Day 08: Loops, loop invariants, and function inputs and outputs

Instructions

Study source: within this activity's existing reading/practice time, use Automate the Boring Stuff with Python, 3rd edition, Chapter 3; Practice Questions; Appendix B answers, then explain one connection to the local task in your own words. Keep the local exercise and acceptance checks; use its original explanation if the external source is unavailable. Spaced-review warm-up (10 minutes): from memory, predict bool(""), bool(" "), and "0" == 0, then check them. Yesterday's review makes this quick; slow answers tell you what to revisit.

New idea: a for loop runs its body once for each item; for a string, each item is one Unicode code point. A loop invariant is a sentence about your variables that is true before the first iteration and stays true after every iteration. For counting, the invariant is: total equals the number of matches among the characters visited so far.

1. Before running, fill in a trace table with columns visited / total / invariant holds?
2. Run the snippet; it prints the same columns so you can check your table.
3. Run the second snippet to see the empty case (the body runs zero times) and a code-point surprise.

text = "été"
target = "é"
total = 0
visited = ""
for ch in text:
    if ch == target:
        total += 1
    visited += ch
    print(repr(visited), total, total == visited.count(target))
print("final", total, visited == text)

text = ""
total = 0
for ch in text:
    total += 1
print("final", total)
print(len("é"), len("e\u0301"), "é" == "e\u0301")

Instructions

Study source: within this activity's existing reading/practice time, use Automate the Boring Stuff with Python, 3rd edition, Chapter 4; Practice Questions; Appendix B answers, then explain one connection to the local task in your own words. Keep the local exercise and acceptance checks; use its original explanation if the external source is unavailable. Spaced-review warm-up: from memory, write what a = [1]; b = a; a.append(2) leaves in b (Day 05), then check.

New idea: a function's contract says what inputs it expects, what it returns, and what it leaves unchanged. count_char expects text (a string) and target (exactly one code point), returns an int, and does not modify its input. return hands a value back to the caller; print only displays text and the function then returns None.

1. Predict all seven outputs of count_char, then run it.
2. Predict the two lines of the shout example, then run it.

def count_char(text, target):
    """Return how many code points in text are equal to target (one code point)."""
    total = 0
    for ch in text:
        if ch == target:
            total += 1
    return total
print(count_char("", "a"))
print(count_char("aaaa", "a"))
print(count_char("猫猫犬", "猫"))
print(count_char("café", "e"))
print(count_char("cafe\u0301", "e"))
original = "été"
print(count_char(original, "é"), original)
print(count_char("abc", "ab"))

def shout(text):
    print(text.upper())
result = shout("hi")
print(result)

Instructions

Spaced review: the Git part reuses Day 07's restore sequence with a Python file, so the commands should feel familiar.

Invariant for two variables: after each iteration, run is the length of the streak of target that ends at the character just visited, and best is the longest streak among all characters visited so far.

1. Predict the four outputs and write a trace table (ch / run / best) for "aabaaa" before running.

def longest_run(text, target):
    best = 0
    run = 0
    for ch in text:
        if ch == target:
            run += 1
        else:
            run = 0
        if run > best:
            best = run
    return best
print(longest_run("", "a"))
print(longest_run("baaab", "a"))
print(longest_run("aabaaa", "a"))
print(longest_run("猫猫犬猫", "猫"))

2. In a new temporary directory, commit the function, add an unwanted debug line, inspect it, and restore. Predict every output first:

cd "$(mktemp -d)"
git init -q
git config user.name "Learner"
git config user.email "[email protected]"
cat > stats.py <<'EOF'
def longest_run(text, target):
    best = 0
    run = 0
    for ch in text:
        if ch == target:
            run += 1
        else:
            run = 0
        if run > best:
            best = run
    return best
print(longest_run("猫猫犬猫", "猫"))
EOF
git add stats.py
git commit -q -m "Add longest_run"
python3 stats.py
printf 'print("debug", longest_run("", "a"))\n' >> stats.py
python3 stats.py
git diff --stat
git restore stats.py
git status --short
python3 stats.py
git log --format=%s

Instructions

Spaced review: the branch inside the loop reuses Day 06 conditions. Do not open the reference answer or use AI until your evidence is submitted.

1. Predict the number printed by this loop, then run it. Also predict the result if the string were empty.

total = 0
for ch in "a猫a":
    if ch != "a":
        total += 10
    else:
        total += 1
print(total)

2. Implement count_in(text, chars): return how many code points of text occur anywhere in chars (for a single character ch, the expression ch in chars is True when ch appears in chars). Write the invariant as a comment or sentence, then predict and test: count_in("", "aeiou"), count_in("banana", "an"), count_in("naïve", "aeiouï"), count_in("cafe\u0301", "é"), count_in("猫と犬", "猫犬").