8 Queens problem

This is a discussion on 8 Queens problem within the C++ Programming forums, part of the General Programming Boards category; i have a homework my homework is to place the remaining 7 queens pieces from the game of chess on ...

  1. #1
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    8 Queens problem

    i have a homework
    my homework is to place the remaining 7 queens pieces from the game of chess
    on a chessboard so that no queen piece is threatening another queen on the board after user had placed the 1st queen on the position of his choice.

    I thought declaring an array like that int queen[8][8] and initially put 0 in entire array thenand put 1 for queen and put 2 to threatining place into this array.

    I wrote this code:

    Code:
    int a,b,check,queen[8][8],A,B,i,j;
    for (i=0;a<=7;a++)
    	{
    		for (i=0;b<=7;b++)
    		{
    			queen[a][b]=0;
    		}
    	}
    a=3;     //suppose user input these co-ordinates
    b=5;
    queen[a][b]=1;
    	//loops for user placed queen
    for (i=a+1;i<=7;i++)   //to lock the corresponding lower column
    	queen[i][b]=2;
    for (i=a-1;i>=0;i--) 	//to lock the corresponding upper column
    	queen[i][b]=2;
    for (i=a-1,j=b+1;i>=0,j<=7;i--,j++)  //to lock the corresponding upper right diagonal
    	queen[i][j]=2;
    for (i=a-1,j=b-1;i>=0,j>=0;i--,j--)    //to lock the corresponding upper left diagonal
    	queen[i][j]=2;
    for (i=a+1,j=b+1;i<=7,j<=7;i++,j++)    //to lock the corresponding lower right diagonal
    	queen[i][j]=2;
    for (i=a+1,j=b-1;i<=7,j>=0;i++,j--)    //to lock the corresponding lower left diagonal
    	queen[i][j]=2;
    
    for (i=0;i<=7;i++)
    {
            for (j=0;j<=7;j++)
            {
                         if (queen[i][j]==0)
                         {
                                      queen[i][j]=1;
                                      // and locked the corresponding column, rows, diagonals for this.
    
                         }
            }
    }
    But the problem is I can't back track to remove the queen if there's no place left for it to place in a particular row.

    I tried to to this but it only increased the size of of my program and still no results.

    Could someone please help me...???

  2. #2
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    I tried to make sense of your code but I had a hard time, so I went ahead and tried to make it look better, I may misunderstand your intentions in my comments so feel free to harp on me if I did.

    Now, what I don't understand is what the problem is?

    Code:
    // Commented and cleared for the sake of clarity
    
    int a, b;		// The input variables, will store the location of the user-defined queen
    int check;		// Declared but not used so far
    int queen[8][8];	// The array that holds the values of where a queen is and can be placed
    int A, B;		// Declared but not used so far
    int i, j;		// Counters
    
    for (i=0;a<=7;a++)			// Loop that runs while a = {0, 1, 2, ..., 7}
    	{
    		for (i=0;b<=7;b++)	// Loop that runs while b = {0, 1, 2, ..., 7}
    		{
    			queen[a][b]=0;	// Initializing the "chest board" array
    		}
    	}
    
    a=3;			//suppose user input these co-ordinates
    b=5;
    
    // If queen[x][y] == 0 then a queen can be placed
    // If queen[x][y] == 1 then a queen is already placed there
    // If queen[x][y] == 2 then a queen can not be placed there, due to risk of death by being in line with another queen
    
    // Set the user-defined queen
    queen[a][b]=1;
    
    /*****************************************************************/
    /*--------NOTE: a corresponds to x and b corresponds to y--------*/
    /*****************************************************************/
    
    // The below is for the user-placed queen --------
    
    for (i=a+1;i<=7;i++)				//to lock the corresponding lower column
    	queen[i][b]=2;
    for (i=a-1;i>=0;i--)				//to lock the corresponding upper column
    	queen[i][b]=2;
    for (i=a-1,j=b+1;i>=0,j<=7;i--,j++)		//to lock the corresponding upper right diagonal
    	queen[i][j]=2;
    for (i=a-1,j=b-1;i>=0,j>=0;i--,j--)		//to lock the corresponding upper left diagonal
    	queen[i][j]=2;
    for (i=a+1,j=b+1;i<=7,j<=7;i++,j++)		//to lock the corresponding lower right diagonal
    	queen[i][j]=2;
    for (i=a+1,j=b-1;i<=7,j>=0;i++,j--)		//to lock the corresponding lower left diagonal
    	queen[i][j]=2;
    
    // The above was for the user-placed queen --------
    
    // The below will cycle through and set queens in the free spaces
    for (i=0;i<=7;i++)
    {
            for (j=0;j<=7;j++)
            {
                         if (queen[i][j]==0)	// In the case we can set a queen
                         {
    				queen[i][j]=1;	// We do
    				// and locked the corresponding column, rows, diagonals for this.
    				// here would be like the above code that we could put in a function
    				// such as "LockQueen(i, j);" that ran the code that was above
    				// in the function for the values i and j being a and b
                         }
            }
    }
    EDIT: Here's the LockQueen function I was talking about, at least I'm pretty sure the int &queen[8][8] part will work, if not just define queen as a global before you declare LockQueen():

    Code:
    void LockQueen(int x, int y, int &queen[8][8])
    {
    	int i; j;
    
    	for (i=x+1;i<=7;i++)				//to lock the corresponding lower column
    		queen[i][y]=2;
    
    	for (i=x-1;i>=0;i--)				//to lock the corresponding upper column
    		queen[i][y]=2;
    
    	for (i=x-1,j=y+1;i>=0,j<=7;i--,j++)		//to lock the corresponding upper right diagonal
    		queen[i][j]=2;
    
    	for (i=x-1,j=y-1;i>=0,j>=0;i--,j--)		//to lock the corresponding upper left diagonal
    		queen[i][j]=2;
    
    	for (i=x+1,j=y+1;i<=7,j<=7;i++,j++)		//to lock the corresponding lower right diagonal
    		queen[i][j]=2;
    
    	for (i=x+1,j=y-1;i<=7,j>=0;i++,j--)		//to lock the corresponding lower left diagonal
    		queen[i][j]=2;
    }
    Last edited by tjpanda; 10-13-2007 at 01:33 PM. Reason: Adding in a function
    "When your work speaks for itself - don't interrupt!"

    -Samantha Ingraham.

  3. #3
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    Yeah, that's what I intended to do. Sorry for not being clear. Actually, the program which I wrote is so messed up now that even I'm having a hard time understanding it.

    Now coming to the problem. The problem is, suppose the function LockQueen worked up to rows 0-2 then it skipped the row 3, the user has inputed by incrementing a variable of the outer loop. But in row 4 there's no suitable place for a queen, so it should go back to row 2 and moves the queen to the next suitable location. THAT's my problem. How could I do that???

    Any help would be greatly appreciated.

    P.S. Please ignore any grammatical errors as English isn't my first language, but I hope I'm lot more clearer this time.

  4. #4
    Cat without Hat CornedBee's Avatar
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    You could make more use of the many values an int can take, e.g.:
    0: No queen, nobody threatening.
    -1: Queen.
    -2: There used to be a queen there, but it was moved.
    > 0: this many pieces are threatening this field.

    Then on placing a queen, you increment every field the queen can reach and that is >= 0. On removing a queen, you decrement every field the queen can reach and that is > 0. (If you encounter a field == 0 on removing, you've got a bug.)
    All the buzzt!
    CornedBee

    "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
    - Flon's Law

  5. #5
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    That's exactly what I initially did.

    0: or no queen
    1: queen
    2: locked path/threatening

    When there's no place go back to the previous iteration and removes the queen with aal its fields and placed 4 where it was initially and then starts the whole process again. Now the problem is, if I place 0 again it would place the queen on the same location again and if I place or 2 it runs into a situation where I can't find a solution.

    Help me please!!!

  6. #6
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    This may not be the best solution but try adding a second 2darry such as:

    Code:
    int badLocation[8][8];		// A mask for "bad spots"
    Then every time you run across a problem like you mentioned you can put a 1 on that location in badLocation and then in your code check to see if the location your about to set a queen in has a 1 in that position on badLocation, if it does then simply don't place it there and move on to the next slot.
    "When your work speaks for itself - don't interrupt!"

    -Samantha Ingraham.

  7. #7
    Cat without Hat CornedBee's Avatar
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    No, it's not the same. Read my post again. My system is more complex.
    All the buzzt!
    CornedBee

    "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
    - Flon's Law

  8. #8
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    Thanks for your response guys.

    tjpanda, I tried what you said but still....the problem is how to remove that particular 1 again if there's no slot for placing a queen after investigating all the previous rows.

    And CornedBee, sorry! I didn't get you. Would you please elaborate it a little more by writing a piece of code for that???

  9. #9
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    Something like this:
    Code:
    bool addQueen(int x, int y)
    {
      if(theField[x][y] != 0) {
        return false; // Field is threatened or already taken or was already taken once.
      }
      for(int i = 0; i < 8; ++i) {
        if(theField[i][y] == -1) {
          return false; // Would threaten on row.
        }
        if(theField[x][i] == -1) {
          return false; // Would threaten on column.
        }
      }
      // Same test for diagonal. Too lazy to implement the coordinate finding.
    
      // OK, we know we can place the queen.
      theField[x][y] = -1;
      for(int i = 0; i < 8; ++i) {
        if(theField[i][y] >= 0) {
          ++theField[i][y]; // One more piece is threatening this field.
        }
        if(theField[x][i] >= 0) {
          ++theField[x][i]; // One more piece is threatening this field.
        }
      }
      // Again, same thing for diagonal moves.
    
      return true;
    }
    
    bool removeQueen(int x, int y)
    {
      if(theField[x][y] != -1) {
        // There's no queen here.
        return false;
      }
    
      theField[x][y] = -2; // There has been a queen here, so don't try placing one again.
    
      for(int i = 0; i < 8; ++i) {
        if(theField[i][y] > 0) {
          --theField[i][y]; // One less piece is threatening this field.
        }
        if(theField[x][i] > 0) {
          --theField[x][i]; // One less piece is threatening this field.
        }
      }
      // Again, same thing for diagonal moves.
    
      return true;
    }
    The nice thing about this is that you don't mark a field unthreatened by removing a queen when in fact a different queen is still threatening it. Which, I guess, was your main problem.
    All the buzzt!
    CornedBee

    "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
    - Flon's Law

  10. #10
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    First of all thanks for your help. By following your advice I wrote this code:

    Code:
    #include "stdafx.h"
    #include <iostream>
    using namespace std;
    int queen[8][8];
    bool addQueen(int x, int y)
    {
    	int j;
    	if(queen[x][y] != 0) 
    	{
    		return false; // Field is threatened or already taken or was already taken once.
    	}
    	for(int i = 0; i < 8; ++i) 
    	{
    		if(queen[i][y] == -1) 
    		{
    			return false; // Would threaten on row.
    		}
    		if(queen[x][i] == -1) 
    		{
    			return false; // Would threaten on column.
    		}
    	}
      // OK, we know we can place the queen.
    	queen[x][y] = -1;
    	for(int i = 0; i < 8; ++i) 
    	{
    		if(queen[i][y] >= 0) 
    		{
    			++queen[i][y]; // One more piece is threatening this field.
    		}
    		if(queen[x][i] >= 0) 
    		{
    			++queen[x][i]; // One more piece is threatening this field.
    		}
    	}
    	for (int i=x-1,j=y+1;i>=0,j<=7;i--,j++)     
    	{
    		if (queen[i][j]>=0)
    			++queen[i][j];
    	}
        for (int i=x-1,j=y-1;i>=0,j>=0;i--,j--)		
    	{
    		if (queen[i][j]>=0)
    			++queen[i][j];
    	}
        for (int i=x+1,j=y+1;i<=7,j<=7;i++,j++)		
    	{
    		if (queen[i][j]>=0)
    			++queen[i][j];
    	}
        for (int i=x+1,j=y-1;i<=7,j>=0;i++,j--)		
    	{
    		if (queen[i][j]>=0)
    			++queen[i][j];
    	}
      
    
    	return true;
    }
    
    bool removeQueen(int x, int y)
    {
    	int j;
    	if(queen[x][y] != -1) 
    	{
        // There's no queen here.
    		return false;
    	}
    
    	queen[x][y] = -2; // There has been a queen here, so don't try placing one again.
    	for(int i = 0; i < 8; ++i) 
    	{
    		if(queen[i][y] > 0) 
    		{
    			--queen[i][y]; // One less piece is threatening this field.
    		}
    		if(queen[x][i] > 0) 
    		{
    			--queen[x][i]; // One less piece is threatening this field.
    		}
    	}
    	
    	for (int i=x-1,j=y+1;i>=0,j<=7;i--,j++)     
    	{
    		if (queen[i][j]>0)
    			--queen[i][j];
    	}
        for (int i=x-1,j=y-1;i>=0,j>=0;i--,j--)		
    	{
    		if (queen[i][j]>0)
    			--queen[i][j];
    	}
        for (int i=x+1,j=y+1;i<=7,j<=7;i++,j++)		
    	{
    		if (queen[i][j]>0)
    			--queen[i][j];
    	}
        for (int i=x+1,j=y-1;i<=7,j>=0;i++,j--)		
    	{
    		if (queen[i][j]>0)
    			--queen[i][j];
    	}
    
      return true;
    }
    
    
    int _tmain(int argc, _TCHAR* argv[])
    {
    	int a, b, i, j; // ab to store the user co-ordinates ans i,j as counters
            bool ans;
        //int queen[8][8];
        for (i = 0; i <= 7; i++)			// Loop that runs while a = {0, 1, 2, ..., 7}
    		{
    			for (j = 0; j <= 7; j++)	// Loop that runs while b = {0, 1, 2, ..., 7}
    			{
    				queen[i][j] = 0;	// Initializing the "chest board" array
                }
            }
        cout<<"Enter Row: ";
        cin>>a;
        cout<<"Enter Column: ";
        cin>>b;
        queen[a][b] = -1;
        for(i = 0; i < 8; ++i) 
    	{
    		++queen[i][b]; // One more piece is threatening this field.
            ++queen[a][i]; // One more piece is threatening this field.
        }
        for (i=a-1,j=b+1;i>=0,j<=7;i--,j++)		//to lock the corresponding upper right diagonal
    		++queen[i][j];
        for (i=a-1,j=b-1;i>=0,j>=0;i--,j--)		//to lock the corresponding upper left diagonal
    	    ++queen[i][j];
        for (i=a+1,j=b+1;i<=7,j<=7;i++,j++)		//to lock the corresponding lower right diagonal
    	    ++queen[i][j];
        for (i=a+1,j=b-1;i<=7,j>=0;i++,j--)		//to lock the corresponding lower left diagonal
    	    ++queen[i][j];
    	for (i=0;i<8;++i)
    	{
    		for (j=0;j<8;++j)
    		{
    			if (i==a)
    				++i;
    			ans = addQueen(i,j);
    			if (!ans)
    				removeQueen(i,j);
    		}
    	}
    	for (i=0;i<8;i++)
    	{
    		for (j=0;j<8;j++)
    		{
    			cout<<"\t"<<queen[i][j];
    		}
    		cout<<"\n";
    	}
    	return 0;
    }
    And the output I got is:

    Enter Row: 3
    Enter Column: 5

    -1 3 3 2 2 4 2 2
    3 3 -1 2 3 2 3 3
    3 4 3 3 3 2 3 -1
    4 3 4 2 3 1 3 3
    2 -1 3 2 4 4 4 2
    2 1 2 2 3 3 2 2
    2 2 3 3 -1 3 3 4
    2 3 3 2 3 3 -1 3

    Enter Row: 0
    Enter Column: 0
    1 3 3 3 3 1 2 2
    3 3 -1 4 2 2 2 1
    3 4 4 3 -1 2 1 1
    3 -1 3 4 4 2 1 1
    3 2 4 -1 3 2 2 1
    1 2 2 2 2 1 1 1
    2 2 1 1 2 1 1 1
    2 1 1 1 1 1 1 1


    Enter Row: 5
    Enter Column: 5
    3 -1 4 3 2 3 1 2
    4 4 3 -1 3 2 1 2
    -1 3 4 4 2 3 1 2
    2 4 -1 3 2 3 2 3
    3 3 3 3 2 3 3 -1
    3 2 2 3 2 1 3 3
    1 1 1 1 2 3 1 2
    1 1 1 2 1 2 1 2

    Any ideas what went wrong or did I misunderstood anything...???
    Last edited by Dashing Boy; 10-14-2007 at 10:35 AM.

  11. #11
    Cat without Hat CornedBee's Avatar
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    Why did you replicate logic from addQueen() in main()?
    All the buzzt!
    CornedBee

    "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
    - Flon's Law

  12. #12
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    To block the path of the user placed queen. Still there's no appropriate solution if I remove that part. .

  13. #13
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    Quote Originally Posted by Dashing Boy View Post
    To block the path of the user placed queen. Still there's no appropriate solution if I remove that part. .
    But surely the function "placeQueen" shouldn't care if it's a user-placed queen or not?

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  14. #14
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    Ooooops........I guess that was the mistake. But still it is not producing the required result after removing that part of code. Here's the result:

    Enter Row: 3
    Enter Column: 5
    -1 3 3 3 2 2 3 1
    3 3 -1 3 3 2 2 2
    3 4 3 4 -1 1 3 1
    3 2 3 3 2 -1 1 0
    2 -1 4 3 4 2 3 1
    3 3 3 -1 3 2 3 2
    2 1 2 2 2 1 2 2
    2 3 2 2 3 2 -1 2



    Enter Row: 0
    Enter Column: 0
    -1 2 3 1 1 2 1 0
    2 -1 3 3 3 3 1 2
    3 3 3 -1 4 2 2 2
    1 3 4 4 3 -1 2 1
    1 3 -1 3 4 4 2 1
    2 3 2 4 -1 3 2 2
    1 1 2 2 2 2 1 1
    0 2 2 1 1 2 1 1

    I even tried after both removing and adding the following part of code but can't find any significant improvement.

    Code:
    if(i==a)           //to skip the row of user placed queen
    {
           i++;
    }

  15. #15
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    Sorry for disturbing your guys again.

    But that's what I did. I made an array of type 'position' to keep track of the queens I placed, like that:

    Code:
    struct position
    {
          int x,y;
    };
    position queen[8];
    And that's the way I used it:

    Code:
    #include "stdafx.h"
    #include <iostream>
    using namespace std;
    int board[8][8],a,b;
    struct position
    {
    	int x,y;
    };
    position queen[8];
    bool addQueen(int x, int y)
    {
    	int j;
    	if(board[x][y] != 0) 
    	{
    		return false; // Field is threatened or already taken or was already taken once.
    	}
    	for(int i = 0; i < 8; ++i) 
    	{
    		if(board[i][y] == -1) 
    		{
    			return false; // Would threaten on row.
    		}
    		if(board[x][i] == -1) 
    		{
    			return false; // Would threaten on column.
    		}
    	}
      // OK, we know we can place the queen.
    	board[x][y] = -1;
    	for(int i = 0; i < 8; ++i) 
    	{
    		if(board[i][y] >= 0) 
    		{
    			++board[i][y]; // One more piece is threatening this field.
    		}
    		if(board[x][i] >= 0) 
    		{
    			++board[x][i]; // One more piece is threatening this field.
    		}
    	}
    	for (int i=x-1,j=y+1;i>=0,j<=7;i--,j++)     
    	{
    		if (board[i][j]>=0)
    			++board[i][j];
    	}
        for (int i=x-1,j=y-1;i>=0,j>=0;i--,j--)		
    	{
    		if (board[i][j]>=0)
    			++board[i][j];
    	}
        for (int i=x+1,j=y+1;i<=7,j<=7;i++,j++)		
    	{
    		if (board[i][j]>=0)
    			++board[i][j];
    	}
        for (int i=x+1,j=y-1;i<=7,j>=0;i++,j--)		
    	{
    		if (board[i][j]>=0)
    			++board[i][j];
    	}
      
    
    	return true;
    }
    
    bool removeQueen(int x, int y)
    {
    	int j;
    	if(board[x][y] != -1) 
    	{
        // There's no queen here.
    		return false;
    	}
    
    	board[x][y] = 0; // There has been a queen here, so don't try placing one again.
    	for(int i = 0; i < 8; ++i) 
    	{
    		if(board[i][y] > 0) 
    		{
    			--board[i][y]; // One less piece is threatening this field.
    		}
    		if(board[x][i] > 0) 
    		{
    			--board[x][i]; // One less piece is threatening this field.
    		}
    	}
    	
    	for (int i=x-1,j=y+1;i>=0,j<=7;i--,j++)     
    	{
    		if (board[i][j]>0)
    			--board[i][j];
    	}
        for (int i=x-1,j=y-1;i>=0,j>=0;i--,j--)		
    	{
    		if (board[i][j]>0)
    			--board[i][j];
    	}
        for (int i=x+1,j=y+1;i<=7,j<=7;i++,j++)		
    	{
    		if (board[i][j]>0)
    			--board[i][j];
    	}
        for (int i=x+1,j=y-1;i<=7,j>=0;i++,j--)		
    	{
    		if (board[i][j]>0)
    			--board[i][j];
    	}
    
      return true;
    }
    
    bool moveQueen (int x, int y)
    {
    	if (x==a)
    		return false;
    	removeQueen(x,y);
    	for (int i=y+1;i<=7;i++)
    	{
    		if (board[x][i]==0)
    		{
    			board[x][i]=-1;
    			return true;
    		}
    	}
    	return false;
    }
    
    int _tmain(int argc, _TCHAR* argv[])
    {
    	int i, j;
    	bool ansAdd, ansRem, ansMov;
        //int queen[8][8];
        for (i = 0; i <= 7; i++)			// Loop that runs while a = {0, 1, 2, ..., 7}
    		{
    			for (j = 0; j <= 7; j++)	// Loop that runs while b = {0, 1, 2, ..., 7}
    			{
    				board[i][j] = 0;	// Initializing the "chest board" array
                }
            }
    	cout<<"Enter Row: ";
    	cin>>a;
    	cout<<"Enter Column: ";
    	cin>>b;
        board[a][b] = -1;
        for (i=0;i<8;++i)
    	{
    		for (j=0;j<8;++j)
    		{
    			if (i==a)                 //if i = user input row
    				++i;
    			if (i==8)
    				break;
    			ansAdd = addQueen(i,j);
    			if (ansAdd)
    			{
    				queen[i].x=i;	     //to keep track of the placed queens
    			        queen[i].y=j;
    				break;
    			}
    			if (!ansAdd && j==7)         // if there's no place for a queen to be placed
    			{
    				ansMov = moveQueen (queen[i].x,queen[i].y);     move queen to the next available slot
    				if (!ansMov)
    					i-=2;
    				
    			}
    			
    		}
    	}
    	for (i=0;i<8;i++)
    	{
    		for (j=0;j<8;j++)
    		{
    			cout<<"\t"<<board[i][j];
    		}
    		cout<<"\n";
    	}
    	return 0;
    }
    And the result is:

    Enter Row: 0
    Enter Column: 0
    0 1 2 -1 0 1 0 -1
    1 -1 3 3 3 3 1 2
    2 3 2 -1 4 2 2 2
    0 3 4 3 3 -1 2 1
    0 3 -1 3 3 4 2 1
    1 3 2 4 -1 2 2 2
    0 1 2 2 2 2 0 1
    0 2 2 1 1 2 1 0

    Enter Row: 3
    Enter Column: 5
    1 2 1 2 -1 1 0 1
    2 2 -1 2 2 1 2 2
    2 2 1 3 0 -1 1 0
    3 2 2 1 2 -1 0 0
    2 -1 3 3 2 2 1 1
    3 2 4 -1 1 1 2 1
    1 2 2 2 1 0 0 1
    -1 3 2 2 2 2 1 1


    Would you please tell me what stupid mistake I'm making this time???

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