How to check if a 8-puzzle is solvable? I have a 8-puzzle
1|2|3
-+-+-
4|5|6
-+-+-
 |8|7

How can be checked if the puzzle is solvable?
Wikipedia states that it is solvable, but does not prove it. Can anybody explain the prove?
 A: If you ignore the gap and just look at the ordered sequence of numbers, any "horizontal" move leaves the sequence unchanged and any "vertical" move of the puzzle has the form
$$(\ldots, x, y, z, \ldots)\to (\ldots, y, z, x, \ldots)$$
or vice versa.
These are even permutations and therefore the group of possible permutations is a subgroup of $A_8$ (alternating group) and cannot be all of $S_8$ (full symmetric group).
The situation in your post corresponds to an odd permutation of the target ordering and therefore is not solvable.
A: Though It's old question but I am trying to answer it.
There is a method to check whether the given state is solvable or not.
Problem State:

1|2|3
-+-+-
4|5|6
-+-+-
 |8|7

Write it in a linear way,
1,2,3,4,5,6,8,7 - Ignore the blank tile
Now find the number of inversion, by counting tiles precedes the another tile with lower number.
In our case,
1,2,3,4,5,6,7 is having 0 inversions, 
and 8 is having 1 inversion as it's preceding the number 7.
Total number of inversion is 1 (odd number) so the puzzle is insolvable.
Let's take another example,

5|2|8
-+-+-
4|1|7
-+-+-
 |3|6

5 precedes 1,2,3,4 - 4 inversions
2 precedes 1 - 1 inversion
8 precedes 1,3,4,6,7 - 5 inversions
4 precedes 1,3 - 2 inversions
1 precedes none - 0 inversions
7 precedes 3,4 - 2 inversions
3 precedes none - 0 inversions
6 precedes none - 0 inversions
total inversions 4+1+5+2+0+2+0+0 = 14 (Even Number) So this puzzle is solvable.
Here is a function which I have used in Flash Game to check whether the given puzzle is solvable or not 
    public function checkSolvable(pList:Array):Boolean{
        trace("checkSolvable called with : " + pList);
        var inversions:Number = 0;

        for(var i:int=0;i<pList.length;i++){
            for(var j:int=i+1;j<pList.length;j++){
                if(pList[j]>pList[i]){
                    inversions++;
                }
            }
        }

        if(inversions%2 == 1){
            trace("It's Unsolvable");
            return false;
        }else{
            trace("It's Solvable");
            return true;
        }
    }

Visit http://www.cs.bham.ac.uk/~mdr/teaching/modules04/java2/TilesSolvability.html for detailed information.
A: Late answer, I know, but I'm expanding on Hagen von Eitzen's answer in a slightly more elementary way, if it's still of interest.
Short answer: This state is not solvable.
First note that every permutation can be represented as a graph of disjoint cycles (see cycle notation). In the usual way, then, we represent a game state as a permutation of the 8 non-blank tiles, flattened to row major order.
Now, we can show that the parity (oddness/evenness) of the number of cycles is invariant under the sliding of the tile. To see why, we only need to consider vertical moves, because horizontal moves preserve the permutation. Here, notice that exactly $3$ elements $x, y, z$ change position to $z, x, y$ or $y, z, x$. The two cases are analogous- just reverse the arrows in the cycle- so we'll just deal with the $z, x, y$ case here. We proceed by focusing on the predecessors of $x, y, z$ in their respective cycles. Now there are $3$ cases to consider:


*

*Same cycle: All elements are in the same cycle. In this case, exactly one cycle remains in the place of the old cycle, so the total number of cycles is preserved. See the image below for this case. Green circles represent predecessors, blue circles represent $x, y, z$. Dotted lines from blue to green denote a sequence of 0 or more elements, including the green i.e. the blue might equal the next green. Similar pictures for the other cases can be drawn.





*

*Two cycles: In this case, there is one cycle containing exactly $2$ elements, and one cycle containing $1$ element. Here, the swap generates $2$ new cycles in place of the old $2$ cycles. So again, the total number of cycles is preserved.

*Three cycles: In this case, all three cycles merge into one cycle. So the total number of cycles decreases by $2$, preserving the parity of the total number of cycles.
Since parity is preserved in each case, it is always preserved. Hence you cannot get from a permutation with an odd number of cycles (like the one in the question) to one with an even number of cycles, by only sliding the blank tile.
