Programming constructs: sequence, selection, loops and functions – IB Computer Science exam questions with mark scheme

1Standard4 marks

Consider the following Python program.

total = 0
for i in range(1, 6):
    if i % 2 == 0:
        total = total + i * i
    else:
        total = total - i
print(total)

(a) Complete a trace table showing the value of total at the end of each iteration, and state the output. [3]

(b) State the output if range(1, 6) is changed to range(1, 5). [1]

i = 1 → total = −1; i = 2 → total = 3(1)
i = 3 → total = 0; i = 4 → total = 16(1)
i = 5 → total = 11, so the output is 11(1)
(b) 16(1)
Answer: (a) −1, 3, 0, 16, 11; output 11 (b) 16
Examiner tip: Odd values are subtracted, even values have their square added; check the sign on every row.

2Standard4 marks

Write a Python function count_vowels(text) that returns the number of vowels (a, e, i, o, u) in the string text. Upper-case and lower-case vowels must both be counted. For example, count_vowels("Education") returns 5.

[4]

Function header with a parameter, e.g. def count_vowels(text):(1)
Initialises a counter to 0 and loops through every character of the string (allow an equivalent built-in approach, e.g. adding text.lower().count(v) for each vowel v)(1)
Converts case (or tests both cases) and tests whether the character is a vowel, incrementing the counter if so (allow ch in "aeiouAEIOU" or a chain of ==/or tests)(1)
Returns the counter (after the loop) rather than printing it(1)

Examiner tip: Return the count after the loop ends; a return inside the loop exits on the first character.

3Core4 marks

In a program, the variables below have been assigned these values.

x = 7
y = 12
flag = False

(a) State whether each of the following conditions evaluates to True or False.

(i) x > 5 and y < 10

(ii) not flag or x == y

(iii) x != 7 or (y >= 12 and not flag) [3]

(b) A youth club accepts members aged 13 to 19 inclusive. Write a single condition, using the variable age, that is true only when age is in this range. [1]

Answers may be written in Python or Java.

(a)(i) False (x > 5 is true but y < 10 is false, so AND gives false)(1)
(a)(ii) True (not flag is true, so OR is true regardless of x == y)(1)
(a)(iii) True (x != 7 is false but y >= 12 and not flag are both true) (accept true/false in any case, e.g. TRUE)(1)
(b) age >= 13 and age <= 19 (accept 13 <= age <= 19 in Python; accept Java age >= 13 && age <= 19; accept age > 12 and age < 20 for integer ages; reject or; reject 13 <= age <= 19 in Java)(1)
Answer: (a) (i) False (ii) True (iii) True (b) age >= 13 and age <= 19
Examiner tip: In (b) students often write or instead of and, which makes the condition true for every age.

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4Core5 marks

A car park calculates the charge, in pounds, using the function below. Drivers with a blue badge park for free.

def charge(hours, blue_badge):
    if blue_badge:
        return 0
    elif hours <= 1:
        return 2
    elif hours <= 4:
        return 5
    else:
        return 5 + (hours - 4) * 3

(a) State the value returned by each of the following calls.

(i) charge(1, False)

(ii) charge(6, False)

(iii) charge(6, True) [3]

(b) A programmer swaps the order of the two conditions so that hours <= 4 is tested before hours <= 1. Explain why this would cause incorrect output. [2]

(a)(i) 2(1)
(a)(ii) 11(1)
(a)(iii) 0(1)
(b) In an if / elif (else if) chain only the first condition that is true is executed / the remaining branches are skipped(1)
(b) Any stay of 1 hour or less also satisfies hours <= 4, so it would be charged 5 instead of 2 / the hours <= 1 branch could never be reached(1)
Answer: (a) (i) 2 (ii) 11 (iii) 0 (b) Only the first true branch runs; short stays also satisfy hours <= 4 so would be charged 5, and the <= 1 branch becomes unreachable.
Examiner tip: For (a)(ii) work through the arithmetic: 5 + (6 − 4) × 3 = 11, not 18.

5Standard5 marks

The program below is meant to read exam marks until the user enters -1, then print the mean of the marks entered. The statements inside the loop are in the wrong order.

total = 0
count = 0
mark = int(input("Mark: "))
while mark != -1:
    mark = int(input("Mark: "))
    total = total + mark
    count = count + 1
print(total / count)

(a) The user enters 60, 70, 80 and then -1. State the output. [1]

(b) Explain why this output is incorrect. [2]

(d) State what would happen if the line mark = int(input("Mark: ")) inside the loop were deleted and the user entered 60 first. [1]

(a) 49.666… (accept 49.67 / 49.7 / 149/3; reject 70)(1)
(b) The first mark (60) is read before the loop but is overwritten before it is added to the total / it is never added(1)
(b) The sentinel value -1 is added to the total and counted as a mark (so total = 149 and count = 3)(1)
(c) total = total + mark, then count = count + 1, then mark = int(input("Mark: ")) as the last statement in the loop (all three in this order; the two updates may be either way round, but the input must come after them)(1)
(d) Infinite loop / the loop never ends because mark never changes and stays 60 (so the condition is always true) (ignore reference to the program crashing later)(1)
Answer: (a) 49.666… (b) 60 is never added, and the sentinel -1 is added and counted. (c) Update total and count first, then read the next mark at the end of the loop body. (d) Infinite loop: mark stays 60, so mark != -1 is always true.
Examiner tip: Say both faults in (b): the lost first value and the sentinel being processed are each worth a mark.

6Standard5 marks

Consider the following Python program.

n = 40213
result = 0
count = 0
while n > 0:
    d = n % 10
    if d % 2 == 1:
        result = result * 10 + d
    else:
        count = count + 1
    n = n // 10
print(result, count)

(a) Copy and complete the trace table below, showing the values of the variables at the end of each pass through the loop, and state the output.

d	result	count	n
	0	0	40213

[4]

(b) State the output if the first line is changed to n = 0, and explain why. [1]

(a) d takes the values 3, 1, 2, 0, 4 in that order(1)
(a) result column: 3, 31, 31, 31, 31(1)
(a) count column: 0, 0, 1, 2, 3 and n column: 4021, 402, 40, 4, 0(1)
(a) Output 31 3 (follow through from their final row of the table)(1)
(b) 0 0, because the condition n > 0 is false at the start so the loop body never executes (both value and reason needed)(1)
Answer: (a) d: 3, 1, 2, 0, 4; result: 3, 31, 31, 31, 31; count: 0, 0, 1, 2, 3; n: 4021, 402, 40, 4, 0. Output: 31 3. (b) 0 0, since the loop condition is checked before the first pass and is false.
Examiner tip: Remember 0 is even: the digit 0 increases count. Students who treat it as neither lose a mark.

7Standard5 marks

Consider the following Python program.

score = 10

def bonus(points):
    score = points * 2
    return score + 1

def add_bonus(points):
    global score
    score = score + bonus(points)

result = bonus(3)
print(score, result)
add_bonus(4)
print(score)

(a) State the output of the program. [3]

(b) Explain why the first print statement does not display 6 for score, even though bonus assigns score = points * 2. [2]

(a) First line shows score as 10(1)
(a) First line shows result as 7 (so the line is 10 7)(1)
(a) Second line is 19(1)
(b) Assigning to score inside bonus creates a local variable (with function scope) because bonus has no global declaration(1)
(b) The local score is separate from (shadows) the global score and is discarded when the function returns, so the global still holds 10(1)
Answer: (a) 10 7 then 19 (b) score inside bonus is a local variable, separate from the global one, so the global score is unchanged at 10.
Examiner tip: Track two different variables called score: only add_bonus, which declares global, changes the one printed.

8Hard6 marks

A cinema booking program has been split into functions. The structure chart shows which function calls which, and the data passed into each function (arrow pointing down) and returned from it (arrow pointing up).

Ticket prices are £9.50 per adult and £6.00 per child. If five or more tickets are bought in one booking, a 10% discount is applied to the whole cost. The function apply_discount(cost, rate) has already been written; for example, apply_discount(40.0, 0.10) returns 36.0.

(a) Outline two benefits of writing this program as separate functions rather than as one block of code. [2]

(b) Construct the function calculate_cost(adults, children), using apply_discount where needed, so that it matches the structure chart. [4]

Answers may be written in Python or Java.

(a) Any two from: each function can be tested / debugged independently; functions such as apply_discount can be reused (called many times / in other programs) without rewriting code; different programmers can work on different functions at the same time; the program is easier to read / maintain because a change (e.g. to prices) is made in one place; local variables reduce unintended side effects (one mark per benefit; two benefits needed)(1)
(a) Second distinct benefit (reject 'makes the program shorter' or 'faster' without justification)(1)
(b) Function header taking two parameters adults and children (e.g. def calculate_cost(adults, children):; Java: static double calculateCost(int adults, int children))(1)
(b) Calculates cost = adults * 9.50 + children * 6.00 (accept 9.5 and 6)(1)
(b) Condition testing adults + children >= 5 and, when true, calling apply_discount(cost, 0.10) and storing the result (accept rate 0.1; reject > 5; reject applying the discount inline instead of calling the function, as the question requires its use)(1)
(b) Returns cost (in all cases, after any discount) (reject printing instead of returning)(1)

Answer: (a) e.g. functions can be tested separately; apply_discount can be reused. (b)

def calculate_cost(adults, children):
    cost = adults * 9.50 + children * 6.00
    if adults + children >= 5:
        cost = apply_discount(cost, 0.10)
    return cost

e.g. calculate_cost(2, 3) returns 33.30.

Examiner tip: The structure chart shows cost going up to main, so the function must return it; printing it inside the function loses the final mark.

9Hard5 marks

A program runs two threads at the same time. Each thread needs exclusive use of a shared printer and a shared log file, which are protected by the locks printer_lock and file_lock. acquire() waits until the lock is free and then takes it.

# Thread A                    # Thread B
printer_lock.acquire()        file_lock.acquire()
file_lock.acquire()           printer_lock.acquire()
print_report()                write_log()
file_lock.release()           printer_lock.release()
printer_lock.release()        file_lock.release()

(a) Explain how this sequence of instructions can lead to deadlock. [3]

(b) Describe how changing the order of instructions in Thread B prevents deadlock. [2]

(a) Thread A acquires printer_lock and Thread B acquires file_lock (before either reaches its second acquire)(1)
(a) Each thread then waits for the lock held by the other (A waits for file_lock, B waits for printer_lock)(1)
(a) Neither can continue to its release statements, so both wait forever / the program halts with no progress (accept 'circular wait')(1)
(b) Thread B should acquire printer_lock first and file_lock second / both threads acquire the locks in the same order(1)
(b) Whichever thread gets printer_lock first can then also get file_lock; the other thread waits for printer_lock while holding nothing, so it can continue once the locks are released (reject 'remove the locks', which allows both threads to use the resources at once)(1)
Answer: (a) A holds the printer lock and B holds the file lock; each waits for the other's lock, so neither reaches its release and both wait forever. (b) Make B acquire printer_lock then file_lock, the same order as A, so the second thread waits while holding nothing.
Examiner tip: Deadlock needs each thread to be holding one resource while waiting for the other's; describe both halves, not just 'they wait for each other'.

10Stretch6 marks

Consider the following Python function.

def grid(n):
    total = 0
    for i in range(1, n + 1):
        line = ""
        for j in range(i, n + 1):
            if (i * j) % 2 == 0:
                line = line + "#"
                total = total + 1
            else:
                line = line + "."
        print(line)
    return total

(a) State the output printed by the call grid(4). [2]

(b) State the value returned by grid(4). [1]

(d) Rewrite the inner for loop as a conditional (while) loop that produces the same result. [2]

(a) Four lines printed, of lengths 4, 3, 2, 1 (in that order)(1)
(a) Exactly .#.#, ###, .#, #(1)
(b) 7(1)
(c) $\frac{n(n+1)}{2}$ (accept $1 + 2 + \dots + n$ or equivalent; accept $\frac{n^2+n}{2}$)(1)
(d) j initialised to i before the loop and loop condition while j <= n: (accept j < n + 1)(1)
(d) Same if/else body inside, and j = j + 1 as the last statement of the loop body (reject an increment placed so that the first value of j is skipped)(1)

Answer: (a) .#.# / ### / .# / # (b) 7 (c) n(n+1)/2 (d)

j = i
while j <= n:
    if (i * j) % 2 == 0:
        line = line + "#"
        total = total + 1
    else:
        line = line + "."
    j = j + 1

Examiner tip: The inner loop starts at i, not 1, so each line is one character shorter; also remember range(i, n + 1) stops at n, not n + 1.

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