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Cambridge International AS & A Level

COMPUTER SCIENCE 9618/23

Paper 2 Fundamental Problem-solving and Programming Skills October/November 2023
MARK SCHEME
Maximum Mark: 75

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the
details of the discussions that took place at an Examiners’ meeting before marking began, which would have
considered the acceptability of alternative answers.

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.

Cambridge International will not enter into discussions about these mark schemes.

Cambridge International is publishing the mark schemes for the October/November 2023 series for most
Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level
components.

This document consists of 11 printed pages.

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PUBLISHED 2023

Generic Marking Principles

These general marking principles must be applied by all examiners when marking candidate answers.
They should be applied alongside the specific content of the mark scheme or generic level descriptors
for a question. Each question paper and mark scheme will also comply with these marking principles.

GENERIC MARKING PRINCIPLE 1:

Marks must be awarded in line with:

• the specific content of the mark scheme or the generic level descriptors for the question
• the specific skills defined in the mark scheme or in the generic level descriptors for the question
• the standard of response required by a candidate as exemplified by the standardisation scripts.

GENERIC MARKING PRINCIPLE 2:

Marks awarded are always whole marks (not half marks, or other fractions).

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:

• marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit
is given for valid answers which go beyond the scope of the syllabus and mark scheme,
referring to your Team Leader as appropriate
• marks are awarded when candidates clearly demonstrate what they know and can do
• marks are not deducted for errors
• marks are not deducted for omissions
• answers should only be judged on the quality of spelling, punctuation and grammar when these
features are specifically assessed by the question as indicated by the mark scheme. The
meaning, however, should be unambiguous.

GENERIC MARKING PRINCIPLE 4:

Rules must be applied consistently, e.g. in situations where candidates have not followed
instructions or in the application of generic level descriptors.

GENERIC MARKING PRINCIPLE 5:

Marks should be awarded using the full range of marks defined in the mark scheme for the question
(however; the use of the full mark range may be limited according to the quality of the candidate
responses seen).

GENERIC MARKING PRINCIPLE 6:

Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should
not be awarded with grade thresholds or grade descriptors in mind.

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Mark scheme abbreviations

/ separates alternative words / phrases within a marking point

// separates alternative answers within a marking point
underline actual word given must be used by the candidate (grammatical variants accepted)
max indicates the maximum number of marks that can be awarded
( ) the word / phrase in brackets is not required but sets the context

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Question Answer Marks

1(a) 4
Assignment statement Data type

MyVar1  Total1 / Total2 REAL

MyVar2  27/10/2023 DATE

MyVar3  "Sum1 / Sum2" STRING

MyVar4  Result1 AND Result2 BOOLEAN

1(b) 4
Expression Evaluates to

Fraction >= 0.2 AND NOT Active FALSE

INT((Fraction1 * 100) + 13.3) 33

STR_TO_NUM(MID(Code, 4, 2)) + 5 28
LENGTH("TRUE" & Code) 11

1(c) The use of a program library (routines) 1

1(d) MP1 Type: Adaptive 2

MP2 Reason: The (user) requirement(s) changes // to accommodate
legislative changes

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Question Answer Marks

2(a) 5

MP1 Initialise Min to first value in Data and MinIndex to 1

MP2 Loop through 29 more values (or 30 values in total)
MP3 Compare element from Data[] with Min
MP4 Set new Min AND save MinIndex when element value < Min in a loop
MP5 Output MinIndex

2(b) MP1 Simplifies the algorithm // easier to write / understand / test / debug 2
MP2 It is possible to iterate through the values // can use a loop // allows the
storage of many values using a single identifier

2(c) One mark per underlined section 2

DECLARE Data : ARRAY[1:120] OF REAL

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Question Answer Marks

3(a) 3

:
MP1 D3, D4, D1, D2 and D5 in question order or reversed - in any five
consecutive locations
MP2 FoQ value matches index with D3
MP3 EoQ value matches index with D5

3(b) MP1 If NumItems is / = 8 // (queue) full then jump to step 6 6

MP2 Increment EndOfQueue
MP3 If EndOfQueue = 9 then set EndOfQueue to 1
MP4 Increment NumItems

MP5 Set the Element at the index …

MP6 stored in EndOfQueue to value/data/item being added

Mark as follows:
Steps 1 to 4: One mark for gaps filled as shown
Step 5: One mark for 'element' and one mark for other two terms

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Question Answer Marks

4(a) PROCEDURE RandList() 6

DECLARE Count, BaseNum, ThisNum : INTEGER
CONSTANT StepVal = 10
BaseNum  0

FOR Count  1 TO 25
ThisNum  BaseNum + INT(RAND(StepVal))
OUTPUT ThisNum
BaseNum  BaseNum + StepVal
NEXT Count

ENDPROCEDURE

MP1 Procedure heading and ending

MP2 Local loop counter Count as integer
MP3 Loop to iterate 25 times or more for each unique number
MP4 ‘Attempt’ to generate a random number including use of INT() in a
loop
MP5 Ensure that number generated is greater than previous and change
‘previous’
MP6 Output random number after an attempt at MP5 in a loop

4(b) One mark for simplified logical expression: 1

MP1 Result[x + 1] <= Result[x]

ALTERNATIVE SOLUTION:

Result[x] >= Result[x + 1]

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Question Answer Marks

5 6

MP1 Row 1 (initialisation)

Each iteration (1 – 5):

MP2 1 – Total 2
MP3 2 – Total 5
MP4 3 – Total 10
MP5 4 – Total 15
MP6 5 – Total 28 and final Mix[1] = 56

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Question Answer Marks

6 6
FUNCTION TestNum(ThisNum : STRING) RETURNS INTEGER

IF LEFT(ThisNum,3) = RIGHT(ThisNum 3) THEN

RETURN 3
ENDIF

IF RIGHT(ThisNum, 3) = "000" THEN

RETURN 2
ENDIF

IF MID(ThisNum, 4, 1) = MID(ThisNum, 5, 1)__

AND MID(ThisNum, 5, 1) = MID(ThisNum, 6, 1) THEN
RETURN 1
ENDIF

RETURN 0

ENDFUNCTION

MP1 Function heading and ending including parameter and return type
MP2 Test for Condition 1
MP3 Test for Condition 2
MP4 Test for Condition 3
MP5 Return the highest value if more than one condition is satisfied
MP6 Return zero if no condition matched

Question Answer Marks

7(a) MP1 iteration / looping 2

MP2 naming all four modules correctly in the correct sequence // e.g.
Module-A repeatedly calls Sub-Y1,then SubY2 then Sub-9

7(b)(i) TYPE MyType 3

DECLARE RA : INTEGER
DECLARE RB : STRING
DECLARE RC : BOOLEAN
ENDTYPE

MP1 TYPE MyType... ENDTYPE

MP2 RA as integer and RB3 as string
MP3 RC as Boolean

7(b)(ii) PROCEDURE Sub-9(BYREF Param : MyType) 2

MP 1 One mark for BYREF

MP 2 One mark for the rest of the statement

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Question Answer Marks

8(a) PROCEDURE ReceiveFile(FileName : STRING) 7

DECLARE FileData : STRING
DECLARE CharCount : INTEGER
CONSTANT Terminator = "****"

OPENFILE FileName FOR WRITE

FileData  GetData()
CharCount  0

WHILE FileData <> Terminator

WRITEFILE FileName, FileData
CharCount  CharCount + LENGTH(FileData)
FileData  GetData()
ENDWHILE

CLOSEFILE FileName

OUTPUT CharCount (, " characters were written to ",

FileName)

ENDPROCEDURE

MP1 OPEN file in WRITE mode and subsequently CLOSE

MP2 Conditional loop until terminator received
MP3 ‘Attempted’ use of GetData() – Ignore CALL …
MP4 Fully correct use GetData()to return the data in a loop
MP5 Maintain CharCount in a loop
MP6 Write each line to the file - except the terminator in a loop
MP7 Final output of message giving number of characters written outside
loop

8(b) Max 3 marks 3

Problem:
• If the file being sent contains a line of the string "****"
• then the file being written by ReceiveFile() will end at this point //
subsequent file lines will be lost

Solution:
• Read the file (at the sending end) to find the number of lines it contains
• Send an initial message which defines the number of lines in the file

ALTERNATIVE SOLUTION:

• (Transmitter program) chooses a different terminator string / character

that doesn't occur in the file
• Transmitter program sends the terminator string / character before first
line of file / before the transfer begins

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Question Answer Marks

8(c) PROCEDURE Chat(Destination : STRING, Port : INTEGER) 7

DECLARE Data : STRING
DECLARE Finished : BOOLEAN
CONSTANT Terminator = "Bye"
CONSTANT STX = CHR(2)
CONSTANT ETX = CHR(3)

Finished  FALSE

REPEAT
Data  GetData()
OUTPUT Data
IF Data = Terminator THEN
Finished  TRUE
ENDIF

IF NOT Finished THEN //about to reply

INPUT Data
Transmit(STX & Destination & MyID & Data & ETX,
Port)
IF Data = Terminator THEN
Finished  TRUE
ENDIF
ENDIF

UNTIL Finished = TRUE

ENDPROCEDURE

Conditional loop
MP1 Conditional loop
MP2 Test for terminator in both cases
MP3 Use GetData() to get the data from the message
MP4 OUTPUT the data in a loop
MP5 INPUT the data reply
MP6 ‘Attempted ‘ use of Transmit to send it in a loop
MP7 Correct formation of parameters to Transmit()

8(d) Note: Max 3 marks (from either limitation or modification list) 3

Limitation:
1 GetData() does not return a value until a message has been received
2 So once a message has been sent the user has to wait for a reply // chat
is half-duplex

Modification:
3 If no response allow the receiver to exit chat at any time …
4 GetData() should immediately return a suitable message // set a time
limit
5 ... which Chat() can detect and respond by allowing the conversation to
continue

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