Tic Tac Toe: Evaluating Heuristic Value of a Node - c++

Pardon me if this question already exists, I've searched a lot but I haven't gotten the answer to the question I want to ask. So, basically, I'm trying to implement a Tic-Tac-Toe AI that uses the Minimax algorithm to make moves.
However, one thing I don't get is, that when Minimax is used on an empty board, the value returned is always 0 (which makes sense because the game always ends in a draw if both players play optimally).
So Minimax always chooses the first tile as the best move when AI is X (since all moves return 0 as value). Same happens for the second move and it always chooses the second tile instead. How can I fix this problem to make my AI pick the move with the higher probability of winning? Here is the evaluation and Minimax function I use (with Alpha-Beta pruning):
int evaluate(char board[3][3], char AI)
{
for (int row = 0; row<3; row++)
{
if (board[row][0] != '_' && board[row][0] == board[row][1] && board[row][1] == board[row][2])
{
if (board[row][0]==AI)
{
return +10;
}
else
{
return -10;
}
}
}
for (int col = 0; col<3; col++)
{
if (board[0][col] != '_' && board[0][col] == board[1][col] && board[1][col] == board[2][col])
{
if (board[0][col]==AI)
{
return +10;
}
else
{
return -10;
}
}
}
if (board[1][1] != '_' && ((board[0][0]==board[1][1] && board[1][1]==board[2][2]) || (board[0][2]==board[1][1] && board[1][1]==board[2][0])))
{
if (board[1][1]==AI)
{
return +10;
}
else
{
return -10;
}
}
return 0;
}
int Minimax(char board[3][3], bool AITurn, char AI, char Player, int depth, int alpha, int beta)
{
bool breakout = false;
int score = evaluate(board, AI);
if(score == 10)
{
return score - depth;
}
else if(score == -10)
{
return score + depth;
}
else if(NoTilesEmpty(board))
{
return 0;
}
if(AITurn == true)
{
int bestvalue = -1024;
for(int i = 0; i < 3; i++)
{
for(int j = 0; j<3; j++)
{
if(board[i][j] == '_')
{
board[i][j] = AI;
bestvalue = max(bestvalue, Minimax(board, false, AI, Player, depth+1, alpha, beta));
alpha = max(bestvalue, alpha);
board[i][j] = '_';
if(beta <= alpha)
{
breakout = true;
break;
}
}
}
if(breakout == true)
{
break;
}
}
return bestvalue;
}
else if(AITurn == false)
{
int bestvalue = +1024;
for(int i = 0; i < 3; i++)
{
for(int j = 0; j<3; j++)
{
if(board[i][j] == '_')
{
board[i][j] = Player;
bestvalue = min(bestvalue, Minimax(board, true, AI, Player, depth+1, alpha, beta));
beta = min(bestvalue, beta);
board[i][j] = '_';
if(beta <= alpha)
{
breakout = true;
break;
}
}
}
if(breakout == true)
{
break;
}
}
return bestvalue;
}
}

Minimax assumes optimal play, so maximizing "probability of winning" is not a meaningful notion: Since the other player can force a draw but cannot force a win, they will always force a draw. If you want to play optimally against a player who is not perfectly rational (which, of course, is one of the only two ways to win*), you'll need to assume some probability distribution over the opponent's moves and use something like ExpectMinimax, where with some probability the opponent's move is overridden by a random mistake. Alternatively, you can deliberately restrict the ply of the minimax search, using a heuristic for the opponent's play beyond a certain depth (but still searching the game tree for your own moves.)
* The other one is not to play.

Organize your code into smaller routines so that it looks tidier and easier to debug. Apart from the recursive minimax function, an all-possible-valid-move generation function and a robust evaluation sub-routine are essential ( which seems lacking here).
For example, at the beginning of the game, the evaluation algorithm should return a non-zero score, every position should have a relative scoring index ( eg middle position may have slightly higher weightage than the corners).
Your minimax boundary condition - return if there is no empty cell positions ; is flawed as it will evaluate even when a winning/losing move occurred in the preceding ply. Such conditions will aggravate in more complex AI games.
If you are new to minimax, you can find plenty of ready to compile sample codes on CodeReview

Related

Test for possible wins in tictactoe game

I'm creating a tictactoe game, and every move I need to test whether a player has won or not, and it's giving me a lot of trouble. I have a 2d vector of all the possible win combinations:
vector<vector<int>> possibleWins {{1,2,3},{4,5,6},{7,8,9},{1,4,7},{2,5,8},{3,6,9},{1,5,9},{3,5,7}};
Every move I loop through the 2d vector and append player1 and player2 vectors with any cells they have marked:
vector<int> totalX {};
vector<int> totalO {};
int count = 1;
for(int i=0; i<3; i++) {
for(int j=0; j<3; j++) {
if(board[i][j] == 'x') {
if(totalX.size() == 0) {
totalX.push_back(count);
}
for(int k=0; k<totalX.size(); k++) {
if(count == totalX[k]) {
break;
}
else {
totalX.push_back(count);
}
}
}
else if(board[i][j] == 'o') {
if(totalO.size() == 0) {
totalO.push_back(count);
}
for(int k=0; k<totalO.size(); k++) {
if(count == totalO[k]) {
break;
}
else {
totalO.push_back(count);
}
}
}
count++;
}
}
I then try to test whether the cells in each players cell vector is a winning combination of cells, and this is proving to be difficult for me:
int xInRow = 0;
for(int x=0; x<totalX.size(); x++) {
for(int y=0; y<possibleWins.size(); y++) {
xInRow = 0;
for(int z=0; z<3; z++) {
if(totalX[x] == possibleWins[y][z]) {
xInRow++;
if(xInRow == 3) {
return X_WON;
}
}
}
}
}
This does not work, and I've tried implementing it in numerous different ways, but I honestly have no idea how to enumerate through all the possible wins and test if a player has one of these combinations.
Is there a way I could structure this better to make it work? I'm pretty lost on this.
There are two approaches. Your code is a bit too complicated for what should be a simple operation so I'm not going to try to understand it.
I agree with YSC that you don't need a std::vector. You know it's a 3x3 grid every time, so a 3x3 enum array should be much better. Something like
enum TTTState {
EMPTY=0,
X,
O
}
TTState board[3][3];
would save you a lot of headache. You can say board[0][0] is the top left, and board[2][2] is the bottom right.
Option 1
I like your idea of possibleWins, so with the new board[3][3] data structure, you can store pair of numbers with int solutions[8][3][2] but this is already kind of messy.
for each solution in solutions
for each triplet in solution
for each pair in triplet
if board[pair[0]][pair[1]] matches all other pair in triplet
then whoever has pieces in the row has won
Option 2
This is probably cleaner. There are 3 possible ways to win. Horizontal, vertical, and diagonal. You can check these three ways separately.
for i = 0 ; i != 3; i++
type = board[i][0]
won = true
for ii = 1; ii != 3; ii++
if board[i][ii] is not type
won = false
if won then you can return the function with who won
for i = 0 ; i != 3; i++
type = board[0][i]
won = true
for ii = 1; ii != 3; ii++
if board[ii][i] is not type
won = false
if won then you can return the function with who won
Diagonal can just be hard coded, since there are only two possible victory positions..
You're complicating things a little bit.
You don't need to first collect which positions a player has used and see how many of those are in a winning position.
Since you already know the winning positions, you only need to check if any one player has occupied all of one.
Assuming that ' ' marks an empty square,
for(const auto& win: possibleWins)
{
if (board[win[0]] == board[win[1]]
&& board[win[1]] == board[win[2]]
&& board[win[0]] != ' ')
{
// if board[win[0]] == 'X' , then X won
// if board[win[0]] == 'O' , then O won
}
}
should do.
| |
---+---+---
| |
---+---+---
| |
is a int myArray[9]; and you put the player moves on it myArray[played_position] = playerValue;, then you could use your vector<vector<int>> possibleWins {{1,2,3},{4,5,6},{7,8,9},{1,4,7},{2,5,8},{3,6,9},{1,5,9},{3,5,7}}; to check after the 3 move if the positions correspond to the same value (player!)...
for(int idx=0; idx<possibleWins.size(); idx++)
{
if(myArray[possibleWins[idx][0]] == myArray[possibleWins[idx][1]] == myArray[possibleWins[idx][2]])
{
return myArray[possibleWins[idx][0];
}
}
It's just an idea, hope it helps you elaborate....

Minimax with alpha-beta pruning problems

I'm making a C++ program for the game chopsticks.
It's a really simple game with only 625 total game states (and it's even lower if you account for symmetry and unreachable states). I have read up minimax and alpha-beta algorithms, mostly for tic tac toe, but the problem I was having was that in tic tac toe it's impossible to loop back to a previous state while that can easily happen in chopsticks. So when running the code it would end up with a stack overflow.
I fixed this by adding flags for previously visited states (I don't know if that's the right way to do it.) so that they can be avoided, but now the problem I have is that the output is not symmetric as expected.
For example in the start state of the game each player has one finger so it's all symmetric. The program tells me that the best move is to hit my right hand with my left but not the opposite.
My source code is -
#include <iostream>
#include <array>
#include <vector>
#include <limits>
std::array<int, 625> t; //Flags for visited states.
std::array<int, 625> f; //Flags for visited states.
int no = 0; //Unused. For debugging.
class gamestate
{
public:
gamestate(int x, bool t) : turn(t) //Constructor.
{
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++) {
val[i][j] = x % 5;
x /= 5;
}
init();
}
void print() //Unused. For debugging.
{
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++)
std::cout << val[i][j] << "\t";
std::cout << "\n";
}
std::cout << "\n";
}
std::array<int, 6> canmove = {{ 1, 1, 1, 1, 1, 1 }}; //List of available moves.
bool isover() //Is the game over.
{
return ended;
}
bool won() //Who won the game.
{
return winner;
}
bool isturn() //Whose turn it is.
{
return turn;
}
std::vector<int> choosemoves() //Choose the best possible moves in the current state.
{
std::vector<int> bestmoves;
if(ended)
return bestmoves;
std::array<int, 6> scores;
int bestscore;
if(turn)
bestscore = std::numeric_limits<int>::min();
else
bestscore = std::numeric_limits<int>::max();
scores.fill(bestscore);
for (int i = 0; i < 6; i++)
if (canmove[i]) {
t.fill(0);
f.fill(0);
gamestate *play = new gamestate(this->playmove(i),!turn);
scores[i] = minimax(play, 0, std::numeric_limits<int>::min(), std::numeric_limits<int>::max());
std::cout<<i<<": "<<scores[i]<<std::endl;
delete play;
if (turn) if (scores[i] > bestscore) bestscore = scores[i];
if (!turn) if (scores[i] < bestscore) bestscore = scores[i];
}
for (int i = 0; i < 6; i++)
if (scores[i] == bestscore)
bestmoves.push_back(i);
return bestmoves;
}
private:
std::array<std::array<int, 2>, 2 > val; //The values of the fingers.
bool turn; //Whose turn it is.
bool ended = false; //Has the game ended.
bool winner; //Who won the game.
void init() //Check if the game has ended and find the available moves.
{
if (!(val[turn][0]) && !(val[turn][1])) {
ended = true;
winner = !turn;
canmove.fill(0);
return;
}
if (!(val[!turn][0]) && !(val[!turn][1])) {
ended = true;
winner = turn;
canmove.fill(0);
return;
}
if (!val[turn][0]) {
canmove[0] = 0;
canmove[1] = 0;
canmove[2] = 0;
if (val[turn][1] % 2)
canmove[5] = 0;
}
if (!val[turn][1]) {
if (val[turn][0] % 2)
canmove[2] = 0;
canmove[3] = 0;
canmove[4] = 0;
canmove[5] = 0;
}
if (!val[!turn][0]) {
canmove[0] = 0;
canmove[3] = 0;
}
if (!val[!turn][1]) {
canmove[1] = 0;
canmove[4] = 0;
}
}
int playmove(int mov) //Play a move to get the next game state.
{
auto newval = val;
switch (mov) {
case 0:
newval[!turn][0] = (newval[turn][0] + newval[!turn][0]);
newval[!turn][0] = (5 > newval[!turn][0]) ? newval[!turn][0] : 0;
break;
case 1:
newval[!turn][1] = (newval[turn][0] + newval[!turn][1]);
newval[!turn][1] = (5 > newval[!turn][1]) ? newval[!turn][1] : 0;
break;
case 2:
if (newval[turn][1]) {
newval[turn][1] = (newval[turn][0] + newval[turn][1]);
newval[turn][1] = (5 > newval[turn][1]) ? newval[turn][1] : 0;
} else {
newval[turn][0] /= 2;
newval[turn][1] = newval[turn][0];
}
break;
case 3:
newval[!turn][0] = (newval[turn][1] + newval[!turn][0]);
newval[!turn][0] = (5 > newval[!turn][0]) ? newval[!turn][0] : 0;
break;
case 4:
newval[!turn][1] = (newval[turn][1] + newval[!turn][1]);
newval[!turn][1] = (5 > newval[!turn][1]) ? newval[!turn][1] : 0;
break;
case 5:
if (newval[turn][0]) {
newval[turn][0] = (newval[turn][1] + newval[turn][0]);
newval[turn][0] = (5 > newval[turn][0]) ? newval[turn][0] : 0;
} else {
newval[turn][1] /= 2;
newval[turn][0] = newval[turn][1];
}
break;
default:
std::cout << "\nInvalid move!\n";
}
int ret = 0;
for (int i = 1; i > -1; i--)
for (int j = 1; j > -1; j--) {
ret+=newval[i][j];
ret*=5;
}
ret/=5;
return ret;
}
static int minimax(gamestate *game, int depth, int alpha, int beta) //Minimax searching function with alpha beta pruning.
{
if (game->isover()) {
if (game->won())
return 1000 - depth;
else
return depth - 1000;
}
if (game->isturn()) {
for (int i = 0; i < 6; i++)
if (game->canmove[i]&&t[game->playmove(i)]!=-1) {
int score;
if(!t[game->playmove(i)]){
t[game->playmove(i)] = -1;
gamestate *play = new gamestate(game->playmove(i),!game->isturn());
score = minimax(play, depth + 1, alpha, beta);
delete play;
t[game->playmove(i)] = score;
}
else
score = t[game->playmove(i)];
if (score > alpha) alpha = score;
if (alpha >= beta) break;
}
return alpha;
} else {
for (int i = 0; i < 6; i++)
if (game->canmove[i]&&f[game->playmove(i)]!=-1) {
int score;
if(!f[game->playmove(i)]){
f[game->playmove(i)] = -1;
gamestate *play = new gamestate(game->playmove(i),!game->isturn());
score = minimax(play, depth + 1, alpha, beta);
delete play;
f[game->playmove(i)] = score;
}
else
score = f[game->playmove(i)];
if (score < beta) beta = score;
if (alpha >= beta) break;
}
return beta;
}
}
};
int main(void)
{
gamestate test(243, true);
auto movelist = test.choosemoves();
for(auto i: movelist)
std::cout<<i<<std::endl;
return 0;
}
I'm passing the moves in a sort of base-5 to decimal system as each hand can have values from 0 to 4.
In the code I have input the state -
3 3
4 1
The output says I should hit my right hand (1) to the opponent's right (3) but it does not say I should hit it to my opponent's left (also 3)
I think the problem is because of the way I handled infinite looping.
What would be the right way to do it? Or if that is the right way, then how do I fix the problem?
Also please let me know how I can improve my code.
Thanks a lot.
Edit:
I have changed my minimax function as follows to ensure that infinite loops are scored above losing but I'm still not getting symmetry. I also made a function to add depth to the score
static float minimax(gamestate *game, int depth, float alpha, float beta) //Minimax searching function with alpha beta pruning.
{
if (game->isover()) {
if (game->won())
return 1000 - std::atan(depth) * 2000 / std::acos(-1);
else
return std::atan(depth) * 2000 / std::acos(-1) - 1000;
}
if (game->isturn()) {
for (int i = 0; i < 6; i++)
if (game->canmove[i]) {
float score;
if(!t[game->playmove(i)]) {
t[game->playmove(i)] = -1001;
gamestate *play = new gamestate(game->playmove(i), !game->isturn());
score = minimax(play, depth + 1, alpha, beta);
delete play;
t[game->playmove(i)] = score;
} else if(t[game->playmove(i)] == -1001)
score = 0;
else
score = adddepth(t[game->playmove(i)], depth);
if (score > alpha) alpha = score;
if (alpha >= beta) break;
}
return alpha;
} else {
for (int i = 0; i < 6; i++)
if (game->canmove[i]) {
float score;
if(!f[game->playmove(i)]) {
f[game->playmove(i)] = -1001;
gamestate *play = new gamestate(game->playmove(i), !game->isturn());
score = minimax(play, depth + 1, alpha, beta);
delete play;
f[game->playmove(i)] = score;
} else if(f[game->playmove(i)] == -1001)
score = 0;
else
score = adddepth(f[game->playmove(i)], depth);
if (score < beta) beta = score;
if (alpha >= beta) break;
}
return beta;
}
}
This is the function to add depth -
float adddepth(float score, int depth) //Add depth to pre-calculated score.
{
int olddepth;
float newscore;
if(score > 0) {
olddepth = std::tan((1000 - score) * std::acos(-1) / 2000);
depth += olddepth;
newscore = 1000 - std::atan(depth) * 2000 / std::acos(-1);
} else {
olddepth = std::tan((1000 + score) * std::acos(-1) / 2000);
depth += olddepth;
newscore = std::atan(depth) * 2000 / std::acos(-1) - 1000;
}
return newscore;
}
Disclaimer: I don't know C++, and I frankly haven't bothered to read the game rules. I have now read the rules, and still stand by what I said...but I still don't know C++. Still, I can present some general knowledge of the algorithm which should set you off in the right direction.
Asymmetry is not in itself a bad thing. If two moves are exactly equivalent, it should choose one of them and not stand helpless like Buridan's ass. You should, in fact, be sure that any agent you write has some method of choosing arbitrarily between policies which it cannot distinguish.
You should think more carefully about the utility scheme implied by refusing to visit previous states. Pursuing an infinite loop is a valid policy, even if your current representation of it will crash the program; maybe the bug is the overflow, not the policy that caused it. If given the choice between losing the game, and refusing to let the game end, which do you want your agent to prefer?
Playing ad infinitum
If you want your agent to avoid losing at all costs -- that is, you want it to prefer indefinite play over loss -- then I would suggest treating any repeated state as a terminal state and assigning it a value somewhere between winning and losing. After all, in a sense it is terminal -- this is the loop the game will enter forever and ever and ever, and the definite result of it is that there is no winner. However, remember that if you are using simple minimax (one utility function, not two), then this implies that your opponent also regards eternal play as a middling result.
It may sound ridiculous, but maybe playing unto infinity is actually a reasonable policy. Remember that minimax assumes the worst case -- a perfectly rational foe whose interests are the exact opposite of yours. But if, for example, you're writing an agent to play against a human, then the human will either err logically, or will eventually decide they would rather end the game by losing -- so your agent will benefit from patiently staying in this Nash equilibrium loop!
Alright, let's end the game already
If you want your agent to prefer that the game end eventually, then I would suggest implementing a living penalty -- a modifier added to your utility which decreases as a function of time (be it asymptotic or without bound). Implemented carefully, this can guarantee that, eventually, any end is preferable to another turn. With this solution as well, you need to be careful about considering what preferences this implies for your opponent.
A third way
Another common solution is to depth-limit your search and implement an evaluation function. This takes the game state as its input and just spits out a utility value which is its best guess at the end result. Is this provably optimal? No, not unless your evaluation function is just completing the minimax, but it means your algorithm will finish within a reasonable time. By burying this rough estimate deep enough in the tree, you wind up with a pretty reasonable model. However, this produces an incomplete policy, which means that it is more useful for a replanning agent than for a standard planning agent. Minimax replanning is the usual approach for complex games (it is, if I'm not mistaken, the basic algorithm followed by Deep Blue), but since this is a very simple game you probably don't need to take this approach.
A side note on abstraction
Note that all of these solutions are conceptualized as either numeric changes to or estimations of the utility function. This is, in general, preferable to arbitrarily throwing away possible policies. After all, that's what your utility function is for -- any time you make a policy decision on the basis of anything except the numeric value of your utility, you are breaking your abstraction and making your code less robust.

Connect Four - Negamax AI evaluation function issue

I'm trying to implement NegaMax ai for Connect 4. The algorithm works well some of the time, and the ai can win. However, sometimes it completely fails to block opponent 3 in a rows, or doesn't take a winning shot when it has three in a row.
The evaluation function iterates through the grid (horizontally, vertically, diagonally up, diagonally down), and takes every set of four squares. It then checks within each of these sets and evaluates based on this.
I've based the function on the evaluation code provided here: http://blogs.skicelab.com/maurizio/connect-four.html
My function is as follows:
//All sets of four tiles are evaluated before this
//and values for the following variables are set.
if (redFoursInARow != 0)
{
redScore = INT_MAX;
}
else
{
redScore = (redThreesInARow * threeWeight) + (redTwosInARow * twoWeight);
}
int yellowScore = 0;
if (yellowFoursInARow != 0)
{
yellowScore = INT_MAX;
}
else
{
yellowScore = (yellowThreesInARow * threeWeight) + (yellowTwosInARow * twoWeight);
}
int finalScore = yellowScore - redScore;
return turn ? finalScore : -finalScore; //If this is an ai turn, return finalScore. Else return -finalScore.
My negamax function looks like this:
inline int NegaMax(char g[6][7], int depth, int &bestMove, int row, int col, bool aiTurn)
{
{
char c = CheckForWinner(g);
if ('E' != c || 0 == depth)
{
return EvaluatePosition(g, aiTurn);
}
}
int bestScore = INT_MIN;
for (int i = 0; i < 7; ++i)
{
if (CanMakeMove(g, i)) //If column i is not full...
{
{
//...then make a move in that column.
//Grid is a 2d char array.
//'E' = empty tile, 'Y' = yellow, 'R' = red.
char newPos[6][7];
memcpy(newPos, g, sizeof(char) * 6 * 7);
int newRow = GetNextEmptyInCol(g, i);
if (aiTurn)
{
UpdateGrid(newPos, i, 'Y');
}
else
{
UpdateGrid(newPos, i, 'R');
}
int newScore = 0; int newMove = 0;
newScore = NegaMax(newPos, depth - 1, newMove, newRow, i, !aiTurn);
newScore = -newScore;
if (newScore > bestScore)
{
bestMove = i;
bestScore = newScore;
}
}
}
}
return bestScore;
}
I'm aware that connect four has been solved are that there are definitely better ways to go about this, but any help or suggestions with fixing/improving this will be greatly appreciated. Thanks!

minimax c++ implementation for tic tac toe

void generate_moves(int gameBoard[9], list<int> &moves)
{
for (int i = 0; i < 9; i++)
{
if (gameBoard[i] == 0){
moves.push_back(i);
}
}
}
int evaluate_position(int gameBoard[9], int playerTurn)
{
state currentGameState = checkWin(gameBoard);
if (currentGameState != PLAYING)
{
if ((playerTurn == 1 && currentGameState == XWIN) || (playerTurn == -1 && currentGameState == OWIN))
return +infinity;
else if ((playerTurn == -1 && currentGameState == XWIN) || (playerTurn == 1 && currentGameState == OWIN))
return -infinity;
else if (currentGameState == DRAW)
return 0;
}
return -1;
}
int MinMove(int gameBoard[9], int playerTurn)
{
//if (checkWin(gameBoard) != PLAYING) { return evaluate_board(gameBoard); };
int pos_val = evaluate_position(gameBoard, playerTurn);
if (pos_val != -1) return pos_val;
int bestScore = +infinity;
list<int> movesList;
generate_moves(gameBoard, movesList);
while (!movesList.empty())
{
gameBoard[movesList.front()] = playerTurn;
int score = MaxMove(gameBoard, playerTurn*-1);
if (score < bestScore)
{
bestScore = score;
}
gameBoard[movesList.front()] = 0;
movesList.pop_front();
}
return bestScore;
}
int MaxMove(int gameBoard[9], int playerTurn)
{
//if (checkWin(gameBoard) != PLAYING) { return evaluate_board(gameBoard); };
int pos_val = evaluate_position(gameBoard, playerTurn);
if (pos_val != -1) return pos_val;
int bestScore = -infinity;
list<int> movesList;
generate_moves(gameBoard, movesList);
while (!movesList.empty())
{
gameBoard[movesList.front()] = playerTurn;
int score = MinMove(gameBoard, playerTurn*-1);
if (score > bestScore)
{
bestScore = score;
}
gameBoard[movesList.front()] = 0;
movesList.pop_front();
}
return bestScore;
}
int MiniMax(int gameBoard[9], int playerTurn)
{
int bestScore = -infinity;
int index = 0;
list<int> movesList;
vector<int> bestMoves;
generate_moves(gameBoard, movesList);
while (!movesList.empty())
{
gameBoard[movesList.front()] = playerTurn;
int score = MinMove(gameBoard, playerTurn);
if (score > bestScore)
{
bestScore = score;
bestMoves.clear();
bestMoves.push_back(movesList.front());
}
else if (score == bestScore)
{
bestMoves.push_back(movesList.front());
}
gameBoard[movesList.front()] = 0;
movesList.pop_front();
}
index = bestMoves.size();
if (index > 0) {
time_t secs;
time(&secs);
srand((uint32_t)secs);
index = rand() % index;
}
return bestMoves[index];
}
I created a tic tac toe program in C++ and tried to implement a MiniMax algorithm with exhaustive search tree.
These are the functions I have written using wiki and with the help of some websites. But the AI just doesn't work right and at times doesn't play its turn at all.
Could someone have a look and please point out if there is anything wrong with the logic?
This is how I think it works:
Minimax : This function starts with very large -ve number as best score and goal is to maximize that number. It calls minMove function. If new score > best score, then best score = new score...
MinMove : This function evaluates game board. If game over then it returns -infinity or +infinity depending on who won. If game is going on this function starts with max +infinity value as best score and goal is to minimize it as much possible. It calls MaxMove with opponent player's turn. (since players alternate turns).
If score < best score then best score = score. ...
MaxMove : This function evaluates game board. If game over then it returns -infinity or +infinity depending on who won. If game is going on this function starts with least -infinity value as best score and goal is to maximize it as much possible. It calls MinMove with opponent player's turn. (since players alternate turns).
If score > best score then best score = score. ...
Minmove and MaxMove call each other mutually recursively, MaxMove maximizing the value and MinMove minimizing it. Finally it returns the best possible moves list.
If there are more than 1 best moves, then a random of them is picked as the computer's move.
In MiniMax, MinMove(gameBoard, playerTurn) should be MinMove(gameBoard, -playerTurn) as you do in MaxMove.
As you use MinMove and MaxMove, your evaluation function should be absolute. I mean +infinity for XWIN
and -infinity for OWIN. And so MinMove can only be use when player == -1 and MaxMove when player == 1, thus the parameter become useless. And so MiniMax can only be used by player == 1.
I have done some changes in your code and it works (https://ideone.com/Ihy1SR).

Maze solver recording backtracked paths

I've gotten my maze solver program to work but it seems to be including back tracked spaces (places it went to and hit a wall so it had to turn around) in the final solution path that it outputs. Here is an example:
How can I prevent this in my current implementation below:
int dir = 4;
bool visited[Max_Maze][Max_Maze][dir];
for (row = 0; row < size; ++ row)
{
for (col = 0; col < size; ++ col)
{
for (dir = 0; dir < 4; ++ dir)
{
visited[row][col][dir]=false;
}
}
}
bool notSolved = true;
int path = 0;
row = 0;
col = 0;
rowStack.push(row);
colStack.push(col);
while (notSolved)
{
//from perspective of person looking at maze on screen
if (((row-1)>=0)&&(maze[row - 1][col] == 0)&&(visited[row][col][0]==false))
{
// if that space is not out of bounds and if you can go up
// and you have not gone in that direction yet, go up
visited[row][col][0] = true;
row--;
rowStack.push(row);
colStack.push(col);
path++;
}
else if (((col+1)<size)&&(maze[row][col + 1] == 0)&&(visited[row][col][1]==false))
{
//else if you can go right etc., go right
visited[row][col][1] = true;
col++;
rowStack.push(row);
colStack.push(col);
path++;
}
else if (((row+1)<size)&&(maze[row + 1][col] == 0)&&(visited[row][col][2]==false))
{
//else if you can go down etc., go down
visited[row][col][2] = true;
row++;
rowStack.push(row);
colStack.push(col);
path++;
}
else if (((col-1)>=0)&&(maze[row][col - 1] == 0)&&(visited[row][col][3]==false))
{
//else if you can go left etc., go left
visited[row][col][3] = true;
col--;
rowStack.push(row);
colStack.push(col);
path++;
}
else
{
//if stuck
if (path == 0)
{
cout << "No Solution Path" << endl;
notSolved = false;
}
else
{
rowStack.pop();
colStack.pop();
row = rowStack.top();
col = colStack.top();
path--;
}
}
if((maze[row][col] == 0) && (row == (size - 1) && col == (size - 1)))
{
//if we reached an exit
cout << "Solution Path:(in reverse)" << endl;
for (int i = 0; i <= path; i++)
{
cout << "row:" << rowStack.top() << " col:" << colStack.top() << endl;
rowStack.pop();
colStack.pop();
}
notSolved = false;
}
}
Simple fix or total restructuring needed?
As the solver goes right into that dead end, it records that it has "visited right from (R, C)", because your visited array is three dimensional. But it never records that it has "visited left from (R, C + 1)". So it thinks it's fine to move to the same position twice, so long as it doesn't make the exact same move twice (which it doesn't, as it's moving left when it backtracks, not right).
It looks like it will work fine as is if you change visited to be a 2-dimensional array and only record positions, not moves. Then every square you've visited before blocks further movement, but that's okay because if the correct solution requires going back to that square you'll eventually hit the else case enough to pop back to it, and from there three must be a never-visited square to explore.
Without commenting on your specific solution, you might consider redoing as a standard depth first search algorithm, which I think you will agree is somewhat more clear:
boolean dfs (State s) {
if (end_state(s)) {
return true;
}
vector<option_t> options = generate_moves(s);
for (vector<option_t>::iterator it = options.begin();
it != options.end(); it++) {
make_move(s, it);
boolean found = dfs(s);
if (found) {
cout << s << endl;
}
undo_move(s, it);
if (found) return true;
}
return false;
}
As you can see, this will work as long as you can create:
some class State that holds your current maze state
end_state function that knows when State is at a solution
generate_moves function that can find all options for current State
make_move that can apply a move against your state
undo_move that can undo a move against state
you define option_t such that it represents a move option
define operator<< for State
I can tell you that certainly the solution I give you does not have problems with printing backtracked spaces (should be clear also from the code).