Computer Science 9618/33 May/June 2024: question paper and mark scheme

Mark scheme

1(a) One mark per mark point (Max 3) 3 MP1 conversion of exponent 001001 to 9 MP2 application of exponent to mantissa to go from 0.100111100 to 100111100 // 256 + 32 + 16 + 8 + 4 seen // 64/128 + 8/128 + 4/128 + 2/128 + 1/128 = 79/128 // 1/2 + 1/16 + 1/32 + 1/64 + 1/128 = 79/128 MP3 correct answer = 316 1(b) One mark per mark point (Max 3) 3 MP1 number converted to binary 10011001.01 // number converted to positive 102.75, reversed bits and 1 added. (0)1100110.11 10011001.00 10011001.01 // -128 + 16 + 8 + 1 + 0.25 = –102.75 MP2 exponent = 7 // Moving binary point the correct number of places MP3 correct answer Mantissa Exponent 1 0 0 1 1 0 0 1 0 1 0 0 0 1 1 1

Mark scheme

2(a) Two marks for all protocols in correct position 2 One mark for at least two protocols in correct position Application Transport Internet Link 2(b) One mark per mark point (Max 2) 2 MP1 The transport layer is responsible for delivery of data from the source host to the destination host MP2 It is where data is broken up into packets and sent to the internet layer MP3 Adds the sequence number to the packet header MP4 It establishes end to end contact MP5 It ensures data arrives error free // It retransmits packets if lost. 2(c) One mark for name of protocol and one mark for expansion (Max 2) 2 HTTP(S) – responsible for correct transfer of files / hypertext documents that make up web pages on the world wide web FTP – used when transferring files from a server to a client on a network POP3 – handles the receiving of emails IMAP – handles the receiving of emails SMTP – handles the sending of emails BitTorrent – provides peer-to-peer file sharing

Mark scheme

3(a) One mark per mark point (Max 2) 3 non-composite data types MP1 Non-composite data types can both be user-defined or primitive MP2 Non-composite data types do not refer to other data types in their definition / contain one data type in their definition MP3 Non-composite data types can be primitive/enumerated/pointer One mark per mark point (Max 2) composite data types MP4 Composite data types can be user-defined or primitive MP5 Composite data types refer to other data types in their definition/contain more than one data type in their definition MP6 Composite data types can be record/set/class 3(b) One mark for TYPE FootballClub and ENDTYPE correct 4 One mark for every two correct declarations Example answer TYPE FootballClub DECLARE TeamName : STRING DECLARE DateOfJoining : DATE DECLARE MainTelephone : STRING DECLARE ManagerName : STRING DECLARE NumberOfMembers : INTEGER DECLARE LeaguePosition : INTEGER ENDTYPE

Mark scheme

4(a) One mark per mark point (Max 2) 2 MP1 Sequential access method searches for records one after the other MP2 … from the physical start of the file until the record is found/the end of file. 4(b) One mark per mark point (Max 3) 3 MP1 For serial files, records are stored in chronological order MP2 … every record needs to be checked until the record is found, or all records have been checked. MP3 For sequential files, records are stored in order of a key field/index, and it is the key field/index that is compared. MP4 … every record is checked until the record is found, or the key field of the current record is greater than the key field of the target record.

Mark scheme

5(a) One mark per correct term (Max 3) 3 (5 + 2) / (9 – 3) * 3 Complete correct answer ((5 + 2) / (9 - 3)) * 3 5(b) One mark 7 3 + 2 One mark 2 8 * - 6 / Complete answer 7 3 + 2 8 * - 6 / 5(c) One mark per ring (Max 4) 4 3 5 7 7 10 10 17 17 12 12 12 12 120 120 12

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 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 1 1 0 1 0 1 1 1 0 0 0 0 1 1 1 1 1 0 1 0 1 1 6(b) Two marks for all five correct terms and no extras 2 One mark for any three correct terms (Z =) A.B.C + A.B.C + A.B.C + A.B.C + A.B.C 6(c)(i) Two marks if all correct 2 One mark if one error present BC A 00 01 11 10 0 1 1 0 0 1 1 1 1 1 6(c)(ii) One mark for each correct loop (Max 2) 2 BC A 00 01 11 10 0 1 1 0 0 1 1 1 1 1 6(c)(iii) A+B 1

Mark scheme

7(a) One mark per point (max 3) 3 MP1 A digital certificate is an electronic/online document. MP2 used to authenticate/prove the identity of a website/the online identity of an individual/organisation MP3 typically issued by a CA MP4 For example: it contains information identifying a website owner/individual and a public key 7(b) One mark per point (max 2) 2 MP1 The digital certificate provides the public key MP2 … that can be used to validate the private key associated with the organisation/website/digital signature

Mark scheme

8(a) One mark for each correctly completed clause (Max 3) 3 (23) feature(sliding_doors). (24) available(sliding_doors, minivan). (25) unavailable(sliding_doors, hatchback). 8(b) (Options =) sunroof, reversing_camera 1 8(c) One mark per mark point (Max 4) 4 MP1 feature(F) MP2 bodystyle(B) MP3 unavailable(F, B) MP4 all correct Boolean operators and punctuation (allow , for AND) and no additional lines of code Example answers may_choose_option(F, B) IF feature(F) AND bodystyle(B) AND NOT unavailable(F, B). feature(F), bodystyle(B), NOT unavailable(F, B).

Mark scheme

9 One mark per mark point (Max 3) 3 MP1 Deep learning learns by finding hidden patterns that are undetectable to humans. MP2 It structures algorithms in layers: input layer, hidden layers and output layer. MP3 … to create an artificial neural network to learn and make intelligent decisions on its own. MP4 It is trained using large quantities of unlabelled data. MP5 Deep learning requires/uses a large number of hidden layers. MP6 … the larger the number of layers, the higher the level of success.

Mark scheme

10(a) One mark 1 The elements are sorted according to the compare function / in ascending / descending order. 10(b) One mark for each correctly completed line (Max 5) 5 DECLARE Names : ARRAY[1:100000] OF STRING DECLARE TopOfList : INTEGER DECLARE EndOfList : INTEGER DECLARE CurrentItem : INTEGER DECLARE ToFind : STRING DECLARE Found : BOOLEAN DECLARE NotInList : BOOLEAN TopOfList ← 1 EndOfList ← 100000 OUTPUT "Which name do you wish to find? " INPUT ToFind Found ← FALSE NotInList ← FALSE WHILE Found = FALSE AND NotInList = FALSE CurrentItem ← (TopOfList + EndOfList) DIV 2 IF ToFind = Names[CurrentItem]// Names[CurrentItem] = ToFind THEN Found ← TRUE ELSE IF TopOfList >= EndOfList THEN NotInList ← TRUE ELSE IF ToFind > Names[CurrentItem] THEN TopOfList ← CurrentItem + 1 ELSE EndOfList ← CurrentItem – 1 ENDIF ENDIF ENDIF ENDWHILE 10(b) IF Found = TRUE THEN OUTPUT "Item found at position ", CurrentItem, " in array" ELSE OUTPUT "Item not in array" ENDIF 10(c) One mark from 2 MP1 Big O for a binary search is O(Log2 n). MP2 Big O notation is used to indicate the time/space complexity of an algorithm. One mark from MP3 The time taken to complete the search increases logarithmically as the number of search items increases linearly MP4 The time taken to complete the search increases linearly as the number of search items increases exponentially MP5 As the search field is repeatedly getting smaller, the number of comparisons made before the item is found, or the number of items runs out, is relatively small.

Mark scheme

11(a) One mark per mark point (Max 2) 2 MP1 Uses hard-wired code/control units MP2 Uses relatively few instructions / simple instructions MP3 Uses relatively few addressing modes MP4 Makes use of a single-cycle for each instruction MP5 Makes use of fixed length / fixed format instructions MP6 Makes use of general-purpose registers MP7 Pipelining is straightforward to apply MP8 The design emphasis is on the software MP9 Processor chips require few transistors. 11(b) One mark per mark point (Max 3) 3 MP1 Once the processor detects an interrupt at the start/end of the fetch-execute cycle MP2 … the current program is temporarily stopped and the status of each register stored on the stack. MP3 After the interrupt has been serviced/the Interrupt Service Routine (ISR) has been executed … MP4 … the registers can be restored to its original status before the interrupt was detected // … the data can be restored from the stack. 11(c) One mark per mark point (Max 3) 3 MP1 Pipelining adds an additional complexity // there could be a number of instructions still in the pipeline when the interrupt is received MP2 All the instructions currently in operation are usually discarded except for the last one/the one at write back MP3 … the interrupt handler routine is applied to the remaining instruction. MP4 Once the interrupt has been serviced the processor can restart with the next instruction in the sequence.