Computer Science 9618/22 May/June 2026: question paper and mark scheme

Mark scheme

1(a)(i) STRING 1 1(a)(ii) IF LENGTH(Pass) < 6 OR LENGTH(Pass) > 8 / >= 9 2 THEN Mark as follows: 1 Mark: Use of LENGTH function twice 1 Mark: Number 8 used in second condition 1(a)(iii) Mark as follows: 7 1 DECLARE LetterFlag, DigitFlag: BOOLEAN 1 mark for BOOLEAN 2 LetterFlag ← FALSE 1 mark for FALSE 3 FOR Index ← 1 TO LENGTH(Pass) 1 mark for LENGTH 4 Character ← MID(Pass, Index, 1) 1 mark for MID, Index and 1 5 IF IS_NUM(Character) = TRUE THEN 1 mark for IS_NUM 6 DigitFlag ← TRUE ELSE LetterFlag ← TRUE 1 mark for both variables being set to TRUE 7 IF NOT(LetterFlag AND DigitFlag) THEN 1 mark for NOT 1(b)(i) One mark per state 2 Example answers: 1 Sensor activated (mode) / / Intruder/movement detected (mode) // Alert Mode 2 Alarm sounding // Light flashing (mode) // Alarm activated 3 Error mode // Power loss detected (mode) // Battery too low mode 4 Tamper detected (mode) // Lockdown (mode) 5 Disarm Mode // Maintenance Mode // Password Change Mode // Service Mode Max 2 Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 1(b)(ii) One mark per event 3 Do not accept repeating Example answers: examples in stem. 1 Movement / pressure (detected) Reject Turning 2 Window opened / broken (detected) system on. Reject 3 Door open (detected) reference to 4 Entering wrong password (repeatedly) creating password. 5 Entering password to enter / exit maintenance mode / silence alarm 6 Tampering / damaging alarm (system) 7 Time delay // Time passed after alarm triggered / before alarm activated 8 Battery charge too low 9 Pressing panic button Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 1(c) Example Solution 3 Do not accept once if lower-case letters or rule being example of test reason for invalid test characters that are tested data data not letters or digits rule 1 1A password is too short used. rule 1 PHRASE345678 Password too long rule 2 ABCDEF Does not contain a digit rule 2 1234567 Does not contain a letter rule 3 PASSWRD2 letters are repeated rule 3 ABCDEF11 Digits / Numbers are repeated rule 3 AAA111 Both digits and letters are repeated Accept Characters repeated Mark as follows: Rule 1 test: 1 mark for one example of test data for rule 1 and appropriate reason Test data A string of more than 8 characters that contains both letters and digits. Reject if test data breaks Rule 2 The letters must be upper case or/and Rule 3 Each character only occurs once Reason Password is too long Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 1(c) Rule 2 test (1 mark): 1 mark for two examples of test data that only breaks rule 2 and appropriate reasons. Test data A string between 6 and 8 characters long Reject if letters that contains no digits repeated Reason Does not contain a digit Test data A string between 6 and 8 characters long that contains no letters Reason Does not contain a letter Reject if digits repeated Rule 3: 1 mark for two examples of test data for rule 3 and appropriate Reject characters reasons repeated Test data A string between 6 and 8 characters long where only digits are repeated Reason Digits are repeated Test data A string between 6 and 8 characters long where both characters and digits are Reject if letters repeated repeated Reject if no letters Reason Both digits and letters are repeated 1(d) One mark per example 3 Example answers: 1 Structure chart // Problem / Program broken into modules // Reject Modular Diagram ‘Decomposition’ by itself 2 Flowchart Reject Module 3 Data Flow Diagram Overview by itself 4 Identifier / Variable table // Data structure / Record structure (designs) 5 File structure // Database Design 6 E-R Diagram 7 HCI Design // Report Design 8 Class Diagram // UML Diagram 9 Programming language to use Max 3 Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 1(e) One mark per stage 2 1 Analysis 2 Coding // Implementation NE Development / Programming 3 Maintenance Max 2 marks 1(f)(i) Integration 1 1(f)(ii) One mark for the name, one for the description 3 Method: Stub (testing) Description: 1 A simple / dummy module is written to replace the module Do not mark (that is not working) description if 2 The simple / dummy module will return an expected value / method incorrect will output a message (to show the module has been called)

Mark scheme

2(a) One mark for either explanation: 1 • To filter out information / data that is not necessary for the task • To keep only information / data that is necessary for the task Max 1 mark 2(b) One mark for each TWO data items 2 Reject Quantity Reject Return date Example answers if Hire date already 1 Tool ID // Tool name // Type of tool given 2 Customer ID // Customer name // Customer address // MP 2 Reject just Account reference ‘Customer’ 3 Hire ID 4 (Hire) Date 5 Hire time 6 Length of hire: Full or Half day 7 Cost of hire 8 Payment details 9 Tool status Max 2 marks Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 2(c) One mark for each process. 2 Examples include: 1 Input customer details / customer ID / payment details / Tool ID 2 Search for required tool // Checking if tool returned // Checking if tool available // Searching for dates when tool is available 3 Record tool booking // Amend Tool booking // Create tool booking // Record when tool returned // Confirm booking 4 Calculating time when tool has to be returned 5 Change / check tool status e.g. hired / available / damaged // Prevent same tool being hired twice 6 Process payment / calculate hire cost Max 2 marks

Mark scheme

3 Example solution 6 MP 1 cannot be in a loop structure Mark points: MP 1 (Step 4) Open WordList.txt / File in read mode (and Do not accept subsequently close) once if MP 2, 3, 4 MP 2 (Step 5) Read the next line from the file is clearly not in a in a loop loop MP 3 (Step 6) Increment line counter Award only one in a loop mark for MP 5 and MP 4 (Step 7) Check if the line read equals the word (input) MP 6 if both in a loop outputs are MP 5 (Step 8) ...If it does output line counter possible MP 6 (Step 9) If word not found in file then output "Not Found" after all iterations MP 7 Can be any MP 7 (Step 10) Otherwise repeat from step 5 loop but it must be clear where it Max 6 marks from 7 possible marks starts and ends and must include at least one of the end conditions. Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/

Mark scheme

4(a) After group 1 6 back stack forward stack www.bbb.home ← TOS www.333.corp www.111.home www.aaa.home ← TOS www.999.corp www.ccc.my www.222.me Web address currently displayed in the browser: www.zzz.corp After group 2 back stack forward stack www.333.corp ← TOS www.zzz.corp ← TOS www.111.home www.aaa.home www.999.corp www.ccc.my www.222.me Website displayed in the browser: www.bbb.home After group 3 back stack forward stack www.111.home ← TOS www.333.corp www.bbb.home www.zzz.corp www.aaa.home www.ccc.my www.222.me ← TOS Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 4(a) Website displayed in the browser: www.999.corp Mark as follows: Group 1 1 1 mark for back stack as shown and TOS position as shown on both stacks 2 1 mark for forward stack as shown and current web address Group 2 3 1 mark for stacks and correct TOS pointers as shown and current web address Group 3 4 1 mark for back stack as shown and TOS as shown 5 1 mark for forward stack as shown and TOS as shown 6 1 mark current web address 4(b) One mark for identifying an ADT. 3 Queue / Linked List One mark per feature Features of a Queue: 1 1 Each queue element contains one data item 2 A front/start pointer and a rear/end pointer 3 Data is added at back / end and removed from front / start // works on a FIFO basis 4 Can be circular or linear Need both MAX 2 marks for features Features of a Linked List: 1 1 Each node / item contains data and a pointer (to the next Reject use of 2 node/item) arrays 2 A pointer to the start of the list 3 Last node in the list has a null pointer 4 Data may be added / removed by manipulating pointers 5 Pointers are followed to move through nodes in the list (in a given sequence) 6 Unused nodes are stored on a free list // a free-list pointer to the Free List 7 Is a dynamic (data structure) MAX 2 marks for features Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/

Mark scheme

5(a) The greater than logic operator / > in the selection statement 1 would need to be changed to a less than operator / < Or by example IF Numbers[Index] > Numbers[Index + 1] THEN would be changed to IF Numbers[Index] < Numbers[Index + 1] THEN OR IF Numbers[Index] <= Numbers[Index + 1] THEN OR IF Numbers[Index + 1] > Numbers[Index] THEN OR IF Numbers[Index + 1] >= Numbers[Index] THEN 5(b)(i) 4 Reject answers 2 marks for inner loop that just repeat the algorithm MP 1: The inner loop iterates one less time (in algorithm B) each time the outer / repeat loop iterates. Reject program stops rather than MP 2: this is because for each iteration of the outer / repeat iteration ends or loop (in algorithm B) the last / one more element in the algorithm ends Numbers array will be in the correct place. 2 marks for outer loop MP 3: The outer / repeat loop (in algorithm B) would stop iterating if no swaps are made / all numbers in array are sorted MP 4: so not as many iterations would be needed (if the Numbers array is not initially sorted) 5(b)(ii) 1 mark: When Numbers (array) / array is already in (ascending) 1 order / sorted 5(c)(i) 1 mark: The parameters / Num1 and Num2 are passed by value 2 / BYVAL // The parameters / Num1 and Num2 are not passed / should be passed by reference / BYREF 1 mark: (The swap made inside) the procedure does not change the arguments used /parameters (passed) / the values stored in the Numbers elements passed // The procedure does not swap / change the values stored in the array Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 5(c)(ii) Solution 2 NE for header mark If ByRef is PROCEDURE Swap(BYREF Num1, Num2 : INTEGER) mentioned in the DECLARE Temp : INTEGER header but syntax Temp ← Num1 is not correct but Num1 ← Num2 allow FT for Num2 ← Temp second mark ENDPROCEDURE 2 marks: code as above 1 mark: just the procedure header as above Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/

Mark scheme

6(a) Mark Schme One – loop used to form return string 8 (See MS Two for solutions using LEFT()/MID() to form return string) Example solution FUNCTION AddPadding() RETURNS STRING CONSTANT Space = ' ' DECLARE Plaintext : STRING DECLARE LenOfPlaintext : INTEGER DECLARE Index : INTEGER REPEAT OUTPUT "Enter plaintext" INPUT Plaintext LenOfPlaintext ← LENGTH(Plaintext) UNTIL LenOfPlaintext >= 1 AND LenOfPlaintext <= 25 FOR Index ← LenOfPlaintext + 1 TO 25 Plaintext ← Plaintext & Space NEXT Index // Alternative way to append spaces using while loop WHILE LENGTH(Plaintext) <> 25 Plaintext ← Plaintext & Space ENDWHILE //Alternative way to append spaces DECLARE SpaceString : STRING SpaceString ← "" FOR Index ← LenOfPlaintext + 1 TO 25 SpaceString ← SpaceString & Space NEXT Index Plaintext ← Plaintext & SpaceString RETURN Plaintext ENDFUNCTION Mark points 1 Function heading, return type and ending 2 Prompt for and input of plaintext MP 4 Must be a 3 Conditional loop that checks one of the boundary values conditional loop 4 Conditional loop that correctly checks both boundaries of If no loop used plaintext then MP3, MP4 5 …Re-input of plaintext in a loop and MP5 can not 6 Loop that iterates for number of spaces to add to end of be given plaintext 7 …Spaces appended to plaintext // Spaces appended to an empty string 8 Correctly padded plaintext returned following a reasonable attempt at MP7 Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. 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PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 6(a) Mark Scheme Two - No Loop used to form return string – use of LEFT()/MID() Function Example Solution: FUNCTION AddPadding() RETURNS STRING CONSTANT Fill = " " //24 Spaces DECLARE Plaintext : STRING DECLARE LenOfPlaintext : INTEGER DECLARE Index : INTEGER REPEAT OUTPUT "Enter plaintext" INPUT Plaintext LenOfPlaintext ← LENGTH(Plaintext) UNTIL LenOfPlaintext >= 1 AND LenOfPlaintext <= 25 Plaintext ← LEFT(Plaintext & Fill, 25) RETURN Plaintext ENDFUNCTION Mark points 1 Function heading, return type and ending 2 Prompt for and input of plaintext 3 Conditional loop that checks one of the boundary values Conditional loop that correctly checks for length of plaintext 4 …Re-input of plaintext in a loop 5 Use of string of at least 24 spaces Correct use of LEFT() / MID() function and & to append padding characters Do not accept use 6 Correctly padded plaintext returned following a reasonable of language attempt at specific string MP 6 and MP 7 manipulation which is not the compatible with the pseudocode used for the syllabus Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 6(b)(i) Solutions: 2 • Grid : ARRAY[1 : 5, 1 : 5] OF CHAR • Grid : ARRAY[0 : 4, 0 : 4] OF CHAR 1 mark for boundaries of both array dimensions: • [1 : 5, 1 : 5] Reject use of • [0 : 4, 0 : 4] different brackets eg ( ) 1 mark for all other parts of declaration Reject CHARACTER instead of CHAR Reject STRING instead of CHAR Must be as exactly as above but accept use of lowercase Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 6(b)(ii) Solution One : Using nested loop and output of each row in a 8 loop following nested loop Example Solution PROCEDURE PopulateGrid(Plaintext : STRING) DECLARE Row, Column, Index : INTEGER Index ← 1 FOR Row ← 1 TO 5 FOR Column ← 1 TO 5 Grid[Row, Column] ← MID(Plaintext, Index, 1) Index ← Index + 1 NEXT Column Next Row FOR Row ← 1 TO 5 OUTPUT Grid[Row, 1], Grid[Row, 2], Grid[Row, 3],_ Grid[Row, 4], Grid[Row, 5] NEXT Row ENDPROCEDURE Mark Points 1 Generate Index, 1 to 25, used to extract all characters from Plaintext 2 Generate both Row and Col in a loop 3 Extract an individual character from Plaintext in a loop Syntax for MID() 4 Attempt to assign the extracted character to Grid in a function must use loop parameters of correct type – 5 Extraction and assignment of all 25 characters in allow other string Plaintext to Grid in a loop handling function combinations if they work. 6 Attempt to identify / isolate 5 characters comprising at least one row Max 7 if Procedure heading/parameter 7 Identify / isolate all 5 characters for all 5 rows type incorrect Penalise incorrect 8 Output all five rows of GRID, each on a separate line syntax for indexes i.e. [Row][Col] instead once. Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 6(b)(ii) Solution Two: Using nested loop and output of each string in a loop following nested loop - Example Solution PROCEDURE PopulateGrid(Plaintext : STRING) DECLARE Row, Column, Index : INTEGER DECLARE RowString : STRING FOR Row ← 1 TO 5 FOR Column ← 1 TO 5 Index ← (Row – 1) * 5 + Column Grid[Row, Column] ← MID(Plaintext, Index, 1) NEXT Column NEXT Row FOR Row ← 1 TO 5 RowString ←"" For Col ← 1 TO 5 RowString ← RowString & Grid[Row, Column] Next Col OUTPUT RowString NEXT Row ENDPROCEDURE Solution Three: Using nested loop where output string is formed in the inner loop and output in string in the outer loop – complex use of Index. No additional loop after nested loop Example Solution PROCEDURE PopulateGrid(Plaintext : STRING) DECLARE Row, Column, Index : INTEGER DECLARE RowString : STRING FOR Row ← 1 TO 5 RowString ← "" FOR Column ← 1 TO 5 Index ← (Row – 1) * 5 + Column Grid[Row, Column] ← MID(Plaintext, Index, 1) RowString ← RowString & MID(Plaintext, Index, 1) NEXT Column OUTPUT RowString NEXT Row ENDPROCEDURE Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 6(b)(ii) Solution Four: No nested loop Example Soution PROCEDURE PopulateGrid(Plaintext : STRING) DECLARE Row, Column, Index : INTEGER Row ← 1 Col ← 0 FOR Index ← 1 TO 25 IF Index MOD 5 = 0 THEN Row ← Row + 1 Col ← 0 ENDIF Col ← Col + 1 Grid[Row, Column] ← MID(Plaintext, Index, 1) NEXT Index FOR Row ← 1 TO 5 OUTPUT Grid[Row, 1], Grid[Row, 2],_ Grid[Row, 3], Grid[Row, 4], Grid[Row, 5] NEXT Row ENDPROCEDURE Trace ID: PC-95C57C5F77 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/