Computer Science 9618/22 May/June 2025: question paper and mark scheme

Mark scheme

1(a)(i) One mark for 1 Choice of program development cycle depends on the approach/method required to produce the program Or One mark for a factor that would influence the choice of program development cycle, e.g. • Rapid development of program/prototypes required • Need for prototypes (at an early stage) • Complexity of problem • Skills / experience of development team / programmer(s) • Budget / Time / Resources available • Size of development team • Developer / programmer can return to earlier stages / any stage • Avoidance of repeating previous stages in development of program • How much involvement clients will have • Allows the (program) requirements to be changed during program development • (Program) requirements agreed at start of development of program Max 1 1(a)(ii) Which programming language would best be suited to control the production 1 line // Which programming language would best suit the problem being solved 1(b)(i) Examples include: 3 • Change to (production line) requirements • New technology / hardware available (to control production line) • Changes made to library modules used • Change in relevant legislation Max 3 1(b)(ii) Perfective // Corrective 1 1(c) 1 mark for each correct row 4 Expression Data type RIGHT(MachineCode, 4) STRING Speed * 2.5 REAL NOT Status BOOLEAN IS_NUM(Check) BOOLEAN 1(d)(i) Two / 2 1 1(d)(ii) 1000 1 1(d)(iii) DECLARE Product : ARRAY [0:99, 0:9] OF INTEGER 2 1 mark for correct upper and lower bound • [0:99, 0:9 1 mark for all other parts of declaration DECLARE Product : ARRAY [0:99, 0:9] OF INTEGER

Mark scheme

2(a) Mark as follows: 2 1. To increase the level of detail of the algorithm // To break the problem / task into smaller steps … 2. … until steps can be directly translated into lines of code // from which it can be programmed 2(b)(i) One mark for each of set test values 2 Hours worked between 1 and 40 inclusive Sales value <= 2000 Bonus Pay: 0 Hours worked above 40 Sales value <= 2000 Bonus Pay: 10 Hours worked between 1 and 40 inclusive Sales value > 2000 Bonus Pay: 50 Hours worked above 40 Sales value above 2000 Bonus Pay: 100 max 2 2(b)(ii) One mark per point 2 • Simple modules are written to replace each of the unfinished modules. • Each simple module will return an expected value / will output a message to show it has been called. 2(b)(iii) One mark for naming type of error and one mark for corresponding 2 description Type of error: Logic (error) Description: Where the program does not behave as expected / Does not give expected result / An error in the logic of the algorithm Or Type of error: Run-time (error) Description: The program performs an illegal operation Max 2

Mark scheme

3 Example solution 8 DECLARE FirstNumber : INTEGER DECLARE SecondNumber : INTEGER DECLARE ThirdNumber : INTEGER FirstNumber ← INT(RAND(21)) – 10 OUTPUT FirstNumber REPEAT SecondNumber ← INT(RAND(21)) – 10 UNTIL FirstNumber <> SecondNumber OUTPUT SecondNumber IF FirstNumber < 0 AND SecondNumber < 0 THEN ThirdNumber ← INT(RAND(6)) + 30 OUTPUT ThirdNumber ENDIF Mark points: 1. Declare all variables used 2. Use RAND() function with any integer parameter 3. Use INT() function with any numeric parameter 4. Conditional loop until two different random numbers are generated 5. Use INT() function using random number generated between −10 and 10 inclusive in a loop 6. Output two different random integers / numbers 7. Check if both random integers / numbers are negative 8. then output a (third) random integer / number between 30 and 35 inclusive

Mark scheme

4 6 One mark for each functional group as listed: 1. Initialise Index used for Number to either 1 or 0 before first loop 2. Loop 100 times 3. Input value 4. Increment array index and store value in Number array at element referenced by array index in a loop 5. Output loop starts at last element in Number array 6. Output Number array element referenced by Index in a loop 7. Output all elements of Number array once in reverse order Max 6

Mark scheme

5(a) 4 Mark as follows: 1 Mark: The three new states drawn and labelled. 1 Mark: Two of the events drawn with labels connecting correct states and correct line direction 1 Mark: Four of the events drawn with labels connecting correct states and correct line direction. 1 Mark: All and only six events drawn with labels connecting correct states and correct line direction. 5(b) Conditional Solution 8 Example Solution DECLARE Count : INTEGER DECLARE Line : STRING DECLARE NextHour, Hour : STRING OPENFILE "TimeTaken.txt" FOR READ Count ← 1 READFILE "TimeTaken.txt", Line Hour ← LEFT(Line, 2) WHILE NOT EOF("TimeTaken.txt") READFILE "TimeTaken.txt", Line NextHour ← LEFT(Line, 2) IF NextHour = Hour THEN Count ← Count + 1 ELSE OUTPUT "Hour : ", Hour, " Total : ", Count Hour ← NextHour Count ← 1 ENDIF ENDWHILE OUTPUT "Hour : ", Hour, " Total : ", Count CLOSEFILE "TimeTaken.txt" Mark as follows: 1. Any Initialisation of count for number of pictures taken each hour 2. Open "TimeTaken.txt" for read and subsequently close 3. Conditional loop until EOF 4. Read a line from "TimeTaken.txt" in a loop 5. Extract Hour from line read in a loop 6. A mechanism to compare current hour extracted from file with last one read from file in a loop 7. If hours same increment count in a loop 8. If not same output count (with a suitable message) and update Hour ← NextHour and set Count to 1 in a loop 9. Output final Count and Hour (with a suitable message) once only Note: max 8 5(b) Alternative solution and mark scheme use of an array to hold count for each hour Also, for use of 24 variables and 24 selection conditions DECLARE HoursArray[0 : 23] OF INTEGER DECLARE Index, Hour : INTEGER DECLARE Line : STRING FOR Index ← 0 TO 23 HoursArray[Index] ← 0 NEXT Index OPENFILE "TimeTaken.txt" FOR READ WHILE NOT EOF("TimeTaken.txt") READFILE "TimeTaken.txt", Line Hour ← STR_TO_NUM(LEFT(Line, 2)) HoursArray[Hour] ← HoursArray[Hour] + 1 ENDWHILE FOR Index ← 0 TO 23 IF HoursArray[Index] <> 0 THEN OUTPUT "Hour : ", Index, " Total : ", HoursArray[Index] ENDIF NEXT Index CLOSEFILE "TimeTaken.txt" Mark as follows: 1. Initialisation of 24 array elements to 0 // Initialisation of 24 Integer variables to 0 2. Open "TimeTaken.txt" for read and subsequently close 3. Conditional loop until EOF 4. Read a line from "TimeTaken.txt" in a loop 5. Extract Hour from line read in a loop 6. Convert Hour to an integer value in a loop 7. Increment appropriate array element / variable in a loop 8. Output of each hour and corresponding count variable (with a suitable message) for all values where count is not zero

Mark scheme

6(a)(i) Value Start Unused New Last Current 4 1043 2 8 8 –1 2 Region 1 2 3 Region 2 3 1 Region 3 1 7 7 4 Region 4 Award 1 mark per region 6(a)(ii) Data Pointer 3 1 1018 1 7 2 1007 2 3 3 1010 3 1 4 1056 4 6 5 1092 5 –1 6 1062 6 5 7 1034 7 8 8 1043 8 4 9 0 9 10 10 0 10 –1 Mark as follows: 1 mark for global array Data row 8 containing the value 1043 1 mark for global array Pointer row 7 containing the value 8 and row 8 containing the value 4 1 mark for all other rows in both arrays 6(b) Example answer: 2 The ADT is a linked list and the procedure Place()inserts / add a new node / value into it / the linked list Mark as follows: 1. 1 mark for identifying the ADT as a linked list 2. 1 mark for identifying operation as inserting / adding a value / node into a linked list

Mark scheme

7(a) Example solution Conditional Loop 8 FUNCTION FindCustomer(CustomerID : INTEGER) RETURNS INTEGER DECLARE Index : INTEGER DECLARE Found : BOOLEAN CONSTANT Unused = 99999 CONSTANT Upper = 1000 Found ← FALSE Index ← 1 IF CustomerID < 10001 OR CustomerID > 11000 THEN RETURN –1 // Out of range value for customer ID ENDIF WHILE Found = FALSE AND Loyalty[Index, 1] <> 99999 IF Loyalty[Index,1] = CustomerID THEN Found ← TRUE ELSE Index ← Index + 1 ENDIF IF Index > Upper THEN RETURN –1 ENDIF ENDWHILE IF Found THEN RETURN Loyalty[Index, 2] ELSE RETURN –1 ENDIF ENDFUNCTION Mark as follows: 1. Create function header and ending with correct parameter and return type 2. Check CustomerID is in range and if not return −1 3. (Conditional) loop iterating through each element in array 4. Check for current element in array contains required customer ID in a loop 5. … If found set value to terminate loop 6. Terminating loop when customer ID found 7. Terminating loop when current customer ID is 99999 8. Return either loyalty points for the customer found or –1 if not found 7(a) Alternative solution using FOR Loop Example solution FUNCTION FindCustomer(CustomerID : INTEGER) RETURNS INTEGER DECLARE Index : INTEGER IF CustomerID < 10001 OR CustomerID > 11000 THEN RETURN –1 // Out of range value for customer ID ENDIF FOR Index ← 1 TO 1000 IF Loyalty[Index, 1] = CustomerID THEN RETURN Loyalty[Index, 2] ENDIF IF Loyalty[Index, 1] = 99999 THEN RETURN –1 ENDIF NEXT Index ENDFUNCTION Mark as follows: 1. Create function header and ending with correct parameter and return type 2. Check CustomerID is within range and if not return –1 3. FOR Loop 4. Loop for elements 1 to 1000 / 0 to 999 5. Check if current element in array contains required Customer ID in a loop 6. … If found correctly return loyalty points // store correct loyalty points and return after loop 7. Check if current element in array is 99999 and return –1 / break loop in a loop 8. return −1 if Customer ID not found Direct access solution Alternative solution subtracting 10000 from CustomerID FUNCTION FindCustomer(CustomerID : INTEGER) RETURNS INTEGER DECLARE Index : INTEGER IF CustomerID < 10001 OR CustomerID > 11000 THEN RETURN –1 // Out of range value for customer ID ENDIF 7(a) Index = CustomerID – 10000 IF Loyalty[Index, 1] = CustomerID THEN RETURN Loyalty[Index, 2] ENDIF RETURN –1 ENDFUNCTION Mark as follows: 1. Create function header and ending with correct parameter and return type 2. Declare Index as Integer 3. Check CustomerID is less than 10001 and return –1 if it is 4. Check CustomerID is greater than 11000 and return –1 if it is 5. Calculate Index by subtracting a 10000 form CustomerID 6. Check if array contains the CusomerID 7. Return Loyalty Points if CustomerID found 8. Return –1 if not found Binary Search Mark Scheme Example Solution FUNCTION FindCustomer(CustomerID : INTEGER) RETURNS INTEGER DECLARE Start : INTEGER DECLARE End : INTEGER DECLARE Mid : INTEGER IF CustomerID < 10001 OR CustomerID > 11000 THEN RETURN –1 // Out of range value for customer ID ENDIF Start ← 1 End ←1000 WHILE Start <= End Mid ← (Start + End) DIV 2 IF Loyalty[Mid, 1] = CustomerID THEN RETURN Loyalty[Mid, 2] ELSE IF Loyalty[Mid, 1] > CustomerID THEN End ← Mid – 1 ELSE Start ← Mid + 1 ENDIF ENDWHILE RETURN –1 ENDFUNCTION 7(a) Mark points 1. Create function header and ending with correct parameter and return type 2. Check CustomerID is within range and if not return –1 3. Conditional loop that halves array search space with each iteration 4. Check if CustomerID is stored in current array element in loop 5. Mechanism to end loop if CustomerID is found in array in a loop 6. Mechanism to end loop if CustomerID is not in array in a loop 7. If CustomerID found in array then return correct loyalty points 8. If CustomerID not in array then return –1 7(b) Example solution 7 PROCEDURE PointsReport() DECLARE Count : INTEGER DECLARE Index : INTEGER DECLARE Sum : INTEGER DECLARE Average : REAL Index ← 1 Count ← 0 Sum ← 0 WHILE Loyalty[Index, 1] <> 99999 AND Index <= 1000 IF Loyalty[Index, 2] >= 11 THEN OUTPUT Loyalty[Index, 1] ENDIF Count ← Count + 1 Sum ← Sum + Loyalty[Index, 2] Index ← Index + 1 ENDWHILE Average ← Sum / Count OUTPUT "The average points of all the customers is ", Average ENDPROCEDURE Mark as follows: 1. Create procedure header and ending 2. Declare and initialise Index, Sum and Count 3. Loop through all elements in Loyalty array 4. … or terminates when column1 of Loyalty array equals 99999 in a loop 5. Check if loyalty points greater than or equal to 11 in a loop 6. If true output customer ID 7 Sum points and Increment Count in a loop 8 Calculate and output average with an appropriate message Max 7 7(c)(i) An array can only store data of the same type // An array cannot store data 1 of different types 7(c)(ii) 1 mark for each point 2 1. a new (composite) data type / record is defined (that consists of both INTEGER and STRING) 2. an array based on this new type is declared Alternative 1 mark for each point 1. Converting loyalty points to string and concatenating / joining / append with Customer ID 2. Storing concatenated string in an array of string