GCSE Computer Science Paper 1: Computational Thinking and Programming Skills (Python)

An algorithm is a sequence of steps that solves a problem or completes a task. It is not tied to any language.

Abstraction keeps only the details that matter. Decomposition breaks a problem into smaller problems.

4 is not greater than 9, so c = 9 - 4 = 5. 12 is greater than 7, so c = 12 - 7 = 5. 6 is not greater than 6, so c = 6 - 6 = 0.

Whichever number is larger, the smaller is subtracted from it, so the output is always the difference.

Goals are whole numbers, an average can have a fractional part, played or not is true or false, and a team name is a string of characters.

COMPUTER has 8 characters. Indexes start at 0, so word[2] is M, and word[1:4] is the characters at indexes 1, 2 and 3: OMP. 17 // 5 is integer division, 3, and 17 % 5 is the remainder, 2.

Indefinite iteration repeats until a condition is met, so the number of repeats is not known in advance. A FOR loop that runs 10 times is definite iteration, and IF is selection.

If the VAT rate changes, the programmer only has to update the constant in one place, rather than finding every 0.2 in the program and risking missing one.

a and not b is True, so the whole OR is True. not a is False and b is False, so not a or b is False. and is worked out before or: a and b is False, and False or c is True.

i = 1: 8 > 3, so largest becomes 8 and total becomes 8. i = 2: 2 is not larger, total becomes 10. i = 3: 7 is not larger than 8, total becomes 17.

The loop starts at index 1, so nums[0], the 3, is never added: the algorithm outputs 17 instead of 20. Setting total to nums[0] instead of 0 at the start fixes it.

A binary search discards half of the list at each step, which only works if the list is sorted.

(0 + 8) DIV 2 = 4: Fig. Lemon comes after Fig, so first becomes 5. (5 + 8) DIV 2 = 6: Kiwi. Lemon comes after Kiwi, so first becomes 7. (7 + 8) DIV 2 = 7: Lemon, found.

Every comparison in a binary search discards half of the items that are left, so even a list of a million items needs only about 20 comparisons. A linear search checks the items one at a time and may have to look at all of them.

Pass 1: 5 and 1 swap, 5 and 4 swap, 5 and 2 swap, giving 1, 4, 2, 5. Pass 2: 1 and 4 stay, 4 and 2 swap, 4 and 5 stay, giving 1, 2, 4, 5.

Merge sort is far more time efficient on large lists. It needs extra memory for the lists it merges, and it is harder to write.

calculate(3, 4) works out 3 * 2 + 4 = 10, which is returned, stored in x and printed.

a and b receive the values passed in, so they are parameters. result is created inside the subroutine, so it is local. x is assigned in the main program, so it is global.

A subroutine can be called whenever it is needed, so the code is written once and reused. Each subroutine is a small, self-contained part that can be tested on its own, or written by a different member of a team, so errors are easier to find and the program is easier to maintain.

Breaking the rules of the language is a syntax error, so the program will not run. A program that runs but gives the wrong result has a logic error.

27 is a typical valid mark, 50 is at the edge of the allowed range, and fifty is the wrong type of data, so it should be rejected.

range(1, 10) stops before 10, so 10 is never added and the total is 45 instead of 55.

for i in range(1, 11):

Check every character for a digit, then test both rules together with and, using >= 8 so that a password of exactly 8 characters counts.

password = input("Enter a password: ")
hasDigit = False
for character in password:
    if character.isdigit():
        hasDigit = True
if len(password) >= 8 and hasDigit:
    print("Strong")
else:
    print("Weak")

algorithm has 9 characters. SUBSTRING(0, 2, word) gives the characters at positions 0 to 2 inclusive: alg. r is at position 4, counting from 0. B comes one after A, so its code is 66.

Loop through the word one letter at a time, converting each to lower case, and add one to a count whenever the letter is in "aeiou".

word = input("Enter a word: ")
count = 0
for letter in word.lower():
    if letter in "aeiou":
        count = count + 1
print("Number of vowels:", count)

randint includes both of its end values, so randint(1, 6) gives 1, 2, 3, 4, 5 or 6.

A record keeps all the related data about one book together under a single name. That makes the program easier to manage: a whole book can be passed to a subroutine in one go, and the library's books can be stored in a single array of Book records.

scores[1] is the row [20, 18, 22], and item 2 of it is 22. scores[2] is [7, 11, 14], and item 0 of it is 7.

Loop through the player's row, scores[player], adding each score to a running total, then return the total.

def playerTotal(scores, player):
    total = 0
    for score in scores[player]:
        total = total + score
    return total

Keep the number of the best player so far, starting with player 0. Loop through the other players and, whenever one has a higher total, remember that player instead. Return the player number, not the total.

def playerTotal(scores, player):
    total = 0
    for score in scores[player]:
        total = total + score
    return total


def bestPlayer(scores):
    best = 0
    for player in range(1, len(scores)):
        if playerTotal(scores, player) > playerTotal(scores, best):
            best = player
    return best

Count the failed attempts in a loop that stops when the details are right or after three attempts, checking the username and the password together with and. After the loop, output one message or the other.

USERNAME = "admin"
PASSWORD = "Secret99"

attempts = 0
granted = False
while attempts < 3 and not granted:
    username = input("Username: ")
    password = input("Password: ")
    if username == USERNAME and password == PASSWORD:
        granted = True
    else:
        attempts = attempts + 1
if granted:
    print("Access granted")
else:
    print("Account locked")

Validate the hours with a while loop. Work out the charge for the first 3 hours and the hours after that separately, cap it at 15, and set it to 0 for a blue badge holder before outputting it.

hours = int(input("Number of hours parked: "))
while hours < 1 or hours > 24:
    hours = int(input("Enter a whole number of hours from 1 to 24: "))
badge = input("Blue badge (Y/N)? ")
if hours <= 3:
    charge = hours * 2.00
else:
    charge = 3 * 2.00 + (hours - 3) * 1.50
if charge > 15:
    charge = 15.00
if badge == "Y":
    charge = 0
print("Charge: £" + format(charge, ".2f"))