ITPEC FE Subject B April 2024 Question 11

A: i + 1, B: <, C: data[i] — each pass scans the unsorted part for its smallest value and swaps it into the first unsorted slot.

On pass i, positions 1 to i - 1 already hold the smallest values in order. minPos starts at i, so the scan only needs the rest of the array: j runs from i + 1 to size (blank A). Ascending order needs the minimum, so a value takes over when data[j] < data[minPos] (blank B). The minimum belongs at position i, so the three-line swap exchanges data[i] with data[minPos] (blank C).

Trace on {12, 11, 13, 5, 6}:

• •i = 1: minimum 5 at position 4, swap → {5, 11, 13, 12, 6}
• •i = 2: minimum 6 at position 5, swap → {5, 6, 13, 12, 11}
• •i = 3: minimum 11 at position 5, swap → {5, 6, 11, 12, 13}
• •i = 4: minimum 12 is already at position 4, so the array is sorted

Answer (e)

Why not others:

• •(a), (b), (c) and (d) — A as 1 lets the scan revisit the sorted prefix, where the smallest value always wins; in (a), pass 2 finds data[1] = 5 and swaps it back out, giving {11, 5, 13, 12, 6}
• •(c), (d), (g) and (h) — B as > selects the maximum on each pass, which sorts in descending order
• •(b), (d), (f) and (h) — C as data[i + 1] puts the minimum into position i + 1 and never fills position i: pass 1 of (f) gives {12, 5, 13, 11, 6}, leaving 12 in front

Key rule: In selection sort, pass i searches only positions i + 1 to the end and swaps the winner into position i; the comparison operator alone decides ascending or descending order.

Sorting · AI-written, may contain errors