Tic-Tac-Toe without AI - c

Question

I'm doing homework for UNI and I got to do a Tic-Tac-Toe without any decision taken by player, the moves are all chosen randomly. So if the character on matrix is ' ' it means it's free, while if it's 'X' or 'O' it should generate another move. This is the code (language C):
if (playerTurn == 1){
playerSymb = 'X';
}
else if (playerTurn == 2){
playerSymb = 'O';
}
if (matrix[rand1][rand2] == ' '){
matrix[rand1][rand2] = playerSymb;
} else if(matrix[rand1][rand2] == 'X' || matrix[rand1][rand2] == 'O'){
do{
randAlt1 = MINRND + rand()%(MAXRND - MINRND +1);
randAlt2 = MINRND + rand()%(MAXRND - MINRND +1);
}while (matrix[randAlt1][randAlt2] != 'X' && matrix[randAlt1][randAlt2] != 'O');
matrix[randAlt1][randAlt2] = playerSymb;
}
I did not copied the whole code because it's not finished at all, i just need help solving this. But if I try to run this, the Symbols can be overwritten, like if I have a 'X' at matrix[1][2], it's possible that it will be a 'O' after some turns. So how can I make moves do not overwrite? (sorry for bad english).

Answer 1

Just put correct condition:
while (matrix[randAlt1][randAlt2] == 'X' || matrix[randAlt1][randAlt2] == 'O')
(i.e. try again if this cell is not empty)
Also it is easy to simplify your code without loosing of anything:
randAlt1 = rand1;
randAlt2 = rand2;
while (matrix[randAlt1][randAlt2] != ' ') {
randAlt1 = MINRND + rand()%(MAXRND - MINRND +1);
randAlt2 = MINRND + rand()%(MAXRND - MINRND +1);
}
matrix[randAlt1][randAlt2] = (playerTurn == 1) ? 'X' : 'O';
And it is better to add loop guard to prevent infinite loop (or to add special checks for this case):
randAlt1 = rand1;
randAlt2 = rand2;
int nbAttempts = 0;
while (matrix[randAlt1][randAlt2] != ' ' && nbAttempts < 100) {
randAlt1 = MINRND + rand()%(MAXRND - MINRND +1);
randAlt2 = MINRND + rand()%(MAXRND - MINRND +1);
nbAttempts++;
}
if (matrix[randAlt1][randAlt2] != ' ') {
// show error message and stop the game
}
matrix[randAlt1][randAlt2] = (playerTurn == 1) ? 'X' : 'O';

Answer 2

You choose an arbitrary position and then test if it is free – possibly multiple times. But you can also choose a number of a free position and then find it.
First set up a turn counter
int turnNo = 0;
then make a loop for alternate moves, which chooses one of 9-turnNo unused positions, finds it, marks is with a player mark and tests if the move made a line of three:
while(turnNo < 9)
{
char currPlayerMark = ...choose 'X' or 'O';
int freePos = 9 - turnNo;
int currPos = rand() % freePos; // 0 .. freePos-1
for(x=0; x<3; x++)
{
for(y=0; y<3; y++)
{
if(matrix[x][y] == ' ') // a free position
if(--currPos < 0) // the sought one
break; // break the inner loop
}
if(currPos < 0)
break; // break the outer loop
}
matrix[x][y] = currPlayerMark;
if(test_for_win_position(x,y))
{
message_a_win_of_player(currPlayerMark);
break; // turnNo < 9 here
}
turnNo ++;
}
Finally test if the loop terminated with no 'win':
if(turnNo == 9)
message_its_a_draw(); // no-one wins
A function to test the win position might look like this:
int test_for_win_position(int x, int y)
{
char mark = matrix[x][y];
// check a column
if(matrix[x][0] == mark && matrix[x][1] == mark && matrix[x][2] == mark)
return 1;
// check a row
if(matrix[0][y] == mark && matrix[1][y] == mark && matrix[2][y] == mark)
return 1;
// check one diagonal
if(x==y)
if(matrix[0][0] == mark && matrix[1][1] == mark && matrix[2][2] == mark)
return 1;
// check another diagonal
if(x+y==2)
if(matrix[0][2] == mark && matrix[1][1] == mark && matrix[2][0] == mark)
return 1;
// current player has not won (yet)
return 0;
}