Coding in the Ontario math curriculum, Grades 1 to 8

# Grade 4 band: a nested loop draws an array (ours)
rows = 3
columns = 4
dots = 0
for r in range(rows):
    line = ""
    for c in range(columns):
        line = line + "o "
        dots = dots + 1
    print(line)
print(rows, "rows of", columns, "is", dots, "dots")

Output: three rows of o o o o, then 3 rows of 4 is 12 dots. The inner loop runs four times for every one turn of the outer loop, which is multiplication made visible: change rows to 5 and the program prints 20 dots.

# Grade 5 band: is a number in the pattern 5, 9, 13, ...? (ours)
target = 47
term = 5
found = False
while term <= target:
    if term == target:
        found = True
    term = term + 4
if found:
    print(target, "is in the pattern")
else:
    print(target, "is not in the pattern")

Output: 47 is not in the pattern. The terms go 5, 9, 13 and on to 45, then 49, which jumps past 47. The if asks a yes-or-no question on every term, which is the binary logic the Grade 5 notes describe.

# Grade 6 band: two ways to add 1 + 2 + ... + 100 (ours)
steps = 0
total = 0
for n in range(1, 101):
    total = total + n
    steps = steps + 1
print("Loop:", total, "after", steps, "additions")

last = 100
quick = last * (last + 1) // 2
print("Pairing rule:", quick, "after 1 calculation")

Output: Loop: 5050 after 100 additions, then Pairing rule: 5050 after 1 calculation. Pair 1 with 100, 2 with 99, and so on: 50 pairs of 101. Same answer, one calculation instead of a hundred, which is what C3.2 means by the efficiency of the code.

# Grade 7 band: a subprogram with a defined count (ours)
def doublings_past(start, limit):
    """How many times must start double before it is more than limit?"""
    count = 0
    value = start
    while value <= limit:
        value = value * 2
        count = count + 1
    return count

print(doublings_past(3, 100))
print(doublings_past(5, 1000))
print(doublings_past(1, 1))

Output: 6, 8 and 1. Starting from 3, five doublings reach 96 and the sixth jumps to 192, past 100; 5 needs eight to pass 1000; 1 needs one. The loop runs until a condition is met, not a fixed number of times, and the function is written once and called three times.

# Grade 8 band: use data to make a decision (ours)
rain_mm = [12, 0, 4, 0, 0, 7, 0]   # one week of rain in the school garden

def mean(values):
    return sum(values) / len(values)

week_mean = mean(rain_mm)
spread = max(rain_mm) - min(rain_mm)
dry_days = 0
for r in rain_mm:
    if r == 0:
        dry_days = dry_days + 1

print("Mean:", round(week_mean, 1), "mm a day")
print("Range:", spread, "mm")
print("Dry days:", dry_days)
if dry_days >= 4:
    print("Decision: water the garden on dry days")
else:
    print("Decision: no extra watering")

Output: Mean: 3.3 mm a day, Range: 12 mm, Dry days: 4, then Decision: water the garden on dry days. The mean alone would hide the dry spell; the count of dry days is what drives the decision. Asking which measure to trust is the Grade 8 skill.