Computer Science 9618/31 October/November 2025: question paper and mark scheme

Mark scheme

1(a)(i) DECLARE Member1 : ClubMember 1 1(a)(ii) One mark for each correct answer 2 Example answer Member1.Code ← 984632 Member1.FeesPaid ← TRUE 1(b)(i) One mark per mark point (Max 2) 2 MP1 TYPE Activity= MP2 (Badminton, Football, Golf, Snooker, Swimming, Tennis) Example answer TYPE Activity = (Badminton, Football, Golf, Snooker, Swimming, Tennis) 1(b)(ii) DECLARE Choice : Activity 1

Mark scheme

2(a) One mark per mark point (Max 2) 2 MP1 Correct mantissa MP2 Correct exponent Mantissa Exponent 0 1 1 1 0 1 0 1 1 0 1 0 1 0 0 1 2(b) One mark per mark point (Max 4) 4 MP1 correct method to find the binary number MP2 additional working towards binary number MP3 correct use of exponent MP4 correct answer in the space provided Two from: e.g. 76.1875 = 64+8+4+0.125+0.0625 (0)1001100.0011 –76.1875 = –128+32+16+2+1+0.5+0.25+0.0625 10110011.11 01 One mark movement of binary point by 7 places // 1.011001111 01 x 27 One mark Mantissa Exponent 1 0 1 1 0 0 1 1 1 1 0 1 0 1 1 1

Mark scheme

3(a) One mark per mark point (Max 2) 2 MP1 Protocols set a standard for communication // Protocols establish a standard set of rules for communication MP2 Protocols enable compatibility between devices from different manufacturers/platforms MP3 Two devices wouldn’t be able to communicate/send messages to each other if they were using different protocols 3(b) One mark for a protocol and one mark for a description (Max 4) 4 Example answers: SMTP [1] a protocol used to send emails between mail servers // a protocol used to send emails from a computer to a mail server [1] IMAP [1] allows users to access/read their emails from any device without removing the message from the mail server // synchronises emails on any device [1] 3(c) Two marks for each description mark as 2 x 2 (Max 4) 4 Example answers: Packets checked at receiving end / on arrival [1] … if packets arrive damaged or don’t arrive at all, a re-send request is sent [1] Packets routed through different paths / sent individually [1] … if a route is blocked, the packet is sent through a different route to ensure it arrives [1] If the packet’s hop count is exceeded [1] … the packet will be retired, which can generate a re-send request [1]

Mark scheme

4(a) One mark per scheduling routine (Max 2) from: 2 • Round robin • Shortest job first • First come first served • Shortest remaining time 4(b) One mark for identification and one mark for a description (Max 4) 4 Two from: Provision of a User Interface // Provision of a Graphical User Interface [1] Allows the user to interact with the computer in a more intuitive way // Icons and menus are used to control devices by simply ‘pointing and clicking’ [1] Use of device drivers [1] Makes it easier to control peripherals such as printers within the operating system of the computer rather than on the separate device itself [1] Device mapping [1] Different devices (physical and virtual) are easy to identify on the network, check their status, or use [1] The user interacts only with the Application / top layer (of the TCP/IP protocol suite) [1] leaving the lower layers and their complexities hidden from the user [1]

Mark scheme

5(a) Two from: 2 • Name of certificate holder // Subject • Serial number • Version number • Expiration date // Start date // Validity (not before/not after) • Certificate holder’s public key // Subject public key • Subject digital signature • Certificate Issuer // Digital signature of CA 5(b) One mark per mark point (Max 3) 3 MP1 A digital certificate provides a public key MP2 … which validates the private key used to create the digital signature MP3 It makes a digital signature virtually impossible to spoof // Provides evidence of signer identity that the document was not altered and the signatures are valid MP4 Non repudiation.

Mark scheme

6(a) One mark for working, (all four columns P, Q, R and S) 3 One mark for first four rows of column Z One mark for second four rows of column Z Working space A B C P Q R S Z 0 0 0 1 1 1 1 0 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 1 0 1 1 1 0 1 1 1 1 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 1 1 0 0 1 1 1 1 1 1 1 0 0 1 1 1 6(b)(i) Two marks if no errors present 2 One mark if only one error present BC 00 01 11 10 A 0 0 0 1 1 1 0 1 0 1 6(b)(ii) One mark for each correct loop (Max 2) 2 BC 00 01 11 10 A 0 0 0 1 1 1 0 1 0 1 6(b)(iii) One mark for each mark point (Max 2) 2 • One correct Boolean term with a + / OR sign • All Boolean terms and operators correct and no other terms present A.B + B.C + A.B.C

Mark scheme

7(a) One from: 1 • A* algorithm • Dijkstra’s algorithm 7(b) One mark for each mark point (Max 5) 5 MP1 Deep learning is a branch of machine learning MP2 It makes use of artificial neural networks MP3 … to work in a similar manner to the human/biological brain // … neural networks can be trained to process data MP4 It is a structure consisting of many interconnected layers MP5 … input, hidden and output // the input layer feeds the first hidden layer, the last hidden layer feeds the output MP6 … gradually breaking down and analysing the data to find (hidden) patterns MP7 Deep leaning’s feature extraction/learning involves how to detect features/categories from data MP8 … and can involve different types of supervision MP9 It works well with large data sets MP10 Deep learning uses back propagation of errors to adjust parameters/weightings to refine its results / improve its accuracy.

Mark scheme

8(a) One mark per mark point (Max 2) 2 MP1 To convert the high-level source code / program into a sequence of tokens MP2 … that can be sent to the parser for syntax analysis MP3 To create a symbol table MP4 To remove the unnecessary white space and comments from the code 8(b) One mark 2 6 – 2 One mark 13 7 + * 5 / Complete answer 2 6 – 13 7 + * 5 / 8(c) One mark per ring (Max 4) 4 12 6 6 6 18 15 15 9 5 5 5 5 90 90 90 90 10

Mark scheme

9(a)(i) One mark for each correctly completed line (Max 5) 5 FUNCTION Pop() RETURNS STRING DECLARE DataItem : STRING DataItem ← "" IF Top > –1 // Top >= Base THEN DataItem ← StackArray[Top] Top ← Top – 1 ELSE DataItem ← "You cannot remove data; the stack is empty" ENDIF RETURN DataItem // StackArray[Top + 1] ENDFUNCTION 9(a)(ii) OUTPUT "The data removed from the stack is ", Pop() 1 9(b) One mark per mark point (Max 3) 3 MP1 A recursive algorithm must call itself / have a general case MP2 It must have a base case / have a stopping condition MP3 It must change its state and move towards the base case

Mark scheme

10 One mark per mark point (Max 2) 3 MP1 Exception handling is a process that responds to unwanted / unexpected events when a program runs MP2 … to prevent the program / computer from stopping unexpectedly One mark for example (Max 1) MP3 Programming errors MP4 User errors MP5 Hardware failure // losing connection to a device such as a printer

Mark scheme

11(a) One mark per mark point (Max 6) 6 MP1 LDM #100 seen MP2 Correct use of STO with labelled address (constant or answer) MP3 Correct use of LDD 632 MP4 Correct use of SUB with labelled address (constant) Opcode Operand LDM #100 STO Constant LDD 632 SUB Constant STO Answer MP5 Storing 100 at a labelled address away from the code MP6 Labelling both addresses away from the code. Label Contents Constant: 100 Answer: 11(b) –55 1