Cambridge AS & A Level Computer Science 9618/32 is the May/June 2026 Paper 3 (Advanced Theory), variant 2. It is an A Level component that lasts 1 hour 30 minutes, is worth 75 marks and is sat by every full A Level candidate. It had eleven compulsory questions and no calculator, and four of them carried 9 marks each: Reverse Polish Notation, Dijkstra’s algorithm, a random-file function and a Node class. This guide goes through the paper question by question and shows what the official mark scheme gave marks for. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of all four 9618 papers from this series, see A Level Computer Science 9618 May/June 2026 past papers: what came up and how it was marked.

9618/32 at a glance

Question	Topic	Marks	Main command word
1	Enumerated data type for zoo animals	5	Write, state, give
2	TCP/IP protocol suite (gap-fill)	4	Complete
3	Floating-point to denary, underflow	5	Calculate, describe
4	Sequential file organisation and access	6	State, describe, outline
5	Virtual machines	4	Explain
6	Reverse Polish Notation and stack evaluation	9	Write, show
7	Karnaugh map, sum-of-products, De Morgan’s laws	8	Complete, draw, simplify
8	Symmetric and asymmetric encryption, verified messages	7	Describe, outline
9	Deep learning, A* and Dijkstra’s algorithm	9	State, outline, calculate
10	Big O of searches, a random-file function	9	Describe, write
11	A Node class and OOP terms	9	Write, complete
	Total	75

Question-by-question walkthrough

Question 1: an enumerated type (5 marks)

• Declaration (2 marks): TYPE Animals = (Alpaca, Chicken, Donkey, Duck, Goat, Pony, Sheep). One mark for the TYPE Animals = part and one for the bracketed list. Only an equals sign was allowed, the values must be in the given order, and quotation marks around the values, or extra punctuation such as underscores between them, lost the mark.
• Type (1 mark): enumerated.
• Two features of the list (2 marks): it is complete and fixed once declared (it holds every possible value and can’t change); it is ordinal, so its values have an order; and its elements are unique. “Chronological order” did not count for the ordinal point.

Question 2: TCP/IP (4 marks)

One mark per gap. The suite has four layers, it sends packets of data across the internet, the lowest layer is the link layer, and that layer interconnects nodes in the network.

Question 3: floating-point (5 marks)

The system used a 12-bit mantissa and a 4-bit exponent, both in two’s complement.

• Convert to denary (3 marks): the exponent 0110 is 6, so move the binary point six places right to get 1011010.00110. Reading it in two’s complement gives −64 + 16 + 8 + 2 + 0.125 + 0.0625 = −37.8125. Two marks were for working (correct use of the exponent and a correct conversion method) and one for the answer. Other methods were allowed, such as working out the mantissa as a fraction and multiplying by 64.
• Underflow (2 marks): a calculation on a normalised floating-point number (a specific example was allowed) gives a result smaller than the smallest value the system can store, because the exponent would be too negative for the bits available, so the result becomes zero.

Question 4: sequential files (6 marks)

• A use (1 mark): batch processing, log files, bills or payroll, backups, or another use where data is stored in a set order. An answer that just described sequential organisation scored nothing.
• How records are organised (2 marks): one after the other, in a given order, usually based on a key field. Chronological order was rejected for the order point, because that describes serial organisation.
• Finding a record (3 marks): each record is read in turn from the start of the file, comparing the record key, until the record is found, or until the key being searched for has been passed or the end of the file is reached. Answers about searching an index, as in indexed sequential files, did not answer the question.

Question 5: virtual machines (4 marks)

Any four of ten points describing what a virtual machine is: a standalone environment running on a physical computer rather than a physical machine; it can emulate another system’s hardware or software and has all the components of a physical computer; the host machine supplies the physical resources such as RAM, under a host operating system; a hypervisor sits between the host and guest operating systems and creates and runs the VMs, allocating host resources; a guest OS runs inside the VM and controls the virtual hardware; and several VMs can run on one host. Benefits, drawbacks and uses scored nothing, because the question asked for a description.

Question 6: Reverse Polish Notation (9 marks)

• Infix to RPN (2 marks): 16 9 7 − 18 6 + / +. One mark for the 16 9 7 − part and one for 18 6 + / +.
• RPN to infix (3 marks): ((6 + 15) − (20 − 3)) / (27 * 9), with the outer brackets around the whole numerator needed for the third mark. Leaving out the brackets around 6 + 15 was also accepted.
• Stack evaluation (4 marks): with c = 10, a = 8, b = 2 and d = 18, the stack grows to 10, 8, 2, then the subtraction leaves 10, 6 and the multiplication leaves 60. Pushing b and c gives 60, 2, 10, the addition gives 60, 12 and the division leaves 5. Pushing d gives 5, 18 and the final multiplication leaves 90. Three marks were for the three stages of the stack and one for showing the interim totals 60 and 5. Operators or letters written in the stack lost marks.

Question 7: Boolean algebra (8 marks)

• Complete the K-map (2 marks): filled from the truth table, with the AB columns in the order 00, 01, 11, 10. The two middle columns (AB = 01 and 11) are all 1s, and the CD = 01 row is all 1s; every other cell is 0. Two marks for no errors, one for one error, and the 0s had to be written in.
• Loops (2 marks): one loop around the eight cells in the two middle columns and one around the four cells in the CD = 01 row. Extra incorrect loops capped this at 1.
• Simplified sum-of-products (1 mark): B + NOT C . D, in either order, with or without the dots.
• De Morgan’s simplification (3 marks): the expression was the NOT of (NOT A . B . C . NOT D), plus A.D with a double bar. Applying De Morgan’s laws to the first term gives A + NOT B + NOT C + D, the double negation law turns the second term into A.D, and the absorption law removes A.D because A is already a term on its own. The answer is X = A + NOT B + NOT C + D: one mark for it and up to two for correct working that uses the laws. Other routes were accepted, and the distributive law also earned a working mark.

Question 8: encryption (7 marks)

• Symmetric against asymmetric (3 marks): symmetric uses one key while asymmetric uses a mathematically linked public and private pair; symmetric uses the same key to encrypt and decrypt, asymmetric uses one key for each; symmetric has the key distribution problem that the matched pair solves; asymmetric keys must be much longer for the same security; and asymmetric allows the sender to be verified. Each point had to mention both types.
• Sending a verified message to a group (4 marks): the business generates a matched key pair, makes its public key available to the group (for example on its website), encrypts the message with its private key and sends it, and anyone in the group decrypts it with the business’s public key, which verifies the sender and gives non-repudiation. The scheme stressed that the steps must be in the right order.

Question 9: AI and graph searches (9 marks)

• Subset of AI using artificial neural networks (1 mark): deep learning. “Machine learning” was not enough, because not all machine learning uses neural networks.
• A compared with Dijkstra (3 marks):* A* uses heuristics (intelligent guesses) to find the shortest path and doesn’t necessarily visit every node; Dijkstra checks each vertex and finds the shortest path to every node, and always finds the shortest path; A* may miss the shortest path if its heuristics are inaccurate. Saying Dijkstra “searches all routes” did not earn the Dijkstra point, because it looks for the shortest path to every node.
• Dijkstra’s algorithm (5 marks): the shortest distances from Home were:

A	B	C	D	E	F
6	13	9	20	17	14

A few of these needed the second route: B is 13 through A and C (6 + 3 + 4), not 14 directly from A; F is 14 through A, not 16 directly from Home; and D is 20 through F, not 21 through C. Up to 3 marks were for working on the diagram or in the space: Home set to 0, every other node marked as infinity at the start, values updated, visited nodes shown, a route calculated and more than one route compared for a node. Two marks were for the answers, one for four or five correct.

Question 10: Big O and a random file (9 marks)

• Search efficiency and Big O (3 marks): a linear search’s time grows linearly with the number of items and its worst case is O(n); a binary search’s time grows logarithmically and its worst case is O(log n). In the live exam, every candidate was given full marks for this part, but it is still worth practising against those four points.
• TempCheck function (6 marks): the file held one temperature per minute, with the position equal to minutes after midnight. The logic: the header takes an INTEGER position as its parameter and RETURNS a STRING; open Temperatures.dat FOR RANDOM; SEEK to the position; GETRECORD into Temperature; use IF … THEN … ELSE … ENDIF to set a message, “Not freezing” above 0 and “Freezing” otherwise; CLOSEFILE; then RETURN the message. One mark each for the header, OPENFILE, SEEK, GETRECORD, a complete selection structure, correct logic between condition and message, CLOSEFILE and RETURN, up to 6. CLOSEFILE only scored if it came after GETRECORD and no RETURN came before it, the parameter name had to match the one used in SEEK, and RETURN and RETURNS were checked carefully.

Question 11: OOP (9 marks)

• Declare the Node class (5 marks): CLASS Node … ENDCLASS; Content declared as a STRING and Pointer as an INTEGER (shown as PRIVATE in the scheme’s example) before the constructor; a PUBLIC PROCEDURE NEW … ENDPROCEDURE; two parameters with suitable names and types, which could be any names except Content and Pointer; and the parameters assigned to the attributes, with setters also allowed.
• OOP terms (4 marks): a method that changes private attribute values is a setter; an occurrence of a class while the program runs is an instance (or object); aggregation is one class or object containing another, a “has a” relationship; and inheritance is deriving a new class from an existing one so it takes its methods and data, an “is a” relationship.

Where the marks are on 9618/32

Topic	Questions	Marks
Data representation (user-defined types, floating point)	1, 3	10
Communication (TCP/IP)	2	4
File organisation	4	6
System software (virtual machines, RPN)	5, 6	13
Boolean algebra and logic	7	8
Security (encryption)	8	7
Artificial intelligence	9	9
Algorithms and file handling (Big O, random files)	10	9
Object-oriented programming	11	9
Total		75

• Writing pseudocode carried 13 marks: the TYPE declaration, TempCheck and the Node class. Each is marked line by line against exact keywords.
• Worked procedures carried 22 marks: floating point, RPN, the K-map and De Morgan simplification, and Dijkstra. All give credit for working, so show every stage.
• Most written answers are short point lists. Questions 4, 5, 8 and 9(b)(i) reward separate, specific points rather than one long paragraph.

How to self-mark 9618/32 with the mark scheme

1. Read the mark points (MP1, MP2 and so on) as a checklist. Tick each one your answer contains and stop at the maximum.
2. Check keywords exactly in the pseudocode. For 1(a), 10(b) and 11(a), look for the precise statements: TYPE with an equals sign, OPENFILE … FOR RANDOM, SEEK, GETRECORD, CLOSEFILE, PUBLIC PROCEDURE NEW. Then check the order of CLOSEFILE and RETURN.
3. Mark working separately from answers. In 3(a), 7(b) and 9(b)(ii), award the working marks even if the final value is wrong.
4. Mark the stack by its stages. For 6(c), check each stage and the interim totals 60 and 5, and remove a mark if you wrote operators or letters in the stack.
5. Use the guidance column for rejects: chronological order in 4(b), machine learning in 9(a), benefits in Question 5 and missing 0s in the K-map.
6. Check both sides of comparisons. In 8(a), each point needs both symmetric and asymmetric encryption.

Sit 9618/32 as a timed mock

The paper allows 90 minutes for 75 marks, about 1 minute 10 seconds per mark. A workable plan:

Block	Marks	Time
Question 1	5	6 minutes
Question 2	4	4 minutes
Question 3	5	6 minutes
Question 4	6	7 minutes
Question 5	4	5 minutes
Question 6	9	11 minutes
Question 7	8	10 minutes
Question 8	7	8 minutes
Question 9	9	11 minutes
Question 10	9	11 minutes
Question 11	9	9 minutes
Check calculations and pseudocode	–	2 minutes
Total	75	90 minutes

For Dijkstra, write the running distance next to each node on the diagram and cross out values as they improve, so the working marks are visible.

Other May/June 2026 9618 papers

• 9618 Paper 12 (Theory Fundamentals)
• 9618 Paper 22 (Fundamental Problem-solving and Programming Skills)
• 9618 Paper 42 (Practical)

Practise and download

• Download 9618/32 and its mark scheme on the Learning Portal →
• All 9618 May/June 2026 papers: overview and marking patterns
• Try a free A Level Computer Science quiz

Need help with a 9618 topic? Book a free trial lesson with an A Level Computer Science tutor →

Frequently asked questions

What topics were on 9618 Paper 32 May/June 2026? +

Eleven questions: an enumerated data type; the TCP/IP protocol suite; floating-point conversion and underflow; sequential files; virtual machines; Reverse Polish Notation; a Karnaugh map and De Morgan's laws; symmetric and asymmetric encryption; deep learning, A* and Dijkstra's algorithm; Big O and a random-file function; and a Node class with OOP terms.

How many marks is 9618/32 and how long is it? +

It is out of 75 marks and lasts 1 hour 30 minutes, with no calculator allowed. Questions 6, 9, 10 and 11 were the largest, at 9 marks each.

What was the floating-point answer on 9618/32 2026? +

The 12-bit mantissa 101101000110 with exponent 0110 (6) is -37.8125. Two marks were for working, using the exponent and converting correctly, and one for the final value.

What were the Dijkstra answers in 9618/32 Question 9? +

The shortest distances from Home were A 6, B 13, C 9, D 20, E 17 and F 14. Up to 3 marks were for working and 2 for the answers, with 1 mark for four or five correct values.

What did the RPN expression in 9618/32 Question 6(c) evaluate to? +

90. The stack holds 60 after c × (a − b), then 5 after dividing by (b + c), and finally 90 after multiplying by d. The interim totals 60 and 5 earned a mark of their own.

Where can I download 9618/32 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge AS & A Level Computer Science 9618 past papers page.

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