Binary Search (Iterative)

On a sorted array, narrow the [lo, hi] window by halving it each step until arr(mid) equals the target or the window is empty. Demonstrates the "discard half the search space" invariant.

Algorithm

Canonical found run uses target = 11 on arr = [1, 3, 5, 7, 9, 11, 13]; it returns index 6 (1-based) after two midpoint probes.

Basic Implementation

basic.f90

program binary_search
    implicit none
    integer :: arr(7) = [1, 3, 5, 7, 9, 11, 13]
    integer :: target, lo, hi, mid, result
    target = 11
    lo = 1
    hi = 7
    result = -1
    do while (lo <= hi)
        mid = lo + (hi - lo) / 2
        if (arr(mid) == target) then
            result = mid
            exit
        end if
        if (arr(mid) < target) then
            lo = mid + 1
        else
            hi = mid - 1
        end if
    end do
    print '(I0)', result
end program binary_search

Complexity

Time: O(log n)
Space: O(1)

Implementation notes

Fortran: integer division on positive (hi - lo) is already truncated, so (hi - lo) / 2 is the midpoint offset. Do not call a library bisect; the goal here is zero external dependencies.
The replay highlights the [lo, hi] window, mid, and which branch (left half / right half / match) the step takes.

inclusive bounds `lo` and `hi` are both inclusive; the loop runs while `lo <= hi`.

overflow-safe midpoint `mid = lo + (hi - lo) / 2` avoids `(lo + hi)` overflow on large inputs.