Algorithm NQueens ( k, n) //Prints all Solution to the n-queens problem
{
for i := 1 to n do
{
if Place (k, i) then
{
x[k] := i;
if ( k = n) then write ( x [1 : n]
else NQueens ( k+1, n);
}
}
}
Algorithm Place (k, i)
{
for j := 1 to k-1 do
if (( x[ j ] = // in the same column
or (Abs( x [ j ] - i) =Abs ( j – k ))) // or in the same diagonal
then return false;
return true;
}
The above code is for solving N Queens problem using backtracking.I think that it can place the first 2 queens of two rows in respective columns and then when it comes to 3rd row queen it can't be placed as no queen needs to be attacking and it will simply exit from Algorithm N queens...So how is this algorithm implements backtracking?
The secret here is the recursion.
Let each level of indentation below indicate a level of recursion.
(not what will actually happen, as the third queen can easily be placed, but it just would've taken quite a bit more writing and/or thinking to get to a case that will actually fail)
try to place first queen
success
try to place second queen
success
try to place third queen
fail
try to place second queen in another position
success
try to place third queen
success
try to place fourth queen
Something more in line with what the code actually does: (still not what will actually happen)
first queen
i = 1
Can place? Yes. Cool, recurse.
second queen
i = 1
Can place? No.
i = 2
Can place? No.
i = 3
Can place? Yes. Cool, recurse.
third queen
i = 1
Can place? No.
i = 2
Can place? No.
... (can be placed at no position)
fail
back to second queen
i = 4
Can place? Yes. Cool, recurse.
third queen
i = 1
Can place? No.
...
I hope that helps.