Sorting

Selection Sort

Repeatedly find the index of the smallest remaining element and swap it into the next "sorted prefix" slot. Unlike bubble sort, only one swap per pass.

Algorithm

Basic Implementation

basic.f90 
program sort_selection
    implicit none
    integer :: arr(5) = [5, 1, 4, 2, 8]
    integer :: n, i, j, min_idx, tmp
    n = 5
    do i = 1, n - 1
        min_idx = i
        do j = i + 1, n
            if (arr(j) < arr(min_idx)) then
                min_idx = j
            end if
        end do
        if (min_idx /= i) then
            tmp = arr(i)
            arr(i) = arr(min_idx)
            arr(min_idx) = tmp
        end if
    end do
    print '(*(I0,1X))', arr
end program sort_selection

The pinned input [5, 1, 4, 2, 8] sorts with two real swaps. The frames keep the running minimum and swap positions visible.

Step 1 - First scan finds 1

In the first pass, min_idx moves from 5 to 1.

First pass over [5, 1, 4, 2, 8]: 1 is the running minimum.i0i1i2i3i451428imin

Step 2 - Swap 5 and 1

The smallest value moves into the first sorted slot.

After swap: [1, 5, 4, 2, 8].i0i1i2i3i415428sorted

Step 3 - Second scan finds 2

In the unsorted suffix, 2 is smaller than 5 and becomes the next minimum.

Second pass: 2 is selected from the suffix.i0i1i2i3i415428sortedimin

Step 4 - Sorted after two swaps

Swapping 5 and 2 gives [1, 2, 4, 5, 8]; later passes find no real swap.

After the second real swap: [1, 2, 4, 5, 8].i0i1i2i3i412458sortedsorted

Complexity

  • Time: O(n^2) regardless of input order
  • Space: O(1)
  • Stable: no
  • Swaps: at most n-1

Implementation notes

  • Fortran: skip the swap when min_idx == i to match the lesson spec's frame counts. Do not delegate to a library sort.
  • The replay highlights j (scanning) versus min_idx (running minimum) distinctly, then animates the per-pass swap.
running minimum Track the index of the smallest value seen during a scan.