2024_AQA: A-level COMPUTER SCIENCE Paper 1 (Merged Question paper and marking scheme): June 2024 A-level

2024_AQA: A-level COMPUTER SCIENCE Paper 1 (Merged Question paper and marking scheme): June 2024 A-level COMPUTER SCIENCE Paper 1 June 2024 Preliminary Material To be opened and issued to candidates on or after 1 September 2023, subject to the instructions given in the Teacher’s Notes (7517/1/TN). Note  The Preliminary Material and Skeleton Program are to be seen by candidates and their teachers only, for use during preparation for the examination on 10 June 2024. They cannot be used by anyone else for any other purpose, other than that stated in the instructions issued, until after the examination date has passed. They must not be provided to third parties. Information  A Skeleton Program is provided separately by your teacher and must be read in conjunction with this Preliminary Material.  You are advised to familiarise yourselves with the Preliminary Material and Skeleton Program before the examination.  A copy of this Preliminary Material, the Skeleton Program and the Data files , , and will be made available to you in hard copy and electronically at the start of the examination.  You must not take any copy of the Preliminary Material, Skeleton Program or any other material into the examination room. A-level Computer Science Paper 1 - Key Areas to Revise Paper 1 of the A-level Computer Science exam primarily focuses on the theoretical aspects of computer science, including algorithms, data structures, programming, and the fundamentals of computer systems. Below are the key areas you should focus on to ensure you're well-prepared for the exam: 1. Programming Fundamentals:  Programming Concepts: Review the core programming concepts such as variables, constants, data types (e.g., integer, string, boolean), operators (arithmetic, logical, relational), and input/output. Be comfortable with writing code using appropriate syntax and operators.  Control Structures: Focus on understanding and implementing control structures such as: o Conditional Statements: if, else, elif, switch, and ternary operators. o Loops: for, while, do-while loops. Understand when to use each loop type effectively.  Functions and Procedures: Be able to define and use functions and procedures, including handling parameters and return values. Understand the difference between procedures and functions.  Recursion: Understand recursive functions, how they work, and when they should be used. Practice solving problems like factorials, Fibonacci series, and recursive algorithms for searching and sorting.  Data Structures: Revise the implementation and manipulation of basic data structures such as arrays, lists, stacks, queues, and linked lists. You should know how to perform operations like insertion, deletion, and searching on these structures. 2. Algorithms and Problem-Solving:  Sorting and Searching Algorithms: Be familiar with common algorithms such as: o Sorting Algorithms: Bubble sort, selection sort, insertion sort, quicksort, and merge sort. Understand their time complexities (Big O notation). o Searching Algorithms: Binary search (for sorted data) and linear search. Understand how to implement these algorithms and their efficiency.  Big O Notation: Understand how to analyze the time and space complexity of algorithms. Be able to calculate the efficiency of algorithms based on their performance with respect to input size.  Algorithm Design: Practice designing algorithms to solve problems, and be able to represent them using pseudocode or flowcharts. Focus on breaking down complex problems into simpler tasks and finding efficient solutions. 3. Data Representation:  Binary and Hexadecimal Representation: Be able to convert between binary, hexadecimal, and decimal number systems. Understand how data is represented in binary (e.g., integers, floating-point numbers).  Character Representation: Study how characters are represented in computers using ASCII, Unicode, and other character encoding schemes.  Data Types: Understand the differences between primitive and non-primitive data types. Be familiar with how various data types are stored and their memory usage.  File Handling: Review how to work with files (reading from and writing to text files). Be familiar with file operations, including opening, closing, reading, and writing files. 4. Computer Systems: 5. Networking and Communication: 6. Databases: 7. Ethical, Legal, and Social Issues in Computing: 7517/1/PM IB/G/Jun24/G4001/E5 2 INSTRUCTIONS FOR CANDIDATES Electronic Answer Document Answers for all questions in all sections must be entered into the word-processed document made available to you at the start of the examination and referred to in the question paper rubrics as the Electronic Answer Document. Preparation for the examination You should ensure that you are familiar with the Preliminary Material and the Skeleton Program for your programming language. IB/G/Jun24/7517/1/PM 3 Symbol Puzzle Symbol Puzzle is a simple puzzle where the user places symbols into a grid, trying to maximise their score as they do so. The user is allowed to place a specified number of symbols (38 in the standard puzzle) into the grid. The puzzle is finished when the last symbol has been placed. To score points, the user must place symbols into cells in the grid so that they match a pattern. In the standard puzzle the user is able to use the symbols Q, T and X. To score 10 points using the symbol Q the user needs to place five Q symbols in a 33 section of the grid so that they match the pattern shown in Figure 1. Figure 1 Q Q Q Q Q To score 10 points using the symbol T the user needs to place five T symbols in a 33 section of the grid so that they match the pattern shown in Figure 2. Figure 2 T T T T T To score 10 points using the symbol X the user needs to place five X symbols in a 33 section of the grid so that they match the pattern shown in Figure 3. Figure 3 X X X X X Turn over ► IB/G/Jun24/7517/1/PM 4 When the user has successfully created one of the allowed patterns within a 33 section of the grid, they are no longer allowed to place the symbol used in that pattern in any of the other cells in that 33 section. However, other symbols can still be placed in empty cells within that 33 section. Some cells are blocked, denoted by the symbol @. The user is not able to place a symbol in a blocked cell. In the standard puzzle, the number of blocked cells and the position of each blocked cell is generated randomly. An example of a starting grid for the standard puzzle is shown in Figure 4. The row and column numbers have been shown. The cell in the bottom right corner of the grid is row 1, column 8 Figure 4 1 8 2 3 @ 4 5 6 7 8 7 6 5 4 3 2 1 @ @ @ @ @ @ Instead of using the standard puzzle, the user may enter the name of a file containing a puzzle. Four text files containing different 55 grid puzzles have been provided: puzzle1, puzzle2, puzzle3 and puzzle4. There are errors in the implementation of the puzzle in the Skeleton Program, which means that it does not work in the way described in this Preliminary Material under all circumstances. END OF PRELIMINARY MATERIAL IB/G/Jun24/7517/1/PM 5 There is no Preliminary Material printed on this page Copyright information For confidentiality purposes, all acknowledgements of third-party copyright material are published in a separate booklet. This booklet is published after each live examination series and is available for free download from Permission to reproduce all copyright material has been applied for. In some cases, efforts to contact copyright-holders may have been unsuccessful and AQA will be happy to rectify any omissions of acknowledgements. If you have any queries please contact the Copyright Team. Copyright © 2024 AQA and its licensors. All rights reserved. IB/G/Jun24/7517/1/PM A-level COMPUTER SCIENCE 7517/1 Paper 1 Mark scheme June 2024 Version: 1.0 Final MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students’ scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. No student should be disadvantaged on the basis of their gender identity and/or how they refer to the gender identity of others in their exam responses. A consistent use of ‘they/them’ as a singular and pronouns beyond ‘she/her’ or ‘he/him’ will be credited in exam responses in line with existing mark scheme criteria. Further copies of this mark scheme are available from Copyright information AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre. Copyright © 2024 AQA and its licensors. All rights reserved. 2 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 Level of response marking instructions Level of response mark schemes are broken down into levels, each of which has a descriptor. The descriptor for the level shows the average performance for the level. There are marks in each level. Before you apply the mark scheme to a student’s answer read through the answer and annotate it (as instructed) to show the qualities that are being looked for. You can then apply the mark scheme. Step 1 Determine a level Start at the lowest level of the mark scheme and use it as a ladder to see whether the answer meets the descriptor for that level. The descriptor for the level indicates the different qualities that might be seen in the student’s answer for that level. If it meets the lowest level then go to the next one and decide if it meets this level, and so on, until you have a match between the level descriptor and the answer. With practice and familiarity you will find that for better answers you will be able to quickly skip through the lower levels of the mark scheme. When assigning a level you should look at the overall quality of the answer and not look to pick holes in small and specific parts of the answer where the student has not performed quite as well as the rest. If the answer covers different aspects of different levels of the mark scheme you should use a best fit approach for defining the level and then use the variability of the response to help decide the mark within the level, ie if the response is predominantly level 3 with a small amount of level 4 material it would be placed in level 3 but be awarded a mark near the top of the level because of the level 4 content. Step 2 Determine a mark Once you have assigned a level you need to decide on the mark. The descriptors on how to allocate marks can help with this. The exemplar materials used during standardisation will help. There will be an answer in the standardising materials which will correspond with each level of the mark scheme. This answer will have been awarded a mark by the Lead Examiner. You can compare the student’s answer with the example to determine if it is the same standard, better or worse than the example. You can then use this to allocate a mark for the answer based on the Lead Examiner’s mark on the example. You may well need to read back through the answer as you apply the mark scheme to clarify points and assure yourself that the level and the mark are appropriate. Indicative content in the mark scheme is provided as a guide for examiners. It is not intended to be exhaustive and you must credit other valid points. Students do not have to cover all of the points mentioned in the Indicative content to reach the highest level of the mark scheme. An answer which contains nothing of relevance to the question must be awarded no marks. 3 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 A-level Computer Science Paper 1 (7517/1) – applicable to all programming languages A, B, C, D and E June 2024 The following annotation is used in the mark scheme: ; – means a single mark // – means an alternative response / – means an alternative word or sub-phrase A. – means an acceptable creditworthy answer R. – means reject answer as not creditworthy NE. – means not enough I. – means ignore DPT. – means ‘Don't penalise twice’. In some questions a specific error made by a candidate, if repeated, could result in the loss of more than one mark. The DPT label indicates that this mistake should only result in a candidate losing one mark, on the first occasion that the error is made. Provided that the answer remains understandable, subsequent marks should be awarded as if the error was not being repeated. 4 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 Examiners are required to assign each of the candidate’s responses to the most appropriate level according to its overall quality, and then allocate a single mark within the level. When deciding upon a mark in a level, examiners should bear in mind the relative weightings of the assessment objectives eg In question 07.1, the marks available for the AO3 elements are as follows: AO3 (design) 4 marks AO3 (programming) 8 marks Where a candidate’s answer only reflects one element of the AO, the maximum mark they can receive will be restricted accordingly. 5 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 Question Marks 01 All marks AO1 (understanding) Example answers Easier to test/debug as each subroutine can be tested separately; Easier to understand the code if sensible identifiers are used for subroutine names; Code can be easily reused as each subroutine is independent of rest of program; Makes it easier to work as a team of programmers as each subroutine can be worked on independently; Can be used as often as needed without having to write the code each time; Makes program easier to maintain/update (in future) as fewer changes will need to be made to make an update; Allows the use of recursive techniques because subroutines can call themselves; Easier to understand the code as each subroutine can be considered in isolation; Less likely to be errors in the code due to reuse of code; Reduces/eliminates side effects (eg unexpected change to value in global variable) through use of local variables; Max 3 Notes for examiners Each advantage must be different. Mark should only be awarded if there is an explanation of how the advantage is achieved. Mark should only be awarded if the explanation is relevant for the stated advantaged. 3 6 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 Question Marks 02 1 All marks AO1 (understanding) When an item is removed from a linear queue; A. by implication that problem arises when items have been deleted all other items need to be shuffled up one // this can result in unusable space in the array // this can result in only being able to add a limited number of items to the queue (eg only ever being able to add five items to a queue which uses an array of size 5); Alternative answer No need to shuffle items up one // no unusable spaces; After deleting an item from a circular queue; 2 02 2 All marks AO1 (understanding) 1. Check that the queue is not already empty // check to see if the current size is 0; R. reference to array instead of queue 2. if it is then deal with (underflow) error // (If it is not then) process/dequeue the front item in the queue; R. reference to array instead of queue (unless reference to array is made using value of front pointer) R. if dequeue would only happen under some of the correct circumstances 3. Reduce the value of the variable used to store the current size by 1; 4. Check if the front is in the last position of the array and if it is set it to the first position in the array (A. 0 or 1 instead of first position in the array); 5. (Else) add 1 to the value of the front pointer; Alternative to mark points 1 and 3 if no current size variable is used: calculate the current size; compare this to 0; Alternative to mark points 1 and 3 if no current size variable is used: check to see if the front/rear is –1;; Alternative to mark points 4 and 5: add 1 to the value of the front pointer; then set it to the remainder from dividing the value of front by the maximum size of the queue; Alternative to mark points 4 and 5: add 1 to the value of the front pointer; then if the pointer has passed the end of the array set it to point to the start of the array. DPT. Rear instead of front Max 4 if any errors 5 7 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 Question Marks 03 1 All marks AO1 (knowledge) (Using a program/algorithm/method to) determine if a program will halt; Max 1 for the following points, but only award mark if 1st mark was awarded: without running the program; 2 for a particular input; 03 2 Mark is for AO1 (understanding) It demonstrates that there are non-computable problems // it demonstrates that there are some problems for which there is no algorithm that can solve them // it demonstrates that there are undecidable problems; 1 8 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 Question Marks 04 1 Mark is for AO1 (knowledge) The number of members of a set // the number of elements in a set; A. the size of a set 1 04 2 Mark is for AO2 (apply) The empty set is also a subset of R // {}/ Ø is also a subset of R; NE. they are not the only subsets of R 1 04 3 Mark is for AO2 (apply) 1 // one; 1 04 4 Mark is for AO1 (understanding) It means either the element (immediately) before or the element (immediately) after // or // alternation; 1 04 5 All marks AO2 (apply) a(bb)*|(bb)+ Mark as follows: 1 mark: expression contains a(bb)* R. bb* 1 mark: expression contains (bb)+ R. bb+ Max 1 mark if any errors Alternative answer a|a?(bb)+ Mark as follows: 1 mark: expression contains a|a? 1 mark: expression contains (bb)+ R. bb+ Max 1 mark if any errors 2 9 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 Alternative answer (a|bb)(bb)* Mark as follows: 1 mark: expression contains a|bb 1 mark: expression contains (bb)* R. bb* Max 1 mark if any errors 04 6 All marks AO2 (apply) (ab|b)(bb)* Mark as follows: 1 mark: expression contains ab|b // (ab)|b 1 mark: expression contains (bb)* R. bb* Max 1 mark if any errors Alternative answer a?b(bb)* Mark as follows: 1 mark: expression contains a?b 1 mark: expression contains (bb)* R. bb* Max 1 mark if any errors Note for examiners Any regular expression that would match with an expression that starts with an optional a followed by a compulsory b should get (at least) one mark. 2 10 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 Question Marks 05 1 Mark is for AO1 (understanding) Reverse Polish (Notation) // RPN; A. Postfix 1 05 2 All marks AO2 (apply) 523*+4+;; Mark as follows: 1 mark: 23* in expression; 1 mark: correct order of operands with + symbols either side of the 4; Max 1 mark if any errors 2 11 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 Question Marks 06 1 All marks AO1 (understanding) Connected; Undirected; A. explanations of connected/undirected 2 06 2 All marks AO2 (apply) Mark as follows: 1. Done set to False, Pos set to –1 and Current set to 0 2. Pos set to 0, Temp[0] set to 0 and Current set to 1 3. Pos set to 1, Temp[1] set to 1 and Current set to 2 4. First output is Y 5. Temp[2] set to 2 with no other values after this, Temp[1] set to 3 with no other values after this and Temp[0] set to 4 with no other values after this; 6. Done column correct 7. Correct order for K, H, U and M in output column with no incorrect outputs I. unnecessary repeated values in a column Max 6 if any errors 7 12 Temp Done Pos [0] [1] [2] Current OUTPUT False -1 0 0 0 1 1 1 2 2 2 -1 Y -1 1 K 3 0 1 3 -1 H -1 0 U 4 -1 0 4 -1 M -1 -1 True MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 06 3 All marks AO2 (analyse) There is no child node to the right of any node // each child node is to the left of its parent node;; Note for examiners: if answer is not correct then max one mark for any of the following points:  each node has, at most, one child node R. each node has one child node  there are no nodes with two child nodes  tree has a depth of five 2 06 4 Mark is for AO2 (analyse) Stack // LIFO (data structure); 1 06 5 Mark is for AO2 (analyse) Change the line Current  Dir1[Current] to Current  Dir2[Current] Change the line Current  Dir2[Temp[Pos]] to Current  Dir1[Temp[Pos]]; // Change Dir1 to Dir2 and change Dir2 to Dir1; 1 13 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/1 – JUNE 2024 Question Marks 07 1 4 marks for AO3 (design) and 8 marks for AO3 (programming) Mark Scheme Level Description Mark Range 4 A line of reasoning has been followed to arrive at a 10–12 logically structured working or almost fully working programmed solution that meets most of the requirements. All of the appropriate design decisions have been taken. To award 12 marks, all of the requirements must be met. 3 There is evidence that a line of reasoning has been 7–9 followed to produce a logically structured program. The program displays relevant prompts, inputs the required data, has at least one iterative structure and at least one selection structure and uses appropriate variables to store most of the needed data. An attempt has been made to test for increasing and decreasing numbers, although this may not work correctly under all circumstances. The solution demonstrates good design work as most of the correct design decisions have been made. 2 A program has been written and some appropriate, 4–6 syntactically correct programming language statements have been written. There is evidence that a line of reasoning has been partially followed as, although the program may not have the required functionality, it can be seen that the response contains some of the statements that would be needed in a working solution. There is evidence of some appropriate design work as the response recognises at least one appropriate technique that could be