Selecting most recent value from a variable

Programming Questions

louiscb

Selecting most recent value from a variable

in Programming Questions • 1 year ago

I am currently in the process of making a tic tac toe machine. I have variables called 'index' and 'index2' that tell me what squares have the value 0 (not selected) and I want to apply them so the computer selects a square that is not selected during its turn. However because 'index' and 'index2' are always updating the computer selects all the squares that are not selected!

void computerlogic (){

for (int i = 0; i < grid.length; i++) {

for (int j = 0; j < grid.length; j++) {

//indicates the squares with a value of 0

if (mousePressed == true) {

if ( grid [i] [j] == 0) {

index = i;

index2 = j;

}

}



//Tells the computer to choose only squares that are not selected

//This is where I am having problems



if (grid [1] [1] == 1 && grid [i] [j] != 2) {

grid [index] [index2] = 2;

}



}

}

}

I want the computer to only choose once per turn however I have no idea how to do this. Also, for bonus points, can someone tell me how to shorten the "mouse press" code? But thats not as important as it works.

Thanks!

All the code:

int cols = 3;

int rows = 3;

int index = -1;

int index2 = -1;

int [] [] grid = new int [cols] [rows];

color noplay = color (255);

color player1 = color(255, 0, 0);

color comp = color(0, 255, 0);

void setup () {

size (300, 300);

for (int i = 0; i < grid.length; i++) {

for (int j = 0; j < grid.length; j++) {

grid [i] [j] = 0;

}

}

}

void draw () {

mousepress (mouseX, mouseY);

computerlogic();

display ();

}

void mousepress (int mX, int mY) {

//First Row

if (mousePressed == true) {

if (mX < width/3 && mY < width/3) {

grid [0] [0] = 1;

}

}

if (mousePressed == true) {

if (mX > width/3 && mY < width/3 && mX < 2*(width/3)) {

grid [1] [0] = 1;

}

}

if (mousePressed == true) {

if (mY < width/3 && mX > 2*(width/3)) {

grid [2] [0] = 1;

}

}

//Second Row

if (mousePressed == true) {

if (mX < width/3 && mY < 2*(width/3) && mY > width/3) {

grid [0] [1] = 1;

}

}

if (mousePressed == true) {

if (mX > width/3 && mY < 2*(width/3) && mY > width/3 && mX < 2*(width/3)) {

grid [1] [1] = 1;

}

}

if (mousePressed == true) {

if (mX > 2*(width/3) && mY > width/3 && mY < 2*(width/3)) {

grid [2] [1] = 1;

}

}

//Third Row

if (mousePressed == true) {

if (mX <width/3 && mY > 2*(width/3)) {

grid [0] [2] = 1;

}

}

if (mousePressed == true) {

if (mX > width/3 && mY > 2*(width/3) && mX < 2*(width/3)) {

grid [1] [2] = 1;

}

}

if (mousePressed == true) {

if (mX > 2*(width/3) && mY > 2*(width/3 )) {

grid [2] [2] = 1;

}

}

}

void computerlogic () {

for (int i = 0; i < grid.length; i++) {

for (int j = 0; j < grid.length; j++) {

//indicates the squares with a value of 0

if (mousePressed == true) {

if ( grid [i] [j] == 0) {

index = i;

index2 = j;

}

}



//Tells the computer to choose only squares that are not selected

//This is where I am having problems



if (grid [1] [1] == 1 && grid [i] [j] != 2) {

grid [index] [index2] = 2;

}



}

}

}

void display () {

for (int i = 0; i < grid.length; i++) {

for (int j = 0; j < grid.length; j++) {

if (grid [i] [j] == 0) {

fill (noplay);

}

else if (grid [i] [j] == 1) {

fill (player1);

}

else if (grid [i] [j] == 2) {

fill (comp);

}

rect (i*100, j*100, width/3, width/3);

}

}

}

Replies(4)

amnon.owed

Re: Selecting most recent value from a variable

1 year ago

Here are some suggestions to shorten your code. Of course still some points to develop left...

Main changes:

Put colors in an array so you can choose the correct color with one line: fill(colors[grid[i][j]]);
Replaced your whole mousepress method with a single line: grid[mouseX/w][mouseY/h] = 1;
No need to initialize grid to zero (this is already the default value)
Use mousePressed (for human player) and mouseReleased (for computer player)
Use mouseX and mouseY directly (that's what they are there for)
Calculate w and h, otherwise things will break at different size or rows/cols numbers
There are different ways to let the comp choose a field. This is my quick (and not very good) attempt that finds the open fields, put's them in a basket and randomly chooses one of them. Just to get some basic interaction going.

Adapted Code

int cols = 3;
int rows = 3;
int w, h;
int[][] grid = new int[cols][rows];
color[] colors = { color(255), color(255, 0, 0), color(0, 255, 0) };
void setup () {
size(500, 500);
w = width/cols;
h = height/rows;
}
void draw () {
for (int i=0; i<cols; i++) {
for (int j=0; j<rows; j++) {
fill(colors[grid[i][j]]);
rect (i*w, j*h, w, h);
}
}
}
void mousePressed() {
grid[mouseX/w][mouseY/h] = 1;
}
void mouseReleased() {
ArrayList <PVector> options = new ArrayList <PVector> ();
for (int i=0; i<cols; i++) {
for (int j=0; j<rows; j++) {
if (grid[i][j] == 0) {
options.add(new PVector(i, j));
}
}
}
if (options.size()>0) {
int randomChoice = int(random(options.size()));
PVector option = options.get(randomChoice);
grid[int(option.x)][int(option.y)] = 2;
}
}

louiscb

Re: Selecting most recent value from a variable

1 year ago

WOW! Thats so smart, I understand most of that, its pure logic. The only thing I didn't get was the use of ArrayLists, PVectors and options.something. I haven't come across any of that in my programming yet, so I didnt get most of mouse released, however I do get the underlying logic. Also, how do you think the best way would be to try to make the computer try to win? I'm pretty sure I wouldn't be able to do it without using different functions that I haven't used yet.

Thanks loads, a really big help :)

amnon.owed

Re: Selecting most recent value from a variable

1 year ago

If you want the computer to win it should consider the best choice, instead of choosing randomly. You could calculate all possible options times a certain number of steps in forwards, but I guess that would be pretty resource intensive. Another route is to have two strategies. One defensive (preventing the human player from scoring) and one offensive (trying to score), which seems a bit easier to program, so would be a worthy next programming task. Of course there are some other things you might want to include now, such as preventing the human from overwriting computer fields. Adding a check if somebody has won and including an end when all fields are taken.

General tip: there are loads of tictactoe examples on OpenProcessing.org

xenophage

Re: Selecting most recent value from a variable

1 year ago

Hi!

Tic Tac Toe was the first time I really attempted some basic AI. It was all done on the TI-83 plus calculator! I'll give you a few tips. Tic tac toe is pretty limited in the choices you have to go if you think about it like a computer. start with move 1. If player one goes in the center (actually the worst place to go first) then if you think about it, there is only 2 choices, side middle, or corner. The computer can choose, and rotate or mirror the board. There is many possibilities, but by mirroring or rotating, theres actually very few places to go. Get some scratch paper out and start with either center, side middle, or corner. From those 3 branches and taking into thought rotation of the board, you'll soon find what moves are best. When I was done with my AI, it was unbeatable. If human went first, computer had a lower chance of winning, but would always at least tie, because I programmed in any counter moves to block human. But when computer went first, it had a high chance of winning because it knew how to create the double win moves. And of course you could let computer play itself, which was always tie. It's not really AI, it just sorts through a small list of possible places to go according to how the board is rotated from different examples. Took me 4 months to perfect it but that was back when I was in high school and didn't spend much time on it. I looked back at it once and thought about adding a computer difficulty level, but I figured out that came down to a matter of risk. The harder the computer would try to win, the more open chances the human had of winning if he/she knew how to play. Hope this helps!

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