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The other day, I wrote a console game of Tic-Tac-Toe in c++ for my son. He wanted me to add a computer, and I ended us using the minimax algorithm for the first time. I did some quick testing, but really just gave my laptop to my son as soon as it was printing stuff, who played with it for a couple minuets. I looked over his sholder once or twice, and noticed that it wasn't playing optimally, iv'e been trying to debug it, but I can't see where it goes wrong. I tried getting rid of alpha beta prunning, but that did not change anything.
For context, on the board the computer is -1, blank is 0, and the player is 1.
Here is the minimax function:
int minimax(int board[9], int depth, int alpha, int beta, bool isMaxizimaizingPlayer)
{
bool found = false;
for (int i = 0; i < 9; i++)
{
if (board[i] == 0)
{
found = true;
}
}
if (!found)
{
return eval(board);
}
if (depth == 0 || eval(board) != 0)
{
return eval(board);
}
if (isMaxizimaizingPlayer)
{
int maxEval = -2;
for (int spot = 0; spot < 9; spot++)
{
if (board[spot] == 0)
{
board[spot] = 1;
int e = minimax(board, depth - 1, alpha, beta, false);
if (e > maxEval)
{
maxEval = e;
}
//if (beta < alpha)
//{
// break;
//}
board[spot] = 0;
}
}
return maxEval;
}
else {
int minEval = 2;
for (int spot = 0; spot < 9; spot++)
{
if (board[spot] == 0)
{
board[spot] = -1;
int e = minimax(board, depth - 1, alpha, beta, true);
if (e < minEval)
{
minEval = e;
}
//if (beta < alpha)
//{
// break;
//}
board[spot] = 0;
}
}
return minEval;
}
}
To be compleate, here is my eval function:
int eval(int board[9])
{
/*horizontial*/
for (int i = 0; i < 3; i++)
{
if (board[i * 3] == board[i * 3 + 1] && board[i * 3 + 2] == board[i * 3] && board[i * 3] != 0)
{
return board[i * 3];
}
}
/*vertical*/
for (int i = 0; i < 3; i++)
{
if (board[i] == board[i + 3] && board[i] == board[i + 6] && board[i] != 0)
{
return board[i];
}
}
/*Both diags*/
if (board[4] != 0) {
if (board[0] == board[4] && board[0] == board[8])
{
return board[4];
}
if (board[2] == board[4] && board[4] == board[6])
{
return board[4];
}
}
return 0;
}
And here is the inital call:
int spot = 0;
int minEval = 2;
for (int i = 0; i < 9; i++)
{
if (board[i] == 0)
{
board[i] = -1;
int score = minimax(board, 3, -2, 2, false);
if (score < minEval) {
minEval = score;
spot = i;
}
board[i] = 0;
}
}
std::cout << "The computer went in spot " << spot + 1 << std::endl;
board[spot] = -1;
printBoard(board);
It looks like you only call minimax with a depth of three, so the algorithm will only look up to three moves ahead, if you want optimal play you need to set the depth to > 9, so that the agent is always looking ahead to the end of the game.
I working on my project this project have a frame to [100] x [25] matrix and i try to add animation but my character is going left but it's not going right.
i tried with "counter" variable but its not work.
int counter = 1;
int counter2 = 1;
int left_border = 1;
int matris1 = sizeof(map)/100;
int matris2 = sizeof(map[0])/4;
int startx = 19;
int starty = 8;
while (true)
...
int right = 0, left = 0;
...
for (int a = 0; a < matris2; a++)
cout << "\n#"; //i have this because i make it square map.
for (int k = 0; k < matris1 - 2; k++)
{
if (left == 1)
{
if (((startx+2)-counter) == left_border)
{
counter = 0;
//cout << "SINIR!!"<< endl ;
}
if (k == (startx-counter) and a == starty)
{
counter += 1;
cout << "O";
}
else {
cout << " ";
}
}
else if (right == 1)
{
if (k == (startx+counter2) and a == starty)
{
counter2 += 1;
cout << "O";
}
its need to be going right but its not.
if you need full code.
https://codeshare.io/UbKVU
[![This is the map and "O" is the character]
https://i.stack.imgur.com/uyGQo.png
The code is very difficult to follow - you should have a coordinate system. I've made a simple example below. Update the player coordinate when a key is pressed and redraw the map x by y position, if the player is there then draw the 'O', otherwise if its a wall draw an 'X' (in this case), otherwise draw a space ' '.
using namespace std;
#include <iostream>
#include <conio.h>
#include <stdlib.h>
#define MAPW 15 // map width
#define MAPH 15 // map height
int map[MAPW][MAPH];
#define WALL 1
#define EMPTY 0
void initmap()
{
// just set the map to have walls around the border
for (int x = 0; x < MAPW; x++)
{
for (int y = 0; y < MAPH; y++)
{
if (x == 0 || y == 0 || x == (MAPW - 1) || y == (MAPH - 1))
map[x][y] = WALL;
else
map[x][y] = EMPTY;
}
}
}
int px = MAPW / 2; // player x
int py = MAPH / 2; // player y
void main()
{
initmap(); // initialize map
cout << "Press A/W/S/D to begin and move";
while (1)
{
if (kbhit()) // key pressed?
{
switch (getch()) // which key?
{
case 'a':
if (px > 0 && map[px - 1][py] != WALL) // can go left?
px--; // update x coordinate
break;
case 'd':
if (px < (MAPW-1) && map[px + 1][py] != WALL) // can go right?
px++; // update x coordinate
break;
case 'w':
if (py > 0 && map[px][py - 1] != WALL) // can go up?
py--; // update y coordinate
break;
case 's':
if (py < MAPH && map[px][py + 1] != WALL) // can go down?
py++; // update y coordinate
break;
}
// update map - clear screen and redraw
system("CLS");
// draw map each line
for (int y = 0; y < MAPH; y++)
{
for (int x = 0; x < MAPW; x++)
{
// its a wall?
if (map[x][y] == WALL)
cout << "X";
else
{
// is the player there?
if (x == px && y == py)
{
// draw the player
cout << "O";
}
else // empty space
cout << " ";
}
}
// next line
cout << "\n";
}
}
}
}
I would like some help understanding why my program is printing a grid of
....................
....................
....................
....................
....................
...............OOOOO
OOOOOOOOOOOOO.......
....................
....................
....................
The correct output would be so:
....................
....................
....................
....................
....................
....................
....................
.............O.O....
..............OO....
..............O.....
The way I wrote it is to create a copy of the old state and manipulate it using the rules of the game. After I check every cell, I store count of the number of neighbors alive for that cell. IF the count is greater than 3 or less than two, the cell will die.
If a cell has a count of 2 or 3 neighbors, it remains alive.
If a dead cell has a count of 3, it becomes alive.
These rules are directly applied to the copy version instead of the old and then print the copy.
I've tried using a debugger but I'm still unsure of how to use it properly. I haven't notice any red flags as of yet.
Here's my code:
#include <iostream>
#include <vector>
using std::vector;
using std::cout;
vector<vector<bool> > world = {
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}
};
void generate(const vector<vector<bool> >&g,vector<vector<bool> >&newworld)
{
int count = 0;
newworld = g;
for(size_t i = 0; i < g.size();i++) {
for(size_t j = 0; j < g[i].size();j++) {
int x = g.size(); //I rows
int y = g[i].size(); //J columns
//wrap the edges with formula (x+n)%n where n = NumOfRows or NumOfCol
if(g[(((i+1)+x)%x)][(((j-1)+y)%y)]==true){//top left
count++;
}
else if(g[(((i+1)+x)%x)][j]==true){//top middle
count++;
}
else if(g[(((i+1)+x)%x)][(((j+1)+y)%y)]==true){//top right
count++;
}
else if(g[i][(((j-1)+y)%y)]==true){//left cell
count++;
}
else if(g[i][(((j+1)+y)%y)]==true){//right cell
count++;
}
else if(g[(((i-1)+x)%x)][(((j-1)+y)%y)]==true){ //bottom left
count++;
}
else if(g[(((i-1)+x)%x)][j]==true){//bottom middle
count++;
}
else if(g[(((i-1)+x)%x)][(((j+1)+y)%y)]==true){//bottom right
count++;
}
if (g[i][j]) {
if(count > 3 || count < 2) {//if alive cell has more than 3 or less than 2, die
newworld[i][j] = false;
}
else if (count == 2 || count == 3) { //remain the same
newworld[i][j] = g[i][j];
}
}
else if (g[i][j] == false) {//dead come alive
if(count == 3) {
newworld[i][j] = true;
}
}
}
}
}
void display(vector<vector<bool> >&a)
{
for(size_t row = 0; row <a.size(); row++) {
for(size_t column = 0; column <a[row].size(); column++){
if (a[row][column]) {
cout << 'O';
}
else {
cout << '.';
}
}
cout << '\n';
}
}
int main()
{
vector<vector<bool> > newworld;
generate(world,newworld);
display(newworld);
return 0;
}
The function generate has (at least) two problem.
count is initialized outside the nested loops, so it's never reset to zero (as it should, for every cell) and keeps growing.
All the conditions are mutually exclusive, so whenever one it's met, the others are skipped. There shouldn't be any else if, but only ifs.
Keeping the data structure you chose, you can rewrite that function as
using gof_t = std::vector<std::vector<bool>>;
void generate(gof_t& g, gof_t& newworld)
{
for(size_t i = 0, x = g.size(); i < x; i++)
{
for(size_t j = 0, y = g[i].size(); j < y; j++)
{
size_t i_prev = (i + x - 1) % x;
size_t i_next = (i + 1) % x;
size_t j_prev = (j + y - 1) % y;
size_t j_next = (j + 1) % y;
int count = g[i_prev][j_prev] + g[i_prev][j] + g[i_prev][j_next]
+ g[i ][j_prev] + g[i ][j_next]
+ g[i_next][j_prev] + g[i_next][j] + g[i_next][j_next];
newworld[i][j] = g[i][j] ? (count == 2 || count == 3) : (count == 3);
}
}
std::swap(g, newworld); // <-- Passing by non const reference, we can swap without copying
}
Live (pun intended), here.
I'm trying to write a maze solver using backtracking. It should see whether it's possible to solve a given puzzle from the starting point S to the end point E. The pseudo code can be seen in this link here. My implementation looks like this:
const int N = 8; // global var
bool exploreMaze(char maze[][N], int x, int y)
{
if(y >= 8 || y < 0 || x >= 7 || x < 0) // null char at end of each 1d array
return false;
if(maze[x][y] == '*')
return false;
if(maze[x][y] == 'E')
return true;
maze[x][y] = '*'; // set grid to '*' as to not loop infinitely
if(exploreMaze(maze, x + 1, y))
{
cout << "up" << ", ";
return true;
}
if(exploreMaze(maze, x - 1, y))
{
cout << "down" << ", ";
return true;
}
if(exploreMaze(maze, x, y - 1))
{
cout << "left" << ", ";
return true;
}
if(exploreMaze(maze, x, y + 1))
{
cout << "right" << ", ";
return true;
}
return false;
}
bool isMazeSolvable(char maze[][N])
{
int startX = -1, startY = -1;
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
if(maze[i][j] == 'S')
startX = i;
startY = j;
}
}
if(startX == -1)
return false;
return exploreMaze(maze, startX, startY);
}
int main()
{
char maze[N][N] = {"*******", " S ", "*******", " E ", "*******",
"*******", "*******", "*******"};
cout << isMazeSolvable(maze);
return 0;
}
The array that I'm testing in main should definitely have no solution, but somehow I'm getting 1(true) as the output. Any ideas?
Your maze '*' is only initialized out in the Y direction by 7 characters, but your maze walker checks out to 8 character. This allows it to walk around the end of your walls.
I added a quick maze print function and changed the exploreMaze to put '.' where it walks. Giving the following output:
Initial maze:
*******
S
*******
E
*******
*******
*******
*******
left, left, left, left, left, up, up, right, right, right, right, right, right,
1
After explore:
*******
.......
*******.
E.....
*******
*******
*******
*******
Soluton: Either change the initializer to use 8 character walls, or change the exploreMaze function to only look 7 characters in the Y direction.
Also note: You are not doing the "backtracking" part of the maze solver, because you mark where you have been but don't clean off your path on your way out of the recursion. Add
maze[x][y] = ' '; // Clear the grid so we can try this spot again in another recursion
to the end of your exploreMaze function
Hey all, working on a C++ little game, "Connect 3." This is just like Connect 4, except we only need a match of 3 to win the game. I am storing my board in a 2D vector, which holds ints.
vector< vector<int> > vector2d;
And I have an "X" stored as a 1, and an "O" stored as a -1, with 0 being an empty space. It seems to be working so far.
So, in my algorithm for playing against the computer, it finds the best move possible. I have the algorithm finished, but it needs to know when a "base case" has been hit. (It's recursive.) A base case is either:
Someone has gotten 3 in a row, or
The board is full
Checking if the board is full is easy. I just iterate through and see if any space is a "0". If it is, the board isn't full. But before I check that, I need to see if anyone has gotten 3 in a row, which is where I'm having issues. The only way I can think of doing this is big and complicated, going through the board 3 different times, looking for Horizontal matches of 3, vertical matches of 3, and Diagonal matches of 3. I'm not even sure where to begin in doing that, and I'm hoping there is a better way to do this. Help would be much appreciated!
Also, not sure I'm allowed to use Boost, I haven't yet so far, and I'd like to not have to use it. (Not sure if the school computers have it).
Edit: The board does not need to be 3 by 3. It could be 1 by 7, 7 by 7, or any size. If it's not a legal size (0,0), my code will tell the user that, but any other board should work. I've used the vector sizes to see how big the board is.
You don't have to check the whole board every time. Only the new piece makes a difference so you only have to check those end conditions that include the new piece. There are 8 different directions you need to check, but every two of them are on the same line and should be checked together. Directions can be defined as (delta_X, delta_Y) pairs: (1,0),(0,1),(1,1),(1,-1). Your code should traverse in each direction (as in code from Leonid) and try to count as many pieces with the same value as new piece. Then it should traverse in opposite direction which is (-x,-y) from current direction, and count those pieces as well. If the number of counted pieces is N-1 (new piece is already counted) then you have a winner.
So lets say you are using an 3x3 board. There are a finite number of winning lines that can be formed.
1 0 0 1 1 1 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0
1 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 0 1 0
1 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 1 1 1 0 1 0
Now if you give each board location a bit assignment as follows:
1 2 4
8 16 32
64 128 256
now you can work out that the 8 winning lines are as follows:
1 | 8 | 64 = 73
1 | 2 | 4 = 7
1 | 16 | 256 = 273
4 | 16 | 64 = 84
4 | 32 | 256 = 292
8 | 16 | 32 = 56
64 | 128 | 256 = 448
2 | 16 | 128 = 146
Now if you store a 1 in any bit position that a given player has you can easily step through each of the "solutions" above and test against the 8 values above.
So suppose the 2 players have the following positions:
1 1 0 0 0 1
1 0 0 0 1 1
1 0 1 0 1 0
If you tot up there values as done for the "solutions" as follows you get
1 | 2 | 8 | 64 | 256 = 331
4 | 16 | 32 | 128 = 180
So we know the winnign line is the 1 | 8 | 64 = 73 line so we can test using a bit wise and as follows
331 & 73 = 73
180 & 73 = 0
So we can easily detect that player 1 has 3 in a row and has one as the result of the "and" is not 0.
This means you can calculate a winner in a maximum of 8 steps (ie checking both players running totals against the 8 possible answers).
Obviously complexity increases as you get larger and it can seem a lot more complicated when you run out of bits (look at std::bitset, for example of how to handle that) but the end game will ALWAYS take less iterations to check than a brute force method. Obviously it takes a bit more time to set up but you only calculate the end game conditions once per board type so that time gets amortised across several plays.
Anyway ... Thats how I'd do it :D
The following C++ O(N*M) solution from algorithmical complexity perspective is the best possible as we need to check in the worst case each cell of the board. It iterates over all cells in the board (i and j), tries to go in 4 directions (k), and from there checks that 3 cells (l) in direction k are occupied and equal.
vector<vector<int> > board(n, vector<int>(m)); // initialize
/* down down-right right up-right */
int di[] = {1, 1, 0, -1 }; // four directions i coordinate
int dj[] = {0, 1, 1, 1 }; // four directions j coordinate
for (int i = 0; i < n; i++) { // for each row
for (int j = 0; j < m; j++) { // for each column
for (int k = 0; k < 4; k++) { // for each direction
int ii = i, jj = j;
bool found = true;
if (board[ii][jj] == 0) continue; // empty space
for (int l = 1; l < 3 && found; l++) { // need 3 in a row
int iii = ii + di[k], jjj = jj + dj[k];
if (iii < 0 || iii >= n) found = false, continue; // off bounds
if (jjj < 0 || jjj >= n) found = false, continue; // off bounds
if (board[iii][jjj] != board[ii][jj]) found = false;
}
if (found) {
printf("Hurray!\n");
return;
}
}
}
}
I made a game like that , the first thing I ever made in C++ actually ( Who needs hello world :P)
And everyone can use it if they want.
Just don't forget it's my first C++ thing and it's definatly not properly coded :P but it has some nice C++ things like that in it. But there's a 100% optimized search algorithm in there that checks the absolute least amount of required permutation to check three in a row win conditions with heavy commenting and ASCII art. That could be quite usefull.
Oh almost forgot the mention, It's a console application thingy (black screen DOS envi ,whatever it's called). It has an AI that (if this is my latest version) Should do pretty well. AND the grid is dynamically built (which was the hard part) U can play 3 in a row, but with a max of 20x20 grid ( lame game I found out, much more fun as 4 in a row with gravity )
Here you go:
// DrieOpEenRij.cpp : Defines the entry point for the console application.
#include "stdafx.h"
#include <iostream>
#include <string>
#include <typeinfo>
using namespace std;
typedef unsigned short USHORT;
//USE ONLY IN A SQUARE GRID
//This method checks a win for the minimimum amount of spaces covering 100% amount of the grid
//It has 100% coverage and close to 0% overhead, discrimination between who to check for is required and
//so currentMove char is required to check for win on 'H' human and 'C' Computer
void CheckForWin(const char* Grid_ptr , const USHORT GridSize , const USHORT GridWidth ,bool &humanWin, bool &computerWin, const char currentMove)
{
//check for an x from 1-end of array
//for all x's check if that makes a 3 line once per linetype
//check for horizontal win (dont get overhead on edges)
//A non square grid will have been detected by now
const USHORT rowStart = 0;
const USHORT rowEnd = GridWidth-1;
USHORT passRowCounter = 1;
const USHORT Side = GridWidth;
const USHORT cond1 = rowEnd-2;
const USHORT cond2 = GridSize-Side*2;
//Check for all human win options ( after a human move )
if (currentMove == 'H')
{
//Check for human win code
//Check all array slots for an occurence of 'X'
for(USHORT i = 0; i < GridSize; i++)
{
//Local stack variables, optimizations for iterations in loops and if statements,
//also for readability, this is (only efficient and) done only when it is guaranteed
//to be used in every for jump.
USHORT iModSide = i % Side;
USHORT SideMinTwo = Side - 2;
USHORT SidePlusTwo = Side + 2;
USHORT iPlusSide = i + Side;
USHORT iPlusSideTimesTwo = i + Side * 2;
USHORT iPlusOne = i + 1;
USHORT iPlusTwo = i + 2;
//If an X is found evaluate a win scenario
if (Grid_ptr[i] == 'X')
{
//For each row -->
if (iModSide < SideMinTwo)
{
//Check horizontal win from left to right
if (Grid_ptr[i + 1] == 'X' && Grid_ptr[i + 2] == 'X')
{
humanWin = true;
break;
}
}
//For the two values under the 'X' (colomn wise) check for 'X''X'
if (iPlusSideTimesTwo < GridSize)
{
if(Grid_ptr[iPlusSide] == 'X' && Grid_ptr[iPlusSideTimesTwo] == 'X')
{
humanWin = true;
break;
}
}
//CHECK FOR DIAGONAL WIN FROM TOP LEFT TO DOWN RIGHT IN ALL POSSIBLE+LEGAL SLOTS!
// [X] [X] [?] [?] This illustration shows that checking only at X will suffice
// [X] [X] [?] [?] for this specific check in screening for all Top Left --> Down Right
// [?] [?] [?] [?] diagonal wins, similarly the Top Right --> Down Left is done mirrored
// [?] [?] [?] [?] All other wins using this vector are impossible!
// Using this amount of conditions to find it saves a lot of searching and with it time
if (iPlusSideTimesTwo < GridSize && iModSide < SideMinTwo)
{
if (Grid_ptr[i+Side+1] == 'X' && Grid_ptr[iPlusSideTimesTwo+2] == 'X')
{
humanWin = true;
break;
}
}
//CHECK FOR DIAGONAL WIN FROM TOP LEFT TO DOWN RIGHT IN ALL POSSIBLE+LEGAL SLOTS!
// [?] [?] [Y] [Y] This illustration shows that checking only at Y will suffice
// [?] [?] [Y] [Y] for this specific check in screening for all Top Right --> Down Left
// [?] [?] [?] [?] diagonal wins, similarly the Top Left --> Down Right is done mirrored
// [?] [?] [?] [?] This because all other wins using this vector are impossible!
// Using this amount of conditions to find it saves a lot of searching and with it time
if (i % Side > 1 && i + Side*2-2 < GridSize)
{
if (Grid_ptr[i+Side-1] == 'X' && Grid_ptr[i+Side*2-2] == 'X')
{
humanWin = true;
break;
}
}
} //end if arrayvalue is 'X'
} //end for each value in array
} //end if currentMove 'H'
else if (currentMove == 'C')
{
//Check for human win code
//Check all array slots for an occurence of 'X'
for(USHORT i = 0; i < GridSize; i++)
{
//Local stack variables, optimizations for iterations in loops and if statements,
//also for readability, this is (only efficient and) done only when it is guaranteed
//to be used in every for jump.
USHORT iModSide = i % Side;
USHORT SideMinTwo = Side - 2;
USHORT SidePlusTwo = Side + 2;
USHORT iPlusSide = i + Side;
USHORT iPlusSideTimesTwo = i + Side * 2;
USHORT iPlusOne = i + 1;
USHORT iPlusTwo = i + 2;
//If an X is found evaluate a win scenario
if (Grid_ptr[i] == 'O')
{
//For each row -->
if (iModSide < SideMinTwo)
{
//Check horizontal win from left to right
if (Grid_ptr[i + 1] == 'O' && Grid_ptr[i + 2] == 'O')
{
computerWin = true;
break;
}
}
//For the two values under the 'O' (colomn wise) check for 'O''O'
if (iPlusSideTimesTwo < GridSize)
{
if(Grid_ptr[iPlusSide] == 'O' && Grid_ptr[iPlusSideTimesTwo] == 'O')
{
computerWin = true;
break;
}
}
//CHECK FOR DIAGONAL WIN FROM TOP LEFT TO DOWN RIGHT IN ALL POSSIBLE+LEGAL SLOTS!
// [X] [X] [?] [?] This illustration shows that checking only at X will suffice
// [X] [X] [?] [?] for this specific check in screening for all Top Left --> Down Right
// [?] [?] [?] [?] diagonal wins, similarly the Top Right --> Down Left is done mirrored
// [?] [?] [?] [?] All other wins using this vector are impossible!
// Using this amount of conditions to find it saves a lot of searching and with it time
if (iPlusSideTimesTwo < GridSize && iModSide < SideMinTwo)
{
if (Grid_ptr[i+Side+1] == 'O' && Grid_ptr[iPlusSideTimesTwo+2] == 'O')
{
computerWin = true;
break;
}
}
//CHECK FOR DIAGONAL WIN FROM TOP LEFT TO DOWN RIGHT IN ALL POSSIBLE+LEGAL SLOTS!
// [?] [?] [Y] [Y] This illustration shows that checking only at Y will suffice
// [?] [?] [Y] [Y] for this specific check in screening for all Top Right --> Down Left
// [?] [?] [?] [?] diagonal wins, similarly the Top Left --> Down Right is done mirrored
// [?] [?] [?] [?] This because all other wins using this vector are impossible!
// Using this amount of conditions to find it saves a lot of searching and with it time
if (iPlusSideTimesTwo+2 < GridSize && iModSide < SidePlusTwo)
{
if (Grid_ptr[i+Side-1] == 'O' && Grid_ptr[i+Side*2-2] == 'O')
{
computerWin = true;
break;
}
}
} //end if arrayvalue is 'O'
} //end for each value in array
}// else if currentMove 'C'
} //end method
//useAI(char* Grid_ptr) { }
//weighGrid (char* Grid_ptr) { for (USHORT i = 0; i < GridSize(find out); i++) {} }
void PrintGrid(char* Grid_ptr, USHORT GridWidth, USHORT GridHeight, USHORT GridSize)
{
//Abort this method if the Grid is not Square
if (GridWidth != GridHeight)
{
cout << "Warning! \n\nGrid is not square. This method will likely fail!" << endl;
cout << "Aborting method!" << endl;
cout << "Press a key to return to program";
}
else
{
//Since this code block's applicable to a square grid
//Width or Height is not relevant, both should work
//I have chosen to stick with Width everywhere.
USHORT rowStart = 0;
USHORT rowEnd = GridWidth-1;
USHORT passRowCounter = 1;
USHORT Side = GridSize / GridHeight;
for(USHORT i = 0; i < Side; i++)
{
//GO TO NEXT ROW CODE
rowEnd = Side * passRowCounter;
passRowCounter++;
//PRINT ALL IN THIS ROW
for (USHORT j = rowStart; j < rowEnd; j++)
{
cout << Grid_ptr[j];
}
rowStart = rowEnd;
cout << "\n";
}
}
}
void useAI(char* Grid_ptr, USHORT GridSize, USHORT GridWidth)
{
//Check all values in the array
//If the value is '?' weigh the priority
//else continue
//Weighing the priority
//If ('O' Present in legal ranges) add prio +1
//The AI Will function on this concept
//All array slots have a weight, the highest weight means the best position
//From top prio to lowest prio that means -->
//WIN IN ONE MOVE (weight + 50)
//NOT LOSE IN ONE MOVE (weight + 15)
//BLOCK ENEMY + LINK UP OWN ( Equal prio but stacks so both matter ) weight +1
//These weights are determined using 8 directional vectors sprouting from all 'X' and 'O' locations in the grid
//In it's path if it encounters on loc 1 'X' loc 2 + weight = 50 , and vice versa, else +1 for all 8 vectors
//Create a weightgrid to store the data
USHORT* WeightGrid_ptr = new USHORT[GridSize];
USHORT* fattest_ptr = new USHORT(0);
USHORT* fattestIndex_ptr = new USHORT(0);
USHORT Side = GridWidth;
//Suggestion for optimization , make a forumula table to play all 8 vectors instead
//Per vector u need Condition for the direction first space and next space. 24 statements in a list
//A bit complex and harder to read so for now went the east 8 vectors copy pasting. But aware of the
//solution none-the-less! Unfortunatly though it seems like a maze of code, it is well documented and
//it's length is over 50% due to optimizations.
for(USHORT i = 0; i < GridSize; i++)
{
if (Grid_ptr[i] == 'X')
{
//CHECK X --> Mid Right Vector
//If within allowed parameters
if(i % Side < Side-2)
{
if(Grid_ptr[i+1] == '?' && Grid_ptr[i+2] == '?')
{
WeightGrid_ptr[i+1] += 1;
WeightGrid_ptr[i+2] += 1;
}
else if(Grid_ptr[i+1] == 'X')
{
WeightGrid_ptr[i+2] += 15;
}
else if (Grid_ptr[i+2] == 'X')
{
WeightGrid_ptr[i+1] += 15;
}
}
//CHECK X --> Down Right Vector
//If within allowed parameters
if (i % Side < Side -2 && i + Side*2 < GridSize)
{
if (Grid_ptr[i+Side+1] == '?' && Grid_ptr[i+Side*2+2] == '?')
{
WeightGrid_ptr[i+Side+1] += 1;
WeightGrid_ptr[i+Side*2+2] += 1;
}
else if(Grid_ptr[i+Side+1] == 'X')
{
WeightGrid_ptr[i+Side*2+2] += 15;
}
else if (Grid_ptr[i+Side*2+2] == 'X')
{
WeightGrid_ptr[i+Side+1] += 15;
}
}
//CHECK X --> Down Mid Vector
//If within allowed paramaters
if (i + Side*2 < GridSize)
{
if (Grid_ptr[i+Side] == '?' && Grid_ptr[i+Side*2] == '?')
{
WeightGrid_ptr[i+Side] += 1;
WeightGrid_ptr[i+Side*2] += 1;
}
else if (Grid_ptr[i+Side] == 'X')
{
WeightGrid_ptr[i+Side*2] += 15;
}
else if (Grid_ptr[i+Side*2] == 'X')
{
WeightGrid_ptr[i+Side] += 15;
}
}
//CHECK X --> Down Left Vector
//If within allowed paramaters
if(i % Side > 1 && i + Side*2 < GridSize)
{
if (Grid_ptr[i + Side*2-1] == '?' && i + Side*2-2 == '?')
{
WeightGrid_ptr[i+Side*2-1] += 1;
WeightGrid_ptr[i+Side*2-2] += 1;
}
else if(Grid_ptr[i + Side*2-2] == 'X')
{
WeightGrid_ptr[i+Side*2-1] += 15;
}
else if(Grid_ptr[i+Side*2-1] == 'X')
{
WeightGrid_ptr[i+Side*2-2] += 15;
}
}
//CHECK X --> Mid Left Vector
//If within allowed parameters
if(i % Side > 1)
{
if (Grid_ptr[i-1] == '?' && Grid_ptr[i-2] == '?')
{
WeightGrid_ptr[i-1] += 1;
WeightGrid_ptr[i-2] += 1;
}
else if(Grid_ptr[i-1] == 'X')
{
WeightGrid_ptr[i-2] += 15;
}
else if(Grid_ptr[i-2] == 'X')
{
WeightGrid_ptr[i-1] += 15;
}
}
//CHECK X --> Top Left Vector
//If within allowed parameters
if( (i) % (Side > 1) && i > Side*2)
{
if (Grid_ptr[i-Side-1] == '?' && Grid_ptr[i-Side*2-2] == '?')
{
WeightGrid_ptr[i-Side-1] += 1;
WeightGrid_ptr[i-Side*2-2] += 1;
}
else if (Grid_ptr[i-Side-1] == 'X')
{
WeightGrid_ptr[i-Side*2-2] += 15;
}
else if (Grid_ptr[i-Side*2-2] == 'X')
{
WeightGrid_ptr[i-Side-1] += 15;
}
}
//CHECK X --> Mid Top Vector
//If within allowed parameters
if (i > Side*2)
{
if(Grid_ptr[i + Side] == '?' && Grid_ptr[i + Side*2] == '?')
{
WeightGrid_ptr[i + Side] += 1;
WeightGrid_ptr[i + Side*2] += 1;
}
else if(Grid_ptr[i + Side] == 'X')
{
WeightGrid_ptr[i + Side*2] += 15;
}
else if (Grid_ptr[i + Side*2] == 'X')
{
WeightGrid_ptr[i + Side] += 15;
}
}
} //end if 'X' detected
else if (Grid_ptr[i] == 'O')
{
//CHECK 8 VECTORS
//Add weights
//CHECK O --> Mid Right Vector
//If within allowed parameters
if(i % Side < Side-2)
{
if(Grid_ptr[i+1] == '?' && Grid_ptr[i+2] == '?')
{
WeightGrid_ptr[i+1] += 1;
WeightGrid_ptr[i+2] += 1;
}
else if(Grid_ptr[i+1] == 'O')
{
WeightGrid_ptr[i+2] += 50;
}
else if (Grid_ptr[i+2] == 'O')
{
WeightGrid_ptr[i+1] += 50;
}
}
//CHECK O --> Down Right Vector
//If within allowed parameters
if (i % Side < Side -2 && i + Side*2 < GridSize)
{
if (Grid_ptr[i+Side+1] == '?' && Grid_ptr[i+Side*2+2] == '?')
{
WeightGrid_ptr[i+Side+1] += 1;
WeightGrid_ptr[i+Side*2+2] += 1;
}
else if(Grid_ptr[i+Side+1] == 'O')
{
WeightGrid_ptr[i+Side*2+2] += 50;
}
else if (Grid_ptr[i+Side*2+2] == 'O')
{
WeightGrid_ptr[i+Side+1] += 50;
}
}
//CHECK O --> Down Mid Vector
//If within allowed paramaters
if (i + Side*2 < GridSize)
{
if (Grid_ptr[i+Side] == '?' && Grid_ptr[i+Side*2] == '?')
{
WeightGrid_ptr[i+Side] += 1;
WeightGrid_ptr[i+Side*2] += 1;
}
else if (Grid_ptr[i+Side] == 'O')
{
WeightGrid_ptr[i+Side*2] += 50;
}
else if (Grid_ptr[i+Side*2] == 'O')
{
WeightGrid_ptr[i+Side] += 50;
}
}
//CHECK O --> Down Left Vector
//If within allowed paramaters
if(i % Side > 1 && i + Side*2 < GridSize)
{
if (Grid_ptr[i + Side*2-1] == '?' && i + Side*2-2 == '?')
{
WeightGrid_ptr[i+Side*2-1] += 1;
WeightGrid_ptr[i+Side*2-2] += 1;
}
else if(Grid_ptr[i + Side*2-2] == 'O')
{
WeightGrid_ptr[i+Side*2-1] += 50;
}
else if(Grid_ptr[i+Side*2-1] == 'O')
{
WeightGrid_ptr[i+Side*2-2] += 50;
}
}
//CHECK O --> Mid Left Vector
//If within allowed parameters
if(i % Side > 1)
{
if (Grid_ptr[i-1] == '?' && Grid_ptr[i-2] == '?')
{
WeightGrid_ptr[i-1] += 1;
WeightGrid_ptr[i-2] += 1;
}
else if(Grid_ptr[i-1] == 'O')
{
WeightGrid_ptr[i-2] += 50;
}
else if(Grid_ptr[i-2] == 'O')
{
WeightGrid_ptr[i-1] += 50;
}
}
//CHECK O --> Top Left Vector
//If within allowed parameters
if( (i) & (Side > 1) && i > Side*2)
{
if (Grid_ptr[i-Side-1] == '?' && Grid_ptr[i-Side*2-2] == '?')
{
WeightGrid_ptr[i-Side-1] += 1;
WeightGrid_ptr[i-Side*2-2] += 1;
}
else if (Grid_ptr[i-Side-1] == 'O')
{
WeightGrid_ptr[i-Side*2-2] += 50;
}
else if (Grid_ptr[i-Side*2-2] == 'O')
{
WeightGrid_ptr[i-Side-1] += 50;
}
}
//CHECK O --> Mid Top Vector
//If within allowed parameters
if (i > Side*2)
{
if(Grid_ptr[i + Side] == '?' && Grid_ptr[i + Side*2] == '?')
{
WeightGrid_ptr[i + Side] += 1;
WeightGrid_ptr[i + Side*2] += 1;
}
else if(Grid_ptr[i + Side] == 'O')
{
WeightGrid_ptr[i + Side*2] += 50;
}
else if (Grid_ptr[i + Side*2] == 'O')
{
WeightGrid_ptr[i + Side] += 50;
}
}
}
} // end for scan 'X' 'O'
//Get highest value from weightgrid, add an 'O' to that position, end method automatically
for (USHORT q = 0; q < GridSize; q++)
{
if (Grid_ptr[q] == '?')
{
//If a better spot is found
if (WeightGrid_ptr[q] > *fattest_ptr)
{
*fattest_ptr = WeightGrid_ptr[q];
*fattestIndex_ptr = q;
}
}
}
Grid_ptr[*fattestIndex_ptr] = 'O';
//SAFE DELETE POINTER WeightGrid_ptr
if (WeightGrid_ptr != NULL)
{
delete[] WeightGrid_ptr;
WeightGrid_ptr = NULL;
}
//SAFE DELETE POINTER fattest_ptr
if (fattest_ptr != NULL)
{
delete fattest_ptr;
fattest_ptr = NULL;
}
//SAFE DELETE POINTER fattestIndex_ptr
if (fattestIndex_ptr != NULL)
{
delete fattestIndex_ptr;
fattestIndex_ptr = NULL;
}
}
int _tmain(int argc, _TCHAR* argv[])
{
//& adress off |-| &x = 0x?
//* value pointed by |-| a = *b
//Make the required variables on the heap
USHORT GridHeight = 0;
USHORT GridWidth = 0;
USHORT GridSize = 0;
USHORT moveCounter = 0;
char currentMove;
USHORT input;
//bool* humanWin_ptr = new bool(false);
//bool* computerWin_ptr = new bool(false);
bool humanWin_ptr = false;
bool computerWin_ptr = false;
bool Draw = false;
cout << "A challanger has arrived!" << endl;
//WARNING FOR THIS BLOCK! Special condition on for loop!
for(;;)
{
cout << "Please state the width for the grid \n";
scanf_s("%hu", &input);
if (input > 2 && input < 20)
{
GridWidth = input;
break; //CRITICAL CODE
}
else
{
cout << "Input was not correct, please state a number between 3 and 20 \n\n";
cout << "Example of correct input '3' (without quotes) \n";
}
}
//WARNING FOR THIS BLOCK! Special condition on for loop!
for(;;)
{
cout << "Please state the height for the grid \n";
scanf_s("%hu", &input);
if (input > 2 && input < 20)
{
GridHeight = input;
break; //CRITICAL CODE
}
else
{
cout << "Input was not correct, please state a number between 3 and 20 \n\n";
cout << "Example of correct input '3' (without quotes) \n";
}
}
cout << "You have succesfully filled in the paperwork to create the Grid" << endl;
GridSize = GridHeight * GridWidth;
cout << "The total GridSize is " << GridSize << " tiles in size" << endl;
//if (GridWidth != GridHeigth)
//{
// cout << "Warning! \n\nGrid is not square. Program may run irregularly!";
// cout << "Close the program or press a key to continue";
// scanf();
//}
//Note: pointer to a Grid object on the heap
char* Grid_ptr = new char[GridSize];
//Initialize Grid as empty
for (USHORT i = 0; i < GridSize; i++)
{
Grid_ptr[i] = '?';
}
//Visualize this step
cout << "Grid created as empty Grid" << endl;
cout << endl;
cout << "Please read the following introduction if you wish for an explanation of the game" << endl;
cout << "You will be reffered to as Player One equally so the opponent as AI" << endl;
cout << "You always start with the first move" << endl;
cout << "The condition for victory is a line of X X X (3 total) in a single line, colomn or a diagonal line across the Grid" << endl;
cout << "Turns are exchanged per move 1 : 1, there are no time limits so use all you need" << endl;
cout << "Player One can not lose this 3x3 Grid game when the best option is always chosen" << endl;
cout << "Consider playing a larger field if you wish to win, Best of luck!" << endl;
cout << "The grid is filled in like this!" << endl;
PrintGrid(Grid_ptr, GridWidth, GridHeight, GridSize);
while(humanWin_ptr == false && computerWin_ptr == false && Draw == false)
{
cout << "Players One's Turn! \n";
cout << "Please fill in the number your X";
currentMove = 'H';
for(;;)
{
scanf_s("%i" , &input);
if (Grid_ptr[input] == 'X' || Grid_ptr[input] == 'O')
{
cout << "That space is already taken ,try another";
}
else
{
Grid_ptr[input] = 'X';
moveCounter++;
break;
}
}
cout << '\n';
PrintGrid(Grid_ptr, GridWidth, GridHeight, GridSize);
CheckForWin(Grid_ptr, GridSize, GridWidth, humanWin_ptr, computerWin_ptr, currentMove);
cout << "AI is making a move!" << endl;
currentMove = 'C';
useAI(Grid_ptr, GridSize, GridWidth);
cout << '\n';
PrintGrid(Grid_ptr, GridWidth, GridHeight, GridSize);
CheckForWin(Grid_ptr, GridSize, GridWidth, humanWin_ptr, computerWin_ptr, currentMove);
if (humanWin_ptr)
{
cout << "Congratulations you have won the game! \n";
char c;
puts ("Enter any text. Include a Space ('.') in a sentence to exit: \n");
do
{
c=getchar();
putchar (c);
}
while (c != ' ');
}
else if (computerWin_ptr)
{
cout << "The computer won this match, better luck next time! \n";
char c;
puts ("Enter any text. Include a Space ('.') in a sentence to exit: \n");
do
{
c=getchar();
putchar (c);
}
while (c != ' ');
}
if (moveCounter >= GridSize)
{
Draw = true;
cout << "The game was a draw, good fighting!";
}
}
//int ch = 0;
//ch = _getch();
//wint_t _getwch( void );
//SAFE DELETE POINTER GRID
if (Grid_ptr != NULL)
{
delete[] Grid_ptr;
Grid_ptr = NULL;
}
/*
//SAFE DELETE POINTER Human Win
if (humanWin_ptr != NULL)
{
delete humanWin_ptr;
humanWin_ptr = NULL;
}
//SAFE DELETE POINTER Computer Win
if (computerWin_ptr != NULL)
{
delete computerWin_ptr;
computerWin_ptr = NULL;
}*/
return 0;
}
What you are asking seams to be about micro optimization. First implement it right, then profile/measure to find bottlenecks, and then think how to improve.
Since the question is so general (and without the example and code), I do not think it is possible to answer differently.