Algorithm of N queens

问题:

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?

回答1:

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.

回答2:

I have code for this without using backtracking, but the nice thing is, it is giving time complexity of big-oh(n).

// when n is even...
for(j=1;j<=n/2;j++)
{
    x[j]=2*j;
};
i=1;
for(j=n/2 +1 ;j<=n;j++)
{
    x[j] =i;
    i=(2*i)+1;
}

// when n is odd..
i=0;
for(j=1;j<=(n/2+1);j++)
{
    x[i] = (2*i)+1;
    i++;
}
i=1;
for(j=(n/2+2);j<=n;j++)
{
    x[j] = 2*i;
    i++;
}

This code works well and gives one solution, but now I am in search of getting all possible solutions using this algorithm.