Exam Pseudocode to Python

Reading exam pseudocode

OCR prints algorithms in its Exam Reference Language (ERL). AQA prints its own pseudocode. Edexcel's Paper 2 is sat on a computer, in Python itself. The ideas are identical. Only the spelling differs, and the exam is easier once you can translate without thinking.

The loop that costs the most marks

# for i = 1 to 5
#     print(i)
# next i
for i in range(1, 6):
    print(i)              # 1 2 3 4 5, one per line

A step is the third value in range:

# for i = 0 to 20 step 5
for i in range(0, 21, 5):
    print(i, end=" ")     # 0 5 10 15 20
print()

Counting down needs a negative step, and the stop value is still one past the end, in the direction you are going:

# for i = 10 to 1 step -3
for i in range(10, 0, -3):
    print(i, end=" ")     # 10 7 4 1
print()

Loop, test last

Python has no do ... until. The body must run at least once and the test comes at the bottom, so you build it from a loop that never ends on its own and a break:

# do
#     total = total + 7
# until total > 30
total = 0
while True:
    total = total + 7
    if total > 30:
        break
print(total)              # 35

The condition is the one you stop on. until total > 30 becomes if total > 30: break. Do not flip it into the loop header unless you also set up the variable first:

total = 7
while total <= 30:
    total = total + 7
print(total)              # 35

That version gives the same answer, but only because the variable was primed to the value the first pass would have produced. The while True form is safer, because it cannot be wrong that way.

Selection with many branches

switch day:
    case "Sat":
        print("Weekend")
    case "Sun":
        print("Weekend")
    default:
        print("Weekday")
endswitch

day = "Sat"
if day == "Sat" or day == "Sun":
    print("Weekend")
else:
    print("Weekday")

Python 3.10 added match, which reads more like the exam's switch. The | means "or" and _ is the default:

day = "Sat"
match day:
    case "Sat" | "Sun":
        print("Weekend")
    case _:
        print("Weekday")

Data types, casting and operators

The exam names five data types. Python has four of them, because it has no separate character type.

print(type(7), type(7.0), type(True), type("7"), type("A"))
# <class 'int'> <class 'float'> <class 'bool'> <class 'str'> <class 'str'>

Casting

Casting converts a value from one type to another. input always gives back a string, so anything you want to calculate with must be cast first.

age_text = "15"
print(int(age_text) + 1)     # 16
print(str(15) + "!")         # 15!
print(float("2.5") * 2)      # 5.0
print(int(7.9))              # 7
print(round(7.9))            # 8

int(7.9) throws the decimal part away, giving 7. round(7.9) rounds, giving 8. They are different operations, and an exam question about a price or an average will care which one you meant.

print(bool(0), bool(1), bool(""), bool("False"))     # False True False True

MOD, DIV and exponent

print(17 / 5, 17 // 5, 17 % 5)     # 3.4 3 2
print(2 ** 3)                      # 8
print(2 ^ 3)                       # 1, which is XOR

MOD and DIV are exam favourites because one pair of lines splits a quantity into units:

total_seconds = 3725
hours = total_seconds // 3600
minutes = (total_seconds % 3600) // 60
seconds = total_seconds % 60
print(hours, "h", minutes, "m", seconds, "s")     # 1 h 2 m 5 s

Learn the five uses of MOD, because the question rarely says "use MOD":

Boolean operators

not binds tightest, then and, then or. Brackets make the intention clear and change the result:

a, b, c = True, False, False
print(a or b and c)         # True, because and goes first: a or (b and c)
print((a or b) and c)       # False
print(not a or b)           # False, because not goes first: (not a) or b

Text handling

word = "Computing"
print(len(word))                     # 9
print(word[2:5])                     # mpu
print(word[:3], word[-3:])           # Com ing
print(word.upper(), word.lower())    # COMPUTING computing
print(ord("A"), chr(66))             # 65 B

Putting ord, chr and MOD together gives the classic Caesar shift, where % 26 is what makes Y wrap round to B:

def shift_letter(letter, shift):
    position = ord(letter) - ord("A")
    new_position = (position + shift) % 26
    return chr(new_position + ord("A"))

print(shift_letter("C", 3), shift_letter("Y", 3))     # F B

Translating whole algorithms

The exam gives you pseudocode to read, to trace, to correct or to complete, and asks you to write your own. In a written paper, an answer in Python is accepted if the logic is unambiguous. In Edexcel Paper 2 the code has to run.

1. A count loop with a running total

total = 0
for i = 1 to 5
    mark = input("Enter mark")
    total = total + int(mark)
next i
average = total / 5
print("Average: " + str(average))

The program below is the same algorithm. A list stands in for the keyboard so that it gives the same output every time you run it.

marks = [60, 72, 85, 41, 90]
total = 0
for i in range(1, 6):
    mark = marks[i - 1]
    total = total + mark
average = total / 5
print("Average: " + str(average))     # Average: 69.6

2. Selection with a compound condition

if mark >= 80 AND attendance >= 90 then
    print("Distinction")
elseif mark >= 50 then
    print("Pass")
else
    print("Fail")
endif

def result(mark, attendance):
    if mark >= 80 and attendance >= 90:
        return "Distinction"
    elif mark >= 50:
        return "Pass"
    else:
        return "Fail"

print(result(85, 95), result(85, 70), result(30, 100))     # Distinction Pass Fail

The second call shows why the order matters. A mark of 85 with 70 per cent attendance fails the first test, so it falls through to the second and earns a Pass.

3. A loop that tests first

secret = 7
guess = 0
tries = 0
while guess != secret
    guess = int(input("Guess"))
    tries = tries + 1
endwhile
print("Found in " + str(tries))

feed = iter([3, 9, 7])
secret = 7
guess = 0
tries = 0
while guess != secret:
    guess = next(feed)
    tries = tries + 1
print("Found in " + str(tries))     # Found in 3

guess is set to 0 before the loop so that the first test has something to compare. A test-first loop needs its variable to exist, and to start at a value that lets the loop begin.

4. Random numbers

The exam's random(1, 6) includes both ends. So does Python's randint, which is unusual, because range does not.

import random
roll = random.randint(1, 6)
print(1 <= roll <= 6)     # True, whichever number came up

The translation checklist

1. Loops. Add one to the stop value of a count loop. Turn until into if ... break.
2. Operators. DIV is //, MOD is %, ^ is **.
3. Text. substring(start, length) is [start:start + length]. ASC is ord and CHR is chr.
4. Types. input gives a string. Cast it before comparing or calculating.
5. Equality. = assigns and == compares, in both the exam and Python.

Key terms you must be able to define

Pseudocode

A way of writing an algorithm in structured, language-neutral English so it can be read without knowing a programming language.

Casting

Converting a value from one data type to another, such as int("7") or str(7).

MOD

The remainder after whole number division. 17 MOD 5 is 2. Written % in Python.

DIV

Whole number division, discarding the remainder. 17 DIV 5 is 3. Written // in Python.

Count-controlled loop

A loop that repeats a set number of times, such as for i = 1 to 5.

Condition-controlled loop

A loop that repeats until or while a condition is met, such as while or do...until.

Concatenation

Joining two strings end to end, using + in the exam and in Python.

Substring

A part of a string, taken by position and length.

ASCII code

The number that stands for a character. ord gives it and chr turns it back into a character.

Flashcards

Say the definition out loud before you flip the card. Retrieving it from memory is what builds recall, reading it again is not.

Reach the top grade on this topic

Full marks on translation questions come from checking the five places where the notations disagree.

• Add one to the stop value of every count loop, and say why in a comment
• Turn do...until into while True with a break on the exit condition
• Know that MOD is the remainder, DIV drops it, and ^ is not power in Python
• Cast the result of input before comparing or calculating with it
• Write the loop for a total or a count rather than reaching for sum or len

Mistakes that cost marks here

total = 0
for i = 1 to 10
    total = total + i * i
next i
print(total)

total = 0
for i in range(1, 11):
    total = total + i * i
print(total)

age = int(input("Age: "))
if age >= 18:
    print("Adult")

do
    guess = int(input("Guess: "))
until guess == 7

while True:
    guess = int(input("Guess: "))
    if guess == 7:
        break