Worked Examples — Lecture 06

Each example below follows the same arc: problem, analysis, algorithm, pseudocode, then Python where the lecture has introduced it. Python blocks are machine-verified; the comment lines # expect: match the actual output. Edge cases and complexity close each example - they are part of the answer, not decoration.

1. Trace the tally: loop reasoning on paper (case cs-021)

Problem. Sum the numbers 1..12, skipping multiples of 3. Predict first, then trace the loop table.

Analysis. The loop has one decision inside: include or skip. A trace table with one row per iteration (n, condition, total) turns prediction into evidence. The condition n mod 3 != 0 does the filtering.

Algorithm.

1. total <- 0
2. for n in 1..12: if n mod 3 != 0 then total <- total + n
3. return total

Pseudocode.

total <- 0 FOR n FROM 1 TO 12 IF n MOD 3 != 0 THEN total <- total + n WRITE total

Trace (dry run). Included: 1, 2, 4, 5, 7, 8, 10, 11. Running total: 1, 3, 7, 12, 19, 27, 37, 48. Skipped at n = 3, 6, 9, 12. Final 48 - matching the reference output.

Expected output. 48

Edge cases. Limit 0: the loop body never runs, total 0. A multiple of 3 exactly at the boundary (12 here) is skipped - the boundary value is the test that catches >= vs > confusion.

Complexity. O(limit). Skipped items cost a condition check but no addition - skipping is not free.

2. Nested loops, traced honestly (case cs-023)

Problem. With n = 4 rows and the rule 'star when column <= row else dot', what does the double loop print? Trace row 2 by hand.

Analysis. Outer loop: rows 1..n. Inner loop: columns 1..2r - 1 - the count grows with r, which is exactly why the trace matters. Character = star if c <= r else dot. Hand-tracing row 2: c runs 1, 2, 3, giving star, star, dot.

Algorithm.

1. for each row r in 1..n
2. line <- empty
3. for c in 1..2r - 1: append star if c <= r else dot
4. print line

Pseudocode.

FOR r FROM 1 TO n line <- "" FOR c FROM 1 TO 2r - 1 IF c <= r THEN line <- line + "" ELSE line <- line + "." WRITE line

Trace (dry run). r = 1: c = 1, giving ''. r = 2: c = 1, 2, 3, giving '.'. r = 3: c = 1..5, giving '..'. r = 4: c = 1..7, giving '****...'. The output matches the reference exactly.

Expected output. four lines: then . then .. then ****...

Edge cases. n = 0 prints nothing. The inner bound 2r - 1 is where most wrong traces go: it depends on r, so each row is longer than the last. Predict the shape before tracing.

Complexity. O(n^2) characters - the sum 1 + 3 + 5 + ... + (2n - 1) = n^2 is the doubling-grid fact from the lecture.

Python companion. The nested-grid count from today's shading exercise, executing:

count = 0
for row in range(4):
    for col in range(3):
        count += 1
print(count)
# expect: 12