2025 Cambridge IAS-Level Computer Science (9618) 模擬試題連答案詳解

Step 1: The subtrahend is 00101100. To find its two's complement, invert all bits to get 11010011 and then add 1, which results in 11010100. Step 2: Add this to the minuend: 01010101 + 11010100 = 100101001. Step 3: Discard the overflow bit outside the 8-bit range, which gives the final result 00101001.

1 mark: Correct two's complement of the subtrahend (11010100). 1 mark: Correct working showing binary addition of the minuend and the calculated two's complement. 1 mark: Correct final 8-bit binary result (00101001).

1. LDD 105 loads the value at address 105 directly into the ACC. Address 105 contains 35, so ACC becomes 35. 2. ADD 106 adds the value at address 106 to the current ACC. 35 + 15 = 50, so ACC becomes 50. 3. STO 107 stores the value of the ACC (50) into address 107. The ACC itself remains unchanged, so ACC is still 50.

1 mark: ACC = 35 after execution of LDD 105. 1 mark: ACC = 50 after execution of ADD 106. 1 mark: ACC = 50 after execution of STO 107.

In tblDepartment, DepartmentID uniquely identifies each department, making it the primary key. In tblEmployee, DepartmentID refers back to the primary key of tblDepartment to link the tables, making it the foreign key. Since one department can employ multiple employees, but each employee belongs to only one department, the relationship is One-to-Many.

1 mark: Correctly identifying DepartmentID as the primary key in tblDepartment. 1 mark: Correctly identifying DepartmentID as the foreign key in tblEmployee. 1 mark: Correctly identifying the relationship as One-to-Many.

Checking row parity: Row 1 has 4 ones (even). Row 2 has 2 ones (even). Row 3 has 5 ones (odd - ERROR!). Row 4 has 2 ones (even). Checking column parity: Col 1 has 2 ones (even). Col 2 has 5 ones (odd - ERROR!). Col 3 has 2 ones (even). Col 4 has 4 ones (even). The intersection of the row and column errors occurs at Row 3, Column 2. Since the received bit value is 1, the corrected value must be 0.

1 mark: Identifying Row 3 as containing the error. 1 mark: Identifying Column 2 as containing the error. 1 mark: Stating that the correct bit value is 0.

1. Normal data: Any integer from 1 to 99 inclusive (e.g., 55) is a valid score within the expected range. 2. Boundary data: 0 or 100 are values on the immediate limits of the acceptable range. 3. Abnormal data: Values outside the accepted boundaries (such as -5 or 105) or data of an incorrect type (such as 'abc' or 45.5) which should be rejected.

1 mark: Providing any valid integer between 1 and 99 inclusive for normal data. 1 mark: Providing exactly 0 or 100 for boundary data. 1 mark: Providing any value less than 0, greater than 100, or a non-integer data type for abnormal data.

When a router receives a packet, it reads the destination IP address contained in the packet's header. It then references its internal routing table, which contains information on network topology and paths. Based on this, the router determines the most efficient path (or next hop) for the packet. Finally, it forwards the packet to the next router or destination interface.

Award 1 mark for each point up to a maximum of 4:
1. Receives a data packet and inspects/reads the destination IP address in the packet header.
2. References/searches its internal routing table.
3. Determines the most efficient or optimum path / next hop for the packet.
4. Forwards/transmits the packet to the next router, node, or destination.

At the end of each Fetch-Execute cycle, the CPU checks the interrupt lines. If an interrupt flag is set and its priority is high enough, the CPU suspends execution of the current program. It saves the contents of current registers (including the Program Counter) onto a stack. The Program Counter is then loaded with the starting address of the Interrupt Service Routine (ISR). Once the ISR completes, the saved register values are popped from the stack, and execution of the original program resumes.

Award 1 mark for each point up to a maximum of 4:
1. Checks for interrupts/interrupt flags at the end of the current Fetch-Execute cycle.
2. Saves current register states / Program Counter (PC) onto the stack.
3. Loads the Program Counter (PC) with the start address of the Interrupt Service Routine (ISR).
4. Restores/pops the register values from the stack once the ISR completes to resume the original program.

Referential integrity ensures that relationships between tables remain consistent. The DBMS prevents orphaned records by ensuring that any foreign key in the child table (ORDER) has a matching primary key in the parent table (CUSTOMER). Data validation involves the DBMS checking inputs against predefined rules (e.g., ensuring order quantity is a positive integer) to prevent incorrect or non-sensical data entry.

Award 1 mark for each point up to a maximum of 4 (Max 2 for referential integrity, Max 2 for data validation):
- Referential Integrity:
1. Ensures that a foreign key value in the child table (ORDER) must match an existing primary key value in the parent table (CUSTOMER).
2. Prevents orphaned records (e.g., cannot delete a customer with active orders / cannot add an order for a non-existent customer).
- Data Validation:
3. Automatic checks performed by the DBMS on input data to ensure it is sensible/conforms to rules.
4. Prevents invalid data types or invalid ranges from being saved (e.g., preventing negative quantities).

Vector graphics store images as mathematical formulas and geometric properties (lines, curves, coordinates) rather than as individual pixels. Because of this, they can be scaled up or down infinitely without any loss of quality or pixelation. Additionally, they have a smaller file size because they only need to store the drawing instructions, whereas bitmaps must store color data for every individual pixel.

Award 1 mark for each point up to a maximum of 4 (Max 2 per advantage fully explained):
- Advantage 1 (Scalability):
1. Vector graphics are scale-invariant / can be resized infinitely.
2. They do not lose quality or become pixelated when scaled because they are recalculated mathematically (unlike bitmaps which stretch pixels).
- Advantage 2 (File Size / Editing):
3. Vector graphics have a smaller file size as they store only mathematical drawing instructions / formulas instead of pixel data.
4. Individual elements/shapes can be easily modified/edited without affecting the rest of the image.

A disk defragmenter utility works by reorganizing files on a hard disk drive so that all segments/blocks of a single file are stored contiguously. This reduces read/write head movement, speeding up file access and system performance. A backup utility creates copies of critical files and stores them on an external drive or cloud storage, enabling recovery in case of hardware failure or malware attack.

Award 1 mark for each point up to a maximum of 4 (Max 2 for defragmenter, Max 2 for backup):
- Disk Defragmenter:
1. Reorganizes fragmented file blocks/sectors on a hard drive to be contiguous.
2. Improves file access speeds / read-write efficiency by reducing physical drive head movement.
- Backup Utility:
3. Creates copies of files and directories to separate storage media (e.g., cloud, external storage).
4. Protects data security/integrity by allowing data recovery in case of system failure, corruption, or malware.

To create a digital signature, the sender inputs the document into a hashing algorithm to produce a unique message digest (hash). The sender then encrypts this hash using their private key. The encrypted hash is the digital signature, which is sent along with the document. To verify, the recipient decrypts the signature using the sender's public key to obtain the decrypted hash. The recipient then runs the received document through the same hashing algorithm. If the decrypted hash and the newly calculated hash are identical, authenticity and integrity are confirmed.

Award 1 mark for each point up to a maximum of 4:
1. Sender calculates a cryptographic hash/message digest of the document.
2. Sender encrypts this hash using their own private key to produce the digital signature.
3. Recipient decrypts the digital signature using the sender's public key to retrieve the original hash.
4. Recipient hashes the received document independently and compares the result with the decrypted hash; if they match, the document has not been altered.

Open-source software gives users access to the underlying source code, allowing them to view, edit, and tailor it to their needs. Its licensing terms generally permit free use, modification, and redistribution. Proprietary software keeps its source code hidden/compiled, preventing any modification. Its license (EULA) restricts copying, distribution, and modification, and usually requires a purchasing fee.

Award 1 mark for each point up to a maximum of 4 (Max 2 for source code, Max 2 for licensing):
- Source Code:
1. Open-source software provides full access to the source code for editing and customization.
2. Proprietary software keeps the source code closed/compiled so users cannot view or modify it.
- Licensing Terms:
3. Open-source licenses (e.g., GPL) allow users to run, share, and redistribute the software freely.
4. Proprietary licenses restrict copying, redistribution, and modification, and typically require a paid user license.

Corrective maintenance is reactive and aims to resolve bugs, logical errors, or coding defects discovered in a system once it is live. For example, fixing an error where a program crashes if a user enters a negative value. Adaptive maintenance is proactive and involves modifying the software so that it remains compatible with an external change, such as hardware upgrades, operating system updates, or new regulatory standards. For example, updating an application to run on the latest version of macOS.

Award 1 mark for each point up to a maximum of 4:
1. Corrective maintenance is defined as identifying and resolving bugs, logical defects, or errors in the existing system.
2. Valid example of corrective maintenance (e.g., fixing a run-time error when a database query fails, resolving a broken link on a form submission).
3. Adaptive maintenance is defined as altering software so that it operates correctly within a changed external environment (OS, hardware, or regulations).
4. Valid example of adaptive maintenance (e.g., rewriting code to support a new API version, updating a payroll module to reflect new tax laws).

During the fetch stage, the CPU retrieves the next instruction from memory:
1. The memory address of the next instruction stored in the Program Counter (PC) is copied into the Memory Address Register (MAR): \(\text{MAR} \leftarrow [\text{PC}]\).
2. The instruction stored at the memory address inside the MAR is loaded via the data bus into the Memory Data Register (MDR): \(\text{MDR} \leftarrow [[\text{MAR}]]\).
3. Simultaneously, the Program Counter (PC) is incremented by 1 to point to the next instruction in sequence: \(\text{PC} \leftarrow [\text{PC}] + 1\).
4. The instruction in the MDR is then copied into the Current Instruction Register (CIR) to be ready for decoding: \(\text{CIR} \leftarrow [\text{MDR}]\).

Award 1 mark for each correct step with its corresponding RTN up to a maximum of 4 marks:
- Copying PC to MAR: \(\text{MAR} \leftarrow [\text{PC}]\) [1 mark]
- Loading instruction from address in MAR to MDR: \(\text{MDR} \leftarrow [[\text{MAR}]]\) [1 mark]
- Incrementing PC: \(\text{PC} \leftarrow [\text{PC}] + 1\) [1 mark]
- Copying MDR to CIR: \(\text{CIR} \leftarrow [\text{MDR}]\) [1 mark]

1. **Carrier Sensing**: Before sending data, a device listens to the shared transmission medium to determine if another device is currently transmitting (if the medium is busy).
2. **Collision Detection**: While transmitting, the device continues to monitor the medium. If two devices transmit at the same time, their signals collide, resulting in an abnormal voltage level which is detected as a collision.
3. **Jam Signal**: Once a collision is detected, the transmitting device immediately stops normal data transmission and broadcasts a 'jam signal' to notify all other devices on the network that a collision has occurred.
4. **Random Backoff**: After the jam signal, each device waiting to transmit generates a random wait time (backoff period) before attempting to sense the medium again. This random delay prevents the devices from immediately colliding again.

Award 1 mark per explanation point up to a maximum of 4:
- 1 mark: Senses/listens to the medium to check if it is idle before transmitting.
- 1 mark: Monitors the medium during transmission to detect overlapping signals/collisions.
- 1 mark: Transmission is aborted and a jam signal is broadcast upon collision detection.
- 1 mark: Devices apply a random backoff time/delay before trying to retransmit.

To retrieve data from both tables, an INNER JOIN is performed on the common key BookID. The fields selected are BookTitle (from tblBook) and DateBorrowed (from tblBorrowing). The filter criteria are applied using a WHERE clause: StudentID must equal 'S1024' and DateReturned must be checked for NULL using the 'IS NULL' operator. Finally, the ORDER BY clause sorts the records by DateBorrowed in descending order (DESC) to show the most recent borrowings first.

1 mark: SELECT BookTitle, DateBorrowed (with or without table prefixes). 1 mark: FROM tblBook INNER JOIN tblBorrowing ON tblBook.BookID = tblBorrowing.BookID (accept implicit join via WHERE). 1 mark: WHERE StudentID = 'S1024'. 1 mark: AND DateReturned IS NULL (reject DateReturned = NULL). 1 mark: ORDER BY DateBorrowed DESC.

Line 1 loads the value at address 108 (12) into ACC. Line 2 subtracts the value at address 109 (4) from ACC, resulting in ACC = 8. Line 3 compares ACC (8) with immediate value 8, setting the comparison flag to Equal. Line 4 jumps to EQUAL (line 8) because the flag is Equal. Line 8 loads immediate value 20 into ACC. Line 9 stores ACC (20) into address 111. Line 10 increments ACC to 21. Line 11 stores ACC (21) into address 110. Line 12 ends execution.

1 mark for each correct value: (i) 8, (ii) Equal, (iii) 20, (iv) 20, (v) 21.

Part (a) uses the CREATE TABLE command. ProjectID uses CHAR(6) because it is a fixed length of 6 characters, and is defined as PRIMARY KEY. ProjectName uses VARCHAR(50) and has the NOT NULL constraint. Budget is specified as DECIMAL or NUMERIC. Part (b) uses UPDATE tblProject, sets the new Budget using SET Budget = Budget * 1.10, and filters the target rows with WHERE Budget < 15000.

Part (a) [3 Marks]: 1 mark for correct CREATE TABLE syntax with tblProject and parentheses. 1 mark for ProjectID CHAR(6) PRIMARY KEY. 1 mark for ProjectName VARCHAR(50) NOT NULL. Part (b) [2 Marks]: 1 mark for UPDATE tblProject SET Budget = Budget * 1.1 (or Budget * 1.10 or Budget + Budget * 0.1). 1 mark for WHERE Budget < 15000.

The loop executes 3 times. Iteration 1: IX is 0. LDX 300 loads Mem[300] (15) into ACC. ADD 310 adds 0, resulting in 15. STO 310 saves 15 in address 310. INC IX sets IX to 1. TXA transfers 1 to ACC. CMP #3 compares 1 with 3 (Not Equal), so JPN jumps back. Iteration 2: IX is 1. LDX 300 loads Mem[301] (22) into ACC. ADD 310 adds 15, resulting in 37. STO 310 saves 37. INC IX sets IX to 2. TXA transfers 2 to ACC. CMP #3 compares 2 with 3 (Not Equal), so JPN jumps back. Iteration 3: IX is 2. LDX 300 loads Mem[302] (8) into ACC. ADD 310 adds 37, resulting in 45. STO 310 saves 45. INC IX sets IX to 3. TXA transfers 3 to ACC. CMP #3 compares 3 with 3 (Equal). Since it is equal, JPN LOOP does not branch, and the program ends. The final value in 310 is 45.

1 mark: Correct number of loop executions (3 times). 1 mark: IX value at end of second iteration (2). 1 mark: ACC value after first TXA (1). 2 marks: Final value in 310 (45) with shown correct addition steps (15 + 22 + 8 = 45). Allow 1 mark if the correct value of 45 is given but working is missing or incorrect.

Stepwise refinement is a top-down design method. The programmer begins with a high-level description of a problem and progressively breaks it down into smaller sub-tasks. This process continues iteratively until each sub-task is simple enough to be translated directly into pseudocode or program code.

1 mark per bullet point: - Decomposition of the main problem into smaller / more manageable sub-tasks. - Process repeated / refined until the sub-tasks can be written as code / individual modules.

A record is a user-defined composite data type that allows grouping of related data items of different types (such as an INTEGER, a STRING, and a REAL together) and refers to them by field names. A 1D array is a built-in data structure that stores a collection of elements of the same data type, accessed via a numeric index.

1 mark per valid difference (max 2): - Record can hold different data types / heterogeneous data, array must hold same data type / homogeneous data. - Record elements are accessed by field names, array elements are accessed by indices. - Array is usually of fixed size, record is a composite user-defined structure.

When a parameter is passed by value (BYVAL), a local copy of the value is created for the procedure. Hence, any modification of this parameter is restricted to the local scope and the calling environment's variable remains unchanged. When passed by reference (BYREF), the procedure receives a pointer to the original memory location, so any modifications directly affect the variable in the calling environment.

1 mark for explaining By Value: By value creates a copy / does not affect the original variable in the calling routine. 1 mark for explaining By Reference: By reference passes the address / memory pointer / directly alters the original variable in the calling routine.

In software development, particularly top-down integration testing, some lower-level modules might not yet be developed or coded. To test the higher-level modules that call these unfinished modules, developers use 'stubs'. A stub is a simple piece of dummy code that simulates the interface of the actual module, usually just displaying a message or returning a fixed hardcoded value, allowing the calling module to be tested without errors.

1 mark per bullet point (max 2): - Acts as a placeholder / dummy module for code that is not yet written / completed. - Allows testing of higher-level modules / program flow to continue by returning predefined mock values.

1. LENGTH(Word) returns 8 since "COMPUTER" has 8 characters. 2. The expression simplifies to MID(Word, 4, 8 - 5) which is MID("COMPUTER", 4, 3). 3. MID("COMPUTER", 4, 3) extracts 3 characters starting from the 4th position (index 4 is 'P'). 4. The extracted characters are 'P', 'U', and 'T', which gives "PUT". 5. The data type of the returned value is STRING.

1 mark for the correct value: "PUT" (accept PUT, reject other lengths or incorrect substrings). 1 mark for the correct data type: STRING (or String).

A structure chart is a design tool used in top-down design to represent the modular structure of a program, displaying how modules are organized hierarchically and how they interact. The flow of data between modules is represented using arrows: a 'data couple' (an arrow with an open circle) represents data being passed, while a 'control flag' (an arrow with a filled circle) represents status signals or control variables (like a boolean flag) being passed.

1 mark for the purpose: To show the hierarchical structure / decomposition / modular breakdown of a program. 1 mark for identifying the arrow: Data couple (accept 'arrow with an open circle') OR control flag / parameter / status flag (accept 'arrow with a filled circle').

Let us trace the algorithm step-by-step with the input "CAMBRIDGE":
1. `OutputStr` is initialized to `""`, and `Count` is initialized to `0`.
2. The loop runs for `Index` from 1 to 9 (length of "CAMBRIDGE"):
- `Index = 1`: `NextChar` is 'C'. It is not a vowel. `OutputStr` becomes `"C"`.
- `Index = 2`: `NextChar` is 'A'. It is a vowel. `Count` becomes `1`.
- `Index = 3`: `NextChar` is 'M'. Not a vowel. `OutputStr` becomes `"CM"`.
- `Index = 4`: `NextChar` is 'B'. Not a vowel. `OutputStr` becomes `"CMB"`.
- `Index = 5`: `NextChar` is 'R'. Not a vowel. `OutputStr` becomes `"CMBR"`.
- `Index = 6`: `NextChar` is 'I'. Vowel. `Count` becomes `2`.
- `Index = 7`: `NextChar` is 'D'. Not a vowel. `OutputStr` becomes `"CMBRD"`.
- `Index = 8`: `NextChar` is 'G'. Not a vowel. `OutputStr` becomes `"CMBRDG"`.
- `Index = 9`: `NextChar` is 'E'. Vowel. `Count` becomes `3`.
3. The loop terminates.
4. `OutputStr` is concatenated with the string representation of `Count` (3), resulting in `"CMBRDG3"`.
5. The value `"CMBRDG3"` is returned.

1 mark: Correctly initializing OutputStr to "" and Count to 0.
1 mark: Correctly writing all values of Index (1 to 9) and NextChar.
1 mark: Correctly updating the Count column only on vowels (to 1, 2, 3).
1 mark: Correctly building up the OutputStr column step-by-step (vowels omitted).
1 mark: Correct final return value "CMBRDG3".

Let us trace the algorithm execution:
1. `Total` is initialized to 0.
2. The loop runs for `Index` from 1 to 4:
- `Index = 1`: `Temp <- Arr[2] - Arr[1]` = `15 - 12 = 3`. Since `3 > 0` is True, `Total <- Total + Temp` = `0 + 3 = 3`.
- `Index = 2`: `Temp <- Arr[3] - Arr[2]` = `10 - 15 = -5`. Since `-5 > 0` is False, `Total <- Total - Temp` = `3 - (-5) = 8`.
- `Index = 3`: `Temp <- Arr[4] - Arr[3]` = `8 - 10 = -2`. Since `-2 > 0` is False, `Total <- Total - Temp` = `8 - (-2) = 10`.
- `Index = 4`: `Temp <- Arr[5] - Arr[4]` = `14 - 8 = 6`. Since `6 > 0` is True, `Total <- Total + Temp` = `10 + 6 = 16`.
3. The loop terminates.
4. `Total` is outputted, which is 16.

1 mark: Correct Index values (1, 2, 3, 4) and corresponding Temp values (3, -5, -2, 6).
1 mark: Correctly updating Total to 3 at Index = 1.
1 mark: Correctly handling negative difference subtraction to yield Total = 8 at Index = 2.
1 mark: Correctly handling negative difference subtraction to yield Total = 10 at Index = 3.
1 mark: Correctly updating Total to 16 at Index = 4 and giving final Output of 16.

Let us trace the logic step-by-step:
1. `Result` is initialized to `""`. `Num` is 59.
2. First iteration of `WHILE Num > 0` (59 > 0 is True):
- `Remainder <- 59 MOD 4` = 3
- `Result <- "3" & ""` = `"3"`
- `Num <- 59 DIV 4` = 14
3. Second iteration of `WHILE Num > 0` (14 > 0 is True):
- `Remainder <- 14 MOD 4` = 2
- `Result <- "2" & "3"` = `"23"`
- `Num <- 14 DIV 4` = 3
4. Third iteration of `WHILE Num > 0` (3 > 0 is True):
- `Remainder <- 3 MOD 4` = 3
- `Result <- "3" & "23"` = `"323"`
- `Num <- 3 DIV 4` = 0
5. The while loop condition `Num > 0` (0 > 0) is False. The loop terminates.
6. The function returns the `Result` string: `"323"`.

1 mark: Correctly tracking values of Num through integer division (59 -> 14 -> 3 -> 0).
1 mark: Correctly calculating and recording all Remainder values (3, 2, 3).
2 marks: Correctly updating Result by prepending the remainder character (award 1 mark for correct characters, 1 mark for correct prepending order: "3" -> "23" -> "323").
1 mark: Correct return value of "323".

PROCEDURE LongestRun(DataArray : ARRAY[1:100] OF INTEGER, NumElements : INTEGER) DECLARE MaxStart, MaxLength, CurrentStart, CurrentLength, i : INTEGER MaxStart <- 1 MaxLength <- 1 CurrentStart <- 1 CurrentLength <- 1 FOR i <- 2 TO NumElements IF DataArray[i] = DataArray[i - 1] THEN CurrentLength <- CurrentLength + 1 ELSE IF CurrentLength > MaxLength THEN MaxLength <- CurrentLength MaxStart <- CurrentStart ENDIF CurrentStart <- i CurrentLength <- 1 ENDIF ENDFOR IF CurrentLength > MaxLength THEN MaxLength <- CurrentLength MaxStart <- CurrentStart ENDIF OUTPUT "Starting index: ", MaxStart OUTPUT "Length: ", MaxLength ENDPROCEDURE

1 mark: Correct procedure header with parameter names and types. 1 mark: Initializing MaxStart, MaxLength, CurrentStart, CurrentLength correctly to 1. 1 mark: Correct loop from index 2 up to NumElements. 1 mark: Checking if the current element is equal to the previous element. 1 mark: Incrementing CurrentLength in the true branch. 1 mark: Updating MaxLength and MaxStart in the false branch if CurrentLength > MaxLength. 1 mark: Resetting CurrentStart to i and CurrentLength to 1 in the false branch. 1 mark: Final comparison check after loop ends and outputting both starting index and maximum length.

FUNCTION ProcessTransactions() RETURNS INTEGER DECLARE Line, Status : STRING DECLARE NonApprovedCount : INTEGER DECLARE Parts : ARRAY[1:3] OF STRING NonApprovedCount <- 0 OPENFILE "Transactions.txt" FOR READ OPENFILE "ApprovedTxns.txt" FOR WRITE WHILE NOT EOF("Transactions.txt") READFILE "Transactions.txt", Line Parts <- SPLIT(Line, ',') Status <- Parts[3] IF Status = "APPROVED" THEN WRITEFILE "ApprovedTxns.txt", Line ELSE NonApprovedCount <- NonApprovedCount + 1 ENDIF ENDWHILE CLOSEFILE "Transactions.txt" CLOSEFILE "ApprovedTxns.txt" RETURN NonApprovedCount ENDFUNCTION

1 mark: Correct function header and RETURN statement with INTEGER return type. 1 mark: Opening Transactions.txt for READ and ApprovedTxns.txt for WRITE. 1 mark: Correct loop using WHILE NOT EOF("Transactions.txt") or equivalent. 1 mark: Reading a line from Transactions.txt inside the loop. 1 mark: Splitting the read line and extracting the Status part correctly (index 3). 1 mark: Conditional check if status is "APPROVED". 1 mark: Writing the line to ApprovedTxns.txt if approved, otherwise incrementing NonApprovedCount. 1 mark: Closing both files correctly.

FUNCTION CountLocalPeaks(Grid : ARRAY[1:10, 1:10] OF INTEGER) RETURNS INTEGER DECLARE r, c, PeakCount : INTEGER PeakCount <- 0 FOR r <- 2 TO 9 FOR c <- 2 TO 9 IF Grid[r, c] > Grid[r - 1, c] AND Grid[r, c] > Grid[r + 1, c] AND Grid[r, c] > Grid[r, c - 1] AND Grid[r, c] > Grid[r, c + 1] THEN PeakCount <- PeakCount + 1 ENDIF ENDFOR ENDFOR RETURN PeakCount ENDFUNCTION

1 mark: Correct function header with 2D array parameter and returns type INTEGER. 1 mark: Initializing PeakCount to 0. 1 mark: Outer loop iterating from index 2 to 9. 1 mark: Inner loop iterating from index 2 to 9 (ensuring boundary elements are skipped). 1 mark: Correct comparison of current element with upper neighbor (Grid[r - 1, c]) and lower neighbor (Grid[r + 1, c]). 1 mark: Correct comparison of current element with left neighbor (Grid[r, c - 1]) and right neighbor (Grid[r, c + 1]). 1 mark: Combining all four comparisons with AND. 1 mark: Incrementing PeakCount and returning the final peak count.

FUNCTION Enqueue(NewItem : INTEGER) RETURNS BOOLEAN DECLARE NextTail : INTEGER NextTail <- (TailPointer MOD 50) + 1 IF NextTail = HeadPointer THEN OUTPUT "Error: Queue is full" RETURN FALSE ELSE QueueArray[TailPointer] <- NewItem TailPointer <- NextTail RETURN TRUE ENDIF ENDFUNCTION

1 mark: Correct function header with parameter NewItem and returns BOOLEAN. 1 mark: Calculating the circular next tail position correctly using MOD 50 + 1 (or equivalent condition). 1 mark: Checking if the next tail position is equal to HeadPointer to detect the queue-full state. 1 mark: Outputting an appropriate error message when full. 1 mark: Returning FALSE when the queue is full. 1 mark: Correctly assigning NewItem to QueueArray at index TailPointer when not full. 1 mark: Updating TailPointer to NextTail (or circular next position) when not full. 1 mark: Returning TRUE when successfully inserted.