR. Extra attributes, tables,

ii) Select SwimmerNo From GalaRaceSwimmer Where (RaceNo = 5); And (GalaNo = 2); Order By TimeRecordedForRace;

iii) Select Swimmer.Surname From Swimmer, GalaRace; Where (GalaNo = 4); And (GalaRace.SwimmerNoOfWinner = Swimmer.SwimmerNo);

3

3

Total 11

I. Brackets surrounding attributes, table names in front of attributes unless incorrectA. Criteria without bracketsR. Extra attributes, tables, criteria in this solution but be careful because candidates may give an alternative involving extra attributes, tables, criteria

A. Asc or Ascending in correct place i.e. after TimeRecordedForRaceR. Asc/Ascending in any other position and/or with other words

A. Select Swimmer.Surname, GalaRace.RaceNo From …OrA. Select Surname, RaceNo From…

Brackets may be omitted.A. GalaRace.GalaNo = 4

And SwimmerNoOfWinner = SwimmerNo is OK

Select Surname From Swimmer Where SwimmerNo In;

(Select SwimmerNoOfWinner From GalaRace; Where GalaNo = 4);

R. = in place of In


Answer Marks

6a i)

ii)

L other entities

1

1

6b i)Select Book.Title From Book ; Where BookISBN = “1-57820-082-2”;

ii)Don’t need Book in SelectSelect Book.AuthorName, Book.ISBN ; From Book, BookCopy ; Where (Book.ISBN = BookCopy.ISBN) ; And )BookCopy.AccessionNumber = 1234) ;

2

4

A. Title

A. ISBN = “1-57820-

Any extra attributes lose mark where extra attributes usedR. 1-57820-082-2 Need quotesA. “1-57820-082-2”R. TblBook – penalise once

Penalise TblBook/TblBookCopy onceR. quotes on 1234Any extra attributes lose mark where extra attributes usedBrackets non-essential. May see

A. BoodCopy.ISBN in place of BOOK.ISBN

A. AccessionNumber in place of BookCopy.AccessionNumbe

C Mail-merge//Mail-merging 1

Total 9


Answer Marks

7a

(NB: don’t allow relationship between Ward and PatientMedicalCondition.)

3

A. In for =

7b (For each extra attribute lose one mark.)

i) Ward(WardName, NurseInCharge, NoOfBeds)

ii)Patient(PatientNo, Surname, Forename, Address, DOB, Gender, WardName)

iii)MedicalConditionID(MedicalConditionNumber, Name, RecommendedStandardTreatment)

iv)PatientMedicalCondition(PatientNo. MedicalConditionNo)

1

2

1

7c (Accept tbl in front of table name.)

Select Patient.Forename, Patient.Surname, PatientMedicalCondition, MedicalConditionNo From Patient, PatientMedicalCondition Where Patient.WardName = ‘Victoria’ And Patient.PatientNo = PatientMedicalCondition.PatientNo

A. Forename, Surname, MedicalConditionNo, WardName

3

Total 12

A. NumberOBeds, NameOfNurseInCharge, NurseInChargeNameR. WardId, Name, NurseName, NameOfNurse, BedNo,

A. Attributes rejected in (ii) and (iii) for PatientNo and MedicalConditionNo R. If attributes used are not consistent with (ii) and (iii)

A. MedicalConditionId, MedicalConditionNumber, MedicalconditionName, ConditionName, StandardTreatment, Treatment, RecommendedTreatmentR. ConditionNumber, ConditionID

A. PatientId, PatientNumber, PatientSurname, PatientForename, PatientAddress, DateOfBirth, PatientDateOfBirth, PatientGender, Sex, PatientSex

6.1End of topic questions1 In serial data transmission, single bits are sent one after another along a single wire2 In parallel data transmission, bits are sent down several wires simultaneously.

a) The transmission of data between computer and printer is controlled by the printer via the Ready/Busy wire. If the printer is ready to receive, the Ready/Busy wire is set by the printer at 5 volts, otherwise it is set to 0 volts. The computer reads the state of Ready/Busy wire. If set to Busy, the computer does not send data. If the signal on this wire is set to Ready, the computer places signals representing the data onto the data wires.

b) A short while afterwards, the computer sets the voltage on the Strobe wire to 5 volts. The printer detects this strobe signal voltage and starts to read the data on the data wires at the same time it sets the Ready/Busy wire status to busy. When the printer is finished reading data, it sets the Ready/Busy wire back to Ready.c) Serial data transmission is used for long-distance communication. The need for only one signal pathway each way is the main reason for this use. This makes it easy to regenerate the signal, which can reduce significantly in strength over long distances. It makes it easier to route through telecommunication switches when routing the signal. It also saves on the cost of cabling.d) Parallel data transmission is used over short distances because it is difficult to keep the voltages on the eight wires ’in-line’ with each other (the problem is known as skew) beyond a certain distance. This can lead to the voltage on each wire being read incorrectly. It is also expensive to run eight or more wires over long distances especially if the signals on the wires need to be switched onto a different path, for example, in digital switching centres such as telephone exchanges. Therefore, parallel communication has been restricted to computer to printer connections and computer buses.3 The Baud rate is the rate at which signals on a wire /line may change. One baud is one

signal change per second. The bit rate is the number of bits transmitted per second. The bit rate can be greater than the baud rate if each signal can code more than one bit. For example, if each signal contains 2 bits and the baud rate is 100 baud (100 signals sent per second) then the bit rate is 2 × 100 bits/s.

4 The greater the bandwidth of the transmission system, the higher is the bit rate that can be transmitted over that system

33 Latency refers to the time delay that can occur between the moment something is initiated and the moment its first effect begins. In a wide area network involving satellites, significant time delay occurs because of the physical distance between the ground stations involved and the geostationary satellite. Requesting and receiving a web page can involve a considerable time delay even though the bit rate of the uplink and downlink to the satellite is high, that is, the bandwidth is large. With a round trip distance in excess of 143,200 kilometres, speed of microwaves 3 × 108 metres/second, the propagation time delay is approximately 0.4 seconds.

34 In asynchronous serial data transmission, the arrival of data cannot be predicted by the receiver, instead a start bit is used to signal the arrival of data and to synchronise the transmitter and receiver temporarily.

a) The arrival of data at the receiver is signalled by a special bit called a start bit.

b) The data bits are sent immediately following the start bit, then a parity bit if the data bits are protected by a parity bit and finally, the transmitter attaches a stop bit. Thus, the start bit and the stop bit frame a transmission of a few data bits encoding a single character, for example, the character ‘A’. The time interval for the stop bit allows the receiver to deal with the received bits, that is, transfer these into the RAM of the computer.

35 A communication protocol is a set of pre-agreed signals, codes and rules to be used for data and information exchange between computers or a computer and a peripheral device such as a printer that ensure that the communication is successful.

36 The handshaking protocol in serial data transmission involves exchange of the following signals between sender and receiver, for example, between computer and a printer.Assuming that the printer is present and switched on, the handshaking protocol for computer serial port to serial port of a printer is as follows:

The sending device, the computer, enquires whether the receiving device is ready to receive. The sending device waits for a response that indicates that the receiving device, the printer, is ready to receive. On receipt of this signal, the sending device coordinates the sending of the data and informs that receiver that it is sending the data. The sender then waits for the receiver to become ready to receive more data.

37 The parity bit is an extra bit added to each 7 or 8 data bits that are transmitted. This parity bit is calculated from the 7 or 8 data bits and is either a 1 or a 0 depending on whether even parity or odd parity is chosen.Even parity:The parity bit is set to either 1 or 0 so that the number of 1s across data bits and parity bit is an even number.

Computer

Computer

Computer

Printer

Printer

Printer

Clear to send pin 8 on printer

Start bit

Clear to send pin 8 on printer

Computer Printer Request to send pin 7 on computerYes, I am

Computer Printer Stop bitThat’s it

Are you ready?

Here it is

I am ready again

Computer Printer Clear to send pin 8 on printerBusy

0 0 1 1 0 0 1 1 0

Parity Bit Data Bits

e.g.Odd parity:The parity bit is set to either 1 or 0 so that the number of 1s across data bits and parity bit is an odd number.e.g.

38 The sending computer generates the correct parity bit and attaches it to the end of the data bits as they are transmitted. The receiving computer regenerates a parity bit from the received data bits, which it checks against the received parity bit. If the two are different, an error during transmission has occurred.

39 Baseband mode of operation dedicates the whole bandwidth of the transmission medium to one data channel at a time (a single channel system).

40 Broadband mode of operation employs the bandwidth of the transmission medium to carry several data streams at the same time (a multi-channel system).

a) Baseband tends to be used over short distances such as found in local area networks where high performance can be achieved with this approach at low cost.

b) Broadband is used for long distance communication because long-distance communications media are expensive to install and maintain. It would be wasteful if each media path could support only a single data stream

Exam practice questionsQuestion 1 – June 2005 CPT1 Q5Question number

Answer Marks

1a Name: Start bit;Purpose: Synchronise receiver;Name: Parity bit;Purpose: Perform parity check// check for errors in transmission;A Prevent errorsName: Stop bit;Purpose: Allow start bit to be recognised// Allow receiver to process received bits;A Indicates end of data

2

2

2

1b the number of signal/voltage changes per second;A rate at which signals are sent; A rate at which voltage changes;

number of bits per second/unit of time;R the rate at which bits are sent (question paraphrased)Range of frequencies a channel can handle;A maximum line speed; A maximum transmission speed

1

1

1

0 0 1 1 0 0 1 1 0


1 0 1 1 0 0 1 1 0


1c A signal can contain one or more bits;Bit rate can be higher than baud rate;bit rate = baud rate * number of bits per signal change

2


Answer Marks

2a Bits are sent along a single wire/line; bits are sent one after another;

R Bits of (the) data

Max 1

2b i Data bit;

Parity (bit);

Signal to start data transfer/strobe;

Signal ‘ready to receive data’/busy;

Signal to acknowledge data transfer/Complete;

Out of paper/ink/error ;

On-line/off-line;

Handshaking//control signal/status signal (BUT only if not by exampleabove);

Ground;ii Transmission over long distances;

When a high data transfer rate is required;

A No driver is available

Max 2

Max 1

2c Data is transmitted intermittently (rather than as a steady stream);

Sender and receiver are only synchronised when data is being sent //start bitsynchronises the receiver;

R Description only of start and stop bits

1


Answer Marks

3a Several bits are transmitted simultaneously/at the same time; down several wires; 1

1

A several bits of data; R dataR down two wires A diagram A a byte/word at a time;

3b Exchange of signals between devices/devices communicating with each other;To establish their readiness to send or receive data – a method of ensuring that both the sender and receiver are ready before transmission begins;

2


Answer Marks

4a) i) Baseband: Single/data signal sent at a time Or single message/packet/frame sent at a time Or uses single channel Or one transmission at a timeA Single stream of data;Over full bandwidth (of the cable) Or occupying full bandwidth (of the cable)Or signal uses all available frequencies;(ignore any additional references which are bits, 0/1 in any part of the above)R Single bits sent at a timeR Only works over short distances

4

4a) ii) Broadband: Several/data signals sent simultaneously Or several messages/packets/frames simultaneously Or more than one signal occupies bandwidth;Each at a different frequency or in a different channel Or in a different time slot;OrMultiple channels used;Each at a different frequency Or in a different time slot;Or signal (or equivalent) uses only one frequency Or signal (or equivalent) uses only part of the bandwidth;;R Fast connectionR Video, sound and text….R ADSL, cable examples, etc

4b) i) Two reasons:Wide area networks expensive to install;Wide area networks expensive to maintain;Wide area networks involve long distances;Can allow multiple data stream to keep costs down;

Max 2

Many channels needed to cope with high volume of trafficOr enables more users to use network without experiencing congestion;R FasterR Can work over longer distancesR More than one user will want to use it simultaneouslyR Cheaper, more efficient


Answer Marks

5a bits transferred simultaneously I concurrently; R data R bytesbits sent down many wires at the same time; A bits of dataA a clear diagram;

5b) i) data get skewed; timing of bits becomes different / out of line;A over longer distances the data may not be correct;A too expensive because of amount of wires/cables/lines;R signal decays R corrupted data

Max 1

5b) ii) use serial transmission

6.2End of topic questions1 a) Local Area Network or LAN consists of linked computers in close proximity.

b) A standalone computer is a computer that is not networked. It therefore requires its own peripherals, such as printer and application software, to be installed locally.

c) A wide area network is a set of links that connect together geographically remote computers.

2 The term ‘topology’, in the context of networking, refers to the shape, configuration or structure of the inter-connections connecting devices to the network.

3 An internet is a collection of local area networks and computers that are inter-connected by a wide area network

4 A computer communicates on the network through a network interface card or network adapter. A network adapter plugs into the motherboard of a computer and into a network cable. Network adapters perform all the functions required to communicate on a network. They convert data from the form stored in the computer to the form transmitted or received on the cable.

5 In the Ethernet protocol, each network card is assigned a unique address called its MAC address. MAC is an acronym for Media Access Control. A MAC address is a 48-bit address, for example 00-02-22-C9-54-13, expressed in hexadecimal and separated into six bytes. Part of the MAC address identifies the manufacturer. Each network card manufacturer has been allocated a block of MAC addresses to assign to their cards.

6 The collision problem occurs in bus networks when two computers transmit onto the bus at the same time. The pulses of voltages from each will eventually collide resulting in higher voltage swings. A computer attempting to read these pulses will fail to do so correctly. When this happens, a collision is said to have occurred and the bus becomes unusable for the duration of the transmissions from both computers.

7 Coaxial cable bus networks are segmented so as to reduce collisions and therefore improve the performance of the network. A network segment in Ethernet is a run of Ethernet cable, for example coaxial cable, to which is attached a number of workstations. Non-switched Ethernet bus networks, for example coaxial cable, are often split into smaller sections, called segments, in order to improve their performance. Performance in a non-switched Ethernet network can drop significantly as more stations are added to it. This is because in Ethernet,

the cable along which data travels is shared by all stations connected to it. If lots of stations have data to transmit, the network gets congested, and many collisions occur. Segmentation is one solution to congestion on an Ethernet network.

8 The central switch in a switched Ethernet local area network ensures that collisions cannot occur by forming a temporary unshared connection between sender and receiver. Frames sent by other computers arrive on dedicated cables and are queued until an unshared temporary connection can be made for these. The switch is thus making and breaking temporary connections between each sender and receiver at high speed. As there is never more than one frame travelling along each connection, collisions cannot occur.

9 In a thin-client network, all processing takes place in a central server, the clients are dumb terminals with little or no processing power or local hard disk storage.

10 A thick- or rich-client network is one in which the applications are run in the client workstations. Therefore, client workstations must have local processing power.

11 Bus and star topologies appear very similar in the way that they are physically wired using the current switch-based hardware. Even thin client systems, which can be considered to resemble a traditional star network, use an Ethernet bus switch to connect a central server to nodes. In a traditional star topology network, each link from node to central computer is an independent link. Each link is therefore secure from eavesdropping by other nodes. If a link to a node goes down, the other links and nodes are unaffected. However, if the central computer goes down, then the whole network will fail. In a true star-based network, the speed of each link to the central computer should remain high because the links are not shared. Traffic between nodes in a switched-based bus network will not be adversely affected if a node goes down unless, of course, the traffic involves the broken node or the node is a domain server that validates users when they attempt to log in. Neither will unplugging a network cable in a switched-based bus network affect the rest of the network. In a coaxial cable bus network, a break in the cable stops the whole network from working. All connected nodes are able to read the frames travelling on the coaxial cable bus network. Therefore, coaxial cable bus networks are not secure against eavesdropping. The frames in a coaxial cable Ethernet bus networks can collide when multiple nodes send at the same time. This results in the network showing a noticeable slowing of performance. Although collisions between frames in switch-based Ethernet bus networks cannot occur, performance can be affected when traffic volumes are high because the buffers in the switches suffer overflow. A wireless network is a broadcast network and so is less secure than a cabled switch-based Ethernet network unless wireless encryption is enabled. With a wireless-based network without encryption, it is possible to eavesdrop on traffic intended for other computers. A wireless network can also suffer congestion because the channels are shared. a) A peer-to-peer network (P2P) is one that has no dedicated servers. All computers are

equal and therefore known as peers.b) A server is a computer that provides shared resources to network users.c) A client is a computer that uses the services provided by a server.d) In server-based networking resource security, administration and other functions are

provided by dedicated servers.e) In Web 2.0, software becomes a service that is accessed over the internet.f) Web services are self-contained, modular applications that can be described, published,

located, and invoked over a network, generally, the web.

g) Software as a Service (SaaS, typically pronounced 'sass') is a model of software deployment where an application is hosted as a service provided to customers across the internet.

h) In web services architecture, all components in the system are services.i) Ajax is a web technology that allows only the part of a web page that needs updating to

be fetched from the web server.j) A wireless network refers to any type of local area network in which the nodes

(computers or computing devices, often portable devices) are not connected by wires but instead use radio waves to transmit data back and forth between the computers.

k) Wi-Fi is the trademark for various IEEE 802.11 technologies that support wireless networking of home and business networks.

l) Bluetooth is a wireless protocol for exchanging data over short distances from fixed and mobile devices.

12 Google’s search engine is an example of Web 2.0. Google began its life as a native web application that has never been sold or packaged for its customers. Instead, the web application, Google’s search engine, is delivered as a service, with customers paying, directly or indirectly, for the use of that service. None of the trappings of the old software industry are present. There are no scheduled software releases, just continuous improvement. There is no licensing or sale, just usage. There is no porting to different platforms so that customers can run the software on their own equipment, just a massively scalable collection of commodity PCs running open-source operating systems plus Google-written applications and utilities that no one outside the company ever gets to see. Google’s data centres where it all happens are closely guarded and visitors are not welcome. The software never needs to be distributed but only performed. Google's service is not a server – though it is delivered by a massive collection of internet servers. Neither is it a browser, though it is experienced by the user within the browser. Nor does its flagship search service even host the content that it enables users to find. This content is located on the web in the web pages. Google’s software finds the links in these web pages and using a very clever algorithm arrives at a page rank index for pages on the web.

13 A router is a device that receives packets or datagrams from one host (computer) or router and uses the destination IP address that they contain to pass on the packets, correctly formatted, to another host (computer) or router.

14 A gateway is a device used to connect networks using different protocols so that information can be passed from one system to another successfully

15 When a computer, X, in the English local area network (LAN) wishes to send to a computer, Y, in the Australian LAN, it knows immediately that Y is on a different network. It therefore sends to the inward facing port of the gateway that it is directly connected to. This gateway then reforms the frame so that it is compatible with the internet and sends this frame on port B to the internet. The internet is a collection of routers. The public IP address of Y’s LAN is used by the routers to route the wide area network frame to the other side of the world. When the frame eventually arrives at the outward facing port of the gateway for the Australian LAN, this gateway reforms the frame before sending it into the LAN via the inward facing port. Both gateways have two network cards, one for the outward facing port and one for the inward facing port.

First Hop

Second Hop

Third Hop

Exam practice questionsQuestion 1 – June 2002 CPT5 Q4 b) ii) and iii)Question number

Answer Marks

1a First mark: (More packet) collisions take place; Stations attempting to send at the same time; Each station broadcasts to every other one; Some stations may be attempting to broadcast at the same time;Second mark: Packets need to be sent again;Station has to retransmit (after a random delay);

Max 2

1b First mark: Exclusive bus connection (made temporarily between sender and receiver)Or data transfers take place in turnsOr switch connects just sender and receiverOr collision domain limited to two stationsOr switch splits bus LAN into several smaller segmentsOr switch allocates a time slot to each transmissionOr each host/computer/station/workstation/node has its own link on which one packet at most can travelOr switch allows a dedicated connection/pathway to be set up when a computer wishes to send information to anotherOr switch separates work stations (in different segments) and only

2

passes packets between segments when necessaryOr switch (ports) act as bridges which segment network;Second mark: therefore collisions cannot occur (between two stations) Or collisions reduced


Answer Marks

2a A set of rules/procedures 12b Bus; R Ethernet on its own 12c Twisted pair//coaxial (cable)//optical fibre//fibre optic; 12d Need first octet or first and second octet or first, second and third octet to be

identical. Also must have four octets.For example:192.168.0.1192.168.0.2One mark for four octets;One mark for same LAN;

3

2e (Use candidates example from D)i 192.168.0;ii 1 or 2;

1

2f a (unique) address/identifier assigned to network card // (unique) hardware address/identifier;

1

2g Any two tasks @ one eachAllocation of port numbers;Routing a packet/frame/segment to correct application/service;Splitting messages/data into packets // Disassembling messages // Assembling packets;Adding TCP headers // Adding sequence nos;Error handling // sets parity bits;Checking that transmission successful;Resending transmission if necessary;A Sets packet size

2

2h Any one of the following applications for one mark;Telnet;Internet browser;http (client) // web server;

1

email;FTP;TFTP;SMTP;R Non-networked applications such as word processor

2i Internet registry // Internet registrar;A I.P. registry/registrar

1

Question 3 – 2010 Specimen question paper COMP3 Q5Question number

Answer Marks

3a A 192.71.0.1; A any valid number instead of 1B 192.71.0.2; A any valid number instead of 2C 192.71.1.1; A any valid number instead of 1

3

6.3End of topic questions1 The common gateway interface or CGI: this is a gateway between the web server and a

web-server extension that tells the server how to send information to a web-server extension, and what the server should do after receiving information from a web-server extension.

2 A web-server extension: this is a program written in native code, that is, an executable or a script that is interpreted by an interpreter running on the web server that extends the functionality of the web server and allows it to generate content at the time of the HTTP request.

3 The common gateway interface or CGI is a gateway between the web server and a web-server extension. The CGI specification tells the server how to send information to a web-server extension, and what the server should do after receiving information from a web server extension. In its simplest form, the gateway consists of two objects, the request object and the response object. An HTTP request sent by a web browser such as Get / is placed in the request object in the gateway. The web-server extension accesses the request object and processes its contents before formulating a response, which is placed in the response object in the gateway. The web server then accesses the response object to pass its contents back to the web browser.

4 Get /webpage.asp?myname=fred&age=65 When the post method is used by the browser or an HTTP application then the data, for

example myname=fred&age=6, is passed in the message body while when the Get method is used the data is passed with the command in the address bar of the browser as shown in the answer to question 4.

6 Dynamic web page content means content that is generated at the time of receipt of the web browser request.

7 A script is executed on the server to determine the parts of the web page that are dynamically created at the time of receipt of the request.

8 www.educational-computing.co.uk?surname=bond&age=69 Perl, php, VBScript10 Internet service providers (ISPs) are able to read an interpreted server-side script because

it is high level and therefore can monitor scripts and prevent scripts that could be harmful from executing. This is not the case with server-side executables which are compiled code presented in low-level form, making it difficult for ISPs to understand whether the executable could be harmful.

11

12Create a connectionSelect a databasePerform a database queryUse returned data, if anyClose connection

13 Delphi, php, VBScript

Exam practice questionsQuestion 1 – January 2003 CPT5 Q7b(i) and (ii)Question number Answer Marks1b) i) A Fred;

B James;11

1b) ii) A Fred;B James;

11


Virus Name

When Type of attack

Type of damage Number of computers affected

Melissa virus

1999 Macro virus

A document was created with the virus in it and anyone who opened it would ‘catch’ the virus. The virus would then send itself by email to the first 50 people in the person’s address book. This made the virus replicate at a fast rate

Thousands of computers worldwide. Melissa shut down internet mail systems that got clogged with infected emailsThe creator of this virus was jailed in 2002

2005 Cross-scripting virus

a malicious website could load another website into another frame or window, then use Javascript to read/write data on the other website

MySpace and Yahoo sites affected

Love Bug 2000 worm flooded the internet with emails with the subject, ILOVEYOU. The body of the deceptive email read, "Kindly check the attached love letter coming from me." When opened, the e-mail wreaked havoc on computers, replicating it automatically, sending copies to everyone in the user’s address book, and damaging computer files, such as MP3s.

affected 80 percent of businesses in Australia and the United States

http://en.wikipedia.org/wiki/Internet

Code Red 2001 worm operated in three stages – scanning, flooding and sleeping. During the scanning phase, the worm searched for vulnerable computers and ran damaging computer code on them. Next, in the flooding phase, the worm sent bogus data packets to the White House website. The White House, however, changed their website’s IP address and was therefore able to avoid the attack.

Blaster 2003 worm wreaked havoc on Microsoft XP, Windows NT 4.0 and Windows Server 2003 users. The worm spread quickly, checking for vulnerable computers and then sending itself to those machines. The worm was intended to attack Microsoft’s update website. Some users found that their computers were sluggish, but otherwise may have been unaware that they had been infected.

2 Here is what one bank says:

- We’ll NEVER send you time sensitive emails to confirm or change your security details, account numbers, card numbers, PINs or expiry dates. - We’ll NEVER ask you to input ALL of your security when logging onto internet banking.- We’ll NEVER send you an email containing a link to a log in page.

You probably get mountains of email every day. The chances are a fair bit of it’s unsolicited but comes with halfway plausible return addresses.

More and more of these emails contain attachments carrying viruses, worms, or Trojan horses which will harm your system if they go undetected.

We’re also seeing a lot more ‘phishing’ emails. This is where a mail is sent from a third party, saying they are from your bank, asking you to click on a link to confirm/update your security information. The page you get directed to usually looks exactly like the bank’s page and when you enter your account and security information, you may get forwarded to the real login page, or have to click another link to login to your account.

What really happens is you actually send your account and security details to someone who wants to get away with your money.

Some junk email uses social engineering to tell you about a contest you’ve won, or details of a product you might fancy. Pretty obviously, the sender wants you to interact with the email in a way that’s financially beneficial to them.

Our emails never ask for personal details. The only time we’ll need those is when you’re securely logged on to our banking or enrolment.

But if you do get mail asking for personal information it’s best to delete it right away. There could be a dodgy dealer on the other end.

Please DON’T respond to any such email. Remember - we would never ask a customer to re-register their security details. And we’ll never ask for your security details until you’re safely logged into your account on our website.

The best advice we can give for emails you’re unsure of is, don’t open them. If you use an email program like Outlook, Eudora or Notes, it’s a good idea to turn off your ‘Preview Pane’ too. Finally, if you receive an attachment you weren’t expecting, or you don’t know whom it’s from, don’t open them. If you haven’t received an email virus or worm yet, you will. Sorry, but it’s only a matter of time.

Every time you receive a message with an attachment, test it with these criteria.

The Know test: Is the email from someone that you know? The Received test: Have you received email from this sender before? The Expect test: Were you expecting email with an attachment from this sender? The Sense test: Does email from the sender make sense? For example, would the sender - let’s

say your mother - send you an email message with the Subject line, "Here you go, ;o)" containing a message with the attachment AnnaKournikova.jpg.vbs? A message like that probably doesn’t make sense. In fact, it’s an instance of the Anna Kournikova worm and reading it can damage your system.

The Virus test: Does this email contain a virus? To determine this, you need to install and use an anti-virus program. Alternatively, use an email service that checks viruses for you.

Some email systems provide additional benefits at no extra cost. Hotmail uses the McAfee scan to look for viruses in your email attachments before they send you it. If the email looks like it contains a virus you will be given a warning before you open it.

3 Strong passwords use combinations of uppercase and lowercase letters, numbers, and punctuation, they aren’t usually found in any dictionary. For example using ‘river’ would be a weak password, whereas ‘r!V3r_78’ would be much stronger.

Exam practice questionsQuestion 1 – January 2007 CPT5 Q5 Question Number

Answer Marks

1a If you send the key with the message, anyone can decrypt the messagekey would need to be sent by means other than email, otherwise anyone could intercept the key and use it do decrypt the message;

1

1b Jill.s public key;Jill.s private key;

11

1c the message data is hashed into a message digest; the message digest is

encrypted; with the sender.s private key;

Jill’s software decrypts the signature; using Jack’s public key; contained in digital certificate sent with message; to verify Jack.s public key; decrypt digital certificate using Certificate Authority.s (trusted third party.s) public key; Jill’s software then hashes the document data into a message digest; If recalculated message digest is the same as the original message digest (decrypted signature); then Jill knows that the signed data has not been changed;

3

Max 4

I decryption of message

Question 2 – January 2006 CPT5 Q7 Question Number

Answer Marks

2a converting/transforming from plain text into ciphertext/secret code;A scrambled; A transposition / conversion / codingthe sender processes the message prior to transmission so that if it is accidentally ordeliberately intercepted while it is being transferred it will be incomprehensible tothe intercepting party;Data coded so that unauthorised users can.t read or access the data;

max 1

2b b) i) B’s public key;ii) B’s private key;

11

2c i) a hashing function is applied to the text of the message;the result/message digest is encrypted;using B’s private key;A the data generated is added to the end of the message;A message/date stamp is used to produce digital signature;

ii) A uses Certificate Authority’s public key;to verify B’s public key;digital signature is decrypted;using B’s public key;the hashing function is applied to the text of the message;the result of the hashing function is compared with the digital signature;if they are the same the message is authentic;

max 3

max 4


Answer Marks

3a E-mail may pass through many computers/servers if it travels over a network, each computer can make a copy/can be accessed;When a message arrives at its destination, it waits until the intended

recipient picks it up. During this time the message is vulnerable to being read or copied by the computer’s operator;Electronic eavesdropping of telephone wires and local area networks is possible;With e-mail alterations lave no trace (no physical damage) wheras with paper alterations leave a physical mark;

Max 13b) i) E-mail encrypted using public key;

Recipient’s private key used to decrypt e-mail;2

ii) E-mail encrypted by sender using private key;Recipient decrypts e-mail using sender’s public key;

2

7.1End of topic questions1 The Humber Bridge solved the problem for commuters of getting from Hull to Grimsby across the Humber River, thus reducing the distance that they had to travel considerably.2 Any three from:

a) The solution must cope with a volume of traffic crossing the bridge per hour up to … b) The solution must cover a distance across water of … c) The solution must be capable of supporting a total weight of … d) The solution must connect to the existing road network in the area.3 a) a ferry between the south and north shores and vice versa

b) a hovercraft between the south and north shores and vice versac) a bridge between the south and north shores and vice versa

4 because the requirements were not precise enough for the bridge designers to work with.5 problem definition

feasibility studyanalysishigh-level designlow-level designimplementation and testingmaintenance

6a) What is the proposed system to do?What are the problems with the current way of doing things?What data/information is recorded in the current system? How much data is recorded at present? What data/information is to be recorded in the proposed system?

How much data will the proposed system record? How frequently will the data need to be updated? Will new records need to be added or old ones deleted? How often? Will the changes come in batches, or in ones and twos? How important is the data/information that is recorded?What processes or functions are performed by the current system? What processes or functions are to be performed by the new system? When should these be done and where? What special algorithms do these processes use, e.g. calculation of compound interest? Which processes should be executed manually?What are the inputs to the current system and what inputs will be required for the proposed system? Ask to see any input documents that are used in the current system.What are the outputs from the current system and what outputs will be required from the proposed system? Establish if hard copy output is required. How often will outputs be required? Ask to see some output from the current system.What computing resources does the end user possess?Is the end user prepared to purchase software/hardware resources?Is security an issue? Should there be limited access to some or all parts of the proposed system?How are exceptions and errors handled in the current system? What errors and exceptions should be reported in the proposed system? How should they be reported? Should anything else be done?Are there any constraints on hardware, software, data, methods of working, cost, time, and so on?Does the user have a particular solution in mind? Do you have some suggestions to make to the user?

b) renewal subscription lettersreminder lettersmembership listsmember’s recordmembership categories list and feespayment records

c) Interviewing staff – facts can be gathered directly from the people who have direct experience of the present system. Full and detailed answers can be obtained by pursuing particular lines of questioning.Examining existing paperwork, documentation, records and procedure manuals for the current system – can be used to identify the data that is used in the current system, the information that is produced by the current system and the procedures that are carried out.

Using a questionnaire to survey opinions of staff enables the same set of questions to be asked to many people. A carefully designed questionnaire can be a very quick and cheap way to obtain specific answers to specific questions from a large number of people.Observation of operation of the current system enables current methods of working to be examined and necessary exceptions to the normal pattern of working to be noted.

7 A prototype is an early or trial working version of the proposed system developed to test possible solutions.Prototyping is used to (any three): clarify requirements

perform risk analysis

find a solution to a particular problem

check whether a solution can handle the workload

test a solution within the proposed environment.

Through a process of refinement to progress, in conjunction with users, to a final working system. This process is called evolutionary, iterative development:

to discover errors in design

to discover problems.

8 Once one phase is complete, it is not revisited. The customer doesn’t get to see the developing product. Therefore, it is more difficult to discover, and expensive to fix, errors in a previous stage. For example, analysis can be incomplete or wrong in places because there is little opportunity to evaluate the consequences of the analysis and also as the project progresses, requirements may change. The customer doesn’t get to see the product until it is ready for delivery. The communication path between the customer and the developers is very long, that is, indirect, because developers talk to analysts who talk to customers. This means that there is greater potential for the developer to misinterpret the customer’s requirements. There is no opportunity for the customer to correct the developer’s error. The waterfall model communicates by written documents forwards through the cycle but no analyst can express the requirements that they know about 100 per cent completely and correctly through writing. Customers express themselves in everyday English whereas developers use technical language. The lack of interaction between customer and developer in the waterfall model may result in a gulf appearing between the developer and the customer which may result in imperfect communication and a product that is far removed from what the customer wanted.

9 In this model, ‘water is allowed to flow up hill’, that is, the analysis, design and implementation phases can be revisited, which is not the case with the waterfall model.

10 The spiral model is more appropriate because you are not yet a professional software developer and you will be working on your own.

11 Volumetrics refers to measuring/assessing the volume of data that a system will be required to process and store. Analysis needs to provide developers with volumetric information in order for the developers to make appropriate design decisions.

12 It is important that the set of objectives is agreed with the end user(s)/customer. However, it is very easy for the analyst and the end user(s) to overlook something important that should be included. A list is not always the best way of visualising what it is that needs to be done. Analysis must also try to represent what needs to be done pictorially. One way that this can be done is using a data flow diagram.Context diagramLevel 1 DFDLevel 2 DFD

13 Business algorithms are algorithms supplied by the end user. These are algorithms used by the existing system or would be used if a system existed. An example is an algorithm for calculating interest on an investment, for example, if a customer invests £100 how much interest would be earned in the first year of the investment and in subsequent years?

14 At least 10:Bubble sortInsertion sortShell sortMerge sortHeap sortQuick sortBucket sortRadix sortDistribution sortShuffle sort

15 a) specificationb) designc) specificationd) designe) specification

16 association, aggregation (composition/containment), inheritance. Inheritance class diagram shows the relationship between two object types in which one object type is a kind of the other object type, for example, a car is a kind of vehicle.Aggregation or composition or containment class diagram shows the relationship between two object types in which one object type has a component which is of the other object type, for example, a car has an engine or contains an engine.Association class diagram shows the relationship when it is not inheritance and it is not aggregation, for example, a snake uses the ground to get from A to B. The relationship is just an association because a snake is not a kind of ground and a snake does not contain ground. It is, in fact, a uses relationship.

17 Analyst doesn’t fully understand the problem to be solved and therefore does not provide a clear statement of the problem to be solved. Consequently, a different problem is solved.

Analyst fails to obtain a complete picture of the customer’s needs/requirements.

Analyst fails to specify in the clearest terms the customer’s requirements to the customer so customer agrees to the development of a solution that is not exactly what they want.Analyst fails to communicate to the developer precisely what the system must do and therefore the developer develops a solution which doesn’t match the customer’s expectations.

Exam practice questionsQuestion number

Answer Marks

1 Problem definition, feasibility study, analysis, design, implementation, testing, maintenance.

3

2a Any three from: Interview, survey, observation, examination of the paperwork.

3

2b 1. InterviewingInterviewing is useful because facts can be gathered directly from the people who have direct experience of the present system. Full and detailed answers can be obtained by pursuing particular lines of questioning

2. Examination of existing paperwork

Examination of the existing paper work, documentation, records and procedure manuals can be used to identify the data that is used in the current system, the information that is produced by the current system and the procedures that are carried out.

3. SurveyQuestionnaires enable the same set of questions to be asked to many people. A carefully designed questionnaire can be a very quick and cheap way to obtain specific answers to specific questions from a large number of people.

4. Observation Observation of the current practice enables current

methods of working to be examined and necessary exceptions to the normal pattern of working to be noted.

3

3a Specification 1

3b Design 1

3c Design 1

3d Specification 1


2

7.3Exam practice questionsQuestion 1 – June 2008 CPT5 Q1eStartDateTime EndDateTime Mileage OverdueHours Normal Erroneous Boundary

01/12/07 06:00 01/12/07 15:30

15 2 1

16/12/07 18:00 12/12/07 09:00

237 3

04/12/07 23:00 04/12/07 08:30

5 2 1

03/12/07 08:00 03/12/07 09:00

0 0 1

01/12/07 06:00 01/12/07 15:30

0 1.5 1 I

01/12/07 06:00 01/12/07 15:30

0 -2 1 I

04/12/07 08:30 05/12/07 23:00

57 0 1 A

01/12/07 06:00

01/12/07 15:30

15 3


Answer Marks

2a normal data: accept valid dates within 120 years before present (i.e. 23/01/2008)birthday sometime before today’s date;birthday sometime after today’s date;reason: to check the routine takes into account whether birthday has already been or not;A 29/2 as a special day to test for

A day/month instead of birthday

The data values should be any dates within 120 years before the present. If the reason is just given as ‘they are normal dates’ when both dates are before or after 23/1 of any year then 2 marks max.

3

2b boundary data:a birthday just before boundary;a birthday just after boundary;

4

a birthday exactly on boundary;A ‘yesterday’, ‘today’ ‘tomorrow’

reason: to check that age is calculated exactly, taking into account whether birthday is past, now or future;// this is the oldest you can be // this is the youngest you can be;

22/1 –24/1 Any year within 1887–20082c R anything that is not valid date format;

erroneous data: a date after today’s date;reason: can’t have an age for someone not born yet;orerroneous data: a date which makes a person over 120 years old;(i.e. Before 24/1/1887)A ‘tomorrow’;

reason: no person expected to be over this age;A outside any expected values // outside range;

Remember that wrong formats or invalid dates are not acceptable answers. Give mark for correct reason even if no acceptable date value given.

2


Answer Marks

3a beta testing: a test prior to commercial release// testing that involves sending the product to beta test sites outside the company for real-world exposure;must convey external testing

A the last stage of testing;

1

8

R error messages3b factors to help maintenance: 1 mark per factor (with some relevant

explanation) to max 3

structured program code // no GOTO statements // using iteration/selection constructs appropriately;

local variables; procedures/functions; with interfaces/parameters;modules/units;

layout/indentation/white space;

meaningful identifier names; self-documenting code // comments/annotation;

object oriented programming // use of classes;

use of pre-tested routines/library routines; error logs;

3

3c types of testing. Any 2 of:

system/integration testing;

functional/black box testing;

structural/white box testing;

acceptance testing;

2

unit/module/subsystem testing;

Alpha testing;

A Dry running/walk-through/inspection testing; R bottom-up/top-down testing.


Answer Marks

4 any normal/typical/valid data;: 1<=Hours ,=35; A 0,= Hours ,= 48boundary/extreme values;: eg Hours = 35, 36 & Hours = 0, 48, 49; A -1test that it correctly calculates premium rate;: Hours >35;any erroneous/invalid data; any negative value; any non-integer data; any value over 48;

1 mark for test data, 1 mark for justification

If justification does not match test data, then 1 mark max.

6


Answer Marks

5 1 mark for test data, 1 mark for justification for 3 sets of test data£1, 50p, 20p, 10p, 5p; simple change of a single coin;£3.85, 15p, 25p, 30p, 35p, 60p, 65p, 70p, 75p, 80p, 85p; change made up of one of several coins;40p, 45p, 90p, 95p; change made up of more than one of the same coins: 2x20p;0p; boundary data: no change / zero coins;5p; minimum change that can be given;£1.95; maximum change that can be given // extreme/boundary value;a negative amount; although the routine that calculates how much change is due should not allow erroneous change, this routine should still test for erroneous input;3p; an amount that is not a multiple of 5p // erroneous data;not all values need to be listedR justifications not referring to scenario

6

R answers which seem to test the coins inserted f the calculation of change

Question 6 – June 2006 CPT5 Q4Question Number

Answer Marks

6 Any two points at 1 mark each:

Bugs/Errors/Mistakes in software/system/code/program/it;Problem NE R data errors (T.O.)Requirements change // adding new tasks;Parameters change e.g. VAR rate, No of users adjusted, No of licences change;Performance needs tuning // buffer size needs adjusting // indexing needs to be switched offOr on // indexes need to be rebuilt;Hardware is changed;Software / system is updated // upgrades;Adaptive/Corrective/Perfective maintenance not enough without explanation

“Keeping up to date” NE

2

Question 7 – January 2006 CPT5 Q1Question Number

Answer Marks

7 Top-down testing;Bottom-up testing;Black Box testing;White Box testing;Dry run/walk-through;Unit/Module testing; A Prototyping;

R Integration/Acceptance/Alpha/Beta/System/Performance/Compatibility testing;R anything clearly late in the development cycle;

Mark first 3 responses onlyBUTbeware of expansion on same line.

3


Answer Marks

8 Any two at 1 mark each:Top-down (testing);Bottom-up (testing);White Box (testing);Integration/Integrated (testing);Interface/Stub (testing);Acceptance (testing);Alpha (testing);Beta (testing);Walkthrough/Dry run/Tracing/Desk checking:Compatibility (testing):Performance (testing):System (testing);Unit/Module (testing);R. Functional (testing)R. Structural (testing)R. Test cases and examples of test cases

Max 2


Answer Marks

9a N.B. Must be an automated system (=control) or a system that provides real time data (=monitoring)Flight control software;Software controlling life support systems;Software controlling hazardous materials with potential for exposure to humans;Software controlling mechanical equipment which could cause death through impact/crushing/cutting;Any software which provides information to operators where an inaccuracy or misinterpretation of the data could result in death/injury through an incorrect decision;A Air traffic control, railway signalling system, traffic lights, heart rate monitor, drip feed controller for administering drugs.

Any two at one mark each2

9b) i) Acceptance testing is specified/performed by customer (against original specifications);

1

9b) ii) Poorly/incorrectly specified system//inadequate/inaccurate systems analysis;Poor training of staff using system//staff use system incorrectly;Situation outside specification occurs or example which relates explicitly to specification or similar, e.g. more users attempt to log on than should;A Virus has entered system//malicious misuseR Design flaws, hardware failure, inadequate testingR Data corrupted

1

9b) iii) N.B. Name requiredSystem testing;

Any one for one

Alpha testingBeta testing;Performance testing;A Black Box testing, White Box testing, integration testing, unit/module testing, top-down testing, interface testingR Phased testing, bottom-up testing

mark1


Answer Marks

10a) i) Strategy: White Box testing;Justification: Need to follow a path through the code (to discover the error);

2

10a) ii) ii) While r >= y 1 mark for identifying(in some way) While statement,1 mark for correct condition r >= yR. q := 0 changed to q := 1

2

10b r never changes or an alternative that describes/implies the same thing;A. gets stuck in an infinite loop

1

10c If y > 0 Then x = y * (result q) + (remainder r);R. y cannot equal zeroA. y must always be greater than zero

1

10d) Acceptance is specified/performed by “customer” ;against original specifications

1

7.4Exam practice questions

Question 1 – January 2006 CPT5 Q5 c) and d)Question number

Answer Marks

1ae) Method1 mark

Justification1 mark

Parallel To check against the pupil’s food diary to ensure the data collected is correct;A if new system fails, they have the old one to fall back on;

Direct It would take too long to check against pupil’s food diaries// not a critical system // cheaper than parallel // cheaper than running two systems;

Pilot The new system may not work very well and should be

2

piloted in one school first (before being rolled out to all schools in the LEA);

Phased Use the data collection module before the diet checking one, in case the system refuses to allow acceptable choices, causing chaos at the checkouts;R explanation of pilot

1b data files will have to be entered into the system; R conversion/transfer of data

users / cafeteria staff / pupils will have to be trained; R printing labels

new hardware has to be installed/purchased; new software has to be installed;

I licences

smart cards will have to be updated; pupils will have to be issued with smart cards;

define/create new operating procedures

A produce documentation/user guide/system guide

3


Answer Marks

2a i. Old system and new system operate alongside each other/in parallel until new system proved; (require time limited trial)

ii. Parts of old system gradually replaced in stages by new system;Do not give mark for a pilot

1

1

2b Information/Data/Files may have to be converted/copied/reformatted/modified so that compatible with new system;

Users/operators will have to be trained so that they can use the new system//Staff needed/hired to maintain new system;

Old data archived;Make full backup before changing to new system;Hardware replaced/upgraded;System software replaced/upgraded;

2

2c NB Emphasis is on changes not performance

Is it possible/How easy is it/How long will it take to correct an error in software;Is it possible/How easy is it/How long will it take to change parameters in system, e.g. VAT rate;Is it possible/How easy is it/How long will it take to change system to cope with more users;Is it possible/How easy it is/How long will it take to change system to cope with more terminals/workstations;Is it possible/How easy is it/How long will it take to change system to cater for more software licences;Is it possible/How easy is it/How long will it take to change system to work with different hardware;Is it possible/How easy is it/How long will it take to update/upgrade system;Ho extensive is support documentation;What is skill level of support staff;What is availability of support staff;Can operators/users configure system/change settings;How long will support be available for;Is source code available;

A One reference only to documentation;R How easy is it to add new features/expand system;R What is standing of developers;

3

7.6End of topic questions1 Evaluation is systematic assessment of whether something meets its objectives or

specifications and how well it meets the latter in terms of effectiveness, useability, maintainability.

2 Effectiveness in meeting requirements: learnability, useability, maintainability. 3 In addition to requiring criteria for judging effectiveness evaluation of effectiveness requires:

a) planningb) planning timec) execution time.4 Monitor number of registered users; monitor number of stored files; monitor frequency of file

uploads/downloads; monitor frequency of use by registered users.

5 Observe users – class of 7-year-olds – using the system: measure improvement in word spelling as software is used; note how quickly 7-year-olds learn to use the system; note how easily the 7-year-olds navigate the software; note how long 7-year-olds take to step through the instructions and to answer.

6 Useability refers to the ease with which a User Interface can be used by its intended audience to achieve defined goals.

7 Target acquisition time Latency Readability Use of metaphors Navigability

Target acquisition time: Latency, e.g. moving the mouse pointer to a position over a button.a) Latency: How long does a user who has initiated an action have to wait before the action is

completed, e.g. a button is clicked to execute an SQL statement that returns a dataset matching the query?

b) Readability: How easy is it for a user to read commands/guidance text?c) Use of metaphors: Visual pictures/icons to enable users to intuitively grasp what can be done.d) Navigability: How easy is it to navigate the user interface?8 Maintainability of software refers to how easy or difficult it is to fix bugs, change parameters

and respond to changing requirements.9 Any four from:a) Is the software modular?

o Procedures/Functions are used that perform a single task and are self-contained.

o Collections of related procedures are placed in independent units that can be compiled separately.

o Information hiding is supported by the use of separate units.

b) Is the code self-documenting?o Are meaningful identifier names used for variable names, procedure/function names, etc?

o Is indentation used to indicate the structure of the code.

c) Blocks of code have one entry point and one exit point.d) Is the use of global variables across units minimised?e) Is good use made of local variables?f) Are parameter values passed in and out of procedures through procedure interfaces?g) Is an object-oriented programming approach used?

o Is inheritance used to allow functionality to be extended?

o Is polymorphism used to offer different functionality depending on object type?

h) Is the database created from a DDL script (which itself has been produced using a CASE tool) so that re-engineering the database can be done using the CASE tool?i) Does querying of the database use SQL?j) Is the code commented where clarification is needed?

k) The documentation that accompanies the software is also essential to the maintenance of this software. Evaluating maintainability of software requires consideration of the following:l) Is the project supported by analysis and design data dictionaries?m) Are entities and entity-relationships documented?n) Is the software accompanied by:

o dataflow diagrams

o system flow charts

o other relevant diagrams?

o) Are algorithms documented?10 Evaluation should take place early in the project when prototyping for the purpose of

clarifying the customer’s requirements. Evaluation should take place when designing so that the effectiveness of different approaches can be made before committing to one particular design. This should include types of interface. When a particular user interface type has been settled on, evaluation of different mock-ups of this user interface should be made. This is still part of the design stage. Evaluation should take place to assess useability and navigability of the user interface for a range of users. This takes place during the later stages of implementation when the design is fairly stable. When the final solution is produced further effectiveness evaluation is necessary to confirm that the design has achieved what it set out to achieve.

Nelson Thornes is responsible for the solution(s) given and they may not constitute the only possible solution(s).

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