COMPUTER SCIENCE 7517/1 Paper 1 Mark scheme June 2023

A-level COMPUTER SCIENCE 7517/1 Paper 1 Mark scheme June 2023 Version: 1.0 Final *236A7517/1/MS* 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. 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 © 2023 AQA and its licensors. All rights reserved. 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. A-level Computer Science Paper 1 (7517/1) – applicable to all programming languages A, B, C, D and E June 2023 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. 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 06.1, 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. Question Marks 01 All marks AO1 (understanding) 1. Check the queue is not already full; 2. Compare the value of the (rear) pointer with the maximum size of the array; 3. If equal then (rear) pointer becomes zero; A. index of the first position in the array instead of zero 4. Otherwise, add one to the (rear) pointer; 5. Insert new item in position indicated by (rear) pointer; Alternative answer 1 1. Check the queue is not already full; 2. Compare the value of the (rear) pointer with the maximum size of the array minus one; 3. If equal then (rear) pointer becomes one; A. index of the first position in the array instead of one 4. Otherwise, add one to the (rear) pointer; 5. Insert new item in position indicated by (rear) pointer; Alternative answer 2 1. Check the queue is not already full; 2. Add one to the (rear) pointer; 3. Compare the value of the (rear) pointer with the maximum size of the array; 4. If equal then (rear) pointer becomes zero; A. index of the first position in the array instead of zero 5. Insert new item in position indicated by (rear) pointer; Alternative answer 3 1. Check the queue is not already full; 2. Add one to the (rear) pointer; 3. Compare the value of the (rear) pointer with the maximum size of the array plus one; 4. If equal then (rear) pointer becomes one; A. index of the first position in the array instead of one 5. Insert new item in position indicated by (rear) pointer; Alternative answer 4 1. Check the queue is not already full; 2. Add one to the (rear) pointer; 3. Use modulus/modulo operator/function with new value of (rear) pointer; 4. Use modulus/modulo operator/function with maximum size of array; 5. Insert new item in position indicated by (rear) pointer; Max 4 if any errors 5 Question Marks 02 1 Mark is for AO2 (analyse) The colour is not yellow // the chosen shape was not the yellow circle // the colour is blue or pink; 1 02 2 Mark is for AO2 (analyse) The shape is not a square // the chosen shape was not the blue square // the shape is a triangle or circle; 1 02 3 Mark is for AO2 (analyse) Pink triangle; 1 Question Marks 03 1 All marks for AO1 (understanding) A root (A. start) node; A. there is a parent-child relationship between nodes Each node has no more than two child nodes; R. has two child nodes 2 03 2 All marks are for AO2 (apply) Note for examiners: answers are in pseudo-code so accept any reasonable representation (including use of string or integer values for rows 3 and 6). Mark as follows: 1 mark: row one correct 1 mark: row two correct 1 mark: rows three and six correct 1 mark: rows four and five correct 4 03 3 Mark is for AO1 (knowledge) O(log2n); I. missing brackets I. missing O I. missing 2 1 03 4 Mark is for AO1 (understanding) Every comparison halves the size of the binary tree to look at; (A. every comparison halves the size of the tree) 1 03 5 Mark is for AO1 (understanding) It does not have exponential (or worse) time complexity; It has a polynomial (or better) time complexity solution; A. It can be solved in a reasonable amount of time regardless of the problem/input size; NE. can be solved in a reasonable amount of time Max 1 1 03 6 All marks AO1 (understanding) Rules/knowledge (about the problem domain); Can be used to find a good/approximate but (probably) not optimal solution to a problem; Can reduce the size of the search/problem space // changing some constraints in the problem; Max 2 2 03 7 All marks AO1 (knowledge) As the size of the input/problem increases; the amount of time taken remains the same; 2 Question Marks 04 1 All marks AO2 (analyse) Language Regular language (Y/N)? Language A N Language B Y Language C Y Language D Y Language E Y Language F Y Mark as follows: 1 mark: any two rows correct 2 marks: any four rows correct 3 marks: all rows correct A. any suitable alternative to N and Y 3 04 2 All marks AO2 (apply) a|ab|b+;; If final answer incorrect award a maximum of 1 mark for any of: • a|ab • b+ Alternative answer ab?|b+;; If final answer incorrect award a maximum of 1 mark for any of: • ab? • b+ 2 04 3 Mark is for AO1 (knowledge) The number of elements/items in a set; A. the size of a set 1 Question Marks 05 1 All marks AO2 (apply) Tape Current State Mark as follows: 1 mark: tape and current state of first row correct. 1 mark: tape and current state of second and third rows correct. 1 mark: read/write head correct for first three rows. 1 mark: tape left of first hash symbol and current state for rows four to ten correct (denoted by blue outline in table above). 1 mark: tape and current state for last row correct. A. any unambiguous way of denoting a blank cell on the tape I. position of read/write head when awarding mark points 1, 2, 4 and 5 Max 4 if any errors 5 05 2 All marks AO2 (analyse) When there are no zeros between the two hash symbols on the tape; When there is a character other than zero (A. a one) between the two hash symbols on the tape; If the machine was not in the start state (S0); Max 2 2