Computational thinking, problem specification and tracing flowcharts – IB Computer Science exam questions with mark scheme

1Core4 marks

A company is designing a mobile app that shows the quickest bus route between two stops in a city.

(a) Explain how abstraction could be used when designing the app's route map. [2]

(b) Explain how decomposition could be used when developing the app. [2]

Abstraction: removes unnecessary detail such as building shapes, road widths and shop names (accept any sensible irrelevant detail for this app)(1)
Keeping only the relevant features, e.g. stops as nodes and journey times between them (a graph), so the route can be calculated more simply / efficiently(1)
Decomposition: break the problem into smaller sub-problems, e.g. finding nearby stops, calculating the route, displaying the map, handling live bus times (accept any sensible sub-problems in context)(1)
Each sub-problem can be designed, coded and tested separately (possibly by different team members) and then combined(1)
Answer: Abstraction: hide irrelevant detail, model stops/times as a graph. Decomposition: split into smaller manageable components.
Examiner tip: Always apply the term to the scenario given; generic definitions alone score at most half marks.

2Core4 marks

The diagram shows five flowchart symbols, labelled A to E.

(a) Identify the letter of the symbol that is used to represent:

(i) a step where the algorithm chooses between two routes; [1]

(ii) reading a value entered by the user. [1]

(b) State the name of symbol E and outline its purpose in a flowchart. [2]

(a)(i) D (decision) (accept the letter alone)(1)
(a)(ii) C (input/output) (do not accept B)(1)
(b) Name: connector (accept on-page / off-page connector; do not accept 'circle' or 'node')(1)
(b) Purpose: joins two parts of a flowchart / shows where the flow continues elsewhere (with matching labels), e.g. to avoid long or crossing flowlines or to continue onto another page (accept: links a flowline to another point with the same label)(1)
Answer: (a)(i) D; (ii) C. (b) Connector: links one point of the flowchart to another point with the same label, avoiding long/crossing flowlines.
Examiner tip: Rectangles (B) are process steps only; any input or output must use the parallelogram.

3Standard5 marks

The flowchart shows an algorithm. div gives the whole-number result of a division and mod gives the remainder.

(a) Copy and complete the trace table for the input N = 3150. [3]

N	D	R	N > 0 ?
3150		0

(b) State the output. [1]

(a) D column: 0, 5, 1, 3 (in this order)(1)
(a) R column: 0, 5, 51, 513 (allow 0 omitted for the first iteration if the initial 0 is shown)(1)
(a) N column: 315, 31, 3, 0 and decision column ending in False/No when N = 0 (accept True × 4 then False)(1)
(b) 513 (do not accept 0513; follow through from their final R)(1)
(c) Reverses the digits of N / outputs the digits of N in reverse order (accept: outputs N written backwards, with any leading zeros lost)(1)
Answer: (a) D = 0, 5, 1, 3; R = 0, 5, 51, 513; N = 315, 31, 3, 0. (b) 513. (c) It reverses the digits of N.
Examiner tip: R is a number, not a string, so the leading 0 of '0513' disappears: write 513.

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4Standard6 marks

A community library wants a system that automatically sends a text message to members whose borrowed books are overdue. The library has 4000 members, a fixed budget of £2000 and wants the system working before the start of the next school term.

(a) Distinguish between a constraint and an objective in a problem specification, using an example of each from this scenario. [2]

(b) Identify one input specification and one output specification for the system. [2]

(a) A constraint is a limit/restriction the solution must work within, e.g. the £2000 budget / the deadline of next term / must handle 4000 members (accept any limit from the scenario)(1)
(a) An objective is a goal the solution aims to achieve, e.g. members are reminded automatically so overdue books are returned sooner / reduce staff time spent chasing loans(1)
(b) Input: e.g. member ID / mobile number, book ID, due date, today's date (from the loan records) (accept any one relevant input with its data, e.g. 'due date as a date')(1)
(b) Output: e.g. a text message to the member naming the book and its due date / a list of members sent reminders (do not accept 'a message' without saying who/what it contains)(1)
(c) A measurable criterion linked to the objectives, e.g. every member with an overdue book receives a text within 24 hours of the due date(1)
(c) A second, different measurable criterion, e.g. total cost stays within £2000 / system live before the term starts / no texts sent to members without overdue books / percentage of overdue books falls (do not award vague criteria such as 'it works well' or 'is user-friendly' without a measure)(1)
Answer: Constraint: a limit such as the £2000 budget; objective: a goal such as reminding members automatically. Input: member phone number and due date; output: reminder text naming the book. Criteria: all overdue members texted within 24 h; cost within £2000.
Examiner tip: Evaluation criteria must be testable: say what will be measured and the target, not just 'it works'.

5Core4 marks

A cinema has 20 rows of seats. Row 1 has 12 seats, row 2 has 14 seats, row 3 has 16 seats, and the pattern continues to the back row.

(a) Describe what is meant by pattern recognition in computational thinking. [2]

(b) Use the pattern to determine the number of seats in row 20. [1]

(c) Outline how recognising this pattern makes it easier to design an algorithm that works out the total number of seats in the cinema. [1]

(a) Identifying similarities / trends / repeated features in data or between problems(1)
(a) So that a common solution (or existing solution) can be reused / predictions can be made, making the problem easier to solve (accept: generalising from the pattern)(1)
(b) 50 (seats in row n = 10 + 2n)(1)
(c) The same rule (add 2 each row / 10 + 2n) can be applied inside a single loop (or formula) instead of storing or calculating every row separately (accept: the algorithm generalises to any number of rows)(1)
Answer: (a) Finding similarities or trends so a general solution can be reused. (b) 50. (c) One rule, 10 + 2n, used in a loop for every row.
Examiner tip: Describe needs two parts: what is recognised (similarities/trends) and why (reuse a solution or generalise).

6Hard7 marks

The flowchart shows an algorithm that reads a list of positive whole numbers. The value −1 is used to end the list.

(a) The values 7, 12, 3, 12, 9, −1 are input in this order. Copy and complete the trace table and state the output. [4]

X	X = −1 ?	MAX	C
		0	0

(b) State the purpose of the algorithm. [1]

(a) X column: 7, 12, 3, 12, 9, −1 with the decision X = −1 False for the first five and True for −1(1)
(a) MAX column: 7, 12 (then unchanged at 12)(1)
(a) C column: 1, 1, (1), 2 (then unchanged at 2) (accept repeated unchanged values or blanks)(1)
(a) Output: 12, 2 (both values required; follow through from their final MAX and C)(1)
(b) Finds the largest value input and counts how many times it occurs (both parts needed)(1)
(c) Output: 0, 0 (both values required)(1)
(c) Explanation: MAX starts at 0, which is larger than every negative input, so X > MAX is never True and the true largest value (−4) is never recorded (accept: MAX should be initialised to the first value input)(1)
Answer: (a) MAX 7, 12; C 1, 1, 2; output 12, 2. (b) Largest value and how many times it appears. (c) 0, 0, because MAX is initialised to 0, above every negative input.
Examiner tip: When tracing, the second 12 is not greater than MAX, so it goes to the 'X = MAX ?' decision and only C changes.

7Stretch8 marks

The flowchart shows an algorithm. div gives the whole-number result of a division and mod gives the remainder.

(a) Trace the flowchart for the input N = 60. Copy and complete the trace table, showing every output. [4]

N	F	N mod F = 0 ?	Output
60	2

(b) The input is now N = 97, which is a prime number. Determine the output and the number of times the decision N mod F = 0 ? is evaluated. [2]

(c) Suggest a change to the algorithm that would greatly reduce the number of decisions evaluated for large prime inputs, and explain why the outputs are still correct. [2]

(a) Outputs 2, 2 in the first two iterations with N becoming 30 then 15(1)
(a) F increasing 2 → 3 when 15 mod 2 ≠ 0, then output 3 with N becoming 5(1)
(a) F increasing 3 → 4 → 5 (5 mod 3 and 5 mod 4 both False), output 5, N becoming 1 and the loop ending (N > 1 False)(1)
(a) Final outputs listed as 2, 2, 3, 5 in this order (the prime factors of 60)(1)
(b) Output: 97(1)
(b) 96 decisions (F takes every value from 2 to 97)(1)
(c) Change: stop testing / leave the loop once F × F > N (F greater than √N), then output N if N > 1(1)
(c) Explanation: if N had a factor larger than √N it would also have one smaller than √N, which would already have been found, so the remaining N must be prime (for 97 only about 8 decisions are needed instead of 96) (accept: test 2 then only odd F, halving the decisions, with a correct reason, for 1 mark only)(1)
Answer: (a) Rows (N, F, decision, output): (60, 2, True, 2), (30, 2, True, 2), (15, 2, False), (15, 3, True, 3), (5, 3, False), (5, 4, False), (5, 5, True, 5), then N = 1 and the loop ends. Outputs 2, 2, 3, 5. (b) 97; 96 decisions. (c) Stop when F × F > N and output N if it is still greater than 1.
Examiner tip: F only increases when the remainder is not 0, so the same factor (2) can be output more than once.

8Standard5 marks

A school's exams officer must produce the timetable for the summer exams by hand, without using any software. There are 35 subjects, 6 exam halls of different sizes and some students take up to 11 subjects.

Explain how the exams officer could apply computational thinking to solve this problem, even though no programming is involved. [5]

Decomposition: break the task into smaller sub-problems, e.g. listing which students take which subjects, allocating dates/sessions, then allocating halls(1)
Abstraction: ignore details that do not affect the timetable (e.g. teachers' names, room decoration) and keep only the essentials such as subject, number of candidates, length of paper and hall capacity(1)
Pattern recognition: notice recurring features, e.g. subjects that share many students (and so must not clash) or that this year's constraints are similar to last year's timetable, so a previous solution can be reused(1)
Algorithmic design: write an ordered set of steps / rules to follow, e.g. schedule the subjects with the most clashes first, then place each paper in the smallest hall that fits(1)
The result is a precise procedure that a person can follow on paper / a checklist, and it can be checked against the requirements (e.g. no student has two exams at once), showing that computational thinking is a problem-solving approach, not just coding(1)
Answer: Apply decomposition, abstraction, pattern recognition and algorithmic design to the timetable, producing a step-by-step manual procedure that can be checked.
Examiner tip: Each concept scores only when it is applied to the exam timetable; a list of definitions gains little.

9Standard5 marks

A teacher wants a flowchart that reads the marks of five students and outputs their mean. The flowchart contains a logic error.

(a) The marks available to be input are 8, 6, 9, 7, 10, 4, in that order. Determine the output of the flowchart as drawn. [2]

(b) Explain the logic error. [1]

(a) Six marks are read (the loop ends when COUNT = 6) so TOTAL = 44 (accept 8+6+9+7+10+4 shown)(1)
(a) Output 8.8 (do not accept 8 or 7.33)(1)
(b) The condition COUNT > 5 is only True when COUNT = 6, so the loop runs once too many / reads six marks but divides by 5 (accept: off-by-one error)(1)
(c) Change the decision to COUNT = 5 ? (accept COUNT ≥ 5 ? or COUNT > 4 ?)(1)
(c) Initialise COUNT to 1 instead of 0 (do not accept changing the average to TOTAL / COUNT, as that still reads six marks)(1)
Answer: (a) 8.8 (six marks read, TOTAL 44). (b) COUNT > 5 needs six iterations. (c) Use COUNT = 5 ? or start COUNT at 1. Corrected output: 8.0.
Examiner tip: Check the decision with the value the counter has after the increment: the loop exits only when COUNT is already 6.

10Hard6 marks

The flowchart shows a search algorithm. The array A holds the following nine values, indexed from 0:

Index	0	1	2	3	4	5	6	7	8
A[Index]	3	8	11	15	21	26	30	34	41

(a) Trace the flowchart when TARGET = 26. Copy and complete the trace table and state the output. [3]

LOW	HIGH	POS
0	8	−1

(b) Determine the output when TARGET = 9, and state how many times the decision A[MID] = TARGET ? is evaluated. [2]

(a) MID values 4, 6, 5 with A[MID] values 21, 30, 26(1)
(a) LOW becomes 5 (after 21 < 26) and HIGH becomes 5 (after 30 > 26), POS set to 5(1)
(a) Output: 5 (accept the index 5; do not accept 26 or 6)(1)
(b) Output: −1 (accept 'not found' only if −1 is stated as the value output)(1)
(b) 3 times (MID = 4, 1, 2; loop then ends because LOW = 2 > HIGH = 1)(1)
(c) The values must be sorted in ascending order (accept 'in order', 'sorted'; do not accept 'no duplicates')(1)
Answer: (a) MID 4, 6, 5; output 5. (b) −1; 3 evaluations. (c) A must be sorted in ascending order.
Examiner tip: MID uses whole-number division: (5 + 8) div 2 = 6, not 6.5 or 7.

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