A Level Computer Science 9618 Paper 31, Oct/Nov 2023

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9618/31 · Oct/Nov 2023

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1

Real numbers are stored in a computer using floating-point representation with: • 12 bits for the mantissa • 4 bits for the exponent • two’s complement form for both the mantissa and exponent.

a

Write the normalised floating-point representation of +65.25 in this system. Show your working.

[3]
b

Explain the problem that will occur in storing the normalised floating-point representation of +65.20 in this system.

[2]
2
a

Draw one line to connect each protocol to its most appropriate use.

[4]
b

Outline the purpose of the Link layer in the TCP/IP protocol suite.

[2]
3

Describe what is meant by enumerated and pointer data types. Enumerated Pointer

[4]
4
a

Describe sequential and random methods of file organisation. Sequential file organisation Random file organisation

[4]
b

Outline the process of sequential access for serial and sequential files.

[2]
5

Describe the features of SISD and MIMD computer architectures. SISD MIMD

[4]
6

This diagram represents a logic circuit.

Figure 1 of 2 for question 6
Figure 2 of 2 for question 6
a

Complete the truth table for the given logic circuit.

[3]
b

Simplify the given Boolean expression using Boolean algebra. Show your working. Y=A‾.B‾.C‾.D‾+A‾.B‾.C.D‾+A‾.B.C‾.D‾+A‾.B.C.D‾

[3]
7
a

A student buys a new computer. State one benefit to the student of a user interface and give an example. Benefit Example

[2]
b

Two of the process states are the running state and the ready state. Identify one other process state.

[1]
c

Outline conditions under which a process could change from the running state to the ready state.

[2]
8
a

A pseudocode algorithm finds a customer account record in a random file and outputs it. The records are stored using the user-defined data type TAccount. TYPE TAccount DECLARE AccountNumber : INTEGER DECLARE LastName : STRING DECLARE FirstName : STRING DECLARE Address : STRING DECLARE ContactNumber : STRING ENDTYPE Complete the file handling pseudocode. The function Hash() takes the customer account number as a parameter, calculates and returns the hash value. DECLARE Customer : TAccount DECLARE Location : INTEGER DECLARE AccountFile : STRING .................................................................................................. "AccountRecords.dat" ............................................................... AccountFile .......................................................... OUTPUT "Please enter an account number" INPUT Customer.AccountNumber Location ← Hash( ..............................................................................................................) SEEK ..................................................................................................................... , Location ............................................................. AccountFile, .......................................................... OUTPUT Customer // output customer record CLOSEFILE AccountFile

[5]
b

Define the term exception handling.

[1]
c

State two possible causes of an exception.

[2]
9
a(i)

Write the infix expression for this Reverse Polish Notation (RPN) expression: 5 2 – 5 4 + * 9 /

[2]
a(ii)

Show how the contents of the following stack will change as the RPN expression in part (a)(i) is evaluated.

[4]
b

Explain how a stack can be used to evaluate RPN expressions.

[3]
10

A stack is to be set up using the information in the table. Identifier | Data type | Description BasePointer | INTEGER | points to the bottom of the stack TopPointer | INTEGER | points to the top of the stack Stack | REAL | 1D array to implement the stack A constant, with identifier Capacity, limits the size of the stack to 25 items.

a

Write the pseudocode for the required declarations.

[3]
b

Complete the pseudocode function Pop() to pop an item from Stack. // popping an item from the stack FUNCTION Pop()............................................................................... DECLARE Item : REAL Item ← 0 ............................................................................... BasePointer THEN Item ← ............................................................................... TopPointer ← ............................................................................... ELSE OUTPUT "The stack is empty – error" ENDIF ............................................................................... ENDFUNCTION

[5]
c

Compare and contrast the queue and stack Abstract Data Types (ADT).

[2]
11

A declarative programming language is used to represent subjects that students can choose to study. Students must choose two subjects. 01 subject(mathematics). 02 subject(physics). 03 subject(chemistry). 04 subject(computer_science). 05 subject(geography). 06 subject(history). 07 subject(english). 08 subject(biology). 09 student(tomaz). 10 student(josephine). 11 student(elspeth). 12 student(nico). 13 student(teresa). 14 student(pietre). 15 choice1(tomaz, mathematics). 16 choice1(teresa, chemistry). 17 choice1(pietre, mathematics). 18 choice1(nico, mathematics). 19 choice1(elspeth, chemistry). 20 choice2(tomaz, computer_science). 21 choice2(nico, geography). These clauses have the meanings: Clause Meaning 01 Mathematics is a subject. 09 Tomaz is a student. 15 Tomaz has chosen mathematics as his first choice. 20 Tomaz has chosen computer science as his second choice.

a

Anthony is a student who would like to study history and geography. Write additional clauses to represent this information. 22 ............................................................................................................................................. 23 ............................................................................................................................................. 24 .............................................................................................................................................

[3]
b

Using the variable X, the goal: choice1(X, chemistry) returns X = teresa, elspeth Write the result returned by the goal: choice1(X, mathematics) X = .....................................................................................................................................

[1]
c

Students must choose two different subjects such that: N may choose S, if N is a student and S is a subject and N has not chosen S as the first choice. Write this as a rule. may_choose_subject(N, S) IF ............................................................................................................................................. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................

[4]
12

Artificial neural networks have played a significant role in the development of machine learning. Explain what is meant by the term artificial neural network.

[4]

Mark scheme

9618/31 · Oct/Nov 2023

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1a · 3 marks

  • 1 mark

    conversion of 65.25 to binary seen e.g. 1000001.01 = 65.25 // 64 + 1 + 0.25 / ¼

  • 1 mark

    correct mantissa: 0 1 0 0 0 0 0 1 0 1 0 0

  • 1 mark

    correct exponent: 0 1 1 1

1b · 2 marks

  • 1 mark

    the decimal fraction 0.20 cannot be represented exactly (the closest is 0.25 / 0.1875)

  • 1 mark

    therefore, there will be a loss of precision due to a rounding error/truncation

2a · 4 marks

  • 1 mark

    HTTP connected to: when transmitting hypertext documents

  • 1 mark

    BitTorrent connected to: to provide peer-to-peer file sharing

  • 1 mark

    SMTP connected to: when sending email messages towards the intended destination

  • 1 mark

    IMAP connected to: when retrieving email messages from a mail server over a TCP/IP connection

2b · 2 marks

  • 1 mark

    To ensure correct network protocols are followed

  • 1 mark

    To enable the upper layers to access the physical medium // enables connection/ communication with the internet / network layer

3 · 4 marks

  • 1 mark

    A user-defined non-composite (data type) (only award once) [enumerated type]

  • 1 mark

    …with a list of all possible values

  • 1 mark

    A user-defined non-composite (data type) (only award once) [pointer type] / …that stores addresses/memory locations only

  • 1 mark

    …and indicates the type of data stored in the memory location.

4a · 4 marks

  • 1 mark

    Records (in the file) are ordered

  • 1 mark

    …based on the key field

  • 1 mark

    Records are stored in no particular order within the file // There is no sequencing in the placement of the records

  • 1 mark

    There is a relationship between the key of the record and its location within the file // a hashing algorithm is used to find the location of the record

4b · 2 marks

  • 1 mark

    Start at the beginning of the file

  • 1 mark

    …check records linearly // …until the desired record is found // … processing / updating records as required //… EOF found.

5 · 4 marks

  • 1 mark

    Single Instruction, Single Data (architecture). // Data is taken from a single source and a single instruction is performed on the data.

  • 1 mark

    Contains one processor, a control unit and a memory unit // …that executes instructions sequentially.

  • 1 mark

    Multiple Instruction, Multiple Data (architecture). // At any time, any processor can execute different instructions on different sets of data.

  • 1 mark

    Contains many processors // …that operate asynchronously / independently.

6a · 3 marks

  • 1 mark

    First shaded block (rows 1 to 4) correct for column Z: 1, 0, 1, 0

  • 1 mark

    Second block (rows 5 to 8) and third block (rows 9 to 12) correct for column Z: 1, 0, 1, 0 and 1, 0, 1, 0

  • 1 mark

    Fourth shaded block (rows 13 to 16) correct for column Z: 1, 0, 0, 0

6b · 3 marks

  • 1 mark

    Working point 1: (Y =) Ā.B̄.C̄.D̄ + Ā.B̄.C.D̄ + Ā.B.C̄.D̄ + Ā.B.C.D̄

  • 1 mark

    Working point 2: Factorising terms e.g. (Y =) Ā.D̄.(B̄.C̄ + B̄.C + B.C̄ + B.C)

  • 1 mark

    (Y =) Ā.D̄

7a · 2 marks

  • 1 mark

    Benefit: The user interface hides the complexities of the computer hardware/operating system from the user // It provides appropriate access systems for users with differing needs // Complex commands involving memory locations/buses/computer hardware/ are avoided

  • 1 mark

    Valid example: Clicking on icon rather than writing code // Using a graphical user interface / icons for navigation

7b · 1 mark

  • 1 mark

    Blocked (state)

7c · 2 marks

  • 1 mark

    When the time slice of the running process expires (round robin).

  • 1 mark

    …and there is a process with a higher priority in the ready queue, the running process is pre-empted // When an interrupt arrives at the CPU, (the process running on the CPU gets pre-empted).

8a · 5 marks

  • 1 mark

    AccountFile  "AccountRecords.dat"

  • 1 mark

    OPENFILE AccountFile FOR RANDOM

  • 1 mark

    Location  Hash(Customer.AccountNumber)

  • 1 mark

    SEEK AccountFile, Location

  • 1 mark

    GETRECORD AccountFile, Customer

8b · 1 mark

  • 1 mark

    (Exception handling is the process of) responding to an unexpected event when the program is running so it does not halt unexpectedly

8c · 2 marks

  • 1 mark

    Programming errors / User errors

  • 1 mark

    Hardware failure / Runtime errors

9a(i) · 2 marks

  • 1 mark

    (5 – 2)

  • 1 mark

    * (5 + 4) / 9

9a(ii) · 4 marks

  • 1 mark

    Ring 1 correct (push 5, push 2, subtract to get 3)

  • 1 mark

    Ring 2 correct (push 5, push 4, add to get 9)

  • 1 mark

    Ring 3 correct (multiply 3 and 9 to get 27)

  • 1 mark

    Ring 4 correct (push 9, divide 27 by 9 to get 3)

9b · 3 marks

  • 1 mark

    Evaluate the RPN expression from left to right

  • 1 mark

    Push each element of the RPN expression onto the stack in order until an operator is reached

  • 1 mark

    Pop the last two elements from the stack and apply the operator // Push the result of the operation onto the stack // Repeat the process until the whole expression is evaluated.

10a · 3 marks

  • 1 mark

    Correct constant declaration

  • 1 mark

    Two correct variable declarations

  • 1 mark

    Correct array declaration

10b · 5 marks

  • 1 mark

    FUNCTION Pop() RETURNS REAL

  • 1 mark

    IF TopPointer >= BasePointer THEN

  • 1 mark

    Item  Stack[TopPointer]

  • 1 mark

    TopPointer  TopPointer – 1

  • 1 mark

    RETURN Item

10c · 2 marks

  • 1 mark

    A queue is a first in first out / FIFO data structure and a stack is a first in last out / FILO / LIFO data structure // Data is removed from a queue in the order it is received and removed from a stack in the reverse order to which it is received

  • 1 mark

    Both ADTs can vary in size / are of indeterminate length // Data is popped and pushed (onto/from a stack) at the same end but it is enqueued and dequeued (to/from a queue) at different/opposite ends // a queue has two accessible ends and a stack has only one // A stack has only one moveable pointer whereas a queue has two.

11a · 3 marks

  • 1 mark

    (22) student(anthony).

  • 1 mark

    (23) choice1(anthony, history).

  • 1 mark

    (24) choice2(anthony, geography).

11b · 1 mark

  • 1 mark

    X = tomaz, pietre, nico

11c · 4 marks

  • 1 mark

    student(N)

  • 1 mark

    subject(S)

  • 1 mark

    choice1(N, S)

  • 1 mark

    all logical operators correct with no additional code (see example answers)

12 · 4 marks

  • 1 mark

    An artificial neural network is the component of artificial intelligence that is meant to simulate the functioning of a biological brain.

  • 1 mark

    Artificial neural networks are a key component of machine learning.

  • 1 mark

    They can solve problems that would prove impossible or difficult for humans // Artificial neural networks have self-learning capabilities that enable them to produce better results as more data becomes available

  • 1 mark

    Artificial neural networks can be layered (input, hidden and output layers) // Artificial neural networks have many interconnected layers, some / many of which are hidden // Weights are assigned between nodes // Weights are adjusted through training to give a more accurate result // More complex learning capabilities / more accurate results are available with larger numbers of hidden layers

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