Consider a 3-puzzle where, like in the usual 8-puzzle game, a tile can only…

Key idea: For a 2x2 sliding puzzle the reachability of a configuration from another is determined by the parity of (number of inversions among tiles) + (row index of the blank counted from the bottom). Configurations with the same parity are reachable from one another.

Initial configuration:

• Top row: 1, 2; bottom row: blank, 3.

Compute parity for the initial state:

• Inversions among tiles 1, 2, 3: 0.

• Blank is on the bottom row, which is row 1 from the bottom.

• Parity = 0 + 1 = 1 (odd). Any reachable configuration must have odd parity.

Now check each candidate configuration by computing (inversions) + (blank row from bottom):

1. Configuration with top row 3, 1 and bottom row blank, 2: tile order 3,1,2 has 2 inversions; blank is on bottom row (1). Sum = 2 + 1 = 3 (odd) → reachable.

2. Configuration with top row blank, 3 and bottom row 2, 1: tile order 3,2,1 has 3 inversions; blank is on the top row (row 2 from bottom). Sum = 3 + 2 = 5 (odd) → reachable.

3. Configuration with top row 1, 3 and bottom row 2, blank: tile order 1,3,2 has 1 inversion; blank is on the bottom row (1). Sum = 1 + 1 = 2 (even) → not reachable.

4. Configuration with top row blank, 2 and bottom row 1, 3: tile order 2,1,3 has 1 inversion; blank is on the top row (row 2 from bottom). Sum = 1 + 2 = 3 (odd) → reachable.

Conclusion: The only configuration that cannot be reached from the given initial state is the one with top row 1 and 3 and bottom row 2 with the blank in the bottom-right.