We place exactly one queen per row. Choosing exactly one queen per row leaves columns and two diagonal identities as the only conflicts.

The row is the recursion depth. A candidate needs three tickets—column, down diagonal, up diagonal. Take them together, explore, then return all three.

What the function promises: Return the number of solutions for n >= 0. The empty board has one empty arrangement.

def count_queens(n):
    if n < 0: raise ValueError("Negative n")
    columns, down, up = set(), set(), set()
    def place(row):
        if row == n: return 1
        total = 0
        for column in range(n):
            marks = column, row-column, row+column
            if marks[0] in columns or marks[1] in down or marks[2] in up: continue
            columns.add(marks[0]); down.add(marks[1]); up.add(marks[2])
            total += place(row+1)
            columns.remove(marks[0]); down.remove(marks[1]); up.remove(marks[2])
        return total
    return place(0)

assert count_queens(0) == 1
assert count_queens(4) == 2
assert count_queens(5) == 10

Work and storage: Worst-case exponential search; sets and recursion use O(n) active space.

When to recognize it: Choosing exactly one queen per row leaves columns and two diagonal identities as the only conflicts.

Check before applying it: Forgetting one undo corrupts sibling branches; symmetry optimizations require careful counting.

Store the chosen column per row and render it only at complete placements.