a level computer science revision pack 02 – algorithms and
TRANSCRIPT
A Level Computer Science Revision Pack
02 – Algorithms and Programming
The mark scheme for each paper follows the questions
Included (in order of appearance) 2019 2018 2017
How to revise Computer Science Practice questions from past papers are one of the best methods of revising topics from the course. This approach, accompanied by creating notes and reading the course textbook as a source for information, has proven successful for many of our previous students. How to revise a particular topic this is generic and by no means a one size fits all approach
1. On a single sheet of A4, write down everything you currently know about the topic. Do this prior to reading the course textbook or seeking help from previous notes.
2. Now consult course textbook for the topic and add to this sheet, anything you did not know that is necessary – once complete, highlight these points – these are the areas you need to learn.
3. Locate questions based around this topic in the past paper pack and attempt to answer them.
4. Confirm with the mark scheme as to your success in answering the question. The end goal of this approach would be that you are comfortably able to produce a piece of A4 for each topic of the course and then apply this information to the past paper questions. Obtaining feedback for answers The students who succeed the best in computer science are those who seek constant feedback from teachers, not just in the scope of a lesson. Any work you produce out of lesson such as past paper question answers or programming challenges, you should want to seek feedback for. This can be achieved by:
1. Taking work to a teacher during school time. 2. Emailing a teacher your answers, questions etc.
Mr Ravenscroft – [email protected] Mr Ebrahim – [email protected]
As your teachers we want to give you feedback!
Study Skills and Support
Exam board: OCR Course length: 2 years How is it assessed? 2 written exams on 01 – Computer Systems and 02 – Algorithms and Programming (each worth 40%) and a programming project worth 20%. Modules covered: 01 – Processors, Input – Output and Storage, Systems software, software development, compression, databases, networks, web technologies, data types, data structures, Boolean algebra, morals and ethics. 02 – Thinking abstractly; ahead; procedurally; logically; concurrently, programing techniques, computational methods, algorithms.
Resource Link Useful For… Requirements Course Textbook
N/A Independent revision & study
Course textbook from the school library.
YouTube YouTube Knowledge booster, second voice
N/A
Past Paper Packs
N/A Exam style question practice, independent study
Past paper pack from class teacher
Departmental resources
All stored within the Microsoft Teams Team for the group.
Accessing departmental materials and lessons
School email and password login.
Mr Fraser www.mrfraser.org Accessing resources and work sheets
mrfraser.org login account (free to create)
Craig n Dave craigndave.org
Resources for topics – broken down by spec
Access is free for most content – school has a paid account
AQA Past Papers
Search ‘AQA A Level Computer Science Past Papers’ on Google.
Different phrasing of exam style questions.
N/A
Class Teachers [email protected] [email protected]
Familiarity with assessment objectives is necessary in exam answers to ensure
application and evaluation
Embed and commit knowledge and understanding to long term memory for
examination recall
Develop awareness and appreciation for the use of computer science in the
wider world; which can be included in extended answers
Develop understanding of programming beyond the curriculum to enable
success in the programming project
Purp
ose
of
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tudy
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OCR is an exempt Charity
* H 4 4 6 0 2 *
Oxford Cambridge and RSA
Tuesday 11 June 2019 – MorningA Level Computer ScienceH446/02 Algorithms and programmingTime allowed: 2 hours 30 minutes
You may use:• a ruler (cm/mm)• an HB pencil
Do not use:• a calculator
Please write clearly in black ink. Do not write in the barcodes.
Centre number Candidate number
First name(s) �
Last name �
INSTRUCTIONS• Use black ink.• Answer all the questions. • Write your answer to each question in the space provided. Additional paper may be
used if required but you must clearly show your candidate number, centre number and question number(s).
INFORMATION• The total mark for this paper is 140.• The marks for each question are shown in brackets [ ].• Quality of extended responses will be assessed in questions marked with an
asterisk (*).• This document consists of 24 pages.
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Section A
1 The temperatures of an ocean are input into a computer system. They are recorded, and will be accessed, in the order in which they arrive. The data for one week is shown:
5, 5.5, 5, 6, 7, 6.5, 6
(a) The data is to be stored in a data structure. The programmer stores the data in a queue.
Explain why a queue is used instead of a stack.
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[2]
(b) The data is processed. After processing, the value for the first day is stored as 0. The value for each following day is stored as an increase, or decrease, from the first day.
For example: if the first day was 7, the second was 6 and the third was 9, after processing it would be stored as 0, –1, 2.
(i) The queue uses dequeue() to return the first element of the queue.
dequeue() is a function.
Explain why dequeue() is a function, not a procedure.
[1]
(ii) Complete the algorithm to process the data in the queue and store the results in an array called processedData.
processedData[0] = 0
firstDay = ………………………………………
for count = 1 to 6
processedData[………………………………] = dequeue() – …………………………………
next count[3]
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(iii) The contents of processedData are shown.
0 0.5 0 1 2 1.5 1
The data needs to be sorted into ascending order.
Explain how a bubble sort algorithm sorts data. Use the current contents of processedData in your explanation.
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(iv) A bubble sort has the following complexities:
Best time O(n)
Average and worst time O(n2)
Worst space O(1)
Describe what each of these complexities mean.
Best time O(n)
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Average and worst time O(n2)
Worst Space O(1)
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2 A program needs to store the names of plants that are in a garden, so they can be easily found and accessed in alphabetical order.
The data is stored in a tree structure. Part of the tree is shown.
Begonia HostaPeony
Daisy Sunflower
Rose
Lily
Fig. 2.1
(a) (i) State the type of tree shown in Fig. 2.1.
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(ii) Show the output of a breadth-first traversal of the tree shown in Fig. 2.1.
[3]
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(iii) Explain how backtracking is used in a depth-first (post-order) traversal. Use the tree in Fig. 2.1 in your explanation.
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(b) The elements in the tree in Fig. 2.1 are read into a linked list producing an alphabetised list.
(i) Complete the following table to show the linked list for the data.
Data item Data NextPointer
0 Begonia
1 Daisy
2 Hosta
3 Lily
4 Peony
5 Rose
6 Sunflower
[2]
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(ii) A new plant, Lavender, needs adding to the linked list. The linked list needs to retain its alphabetical order.
Complete the table to show the linked list after Lavender is added.
Data item Data NextPointer
0 Begonia
1 Daisy
2 Hosta
3 Lily
4 Peony
5 Rose
6 Sunflower
[3]
(iii) Hosta needs removing from the linked list.
Explain how a data item is removed from a linked list. Use the removal of Hosta in your answer.
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(iv) The linked list is stored as a 2D array with the identifier plantList. The index of the first element of the linked list is stored in the identifier firstElement.
All contents of the linked list need to be output in alphabetical order.
Write an algorithm to follow the pointers to output the contents of the linked list in alphabetical order.
Add comments to explain your code.
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3 A recursive function, GCD, is given in pseudocode.
function GCD(num1, num2)
if num2 == 0 then
return num1
else
return GCD(num2, num1 MOD num2)
endif
endfunction
(a) The function uses branching.
(i) Identify the type of branching statement used in the function.
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(ii) Explain the difference between branching and iteration.
[2]
(iii) Identify the two parameters in the function.
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(iv) State whether the parameters should be passed by value, or by reference. Justify your answer.
[2]
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(v) Describe the arithmetic operation of MOD. Use an example in your answer.
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(b) Trace the recursive function when it is called by the statement GCD(250, 20). Give the final value returned.
Final return value: .[3]
(c) The function has been rewritten using iteration instead of recursion.
(i) State one benefit and one drawback of using iteration instead of recursion.
Benefit
Drawback
[2]
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(ii) Complete the missing statements in this iterative version of the function.
function newGCD(num1, num2)
temp = 0
while (num2 != ………………………………
………………………………
………………………………
………………………………
)
= num2
num2 = num1 MOD
num1 = temp
endwhile
return
endfunction[4]
4 Mabel is a software engineer. She is writing a computer game for a client. In the game the main character has to avoid their enemies. This becomes more difficult as the levels of the game increase.
(a) Mabel uses decomposition to design the program.
Explain how decomposition can aid the design of this program.
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(b) The computer game allows a user to select a character (e.g. name, gender). They can then choose a level for the game (easy, normal, challenging). The user controls their character by moving it left or right. The character can jump using space bar as an input. If the character touches one of the enemies then it loses a life. The character has to make it to the end of the level without losing all their lives.
The game is designed in a modular way.
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(i) One sub-procedure will handle the user input.
Describe three other sub-procedures Mabel could create for the given game description.
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(ii) Describe the decision that the program will need to make within the user input sub-procedure and the result of this decision.
[2]
(iii) Define pipelining and give an example of how it could be applied in the program.
[2]
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(c) The game’s ‘challenging’ level has intelligent enemies that hunt down the character in an attempt to stop the user from winning. The program plans the enemies’ moves in advance to identify the most efficient way to stop the user from winning the game.
The possible moves are shown in a graph. Each node represents a different state in the game. The lines represent the number of moves it will take to get to that state.
AB E
F G
HD
C
2
14
1
3
12
2
4
Show how Dijkstra’s algorithm would find the shortest path from A to H.
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(d)* Mabel has been told that true programmers write programs in a text editor, and do not use IDEs. Mabel does not agree with this statement.
Discuss the use of an IDE in the development of this program.
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5 A 1-dimensional array stores the following data:
Index 0 1 2 3 4 5
Data 2 18 6 4 12 3
(a) The array needs sorting into descending order.
Describe how a merge sort would sort the given array into descending order.
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(b) An insertion sort can be used to sort the array instead of a merge sort.
Explain why an insertion sort might use less memory than a merge sort.
[2]
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6* Benedict runs a social networking website. He has been told he should use data mining to help him enhance and improve his website.
Evaluate the use of data mining to help Benedict enhance and improve his social networking website.
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Section B
7 A program is needed to plan the layout of a garden.
The program will allow the user to create an image of the garden, for example:
Vegetable bed
Flower bed
Path Patio
Peartree
PeartreeApple
tree
(a) The programmer will use abstraction to produce the program interface to represent the garden.
(i) Give two different examples of how abstraction has been used to produce the layout of the garden.
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(ii) Explain the need for abstraction in the production of this program.
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(iii) The user needs to input data into the program to set up their garden layout.
Identify three pieces of data that the user may input into this program.
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(b) The program is to be built using object oriented programming.
All items that can be added to the garden are declared as instances of the class GardenItem.
The class has the following attributes:
Attribute Description Example
itemName The name of the item Flowerbed
length The length of the item in metres 2
width The width of the item in metres 1
(i) The constructor method sets the attributes to values that are passed as parameters.
Write pseudocode or program code to declare the class GardenItem and its constructor. All attributes should be private and initialised through the constructor (e.g. daisies = new GardenItem("Flowerbed",2,1) ). [4]
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(ii) The trees in the garden layouts are defined by the class Tree. This class inherits from GardenItem.
The class Tree has the additional attributes: height, sun, shade.
If sun is true then the tree can grow in full sun, if it is false then it cannot.
If shade is true then the tree can grow in full shade, if it is false then it cannot.
The length and width of a tree are the same. Only one value for these measurements is passed to the constructor.
Write an algorithm, using pseudocode or program code, to declare the class Tree. Declare all attributes as private.
[5]
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(iii) The Common Oak is a type of tree. It has a maximum height, length and width of 40 m. It can grow in sun and shade.
Write a statement, using pseudocode or program code, to declare an instance of tree for the Common Oak. Give the object the identifier firstTree.
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(iv) The classes GardenItem and Tree use get and set methods to access and alter their private attributes.
Write the get method getItemName and set method setItemName for class GardenItem. The set method takes the new value as a parameter.
Do not write any other methods, or re-declare the class.
[4]
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(v) The trees in the garden layouts are stored in a 1-dimensional array, treeArray. The array can store a maximum of 1000 items. The array has global scope.
A procedure, findTree, takes as parameters:
• The maximum height of a tree• The maximum width of a tree• Whether the tree can live in full sun• Whether the tree can live in full shade.
It searches the array, treeArray, for all trees that do not exceed the maximum height and width, and that can grow in the conditions available. If there are no suitable trees, a suitable message is output.
It outputs the name and details of the trees found in an appropriate message.
Call the get methods, getItemName, getHeight, getWidth, getSun, getShade, to access the attributes.
Write, using pseudocode or program code, the procedure findTree.
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(c)* The programmer is designing the program to make use of caching and re-useable components.
Explain and evaluate the use of caching and re-useable components in the design of the garden program.
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Oxford Cambridge and RSA
Copyright Information
OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download from our public website (www.ocr.org.uk) after the live examination series.
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Oxford Cambridge and RSA Examinations
GCE Computer Science
H446/02: Algorithms and programming
Advanced GCE
Mark Scheme for June 2019
OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. © OCR 2019
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er a
nd
e
nd
s w
ith th
e a
wa
rdin
g o
f gra
de
s. Q
uestio
n p
ap
ers
and
Ma
rk S
ch
em
es a
re d
eve
lope
d in
asso
cia
tion
with
ea
ch
oth
er s
o th
at is
su
es o
f d
iffere
ntia
tion
an
d p
ositiv
e a
ch
ieve
me
nt c
an b
e a
ddre
ssed
from
the v
ery
sta
rt. T
his
Ma
rk S
ch
em
e is
a w
ork
ing d
ocu
men
t; it is n
ot e
xh
au
stiv
e; it d
oe
s n
ot p
rovid
e ‘c
orre
ct’ a
nsw
ers
. Th
e M
ark
Sch
em
e c
an o
nly
pro
vid
e ‘b
est
gu
esses’ a
bo
ut h
ow
the q
uestio
n w
ill wo
rk o
ut, a
nd it is
su
bje
ct to
revis
ion a
fter w
e h
ave
loo
ke
d a
t a w
ide ra
ng
e o
f scrip
ts.
Th
e E
xa
min
ers
’ Sta
nd
ard
isa
tion
Me
etin
g w
ill ensu
re th
at th
e M
ark
Sch
em
e c
ove
rs th
e ra
ng
e o
f can
did
ate
s’ re
spo
nses to
the
qu
estio
ns, a
nd th
at a
ll E
xa
min
ers
und
ers
tand
an
d a
pp
ly th
e M
ark
Sch
em
e in
the s
am
e w
ay. T
he
Ma
rk S
ch
em
e w
ill be d
iscu
ssed
an
d a
me
nd
ed
at th
e m
eetin
g, a
nd
adm
inis
trativ
e p
roced
ure
s w
ill be c
onfirm
ed. C
o-o
rdin
atio
n s
crip
ts w
ill be is
su
ed
at th
e m
eetin
g to
exe
mp
lify a
spe
cts
of c
an
did
ate
s’ re
spo
nses a
nd
achie
ve
me
nts
; the c
o-o
rdin
atio
n s
crip
ts th
en
be
co
me
part o
f this
Ma
rk S
ch
em
e.
Be
fore
the S
tand
ard
isatio
n M
eetin
g, y
ou s
hou
ld re
ad
an
d m
ark
in p
en
cil a
num
ber o
f scrip
ts, in
ord
er to
ga
in a
n im
pre
ssio
n o
f the ra
ng
e o
f re
spo
nse
s a
nd
achie
ve
me
nt th
at m
ay b
e e
xp
ecte
d.
In y
our m
ark
ing
, yo
u w
ill enco
un
ter v
alid
respo
nse
s w
hic
h a
re n
ot c
ove
red b
y th
e M
ark
Sch
em
e: th
ese re
spo
nse
s m
ust b
e c
redite
d. Y
ou w
ill e
nco
un
ter a
nsw
ers
wh
ich
fall o
uts
ide th
e ‘ta
rge
t rang
e’ o
f Ba
nd
s fo
r the p
ape
r wh
ich
yo
u a
re m
ark
ing
. Ple
ase
ma
rk th
ese
an
sw
ers
accord
ing
to th
e
ma
rkin
g c
riteria
. P
lease
read
care
fully
all th
e s
crip
ts in
yo
ur a
llocatio
n a
nd
make
eve
ry e
ffort to
look p
ositiv
ely
for a
ch
ieve
me
nt th
roug
hou
t the
ab
ility ra
ng
e. A
lwa
ys
be p
rep
are
d to
use th
e fu
ll rang
e o
f ma
rks.
LE
VE
LS
OF
RE
SP
ON
SE
QU
ES
TIO
NS
: T
he in
dic
ativ
e c
on
tent in
dic
ate
s th
e e
xp
ecte
d p
ara
me
ters
for c
and
idate
s’ a
nsw
ers
, but b
e p
repa
red
to re
cog
nis
e a
nd
cre
dit u
ne
xp
ecte
d a
ppro
ach
es
wh
ere
they s
how
rele
va
nce
. U
sin
g ‘b
est-fit’, d
ecid
e firs
t wh
ich
se
t of B
AN
D D
ES
CR
IPT
OR
S b
est d
escrib
es th
e o
ve
rall q
uality
of th
e a
nsw
er. O
nce
the b
an
d is
locate
d, a
dju
st
the m
ark
co
nce
ntra
ting
on
featu
res o
f the a
nsw
er w
hic
h m
ake it s
trong
er o
r we
ake
r follo
win
g th
e g
uid
elin
es fo
r refin
em
ent.
H
igh
est m
ark
: If cle
ar e
vid
en
ce o
f all th
e q
ualitie
s in
the b
an
d d
escrip
tors
is s
how
n, th
e H
IGH
ES
T M
ark
sh
ou
ld b
e a
wa
rde
d.
L
ow
est m
ark
: If the a
nsw
er s
how
s th
e c
and
idate
to b
e b
ord
erlin
e (i.e
. they h
ave
ach
ieve
d a
ll the q
ua
lities o
f the b
an
ds b
elo
w a
nd
sh
ow
lim
ited e
vid
ence
of m
eetin
g th
e c
riteria
of th
e b
and
in q
uestio
n) th
e L
OW
ES
T m
ark
sh
ou
ld b
e a
wa
rde
d.
M
idd
le m
ark
: Th
is m
ark
sh
ou
ld b
e u
sed fo
r ca
nd
idate
s w
ho a
re s
ecure
in th
e b
an
d. T
hey a
re n
ot ‘b
ord
erlin
e’ b
ut th
ey h
ave
only
achie
ve
d
so
me
of th
e q
ualitie
s in
the b
an
d d
escrip
tors
. B
e p
repa
red to
use th
e fu
ll rang
e o
f ma
rks. D
o n
ot re
serv
e (e
.g.) h
igh B
an
d 3
ma
rks ‘in
ca
se
’ so
me
thin
g tu
rns u
p o
f a q
uality
yo
u h
ave
no
t ye
t se
en.
If an a
nsw
er g
ive
s c
lea
r evid
en
ce o
f the q
ualitie
s d
escrib
ed
in th
e b
an
d d
escrip
tors
, rew
ard
appro
pria
tely
.
A
O1
A
O2
A
O3
Hig
h (th
oro
ug
h)
Pre
cis
ion in
the u
se o
f que
stio
n
term
inolo
gy. K
now
ledg
e s
how
n is
co
nsis
tent a
nd
we
ll-deve
lope
d.
Cle
ar a
pp
recia
tion o
f the q
uestio
n
from
a ra
ng
e o
f diffe
rent
pers
pectiv
es m
akin
g e
xte
nsiv
e
use o
f acq
uire
d k
now
led
ge a
nd
u
nd
ers
tand
ing
.
Kn
ow
ledg
e a
nd
un
de
rsta
ndin
g
sh
ow
n is
co
nsis
tently
app
lied to
co
nte
xt e
na
blin
g a
log
ical a
nd
su
sta
ined
arg
um
ent to
de
ve
lop.
Exa
mp
les u
se
d e
nh
an
ce ra
ther
than
detra
ct fro
m re
spo
nse
.
Co
nce
rted e
ffort is
ma
de to
co
nsid
er a
ll aspe
cts
of a
syste
m /
pro
ble
m o
r we
igh
up b
oth
sid
es to
a
n a
rgum
ent b
efo
re fo
rmin
g a
n
ove
rall c
onclu
sio
n. J
udg
em
ents
m
ade
are
ba
se
d o
n a
pp
ropria
te
and
co
ncis
e a
rgu
me
nts
that h
ave
b
ee
n d
eve
lope
d in
respon
se
resultin
g in
them
bein
g b
oth
su
pp
orte
d a
nd
realis
tic.
Mid
dle
(rea
so
nab
le)
Aw
are
ne
ss o
f the m
ean
ing o
f the
term
s in
the q
uestio
n. K
no
wle
dg
e
is s
oun
d a
nd e
ffectiv
ely
d
em
onstra
ted. D
em
and
s o
f q
uestio
n u
nd
ers
tood a
ltho
ug
h a
t tim
es o
pp
ortu
nitie
s to
ma
ke u
se o
f a
cq
uire
d k
now
ledg
e a
nd
und
ers
tand
ing
not a
lwa
ys ta
ke
n.
Kn
ow
ledg
e a
nd
un
de
rsta
ndin
g
app
lied to
co
nte
xt. W
hils
t cle
ar
evid
en
ce th
at a
n a
rgum
en
t build
s
and
de
ve
lops th
roug
h re
sp
on
se
there
are
times w
hen
o
pp
ortu
nitie
s a
re m
isse
d to
use a
n
exa
mp
le o
r rela
te a
n a
spe
ct o
f kn
ow
ledg
e o
r und
ers
tand
ing
to
the c
onte
xt p
rovid
ed
.
Th
ere
is a
reaso
na
ble
atte
mp
t to
reach
a c
onclu
sio
n c
onsid
erin
g
aspe
cts
of a
syste
m / p
rob
lem
or
we
igh
ing u
p b
oth
sid
es o
f an
arg
um
ent. H
ow
eve
r the im
pact o
f th
e c
onclu
sio
n is
ofte
n le
sse
ne
d
by a
lack o
f su
pp
orte
d ju
dg
em
ents
w
hic
h a
cco
mp
an
y it. T
his
inab
ility
to b
uild
on
an
d d
eve
lop lin
es o
f a
rgu
me
nt a
s d
eve
lope
d in
the
respo
nse
ca
n d
etra
ct fro
m th
e
ove
rall q
uality
of th
e re
spo
nse
.
Lo
w (b
as
ic)
Co
nfu
sio
n a
nd
ina
bility
to
deco
nstru
ct te
rmin
olo
gy a
s u
se
d
in th
e q
uestio
n. K
now
led
ge p
artia
l a
nd
su
pe
rficia
l. Focu
s o
n q
uestio
n
narro
w a
nd
ofte
n o
ne
-d
ime
nsio
na
l.
Inab
ility to
ap
ply
kn
ow
ledg
e a
nd
u
nd
ers
tand
ing
in a
ny s
usta
ined
w
ay to
co
nte
xt re
sultin
g in
te
nu
ou
s a
nd u
nsu
pp
orte
d
sta
tem
ents
bein
g m
ade.
Exa
mp
les if u
sed
are
for th
e m
ost
part irre
leva
nt a
nd
unsu
bsta
ntia
ted.
Little
or n
o a
ttem
pt to
prio
ritise o
r w
eig
h u
p fa
cto
rs d
urin
g c
ours
e o
f a
nsw
er. C
onclu
sio
n is
ofte
n
dis
locate
d fro
m re
sp
on
se
an
d a
ny
judg
em
ents
lack s
ubsta
nce
du
e in
p
art to
the b
asic
leve
l of a
rgu
me
nt
that h
as b
ee
n d
em
onstra
ted
thro
ug
hout re
spo
nse.
A
ssessme
nt O
bje
ctive
AO
1
Dem
on
strate kno
wled
ge and
un
derstan
din
g of th
e prin
ciples an
d co
ncep
ts of co
mp
ute
r science, in
clud
ing ab
straction
, logic, algo
rithm
s and
data
represe
ntatio
n.
AO
1.1
D
emo
nstrate kn
ow
led
ge o
f the p
rincip
les and
con
cepts o
f abstractio
n, lo
gic, algorith
ms, d
ata represe
ntatio
n o
r oth
er as app
rop
riate.
AO
1.2
D
emo
nstrate u
nd
erstan
din
g of th
e prin
ciples an
d co
ncep
ts of ab
straction
, logic, algo
rithm
s, data rep
resentatio
n o
r oth
er as app
rop
riate.
AO
2
Ap
ply kn
ow
ledge an
d u
nd
erstand
ing o
f the p
rincip
les and
con
cepts o
f com
pu
ter scien
ce inclu
din
g to an
alyse pro
blem
s in co
mp
utatio
nal te
rms.
AO
2.1
A
pp
ly kno
wled
ge and
un
derstan
din
g of th
e prin
ciples an
d co
ncep
ts of co
mp
ute
r science.
AO
2.2
A
nalyse
pro
blem
s in co
mp
utatio
nal te
rms.
AO
3
Design
, pro
gram an
d evalu
ate co
mp
ute
r systems th
at solve p
rob
lems, m
aking reaso
ned
jud
gemen
ts abo
ut th
ese an
d p
resentin
g con
clusio
ns.
AO
3.1
D
esign co
mp
ute
r systems th
at solve p
rob
lems.
AO
3.2
P
rogram
com
pu
ter system
s that so
lve pro
blem
s.
AO
3.3
Evalu
ate com
pu
ter systems th
at solve p
rob
lems, m
aking re
ason
ed ju
dgem
ents ab
ou
t these an
d p
resen
ting co
nclu
sion
s.
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
1a
1 m
ark
per b
ulle
t
Q
ueu
e o
utp
uts
data
in a
Firs
t In F
irst O
ut fa
shio
n
It w
ill retrie
ve
the te
mp
era
ture
va
lues in
the o
rder th
ey w
ere
re
cord
ed
or
S
tack o
utp
uts
the d
ata
in a
Last In
Firs
t Out fa
shio
n
It w
ill retrie
ve
the te
mp
era
ture
va
lues in
the re
ve
rse
of th
e o
rder
they w
ere
record
ed
2
AO
1.2
(1
) A
O2
.2
(1)
Ma
rk P
oin
t 1 is
the d
efin
ition
Ma
rk P
oin
t 2 is
for c
onte
xt o
f the te
mp
era
ture
va
lues
1bi
It retu
rns a
va
lue
1
AO
2.1
(1
)
1bii
1 m
ark
per c
om
ple
ted w
ord
processedData[0] = 0
firstDay = dequeue()
for count = 1 to 6
processedData[count] = dequeue() - firstDay
next count
3
AO
2.2
(1
) A
O3
.2
(2)
Exa
ct a
nsw
ers
only
1biii
1 m
ark
per b
ulle
t to m
ax 5
.
Ma
x 3
if no a
pp
lica
tion
to d
ata
in processedData
C
om
pa
res e
ach
pa
ir of d
ata
e.g
. 0 a
nd 0
.5
If th
ey a
re in
the c
orre
ct o
rde
r it mo
ve
s to
the n
ext p
air e
.g. 0
.5
and
0
If th
ey a
re in
the w
rong
ord
er it s
wa
ps th
em
e.g
. 0.5
an
d 0
b
eco
me
s 0
an
d 0
.5
C
ontin
ue
s to
the e
nd
of th
e a
rray e
.g. P
ass 1
com
ple
te
If th
ere
ha
s b
ee
n a
sw
ap it c
hecks a
ga
in e
.g. P
ass 2
com
ple
te
If th
ere
ha
ve
be
en n
o s
wa
ps it is
so
rted
0
0.5
0
1
2
1.5
1
Pa
ss 1
0
0
0.5
1
2
1.5
1
0
0
0.5
1
1.5
2
1
0
0
0.5
1
1.5
1
2
0
0
0.5
1
1
1.5
2
Pa
ss 2
0
0
0.5
1
1
1.5
2
No
sw
aps
5
AO
1.1
(2
) A
O1
.2
(1)
AO
2.1
(1
) A
O2
.2
(1)
Allo
w (fu
ll) cre
dit fo
r table
s s
ho
win
g th
e
bub
ble
sort b
ein
g c
om
ple
ted.
1biv
1 m
ark
per b
ulle
t O
(n)
Lin
ea
r
B
est tim
e g
row
s a
t the s
am
e ra
te a
s th
e n
um
ber o
f ele
me
nts
T
his
is th
e c
ase w
hen
the
da
ta is
alre
ad
y in
ord
er
O(n
2)
P
oly
nom
ial / Q
uad
ratic
W
ors
t and a
ve
rag
e tim
e is
pro
portio
na
l to th
e s
qua
re (p
oly
nom
ial)
of th
e n
um
ber o
f ele
me
nts
W
ors
t case
is w
hen
the d
ata
is in
itially
in th
e re
ve
rse
ord
er
O(1
) C
onsta
nt
W
ill alw
ays ta
ke
the s
am
e a
mo
unt o
f mem
ory
(in a
dd
ition to
the
list its
elf).
6
AO
1.1
(3
) A
O1
.2
(3)
No
te: F
irst M
ark
Po
int is
for th
e id
en
tifica
tion,
se
con
d M
ark
Po
int is
for th
e d
escrip
tion
No
te: D
o n
ot a
llow
descrip
tion
s re
latin
g to
time
co
mp
lexity
for 'W
ors
t Spa
ce O
(1)'
No
te: D
o n
ot a
llow
‘eq
ual to
’ in d
escrip
tion
s,
O(n
) and O
(n2) g
row
in p
ropo
rtion to
the n
um
ber
of ite
ms
2ai
Bin
ary
Tre
e / B
inary
Se
arc
h T
ree
1
AO
2.1
(1
)
2a
ii
1 m
ark
per b
ulle
t
1
st la
ye
r: Lily
2
nd la
ye
r: Da
isy, S
unflo
we
r
3
rd laye
r: Beg
onia
, Ho
sta
, Pe
on
y
4th la
ye
r: Ro
se
3
AO
1.2
(1
) A
O2
.1
(1)
AO
2.2
(1
)
2a
iii
1 m
ark
per b
ulle
t to m
ax 4
. Ma
x 2
mark
s fo
r no a
pplic
atio
n to
the tre
e.
D
epth
first s
tarts
at th
e ro
ot (L
ily)
…
and
go
es a
ll the w
ay d
ow
n o
ne
bra
nch
to th
e b
otto
m (B
eg
onia
)
It s
tore
s w
hic
h n
od
es it h
as v
isite
d / p
ush
es n
od
es v
isite
d o
nto
a
sta
ck
W
hen
it ca
nn
ot g
o a
ny fu
rther
…
It then b
acktra
cks/re
turn
s to
the p
revio
us n
od
e
A
nd c
ontin
ue
s to
backtra
ck u
ntil a
nod
e is
reach
ed
with
un
vis
ited
ch
ildre
n.
…
and
ch
ecks d
ow
n th
at b
ran
ch
In
the tre
e s
how
n, a
fter v
isitin
g B
eg
onia
, the a
lgo
rithm
wo
uld
b
acktra
ck to
Dais
y…
…
and
wo
uld
then
vis
it Ho
sta
(Acce
pt a
ny o
ther e
xa
mp
le)
4
AO
1.1
(1
) A
O1
.2
(1)
AO
2.1
(1
) A
O2
.2
(1)
2bi
1 m
ark
per b
ulle
t
C
orre
ct N
extP
oin
ter v
alu
es
S
uita
ble
en
d/n
ull p
oin
ter
Da
ta
item
D
ata
N
extP
oin
ter
0
Be
go
nia
1
1
Da
isy
2
2
Ho
sta
3
3
Lily
4
4
Pe
on
y
5
5
Ro
se
6
6
Su
nflo
we
r n
ull
2
AO
2.1
(2
)
Exa
ct v
alu
es o
nly
. Allo
w -1
for n
ull p
oin
ter o
r e
qu
iva
lent s
uch a
s Φ
. D
o n
ot a
llow
a b
lank o
r 0.
2b
ii
1 m
ark
per b
ulle
t
L
ave
nd
er a
dd
ed
in p
ositio
n 7
…
Ho
sta
poin
ts to
7
…
Lave
nd
er p
oin
ts to
3
Da
ta
item
D
ata
N
extP
oin
ter
0
Be
go
nia
1
1
Da
isy
2
2
Ho
sta
7
3
Lily
4
4
Pe
on
y
5
5
Ro
se
6
6
Su
nflo
we
r n
ull
7
Lave
nd
er
3
3
AO
1.2
(1
) A
O2
.2
(2)
Do
no
t cre
dit a
nsw
ers
tha
t do n
ot p
lace la
ve
nd
er
in p
ositio
n 7
an
d th
en u
pd
ate
po
inte
r positio
ns
2b
iii
1 m
ark
per b
ulle
t
T
rave
rse th
e lis
t to th
e ite
m im
me
dia
tely
prio
r to th
e ite
m to
be
rem
oved (1
)
…
wh
ich
is D
ata
Item
1 - D
ais
y
F
ind th
e v
alu
e o
f the N
extP
oin
ter o
f the ite
m to
be re
mo
ve
d
…
wh
ich
is th
e N
extP
oin
ter o
f Da
taIte
m 2
- Ho
sta
, va
lue 7
S
et th
e n
extP
oin
ter o
f the
item
prio
r to th
e ite
m to
be re
mo
ve
d to
th
e N
extP
oin
ter v
alu
e o
f the D
ata
Item
to b
e re
move
d
…
upd
ate
the N
extP
oin
ter o
f Da
taIte
m 1
- Da
isy fro
m 2
to 7
(L
ave
nd
er)
4
AO
1.1
(1
) A
O1
.2
(1)
AO
2.2
(2
)
Fin
d th
e ite
m b
efo
re ite
m to
be d
ele
ted (D
ais
y)
Fin
d n
extP
tr of ite
m to
be
de
lete
d (H
osta
) U
pda
te n
extP
tr of th
e ite
m b
efo
re (D
ais
y) to
the
nextP
tr of ite
m to
be d
ele
ted (H
osta
) i.e
. Da
isy 2
is u
pd
ate
d to
Da
isy 7
A
llow
FT
from
2b(ii/iii) if c
and
idate
ha
s u
sed
ta
ble
in fig
2.1
(e.g
. Da
isy w
ould
no
w p
oin
t to
Lily
at p
ositio
n 3
)
2b
iv
1 m
ark
per b
ulle
t
S
tart a
t the firstElement
in th
e lis
t
C
orre
ctly
loop
ing u
ntil n
ull p
oin
ter fo
un
d / e
nd o
f list
O
utp
uttin
g th
e d
ata
ele
me
nt
A
cce
ssin
g th
e p
oin
ter to
the n
ext e
lem
ent
A
ppro
pria
te c
om
me
nt(s
) e
.g.
currentElement = firstElement
while(currentElement != null) //Continue until last
node
print(plantList[currentElement,0])
currentElement = plantList[currentElement,1]
endwhile
5
AO
2.1
(1
) A
O2
.2
(2)
AO
3.2
(2
)
No
te: S
olu
tion m
ust u
tilise
po
inte
rs in
a lin
ke
d
list; it c
ann
ot u
se a
FO
R lo
op
as th
e n
um
ber o
f e
lem
en
ts is
not k
now
n a
nd th
e d
ata
is n
ot in
o
rde
r by in
de
x n
um
ber
No
te: id
entifie
rs g
ive
n in
the q
uestio
n a
s
plantList
and firstElement
sh
ou
ld b
e
used
accura
tely
in th
e s
olu
tion
N
ote
: allo
w c
redit fo
r answ
ers
that in
terp
ret th
e
data
stru
ctu
re a
s a
n a
rray o
f record
s/s
tructu
res
with
data
/poin
ter fie
lds
3a
i if
1
AO
1.1
(1
)
3aii
1 m
ark
per b
ulle
t
B
ranch
ing d
ecid
es w
hic
h c
od
e is
run / o
nly
runs c
ode
on
ce
Ite
ratio
n re
pe
ate
dly
runs th
e s
am
e c
ode
in th
e s
am
e s
eq
uence
2
AO
1.2
(2
)
3a
iii n
um
1, n
um
2
1
AO
2.1
(1
)
Exa
ct id
en
tifier n
am
es re
qu
ired
3aiv
1 m
ark
per b
ulle
t
B
y V
alu
e
…
the o
rigin
al v
alu
es d
o n
ot n
ee
d to
be m
odifie
d
…
byR
ef w
ould
no
t wo
rk / w
ould
ca
use
the ro
utin
e to
cra
sh
2
AO
2.2
(2
)
3av
1 m
ark
per b
ulle
t
G
ive
s th
e re
ma
inder a
fter d
ivis
ion
E
.g. 1
0 M
OD
3 =
1
2
AO
1.1
(1
) A
O1
.2
(1)
3b
1 m
ark
per b
ulle
t to m
ax 3
N
um
2 !=
0 th
ere
fore
retu
rn G
CD
(20,1
0)
N
um
2 !=
0 th
ere
fore
retu
rn G
CD
(10,0
)
F
inal re
turn
va
lue =
10
3
AO
2.1
(1
) A
O2
.2
(2)
Allo
w F
T fo
r num
eric
al e
rrors
3ci
1 m
ark
for b
en
efit, 1
ma
rk fo
r dra
wb
ack
Be
ne
fit:
T
he p
rog
ram
ca
n/m
igh
t run fa
ste
r
C
ann
ot ru
n o
ut o
f sta
ck s
pace
/mem
ory
E
asie
r to tra
ce/fo
llow
D
raw
back:
Ite
ratio
n c
an le
ad
to le
ng
thie
r co
de
Ite
ratio
n c
an le
ad
to c
ode
that lo
oks m
ore
co
mp
lex / is
hard
er to
u
nd
ers
tand
so
me
pro
ble
ms a
re m
ore
ele
ga
ntly
co
de
d w
ith a
recurs
ive
so
lutio
n
2
AO
1.1
(1
)
3cii
1 m
ark
for e
ach c
orre
ct s
tate
me
nt
function newGCD(num1, num2)
temp = 0
while (num2 != 0)
temp = num2
num2 = num1 MOD num2
num1 = temp
endwhile
return num1
endfunction
4
AO
2.2
(2
) A
O3
.2
(2)
4a
1 m
ark
per b
ulle
t
S
he c
an s
plit th
e p
rob
lem
dow
n in
to s
ub p
rob
lem
s
It w
ill cre
ate
s a
mo
re m
an
ag
eab
le p
roble
m / s
imple
r to
und
ers
tand / m
ain
tain
ca
n ta
ckle
ea
ch s
ub p
roble
m in
de
pe
nd
en
tly
2
AO
1.2
(1
) A
O2
.2
(1)
4bi
1 m
ark
per b
ulle
t, ma
x 2
per s
ub-p
roce
dure
e
.g.
Se
lect c
hara
cte
r (nam
e, g
end
er)
G
ive
s th
e u
ser o
ptio
ns fo
r ch
oo
sin
g a
ch
ara
cte
r
C
hoo
se le
ve
l
G
ive
the u
ser th
e c
hoic
e o
f leve
l (easy, n
orm
al, c
ha
lleng
ing
) and
take
the u
se
r inpu
t
T
ouch e
ne
my
C
alle
d to
de
term
ine if th
e c
hara
cte
r touch
es a
n e
ne
my
L
ose
life
R
em
ove
a life
, if <0
then
gam
e o
ve
r
E
nd le
ve
l
M
ove
on
to n
ext le
ve
l O
ne m
ark
for id
en
tifyin
g s
ensib
le s
ub
routin
e, 1
ma
rk fo
r descrip
tion
6
AO
2.1
(2
) A
O2
.2
(2)
AO
3.2
(2
)
Do
no
t aw
ard
any u
ser in
put re
late
d p
roced
ure
s
e.g
. Left/R
igh
t inpu
t (but c
hara
cte
r mo
vem
ent
outp
ut o
n s
cre
en le
ft/right w
ould
be
va
lid)
Allo
w o
the
r reaso
na
ble
respo
nse
s fro
m th
e
sce
na
rio e
.g. g
ene
rate
en
em
y()
4bii
1 m
ark
per b
ulle
t
D
ecis
ion b
ase
d o
n w
hat th
e u
ser h
as in
pu
t
E
.g. If th
ey c
lick le
ft mo
ve
the c
hara
cte
r left // if th
ey c
lick rig
ht
mo
ve
the c
hara
cte
r right // if th
ey c
lick s
pace b
ar m
ake
the
ch
ara
cte
r jum
p
2
AO
2.1
(1
) A
O2
.2
(1)
4biii
1 m
ark
per b
ulle
t
T
he re
sult fro
m o
ne p
roce
ss / p
roce
du
re fe
ed
s in
to th
e n
ext
E
.g. th
e re
sult o
f dete
ctin
g a
ch
ara
cte
r touch
ing
an
en
em
y fe
eds
into
redu
cin
g th
e n
um
ber o
f live
s
2
AO
1.2
(1
) A
O2
.2
(1)
No
te: 1
Ma
rk M
ax fo
r a g
en
eric
descrip
tion
of
pip
elin
ing
4c
1 m
ark
for fin
al s
olu
tion, m
ax 5
for s
how
ing
the s
tag
es
M
ark
A a
s th
e c
urre
nt n
od
e in
itially
R
ecord
B =
1, C
= 2
(mark
A a
s v
isite
d)
R
ecord
E =
5 (a
nd m
ark
B a
s v
isite
d)
(R
ecord
D =
3, F
= 5
(and
mark
B a
s v
isite
d)
C
han
ge
E to
4 (o
ve
rridin
g p
revio
us v
alu
e, a
nd
ma
rk D
as v
isite
d)
R
ecord
G =
6 (a
nd m
ark
E a
s v
isite
d)
…
Do
no
t ch
ang
e G
as g
reate
r than c
urre
nt (m
ark
F a
s v
isite
d)
(G
as v
isite
d) H
= 1
0 (M
ark
G a
s v
isite
d)
S
olu
tion
: A-C
-D-E
-G-H
pa
th le
ng
th 1
0
No
de
Vis
ited
Fro
m A
P
revio
us N
ode
A
0
- 1
Ma
rk
B
1
A
1 M
ark
C
2
A
D
3
C
1 M
ark
F
5
C
E
5 4
B
D
2 M
ark
s
Initia
l vis
it, plu
s
ove
rride
va
lues
G
6
E
1 M
ark
H
10
G
1 M
ark
6
AO
1.2
(1
) A
O2
.2
(3)
AO
2.2
(2
)
Gu
idan
ce –
1 m
ark
only
for s
tatin
g th
e s
olu
tion
of A
-C-D
-E-G
-H le
ng
th 1
0
4d
Mark
Ban
d 3
– H
igh
lev
el
(7-9
mark
s)
The c
andid
ate
dem
onstra
tes a
thoro
ugh k
no
wle
dge a
nd u
nders
tan
din
g o
f IDE
s;
the m
ate
rial is
genera
lly a
ccura
te a
nd d
eta
iled.
The c
andid
ate
is a
ble
to a
pply
their k
no
wle
dge
an
d u
nders
tand
ing d
irectly
and
consis
tently
to th
e c
onte
xt p
rovid
ed.
Evid
ence/e
xam
ple
s w
ill be
explic
itly re
levan
t to th
e e
xpla
natio
n.
The c
andid
ate
is a
ble
to w
eig
h u
p th
e c
onte
xt w
hic
h re
sults
in a
sup
porte
d a
nd
realis
tic ju
dgm
ent a
s to
wh
eth
er ID
Es a
re u
sefu
l in th
is c
onte
xt.
There
is a
we
ll-develo
pe
d lin
e o
f reason
ing w
hic
h is
cle
ar a
nd lo
gic
ally
stru
ctu
red. T
he
info
rmatio
n p
resente
d is
rele
va
nt a
nd s
ubsta
ntia
ted.
Mark
Ban
d 2
– M
id le
vel
(4-6
mark
s)
The c
andid
ate
dem
onstra
tes re
asonab
le k
nole
dg
e a
nd
und
ers
tand
ing
of ID
Es;
the m
ate
rial is
genera
lly a
ccura
te b
ut a
t times u
nderd
eve
lop
ed.
The c
andid
ate
is a
ble
to a
pply
their k
no
wle
dge
an
d u
nders
tand
ing d
irectly
to th
e
conte
xt p
rovid
ed a
ltho
ugh
one o
r two o
pp
ortu
nitie
s a
re m
issed.
Evid
ence/e
xam
ple
s a
re fo
r the m
ost p
art im
plic
itly re
levan
t to th
e e
xpla
natio
n.
The c
andid
ate
makes a
reasona
ble
atte
mpt to
com
e to
a c
onclu
sio
n s
ho
win
g
som
e re
cognitio
n o
f influ
en
cin
g fa
cto
rs th
at w
ou
ld d
ete
rmin
e w
he
ther ID
Es a
re
usefu
l in th
is c
onte
xt.
There
is a
line
of re
ason
ing
pre
sente
d w
ith s
om
e s
tructu
re. T
he in
form
atio
n
pre
sente
d is
in th
e m
ost p
art re
levan
t an
d s
up
porte
d b
y s
om
e e
vid
ence
M
ark
Ban
d 1
– L
ow
Lev
el
(1-3
mark
s)
The c
andid
ate
dem
onstra
tes a
basic
kno
wle
dg
e o
f IDE
s w
ith lim
ited
unders
tand
ing s
ho
wn; th
e m
ate
rial is
basic
an
d c
onta
ins s
om
e in
accura
cie
s.
The c
andid
ate
s m
akes a
limite
d a
ttem
pt to
ap
ply
acq
uire
d k
no
wle
dge a
nd
unders
tand
ing to
the c
on
text p
rovid
ed.
The c
andid
ate
pro
vid
es n
oth
ing m
ore
than a
n u
nsupp
orte
d a
ssertio
n.
The in
form
atio
n is
basic
an
d c
om
unic
ate
d in
an u
nstru
ctu
red w
ay. T
he
info
rmatio
n is
supp
orte
d b
y lim
ited e
vid
ence a
nd th
e re
latio
nsh
ip to
the e
vid
ence
may n
ot b
e c
lear.
0 m
ark
s
No a
ttem
pt to
answ
er th
e q
uestio
n o
r resp
onse is
not w
orth
y o
f cre
dit.
9
AO
1.1
(2
) A
O1
.2
(2)
AO
2.1
(2
) A
O3
.3
(3)
AO
1: K
no
wle
dg
e a
nd
Un
de
rsta
nd
ing
In
dic
ativ
e c
on
ten
t T
ools
to a
id w
riting
C
olo
ure
d fo
nt
P
redic
tive
text
A
uto
-corre
ct
To
ols
to a
id d
e-b
ug
gin
g
S
tepp
ing
B
reak p
oin
ts
V
aria
ble
wa
tch
win
do
w
AO
2: A
pp
licatio
n
e.g
.
C
an w
rite s
ub
routin
es fo
r the p
rog
ram
and
it w
ill tell y
ou w
hat p
ara
mete
rs a
re n
ee
de
d
A
llow
yo
u to
run th
e p
rog
ram
with
ou
t exitin
g
the s
oftw
are
/ havin
g to
load
a s
epara
te
co
mp
iler
In
teg
rate
s o
ther to
ols
such
as v
ers
ion
co
ntro
l.
C
an re
du
ce s
pellin
g e
rrors
C
an u
se to
fix e
rrors
that m
igh
t occur /
debug
A
O3
: Eva
lua
tion
e.g
.
U
se
r friend
ly fo
r novic
es
In
cre
ase s
pee
d o
f writin
g
F
ew
er m
ista
ke
s
In
cre
ase s
pee
d o
f testin
g / fin
din
g e
rrors
C
olla
bo
rativ
e te
am
wo
rkin
g fa
cilita
ted
5a
1 m
ark
per b
ulle
t to m
ax 6
. M
ax 4
if ge
ne
ric d
escrip
tion g
ive
n w
ith n
o a
pp
lica
tion
M
ax 4
if a d
iag
ram
ma
tic s
olu
tion
is g
ive
n w
ith n
o d
escrip
tion
S
plits
the lis
t in h
alf re
pea
ted
ly…
…
until it is
in in
dep
en
den
t arra
ys / e
lem
ents
e.g
. 2, 1
8, 6
, 4, 1
2,
3
C
om
pa
re th
e firs
t two
item
s (in
de
x 0
an
d 1
) e.g
. 2, 1
8
…
and
co
mb
ine to
cre
ate
a n
ew
arra
y in
de
sce
nd
ing
ord
er i.e
. 18,
2
R
epe
at w
ith in
dexe
s 2
an
d 3
(6, 4
), then
4 a
nd
5 (1
2, 3
)
C
om
pa
re th
e firs
t ele
men
t in th
e firs
t two
ne
w a
rrays
…
Ch
oo
se th
e la
rge
st e
lem
ent, w
riting th
is to
the n
ew
arra
y firs
t
…
repe
at u
ntil n
o e
lem
ents
left
C
om
bin
e th
e tw
o re
ma
inin
g lis
ts in
to o
ne lis
t
e.g
. [2
, 18, 6
, 4, 1
2, 3
] [2
, 18, 6
] [4, 1
2, 3
] [2
, 18] [6
] [4, 1
2] [3
] [2
] [18] [6
] [4] [1
2] [3
] [1
8, 2
] [6, 4
] [12, 3
] [1
8, 6
, 4, 2
] [12, 3
] [1
8, 1
2, 6
, 4, 3
, 2]
e.g
. [2
, 18, 6
, 4, 1
2, 3
] [2
, 18, 6
] [4, 1
2, 3
] [2
, 18] [6
] [4, 1
2] [3
] [2
] [18] [6
] [4] [1
2] [3
] [1
8 2
] [6] [1
2 4
] [3]
[18 6
2] [1
2 4
3]
[18, 1
2, 6
, 4, 3
, 2]
6
AO
1.2
(3
) A
O2
.2
(3)
Allo
w m
ax 5
if corre
ct d
escrip
tion
but in
a
scen
din
g o
rder.
5b
1 m
ark
per b
ulle
t
M
erg
e s
ort m
igh
t cre
ate
a n
ew
arra
y e
ach
time it s
plits
and
m
erg
es / o
ften im
ple
me
nte
d re
curs
ive
ly w
hic
h p
lace
s a
dd
itiona
l d
ata
on th
e s
tack
In
sertio
n s
ort d
oe
s n
ot u
se
an
y a
dd
itiona
l arra
ys//In
sertio
n s
ort is
a
n in
-pla
ce a
lgo
rithm
.
2
AO
1.2
(2
)
6
Mark
Ban
d 3
– H
igh
lev
el
(7-9
mark
s)
The c
andid
ate
dem
onstra
tes a
thoro
ugh k
no
wle
dge a
nd u
nders
tan
din
g o
f data
m
inin
g; th
e m
ate
rial is
ge
ne
rally
accura
te a
nd d
eta
iled.
The c
andid
ate
is a
ble
to a
pply
their k
no
wle
dge
an
d u
nders
tand
ing d
irectly
and
consis
tently
to th
e c
onte
xt p
rovid
ed.
Evid
ence/e
xam
ple
s w
ill be
explic
itly re
levan
t to th
e e
xpla
natio
n.
The c
andid
ate
is a
ble
to w
eig
h u
p th
e c
onte
xt w
hic
h re
sults
in a
sup
porte
d a
nd
realis
tic ju
dgm
ent a
s to
wh
eth
er it is
possib
le to
use d
ata
min
ing in
this
conte
xt.
There
is a
we
ll-develo
pe
d lin
e o
f reason
ing w
hic
h is
cle
ar a
nd lo
gic
ally
stru
ctu
red. T
he
info
rmatio
n p
resente
d is
rele
va
nt a
nd s
ubsta
ntia
ted.
Mark
Ban
d 2
– M
id le
vel
(4-6
mark
s)
The c
andid
ate
dem
onstra
tes re
asonab
le k
no
wle
dg
e a
nd u
nders
tan
din
g o
f data
m
inin
g; th
e m
ate
rial is
ge
ne
rally
accura
te b
ut a
t times u
nderd
evelo
ped
. T
he c
andid
ate
is a
ble
to a
pply
their k
no
wle
dge
an
d u
nders
tand
ing d
irectly
to th
e
conte
xt p
rovid
ed a
ltho
ugh
one o
r two o
pp
ortu
nitie
s a
re m
issed.
Evid
ence/e
xam
ple
s a
re fo
r the m
ost p
art im
plic
itly re
levan
t to th
e e
xpla
natio
n.
Th
e c
and
idate
makes a
reaso
na
ble
atte
mpt to
com
e to
a c
onclu
sio
n
sh
ow
ing
so
me re
cog
nitio
n o
f influ
en
cin
g fa
cto
rs th
at w
ould
de
term
ine
wh
eth
er it is
possib
le to
use d
ata
min
ing
. T
here
is a
line
of re
ason
ing
pre
sente
d w
ith s
om
e s
tructu
re. T
he in
form
atio
n
pre
sente
d is
in th
e m
ost p
art re
levan
t an
d s
up
porte
d b
y s
om
e e
vid
ence
. M
ark
Ban
d 1
– L
ow
Lev
el
(1-3
mark
s)
The c
andid
ate
dem
onstra
tes a
basic
kno
wle
dg
e o
f data
min
ing w
ith lim
ited
unders
tand
ing s
ho
wn; th
e m
ate
rial is
basic
an
d c
onta
ins s
om
e in
accura
cie
s.
The c
andid
ate
s m
akes a
limite
d a
ttem
pt to
ap
ply
acq
uire
d k
no
wle
dge a
nd
unders
tand
ing to
the c
on
text p
rovid
ed.
The c
andid
ate
pro
vid
es n
oth
ing m
ore
than a
n u
nsupp
orte
d a
ssertio
n.
The in
form
atio
n is
basic
an
d c
om
unic
ate
d in
an u
nstru
ctu
red w
ay. T
he
info
rmatio
n is
supp
orte
d b
y lim
ited e
vid
ence a
nd th
e re
latio
nsh
ip to
the e
vid
ence
may n
ot b
e c
lear.
0 m
ark
s
No a
ttem
pt to
answ
er th
e q
uestio
n o
r resp
onse is
not w
orth
y o
f cre
dit.
9
AO
1.1
(2
) A
O1
.2
(2)
AO
2.1
(2
) A
O3
.3
(3)
AO
1: K
no
wle
dg
e a
nd
Un
de
rsta
nd
ing
In
dic
ativ
e c
on
ten
t
E
xtra
ctin
g d
ata
from
data
ba
ses
U
sin
g la
rge d
ata
sets
L
oo
kin
g fo
r patte
rns/s
pecific
occurre
nce
s o
f d
ata
G
ath
erin
g d
ata
that c
an b
e a
na
lyse
d a
nd
u
sed
to in
form
decis
ions
A
O2
: Ap
plic
atio
n
e.g
.
U
se
to fin
d o
ut w
hat h
is u
sers
do
F
ind fe
atu
res th
at a
re u
se
d m
ost o
ften
F
ind fe
atu
res th
at a
re n
ot u
sed
F
ind o
ut w
hat p
eo
ple
in h
is ta
rget a
ge g
rou
p
do o
n o
ther s
ites
F
ind o
ut c
hara
cte
ristic
s o
f peo
ple
wh
o u
se
the s
ite
AO
3: E
va
lua
tion
e
.g.
C
an id
en
tify a
rea
s to
focu
s a
ttentio
n
S
ave
time
an
d m
one
y b
y id
en
tifyin
g a
reas
that a
re n
ot p
op
ula
r/use
d
N
ew
featu
res ta
rgete
d a
t sp
ecific
gro
up
s
co
uld
brin
g in
ne
w b
usin
ess e
.g. a
dve
rtisin
g
B
ut c
are
wo
uld
ne
ed to
be
ap
plie
d to
priv
acy
issu
es / G
DP
R a
nd p
ote
ntia
l imp
act o
n th
e
users
7ai
1 m
ark
per e
xa
mp
le
e.g
. N
o a
ctu
al im
ag
es s
ho
wn
Ite
ms a
re n
am
ed / la
be
lled
S
imp
lified la
yo
ut w
ith s
ha
pe
s
2
AO
2.1
(2
)
Allo
w a
ny re
aso
na
ble
exa
mp
les, b
ut th
ey m
ust
be fo
r diffe
rent a
sp
ects
7aii
1 m
ark
per b
ulle
t to m
ax 3
e
.g.
Re
du
ces c
om
ple
xity
of d
esig
n
R
edu
ces c
om
ple
xity
of p
rog
ram
min
g
R
edu
ce m
em
ory
/pro
cessin
g re
qu
irem
ents
C
ould
invo
lve
a la
rge
num
ber o
f imag
es th
at w
ould
take
e
xce
ssiv
e m
em
ory
R
eality
co
nta
ins th
ing
s th
at a
ren’t re
leva
nt to
a c
om
pute
r pro
gra
m
3
AO
1.1
(1
) A
O1
.2
(1)
AO
2.1
(1
)
No
te: d
o n
ot a
llow
answ
ers
rela
ted
to th
e u
ser
exp
erie
nce
/ user in
terp
reta
tion
, the q
uestio
n is
a
bo
ut th
e p
rodu
ctio
n o
f the s
yste
m
7aiii
1 m
ark
per e
xa
mp
le
e.g
. G
ard
en d
ime
nsio
ns/w
idth
/leng
th
N
um
be
r of ite
ms in
the g
ard
en
N
am
e o
f item
s in
the g
ard
en
L
oca
tion
of ite
ms in
the g
ard
en
3
AO
2.1
(3
)
7bi
1 m
ark
per b
ulle
t to m
ax 4
C
lass d
ecla
ratio
n
3
attrib
ute
s d
ecla
red
C
onstru
cto
r
…
takin
g p
ara
mete
rs
…
se
tting th
e a
ttribute
s to
the p
ara
me
ters
e.g.
class GardenItem
private itemName
private length
private width
public procedure new(pItemName, pLength, pWidth)
itemName = pItemName
length = pLength
width = pWidth
endprocedure
endclass
4
AO
1.1
(1
) A
O2
.1
(1)
AO
2.2
(1
) A
O3
.2
(1)
No
te th
at e
xa
mp
le a
nsw
ers
are
giv
en in
the
sp
ecific
atio
n p
se
udo
code
. An
y p
seu
do
cod
e
answ
er th
at c
an b
e u
nd
ers
tood b
y a
‘com
pete
nt
pro
gra
mm
er’ s
hou
ld b
e a
cce
pte
d
7bii
1 m
ark
per b
ulle
t
C
lass d
ecla
ratio
n in
heritin
g fro
m gardenItem
A
dditio
na
l 3 p
rope
rties d
ecla
red
as p
riva
te
C
onstru
cto
r takes a
ll 5 p
ara
mete
rs
U
se
of s
upe
r (or e
qu
iva
lent) to
se
t su
per c
lass p
ara
mete
rs
R
em
ain
de
r of p
rope
rties s
et to
para
mete
rs
e.g
. class Tree inherits GardenItem
private height
private sun
private shade
public procedure new(pName, pHeight, pLenWidth,
pSun, pShade)
super.itemName = pName
super.length = pLenWidth
super.width = pLenWidth
height = pHeight
sun = pSun
shade = pShade
endprocedure
endclass
5
AO
1.1
(1
) A
O2
.2
(1)
AO
3.2
(3
)
Acce
pt s
olu
tions th
at c
all th
e p
are
nt’s
co
nstru
cto
r. class Tree inherits GardenItem
private height
private sun
private shade
public procedure new(pName,
pHeight, pLenWidth, pSun, pShade)
height = pHeight
sun = pSun
shade = pShade
super.new(pName,pLenWidth,pLenWidth)
endprocedure
endclass
7biii
1 m
ark
per b
ulle
t
D
ecla
ratio
n o
f insta
nce o
f tree (i.e
. new Tree
)
S
torin
g re
sult in
firstTree
A
ll para
me
ters
inclu
de
d a
nd in
the s
am
e o
rder a
s 7
bii
…
with
ap
pro
pria
te d
ata
typ
es
e.g
. firstTree = new Tree("Common Oak",40,40,true, true)
4
AO
1.1
(1
) A
O2
.2
(1)
AO
3.2
(2
)
7biv
1 m
ark
per b
ulle
t
G
et m
eth
od d
ecla
ratio
n
R
etu
rns itemName
S
et m
eth
od d
ecla
ratio
n
…
take
s v
alu
e a
s a
para
me
ter
…
assig
ns p
ara
mete
r to itemName
e.g
. function getItemName()
return itemName
endfunction
procedure setItemName(newname)
itemName = newname
endprocedure
4
AO
1.2
(2
) A
O2
.2
(2)
7bv
1 m
ark
per b
ulle
t to m
ax 6
P
roced
ure
decla
ratio
n w
ith a
ll four p
ara
me
ters
L
oo
pin
g 1
00
0 tim
es / to
end
of a
rray
C
heckin
g if h
eig
ht a
nd
wid
th a
re le
ss th
an o
r eq
ual to
the
ma
xim
um
heig
ht a
nd w
idth
C
heckin
g if s
un a
nd
sh
ad
e m
atc
h
O
utp
uttin
g v
alu
e(s
) usin
g g
et m
eth
od
s
O
utp
ut a
n a
pp
ropria
te m
essag
e
O
utp
uts
a m
essag
e if n
o m
atc
hin
g tre
e is
foun
d
e.g
. procedure findTree(pHeight, pWidth, pSun, pShade)
flag = false
for i = 0 to 999
if treeArray[i].getHeight() <= pHeight then
if treeArray[i].getWidth() <= pWidth then
if treeArray[i].getSun() == pSun then
if treeArray[i].getShade() == pShade then
flag = true
print(treeArray[i].getItemName() + “
height: “ + treeArray[i].getHeight + “ width: “ +
treeArray[i].getWidth() + “ Sun?: “ +
treeArray[i].getSun() + “ Shade?: “ +
treeArray[i].getShade()
endif
endif
endif
endif
next count
if flag == false then
print(“No suitable trees”)
endif
endprocedure
6
AO
1.2
(1
) A
O2
.2
(2)
AO
3.2
(3
)
7c
Ma
rk B
an
d 3
– H
igh
leve
l (7
-9 m
ark
s)
Th
e c
an
did
ate
dem
onstra
tes a
tho
roug
h k
no
wle
dge
and
unde
rsta
nd
ing
of c
ach
ing
an
d
reu
sea
ble
co
mpo
ne
nts
; the
ma
teria
l is g
en
era
lly a
ccu
rate
and
de
taile
d.
Th
e c
an
did
ate
is a
ble
to a
pp
ly th
eir k
no
wle
dge
and
und
ers
tan
din
g d
irectly
an
d
co
nsis
ten
tly to
the
co
nte
xt p
rovid
ed
. E
vid
en
ce/e
xa
mp
les w
ill be
exp
licitly
rele
va
nt to
the
exp
lan
atio
n.
Th
e c
an
did
ate
is a
ble
to w
eig
h u
p th
e u
se
of b
oth
ca
chin
g a
nd
reusa
ble
co
mpo
nen
ts
wh
ich
results
in a
su
ppo
rted
an
d re
alis
tic ju
dg
me
nt a
s to
wh
eth
er it is
possib
le to
use
the
m in
this
con
text.
Th
ere
is a
we
ll-deve
lop
ed
line
of re
aso
nin
g w
hic
h is
cle
ar a
nd
logic
ally
stru
ctu
red
. Th
e
info
rma
tion
pre
sen
ted
is re
leva
nt a
nd
su
bsta
ntia
ted
. M
ark
Ban
d 2
– M
id le
ve
l (4
-6 m
ark
s)
Th
e c
an
did
ate
dem
onstra
tes re
ason
able
kn
ole
dge
and
und
ers
tand
ing o
f ca
ch
ing
and
re
usea
ble
co
mpo
ne
nts
; the
ma
teria
l is g
en
era
lly a
ccu
rate
but a
t times u
nd
erd
eve
lope
d.
Th
e c
an
did
ate
is a
ble
to a
pp
ly th
eir k
no
wle
dge
and
und
ers
tan
din
g d
irectly
to th
e c
on
text
pro
vid
ed
alth
ou
gh
one
or tw
o o
pp
ortu
nitie
s a
re m
isse
d. E
vid
en
ce
/exa
mp
les a
re fo
r the
m
ost p
art im
plic
itly re
leva
nt to
the
exp
lan
atio
n.
Th
e c
an
did
ate
ma
kes a
reason
ab
le a
ttem
pt to
com
e to
a c
on
clu
sio
n s
ho
win
g s
om
e
reco
gn
ition
of in
flue
ncin
g fa
cto
rs th
at w
ou
ld d
ete
rmin
e w
he
the
r it is p
ossib
le to
use
ca
ch
ing a
nd re
usa
ble
co
mp
onm
en
ts in
this
co
nte
xt.
Th
ere
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A Level Computer ScienceH446/02 Algorithms and programming
Friday 15 June 2018 – MorningTime allowed: 2 hours 30 minutes
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INSTRUCTIONS• Use black ink.• Complete the boxes above with your name, centre number and candidate number.• Answer all the questions. • Write your answer to each question in the space provided. Additional paper may be
used if required but you must clearly show your candidate number, centre number and question number(s).
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INFORMATION• The total mark for this paper is 140.• The marks for each question are shown in brackets [ ].• Quality of extended responses will be assessed in questions marked with an
asterisk (*).• This document consists of 28 pages.
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2Section A
Answer all the questions.
1 A program stores entered data in a binary search tree.
The current contents of the tree are shown:
Spain
NorwayGermanyAustria
Italy
France
(a) Complete the diagram to show the contents of the tree after the following data is added:
England, Scotland, Wales, Australia [3]
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(b) Show the order of the nodes visited in a breadth first traversal on the following tree.
Spain
NorwayGermanyAustria
Italy
France
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4 (c) A pseudocode algorithm is written to search the tree to determine if the data item “Sweden” is
in the tree.
The function currentNode.left()returns the node positioned to the left of currentNode.
The function currentNode.right()returns the node positioned to the right of currentNode.
function searchForData(currentNode:byVal, searchValue:byVal)
thisNode = getData(……………………………………………………………………………………………………………)
if thisNode == …………………………………………………………………………………………………………… then
return ……………………………………………………………………………………………………………
elseif thisNode < searchValue then
if currentNode.left() != null then
return (searchForData(currentNode.left(), searchValue))
else
return ……………………………………………………………………………………………………………
endif
else
if …………………………………………………………………………………………………………… != null then
return (searchForData(currentNode.right(), searchValue))
else
return false
endif
endif
endfunction
(i) Complete the algorithm.
[5]
(ii) The algorithm needs to be used in different scenarios, with a range of different trees.
Identify two preconditions needed of a tree for this algorithm to work.
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62 A company merger is joining five e-commerce retailers under one company, OCRRetail. Each
retailer has a different sales system and OCRRetail wants to develop one computer system that can be used by all the retailers.
Mary’s software development company has been employed to analyse and design a solution for
the company.
(a) (i) Two computational methods (techniques used to solve a problem using computational thinking) that Mary will use are problem recognition and decomposition.
State what is meant by problem recognition and decomposition.
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(ii) State one additional computational method.
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(b) Mary plans to use data mining to generate information about OCRRetail’s customers. Mary will use this information to benefit the company.
(i) Define the term ‘data mining’.
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(ii) Identify two pieces of information that data mining could provide OCRRetail about sales, and state how OCRRetail could make use of this information.
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(c) Mary has developed the program and is considering using performance modelling before installing the system.
(i) Define the term ‘performance modelling’.
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(ii) Identify one way performance modelling could be used to test the new system.
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(d) Mary created the program as a series of sub-programs that can be reused.
Describe one benefit of Mary creating reusable program components.
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83 A puzzle has multiple ways of reaching the end solution. Fig. 3 shows a graph that represents all
possible routes to the solution. The starting point of the game is represented by A, the solution is represented by J. The other points in the graph are possible intermediary stages.
Fig. 3
(a) The graph in Fig. 3 is a visualisation of the problem.
(i) Identify one difference between a graph and a tree.
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(ii) Explain how the graph is an abstraction of the problem.
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(iii) Identify two advantages of using a visualisation such as the one shown in Fig. 3.
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(b) Demonstrate how Dijkstra’s algorithm would find the shortest path to the solution in Fig. 3.
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10 (c)* The creator of the puzzle has been told that the A* algorithm is more efficient at finding the
shortest path because it uses heuristics.
Compare the performance of Dijkstra’s algorithm and the A* search algorithm, making reference to heuristics, to find the shortest path to the problem.
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(d) A computer program version of the puzzle is to be developed. A programmer will use an IDE to debug the program during development.
Describe three features of an IDE that help debug the program.
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4 A recursive function, generate, is shown.
function generate(num1:byval) if num1 > 10 then return 10 else return num1 + (generate(num1 + 1) DIV 2) endif endfunction
(a) Trace the algorithm to show the value returned when generate(7) is called. Show each step of your working.
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(b) The parameter, num1, is passed by value.
Explain why the parameter was passed by value instead of by reference.
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(c)* Parameters can be used to reduce the use of global variables.
Compare the use of parameters to global variables in recursive functions.
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14 (d) A student called Jason writes a recursive algorithm. The recursive algorithm uses more
memory than if Jason had written it as an iterative algorithm.
Explain why the recursive algorithm uses more memory than the iterative algorithm.
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5 A computer program stores data input on a stack named dataItems. The stack has two sub-programs to add and remove data items from the stack. The stack is implemented as a 1D array, dataArray.
Sub-program Description
push() The parameter is added to the top of the stack
pop() The element at the top of the stack is removed
The current contents of dataItems are shown:
6
15
100
23
(a) Show the contents of the stack dataItems after each line of the following lines of code are run
01 push(13) 02 pop() 03 push(10) 04 push(20)
Line 01 Line 02 Line 03 Line 04 Line 01 Line 02 Line 03 Line 04
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100
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16 (b) The main program asks a user to push or pop an item from the stack. If the user chooses
‘push’, the data item is added to the stack. If the user chooses “pop”, the next item is removed from the stack, multiplied by 3 and output.
The main program is shown:
01 userAnswer = input("Would you like to push or pop an item?") 02 if userAnswer == "push" then 03 push(input("Enter data item")) 04 else 05 print(pop() * 3) 06 endif
(i) Before the sub-programs, push() and pop(), can add or remove items from the stack, a selection statement is used to decide if each action is possible.
Describe the decision that needs to be made in each sub-program and how this impacts the next process.
push() ..............................................................................................................................
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(ii) The algorithm does not work when the user enters "PUSH" or "Push". The algorithm needs to be changed in order to accept these inputs.
Identify the line number to be changed and state the change that should be made.
Line number .......................................................................................................................
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(c) The stack is implemented as a 1D array, dataArray.
Describe how a 1D array can be set up and used to push and pop items as a stack.
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18 (d) As an array, the data in dataArray is sorted and then searched for a specific value.
(i) The data in dataArray is sorted into ascending order using an insertion sort.
The current contents of dataArray are shown:
100 22 5 36 999 12
Show the steps of an insertion sort on the current contents of the array dataArray.
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(ii) The array dataArray can now be searched using a binary search.
Describe the stages of a binary search on an array of size n.
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20 (iii) The array has 50 items.
The function, searchItem(), performs a linear search for a data item.
function searchItem(dataItem) for count = 0 to 49 if dataArray[count] == dataItem then return(count) endif next count return(-1) endfunction
Rewrite the function using a while loop.
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Section B
Answer all questions.
6 Kamran is writing a program to manipulate the data for a set of items.
For each item, the program needs to store:• Item name (e.g. Box)• Cost (e.g. 22.58)• Date of arrival (e.g. 1/5/2018)• Transferred (e.g. true)
The items are added to a queue for processing.
The queue is defined as a class, itemQueue.
itemQueue
theItems[10] : Itemshead : Integertail : IntegernumItems : Integer
constructorenqueuer()dequeuer()setnumItems()getnumItems()
The head attribute points to the first element in the queue. The tail attribute points to the next available space in the queue. The numItems attribute states how many items are currently in the queue.
(a) The data about the items can be stored using either a record structure, or as objects of a class.
(i) Explain the similarities and differences between a record and a class.
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22 (ii) Kamran chooses to use a record structure to store the data about the items.
Record structures may be declared using the following syntax:
recordStructure recordstructurename fieldname : datatype … endRecordStructure
Complete the pseudocode to declare a record called items.
recordStructure …………………………………………………………………………….
itemName : …………………………………………………………………
…………………………………………………….……………: Currency
…………………………………………………….……………: Date
transferred : …………………………………………………….……………
endRecordStructure
[5]
(iii) New records may be created using the following syntax:
recordidentifier : recordstructurename recordidentifier.fieldname = data …
Write a programming statement to create a new item, using the identifier ‘box1’, with the item name “Box”, the cost 22.58, date of arrival 1/5/2018 and transferred true.
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(b) The array, theItems, stores the items in the queue. When the tail of the queue exceeds the last element in the array, it adds a new item to the first element if it is vacant.
For example, in the following queue, the next item to be added would be placed at index 0.
Index 0 1 2 3 4 5 6 7 8 9
Element Data Data Data Data Data Data Data
(i) Define the term ‘queue’.
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(ii) The attributes in itemQueue are all declared as private.
Explain how a private attribute improves the integrity of the data.
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(iii) The constructor method creates a new instance of itemQueue and sets the head, tail and numItems attributes to 0.
Write an algorithm, using pseudocode or program code, for the constructor including the initialisation for all attributes.
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24 (iv) The enqueue method:
• takes as a parameter the item to insert in the queue• checks if the queue is full• reports an error and returns false if the queue is full• does the following if the queue is not full:
o adds the item to the array at the tail position and adjusts the pointer(s)o returns true
The attribute numItems stores the number of items currently in the queue.
Write an algorithm, using pseudocode or program code, for the enqueue method.
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(v) Write a programming statement to declare an instance of itemQueue called myItems.
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(vi) Write a procedure, insertItems(), to ask the user to input the data for an item. The item is then added to the queue myItems. The user is continually asked to input data items until the queue is full.
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26 (vii) When the main program ends, the items and the queue no longer exist.
Describe how Kamran could amend the program to make sure the items and queue still exist and are used the next time the program is run.
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(c)* Kamran wants to expand the program to allow it to handle up to 100,000,000 items and to allow him to search for data about items. Kamran is worried that the increase in the number of items will cause a decrease in the performance of the program. He decides to investigate the benefits of caching and concurrent processing.
Evaluate the use of caching and concurrent processing in this scenario and make a recommendation to Kamran.
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END OF QUESTION PAPER
28
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Oxford Cambridge and RSA Examinations
GCE
Computer Science
Unit H446/02: Algorithms and programming
Advanced GCE
Mark Scheme for June 2018
OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. © OCR 2018
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tion
M
ean
ing
Om
issio
n m
ark
Be
ne
fit of th
e d
oubt
Inco
rrect p
oin
t
Follo
w th
rou
gh
No
t an
sw
ere
d q
uestio
n
No b
en
efit o
f doubt g
ive
n
Re
pe
at
Co
rrect p
oin
t
To
o v
ag
ue
Bla
nk P
ag
e –
this
annota
tion m
ust b
e u
sed
on a
ll bla
nk p
ag
es w
ithin
an a
nsw
er b
ookle
t (stru
ctu
red o
r unstru
ctu
red) a
nd o
n e
ach p
ag
e o
f an a
dd
itional o
bje
ct w
he
re th
ere
is n
o c
an
did
ate
response.
Leve
l 1
Leve
l 2
Leve
l 3
H4
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/02
M
ark
Sc
he
me
Ju
ne
20
18
4
Su
bje
ct-s
pe
cific
Ma
rkin
g In
stru
ctio
ns
IN
TR
OD
UC
TIO
N
Yo
ur firs
t task a
s a
n E
xa
min
er is
to b
ecom
e th
oro
ug
hly
fam
iliar w
ith th
e m
ate
rial o
n w
hic
h th
e e
xa
min
atio
n d
ep
en
ds. T
his
ma
teria
l inclu
de
s:
th
e s
pecific
atio
n, e
spe
cia
lly th
e a
ssessm
ent o
bje
ctiv
es
th
e q
uestio
n p
ap
er a
nd its
rubric
s
th
e m
ark
sch
em
e.
Yo
u s
hou
ld e
nsu
re th
at y
ou h
ave
co
pie
s o
f these
ma
teria
ls.
Yo
u s
hou
ld e
nsure
als
o th
at y
ou a
re fa
milia
r with
the a
dm
inis
trativ
e p
rocedu
res re
late
d to
the m
ark
ing p
roce
ss. T
hese
are
se
t out in
the O
CR
b
oo
kle
t Instru
ctio
ns fo
r Ex
am
iners
. If yo
u a
re e
xa
min
ing
for th
e firs
t time
, ple
ase
read c
are
fully
Ap
pen
dix
5 In
trod
uc
tion
to S
crip
t Ma
rkin
g:
No
tes
for N
ew
Ex
am
iners
. P
lease
ask fo
r help
or g
uid
an
ce w
hen
eve
r yo
u n
ee
d it. Y
our firs
t poin
t of c
onta
ct is
yo
ur T
eam
Lead
er.
H4
46
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M
ark
Sc
he
me
Ju
ne
20
18
5
US
ING
TH
E M
AR
K S
CH
EM
E
Ple
ase
stu
dy th
is M
ark
Sch
em
e c
are
fully
. Th
e M
ark
Sch
em
e is
an in
teg
ral p
art o
f the p
rocess th
at b
eg
ins w
ith th
e s
ettin
g o
f the q
uestio
n p
ap
er
and
en
ds w
ith th
e a
wa
rdin
g o
f gra
de
s. Q
uestio
n p
ape
rs a
nd M
ark
Sch
em
es a
re d
eve
lope
d in
asso
cia
tion
with
ea
ch
oth
er s
o th
at is
su
es o
f d
iffere
ntia
tion
an
d p
ositiv
e a
ch
ieve
me
nt c
an b
e a
ddre
ssed
from
the v
ery
sta
rt. T
his
Ma
rk S
ch
em
e is
a w
ork
ing d
ocu
men
t; it is n
ot e
xh
au
stiv
e; it d
oe
s n
ot p
rovid
e ‘c
orre
ct’ a
nsw
ers
. Th
e M
ark
Sch
em
e c
an o
nly
pro
vid
e ‘b
est
gu
esses’ a
bo
ut h
ow
the q
uestio
n w
ill wo
rk o
ut, a
nd it is
su
bje
ct to
revis
ion a
fter w
e h
ave
loo
ke
d a
t a w
ide ra
ng
e o
f scrip
ts.
Th
e E
xa
min
ers
’ Sta
nd
ard
isa
tion
Me
etin
g w
ill ensu
re th
at th
e M
ark
Sch
em
e c
ove
rs th
e ra
ng
e o
f can
did
ate
s’ re
spo
nses to
the
qu
estio
ns, a
nd th
at a
ll E
xa
min
ers
und
ers
tand
an
d a
pp
ly th
e M
ark
Sch
em
e in
the s
am
e w
ay. T
he
Ma
rk S
ch
em
e w
ill be d
iscu
ssed
an
d a
me
nd
ed
at th
e m
eetin
g, a
nd
adm
inis
trativ
e p
roced
ure
s w
ill be c
onfirm
ed. C
o-o
rdin
atio
n s
crip
ts w
ill be is
su
ed
at th
e m
eetin
g to
exe
mp
lify a
spe
cts
of c
andid
ate
s’ re
spo
nses a
nd
achie
ve
me
nts
; the c
o-o
rdin
atio
n s
crip
ts th
en
be
co
me
part o
f this
Ma
rk S
ch
em
e.
Be
fore
the S
tand
ard
isatio
n M
eetin
g, y
ou s
hou
ld re
ad
an
d m
ark
in p
en
cil a
num
ber o
f scrip
ts, in
ord
er to
ga
in a
n im
pre
ssio
n o
f the ra
ng
e o
f re
spo
nse
s a
nd
ach
ieve
me
nt th
at m
ay b
e e
xp
ecte
d.
In y
our m
ark
ing
, yo
u w
ill enco
un
ter v
alid
respo
nse
s w
hic
h a
re n
ot c
ove
red b
y th
e M
ark
Sch
em
e: th
ese re
spo
nse
s m
ust b
e c
redite
d. Y
ou w
ill e
nco
un
ter a
nsw
ers
wh
ich
fall o
uts
ide th
e ‘ta
rge
t rang
e’ o
f Ba
nd
s fo
r the p
ape
r wh
ich
yo
u a
re m
ark
ing
. Ple
ase
ma
rk th
ese
an
sw
ers
accord
ing
to th
e
ma
rkin
g c
riteria
. P
lea
se re
ad
care
fully
all th
e s
crip
ts in
yo
ur a
llocatio
n a
nd
make
eve
ry e
ffort to
look p
ositiv
ely
for a
ch
ieve
me
nt th
roug
hou
t the
ability
rang
e. A
lwa
ys
be p
rep
are
d to
use th
e fu
ll rang
e o
f ma
rks.
H4
46
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M
ark
Sc
he
me
Ju
ne
20
18
6
LE
VE
LS
OF
RE
SP
ON
SE
QU
ES
TIO
NS
: T
he in
dic
ativ
e c
on
tent in
dic
ate
s th
e e
xp
ecte
d p
ara
me
ters
for c
and
idate
s’ a
nsw
ers
, but b
e p
repa
red
to re
cog
nis
e a
nd
cre
dit u
ne
xp
ecte
d
app
roa
ch
es w
here
they s
how
rele
va
nce
. U
sin
g ‘b
est-fit’, d
ecid
e firs
t wh
ich
se
t of B
AN
D D
ES
CR
IPT
OR
S b
est d
escrib
es th
e o
ve
rall q
uality
of th
e a
nsw
er. O
nce
the b
an
d is
locate
d, a
dju
st
the m
ark
co
nce
ntra
ting
on
featu
res o
f the a
nsw
er w
hic
h m
ake it s
trong
er o
r we
ake
r follo
win
g th
e g
uid
elin
es fo
r refin
em
ent.
H
igh
est m
ark
: If cle
ar e
vid
en
ce o
f all th
e q
ualitie
s in
the b
an
d d
escrip
tors
is s
how
n, th
e H
IGH
ES
T M
ark
sh
ou
ld b
e a
wa
rde
d.
L
ow
est m
ark
: If the a
nsw
er s
how
s th
e c
and
idate
to b
e b
ord
erlin
e (i.e
. they h
ave
ach
ieve
d a
ll the q
ua
lities o
f the b
an
ds b
elo
w a
nd
sh
ow
lim
ited e
vid
en
ce o
f me
etin
g th
e c
riteria
of th
e b
and
in q
uestio
n) th
e L
OW
ES
T m
ark
sh
ou
ld b
e a
wa
rde
d.
M
idd
le m
ark
: Th
is m
ark
sh
ou
ld b
e u
sed fo
r ca
nd
idate
s w
ho a
re s
ecure
in th
e b
an
d. T
hey a
re n
ot ‘b
ord
erlin
e’ b
ut th
ey h
ave
only
achie
ve
d
so
me
of th
e q
ualitie
s in
the b
an
d d
escrip
tors
. B
e p
repa
red to
use th
e fu
ll rang
e o
f ma
rks. D
o n
ot re
serv
e (e
.g.) h
igh B
an
d 3
ma
rks ‘in
ca
se
’ so
me
thin
g tu
rns u
p o
f a q
uality
yo
u h
ave
no
t ye
t se
en
. If an a
nsw
er g
ive
s c
lear e
vid
en
ce o
f the q
ua
lities d
escrib
ed
in th
e b
and
de
scrip
tors
, rew
ard
app
rop
riate
ly.
H4
46
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M
ark
Sc
he
me
Ju
ne
20
18
7
A
O1
A
O2
A
O3
Hig
h (th
oro
ug
h)
Pre
cis
ion in
the u
se o
f que
stio
n
term
inolo
gy. K
now
ledg
e s
how
n is
co
nsis
tent a
nd
we
ll-deve
lope
d.
Cle
ar a
pp
recia
tion o
f the q
uestio
n
from
a ra
ng
e o
f diffe
rent
pers
pectiv
es m
akin
g e
xte
nsiv
e
use o
f acq
uire
d k
now
led
ge a
nd
u
nd
ers
tand
ing
.
Kn
ow
ledg
e a
nd
un
de
rsta
ndin
g
sh
ow
n is
co
nsis
tently
app
lied to
co
nte
xt e
na
blin
g a
log
ical a
nd
su
sta
ined
arg
um
ent to
de
ve
lop.
Exa
mp
les u
se
d e
nh
an
ce ra
ther
than
detra
ct fro
m re
spo
nse
.
Co
nce
rted e
ffort is
ma
de to
co
nsid
er a
ll aspe
cts
of a
syste
m /
pro
ble
m o
r we
igh
up b
oth
sid
es to
a
n a
rgum
ent b
efo
re fo
rmin
g a
n
ove
rall c
on
clu
sio
n. J
udg
em
ents
m
ade
are
ba
se
d o
n a
pp
ropria
te
and
co
ncis
e a
rgu
me
nts
that h
ave
b
ee
n d
eve
lope
d in
respon
se
resultin
g in
them
bein
g b
oth
su
pp
orte
d a
nd
realis
tic.
Mid
dle
(rea
so
nab
le)
Aw
are
ne
ss o
f the m
ean
ing o
f the
term
s in
the q
uestio
n. K
no
wle
dg
e
is s
oun
d a
nd e
ffectiv
ely
d
em
onstra
ted. D
em
and
s o
f q
uestio
n u
nd
ers
tood a
ltho
ug
h a
t tim
es o
pp
ortu
nitie
s to
ma
ke u
se o
f a
cq
uire
d k
now
ledg
e a
nd
und
ers
tand
ing
not a
lwa
ys ta
ke
n.
Kn
ow
ledg
e a
nd
un
de
rsta
ndin
g
app
lied to
co
nte
xt. W
hils
t cle
ar
evid
en
ce th
at a
n a
rgum
en
t build
s
and
de
ve
lops th
roug
h re
sp
on
se
there
are
times w
hen
opp
ortu
nitie
s
are
mis
se
d to
use a
n e
xa
mp
le o
r re
late
an a
spe
ct o
f kn
ow
ledg
e o
r u
nd
ers
tand
ing
to th
e c
onte
xt
pro
vid
ed
.
Th
ere
is a
reaso
na
ble
atte
mp
t to
reach
a c
onclu
sio
n c
onsid
erin
g
aspe
cts
of a
syste
m / p
rob
lem
or
we
igh
ing u
p b
oth
sid
es o
f an
arg
um
ent. H
ow
eve
r the im
pact o
f th
e c
onclu
sio
n is
ofte
n le
sse
ne
d
by a
lack o
f su
pp
orte
d ju
dg
em
ents
w
hic
h a
cco
mp
an
y it. T
his
inab
ility
to b
uild
on
an
d d
eve
lop lin
es o
f a
rgu
me
nt a
s d
eve
lope
d in
the
respo
nse
ca
n d
etra
ct fro
m th
e
ove
rall q
uality
of th
e re
spo
nse
.
Lo
w (b
as
ic)
Co
nfu
sio
n a
nd
ina
bility
to
deco
nstru
ct te
rmin
olo
gy a
s u
se
d in
th
e q
uestio
n. K
now
ledg
e p
artia
l a
nd
su
pe
rficia
l. Focu
s o
n q
uestio
n
narro
w a
nd
ofte
n o
ne
-dim
ensio
na
l.
Inab
ility to
ap
ply
kn
ow
ledg
e a
nd
und
ers
tand
ing
in a
ny s
usta
ined
w
ay to
co
nte
xt re
sultin
g in
tenu
ou
s
and
un
su
pp
orte
d s
tate
me
nts
bein
g
ma
de
. Exa
mp
les if u
se
d a
re fo
r th
e m
ost p
art irre
leva
nt a
nd
unsu
bsta
ntia
ted.
Little
or n
o a
ttem
pt to
prio
ritise o
r w
eig
h u
p fa
cto
rs d
urin
g c
ours
e o
f a
nsw
er. C
onclu
sio
n is
ofte
n
dis
locate
d fro
m re
sp
on
se
an
d a
ny
judg
em
ents
lack s
ubsta
nce
du
e in
p
art to
the b
asic
leve
l of a
rgu
me
nt
that h
as b
ee
n d
em
onstra
ted
thro
ug
hout re
spo
nse.
H4
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M
ark
Sc
he
me
Ju
ne
20
18
8
A
ss
es
sm
en
t Ob
jec
tive
AO
1
De
mo
nstra
te k
now
ledg
e a
nd
un
ders
tand
ing
of th
e p
rincip
les a
nd
co
nce
pts
of c
om
pute
r scie
nce
, inclu
din
g a
bstra
ctio
n, lo
gic
, alg
orith
ms
and
da
ta re
pre
sen
tatio
n.
AO
1.1
D
em
on
stra
te k
no
wle
dg
e o
f the p
rincip
les a
nd c
on
cep
ts o
f abstra
ctio
n, lo
gic
, alg
orith
ms, d
ata
repre
se
nta
tion
or o
ther a
s a
ppro
pria
te.
AO
1.2
D
em
on
stra
te u
nd
ers
tan
din
g o
f the p
rincip
les a
nd
co
nce
pts
of a
bstra
ctio
n, lo
gic
, alg
orith
ms, d
ata
repre
sen
tatio
n o
r oth
er a
s a
pp
ropria
te.
AO
2
Ap
ply
kn
ow
ledg
e a
nd
und
ers
tand
ing
of th
e p
rincip
les a
nd
co
nce
pts
of c
om
pute
r scie
nce
inclu
din
g to
ana
lyse
pro
ble
ms in
co
mp
uta
tiona
l te
rms.
AO
2.1
A
pply
kn
ow
ledg
e a
nd
und
ers
tand
ing
of th
e p
rincip
les a
nd
co
nce
pts
of c
om
pute
r scie
nce
.
AO
2.2
A
naly
se
pro
ble
ms in
com
puta
tiona
l term
s.
AO
3
De
sig
n, p
rog
ram
and
eva
luate
com
pute
r syste
ms th
at s
olv
e p
roble
ms, m
akin
g re
aso
ne
d ju
dg
em
ents
abo
ut th
ese a
nd
pre
sen
ting
co
nclu
sio
ns.
AO
3.1
D
esig
n c
om
pute
r syste
ms th
at s
olv
e p
roble
ms.
AO
3.2
P
rogra
m c
om
pute
r syste
ms th
at s
olv
e p
roble
ms.
AO
3.3
E
va
luate
co
mp
ute
r syste
ms th
at s
olv
e p
roble
ms, m
akin
g re
ason
ed
judg
em
ents
abo
ut th
ese
an
d p
resen
ting
co
nclu
sio
ns.
H4
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Ju
ne
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9
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
1
(a)
1
mark
for e
ach o
f: -
Sco
tlan
d in
corre
ct p
lace
- W
ale
s in
corre
ct p
lace
- A
ustra
lia a
nd
Eng
land b
oth
in c
orre
ct p
lace
3
AO
2.2
(3)
1
(b)
1
mark
per b
ulle
t to m
ax
Ita
ly
F
ran
ce, S
pain
A
ustria
, Germ
any, N
orw
ay
3
AO
1.1
(1
) A
O2
.1 (1
) A
O2
.2 (1
)
H4
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M
ark
Sc
he
me
Ju
ne
20
18
10
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
1
(c)
(i) 1
mark
per b
ulle
t to m
ax 5
function searchForData(currentNode:byVal, searchValue:byVal)
thisNode = getData(currentNode)
if thisNode == searchValue then
return true
elseif thisNode < searchValue then
if currentNode.left () != null then
return (searchForData(currentNode.left (),
searchValue))
else
return false
endif
else
if currentNode.right() != null then
return (searchForData(currentNode.right (),
searchValue))
else
return false
endif
endif
endfunction
5
AO
2.2
(2)
AO
3.2
(3)
Th
e lin
e elseif thisNode <
searchValue then
sh
ou
ld
have
read elseif thisNode
> searchValue then
If ca
nd
idate
s a
ttem
pt to
co
rrect
the c
ode a
nd
their a
nsw
ers
are
co
nsis
tent w
ith, a
nd
wo
rk w
ith
their a
me
ndm
en
t, su
ch
an
sw
ers
sh
ou
ld b
e c
redite
d.
1
(c)
(ii)
It’s a
bin
ary
tree
It’s
ord
ere
d / s
orte
d
2
AO
2.2
(2)
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
11
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
2
(a)
(i) R
ecog
nitio
n
Id
en
tify th
ere
is a
pro
ble
m to
be s
olv
ed // w
hat th
e
pro
ble
m is
D
ecom
positio
n
S
plittin
g d
ow
n a
pro
ble
m in
to s
ub-p
rob
lem
s
2
AO
1.1
(2
)
2
(a)
(ii) e
.g.
D
ivid
e a
nd
co
nq
uer
A
bstra
ctio
n
1
AO
1.1
(1
)
Acce
pt o
ther c
red
ible
an
sw
ers
e.g
.: C
ritica
l thin
kin
g, M
ode
lling
, He
uris
tics, C
oncu
rrency,
Vis
ualis
atio
n, B
acktra
ckin
g
2
(b)
(i)
Tu
rnin
g la
rge
qu
antitie
s o
f data
into
usefu
l info
rma
tion
/ F
indin
g p
atte
rns w
ithin
larg
e q
uantitie
s o
f in
form
atio
n
1
AO
1.1
(1)
Mu
st re
fer to
larg
e q
uan
tities o
f data
2
(b)
(ii) 1
mark
per id
entify
ing
data
, 1 fo
r use
e.g
.
Id
en
tify c
usto
mer tre
nd
s
T
o id
en
tify ite
ms to
se
ll/offe
rs to
sen
d c
usto
me
rs
Id
en
tify w
hic
h s
tore
s a
re m
akin
g th
e m
ost p
rofit
T
o id
en
tify w
hat th
e o
ther s
tore
s a
re d
oin
g w
ell
W
hic
h ite
ms a
re n
ot s
ellin
g w
ell
T
o re
pla
ce th
em
with
oth
er ite
ms
4
AO
2.2
(4
) A
cce
pt a
ny v
alid
respo
nse
s
2
(c)
(i) S
imu
late
/test th
e b
eh
avio
ur o
f the s
yste
m b
efo
re it is
used
1
AO
1.1
(1)
2
(c)
(ii) e
.g.
T
estin
g it w
ith a
larg
e n
um
ber o
f sim
ulta
ne
ous o
rders
(s
tress te
stin
g)
T
estin
g it w
ith a
larg
e n
um
ber o
f cu
sto
me
rs/ite
ms/o
rders
1
AO
2.2
(1)
2
(d)
1
mark
per b
ulle
t to m
ax 2
e.g
.
th
e c
om
pon
ents
ca
n b
e u
sed
in a
futu
re p
rog
ram
…
th
ey d
o n
ot n
ee
d to
be re
writte
n / s
ave
s tim
e
2
AO
1.1
(1)
AO
2.1
(1)
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
12
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
th
ey h
ave
alre
ad
y b
ee
n te
ste
d…
…
it will s
ave
time
3
(a)
(i) A
ny o
ne
from
: -
A g
rap
h h
as c
ycle
s
- A
gra
ph
ca
n b
e d
irecte
d/u
nd
irecte
d
- A
tree h
as a
hie
rarc
hy (e
.g. P
are
nt/C
hild
)
1
AO
1.2
(1)
Allo
w a
ny a
pp
rop
riate
descrip
tion
e.g
. gra
ph c
an b
e
we
igh
ted, tre
e h
as a
root
3
(a)
(ii) 1
mark
per b
ulle
t to m
ax 2
T
he p
uzzle
is n
ot s
how
n in
the d
iag
ram
T
he g
raph s
how
s d
iffere
nt s
eq
uen
ces o
f su
b p
rob
lem
s
in th
e p
uzzle
that c
an b
e s
olv
ed to
ge
t to th
e fin
al
so
lutio
n
T
he p
uzzle
do
es n
ot h
ave
all s
tate
s v
isib
le a
t once
2
AO
1.2
(1)
AO
2.1
(1)
An
sw
ers
mu
st b
e in
co
nte
xt o
f the p
uzzle
3
(a)
(iii) 1
mark
per b
ulle
t to m
ax 2
e
.g.
V
isu
alis
atio
ns b
en
efit h
um
ans ra
ther th
an c
om
pute
rs
V
isu
alis
atio
ns p
resen
t the
info
rma
tion in
a s
imp
ler
form
to u
nd
ers
tan
d
V
isu
alis
atio
ns c
an
best e
xp
lain
co
mp
lex s
ituatio
ns
2
AO
1.1
(1)
AO
2.1
(1
)
3
(b)
1
mark
per b
ulle
t
M
ark
A a
s th
e in
itial n
ode
an
d th
en
vis
it B (5
)
N
ode
E (8
) is th
en v
isite
d (c
hose
n fro
m C
(13), D
(14), E
(8
))
N
ode
I (12) is
then
vis
ited
afte
r E
N
ode
J (1
4) is
then v
isite
d a
fter I
:
V
isitin
g G
(18) fro
m I;
V
isitin
g G
(15) fro
m C
– o
ve
rridin
g th
e p
revio
us v
alu
e o
f 18
so
lutio
n A
-B-E
-I-J p
ath
len
gth
14
7
AO
1.2
(3)
AO
2.1
(2)
AO
2.2
(2)
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
13
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
3
(c)
M
ark
Ba
nd
3 –
Hig
h le
ve
l (7
-9 m
ark
s)
Th
e c
and
idate
dem
onstra
tes a
thoro
ug
h k
now
led
ge a
nd
u
nd
ers
tand
ing
of D
ijkstra
’s a
nd
A*; th
e m
ate
rial is
ge
nera
lly
accura
te a
nd
deta
iled.
Th
e c
and
idate
is a
ble
to a
pp
ly th
eir k
now
ledg
e a
nd
u
nd
ers
tand
ing
dire
ctly
an
d c
onsis
tently
to th
e c
on
text
pro
vid
ed
. Evid
en
ce/e
xa
mp
les w
ill be e
xp
licitly
rele
va
nt to
th
e e
xp
lana
tion
. T
he
re is
a w
ell-d
eve
lope
d lin
e o
f reaso
nin
g w
hic
h is
cle
ar
and
log
ica
lly s
tructu
red. T
he
info
rma
tion p
rese
nte
d is
re
levan
t and s
ubsta
ntia
ted.
Ma
rk B
an
d 2
– M
id le
ve
l (4
-6 m
ark
s)
Th
e c
and
idate
dem
onstra
tes re
aso
na
ble
kn
ow
ledg
e a
nd
u
nd
ers
tand
ing
of D
ijkstra
’s a
nd
A*; th
e m
ate
rial is
ge
nera
lly
accura
te b
ut a
t time
s u
nd
erd
eve
lope
d.
Th
e c
and
idate
is a
ble
to a
pp
ly th
eir k
now
ledg
e a
nd
u
nd
ers
tand
ing
dire
ctly
to th
e c
onte
xt p
rovid
ed
alth
ou
gh o
ne
o
r two
op
portu
nitie
s a
re m
isse
d. E
vid
en
ce/e
xa
mp
les a
re fo
r th
e m
ost p
art im
plic
itly re
leva
nt to
the e
xp
lana
tion
. T
he c
and
idate
pro
vid
es a
reaso
na
ble
dis
cussio
n, th
e
ma
jority
of w
hic
h is
focuse
d. E
va
luativ
e c
om
me
nts
are
, for
the m
ost p
art a
pp
ropria
te, a
lthoug
h o
ne
or tw
o o
pp
ortu
nitie
s
for d
eve
lopm
ent a
re m
isse
d.
The
re is
a lin
e o
f reaso
nin
g p
rese
nte
d w
ith s
om
e s
tructu
re.
The
info
rma
tion p
rese
nte
d is
in th
e m
ost p
art re
leva
nt a
nd
su
pp
orte
d b
y s
om
e e
vid
ence
. M
ark
Ba
nd
1 –
Lo
w L
eve
l (1
-3 m
ark
s)
Th
e c
and
idate
dem
onstra
tes a
basic
kno
wle
dg
e o
f Dijk
stra
’s
and
A* w
ith lim
ited u
nd
ers
tand
ing s
how
n; th
e m
ate
rial is
b
asic
and
co
nta
ins s
om
e in
accu
racie
s. T
he c
and
idate
s
make
s a
limite
d a
ttem
pt to
ap
ply
acq
uire
d k
now
led
ge a
nd
9
AO
1.1
(2)
AO
1.2
(2)
AO
2.1
(2)
AO
3.3
(3)
AO
1: K
no
wle
dg
e a
nd
Un
de
rsta
nd
ing
In
dic
ativ
e c
on
ten
t
H
euris
tic h
elp
s p
rodu
ce a
so
lutio
n in
a fa
ste
r time
A
* uses e
stim
ate
d d
ista
nce
from
final n
od
e
D
ijkstra
uses a
we
igh
t/dis
tance
A
* ch
oo
ses w
hic
h p
ath
to ta
ke n
ext b
ase
d o
n lo
we
st
cu
rrent d
ista
nce
trave
lled
A
O2
: Ap
plic
atio
n
D
escrip
tion o
f how
A* w
ill diffe
r from
Dijk
stra
, e.g
. ta
kin
g th
e s
horte
r route
A-B
-E-I b
efo
re e
xp
lorin
g
nod
es fro
m D
and E
D
escrip
tion o
f the d
iffere
nt n
um
ber o
f com
paris
ons
that w
ould
be
nee
de
d in
this
pro
ble
m
A
* doe
sn’t n
ee
d to
find a
ll possib
le s
olu
tion
s (s
ave
s
time
) A
O3
: Eva
lua
tion
C
and
idate
s w
ill nee
d to
eva
luate
the b
en
efits
and
d
raw
backs o
f each
alg
orith
m
S
ma
ll-sca
le p
roble
m
Q
uic
k to
find a
so
lutio
n u
sin
g e
ither m
eth
od
D
iffere
nce in
pro
gra
mm
ing c
om
ple
xity
is m
inim
al
D
on’t k
now
if this
pro
ble
m n
ee
ds to
sca
le
M
ost e
fficie
nt ro
ute
ne
ede
d
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
14
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
und
ers
tand
ing
to th
e c
onte
xt p
rovid
ed
. T
he c
and
idate
pro
vid
es a
limite
d d
iscu
ssio
n w
hic
h is
na
rrow
in
focus. J
udg
em
ents
if ma
de
are
we
ak a
nd
u
nsu
bsta
ntia
ted.
The
info
rma
tion is
basic
and
co
mu
nic
ate
d in
an
unstru
ctu
red w
ay. T
he
info
rma
tion
is s
upp
orte
d b
y lim
ited
evid
en
ce a
nd
the re
latio
nship
to th
e e
vid
en
ce m
ay n
ot b
e
cle
ar.
0 m
ark
s
No
atte
mpt to
an
sw
er th
e q
uestio
n o
r respo
nse is
not w
orth
y
of c
redit.
3
(d)
1
mark
per b
ulle
t to m
ax 4
e
.g.
U
nde
rlines s
yn
tax e
rrors
dyn
am
ica
lly
C
an b
e c
orre
cte
d b
efo
re ru
nn
ing // s
ave
s tim
es
W
atc
h w
indo
w
V
iew
how
va
riab
les c
ha
ng
e d
urin
g ru
nn
ing o
f the
pro
gra
m
B
reak p
oin
ts
S
top th
e p
rog
ram
at s
et p
oin
ts to
ch
eck th
e v
alu
es o
f va
riab
les
E
rror m
essag
e lis
t
T
ells
yo
u w
here
erro
rs a
re a
nd
sug
gests
co
rrectio
ns
S
tep-m
ode
E
xe
cute
s p
rog
ram
one
sta
tem
ent a
t a tim
e to
wa
tch
va
riab
le v
alu
es a
nd
pro
gra
m p
ath
wa
ys
T
race
s
P
rint-o
uts
of v
aria
ble
va
lues fo
r each s
tate
me
nt
6
AO
1.1
(3)
AO
1.2
(3)
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
15
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
exe
cutio
n w
ithin
a p
rog
ram
Cra
sh-d
um
p/p
ost-m
orte
m ro
utin
e
S
how
s th
e s
tate
of v
aria
ble
s w
here
an
erro
r occurs
S
tack c
onte
nts
S
how
s s
eq
uen
cin
g th
roug
h p
roce
dure
s/m
odu
les
C
ross-re
fere
nce
rs
Id
en
tifies w
here
va
riab
les/c
onsta
nts
are
used in
a
pro
gra
m to
avo
id d
up
lica
tion
s
4
(a)
1
mark
per b
ulle
t for w
ork
ing
to m
ax 6
g
enera
te(7
) re
turn
7 +
(ge
ne
rate
(8) D
IV 2
)
g
enera
te(8
) re
turn
8 +
(ge
nera
te(9
) DIV
2)
ge
nera
te(9
) re
turn
9 +
(ge
nera
te(1
0) D
IV 2
) g
enera
te(1
0)
retu
rn 1
0 +
(ge
nera
te(1
1) D
IV 2
)
g
enera
te(1
1)
retu
rn 1
0
R
ew
indin
g: re
turn
10 +
(10 D
IV 2
) = 1
0 +
5 =
15
re
turn
9 +
(15 D
IV 2
) = 9
+ 7
= 1
6
retu
rn 8
+ (1
6 D
IV 2
) = 8
+ 8
= 1
6
re
turn
7 +
(16 D
IV 2
) = 7
+ 8
= 1
5
6
AO
1.2
(1)
AO
2.2
(5)
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
16
Qu
es
tion
A
nsw
er
Ma
rks
Gu
idan
ce
4
(b)
If the v
alu
e is
se
nt b
y v
alu
e, n
um
1 w
ill not b
e o
ve
rridde
n
/ it is a
co
py o
f the p
ara
me
ter th
at is
used (1
) and
this
w
ill pro
du
ce th
e c
orre
ct o
utp
ut (1
)
if th
e p
ara
me
ter h
ad
be
en
pa
sse
d b
y re
fere
nce
it wo
uld
n
ot p
rodu
ce th
e c
orre
ct re
sult (1
) as n
um
1 w
ould
be
o
ve
rridde
n / b
eca
use
it is a
po
inte
r to th
e a
dd
ress o
f the
va
riab
le (1
)
2
AO
2.1
(1)
AO
2.2
(1)
4
(c)
M
ark
Ba
nd
3 –
Hig
h le
ve
l (7
-9 m
ark
s)
Th
e c
and
idate
dem
onstra
tes a
thoro
ug
h k
now
led
ge a
nd
u
nd
ers
tand
ing
of p
ara
mete
rs a
nd
glo
ba
l va
riable
s; th
e
ma
teria
l is g
ene
rally
accu
rate
and
deta
iled.
Th
e c
and
idate
is a
ble
to a
pp
ly th
eir k
now
ledg
e a
nd
u
nd
ers
tand
ing
dire
ctly
an
d c
onsis
tently
to th
e c
on
text
pro
vid
ed
. Evid
en
ce/e
xa
mp
les w
ill be e
xp
licitly
rele
va
nt to
th
e e
xp
lana
tion
. T
he
re is
a w
ell-d
evelo
pe
d lin
e o
f reaso
nin
g w
hic
h is
cle
ar
and
log
ica
lly s
tructu
red. T
he
info
rma
tion p
rese
nte
d is
re
levan
t and s
ubsta
ntia
ted.
Ma
rk B
an
d 2
– M
id le
ve
l (4
-6 m
ark
s)
Th
e c
and
idate
dem
onstra
tes re
aso
na
ble
kn
ow
ledg
e a
nd
u
nd
ers
tand
ing
of p
ara
mete
rs a
nd
glo
ba
l va
riable
s; th
e
ma
teria
l is g
ene
rally
accu
rate
but a
t time
s u
nd
erd
eve
lope
d.
Th
e c
and
idate
is a
ble
to a
pp
ly th
eir k
now
ledg
e a
nd
u
nd
ers
tand
ing
dire
ctly
to th
e c
onte
xt p
rovid
ed
alth
ou
gh o
ne
o
r two
op
portu
nitie
s a
re m
isse
d. E
vid
en
ce/e
xa
mp
les a
re fo
r th
e m
ost p
art im
plic
itly re
leva
nt to
the e
xp
lana
tion
. T
he c
and
idate
pro
vid
es a
reaso
na
ble
dis
cussio
n, th
e
ma
jority
of w
hic
h is
focuse
d. E
va
luativ
e c
om
me
nts
are
, for
the m
ost p
art a
pp
ropria
te, a
lthoug
h o
ne
or tw
o o
pp
ortu
nitie
s
for d
eve
lopm
ent a
re m
isse
d.
The
re is
a lin
e o
f reaso
nin
g p
rese
nte
d w
ith s
om
e s
tructu
re.
The
info
rma
tion p
rese
nte
d is
in th
e m
ost p
art re
leva
nt a
nd
9
AO
1.1
(2)
AO
1.2
(2)
AO
2.1
(2)
AO
3.3
(3)
AO
1: K
no
wle
dg
e a
nd
Un
de
rsta
nd
ing
In
dic
ativ
e c
on
ten
t
P
ara
mete
r allo
ws a
va
lue
to b
e s
ent to
a s
ub-
pro
gra
m
G
loba
l va
riab
les c
an
be a
ccesse
d th
roug
hout th
e
sco
pe
of th
e p
rog
ram
L
oca
l va
riab
les c
an
only
be a
ccesse
d w
ithin
the
sco
pe
of th
e s
ub-p
rog
ram
it's d
efin
ed
with
in –
a
para
mete
r becom
es a
local v
aria
ble
in th
e fu
nctio
n
AO
2: A
pp
licatio
n
If g
loba
l, eq
uiv
ale
nt o
f by re
fere
nce -v
alu
e w
ould
be
o
ve
r-ridde
n
G
loba
l varia
ble
take
s m
ore
me
mory
than a
loca
l va
riab
le/p
ara
me
ter
In
recurs
ion, e
ach c
all p
rodu
ces a
ne
w lo
cal
va
riab
le fo
r num
1
AO
3: E
va
lua
tion
C
and
idate
s w
ill nee
d to
eva
luate
the b
en
efits
and
d
raw
backs o
f each
alg
orith
m
G
loba
l wo
uld
req
uire
alte
ring th
e a
lgorith
m a
s th
e
va
lue w
ould
be
ove
r-ridde
n o
n e
ach c
all
G
loba
l wo
uld
me
an
that m
em
ory
sp
ace
is k
ept
thro
ug
hout th
e ru
nn
ing o
f the p
rog
ram
, not ju
st th
e
su
b-p
rog
ram
P
ara
mete
r ena
ble
s m
em
ory
to b
e re
allo
cate
d
M
an
y m
ore
me
mory
sp
ace
s n
ee
de
d fo
r para
mete
r
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
17
Qu
es
tion
A
nsw
er
Ma
rks
Gu
idan
ce
su
pp
orte
d b
y s
om
e e
vid
ence
. M
ark
Ba
nd
1 –
Lo
w L
eve
l (1
-3 m
ark
s)
Th
e c
and
idate
dem
onstra
tes a
basic
kno
wle
dg
e o
f p
ara
mete
rs a
nd
glo
ba
l va
riab
les w
ith lim
ited u
nde
rsta
nd
ing
sh
ow
n; th
e m
ate
rial is
basic
and
co
nta
ins s
om
e
inaccu
racie
s. T
he c
and
idate
s m
ake
s a
limite
d a
ttem
pt to
a
pp
ly a
cq
uire
d k
now
ledg
e a
nd
un
ders
tand
ing
to th
e c
onte
xt
pro
vid
ed
. T
he c
and
idate
pro
vid
es a
limite
d d
iscu
ssio
n w
hic
h is
na
rrow
in
focus. J
udg
em
ents
if ma
de
are
we
ak a
nd
u
nsu
bsta
ntia
ted.
The
info
rma
tion is
basic
and
co
mu
nic
ate
d in
an
unstru
ctu
red w
ay. T
he
info
rma
tion
is s
upp
orte
d b
y lim
ited
evid
en
ce a
nd
the re
latio
nship
to th
e e
vid
en
ce m
ay n
ot b
e
cle
ar.
0 m
ark
s
No
atte
mpt to
an
sw
er th
e q
uestio
n o
r respo
nse is
not w
orth
y
of c
redit.
in re
curs
ion, 1
for e
ach
ca
ll
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
18
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
4
(d)
1
mark
per b
ulle
t
E
ach re
curs
ive
ca
ll sto
res th
e c
urre
nt s
tate
on th
e s
tack // c
reate
s n
ew
va
riable
s
Ite
ratio
n re
use
s th
e s
am
e v
aria
ble
s
2
AO
1.2
(1)
AO
2.1
(1)
5
(a)
1
mark
for e
ach c
orre
ct s
tack
20
13
10
10
6
6
6
6
15
15
15
15
100
100
100
100
23
23
23
23
4
AO
1.2
(2)
AO
2.2
(2)
5
(b)
(i) 1
mark
per b
ulle
t, ma
x 2
for in
sert, m
ax 2
for re
mo
ve
p
ush
C
heck if th
e s
tack is
full (p
oin
ter =
arra
y.le
ng
th/a
rray.le
ng
th+
1)
If it is
not –
insert th
e ite
m
If it is
– re
turn
/erro
r that th
e s
tack is
full
pop
C
heck if th
e s
tack is
em
pty
(poin
ter =
0/1
)
If it is
– re
turn
/erro
r that th
e s
tack is
em
pty
If it is
not –
retu
rn th
e ite
m
4
AO
1.2
(2)
AO
2.2
(2)
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
19
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
5
(b)
(ii) 1
mark
per lin
e, 1
for c
han
ge
lin
e 0
2
In
clu
de
an
OR
with
va
riatio
ns (e
.g. userAnswer = "PUSH" OR userAnswer =
"Push"
etc
.)/Co
nve
rt inp
ut to
up
perc
ase/lo
we
rca
se
and
just c
om
pare
to e
qu
iva
lent
2
AO
2.2
(2)
5
(c)
1
mark
per b
ulle
t to m
ax 3
A
rray s
ize
defin
ed
A
sta
ck p
oin
ter is
used
to p
oin
t to th
e to
p o
f the s
tack
W
hen
an ite
m is
pushe
d th
e s
tack p
oin
ter is
incre
me
nte
d
W
hen
an ite
m is
pop
pe
d th
e s
tack p
oin
ter is
decre
me
nte
d
3
AO
1.2
(1)
AO
2.1
(1)
AO
2.2
(1)
5
(d)
(i) 1
mark
per ro
w (a
fter firs
t row
)
100
22
5
36
999
12
22
100
5
36
999
12
1 m
ark
5
22
100
36
999
12
1 m
ark
5
22
36
100
999
12
1 m
ark
5
22
36
100
999
12
1 m
ark
5
12
22
36
100
999
1 m
ark
5
AO
2.2
(5)
5
(d)
(ii) 1
mark
per b
ulle
t to m
ax 7
R
epe
at
C
alc
ula
ting
an a
rray m
idp
oin
t…
…
by a
dd
ing
the a
rray lo
we
r bou
nd
to th
e a
rray u
pp
er b
ou
nd, d
ivid
ing
by 2
an
d ro
un
din
g
C
om
pa
re a
rray m
idpo
int w
ith v
alu
e to
se
arc
h fo
r…
…
if eq
ual s
et fo
un
d fla
g to
true
…
if arra
y m
idpo
int <
va
lue to
se
arc
h fo
r, ch
ang
e lo
we
rbo
und
to e
qu
al m
idpo
int +
1
…
if arra
y m
idpo
int >
va
lue to
se
arc
h fo
r, ch
ang
e u
ppe
rbo
und
to e
qu
al m
idpo
int –
1
U
ntil lo
we
rbo
und
is g
reate
r than
or e
qu
al to
up
pe
rbou
nd
R
etu
rn/o
utp
ut fo
un
d fla
g
7
AO
1.1
(2)
AO
1.2
(3)
AO
2.1
(1)
AO
2.2
(1)
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
20
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
5
(d)
(iii) 1
mark
per b
ulle
t
S
ettin
g v
aria
ble
to s
tart a
t 0
S
uita
ble
wh
ile s
tructu
re (e
nd
wh
ile o
r cle
ar in
de
nta
tion)
lo
op
ing
50 tim
es
In
cre
me
ntin
g th
e v
aria
ble
with
in th
e lo
op
e
.g. 1
function searchItem(dataItem)
count = 0
while count < 50
if dataArray(count) == dataItem then
return(count)
endif
count = count + 1
endwhile
return(-1)
endfunction
e.g
. 2
function searchItem(dataItem)
count = 0
while count < 50 and dataArray[count]!=dataItem
count = count + 1
endwhile
if count==50
count=-1
endif
return(count)
endfunction
4
AO
1.2
(1)
AO
3.1
(1)
AO
3.2
(2)
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
21
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
6
(a)
(i) 1
mark
per b
ulle
t to m
ax 3
R
ecord
is a
data
stru
ctu
re…
…
A c
lass is
a te
mp
late
for m
akin
g d
ata
stru
ctu
res (o
bje
cts
)
C
lass a
lso
ha
s m
eth
od
s (w
hic
h d
escrib
es fu
nctio
nality
)
B
oth
sto
re d
ata
of d
iffere
nt ty
pes
W
hic
h c
an b
e a
ccesse
d b
y th
eir n
am
es
B
ut c
lasses c
an m
ake th
em
acce
ssib
le v
ia m
eth
od
s
B
oth
ca
n h
ave
mu
ltiple
‘insta
nce
s’
C
lass c
an
inclu
de
vis
ibility
of p
rop
ertie
s / p
riva
te
3
AO
1.2
(3)
6
(a)
(ii) 1
mark
per s
pace
recordStructure items
itemName : String
cost : Currency
dateArrival : Date
transferred : Boolean
endRecordStructure
5
AO
2.2
(2)
AO
3.2
(3)
6
(a)
(iii) 1
mark
per b
ulle
t to m
ax 3
D
ecla
ring
box1
as a
n ite
m
U
sin
g B
ox1
. (or e
qu
iva
len
t) for e
ach v
aria
ble
S
ettin
g e
ach v
aria
ble
(ma
tch
ing 6
aii) c
orre
ctly
e
.g.
Box1 : Items
Box1.itemName = “Box”
Box1.cost = 22.58
Box1.dateArrival = “1/5/2018”
Box1.transfered = True
3
AO
2.2
(2)
AO
3.2
(1)
En
sure
va
riab
le
nam
es fo
r co
st
and
da
teA
rriva
l are
co
nsis
tent
with
va
riab
le
nam
es g
ive
n in
a
(ii)
6
(b)
(i) 1
mark
per b
ulle
t to m
ax 2
A
data
stru
ctu
re
F
IFO
(first in
first o
ut)
2
AO
1.1
(2)
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
22
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
6
(b)
(ii) 1
mark
per b
ulle
t to m
ax 2
P
rope
rties (a
re e
nca
psu
late
d) a
nd
ca
n o
nly
be a
cce
ssed
thro
ug
h th
eir m
eth
od
s
E
nfo
rce v
alid
atio
n th
roug
h th
e m
eth
od
// inap
pro
pria
te d
ata
ca
n b
e c
aug
ht b
efo
re e
nte
red
C
ann
ot b
e c
hang
ed/a
ccesse
d a
ccid
en
tally
2
AO
1.2
(2)
6
(b)
(iii) 1
mark
per b
ulle
t to m
ax
C
onstru
cto
r meth
od
/new
S
ettin
g head
and tail
to 0
with
in c
on
stru
cto
r meth
od
e.g
. public procedure new()
head = 0
tail = 0
numItems = 0
endprocedure
2
AO
2.2
(1)
AO
3.2
(1)
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
23
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
6
(b)
(iv)
1 m
ark
per b
ulle
t to m
ax 6
F
un
ctio
n d
ecla
ratio
n, ta
kin
g ite
m a
s a
para
mete
r
C
heckin
g if th
e q
ueu
e is
full…
…
outp
uttin
g/re
po
rting e
rror a
nd re
turn
ing
false
A
ddin
g th
e ite
m to
the tail
positio
n
C
orre
ctly
upd
atin
g th
e tail
poin
ter (e
ither b
efo
re o
r afte
r add
ition)
In
cre
me
ntin
g numItems
and
retu
rnin
g true
if succe
ssfu
l
e.g
. public function enqueue(newItem : items) : boolean
if numItems = 10 then
print("Error: The queue is full")
return false
else
theItems[tail] = newItem
if tail = 9 then
tail = 0
else
tail += 1
endif
numItems += 1
return true
endif
endprocedure
6
AO
2.2
(3)
AO
3.1
(1
) A
O3
.2 (2
)
6
(b)
(v)
e.g
.
myItems =
(new
) itemQueue()
1
AO
2.1
(1)
Allo
w fo
llow
th
roug
h if th
ey
have
pa
ram
ete
rs
in 6
(b)(iii)
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
24
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
6
(b)
(vi)
1 m
ark
per b
ulle
t to m
ax 5
P
roced
ure
decla
ratio
n fo
r insertItems
A
skin
g fo
r inpu
t of d
ata
item
s fo
r a n
ew
item
…..
…
usin
g re
cord
stru
ctu
re c
orre
ctly
U
se
of myItems.enqueue
L
oo
pin
g w
hile
the q
ueu
e is
not fu
ll e
.g.
procedure insertItems()
newItem : Items
itemCount = myItems.getnumItems()
while itemCount < 10
newItem.itemName = input("Enter the item name")
newItem.cost = input("Enter the item cost")
newItem.dateArrival = input("Enter the date of arrival")
newItem.transferred = input("Has it been transferred?")
myItems.enqueue(newItem)
itemCount = itemCount + 1
endwhile
myItems.setnumItems(itemCount)
endprocedure
5
AO
2.2
(2)
AO
3.1
(1
) A
O3
.2 (2
)
6
(b)
(vii)
1 m
ark
per b
ulle
t to m
ax 2
S
tore
the ite
ms a
nd q
ueu
e to
an e
xte
rna
l file (w
hen
the p
rog
ram
clo
ses)
L
oa
d th
e ite
ms a
nd q
ueue
from
the file
wh
en
it sta
rts
2
AO
2.1
(1)
AO
2.2
(1)
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
25
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
6
(c)
Mark
Ban
d 3
– H
igh
lev
el
(7-9
ma
rks)
Th
e c
and
idate
dem
onstra
tes a
thoro
ug
h k
now
led
ge a
nd
u
nd
ers
tand
ing
of c
achin
g a
nd
co
ncurre
nt p
roce
ssin
g; th
e
ma
teria
l is g
ene
rally
accu
rate
and
deta
iled.
Th
e c
and
idate
is a
ble
to a
pp
ly th
eir k
no
wle
dg
e a
nd
u
nd
ers
tand
ing
dire
ctly
an
d c
onsis
tently
to th
e c
on
text
pro
vid
ed
. Evid
en
ce/e
xa
mp
les w
ill be e
xp
licitly
rele
va
nt to
th
e e
xp
lana
tion
. T
he
re is
a w
ell-d
eve
lope
d lin
e o
f reaso
nin
g w
hic
h is
cle
ar
and
log
ica
lly s
tructu
red. T
he
info
rma
tion p
rese
nte
d is
re
levan
t and s
ubsta
ntia
ted.
Ma
rk B
an
d 2
– M
id le
ve
l (4
-6 m
ark
s)
Th
e c
and
idate
dem
onstra
tes re
aso
na
ble
kn
ow
ledg
e a
nd
u
nd
ers
tand
ing
of c
achin
g a
nd
co
ncurre
nt p
roce
ssin
g; th
e
ma
teria
l is g
ene
rally
accu
rate
but a
t time
s u
nd
erd
eve
lope
d.
Th
e c
and
idate
is a
ble
to a
pp
ly th
eir k
now
ledg
e a
nd
u
nd
ers
tand
ing
dire
ctly
to th
e c
onte
xt p
rovid
ed
alth
ou
gh
one
or tw
o o
pp
ortu
nitie
s a
re m
isse
d. E
vid
en
ce/e
xam
ple
s
are
for th
e m
ost p
art im
plic
itly re
leva
nt to
the e
xp
lana
tion
. T
he c
and
idate
pro
vid
es a
reaso
na
ble
dis
cussio
n, th
e
ma
jority
of w
hic
h is
focuse
d. E
va
luativ
e c
om
me
nts
are
, for
the m
ost p
art a
pp
ropria
te, a
lthoug
h o
ne
or tw
o
opp
ortu
nitie
s fo
r deve
lopm
ent a
re m
isse
d.
The
re is
a lin
e o
f reaso
nin
g p
rese
nte
d w
ith s
om
e s
tructu
re.
The
info
rma
tion p
rese
nte
d is
in th
e m
ost p
art re
leva
nt a
nd
su
pp
orte
d b
y s
om
e e
vid
ence
. M
ark
Ba
nd
1 –
Lo
w L
eve
l (1
-3 m
ark
s)
Th
e c
and
idate
dem
onstra
tes a
basic
kno
wle
dg
e o
f ca
chin
g
and
co
ncu
rrent p
roce
ssin
g w
ith lim
ited u
nd
ers
tand
ing
sh
ow
n; th
e m
ate
rial is
basic
and
co
nta
ins s
om
e
inaccu
racie
s. T
he c
an
did
ate
s m
ake
s a
limite
d a
ttem
pt to
9
AO
1.1
(2)
AO
1.2
(2)
AO
2.1
(2)
AO
3.3
(3)
AO
1: K
no
wle
dg
e a
nd
Un
de
rsta
nd
ing
In
dic
ativ
e c
on
ten
t C
achin
g
P
revio
usly
used
da
ta is
sto
red in
a lo
catio
n…
th
at c
an b
e q
uic
kly
accesse
d …
to
sp
ee
d u
p re
trieva
l if ne
ed
ed
in fu
ture
C
oncu
rrent P
roce
ssin
g
se
ve
ral p
roce
sse
s w
ork
sim
ulta
ne
ou
sly
to s
olv
e a
p
rob
lem
A
O2
: Ap
plic
atio
n
Ca
chin
g
se
arc
h fo
r pre
vio
usly
se
arc
hed
for d
ata
item
s in
a
faste
r seco
nd
ary
sto
rag
e d
evic
e/R
AM
S
pee
d u
p a
ccess fo
r that ite
m
…R
elie
s o
n s
am
e ite
m b
ein
g s
earc
hed fo
r mu
ltiple
tim
es
…K
am
ran n
ee
ds to
decid
e h
ow
feasib
le th
is is
b
ase
d o
n th
e n
um
ber o
f item
C
oncu
rrent
C
om
pu
ter w
ould
ha
ve
mu
ltiple
pro
cesso
rs…
E
ach s
earc
hin
g p
art o
f the
da
ta s
tructu
re a
t one
tim
e…
T
his
wo
uld
be
limite
d b
y b
ottle
ne
cks s
uch a
s
accessin
g th
e s
tora
ge d
evic
e
T
he n
pro
cessors
co
uld
pote
ntia
lly m
ean
an
incre
ase
of u
p to
1/n
of tim
e…
rea
listic
ally
sp
ee
d
incre
ase
is lik
ely
to b
e le
ss th
an th
at
O
nly
usefu
l if usin
g lin
ea
r se
arc
h // b
inary
se
arc
h
ca
nn
ot b
e p
erfo
rme
d c
oncu
rrently
A
O3
: Eva
lua
tion
C
and
idate
s w
ill nee
d to
eva
luate
the b
en
efits
and
d
raw
backs o
f ca
chin
g a
nd
co
ncu
rrent p
roce
ssin
g
Allo
w a
ny p
oin
t of v
iew
(ca
chin
g / c
oncu
rrent / b
oth
) as
long
as a
rgum
ent is
pre
se
nte
d s
uita
bly
.
H4
46
/02
M
ark
Sc
he
me
Ju
ne
20
18
26
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
app
ly a
cq
uire
d k
now
ledg
e a
nd
un
ders
tand
ing
to th
e
co
nte
xt p
rovid
ed
. T
he c
and
idate
pro
vid
es a
limite
d d
iscu
ssio
n w
hic
h is
n
arro
w in
focus. J
udg
em
ents
if ma
de a
re w
eak a
nd
u
nsu
bsta
ntia
ted.
The
info
rma
tion is
basic
and
co
mu
nic
ate
d in
an
unstru
ctu
red w
ay. T
he
info
rma
tion
is s
upp
orte
d b
y lim
ited
evid
en
ce a
nd
the re
latio
nship
to th
e e
vid
en
ce m
ay n
ot b
e
cle
ar.
0 m
ark
s
No
atte
mpt to
an
sw
er th
e q
uestio
n o
r respo
nse is
not
wo
rthy o
f cre
dit.
Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA Registered Company Number: 3484466 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: 01223 552552 Facsimile: 01223 552553 © OCR 2018
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INSTRUCTIONS• Use black ink.• Complete the boxes above with your name, centre number and candidate number.• Answer all the questions. • Write your answer to each question in the space provided. Additional paper may be
used if required but you must clearly show your candidate number, centre number and question number(s).
• Do not write in the barcodes.
INFORMATION• The total mark for this paper is 140.• The marks for each question are shown in brackets [ ].• Quality of extended responses will be assessed in questions marked with an
asterisk (*).• This document consists of 28 pages.
Turn over© OCR 2017 [601/4911/5]DC (SC/SW) 155854/7 R
Last name
First name
Candidatenumber
Centrenumber
Oxford Cambridge and RSA
A Level Computer ScienceH446/02 Algorithms and Programming
Thursday 22 June 2017 – MorningTime allowed: 2 hours 30 minutes
Do not use:• A calculator
*6827705853*
OCR is an exempt Charity
* H 4 4 6 0 2 *
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© OCR 2017
Answer all the questions.
Section A
1 A programmer needs to sort an array of numeric data using an insertion sort.
(a) (i) The following, incomplete, algorithm performs an insertion sort.
Complete the algorithm.
procedure sortit(dataArray, lastIndex)
for x = 1 to lastIndex
currentData = dataArray[……………………]
position = x
while (position > 0 AND dataArray[position-1] > currentData)
dataArray[position] = dataArray[………………………]
position = position - 1
endwhile
dataArray[position] = …………………………
next x
endprocedure
[3]
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(ii) Show how an insertion sort would sort the following data:
6 1 15 12 5 6 9
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(b) (i) Using Big-O notation state the best case complexity of insertion sort.
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(ii) Explain what your answer to part (b)(i) means.
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(c*) The number of data items in the array is continually increasing.
Insertion sort has a worst case time complexity of O(n2) and space complexity of O(1).
An alternative sorting algorithm that could be used is bubble sort which also has a worst case time complexity of O(n2) and space complexity of O(1).
Briefly outline how the bubble sort algorithm works. Discuss the relationship between the complexities and the two sorting algorithms and justify which of the two algorithms is best suited to sorting the array. [9]
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2 A programmer is developing an ordering system for a fast food restaurant. When a member of staff inputs an order, it is added to a linked list for completion by the chefs.
(a) Explain why a linked list is being used for the ordering system.
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(b) Each element in a linked list has:
• a pointer, nodeNo, which gives the number of that node • the order number, orderNo • a pointer, next, that points to the next node in the list
Fig. 2.1 shows the current contents of the linked list, orders.
nodeNo orderNo next
0 154 1
1 157 2
2 155 3
3 156 ∅
Fig. 2.1
∅ represents a null pointer.
(i) Order 158 has been made, and needs adding to the end of the linked list.
Add the order, 158, to the linked list as shown in Fig. 2.1. Show the contents of the linked list in the following table.
nodeNo orderNo next
[2]
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(ii) Order 159 has been made. This order has a high priority and needs to be the second order in the linked list.
Add the order, 159, to the original linked list as shown in Fig. 2.1. Show the contents of the linked list in the following table.
nodeNo orderNo next
[3]
(c) The linked list is implemented using a 2D array, theOrders:
• Row 0 stores orderNo • Row 1 stores next
The data now stored in theOrders is shown in Fig. 2.2.
184 186 185 187
1 2 3
Fig. 2.2
theOrders[1,0] would return 1
The following algorithm is written:
procedure x() finished = false count = 0 while NOT(finished) if theOrders[1,count] == null then finished = true else output = theOrders[0,count] print(output) count = theOrders[1,count] endif endwhile output = theOrders[0,count] print(output)
endprocedure
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(i) Outline why nodeNo does not need to be stored in the array.
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(ii) Complete the trace table for procedure x, for the data shown in Fig. 2.2.
finished count output
[3]
(iii) Describe the purpose of procedure x.
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(iv) A new order, 190, is to be added to theOrders. It needs to be the third element in the list.
The current contents of the array are repeated here for reference:
184 186 185 187
1 2 3
Describe how the new order, 190, can be added to the array, so the linked list is read in the correct order, without rearranging the array elements.
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(d) The user needs to be able to search for, and find, a specific order number.
State an appropriate search algorithm that could be used, and justify your choice against an alternative Search algorithm.
Appropriate Search Algorithm ....................................................................................................
Justification ................................................................................................................................
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(e) The programmer is writing the program using an IDE.
Identify three features of an IDE that the programmer would use when writing the code and describe how the features benefit the programmer.
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(f*) The programmer is considering using concurrent programming.
Discuss how concurrent programming can be applied to the food ordering system and the benefits and limitations of doing so. [9]
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3 An encryption routine reads a line of text from a file, reverses the order of the characters in the string and subtracts 10 from the ASCII value of each letter, then saves the new string into the same file.
The program is split into sub-procedures. Three sub-procedures are described as follows:
• Read string from file • Push each character of the string onto a stack • Read and encrypt each character message
(a) (i) Identify one further sub-procedure that could be used in the program.
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(ii) Describe two advantages of splitting the problem into sub-procedures.
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(b) A function, readMessage:
• takes the file name as a parameter • reads and returns the line of text
Complete the pseudocode algorithm for readMessage:
function …………………………………………………………(fileName)
messageFile = openRead(…………………………………………………………)
message = messageFile.readLine()
messageFile. …………………………………………………………
return …………………………………………………………
endfunction [4]
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(c) A function, push, can be used to add a character to a stack. For example:
theStack.push("H")
places the character H onto the stack, theStack.
A procedure, pushToStack, takes a string as a parameter and pushes each character of the message onto the stack, messageStack.
Complete the procedure below.
Add comments to explain how your code works.
procedure pushToStack(message)
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endprocedure [5]
(d) Describe the steps that the program would have to take in order to encrypt the characters stored in the stack, and save them in a single variable.
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4 A data structure is shown below in Fig. 4.1.
D X
HG
I
M
J
NL
K
E F
B A C
Fig. 4.1
(a) Identify the data structure shown in Fig. 4.1.
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(b) The programmer is considering using a depth-first (post-order) traversal, or a breadth-first traversal to find the path between node A and node X.
(i) Explain the difference between a depth-first (post-order) and breadth-first traversal.
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(ii) Show how a depth-first (post-order) traversal would find the path between node A and node X for the structure shown in Fig. 4.1.
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(iii) Explain how you used backtracking in your answer to part (b)(ii).
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5 A recursive function, calculate, is shown below:
01 function calculate(num1, num2)02 if num1 == num2 then03 return num104 elseif num1 < num2 then05 return calculate(num1, (num2-num1))06 else07 return calculate(num2, (num1-num2))08 endif09 endfunction
(a) Identify the lines where recursion is used.
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(b) Trace the algorithm, showing the steps and result when the following line is run:
print(calculate(4,10))
[5]
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(c) Re-write the function so it uses iteration instead of recursion.
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Section B
Answer all questions.
6 A software developer is creating a Virtual Pet game.
The user can choose the type of animal they would like as their pet, give it a name and then they are responsible for caring for that animal. The user will need to feed, play with, and educate their pet.
The aim is to keep the animal alive and happy, for example if the animal is not fed over a set period of time then the pet will die.
• The game tells the user how hungry or bored the animal is as a percentage (%) and the animal’s intelligence is ranked as a number between 0 and 150 (inclusive).
• Hunger and boredom increase by 1% with every tick of a timer. • When the feed option is selected, hunger is reduced to 0. • When the play option is selected, bored is reduced to 0. • When the read option is selected, the intelligence is increased by 0.6% of its current
value.
An example of the game is shown:
What type of pet would you like? Fox or Elephant?
Fox
What would you like to name your Fox?
Joanne
Joanne’s stats are
Hunger: 56%
Bored: 85%
Intelligence: 20
What would you like to do with your pet? Play, Read or Feed?
Fig. 1.1
(a) Identify three inputs that the user will have to enter to start, and/or play the game.
1 .................................................................................................................................................
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3 ................................................................................................................................................. [3]
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(b) The developer is using decomposition to design the game.
(i) Describe the process of decomposition.
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(ii) The developer has produced the following structure diagram for the game:
Virtual Pet
Choose Pet
Start game Play game
ReadFeed
ReduceHunger to 0
Timer
Complete the structure diagram for the Virtual Pet game by filling in the empty boxes. [6]
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(c) The developer needs to write procedures for the options play and read. Each of the options changes its corresponding value, and outputs the results to the screen.
(i) Write a procedure, using pseudocode, to reset bored and output the new value in an appropriate message.
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(ii) Write a procedure, using pseudocode, to increase intelligence by 0.6% and output the new intelligence in an appropriate message.
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(d) The developer is extending the game to allow users to have multiple pets of different types. The developer has written a class, Pet.
The attributes and methods in the class are described in the table:
Identifier Attribute/Method Description
petName Attribute Stores the pet’s name
bored Attribute Stores the % bored
hunger Attribute Stores the % hunger
intelligence Attribute Stores the intelligence
type Attribute Stores the type of animal
new Method Creates a new instance of pet
feed Method Reduces hunger to 0 and outputs hunger
play Method Reduces bored to 0 and outputs bored
read Method Increases intelligence by a set value
outputGreeting Method Outputs a message to the user
Part of the class declaration is given:
class Pet private petName private bored private hunger private intelligence private type … …
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(i) After a user enters the pet name, and chooses a type, the constructor method of Pet is called to create a new instance. The method needs to set petName, as well as hunger, bored and intelligence to starting values of 0.
Write, using pseudocode, the constructor method for this class.
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(ii) Write a line of code that creates a new instance of Pet for a Tiger called “Springy”.
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(iii) The method outputGreeting for the superclass is written as follows:
public procedure outputGreeting() print("Hello, I'm " + petName + ", I'm a " + type) endprocedure
A class is needed for Tiger. The class needs to:
• inherit the methods and attributes from pet • in the constructor, set type to Tiger, intelligence to 10, hunger to 50 and
bored to 10 • extend the method outputGreeting, by outputting an additional line that says “I like to eat meat and roar”
Write, using pseudocode, the class Tiger.
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(e*) The developer made use of abstraction when creating the Virtual Pet game.
Discuss the need for and purpose of abstraction and how abstraction will be used in the development of the game. [9]
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© OCR 2017
(f) The developer is storing the user’s pets in a 1-dimensional array. At each timer interval, the array is searched, using a linear search, to check if any pets’ hunger or bored values are greater than 90%. If they are, an alert is displayed to the user.
(i) State the complexity of searching the pets in Big-O notation.
...................................................................................................................................... [1]
(ii) A given computer takes 4 milliseconds (ms) to search an array of 20 pets. Calculate an estimate of how long the computer will take to search an array of 100 pets.
Show your working.
...........................................................................................................................................
...........................................................................................................................................
...................................................................................................................................... [2]
END OF QUESTION PAPER
28
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Oxford Cambridge and RSA Examinations
GCE
Computer Science
Unit H446A/02: Algorithms and programming
Advanced GCE
Mark Scheme for June 2017
OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2017
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
3
An
no
tatio
ns
An
no
tatio
n
Mea
nin
g
Om
issio
n m
ark
Be
nefit o
f the d
oubt
Inco
rrect p
oin
t
Follo
w th
rou
gh
No
t answ
ere
d q
uestio
n
No b
en
efit o
f doubt g
ive
n
Re
pe
at
Co
rrect p
oin
t
To
o v
ag
ue
Bla
nk P
ag
e –
this
annota
tion m
ust b
e u
sed o
n a
ll bla
nk p
ag
es w
ithin
an a
nsw
er b
ookle
t (stru
ctu
red o
r unstru
ctu
red) a
nd o
n
each p
ag
e o
f an a
dd
itional o
bje
ct w
he
re th
ere
is n
o c
andid
ate
response.
Leve
l 1
Leve
l 2
Leve
l 3
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
4
Su
bje
ct-s
pe
cific
Ma
rkin
g In
stru
ctio
ns
IN
TR
OD
UC
TIO
N
Yo
ur firs
t task a
s a
n E
xa
min
er is
to b
ecom
e th
oro
ug
hly
fam
iliar w
ith th
e m
ate
rial o
n w
hic
h th
e e
xa
min
atio
n d
ep
en
ds. T
his
ma
teria
l inclu
de
s:
th
e s
pecific
atio
n, e
spe
cia
lly th
e a
ssessm
ent o
bje
ctiv
es
th
e q
uestio
n p
ap
er a
nd its
rubric
s
th
e m
ark
sch
em
e.
Yo
u s
hou
ld e
nsu
re th
at y
ou h
ave
co
pie
s o
f these
ma
teria
ls.
Yo
u s
hou
ld e
nsu
re a
lso
that y
ou a
re fa
milia
r with
the a
dm
inis
trativ
e p
rocedu
res re
late
d to
the m
ark
ing p
roce
ss. T
hese
are
se
t out in
the O
CR
b
oo
kle
t Instru
ctio
ns fo
r Ex
am
iners
. If yo
u a
re e
xa
min
ing
for th
e firs
t time
, ple
ase
read c
are
fully
Ap
pen
dix
5 In
trod
uc
tion
to S
crip
t Ma
rkin
g:
No
tes
for N
ew
Ex
am
iners
. P
lease
ask fo
r help
or g
uid
an
ce w
hen
eve
r yo
u n
ee
d it. Y
our firs
t poin
t of c
onta
ct is
yo
ur T
eam
Lead
er.
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
5
US
ING
TH
E M
AR
K S
CH
EM
E
Ple
ase
stu
dy th
is M
ark
Sch
em
e c
are
fully
. Th
e M
ark
Sch
em
e is
an in
teg
ral p
art o
f the p
rocess th
at b
eg
ins w
ith th
e s
ettin
g o
f the q
uestio
n p
ap
er
and
en
ds w
ith th
e a
wa
rdin
g o
f gra
de
s. Q
uestio
n p
ape
rs a
nd M
ark
Sch
em
es a
re d
eve
lope
d in
asso
cia
tion
with
ea
ch
oth
er s
o th
at is
su
es o
f d
iffere
ntia
tion
an
d p
ositiv
e a
ch
ieve
me
nt c
an b
e a
ddre
ssed
from
the v
ery
sta
rt. T
his
Ma
rk S
ch
em
e is
a w
ork
ing d
ocu
men
t; it is n
ot e
xh
au
stiv
e; it d
oe
s n
ot p
rovid
e ‘c
orre
ct’ a
nsw
ers
. Th
e M
ark
Sch
em
e c
an o
nly
pro
vid
e ‘b
est
gu
esses’ a
bo
ut h
ow
the q
uestio
n w
ill wo
rk o
ut, a
nd it is
su
bje
ct to
revis
ion a
fter w
e h
ave
loo
ke
d a
t a w
ide ra
ng
e o
f scrip
ts.
Th
e E
xa
min
ers
’ Sta
nd
ard
isa
tion
Me
etin
g w
ill ensu
re th
at th
e M
ark
Sch
em
e c
ove
rs th
e ra
ng
e o
f can
did
ate
s’ re
spo
nses to
the
qu
estio
ns, a
nd th
at a
ll E
xa
min
ers
und
ers
tand
an
d a
pp
ly th
e M
ark
Sch
em
e in
the s
am
e w
ay. T
he
Ma
rk S
ch
em
e w
ill be d
iscu
ssed
an
d a
me
nd
ed
at th
e m
eetin
g, a
nd
adm
inis
trativ
e p
roced
ure
s w
ill be c
onfirm
ed. C
o-o
rdin
atio
n s
crip
ts w
ill be is
su
ed
at th
e m
eetin
g to
exe
mp
lify a
spe
cts
of c
andid
ate
s’ re
spo
nses a
nd
achie
ve
me
nts
; the c
o-o
rdin
atio
n s
crip
ts th
en
be
co
me
part o
f this
Ma
rk S
ch
em
e.
Be
fore
the S
tand
ard
isatio
n M
eetin
g, y
ou s
hou
ld re
ad
an
d m
ark
in p
en
cil a
num
ber o
f scrip
ts, in
ord
er to
ga
in a
n im
pre
ssio
n o
f the ra
ng
e o
f re
spo
nse
s a
nd
ach
ieve
me
nt th
at m
ay b
e e
xp
ecte
d.
In y
our m
ark
ing
, yo
u w
ill enco
un
ter v
alid
respo
nse
s w
hic
h a
re n
ot c
ove
red b
y th
e M
ark
Sch
em
e: th
ese re
spo
nse
s m
ust b
e c
redite
d. Y
ou w
ill e
nco
un
ter a
nsw
ers
wh
ich
fall o
uts
ide th
e ‘ta
rge
t rang
e’ o
f Ba
nd
s fo
r the p
ape
r wh
ich
yo
u a
re m
ark
ing
. Ple
ase
ma
rk th
ese
an
sw
ers
accord
ing
to th
e
ma
rkin
g c
riteria
. P
lease
read
care
fully
all th
e s
crip
ts in
yo
ur a
llocatio
n a
nd
make
eve
ry e
ffort to
look p
ositiv
ely
for a
ch
ieve
me
nt th
roug
hou
t the
ab
ility ra
ng
e. A
lwa
ys
be p
rep
are
d to
use th
e fu
ll rang
e o
f ma
rks.
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
6
LE
VE
LS
OF
RE
SP
ON
SE
QU
ES
TIO
NS
: T
he in
dic
ativ
e c
on
tent in
dic
ate
s th
e e
xp
ecte
d p
ara
me
ters
for c
and
idate
s’ a
nsw
ers
, but b
e p
repa
red
to re
cog
nis
e a
nd
cre
dit u
ne
xp
ecte
d
app
roa
ch
es w
here
they s
how
rele
va
nce
. U
sin
g ‘b
est-fit’, d
ecid
e firs
t wh
ich
se
t of B
AN
D D
ES
CR
IPT
OR
S b
est d
escrib
es th
e o
ve
rall q
uality
of th
e a
nsw
er. O
nce
the b
an
d is
locate
d, a
dju
st
the m
ark
co
nce
ntra
ting
on
featu
res o
f the a
nsw
er w
hic
h m
ake it s
trong
er o
r we
ake
r follo
win
g th
e g
uid
elin
es fo
r refin
em
ent.
H
igh
est m
ark
: If cle
ar e
vid
en
ce o
f all th
e q
ualitie
s in
the b
an
d d
escrip
tors
is s
how
n, th
e H
IGH
ES
T M
ark
sh
ou
ld b
e a
wa
rde
d.
L
ow
est m
ark
: If the a
nsw
er s
how
s th
e c
and
idate
to b
e b
ord
erlin
e (i.e
. they h
ave
ach
ieve
d a
ll the q
ua
lities o
f the b
an
ds b
elo
w a
nd
sh
ow
limite
d e
vid
en
ce o
f me
etin
g th
e c
riteria
of th
e b
and
in q
uestio
n) th
e L
OW
ES
T m
ark
sh
ou
ld b
e a
wa
rde
d.
M
idd
le m
ark
: Th
is m
ark
sh
ou
ld b
e u
sed fo
r ca
nd
idate
s w
ho a
re s
ecure
in th
e b
an
d. T
hey a
re n
ot ‘b
ord
erlin
e’ b
ut th
ey h
ave
only
achie
ve
d
so
me
of th
e q
ualitie
s in
the b
an
d d
escrip
tors
.
Be
pre
pa
red to
use th
e fu
ll rang
e o
f ma
rks. D
o n
ot re
serv
e (e
.g.) h
igh B
an
d 3
ma
rks ‘in
ca
se
’ so
me
thin
g tu
rns u
p o
f a q
uality
yo
u h
ave
no
t ye
t se
en
. If an a
nsw
er g
ive
s c
lear e
vid
en
ce o
f the q
ua
lities d
escrib
ed
in th
e b
and
de
scrip
tors
, rew
ard
app
rop
riate
ly.
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
7
A
O1
A
O2
A
O3
Hig
h (th
oro
ug
h)
Pre
cis
ion in
the u
se o
f que
stio
n
term
inolo
gy. K
now
ledg
e s
how
n is
co
nsis
tent a
nd
we
ll-deve
lope
d.
Cle
ar a
pp
recia
tion o
f the q
uestio
n
from
a ra
ng
e o
f diffe
rent
pers
pectiv
es m
akin
g e
xte
nsiv
e
use o
f acq
uire
d k
now
led
ge a
nd
u
nd
ers
tand
ing
.
Kn
ow
ledg
e a
nd
un
de
rsta
ndin
g
sh
ow
n is
co
nsis
tently
app
lied to
co
nte
xt e
na
blin
g a
log
ical a
nd
su
sta
ined
arg
um
ent to
de
ve
lop.
Exa
mp
les u
se
d e
nh
an
ce ra
ther
than
detra
ct fro
m re
spo
nse
.
Co
nce
rted e
ffort is
ma
de to
co
nsid
er a
ll aspe
cts
of a
syste
m /
pro
ble
m o
r we
igh
up b
oth
sid
es to
a
n a
rgum
ent b
efo
re fo
rmin
g a
n
ove
rall c
onclu
sio
n. J
udg
em
ents
m
ade
are
ba
se
d o
n a
pp
ropria
te
and
co
ncis
e a
rgu
me
nts
that h
ave
b
ee
n d
eve
lope
d in
respon
se
resultin
g in
them
bein
g b
oth
su
pp
orte
d a
nd
realis
tic.
Mid
dle
(rea
so
nab
le)
Aw
are
ne
ss o
f the m
ean
ing o
f the
term
s in
the q
uestio
n. K
no
wle
dg
e
is s
oun
d a
nd e
ffectiv
ely
d
em
onstra
ted. D
em
and
s o
f q
uestio
n u
nd
ers
tood a
ltho
ug
h a
t tim
es o
pp
ortu
nitie
s to
ma
ke u
se o
f a
cq
uire
d k
now
ledg
e a
nd
und
ers
tand
ing
not a
lwa
ys ta
ke
n.
Kn
ow
ledg
e a
nd
un
de
rsta
ndin
g
app
lied to
co
nte
xt. W
hils
t cle
ar
evid
en
ce th
at a
n a
rgum
en
t build
s
and
de
ve
lops th
roug
h re
sp
on
se
there
are
times w
hen
opp
ortu
nitie
s
are
mis
se
d to
use a
n e
xa
mp
le o
r re
late
an a
spe
ct o
f kn
ow
ledg
e o
r u
nd
ers
tand
ing
to th
e c
onte
xt
pro
vid
ed
.
Th
ere
is a
reaso
na
ble
atte
mp
t to
reach
a c
onclu
sio
n c
onsid
erin
g
aspe
cts
of a
syste
m / p
rob
lem
or
we
igh
ing u
p b
oth
sid
es o
f an
arg
um
ent. H
ow
eve
r the im
pact o
f th
e c
onclu
sio
n is
ofte
n le
sse
ne
d
by a
lack o
f su
pp
orte
d ju
dg
em
ents
w
hic
h a
cco
mp
an
y it. T
his
inab
ility
to b
uild
on
an
d d
eve
lop lin
es o
f a
rgu
me
nt a
s d
eve
lope
d in
the
respo
nse
ca
n d
etra
ct fro
m th
e
ove
rall q
uality
of th
e re
spo
nse
.
Lo
w (b
as
ic)
Co
nfu
sio
n a
nd
ina
bility
to
deco
nstru
ct te
rmin
olo
gy a
s u
se
d in
th
e q
uestio
n. K
now
ledg
e p
artia
l a
nd
su
pe
rficia
l. Focu
s o
n q
uestio
n
narro
w a
nd
ofte
n o
ne
-dim
ensio
na
l.
Inab
ility to
ap
ply
kn
ow
ledg
e a
nd
u
nd
ers
tand
ing
in a
ny s
usta
ined
wa
y to
co
nte
xt re
sultin
g in
tenu
ou
s
and
un
su
pp
orte
d s
tate
me
nts
bein
g
ma
de
. Exa
mp
les if u
se
d a
re fo
r th
e m
ost p
art irre
leva
nt a
nd
unsu
bsta
ntia
ted.
Little
or n
o a
ttem
pt to
prio
ritise o
r w
eig
h u
p fa
cto
rs d
urin
g c
ours
e o
f a
nsw
er. C
onclu
sio
n is
ofte
n
dis
loca
ted
from
resp
on
se
an
d a
ny
judg
em
ents
lack s
ubsta
nce
du
e in
p
art to
the b
asic
leve
l of a
rgu
me
nt
that h
as b
ee
n d
em
onstra
ted
thro
ug
hout re
spo
nse.
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
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8
A
ssessme
nt O
bje
ctive
AO
1
Dem
on
strate kno
wled
ge and
un
derstan
din
g of th
e prin
ciples an
d co
ncep
ts of co
mp
ute
r science, in
clud
ing ab
straction
, logic, algo
rithm
s and
data
represe
ntatio
n.
AO
1.1
D
emo
nstrate kn
ow
led
ge o
f the p
rincip
les and
con
cepts o
f abstractio
n, lo
gic, algorith
ms, d
ata represe
ntatio
n o
r oth
er as app
rop
riate.
AO
1.2
D
emo
nstrate u
nd
erstand
ing o
f the p
rincip
les and
con
cepts o
f abstractio
n, lo
gic, algorith
ms, d
ata represen
tation
or o
ther as ap
pro
priate.
AO
2
Ap
ply kn
ow
ledge an
d u
nd
erstand
ing o
f the p
rincip
les and
con
cepts o
f com
pu
ter scien
ce inclu
din
g to an
alyse pro
blem
s in co
mp
utatio
nal te
rms.
AO
2.1
A
pp
ly kno
wled
ge and
un
derstan
din
g of th
e prin
ciples an
d co
ncep
ts of co
mp
ute
r science.
AO
2.2
A
nalyse p
rob
lems in
com
pu
tation
al term
s.
AO
3
Design
, pro
gram an
d evalu
ate com
pu
ter system
s that so
lve pro
blem
s, makin
g reason
ed ju
dgem
ents ab
ou
t these
and
prese
ntin
g con
clusio
ns.
AO
3.1
D
esign co
mp
ute
r systems th
at solve p
rob
lems.
AO
3.2
P
rogram
com
pu
ter system
s that so
lve pro
blem
s.
AO
3.3
Evalu
ate com
pu
ter systems th
at solve p
rob
lems, m
aking re
ason
ed ju
dgem
ents ab
ou
t these an
d p
resen
ting co
nclu
sion
s.
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
9
Qu
es
tion
A
nsw
er
Ma
rks
Gu
ida
nc
e
1
a
i 1
mark
for e
ach c
orre
ct ite
m in
bo
ld
procedure sortit(dataArray, lastIndex)
for x = 1 to lastIndex
currentData = dataArray[x]
position = x
while (position > 0 AND dataArray[x – 1] > currentData)
dataArray[position] = dataArray[position-1]
position = position – 1
endwhile
dataArray[position] = currentData
next x
endprocedure
3
AO
1.1
(3
)
answ
ers
mu
st b
e in
the c
orre
ct c
ase a
s
giv
en
e.g
. currentData
1
a
ii 1
mark
for c
onte
nts
of e
ach
row
in ta
ble
6
1
15
12
5
6
9
1
6
15
12
5
6
9
6 is
the s
orte
d lis
t 1
is th
e c
om
pare
d to
sorte
d lis
t 1
is p
ut in
pla
ce in
so
rted lis
t
1
1
6
15
12
5
6
9
15 is
com
pare
d
15 is
in p
lace in
so
rted lis
t 1
1
6
12
15
5
6
9
12 is
com
pare
d
12 is
in p
lace in
so
rted lis
t 1
1
5
6
12
15
6
9
5 is
com
pare
d
5 is
in p
lace in
so
rted lis
t 1
1
5
6
6
12
15
9
6 is
com
pare
d
6 is
in p
lace in
so
rted lis
t 1
1
5
6
6
9
12
15
9 is
com
pare
d a
nd
put in
pla
ce
1
6
AO
2.1
(6
)
… e
ach ro
w is
dep
en
de
nt u
po
n th
e
pre
ce
din
g ro
w b
ein
g c
orre
ct
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
10
1
b
i O
(n)
1
AO
1.1
(1
)
1
b
ii 1
mark
per b
ulle
t to m
ax 3
T
he b
est c
ase is
for a
sorte
d lis
t (O(n
))
A
s th
e n
um
ber o
f ele
men
ts in
cre
ase
s
…
the n
um
ber o
f ste
ps in
cre
ase
s in
a lin
ea
r fashio
n
3
AO
1.2
(3
)
B(ii) d
ep
en
de
nt u
po
n b
(i) bein
g c
orre
ct
i.e. a
nsw
ers
for O
(n) o
nly
A
cce
pt a
ppro
pria
te g
raph
for b
ulle
t p
oin
ts 2
an
d 3
1
c
M
ark
Ba
nd
3 –
Hig
h le
ve
l (7
-9 m
ark
s)
Th
e c
and
idate
dem
onstra
tes a
tho
rou
gh
kn
ow
ledg
e a
nd
un
ders
tand
ing
of h
ow
b
ub
ble
so
rt wo
rks a
nd B
ig O
com
ple
xity
; the m
ate
rial is
ge
ne
rally
accura
te a
nd
d
eta
iled.
Th
e c
and
idate
is a
ble
to a
pp
ly th
eir k
now
ledg
e a
nd
un
ders
tand
ing d
irectly
and
co
nsis
tently
to th
e c
onte
xt p
rovid
ed
. Evid
en
ce/e
xa
mp
les w
ill be e
xp
licitly
re
leva
nt to
the e
xp
lana
tion
. T
he
re is
a w
ell-d
eve
lope
d lin
e o
f reaso
nin
g w
hic
h is
cle
ar a
nd
log
ica
lly
stru
ctu
red. T
he in
form
atio
n p
rese
nte
d is
rele
vant a
nd
su
bsta
ntia
ted.
Ma
rk B
an
d 2
– M
id le
ve
l (4
-6 m
ark
s)
Th
e c
and
idate
dem
onstra
tes re
as
on
ab
le k
now
led
ge a
nd
un
ders
tand
ing
of o
f h
ow
bub
ble
so
rt wo
rks a
nd B
ig O
com
ple
xity
; the m
ate
rial is
ge
ne
rally
accu
rate
b
ut a
t time
s u
nd
erd
eve
lope
d.
Th
e c
and
idate
is a
ble
to a
pp
ly th
eir k
now
ledg
e a
nd
un
ders
tand
ing d
irectly
to
the c
onte
xt p
rovid
ed
alth
ou
gh o
ne
or tw
o o
pp
ortu
nitie
s a
re m
isse
d.
Evid
en
ce/e
xa
mp
les a
re fo
r the m
ost p
art im
plic
itly re
leva
nt to
the e
xp
lana
tion
. T
he c
and
idate
pro
vid
es a
reaso
na
ble
dis
cussio
n, th
e m
ajo
rity o
f wh
ich
is
focuse
d. E
va
luativ
e c
om
me
nts
are
, for th
e m
ost p
art a
pp
ropria
te, a
lthoug
h
one
or tw
o o
pp
ortu
nitie
s fo
r deve
lopm
ent a
re m
isse
d.
The
re is
a lin
e o
f reaso
nin
g p
rese
nte
d w
ith s
om
e s
tructu
re. T
he in
form
atio
n
pre
se
nte
d is
in th
e m
ost p
art re
levan
t and s
upp
orte
d b
y s
om
e e
vid
en
ce.
Ma
rk B
an
d 1
– L
ow
Leve
l
9
AO
1.1
(2
) A
O1
.2
(2)
AO
2.1
(2
) A
O3
.3
(3)
AO
1: K
no
wle
dg
e a
nd
Un
de
rsta
nd
ing
In
dic
ativ
e c
on
ten
t
D
escrip
tion o
f bub
ble
sort:
S
tartin
g a
t the b
eg
innin
g o
f the
list Ite
ms a
re s
wa
pp
ed
with
their n
eig
hbo
ur if th
ey a
re o
ut o
f
ord
er.
E
ach p
air o
f neig
hbo
urs
is
ch
ecke
d in
ord
er.
W
hen
a s
wa
p is
ma
de a
flag
is
se
t.
If a
t the e
nd
of th
e lis
t the
flag
has b
ee
n s
et th
e fla
g is
un
set
and
the a
lgorith
m s
tarts
from
the b
eg
innin
g o
f the lis
t ag
ain
.
W
hen
the a
lgo
rithm
ge
ts to
the
end
of th
e lis
t and th
e fla
g is
unse
t the lis
t is s
orte
d a
nd
the
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
11
(1-3
ma
rks)
Th
e c
and
idate
dem
onstra
tes a
bas
ic k
now
ledg
e o
f of h
ow
bub
ble
so
rt wo
rks
and
Big
O c
om
ple
xity
with
limite
d u
nd
ers
tand
ing
sh
ow
n; th
e m
ate
rial is
basic
a
nd
co
nta
ins s
om
e in
accu
racie
s. T
he c
and
idate
s m
akes a
limite
d a
ttem
pt to
a
pp
ly a
cq
uire
d k
now
ledg
e a
nd
un
ders
tand
ing
to th
e c
onte
xt p
rovid
ed
. T
he c
and
idate
pro
vid
es a
limite
d d
iscu
ssio
n w
hic
h is
na
rrow
in fo
cus.
Ju
dg
em
ents
if ma
de a
re w
eak a
nd
un
su
bsta
ntia
ted
. T
he
info
rma
tion is
basic
and
co
mu
nic
ate
d in
an u
nstru
ctu
red w
ay. T
he
in
form
atio
n is
su
pp
orte
d b
y lim
ited e
vid
en
ce a
nd
the re
latio
nship
to th
e
evid
en
ce m
ay n
ot b
e c
lear.
0 m
ark
s
No
atte
mpt to
an
sw
er th
e q
uestio
n o
r respo
nse is
not w
orth
y o
f cre
dit.
alg
orith
m fin
ish
es.
O
(n2) d
en
ote
s a
s th
e d
ata
siz
e
incre
ase
s th
e tim
e th
e lis
t take
s
to s
ort in
cre
ase
s in
a q
ua
dra
tic
ma
nn
er.
O
(1) d
en
ote
s th
e s
pace
used
is
co
nsta
nt
AO
2: A
pp
licatio
n
A
s d
ata
se
t ge
ts b
igg
er, b
ub
ble
so
rt’s tim
e g
ets
larg
er a
t an
incre
asin
g ra
te..
C
om
ple
xity
doe
sn’t d
en
ote
the a
ctu
al
time
but th
e o
rder w
ith w
hic
h th
e
time
/sp
ace
gro
ws.
O
(1) sp
ace com
plexity m
eans n
o m
atter
ho
w b
ig the d
ata set b
ecom
es the
amo
un
t of sp
ace (extra to th
e data
itself) remain
s the sam
e.
O
(n2) tim
e c
om
ple
xity
me
an
s a
s n
incre
ase
s tim
e in
cre
ase
s b
y n
2 / if n
dou
ble
s th
e tim
e ta
ke
n is
sq
uare
d.
B
ubb
le s
ort c
an b
e tw
ea
ke
d w
ith
imp
rove
me
nts
(e.g
. ch
eckin
g o
ne
less ite
m p
er ite
ratio
n a
nd
alte
rnatin
g
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
12
so
rting
dire
ctio
ns).
T
hese o
ptim
isa
tion
s d
on
’t ch
ang
e
the c
om
ple
xity
. IT w
ill run a
little
qu
icke
r on s
ma
ller s
ets
bu
t time
take
n in
cre
ase
s ra
pid
ly w
ith d
ata
siz
e.
W
hen
ch
oo
sin
g a
n a
lgo
rithm
we
ma
y
als
o w
ant to
take in
to a
cco
unt th
e
ave
rag
e a
nd b
est c
ase s
ce
na
rios.
(in th
is c
ase th
ey a
re a
lso
the s
am
e
for b
oth
alg
orith
ms.)
A
O3
: Eva
lua
tion
T
he a
lgorith
ms m
ay h
ave
the
sa
me
time c
om
ple
xity
bu
t this
doe
s n
ot m
ean
they ta
ke th
e
sa
me
time to
exe
cute
on th
e
sa
me
data
se
t.
In
sertio
n s
ort g
ene
rally
perfo
rms
qu
icke
r than
bu
bble
so
rt and
is
there
fore
pre
fera
ble
. (Neith
er
sca
le w
ell h
ow
eve
r.)
B
oth
alg
orith
ms h
ave
a s
pace
co
mp
lexity
of O
(1). T
his
is
beca
use
bo
th a
lgo
rithm
s a
re in
-
pla
ce (i.e
. all s
ortin
g ta
kes p
lace
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
13
with
in th
e a
ctu
al d
ata
).
B
oth
have
a tim
e c
om
ple
xity
of
O(n
2) as a
co
nseq
uence
of th
eir
neste
d lo
op
s.
(NB
last tw
o p
oin
ts a
re o
nly
like
ly
to a
pp
ear in
the v
ery
hig
he
st
ma
rk a
nsw
ers
.)
2
a
1
mark
per b
ulle
t, to m
ax 2
, e.g
.
O
rders
ca
n b
e p
roce
ssed
in th
e o
rder th
ey a
re in
the q
ueu
e
O
rders
ca
n b
e in
serte
d a
t any p
lace in
the lis
t e.g
. hig
h p
riority
item
inserte
d e
arlie
r in th
e lis
t
O
rders
ca
n b
e d
ele
ted
from
any p
ositio
n in
the lis
t once th
ey a
re
co
mp
lete
L
ist is
dyn
am
ic…
…
to a
llow
ord
ers
to b
e a
dde
d / d
ele
ted
2
AO
1.2
(1
) A
O2
.1
(1)
2
b
i 1
mark
per b
ulle
t
nodeNo
and next
co
lum
ns a
re b
oth
co
rrect
orderNo
co
lum
n is
co
rrect
nodeNo
orderNo
next
2
AO
1.2
(2
)
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
14
154
1
1
157
2
2
155
3
3
156
4
4
158
2
b
ii 1
mark
per c
orre
ct c
olu
mn
nodeNo
orderNo
next
154
4
1
157
2
2
155
3
3
156
4
159
1
3
AO
1.2
(3
)
2
c
i
Th
e in
de
x/s
ubscrip
t of th
e a
rray a
cts
as th
e n
od
eN
o
1
AO
1.2
(1
)
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
15
2
c
ii 1
mark
for e
ach c
orre
ctly
co
mp
lete
d c
olu
mn
Finished
Count
output
Fals
e
0
184
(Fals
e)
1
186
(Fals
e)
2
185
Tru
e
3
187
3
AO
1.2
(1
) A
O2
.2
(2)
2
c
iii 1
mark
per b
ulle
t to m
ax 2
O
utp
ut th
e o
rder n
um
bers
...
...in
the o
rder th
ey a
re in
the lin
ke
d lis
t
2
AO
2.2
(2
)
2
c
iv
1 m
ark
per b
ulle
t to m
ax
O
rder 1
90 is
add
ed to
the
en
d
P
oin
ters
are
up
date
d
1
86
will p
oin
t to 4
1
90
will p
oin
t to 2
OR
Inde
x
0
1
2
3
4
Da
ta
184
186
185
187
190
Po
inte
r 1
4
3
2
4
AO
1.2
(2
) A
O2
.1
(2)
If a d
iag
ram
is g
ive
n th
en
the m
ark
for
upd
atin
g th
e p
oin
ters
is im
plic
it
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
16
2
d
A
lgo
rithm
, ma
x 1
lin
ea
r
Ju
stific
atio
n, 1
mark
per b
ulle
t to m
ax 2
Ite
ms d
o n
ot h
ave
to b
e in
a s
pecific
ord
er
B
inary
nee
ds ite
ms in
ord
er
3
AO
1.1
(1
) A
O2
.1
(2)
No
ma
rks fo
r justific
atio
n if lin
ea
r has n
ot
bee
n id
en
tified
2
e
1
mark
for fe
atu
re, 1
for b
en
efit. M
ax 2
pe
r featu
re.
e.g
. A
uto
-com
ple
te
C
an v
iew
iden
tifiers
/avo
id s
pe
lling
mis
takes
C
olo
ur c
odin
g te
xt/s
yn
tax h
igh
ligh
ting
C
an id
en
tify fe
atu
res q
uic
kly
/use to
ch
eck c
ode
is c
orre
ct
S
tepp
ing
R
un o
ne
line a
t a tim
e a
nd
ch
eck re
sult
B
reakp
oin
ts
S
top th
e c
od
e a
t a s
et p
oin
t to c
heck v
alu
e o
f va
riable
(s)
V
aria
ble
wa
tch
/wa
tch
win
do
w
6
AO
1.1
(3
) A
O1
.2
(3)
Qu
estio
n s
tate
s w
hen
writin
g th
e c
ode
, th
ere
fore
use o
f co
mp
iler/p
rod
ucin
g .e
xe
etc
. are
not a
wa
rde
d m
ark
s
Acce
pt a
ny s
uita
ble
featu
res e
.g. tra
ces,
cra
sh d
um
p, s
tack c
onte
nts
, cro
ss-
refe
rence
s, lin
e n
um
bers
, auto
-inde
nt
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
17
C
heck v
alu
es o
f va
riab
les a
nd
ho
w th
ey c
hang
e d
urin
g th
e e
xe
cutio
n
E
rror d
iagn
ostic
s
L
oca
te a
nd re
port e
rrors
/giv
e d
eta
il on e
rrors
2
f
Ma
rk B
an
d 3
– H
igh
leve
l (7
-9 m
ark
s)
Th
e c
and
idate
dem
onstra
tes a
tho
rou
gh
kn
ow
ledg
e a
nd
un
ders
tand
ing
of
co
ncu
rrent p
rog
ram
min
g; th
e m
ate
rial is
ge
ne
rally
accura
te a
nd d
eta
iled.
Th
e c
and
idate
is a
ble
to a
pp
ly th
eir k
now
ledg
e a
nd
un
ders
tand
ing d
irectly
and
co
nsis
tently
to th
e c
onte
xt p
rovid
ed
. Evid
en
ce/e
xa
mp
les w
ill be e
xp
licitly
re
leva
nt to
the e
xp
lana
tion
. T
he
re is
a w
ell-d
eve
lope
d lin
e o
f reaso
nin
g w
hic
h is
cle
ar a
nd
log
ica
lly
stru
ctu
red. T
he in
form
atio
n p
rese
nte
d is
rele
vant a
nd
su
bsta
ntia
ted.
Ma
rk B
an
d 2
– M
id le
ve
l (4
-6 m
ark
s)
Th
e c
and
idate
dem
onstra
tes re
as
on
ab
le k
now
led
ge a
nd
un
ders
tand
ing
of
co
ncu
rrent p
rog
ram
min
g; th
e m
ate
rial is
ge
ne
rally
accura
te b
ut a
t times
und
erd
eve
lope
d.
Th
e c
and
idate
is a
ble
to a
pp
ly th
eir k
now
ledg
e a
nd
un
ders
tand
ing d
irectly
to
the c
onte
xt p
rovid
ed
alth
ou
gh o
ne
or tw
o o
pp
ortu
nitie
s a
re m
isse
d.
Evid
en
ce/e
xa
mp
les a
re fo
r the m
ost p
art im
plic
itly re
leva
nt to
the e
xp
lana
tion
. T
he c
and
idate
pro
vid
es a
reaso
na
ble
dis
cussio
n, th
e m
ajo
rity o
f wh
ich
is
focuse
d. E
va
luativ
e c
om
me
nts
are
, for th
e m
ost p
art a
pp
ropria
te, a
lthoug
h
one
or tw
o o
pp
ortu
nitie
s fo
r deve
lopm
ent a
re m
isse
d.
There
is a
line o
f reaso
nin
g p
rese
nte
d w
ith s
om
e s
tructu
re. T
he in
form
atio
n
pre
se
nte
d is
in th
e m
ost p
art re
levan
t and s
upp
orte
d b
y s
om
e e
vid
en
ce.
Ma
rk B
an
d 1
– L
ow
Leve
l (1
-3 m
ark
s)
Th
e c
and
idate
dem
onstra
tes a
bas
ic k
now
ledg
e o
f co
ncurre
nt p
rog
ram
min
g
with
limite
d u
nd
ers
tand
ing
sh
ow
n; th
e m
ate
rial is
basic
and
co
nta
ins s
om
e
9
AO
1.1
(2
) A
O1
.2
(2)
AO
2.1
(2
) A
O3
.3
(3)
AO
1: K
no
wle
dg
e a
nd
Un
de
rsta
nd
ing
In
dic
ativ
e c
on
ten
t
P
rocesse
s a
re h
ap
pe
nin
g a
t the
sa
me
time/a
t ove
rlap
pin
g tim
es
O
nly
1 p
rocess c
an a
ctu
ally
hap
pe
n
at a
time o
n a
sin
gle
core
pro
cesso
r,
co
ncu
rrent trie
s to
sim
ula
te m
ultip
le
pro
ce
sse
s
O
ne p
rocess m
ay n
ee
d to
sta
rt
befo
re a
se
con
d h
as fin
ish
ed
In
div
idua
l pro
ce
sse
s a
re th
rea
ds,
each
thre
ad
has a
life lin
e
AO
2: A
pp
licatio
n
M
ultip
le o
rders
ca
n b
e m
ade
an
d
add
ed
to th
e lis
t at th
e s
am
e tim
e
P
rogra
mm
ing
will n
ee
d to
allo
w
mu
ltiple
thre
ad
s to
ma
nip
ula
te a
sin
gle
list
W
ill allo
w th
ose
read
ing
and
writin
g
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
18
inaccu
racie
s. T
he c
and
idate
s m
ake
s a
limite
d a
ttem
pt to
app
ly a
cq
uire
d
kn
ow
ledg
e a
nd u
nd
ers
tan
din
g to
the c
onte
xt p
rovid
ed
. T
he c
and
idate
pro
vid
es a
limite
d d
iscu
ssio
n w
hic
h is
na
rrow
in fo
cus.
Ju
dg
em
ents
if ma
de a
re w
eak a
nd
un
su
bsta
ntia
ted
. T
he
info
rma
tion is
basic
and
co
mu
nic
ate
d in
an u
nstru
ctu
red w
ay. T
he
in
form
atio
n is
su
pp
orte
d b
y lim
ited e
vid
en
ce a
nd
the re
latio
nship
to th
e
evid
en
ce m
ay n
ot b
e c
lear.
0 m
ark
s
No
atte
mpt to
an
sw
er th
e q
uestio
n o
r respo
nse is
not w
orth
y o
f cre
dit.
to m
anip
ula
te a
t the s
am
e tim
e
L
ockin
g w
ill nee
d im
ple
me
ntin
g –
mo
re c
om
ple
x p
rog
ram
min
g
AO
3: E
va
lua
tion
W
ill allo
w fo
r mu
ltiple
ord
ers
at th
e
sa
me
time –
as it w
ould
ha
ppe
n in
real life
A
cce
ss to
the lin
ke
d lis
t will n
ee
d to
be lim
ited s
o it c
ann
ot b
e
accesse
d/o
ve
rwritte
n b
y tw
o th
rea
ds
tryin
g to
do d
iffere
nt o
pera
tion
s
N
ot a
ll of th
e p
roce
ss w
ill be
para
llelis
eab
le. X
pro
cesso
rs d
oe
s
not m
ean
it will ru
n in
1/x
th o
f the
time
of o
ne p
roce
ssor.
H446/0
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Ma
rk S
ch
em
e
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ne 2
017
19
3
a
i S
ave
strin
g in
file
1
AO
2.2
(1
)
3
a
ii 1
mark
per b
ulle
t, ma
x 2
per a
dva
nta
ge
to m
ax 4
e
.g.
Pro
ced
ure
s c
an b
e re
-use
d
N
o n
ee
d to
rep
rog
ram
/save
s tim
e
P
rogra
m c
an b
e s
plit b
etw
een
pro
gra
mm
ers
C
an s
pecia
lise
in th
eir a
rea
S
pee
d u
p c
om
ple
tion
time
A
s m
ultip
le p
roced
ure
s w
ork
ed o
n c
oncu
rrently
E
asy to
test/d
eb
ug
A
s e
ach
mo
du
le c
an b
e te
ste
d o
n its
ow
n th
en
com
bin
ed
.
4
AO
1.2
(4
)
Allo
w a
ny a
pp
rop
riate
adva
nta
ge
s
3
b
1
mark
each
function readMessage(fileName)
messageFile = openRead(fileName)
message = messageFile.readLine()
messageFile.close()
return message
endfunction
4
AO
2.1
(4
)
We a
re n
ot te
stin
g p
se
udo
cod
e
kn
ow
ledg
e –
answ
ers
tha
t wo
rk b
ut d
o
not m
atc
h th
e p
seu
do c
od
e g
ive
n s
hou
ld
still b
e c
redite
d fu
ll mark
s.
readMessage
and fileName
and message
are
ca
se s
ensitiv
e
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
20
3
c
1
mark
per b
ulle
t to m
ax 5
U
se
of a
pp
ropria
te lo
op
C
orre
ct e
nd c
ond
ition (le
ng
th o
f message
)
C
orre
ct u
se o
f .push
with
messageStack
A
cce
ssin
g substring
(or e
qu
iva
lent) c
orre
ctly
A
ppro
pria
te c
om
me
nt(s
)
procedure pushToStack(message)
for x = 0 to message.length() //loop through each
//letter
messageStack.push(message.substring(x,1)) //take
//each character and push onto stack
next x //move to next letter
endprocedure
5
AO
2.1
(2
) A
O3
.2
(3)
3
d
1
mark
per b
ulle
t to m
ax 5
P
op e
lem
en
t from
sta
ck
C
onve
rt to A
SC
II va
lue
S
ubtra
ct 1
0 fro
m A
SC
II va
lue
C
onve
rt back to
ch
ara
cte
r
A
ppe
nd
/co
nca
ten
ate
with
va
riab
le
5
AO
1.2
(2
) A
O2
.2
(3)
Acce
pt p
seu
do
cod
e e
qu
iva
lent.
H446/0
2
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rk S
ch
em
e
Ju
ne 2
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21
4
a
T
ree
// Gra
ph
(und
irecte
d)
1
AO
1.2
(1
)
Do
no
t acce
pt b
inary
tree
4
b
i 1
mark
per b
ulle
t to m
ax 4
D
epth
-first g
oes to
left c
hild
no
de w
hen
it ca
n...
If th
ere
is n
o le
ft ch
ild it g
oe
s to
the rig
ht c
hild
w
hen
there
are
no c
hild
nod
es th
e a
lgo
rithm
‘vis
its’ it’ a
nd
ba
cktra
cks to
the p
are
nt n
od
e.
B
read
th-firs
t vis
its a
ll nod
es c
onn
ecte
d d
irectly
to s
tart n
od
e...
T
hen v
isits
all n
od
es d
irectly
co
nn
ecte
d to
each
of th
ose n
od
es (a
nd th
en
all n
od
es d
irectly
co
nn
ecte
d to
those
no
des a
nd s
o o
n…
)
D
epth
-first u
ses a
sta
ck
B
read
th-firs
t uses a
qu
eu
e
4
AO
1.2
(4
)
4
b
ii 1
mark
per n
od
e in
corre
ct o
rder
D
K
L
H
B
G
(X)
6
AO
2.1
(6
)
4
b
iii M
ax 3
e.g
.
W
hen
a n
od
e d
oe
s n
ot h
ave
any n
od
e to
vis
it e.g
. D
T
he a
lgorith
m g
oes b
ack to
the p
revio
us v
isite
d n
od
e e
.g. B
T
o c
heck fo
r furth
er n
ode
s to
vis
it e.g
. H
3
AO
1.2
(2
) A
O2
.1
(1)
H446/0
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rk S
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e
Ju
ne 2
017
22
T
his
repe
ats
until a
ne
w n
od
e c
an b
e v
isite
d, o
r all n
od
es h
ave
be
en
vis
ited
5
a
05 and 07
1
AO
2.1
(1
)
5
b
1
mark
for e
ach h
igh
ligh
ted e
lem
en
t
calculate(4,10)
if 4 == 10 FALSE
elseif 4 < 10 TRUE
return calculate(4, (10-4))
return calculate(4, 6)
if 4 == 6 FALSE
elseif 4 < 6 TRUE
return calculate(4, 6-4)
return calculate(4, 2)
if 4 == 2 FALSE
elseif 4 < 2 FALSE
else
return calculate(2, 4-2)
return calculate(2,2)
if 2 == 2 TRUE
return 2
return 2
return 2
return 2
5
AO
2.1
(5
)
Allo
w tra
ce ta
ble
or a
ny s
ensib
le
eq
uiv
ale
nt.
H446/0
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Ma
rk S
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em
e
Ju
ne 2
017
23
output(2)
5
c
1
mark
per b
ulle
t to m
ax 4
S
uita
ble
loo
p w
ith c
orre
ct c
ond
ition
In
IF: O
ve
rwritin
g n
um
2 w
ith n
um
2 –
num
1
In
EL
SE
: Ove
rwritin
g n
um
1 w
ith n
um
2...
... O
ve
rwritin
g n
um
2 w
ith n
um
1-n
um
2 c
orre
ctly
(usin
g a
tem
p v
aria
ble
)
e.g
. while num1 != num2
if num1 < num2 then
num2 = num2 – num1
else
temp = num1 – num2
num1 = num2
num2 = temp
endif
endwhile
4
AO
2.1
(1
) A
O2
.2
(1)
AO
3.2
(2
)
Alte
rna
tive
ly s
wa
pp
ing
va
lues b
y:
temp = num1
num1 = num2
num2 = temp – num2
H446/0
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rk S
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Ju
ne 2
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24
6
a
1
mark
per in
put to
ma
x 3
C
hoic
e o
f pet
P
et n
am
e
F
eed
P
lay
R
ead
3
AO
2.1
(3
)
Allo
w a
ny re
aso
na
ble
inp
ut to
this
syste
m
6
b
i 1
mark
per b
ulle
t to m
ax 2
S
plittin
g a
pro
ble
m d
ow
n
In
to its
co
mp
on
ent p
arts
/su
b-p
roce
dure
s/m
odu
les
2
AO
1.1
(2
)
6
b
ii 1
mark
per b
ox
6
AO
2.2
(6
)
Ca
lcu
latio
ns m
ust b
e c
orre
ct
H446/0
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rk S
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25
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
26
6
c
i 1
mark
per b
ulle
t to m
ax 3
D
efin
ing
pro
ced
ure
pla
y
o
Re
settin
g b
ore
d to
0
o
Ou
tputtin
g re
sult
e.g
. procedure play()
bored = 0
print("bored: " + bored + “%”)
endprocedure
3
AO
3.2
(3
)
6
c
ii 1
mark
per b
ulle
t to m
ax 3
De
finin
g p
roced
ure
read
o
Co
rrect c
alc
ula
tion
o
Ou
tputtin
g re
sult
e.g
. procedure read()
intelligence = intelligence * 1.006
print("intelligence: " + intelligence)
endprocedure
3
AO
2.2
(1
) A
O3
.2
(2)
6
d
i 1
mark
per b
ulle
t to m
ax 4
C
orre
ct d
ecla
ratio
n, a
pp
ropria
te n
am
e (e
.g. new
)
T
akin
g name
and theType
as a
para
me
ter
S
ettin
g petName
to p
ara
me
ter
4
AO
2.2
(1
) A
O3
.2
(3)
H446/0
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rk S
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Ju
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27
S
ettin
g bored
, hunger
and intelligence
to 0
e.g
. public procedure new(name, theType)
petName = name
bored = 0
hunger = 0
intelligence = 0
type = theType
endprocedure
6
d
ii 1
mark
per b
ulle
t to m
ax 2
myPet
/app
rop
riate
= new pet
Springy
and Tiger
, in ""
, in s
am
e o
rder a
s c
onstru
cto
r decla
ratio
n
e.g
. myPet = new pet("Springy", "Tiger")
2
AO
2.1
(2
)
6
d
iii 1
mark
per b
ulle
t to m
ax
C
lass d
ecla
ratio
n in
clu
din
g in
he
rit (or e
qu
iva
lent e
.g. T
ige
r exte
nd
s P
et,
Tig
er::P
et, T
iger(P
et))
C
onstru
cto
r pro
ce
dure
(ne
w) w
ith a
ll attrib
ute
s p
rese
nt
b
ore
d =
10, h
ung
er =
50, in
tellig
ence
= 1
0, ty
pe =
“Tig
er”
o
utp
utG
reetin
g p
roced
ure
O
utp
uttin
g o
rigin
al a
nd
ne
w m
essag
es c
orre
ctly
e.g
. class Tiger inherits Pet
5
AO
2.2
(2
) A
O3
.2
(3)
H446/0
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Ma
rk S
ch
em
e
Ju
ne 2
017
28
public procedure new(name)
petName = name
bored = 10
hunger = 50
intelligence = 10
type = "Tiger"
endprocedure
public procedure outputGreeting()
print("Hello, I'm " + petName + ", I'm a " + type)
print("I like to eat meat and roar")
endprocedure
endclass
Acce
pt
super.outputGreeting()
In p
lace o
f first print
sta
tem
ent
H446/0
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rk S
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29
6
e
M
ark
Ba
nd
3 –
Hig
h le
ve
l (7
-9 m
ark
s)
Th
e c
and
idate
dem
onstra
tes a
tho
rou
gh
kn
ow
ledg
e a
nd
un
ders
tand
ing
of
abstra
ctio
n; th
e m
ate
rial is
ge
nera
lly a
ccura
te a
nd d
eta
iled.
Th
e c
and
idate
is a
ble
to a
pp
ly th
eir k
now
ledg
e a
nd
un
ders
tand
ing d
irectly
and
co
nsis
tently
to th
e c
on
text p
rovid
ed
. Evid
en
ce/e
xa
mp
les w
ill be e
xp
licitly
re
leva
nt to
the e
xp
lana
tion
. T
he
re is
a w
ell-d
eve
lope
d lin
e o
f reaso
nin
g w
hic
h is
cle
ar a
nd
log
ica
lly
stru
ctu
red. T
he in
form
atio
n p
rese
nte
d is
rele
vant a
nd
su
bsta
ntia
ted.
Ma
rk B
an
d 2
– M
id le
ve
l (4
-6 m
ark
s)
Th
e c
and
idate
dem
onstra
tes re
as
on
ab
le k
now
led
ge a
nd
un
ders
tand
ing
of
abstra
ctio
n; th
e m
ate
rial is
ge
nera
lly a
ccura
te b
ut a
t time
s u
nd
erd
eve
lope
d.
Th
e c
and
idate
is a
ble
to a
pp
ly th
eir k
now
ledg
e a
nd
un
ders
tand
ing d
irectly
to
the c
onte
xt p
rovid
ed
alth
ou
gh o
ne
or tw
o o
pp
ortu
nitie
s a
re m
isse
d.
Evid
en
ce/e
xa
mp
les a
re fo
r the m
ost p
art im
plic
itly re
leva
nt to
the e
xp
lana
tion
. T
he c
and
idate
pro
vid
es a
reaso
na
ble
dis
cussio
n, th
e m
ajo
rity o
f wh
ich
is
focuse
d. E
va
luativ
e c
om
me
nts
are
, for th
e m
ost p
art a
pp
ropria
te, a
lthoug
h
one
or tw
o o
pp
ortu
nitie
s fo
r deve
lopm
ent a
re m
isse
d.
The
re is
a lin
e o
f reaso
nin
g p
rese
nte
d w
ith s
om
e s
tructu
re. T
he in
form
atio
n
pre
se
nte
d is
in th
e m
ost p
art re
levan
t and s
upp
orte
d b
y s
om
e e
vid
en
ce.
Ma
rk B
an
d 1
– L
ow
Leve
l (1
-3 m
ark
s)
Th
e c
and
idate
dem
onstra
tes a
bas
ic k
now
ledg
e o
f abstra
ctio
n w
ith lim
ited
u
nd
ers
tand
ing
sh
ow
n; th
e m
ate
rial is
basic
and
co
nta
ins s
om
e in
accura
cie
s.
Th
e c
and
idate
s m
ake
s a
limite
d a
ttem
pt to
ap
ply
acq
uire
d k
now
ledg
e a
nd
und
ers
tand
ing
to th
e c
onte
xt p
rovid
ed
. T
he c
and
idate
pro
vid
es a
limite
d d
iscu
ssio
n w
hic
h is
na
rrow
in fo
cus.
Ju
dg
em
ents
if ma
de a
re w
eak a
nd
un
su
bsta
ntia
ted
. T
he
info
rma
tion is
basic
and
co
mu
nic
ate
d in
an u
nstru
ctu
red w
ay. T
he
in
form
atio
n is
su
pp
orte
d b
y lim
ited e
vid
en
ce a
nd
the re
latio
nship
to th
e
evid
en
ce m
ay n
ot b
e c
lear.
9
AO
1.1
(2
) A
O1
.2
(2)
AO
2.1
(2
) A
O3
.3
(3)
AO
1: K
no
wle
dg
e a
nd
Un
de
rsta
nd
ing
In
dic
ativ
e c
on
ten
t
R
em
ova
l of u
nn
ece
ssary
ele
me
nts
U
se
s s
ym
bols
to re
pre
sen
t ele
me
nts
of
the p
roble
m
In
cre
ase c
han
ce o
f cre
atin
g th
e p
rog
ram
su
ccessfu
lly
R
edu
ces p
rog
ram
min
g tim
e a
nd fa
cto
rs
that c
an d
etra
ct fro
m th
e p
rog
ram
AO
2: A
pp
licatio
n
E
xa
mp
les o
f use in
this
syste
m e
.g.
o
En
viro
nm
ent is
not s
ho
wn
o
Mo
ve
me
nts
redu
ced/re
mo
ve
d
o
Oth
er fa
cto
rs th
at c
an b
e
don
e/a
ffect th
e 'p
et' a
re re
mo
ve
d
o
Tim
e m
ay n
ot re
pre
sente
d a
s
min
ute
s, s
econ
ds
AO
3: E
va
lua
tion
R
edu
ces c
om
ple
xity
of p
rog
ram
min
g
R
eq
uire
s le
ss c
om
puta
tiona
l pow
er, s
o
the g
am
e c
an b
e p
laye
d o
n lo
we
r sp
ec
H446/0
2
Ma
rk S
ch
em
e
Ju
ne 2
017
30
0 m
ark
s
No
atte
mpt to
an
sw
er th
e q
uestio
n o
r respo
nse is
not w
orth
y o
f cre
dit.
devic
es e
.g. p
ho
nes
F
ocu
s is
on th
e c
ore
aspe
cts
of th
e
pro
gra
m ra
ther th
an
the e
xtra
s
T
oo m
uch a
bstra
ctio
n c
an
de
tract fro
m
the a
pp
ea
l of th
e g
am
e, m
ay b
e to
o
sim
plis
tic/n
ot re
alis
tic e
no
ug
h, m
ay n
ot
have
en
oug
h s
co
pe
to e
ng
ag
e u
sers
6
f i
O(n
) 1
AO
1.1
(1
)
6
f ii
1 m
ark
per b
ulle
t to m
ax 2
2
0(m
s)
…
sh
ow
ing
wo
rkin
g
2
AO
1.2
(1
) A
O2
.1
(1)
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