No. All blanks share constraints; choosing minimum remaining values exposes contradictions earlier.

Each blank asks for the intersection of three allowed sets. Pick the smallest intersection first; if it is empty, fail immediately. Every tentative digit owns three marks to undo.

What the function promises: Solve a 9×9 board in place using digits 1..9 and 0 for blanks. Return False for an invalid or unsatisfiable board.

def solve_sudoku(board):
    digits = set(range(1,10))
    rows, cols, boxes = [set() for _ in range(9)], [set() for _ in range(9)], [set() for _ in range(9)]
    blanks = []
    for r in range(9):
        for c in range(9):
            value = board[r][c]
            box = (r//3)*3+c//3
            if value == 0: blanks.append((r,c))
            elif value not in digits or value in rows[r] or value in cols[c] or value in boxes[box]:
                return False
            else: rows[r].add(value); cols[c].add(value); boxes[box].add(value)
    def search():
        if not blanks: return True
        best = min(range(len(blanks)), key=lambda i: len(digits-rows[blanks[i][0]]-cols[blanks[i][1]]-boxes[(blanks[i][0]//3)*3+blanks[i][1]//3]))
        r, c = blanks.pop(best); box = (r//3)*3+c//3
        for value in sorted(digits-rows[r]-cols[c]-boxes[box]):
            board[r][c] = value; rows[r].add(value); cols[c].add(value); boxes[box].add(value)
            if search(): return True
            rows[r].remove(value); cols[c].remove(value); boxes[box].remove(value)
        board[r][c] = 0; blanks.insert(best,(r,c))
        return False
    return search()

puzzle = [[5,3,0,0,7,0,0,0,0],[6,0,0,1,9,5,0,0,0],[0,9,8,0,0,0,0,6,0],[8,0,0,0,6,0,0,0,3],[4,0,0,8,0,3,0,0,1],[7,0,0,0,2,0,0,0,6],[0,6,0,0,0,0,2,8,0],[0,0,0,4,1,9,0,0,5],[0,0,0,0,8,0,0,7,9]]
assert solve_sudoku(puzzle)
assert all(set(row) == set(range(1,10)) for row in puzzle)

Work and storage: Exponential worst case; constraint sets use O(81) space and recursion has at most 81 blanks.

When to recognize it: All blanks share constraints; choosing minimum remaining values exposes contradictions earlier.

Check before applying it: A solver must validate the initial board and define behavior for zero or multiple solutions.

After finding one solution, continue searching until a second appears; do not stop at the first.