N Queens Puzzle - 此解决方案中的回溯位置在哪里？

Question

While studying the well known N Queens puzzle , I came across this straightforward and easy to understand implementation in C:

#include<stdio.h>
#include<math.h>

int board[20],count;

int main()
{
 int n,i,j;
 void queen(int row,int n);

 printf(" - N Queens Problem Using Backtracking -");
 printf("\n\nEnter number of Queens:");
 scanf("%d",&n);
 queen(1,n);
 return 0;
}

//function for printing the solution
void print(int n)
{
 int i,j;
 printf("\n\nSolution %d:\n\n",++count);

 for(i=1;i<=n;++i)
  printf("\t%d",i);

 for(i=1;i<=n;++i)
 {
  printf("\n\n%d",i);
  for(j=1;j<=n;++j) //for nxn board
  {
   if(board[i]==j)
    printf("\tQ"); //queen at i,j position
   else
    printf("\t-"); //empty slot
  }
 }
}

/*funtion to check conflicts
If no conflict for desired postion returns 1 otherwise returns 0*/
int place(int row,int column)
{
 int i;
 for(i=1;i<=row-1;++i)
 {
  //checking column and digonal conflicts
  if(board[i]==column)
   return 0;
  else
   if(abs(board[i]-column)==abs(i-row))
    return 0;
 }

 return 1; //no conflicts
}

//function to check for proper positioning of queen
void queen(int row,int n)
{
 int column;
 for(column=1;column<=n;++column)
 {
  if(place(row,column))
  {
   board[row]=column; //no conflicts so place queen
   if(row==n) //dead end
    print(n); //printing the board configuration
   else //try queen with next position
    queen(row+1,n);
  }
 }
}

However, as much as most of it looks correct to me, I cannot see the backtracking in it. What am I missing?

In my opinion, in the queen() function, there should be a check after the for loop to see whether the search exhausted without success for that particular row/queen, and if so, backtrack by simply calling itself with row-1. Is this assumption correct?

Answer 1

let's get deeper look in this code:

void queen(int row,int n)
{
 int column;
 for(column=1;column<=n;++column)
 {
  if(place(row,column))
  {
   board[row]=column; //no conflicts so place queen
   if(row==n) //dead end
    print(n); //printing the board configuration
   else //try queen with next position
    queen(row+1,n);
  }
 }
}

Yes, it's backtracking . Since it'll try every possible solution candidate until the some finish condition. on some row value, for(column=1;column<=n;++column) will ensure to try every possible value of column and check if it feasible with place(row,column) , then go deeper to row +1. After finishing this, this algorithm will resume to next column.

In other word, this algorithm will print every possible solution, of n -queen.

Answer 2

The backtracking is hidden in the recursive call of the function queen() . It checks column for column by try and error and places the queen if found a hit. Then it recursively calls the function queen() with the next row and traverses this next row column for column until it finds a place where the queen cannot be beaten. And this recursive call is repeated until all queens are placed.

The main idea of backtracking is the try and error approach. If not all queens can be placed, the algorithm jumps back one tree branch and begins again with the next column.

Answer 3

The backtracking is done by the implucit return statement at the end of queens function. It comes after all columns are checked, ie the exhaustive search you mentioned.

To see it more clearly, rewrite the function to use an explicit stack data structure in lieu of the implicit call stack. Then the backtracking will be expressed ecplicitly as pop(stack) .