Computer Programming 11 and 12 in BC, and Computer Science 11

Sources: BC Applied Design, Skills, and Technologies and BC Mathematics course lists, and the course pages linked below, in copies fetched raw on 22 October 2026. A second visit that day could not connect to the curriculum site, so the Computer Science 12 content page was never read and nothing on this page describes it. Neither the ministry nor any school district has a connection with Modern Age Coders.

Sources: Computer Programming 11 and Computer Programming 12 on curriculum.gov.bc.ca, research copies fetched raw on 22 October 2026. The Grade 12 page repeats the three big ideas of Grade 11 word for word.

1. Write the requirements first. Input: a 24-hour time as HH:MM in BC. Output: the time in India and whether it is the same day or the next. BC has been on UTC-7 all year since March 2026 and India is on UTC+5:30, so the gap is 12 hours 30 minutes, always.
2. Anticipate. List what a user could type wrongly: a missing colon, single-digit minutes, letters, an hour of 24, minutes of 60.
3. Detect. Check the shape of the input, then the ranges. Let Python’s own int() detect letters, because it already raises an error for them.
4. Resolve. Catch each error where the program is used and print a message that tells the user how to fix it, instead of a traceback.
5. Fill in the expected column before running. Then run every case and compare.

# Pacific time to India time, with error handling (ours)
# BC is UTC-7 all year from 8 March 2026; India is UTC+5:30, so India is 12 h 30 min ahead.
GAP_MINUTES = 12 * 60 + 30

def parse_clock(text):
    """Turn 'HH:MM' (24-hour) into minutes after midnight, or raise ValueError."""
    parts = text.strip().split(":")
    if len(parts) != 2 or len(parts[1]) != 2:
        raise ValueError("use the form HH:MM, for example 16:30")
    hours, minutes = int(parts[0]), int(parts[1])   # int() raises ValueError on letters
    if not (0 <= hours <= 23 and 0 <= minutes <= 59):
        raise ValueError("hours run 0 to 23 and minutes 0 to 59")
    return hours * 60 + minutes

def to_india(text):
    total = parse_clock(text) + GAP_MINUTES
    day = "next day" if total >= 24 * 60 else "same day"
    total %= 24 * 60
    return f"{total // 60:02d}:{total % 60:02d} ({day})"

for entry in ["16:30", "09:00", "11:29", "11:30", "23:59", "24:00", "7:5", "ab:30", "four"]:
    try:
        print(entry, "->", to_india(entry))
    except ValueError as err:
        print(entry, "-> rejected:", err)

1. 0b1101 11 0xff. Thirteen is 8 + 4 + 1, so binary 1101, and Python prefixes binary with 0b. The string 1011 read in base 2 is 8 + 2 + 1 = 11. Two hundred fifty-five is 15 × 16 + 15, so hexadecimal ff, prefixed 0x.
2. 2 5. The count inside bump is a local variable, created by the assignment on its first line. It starts at 1, becomes 2 and is returned. The global count outside was never touched, so it is still 5: local versus global, exactly as the CS 11 content names it.
3. The two columns are identical: True, True, True, False for the rows FF, FT, TF, TT. That is one of De Morgan’s laws: not (a and b) equals (not a) or (not b). The other law gives not (raining or windy) = (not raining) and (not windy).
4. 4. When i is 1, j takes only 0. When i is 2, j takes 0 and 1, adding 1. When i is 3, j takes 0, 1 and 2, adding 3. The total is 0 + 1 + 3 = 4. The inner loop runs i times, because range(i) stops before i.
5. For 31: indexes 3, then 5. The first middle is (0 + 6) // 2 = 3, holding 19, which is too small, so lo becomes 4; the next middle is (4 + 6) // 2 = 5, holding 31. For 10: indexes 3, 1, 2, then lo passes hi and the search reports not found. A linear search checks 3, 8, 12, 19, 25 and 31: six checks to find 31.
6. It prints 18. The line a = b overwrites a before the remainder is taken, so a % b becomes 18 % 18 = 0 and the loop ends at once. Both new values must come from the old pair: a, b = b, a % b. Then 48, 18 becomes 18, 12, then 12, 6, then 6, 0, and the answer is 6. That is Euclid’s algorithm.
7. It prints [4, -5, 7]. Removing an item while looping over the same list shifts everything after it one place left, so the loop skips the item that slid into the removed one’s place. After -2 is removed, -5 slides into its index and is never examined. Build a new list instead: [t for t in temps if t >= 0].
8. 12 and 42 (see the code below). The least common multiple is a × b divided by the greatest common factor. Dividing first keeps the intermediate number small, which matters little in Python but prevents overflow in Java with large ints.
9. See the code below. Our six tests and results: V6B 1A1 True; K1A 0B1 True; v6b 1a1 False (lower case); V6B1A1 False (no space); V6B 1AA False (letter where a digit belongs); the empty string False. The length check comes first so that a short string never reaches the loop.
10. One good set: (1) store member names and meeting dates; (2) for each meeting, record present or absent for every member; (3) a member is reported when they are absent at three meetings in a row, in date order; (4) the report lists each such member once, with the dates missed. For the hour: the driver codes the data entry while the navigator writes the test table for requirement 3, then they swap. Writing requirement 3 precisely (is “three in a row” three meetings or three weeks?) is the real lesson.

def gcf(a, b):
    while b != 0:
        a, b = b, a % b      # both new values come from the old pair
    return a

def lcm(a, b):
    return a // gcf(a, b) * b   # divide first so the product stays small

print(gcf(48, 18), lcm(4, 6), lcm(21, 6))   # 6 12 42

def is_postal_code(text):
    """True for letter digit letter, space, digit letter digit, e.g. V6B 1A1."""
    if len(text) != 7 or text[3] != " ":
        return False
    pattern = "LDL DLD"
    for ch, kind in zip(text, pattern):
        if kind == "L" and not (ch.isalpha() and ch.isupper()):
            return False
        if kind == "D" and not ch.isdigit():
            return False
    return True

Sources: the UBC Computer Science (Vancouver) program page and UBC’s requirements tool (Science, applicants from British Columbia), its response kept as a raw file on 22 October 2026; CEMC’s CCC page and How contests work, read 22 October 2026. We are not connected with UBC or CEMC.