Thread: tick- tack- toe

return (o) or return(0);???

You can't do this:
A[i][j]==A[i+1][j+1]==A[i+2][j+2]
What's happenign is:
A[i][j]==A[i+1][j+1] return 1 if true or 0 if false then:
1:0==A[i+2][j+2]
which I think is always false
do instead
A[i][j]==A[i+1][j+1] && A[i+1][j+1]==A[i+2][j+2]

More

Code:

for (i=j=0;i<n;i++,j++)
    if (A[i][j]==A[i+1][j+1]==A[i+2][j+2])
        break;

... why do you do this?? Should you place return 1 instead break? Like it is now it simply runs a cicle and breaks. The i and j variables are then re-defined, and nothing is done regarding the previous cicle.

But if you have one cicle after another the values of i and j are redefined... if the break statment is reached, no more for()s should be parsed. Instead of break write return A[i][j]==x.
Basicly it'll be like this:

Code:

int TickTackToe(char A[N][N],int n)
{
	int i,j;

	for (i=j=0;i<n;i++,j++)//check diagonal up,left->right,down
		if (A[i][j]==A[i+1][j+1]==A[i+2][j+2])
 			return A[i][j]=='x';

	for (i=n-1,j=0;i>=0;i--,j++)//check diagonal right,up->down,left
   		if (A[i][j]==A[i-1][j+1]==A[i-2][j+2])
  			return A[i][j]=='x';

	for(i=0;i<n;i++)//check vertical lines
	   	for(j=0;j<n;j++)
			if (A[i][j]==A[i+1][j]==A[i+2][j])
				return A[i][j]=='x';

	for(j=0;j<n;j++)//check horizontal lines
		for(i=0;i<n;i++)
			if (A[i][j]==A[i][j+1]==A[i][j+2])
				return A[i][j]=='x';
	//return statment should have been reached... if not no line was detected
	return -1;
}

Hope this one helps.
At the end you had:
if (flag==x)
return (1);
else if (flag==o)
return (0);

return A[i][j]=='x'; does the same stuff: if that char is 'x' return 1, else 0. If no line is detected the function at the end return -1.

This is also works...
Thank you, xErath.

Code:

int TickTackToe(char A[N][N],int n)
{
	int i,j;
	char x='x',o='o',flag;
	
	for (i=j=0;i<n;i++,j++)
		if (A[i][j]==A[i+1][j+1]&&A[i+1][j+1]==A[i+2][j+2])
		 break;

	for (i=n-1,j=n-1-i;i>=0;i--,j++)
	   if (A[i][j]==A[i-1][j+1]&&A[i-1][j+1]==A[i-2][j+2])
	                 break;

	for(i=0;i<n;i++)
	   for(j=0;j<n;j++)
	       if (A[i][j]==A[i+1][j]&&A[i+1][j]==A[i+2][j])		       break;

			
	 for(j=0;j<n;j++)
	   for(i=0;i<n;i++)
	       if (A[i][j]==A[i][j+1]&&A[i][j+1]==A[i][j+2])

	flag=A[i][j];
                	if (flag==x)
		    return (1);
	     else 
		if (flag==o)
		   return (0);
                     else
		return (-1);
}

Ronen