If I successfully arrange all the blocks of every row of a sliding puzzle except the last row in proper order, will the n-th row be automatically resolved? (Assuming that the puzzle is solvable in the 1st place, and the hole is in the last row).
The answer is
In a 2x2 pattern, if you solve a row then you automatically solved the other too. Same for a 3x3 pattern, assuming that the hole is positioned in the last row, because swapping two blocks is an impossible move (it has been proven, only half configurations are solvable, as stated here). In a 4x4 or bigger pattern, there are valid permutations for the last row, so you can't assure that solving the other rows automatically completes the game.
As mentioned before. If you've done these kind of puzzles before a common strategy is to solve row by row but the final two rows together just because it won't always work out.
Consider this bottom right segment of any puzzle larger than 2x4 in correct order:
you can now do this to make the top row still correct but the bottom not.
1234 567 234 1567 2 34 1567 2534 1 67 2534 167 253 1674 253 1674 1253 674 1253 67 4 12 3 6754 1234 675
The Answer is
No it's not automatically resolved (for N>3)
We can minimize the Problem to "can we mix all solved NxN sliding puzzle (with N>3) that way that the last row with the hole is not sorted!" and there is a way.
To make things easy: We shift the buttom row so that the hole is in the last column
We only look at the bottom 4x2 (the rest stays sorted):
a = (n-3,n-1)
b = (n-2, n-1)
c = (n-1, n-1)
d = (n, n-1)
e = (n-3,n)
f = (n-2, n)
g = (n-1,n)
_ = hole(n,n)
Lets begin sliding:
And that's it.
Now the n-1 row is sorted again and the n row is not. So in conclusion there is a way to slide the puzzle that n-1 rows are sorted but the last is not. Perhaps even fewer moves are necessary but that wasn't the question.
edited: I needed more than 23 Minutes to write this??? I didn't saw Ivo's solution...