确定Tic Tac Toe游戏的算法
我在Java中编写了一个井字游戏，我目前确定游戏结束的方法考虑了游戏结束的以下可能情况：
- 董事会已经满员，尚未宣布获胜者：比赛是平局。
- Cross赢了。
- Circle赢了。
不幸的是，要这样做，它会从表中读取这些方案的预定义集合。考虑到电路板上只有9个空格，这并不一定是坏的，因此表格有点小，但有没有更好的算法来确定游戏是否结束？确定某人是否赢了是问题的关键，因为检查9个空格是否已满是微不足道的。
表方法可能是解决方案，但如果没有，那是什么？另外，如果电路板的尺寸不是n=9怎么办？如果它是一个更大的电路板，比如说n=16，n=25等，会导致连续放置的项目数量为x=4，x=5等等。 ？用于所有n = { 9, 16, 25, 36 ... }的一般算法？
25 个答案:
答案 0 :(得分：119)
您知道获胜的举动只能在X或O进行最近的移动之后发生，因此您只能搜索包含在该移动中的可选诊断的行/列，以在尝试确定获胜时限制您的搜索空间板。此外，由于在最后一次移动中，如果它不是一个获胜的移动，那么在平局游戏中有一定数量的移动它默认为抽奖游戏。
编辑：此代码用于n个n板，连续n个赢（3x3板请求连续3个等）
编辑：添加代码来检查反诊断，我无法找出一个非循环方式来确定该点是否在反诊断上，这就是为什么缺少该步骤
public class TripleT {
enum State{Blank, X, O};
int n = 3;
State[][] board = new State[n][n];
int moveCount;
void Move(int x, int y, State s){
if(board[x][y] == State.Blank){
board[x][y] = s;
}
moveCount++;
//check end conditions
//check col
for(int i = 0; i < n; i++){
if(board[x][i] != s)
break;
if(i == n-1){
//report win for s
}
}
//check row
for(int i = 0; i < n; i++){
if(board[i][y] != s)
break;
if(i == n-1){
//report win for s
}
}
//check diag
if(x == y){
//we're on a diagonal
for(int i = 0; i < n; i++){
if(board[i][i] != s)
break;
if(i == n-1){
//report win for s
}
}
}
//check anti diag (thanks rampion)
if(x + y == n - 1){
for(int i = 0; i < n; i++){
if(board[i][(n-1)-i] != s)
break;
if(i == n-1){
//report win for s
}
}
}
//check draw
if(moveCount == (Math.pow(n, 2) - 1)){
//report draw
}
}
}
答案 1 :(得分：32)
你可以使用魔方（{3}}，如果任何行，列或诊断加起来，那么玩家就赢了。
答案 2 :(得分：19)
这类似于Osama ALASSIRY's answer，但它为线性空间和恒定时间交换恒定空间和线性时间。也就是说，初始化后没有循环。
为每行，每列和两个对角线（对角线和反对角线）初始化一对(0,0)。这些对表示相应行，列或对角线中片段的累积(sum,sum)，其中
A piece from player A has value (1,0) A piece from player B has value (0,1)
当玩家放置一块时，更新相应的行对，列对和对角线对（如果在对角线上）。如果任何新更新的行，列或对角线对等于(n,0)或(0,n)，则A或B分别获胜。
渐近分析：
O(1) time (per move) O(n) space (overall)
对于内存使用，使用4*(n+1)整数。
two_elements*n_rows + two_elements*n_columns + two_elements*two_diagonals = 4*n + 4 integers = 4(n+1) integers
练习：你能看到如何在O（1）时间内进行抽签测试吗？如果是这样，你可以在平局的早期结束比赛。
答案 3 :(得分：18)
这个伪代码怎么样：
玩家在位置（x，y）放下一块之后：
col=row=diag=rdiag=0
winner=false
for i=1 to n
if cell[x,i]=player then col++
if cell[i,y]=player then row++
if cell[i,i]=player then diag++
if cell[i,n-i+1]=player then rdiag++
if row=n or col=n or diag=n or rdiag=n then winner=true
我使用char [n，n]数组，其中O，X和空格为空。
- 简单。
- 一个循环。
- 五个简单变量：4个整数和一个布尔值。
- 缩放到任何大小的n。
- 仅检查当前作品。
- 没有魔力。 ：）
答案 4 :(得分：11)
这是我为一个我在javascript中工作的项目写的解决方案。如果你不介意几个阵列的内存成本，它可能是你能找到的最快最简单的解决方案。它假设你知道最后一步的位置。
/*
* Determines if the last move resulted in a win for either player
* board: is an array representing the board
* lastMove: is the boardIndex of the last (most recent) move
* these are the boardIndexes:
*
* 0 | 1 | 2
* ---+---+---
* 3 | 4 | 5
* ---+---+---
* 6 | 7 | 8
*
* returns true if there was a win
*/
var winLines = [
[[1, 2], [4, 8], [3, 6]],
[[0, 2], [4, 7]],
[[0, 1], [4, 6], [5, 8]],
[[4, 5], [0, 6]],
[[3, 5], [0, 8], [2, 6], [1, 7]],
[[3, 4], [2, 8]],
[[7, 8], [2, 4], [0, 3]],
[[6, 8], [1, 4]],
[[6, 7], [0, 4], [2, 5]]
];
function isWinningMove(board, lastMove) {
var player = board[lastMove];
for (var i = 0; i < winLines[lastMove].length; i++) {
var line = winLines[lastMove][i];
if(player === board[line[0]] && player === board[line[1]]) {
return true;
}
}
return false;
}
答案 5 :(得分：6)
我刚为C编程课写过这篇文章。
我发布它是因为此处的其他示例都不适用于任何大小的矩形网格，并且任何数字 N 连续标记都可以获胜
你会在checkWinner()函数中找到我的算法，例如它。它没有使用魔术数字或任何想要检查胜利者的东西，它只是使用四个for循环 - 代码得到很好的评论，所以我会让它自己说话我猜。
// This program will work with any whole number sized rectangular gameBoard.
// It checks for N marks in straight lines (rows, columns, and diagonals).
// It is prettiest when ROWS and COLS are single digit numbers.
// Try altering the constants for ROWS, COLS, and N for great fun!
// PPDs come first
#include <stdio.h>
#define ROWS 9 // The number of rows our gameBoard array will have
#define COLS 9 // The number of columns of the same - Single digit numbers will be prettier!
#define N 3 // This is the number of contiguous marks a player must have to win
#define INITCHAR ' ' // This changes the character displayed (a ' ' here probably looks the best)
#define PLAYER1CHAR 'X' // Some marks are more aesthetically pleasing than others
#define PLAYER2CHAR 'O' // Change these lines if you care to experiment with them
// Function prototypes are next
int playGame (char gameBoard[ROWS][COLS]); // This function allows the game to be replayed easily, as desired
void initBoard (char gameBoard[ROWS][COLS]); // Fills the ROWSxCOLS character array with the INITCHAR character
void printBoard (char gameBoard[ROWS][COLS]); // Prints out the current board, now with pretty formatting and #s!
void makeMove (char gameBoard[ROWS][COLS], int player); // Prompts for (and validates!) a move and stores it into the array
int checkWinner (char gameBoard[ROWS][COLS], int player); // Checks the current state of the board to see if anyone has won
// The starting line
int main (void)
{
// Inits
char gameBoard[ROWS][COLS]; // Our gameBoard is declared as a character array, ROWS x COLS in size
int winner = 0;
char replay;
//Code
do // This loop plays through the game until the user elects not to
{
winner = playGame(gameBoard);
printf("\nWould you like to play again? Y for yes, anything else exits: ");
scanf("%c",&replay); // I have to use both a scanf() and a getchar() in
replay = getchar(); // order to clear the input buffer of a newline char
// (http://cboard.cprogramming.com/c-programming/121190-problem-do-while-loop-char.html)
} while ( replay == 'y' || replay == 'Y' );
// Housekeeping
printf("\n");
return winner;
}
int playGame(char gameBoard[ROWS][COLS])
{
int turn = 0, player = 0, winner = 0, i = 0;
initBoard(gameBoard);
do
{
turn++; // Every time this loop executes, a unique turn is about to be made
player = (turn+1)%2+1; // This mod function alternates the player variable between 1 & 2 each turn
makeMove(gameBoard,player);
printBoard(gameBoard);
winner = checkWinner(gameBoard,player);
if (winner != 0)
{
printBoard(gameBoard);
for (i=0;i<19-2*ROWS;i++) // Formatting - works with the default shell height on my machine
printf("\n"); // Hopefully I can replace these with something that clears the screen for me
printf("\n\nCongratulations Player %i, you've won with %i in a row!\n\n",winner,N);
return winner;
}
} while ( turn < ROWS*COLS ); // Once ROWS*COLS turns have elapsed
printf("\n\nGame Over!\n\nThere was no Winner :-(\n"); // The board is full and the game is over
return winner;
}
void initBoard (char gameBoard[ROWS][COLS])
{
int row = 0, col = 0;
for (row=0;row<ROWS;row++)
{
for (col=0;col<COLS;col++)
{
gameBoard[row][col] = INITCHAR; // Fill the gameBoard with INITCHAR characters
}
}
printBoard(gameBoard); // Having this here prints out the board before
return; // the playGame function asks for the first move
}
void printBoard (char gameBoard[ROWS][COLS]) // There is a ton of formatting in here
{ // That I don't feel like commenting :P
int row = 0, col = 0, i=0; // It took a while to fine tune
// But now the output is something like:
printf("\n"); //
// 1 2 3
for (row=0;row<ROWS;row++) // 1 | |
{ // -----------
if (row == 0) // 2 | |
{ // -----------
printf(" "); // 3 | |
for (i=0;i<COLS;i++)
{
printf(" %i ",i+1);
}
printf("\n\n");
}
for (col=0;col<COLS;col++)
{
if (col==0)
printf("%i ",row+1);
printf(" %c ",gameBoard[row][col]);
if (col<COLS-1)
printf("|");
}
printf("\n");
if (row < ROWS-1)
{
for(i=0;i<COLS-1;i++)
{
if(i==0)
printf(" ----");
else
printf("----");
}
printf("---\n");
}
}
return;
}
void makeMove (char gameBoard[ROWS][COLS],int player)
{
int row = 0, col = 0, i=0;
char currentChar;
if (player == 1) // This gets the correct player's mark
currentChar = PLAYER1CHAR;
else
currentChar = PLAYER2CHAR;
for (i=0;i<21-2*ROWS;i++) // Newline formatting again :-(
printf("\n");
printf("\nPlayer %i, please enter the column of your move: ",player);
scanf("%i",&col);
printf("Please enter the row of your move: ");
scanf("%i",&row);
row--; // These lines translate the user's rows and columns numbering
col--; // (starting with 1) to the computer's (starting with 0)
while(gameBoard[row][col] != INITCHAR || row > ROWS-1 || col > COLS-1) // We are not using a do... while because
{ // I wanted the prompt to change
printBoard(gameBoard);
for (i=0;i<20-2*ROWS;i++)
printf("\n");
printf("\nPlayer %i, please enter a valid move! Column first, then row.\n",player);
scanf("%i %i",&col,&row);
row--; // See above ^^^
col--;
}
gameBoard[row][col] = currentChar; // Finally, we store the correct mark into the given location
return; // And pop back out of this function
}
int checkWinner(char gameBoard[ROWS][COLS], int player) // I've commented the last (and the hardest, for me anyway)
{ // check, which checks for backwards diagonal runs below >>>
int row = 0, col = 0, i = 0;
char currentChar;
if (player == 1)
currentChar = PLAYER1CHAR;
else
currentChar = PLAYER2CHAR;
for ( row = 0; row < ROWS; row++) // This first for loop checks every row
{
for ( col = 0; col < (COLS-(N-1)); col++) // And all columns until N away from the end
{
while (gameBoard[row][col] == currentChar) // For consecutive rows of the current player's mark
{
col++;
i++;
if (i == N)
{
return player;
}
}
i = 0;
}
}
for ( col = 0; col < COLS; col++) // This one checks for columns of consecutive marks
{
for ( row = 0; row < (ROWS-(N-1)); row++)
{
while (gameBoard[row][col] == currentChar)
{
row++;
i++;
if (i == N)
{
return player;
}
}
i = 0;
}
}
for ( col = 0; col < (COLS - (N-1)); col++) // This one checks for "forwards" diagonal runs
{
for ( row = 0; row < (ROWS-(N-1)); row++)
{
while (gameBoard[row][col] == currentChar)
{
row++;
col++;
i++;
if (i == N)
{
return player;
}
}
i = 0;
}
}
// Finally, the backwards diagonals:
for ( col = COLS-1; col > 0+(N-2); col--) // Start from the last column and go until N columns from the first
{ // The math seems strange here but the numbers work out when you trace them
for ( row = 0; row < (ROWS-(N-1)); row++) // Start from the first row and go until N rows from the last
{
while (gameBoard[row][col] == currentChar) // If the current player's character is there
{
row++; // Go down a row
col--; // And back a column
i++; // The i variable tracks how many consecutive marks have been found
if (i == N) // Once i == N
{
return player; // Return the current player number to the
} // winnner variable in the playGame function
} // If it breaks out of the while loop, there weren't N consecutive marks
i = 0; // So make i = 0 again
} // And go back into the for loop, incrementing the row to check from
}
return 0; // If we got to here, no winner has been detected,
} // so we pop back up into the playGame function
// The end!
// Well, almost.
// Eventually I hope to get this thing going
// with a dynamically sized array. I'll make
// the CONSTANTS into variables in an initGame
// function and allow the user to define them.
答案 6 :(得分：5)
如果电路板 n × n ，那么 n 行， n 列和2个对角线。检查所有X或全O的每一个以找到胜利者。
如果只需要 x ＆lt; n 连续的方块赢了，然后它有点复杂了。最明显的解决方案是检查每个 x × x 方块以获得胜利者。这里有一些代码可以证明这一点。
（我实际上没有测试过这个*咳嗽*，但在第一次尝试时编译了，伙计我！）
public class TicTacToe
{
public enum Square { X, O, NONE }
/**
* Returns the winning player, or NONE if the game has
* finished without a winner, or null if the game is unfinished.
*/
public Square findWinner(Square[][] board, int lengthToWin) {
// Check each lengthToWin x lengthToWin board for a winner.
for (int top = 0; top <= board.length - lengthToWin; ++top) {
int bottom = top + lengthToWin - 1;
for (int left = 0; left <= board.length - lengthToWin; ++left) {
int right = left + lengthToWin - 1;
// Check each row.
nextRow: for (int row = top; row <= bottom; ++row) {
if (board[row][left] == Square.NONE) {
continue;
}
for (int col = left; col <= right; ++col) {
if (board[row][col] != board[row][left]) {
continue nextRow;
}
}
return board[row][left];
}
// Check each column.
nextCol: for (int col = left; col <= right; ++col) {
if (board[top][col] == Square.NONE) {
continue;
}
for (int row = top; row <= bottom; ++row) {
if (board[row][col] != board[top][col]) {
continue nextCol;
}
}
return board[top][col];
}
// Check top-left to bottom-right diagonal.
diag1: if (board[top][left] != Square.NONE) {
for (int i = 1; i < lengthToWin; ++i) {
if (board[top+i][left+i] != board[top][left]) {
break diag1;
}
}
return board[top][left];
}
// Check top-right to bottom-left diagonal.
diag2: if (board[top][right] != Square.NONE) {
for (int i = 1; i < lengthToWin; ++i) {
if (board[top+i][right-i] != board[top][right]) {
break diag2;
}
}
return board[top][right];
}
}
}
// Check for a completely full board.
boolean isFull = true;
full: for (int row = 0; row < board.length; ++row) {
for (int col = 0; col < board.length; ++col) {
if (board[row][col] == Square.NONE) {
isFull = false;
break full;
}
}
}
// The board is full.
if (isFull) {
return Square.NONE;
}
// The board is not full and we didn't find a solution.
else {
return null;
}
}
}
答案 7 :(得分：3)
我不太了解Java，但我知道C，所以我尝试了adk's magic square idea（以及Hardwareguy's search restriction）。
// tic-tac-toe.c
// to compile:
// % gcc -o tic-tac-toe tic-tac-toe.c
// to run:
// % ./tic-tac-toe
#include <stdio.h>
// the two types of marks available
typedef enum { Empty=2, X=0, O=1, NumMarks=2 } Mark;
char const MarkToChar[] = "XO ";
// a structure to hold the sums of each kind of mark
typedef struct { unsigned char of[NumMarks]; } Sum;
// a cell in the board, which has a particular value
#define MAGIC_NUMBER 15
typedef struct {
Mark mark;
unsigned char const value;
size_t const num_sums;
Sum * const sums[4];
} Cell;
#define NUM_ROWS 3
#define NUM_COLS 3
// create a sum for each possible tic-tac-toe
Sum row[NUM_ROWS] = {0};
Sum col[NUM_COLS] = {0};
Sum nw_diag = {0};
Sum ne_diag = {0};
// initialize the board values so any row, column, or diagonal adds to
// MAGIC_NUMBER, and so they each record their sums in the proper rows, columns,
// and diagonals
Cell board[NUM_ROWS][NUM_COLS] = {
{
{ Empty, 8, 3, { &row[0], &col[0], &nw_diag } },
{ Empty, 1, 2, { &row[0], &col[1] } },
{ Empty, 6, 3, { &row[0], &col[2], &ne_diag } },
},
{
{ Empty, 3, 2, { &row[1], &col[0] } },
{ Empty, 5, 4, { &row[1], &col[1], &nw_diag, &ne_diag } },
{ Empty, 7, 2, { &row[1], &col[2] } },
},
{
{ Empty, 4, 3, { &row[2], &col[0], &ne_diag } },
{ Empty, 9, 2, { &row[2], &col[1] } },
{ Empty, 2, 3, { &row[2], &col[2], &nw_diag } },
}
};
// print the board
void show_board(void)
{
size_t r, c;
for (r = 0; r < NUM_ROWS; r++)
{
if (r > 0) { printf("---+---+---\n"); }
for (c = 0; c < NUM_COLS; c++)
{
if (c > 0) { printf("|"); }
printf(" %c ", MarkToChar[board[r][c].mark]);
}
printf("\n");
}
}
// run the game, asking the player for inputs for each side
int main(int argc, char * argv[])
{
size_t m;
show_board();
printf("Enter moves as \"<row> <col>\" (no quotes, zero indexed)\n");
for( m = 0; m < NUM_ROWS * NUM_COLS; m++ )
{
Mark const mark = (Mark) (m % NumMarks);
size_t c, r;
// read the player's move
do
{
printf("%c's move: ", MarkToChar[mark]);
fflush(stdout);
scanf("%d %d", &r, &c);
if (r >= NUM_ROWS || c >= NUM_COLS)
{
printf("illegal move (off the board), try again\n");
}
else if (board[r][c].mark != Empty)
{
printf("illegal move (already taken), try again\n");
}
else
{
break;
}
}
while (1);
{
Cell * const cell = &(board[r][c]);
size_t s;
// update the board state
cell->mark = mark;
show_board();
// check for tic-tac-toe
for (s = 0; s < cell->num_sums; s++)
{
cell->sums[s]->of[mark] += cell->value;
if (cell->sums[s]->of[mark] == MAGIC_NUMBER)
{
printf("tic-tac-toe! %c wins!\n", MarkToChar[mark]);
goto done;
}
}
}
}
printf("stalemate... nobody wins :(\n");
done:
return 0;
}
它编译和测试很好。
% gcc -o tic-tac-toe tic-tac-toe.c
% ./tic-tac-toe
| |
---+---+---
| |
---+---+---
| |
Enter moves as " " (no quotes, zero indexed)
X's move: 1 2
| |
---+---+---
| | X
---+---+---
| |
O's move: 1 2
illegal move (already taken), try again
O's move: 3 3
illegal move (off the board), try again
O's move: 2 2
| |
---+---+---
| | X
---+---+---
| | O
X's move: 1 0
| |
---+---+---
X | | X
---+---+---
| | O
O's move: 1 1
| |
---+---+---
X | O | X
---+---+---
| | O
X's move: 0 0
X | |
---+---+---
X | O | X
---+---+---
| | O
O's move: 2 0
X | |
---+---+---
X | O | X
---+---+---
O | | O
X's move: 2 1
X | |
---+---+---
X | O | X
---+---+---
O | X | O
O's move: 0 2
X | | O
---+---+---
X | O | X
---+---+---
O | X | O
tic-tac-toe! O wins!
% ./tic-tac-toe
| |
---+---+---
| |
---+---+---
| |
Enter moves as " " (no quotes, zero indexed)
X's move: 0 0
X | |
---+---+---
| |
---+---+---
| |
O's move: 0 1
X | O |
---+---+---
| |
---+---+---
| |
X's move: 0 2
X | O | X
---+---+---
| |
---+---+---
| |
O's move: 1 0
X | O | X
---+---+---
O | |
---+---+---
| |
X's move: 1 1
X | O | X
---+---+---
O | X |
---+---+---
| |
O's move: 2 0
X | O | X
---+---+---
O | X |
---+---+---
O | |
X's move: 2 1
X | O | X
---+---+---
O | X |
---+---+---
O | X |
O's move: 2 2
X | O | X
---+---+---
O | X |
---+---+---
O | X | O
X's move: 1 2
X | O | X
---+---+---
O | X | X
---+---+---
O | X | O
stalemate... nobody wins :(
%
这很有趣，谢谢！
实际上，考虑到它，你不需要一个魔方，只需要每行/每列/对角线的计数。这比将{＃1}}×n矩阵概括为方形要容易一些，因为您只需要计算到n。
答案 8 :(得分：3)
在我的一次采访中，我被问到了同样的问题。 我的想法： 用0初始化矩阵。 保留3个阵列 1）sum_row（大小n） 2）sum_column（大小n） 3）对角线（尺寸2）
对于每次移动，通过（X）将框值减1，对于每次移动，按（0）将其增加1。 在任何时候，如果在当前移动中被修改的行/列/对角线具有-3或+3的总和意味着有人赢得了游戏。 对于平局，我们可以使用上述方法来保持moveCount变量。
你认为我错过了什么吗？
编辑：相同可用于nxn矩阵。总和应该是+3或-3。
答案 9 :(得分：2)
一种非循环方式来确定该点是否在抗诊断上：
`if (x + y == n - 1)`
答案 10 :(得分：1)
我在行，对角线，对角线检查中进行了一些优化。如果我们需要检查特定的列或对角线，它主要在第一个嵌套循环中决定。因此，我们避免检查列或对角线，以节省时间。当电路板尺寸更大且未填充大量电池时，这会产生很大的影响。
这是java代码。
int gameState(int values[][], int boardSz) {
boolean colCheckNotRequired[] = new boolean[boardSz];//default is false
boolean diag1CheckNotRequired = false;
boolean diag2CheckNotRequired = false;
boolean allFilled = true;
int x_count = 0;
int o_count = 0;
/* Check rows */
for (int i = 0; i < boardSz; i++) {
x_count = o_count = 0;
for (int j = 0; j < boardSz; j++) {
if(values[i][j] == x_val)x_count++;
if(values[i][j] == o_val)o_count++;
if(values[i][j] == 0)
{
colCheckNotRequired[j] = true;
if(i==j)diag1CheckNotRequired = true;
if(i + j == boardSz - 1)diag2CheckNotRequired = true;
allFilled = false;
//No need check further
break;
}
}
if(x_count == boardSz)return X_WIN;
if(o_count == boardSz)return O_WIN;
}
/* check cols */
for (int i = 0; i < boardSz; i++) {
x_count = o_count = 0;
if(colCheckNotRequired[i] == false)
{
for (int j = 0; j < boardSz; j++) {
if(values[j][i] == x_val)x_count++;
if(values[j][i] == o_val)o_count++;
//No need check further
if(values[i][j] == 0)break;
}
if(x_count == boardSz)return X_WIN;
if(o_count == boardSz)return O_WIN;
}
}
x_count = o_count = 0;
/* check diagonal 1 */
if(diag1CheckNotRequired == false)
{
for (int i = 0; i < boardSz; i++) {
if(values[i][i] == x_val)x_count++;
if(values[i][i] == o_val)o_count++;
if(values[i][i] == 0)break;
}
if(x_count == boardSz)return X_WIN;
if(o_count == boardSz)return O_WIN;
}
x_count = o_count = 0;
/* check diagonal 2 */
if( diag2CheckNotRequired == false)
{
for (int i = boardSz - 1,j = 0; i >= 0 && j < boardSz; i--,j++) {
if(values[j][i] == x_val)x_count++;
if(values[j][i] == o_val)o_count++;
if(values[j][i] == 0)break;
}
if(x_count == boardSz)return X_WIN;
if(o_count == boardSz)return O_WIN;
x_count = o_count = 0;
}
if( allFilled == true)
{
for (int i = 0; i < boardSz; i++) {
for (int j = 0; j < boardSz; j++) {
if (values[i][j] == 0) {
allFilled = false;
break;
}
}
if (allFilled == false) {
break;
}
}
}
if (allFilled)
return DRAW;
return INPROGRESS;
}
答案 11 :(得分：1)
我喜欢这个算法，因为它使用了1x9对3x3的电路板表示。
private int[] board = new int[9];
private static final int[] START = new int[] { 0, 3, 6, 0, 1, 2, 0, 2 };
private static final int[] INCR = new int[] { 1, 1, 1, 3, 3, 3, 4, 2 };
private static int SIZE = 3;
/**
* Determines if there is a winner in tic-tac-toe board.
* @return {@code 0} for draw, {@code 1} for 'X', {@code -1} for 'Y'
*/
public int hasWinner() {
for (int i = 0; i < START.length; i++) {
int sum = 0;
for (int j = 0; j < SIZE; j++) {
sum += board[START[i] + j * INCR[i]];
}
if (Math.abs(sum) == SIZE) {
return sum / SIZE;
}
}
return 0;
}
答案 12 :(得分：1)
我迟到了，但是我想指出我发现使用magic square的一个好处，即它可以用来获得引起胜负的广场的引用。下一回合，而不仅仅是用于计算游戏何时结束。
采取这个神奇的方块：
4 9 2
3 5 7
8 1 6
首先，设置一个scores数组，每次移动时该数组都会递增。有关详细信息，请参阅this answer。现在如果我们在[0,0]和[0,1]连续两次非法播放X，那么scores数组看起来像这样：
[7, 0, 0, 4, 3, 0, 4, 0];
董事会看起来像这样：
X . .
X . .
. . .
然后，我们要做的就是获得对哪个方块获胜/阻止的引用：
get_winning_move = function() {
for (var i = 0, i < scores.length; i++) {
// keep track of the number of times pieces were added to the row
// subtract when the opposite team adds a piece
if (scores[i].inc === 2) {
return 15 - state[i].val; // 8
}
}
}
实际上，实现需要一些额外的技巧，比如处理数字键（在JavaScript中），但我发现它非常简单并且喜欢休闲数学。
答案 13 :(得分：1)
超高效的位板
让我们将游戏存储在一个二进制整数中，然后只需一步即可评估所有内容！
- 我们知道 X 的移动占用了 9 位：
xxx xxx xxx - 我们知道 O 的移动占用了 9 位：
ooo ooo ooo
因此，棋盘位置可以仅用 18 位表示：xoxoxo xoxoxo xoxoxo
但是，虽然这看起来很有效，但它并不能帮助我们确定胜利。我们需要一种更有用的位模式……一种不仅可以对移动进行编码，而且还可以以合理的方式对行、列和对角线进行编码的位模式。
我会通过为每个棋盘位置使用一个巧妙的整数值来做到这一点。
选择更有用的表示
首先，我们需要一个棋盘符号，以便我们可以讨论这个问题。所以，类似于国际象棋，让我们用字母对行进行编号，用数字对列进行编号 - 这样我们就知道我们在谈论哪个方块
| 1 | 2 | 3 | |
|---|---|---|---|
| A | a1 | a2 | a3 |
| B | b1 | b2 | b3 |
| C | c1 | c2 | c3 |
让我们给每一个二进制值。
a1 = 100 000 000 100 000 000 100 000 ; Row A Col 1 (top left corner)
a2 = 010 000 000 000 100 000 000 000 ; Row A Col 2 (top edge)
a3 = 001 000 000 000 000 100 000 100 ; Row A Col 3 (top right corner)
b1 = 000 100 000 010 000 000 000 000 ; Row B Col 1 (left edge)
b2 = 000 010 000 000 010 000 010 010 ; Row B Col 2 (middle square)
b3 = 000 001 000 000 000 010 000 000 ; Row B Col 4 (right edge)
c1 = 000 000 100 001 000 000 000 001 ; Row C Col 1 (bottom left corner)
c2 = 000 000 010 000 001 000 000 000 ; Row C Col 2 (bottom edge)
c3 = 000 000 001 000 000 001 001 000 ; Row C Col 3 (bottom right corner)
...其中，二进制值对位置出现在哪些行、列和对角线上进行编码。（稍后我们将看看这是如何工作的）
我们将使用这些值来构建游戏的两种表示形式，一种用于 X，一种用于 O
- X 以一个空板开始：
000 000 000 000 000 000 000 000 - O 以一块空板开始：
000 000 000 000 000 000 000 000
我们跟着X走 （O应该是一样的原理）
- X 玩 A1...所以我们或（X 板）与值 A1
- X 玩 A2...所以我们用值 A2 进行 OR
- X 玩 A3...所以我们用值 A3 进行 OR
这对 X 的董事会价值有何影响：
a1 = 100 000 000 100 000 000 100 000... ORed witha2 = 010 000 000 000 100 000 000 000... ORed witha3 = 001 000 000 000 000 100 000 100... 等于：
XB = 111 000 000 100 100 100 100 100
从左到右阅读我们看到 X 有：
111（所有位置）在第 1 行 （\o/ 赢了，耶！）000（没有位置）在第 2 行000（没有位置）在第 3 行100（一个位置）仅第 1 列的第一个位置100（一个位置）仅第 1 列的第一个位置100（一个位置）仅第 1 列的第一个位置100(One position) 仅对角线 1 的第一个位置100(One position) 仅对角线 2 的第一个位置
您会注意到，每当 X（或 O）有获胜线时，他的棋盘值中也会有三个连续的位。正是这三个位的位置，决定了他赢得了哪一行/列/对角线。
所以，现在的技巧是找到一种方法来在单个操作中检查此（设置了三个连续位）条件。
修改值使检测更容易
为了帮助解决这个问题，让我们改变我们的位表示，以便三组之间总是有零（因为 001 110 也是三个连续的位 - 但它们不是有效的胜利......所以，一个固定的零间隔会破坏这些：0 001 0 110)
因此，在添加一些间距零之后，我们可以确信 X 或 O 的棋盘值中的任何三个连续设置位都表示获胜！
因此，我们的新二进制值（带零填充）如下所示：
a1 = 100 0 000 0 000 0 100 0 000 0 000 0 100 0 000 0; 0x80080080（十六进制）a2 = 010 0 000 0 000 0 000 0 100 0 000 0 000 0 000 0; 0x40008000a3 = 001 0 000 0 000 0 000 0 000 0 100 0 000 0 100 0; 0x20000808b1 = 000 0 100 0 000 0 010 0 000 0 000 0 000 0 000 0; 0x08040000b2 = 000 0 010 0 000 0 000 0 010 0 000 0 010 0 010 0; 0x04004044b3 = 000 0 001 0 000 0 000 0 000 0 010 0 000 0 000 0; 0x02000400c1 = 000 0 000 0 100 0 001 0 000 0 000 0 000 0 001 0; 0x00820002c2 = 000 0 000 0 010 0 000 0 001 0 000 0 000 0 000 0; 0x00402000c3 = 000 0 000 0 001 0 000 0 000 0 001 0 001 0 000 0; 0x00200220
您会注意到板子的每个“赢线”现在需要 4 位。
8 个 winlines x 每个 4 位 = 32 位！是不是很方便:))))))
解析
我们可以移动所有位以寻找三个连续的位，但这需要 32 次移动 x 2 个玩家......以及一个计数器来跟踪。很慢！
我们可以与 0xF 进行 AND 运算，寻找值 8+4+2=14。这将允许我们一次检查 4 位。将轮班次数减少四分之一。但同样，这很慢！
所以，相反，让我们一次检查所有的可能性......
超高效的胜利检测
假设我们想要评估 A3+A1+B2+C3 的情况（在对角线上获胜）
a1 = 100 0 000 0 000 0 100 0 000 0 000 0 100 0 000 0, OR
a3 = 001 0 000 0 000 0 000 0 000 0 100 0 000 0 100 0, OR
b2 = 000 0 010 0 000 0 000 0 010 0 000 0 010 0 010 0, OR
c3 = 000 0 000 0 001 0 000 0 000 0 001 0 001 0 000 0, =
XB = 101 0 010 0 001 0 100 0 010 0 101 0 111 0 110 0 (See the win, on Diagonal 1?)
现在，让我们通过有效地将三位合并为一位来检查它是否获胜......
只需使用：XB AND (XB << 1) AND (XB >> 1)
换句话说：XB与（XB左移）AND（XB右移）
让我们尝试一个例子...
10100100001010000100101011101100 ; whitespaces removed for easy shifting
(AND)
01001000010100001001010111011000 ; XB shifted left
(AND)
01010010000101000010010101110110 ; XB shifted left
(Equals)
00000000000000000000000001000000
看到了吗？任何非零结果都意味着胜利！
但是，他们在哪里赢了
想知道他们在哪里获胜？好吧，你可以使用第二个表：
0x40000000 = RowA
0x04000000 = RowB
0x00400000 = RowC
0x00040000 = Col1
0x00004000 = Col2
0x00000400 = Col3
0x00000040 = Diag1
0x00000004 = Diag2
然而，我们可以比这更聪明，因为模式非常规则！
例如，在汇编中，您可以使用 BSF (Bit Scan Forward) 来查找前导零的数量。然后减去 2，然后减去 /4（右移 2） - 得到一个介于 0 和 8 之间的数字......您可以将其用作索引来查找获胜字符串数组：
{"wins the top row", "takes the middle row!", ... "steals the diagonal!" }
这使得整个游戏逻辑......从移动检查到棋盘更新，再到赢/输检测和适当的成功消息，所有这些都包含在少数 ASM 指令中。
...它小巧、高效且超快！
检查移动是否可玩
显然，“X's board”与“O's board”的 ORing = ALL POSITIONS
因此，您可以很容易地检查移动是否有效。如果用户选择 UpperLeft，则此位置为整数值。只需使用（XB OR OB）检查此值的“与”...
...如果结果非零，则该位置已在使用中。
结论
如果您正在寻找处理板的有效方法，请不要从板对象开始。尝试发现一些有用的抽象。
看看状态是否适合一个整数，并考虑一个“简单”的位掩码来处理会是什么样子。通过一些巧妙的整数选择来表示移动、位置或棋盘……您可能会发现可以非常有效地玩、评估和评分整个游戏 - 使用简单的按位逻辑。
致歉
顺便说一句，我不是 StackOverflow 上的常客，所以我希望这篇文章不会太混乱而无法遵循。另外，请善待......“人类”是我的第二语言，我还不太流利;)
无论如何，我希望这对某人有所帮助。
答案 14 :(得分：0)
不确定是否已发布此方法。这应该适用于任何m * n板，并且玩家应该连续填补“ winnerPos ”位置。这个想法是基于运行窗口的。
private boolean validateWinner(int x, int y, int player) {
//same col
int low = x-winnerPos-1;
int high = low;
while(high <= x+winnerPos-1) {
if(isValidPos(high, y) && isFilledPos(high, y, player)) {
high++;
if(high - low == winnerPos) {
return true;
}
} else {
low = high + 1;
high = low;
}
}
//same row
low = y-winnerPos-1;
high = low;
while(high <= y+winnerPos-1) {
if(isValidPos(x, high) && isFilledPos(x, high, player)) {
high++;
if(high - low == winnerPos) {
return true;
}
} else {
low = high + 1;
high = low;
}
}
if(high - low == winnerPos) {
return true;
}
//diagonal 1
int lowY = y-winnerPos-1;
int highY = lowY;
int lowX = x-winnerPos-1;
int highX = lowX;
while(highX <= x+winnerPos-1 && highY <= y+winnerPos-1) {
if(isValidPos(highX, highY) && isFilledPos(highX, highY, player)) {
highX++;
highY++;
if(highX - lowX == winnerPos) {
return true;
}
} else {
lowX = highX + 1;
lowY = highY + 1;
highX = lowX;
highY = lowY;
}
}
//diagonal 2
lowY = y+winnerPos-1;
highY = lowY;
lowX = x-winnerPos+1;
highX = lowX;
while(highX <= x+winnerPos-1 && highY <= y+winnerPos-1) {
if(isValidPos(highX, highY) && isFilledPos(highX, highY, player)) {
highX++;
highY--;
if(highX - lowX == winnerPos) {
return true;
}
} else {
lowX = highX + 1;
lowY = highY + 1;
highX = lowX;
highY = lowY;
}
}
if(highX - lowX == winnerPos) {
return true;
}
return false;
}
private boolean isValidPos(int x, int y) {
return x >= 0 && x < row && y >= 0 && y< col;
}
public boolean isFilledPos(int x, int y, int p) throws IndexOutOfBoundsException {
return arena[x][y] == p;
}
答案 15 :(得分：0)
我只想分享我在Javascript中所做的事情。我的想法是要找到搜索方向；在网格中，它可能是8个方向，但搜索应该是双向的，因此8/2 = 4个方向。当玩家移动时，搜索将从该位置开始。它会搜索4个不同的双向，直到其值与玩家的石头（O或X）不同为止。
每次进行双向搜索时，可以添加两个值，但由于起点重复，因此需要减去一个值。
getWin(x,y,value,searchvector) {
if (arguments.length==2) {
var checkTurn = this.state.squares[y][x];
var searchdirections = [[-1,-1],[0,-1],[1,-1],[-1,0]];
return searchdirections.reduce((maxinrow,searchdirection)=>Math.max(this.getWin(x,y,checkTurn,searchdirection)+this.getWin(x,y,checkTurn,[-searchdirection[0],-searchdirection[1]]),maxinrow),0);
} else {
if (this.state.squares[y][x]===value) {
var result = 1;
if (
x+searchvector[0] >= 0 && x+searchvector[0] < 3 &&
y+searchvector[1] >= 0 && y+searchvector[1] < 3
) result += this.getWin(x+searchvector[0],y+searchvector[1],value,searchvector);
return result;
} else {
return 0;
}
}
}
此函数可以与两个参数（x，y）一起使用，它们是最后一个移动的坐标。在初始执行中，它使用4个参数递归调用四个双向搜索。所有结果均以长度形式返回，该函数最终在4个搜索双向中选择最大长度。
class Square extends React.Component {
constructor(props) {
super(props);
this.state = {value:null};
}
render() {
return (
<button className="square" onClick={() => this.props.onClick()}>
{this.props.value}
</button>
);
}
}
class Board extends React.Component {
renderSquare(x,y) {
return <Square value={this.state.squares[y][x]} onClick={() => this.handleClick(x,y)} />;
}
handleClick(x,y) {
const squares = JSON.parse(JSON.stringify(this.state.squares));
if (!squares[y][x] && !this.state.winner) {
squares[y][x] = this.setTurn();
this.setState({squares: squares},()=>{
console.log(`Max in a row made by last move(${squares[y][x]}): ${this.getWin(x,y)-1}`);
if (this.getWin(x,y)==4) this.setState({winner:squares[y][x]});
});
}
}
setTurn() {
var prevTurn = this.state.turn;
this.setState({turn:prevTurn == 'X' ? 'O':'X'});
return prevTurn;
}
getWin(x,y,value,searchvector) {
if (arguments.length==2) {
var checkTurn = this.state.squares[y][x];
var searchdirections = [[-1,-1],[0,-1],[1,-1],[-1,0]];
return searchdirections.reduce((maxinrow,searchdirection)=>Math.max(this.getWin(x,y,checkTurn,searchdirection)+this.getWin(x,y,checkTurn,[-searchdirection[0],-searchdirection[1]]),maxinrow),0);
} else {
if (this.state.squares[y][x]===value) {
var result = 1;
if (
x+searchvector[0] >= 0 && x+searchvector[0] < 3 &&
y+searchvector[1] >= 0 && y+searchvector[1] < 3
) result += this.getWin(x+searchvector[0],y+searchvector[1],value,searchvector);
return result;
} else {
return 0;
}
}
}
constructor(props) {
super(props);
this.state = {
squares: Array(3).fill(Array(3).fill(null)),
turn: 'X',
winner: null
};
}
render() {
const status = !this.state.winner?`Next player: ${this.state.turn}`:`${this.state.winner} won!`;
return (
<div>
<div className="status">{status}</div>
<div className="board-row">
{this.renderSquare(0,0)}
{this.renderSquare(0,1)}
{this.renderSquare(0,2)}
</div>
<div className="board-row">
{this.renderSquare(1,0)}
{this.renderSquare(1,1)}
{this.renderSquare(1,2)}
</div>
<div className="board-row">
{this.renderSquare(2,0)}
{this.renderSquare(2,1)}
{this.renderSquare(2,2)}
</div>
</div>
);
}
}
class Game extends React.Component {
render() {
return (
<div className="game">
<div className="game-board">
<Board />
</div>
<div className="game-info">
<div>{/* status */}</div>
<ol>{/* TODO */}</ol>
</div>
</div>
);
}
}
// ========================================
ReactDOM.render(
<Game />,
document.getElementById('root')
);body {
font: 14px "Century Gothic", Futura, sans-serif;
margin: 20px;
}
ol, ul {
padding-left: 30px;
}
.board-row:after {
clear: both;
content: "";
display: table;
}
.status {
margin-bottom: 10px;
}
.square {
background: #fff;
border: 1px solid #999;
float: left;
font-size: 24px;
font-weight: bold;
line-height: 34px;
height: 34px;
margin-right: -1px;
margin-top: -1px;
padding: 0;
text-align: center;
width: 34px;
}
.square:focus {
outline: none;
}
.kbd-navigation .square:focus {
background: #ddd;
}
.game {
display: flex;
flex-direction: row;
}
.game-info {
margin-left: 20px;
}<script src="https://cdnjs.cloudflare.com/ajax/libs/react/16.6.3/umd/react.production.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/16.6.3/umd/react-dom.production.min.js"></script>
<div id="errors" style="
background: #c00;
color: #fff;
display: none;
margin: -20px -20px 20px;
padding: 20px;
white-space: pre-wrap;
"></div>
<div id="root"></div>
<script>
window.addEventListener('mousedown', function(e) {
document.body.classList.add('mouse-navigation');
document.body.classList.remove('kbd-navigation');
});
window.addEventListener('keydown', function(e) {
if (e.keyCode === 9) {
document.body.classList.add('kbd-navigation');
document.body.classList.remove('mouse-navigation');
}
});
window.addEventListener('click', function(e) {
if (e.target.tagName === 'A' && e.target.getAttribute('href') === '#') {
e.preventDefault();
}
});
window.onerror = function(message, source, line, col, error) {
var text = error ? error.stack || error : message + ' (at ' + source + ':' + line + ':' + col + ')';
errors.textContent += text + '\n';
errors.style.display = '';
};
console.error = (function(old) {
return function error() {
errors.textContent += Array.prototype.slice.call(arguments).join(' ') + '\n';
errors.style.display = '';
old.apply(this, arguments);
}
})(console.error);
</script>
答案 16 :(得分：0)
这是 React 中的示例实现：CodeSandbox Demo。
算法非常简单：
For every move:
checkDiagonals()
checkVerticals()
checkHorizontals()
为“O”输入分配 0，为“X”输入分配 1。检查函数的结果可以是0，或1，或2（平局）。
这是实现：
const checkDiagonals = () => {
if ((state[0][0] === val && state[1][1] === val && state[2][2] === val) ||
(state[0][2] === val && state[1][1] === val && state[2][0] === val)) {
return val;
}
return -1;
}
const checkVerticals = () => {
for (let i = 0; i <= 2; i++) {
if (state[0][i] === val && state[1][i] === val && state[2][i] === val) {
return val;
}
}
return -1;
}
const checkHorizontals = () => {
for (let i = 0; i <= 2; i++) {
if (state[i][0] === val && state[i][1] === val && state[i][2] === val) {
return val;
}
}
return -1;
}
剩下的就是有一个函数可以在每个用户输入时触发：
const updateWinningPlayer = () => {
const diagonals = checkDiagonals();
const verticals = checkVerticals();
const horizontals = checkHorizontals();
if (diagonals !== -1) {
setWinner(diagonals)
return;
}
if (verticals !== -1) {
setWinner(verticals);
return;
}
if (horizontals !== -1) {
setWinner(horizontals);
return;
}
if (isDraw()) {
setWinner(2);
}
}
这就给你了！
GitHub 存储库链接：https://github.com/mkotsollaris/tic-tac-toe
答案 17 :(得分：0)
这是我使用二维数组的解决方案：
private static final int dimension = 3;
private static final int[][] board = new int[dimension][dimension];
private static final int xwins = dimension * 1;
private static final int owins = dimension * -1;
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int count = 0;
boolean keepPlaying = true;
boolean xsTurn = true;
while (keepPlaying) {
xsTurn = (count % 2 == 0);
System.out.print("Enter i-j in the format:");
if (xsTurn) {
System.out.println(" X plays: ");
} else {
System.out.println(" O plays: ");
}
String result = null;
while (result == null) {
result = parseInput(scanner, xsTurn);
}
String[] xy = result.split(",");
int x = Integer.parseInt(xy[0]);
int y = Integer.parseInt(xy[1]);
keepPlaying = makeMove(xsTurn, x, y);
count++;
}
if (xsTurn) {
System.out.print("X");
} else {
System.out.print("O");
}
System.out.println(" WON");
printArrayBoard(board);
}
private static String parseInput(Scanner scanner, boolean xsTurn) {
String line = scanner.nextLine();
String[] values = line.split("-");
int x = Integer.parseInt(values[0]);
int y = Integer.parseInt(values[1]);
boolean alreadyPlayed = alreadyPlayed(x, y);
String result = null;
if (alreadyPlayed) {
System.out.println("Already played in this x-y. Retry");
} else {
result = "" + x + "," + y;
}
return result;
}
private static boolean alreadyPlayed(int x, int y) {
System.out.println("x-y: " + x + "-" + y + " board[x][y]: " + board[x][y]);
if (board[x][y] != 0) {
return true;
}
return false;
}
private static void printArrayBoard(int[][] board) {
for (int i = 0; i < dimension; i++) {
int[] height = board[i];
for (int j = 0; j < dimension; j++) {
System.out.print(height[j] + " ");
}
System.out.println();
}
}
private static boolean makeMove(boolean xo, int x, int y) {
if (xo) {
board[x][y] = 1;
} else {
board[x][y] = -1;
}
boolean didWin = checkBoard();
if (didWin) {
System.out.println("keep playing");
}
return didWin;
}
private static boolean checkBoard() {
//check horizontal
int[] horizontalTotal = new int[dimension];
for (int i = 0; i < dimension; i++) {
int[] height = board[i];
int total = 0;
for (int j = 0; j < dimension; j++) {
total += height[j];
}
horizontalTotal[i] = total;
}
for (int a = 0; a < horizontalTotal.length; a++) {
if (horizontalTotal[a] == xwins || horizontalTotal[a] == owins) {
System.out.println("horizontal");
return false;
}
}
//check vertical
int[] verticalTotal = new int[dimension];
for (int j = 0; j < dimension; j++) {
int total = 0;
for (int i = 0; i < dimension; i++) {
total += board[i][j];
}
verticalTotal[j] = total;
}
for (int a = 0; a < verticalTotal.length; a++) {
if (verticalTotal[a] == xwins || verticalTotal[a] == owins) {
System.out.println("vertical");
return false;
}
}
//check diagonal
int total1 = 0;
int total2 = 0;
for (int i = 0; i < dimension; i++) {
for (int j = 0; j < dimension; j++) {
if (i == j) {
total1 += board[i][j];
}
if (i == (dimension - 1 - j)) {
total2 += board[i][j];
}
}
}
if (total1 == xwins || total1 == owins) {
System.out.println("diagonal 1");
return false;
}
if (total2 == xwins || total2 == owins) {
System.out.println("diagonal 2");
return false;
}
return true;
}
答案 18 :(得分：0)
如果你有考试的寄宿家庭5 * 5，我使用下一种检查方法：
public static boolean checkWin(char symb) {
int SIZE = 5;
for (int i = 0; i < SIZE-1; i++) {
for (int j = 0; j <SIZE-1 ; j++) {
//vertical checking
if (map[0][j] == symb && map[1][j] == symb && map[2][j] == symb && map[3][j] == symb && map[4][j] == symb) return true; // j=0
}
//horisontal checking
if(map[i][0] == symb && map[i][1] == symb && map[i][2] == symb && map[i][3] == symb && map[i][4] == symb) return true; // i=0
}
//diagonal checking (5*5)
if (map[0][0] == symb && map[1][1] == symb && map[2][2] == symb && map[3][3] == symb && map[4][4] == symb) return true;
if (map[4][0] == symb && map[3][1] == symb && map[2][2] == symb && map[1][3] == symb && map[0][4] == symb) return true;
return false;
}
我认为，它更清楚，但可能不是最佳方式。
答案 19 :(得分：0)
这是一种非常简单的检查方法。
public class Game() {
Game player1 = new Game('x');
Game player2 = new Game('o');
char piece;
Game(char piece) {
this.piece = piece;
}
public void checkWin(Game player) {
// check horizontal win
for (int i = 0; i <= 6; i += 3) {
if (board[i] == player.piece &&
board[i + 1] == player.piece &&
board[i + 2] == player.piece)
endGame(player);
}
// check vertical win
for (int i = 0; i <= 2; i++) {
if (board[i] == player.piece &&
board[i + 3] == player.piece &&
board[i + 6] == player.piece)
endGame(player);
}
// check diagonal win
if ((board[0] == player.piece &&
board[4] == player.piece &&
board[8] == player.piece) ||
board[2] == player.piece &&
board[4] == player.piece &&
board[6] == player.piece)
endGame(player);
}
}
答案 20 :(得分：0)
9个插槽的后续方法怎么样？为3x3矩阵（a1，a2 ...... a9）声明9个整数变量，其中a1，a2，a3表示第1行，a1，a4，a7将形成第1列（你明白了）。使用“1”表示Player-1，使用“2”表示Player-2。
有8种可能的胜利组合： Win-1：a1 + a2 + a3（根据哪位玩家获胜，答案可能是3或6） Win-2：a4 + a5 + a6 Win-3：a7 + a8 + a9 Win-4：a1 + a4 + a7 .... Win-7：a1 + a5 + a9 Win-8：a3 + a5 + a7
现在我们知道，如果玩家1越过a1，那么我们需要重新评估3个变量的总和：Win-1，Win-4和Win-7。无论哪个'赢 - ？'变量达到3或6首先赢得比赛。如果Win-1变量首先达到6，则Player-2获胜。
我明白这个解决方案不易扩展。
答案 21 :(得分：0)
这是我提出的解决方案，它将符号存储为字符并使用char的int值来确定X或O是否已赢（查看裁判的代码）
public class TicTacToe {
public static final char BLANK = '\u0000';
private final char[][] board;
private int moveCount;
private Referee referee;
public TicTacToe(int gridSize) {
if (gridSize < 3)
throw new IllegalArgumentException("TicTacToe board size has to be minimum 3x3 grid");
board = new char[gridSize][gridSize];
referee = new Referee(gridSize);
}
public char[][] displayBoard() {
return board.clone();
}
public String move(int x, int y) {
if (board[x][y] != BLANK)
return "(" + x + "," + y + ") is already occupied";
board[x][y] = whoseTurn();
return referee.isGameOver(x, y, board[x][y], ++moveCount);
}
private char whoseTurn() {
return moveCount % 2 == 0 ? 'X' : 'O';
}
private class Referee {
private static final int NO_OF_DIAGONALS = 2;
private static final int MINOR = 1;
private static final int PRINCIPAL = 0;
private final int gridSize;
private final int[] rowTotal;
private final int[] colTotal;
private final int[] diagonalTotal;
private Referee(int size) {
gridSize = size;
rowTotal = new int[size];
colTotal = new int[size];
diagonalTotal = new int[NO_OF_DIAGONALS];
}
private String isGameOver(int x, int y, char symbol, int moveCount) {
if (isWinningMove(x, y, symbol))
return symbol + " won the game!";
if (isBoardCompletelyFilled(moveCount))
return "Its a Draw!";
return "continue";
}
private boolean isBoardCompletelyFilled(int moveCount) {
return moveCount == gridSize * gridSize;
}
private boolean isWinningMove(int x, int y, char symbol) {
if (isPrincipalDiagonal(x, y) && allSymbolsMatch(symbol, diagonalTotal, PRINCIPAL))
return true;
if (isMinorDiagonal(x, y) && allSymbolsMatch(symbol, diagonalTotal, MINOR))
return true;
return allSymbolsMatch(symbol, rowTotal, x) || allSymbolsMatch(symbol, colTotal, y);
}
private boolean allSymbolsMatch(char symbol, int[] total, int index) {
total[index] += symbol;
return total[index] / gridSize == symbol;
}
private boolean isPrincipalDiagonal(int x, int y) {
return x == y;
}
private boolean isMinorDiagonal(int x, int y) {
return x + y == gridSize - 1;
}
}
}
此外，我的单元测试验证它确实有效
import static com.agilefaqs.tdd.demo.TicTacToe.BLANK;
import static org.junit.Assert.assertArrayEquals;
import static org.junit.Assert.assertEquals;
import org.junit.Test;
public class TicTacToeTest {
private TicTacToe game = new TicTacToe(3);
@Test
public void allCellsAreEmptyInANewGame() {
assertBoardIs(new char[][] { { BLANK, BLANK, BLANK },
{ BLANK, BLANK, BLANK },
{ BLANK, BLANK, BLANK } });
}
@Test(expected = IllegalArgumentException.class)
public void boardHasToBeMinimum3x3Grid() {
new TicTacToe(2);
}
@Test
public void firstPlayersMoveMarks_X_OnTheBoard() {
assertEquals("continue", game.move(1, 1));
assertBoardIs(new char[][] { { BLANK, BLANK, BLANK },
{ BLANK, 'X', BLANK },
{ BLANK, BLANK, BLANK } });
}
@Test
public void secondPlayersMoveMarks_O_OnTheBoard() {
game.move(1, 1);
assertEquals("continue", game.move(2, 2));
assertBoardIs(new char[][] { { BLANK, BLANK, BLANK },
{ BLANK, 'X', BLANK },
{ BLANK, BLANK, 'O' } });
}
@Test
public void playerCanOnlyMoveToAnEmptyCell() {
game.move(1, 1);
assertEquals("(1,1) is already occupied", game.move(1, 1));
}
@Test
public void firstPlayerWithAllSymbolsInOneRowWins() {
game.move(0, 0);
game.move(1, 0);
game.move(0, 1);
game.move(2, 1);
assertEquals("X won the game!", game.move(0, 2));
}
@Test
public void firstPlayerWithAllSymbolsInOneColumnWins() {
game.move(1, 1);
game.move(0, 0);
game.move(2, 1);
game.move(1, 0);
game.move(2, 2);
assertEquals("O won the game!", game.move(2, 0));
}
@Test
public void firstPlayerWithAllSymbolsInPrincipalDiagonalWins() {
game.move(0, 0);
game.move(1, 0);
game.move(1, 1);
game.move(2, 1);
assertEquals("X won the game!", game.move(2, 2));
}
@Test
public void firstPlayerWithAllSymbolsInMinorDiagonalWins() {
game.move(0, 2);
game.move(1, 0);
game.move(1, 1);
game.move(2, 1);
assertEquals("X won the game!", game.move(2, 0));
}
@Test
public void whenAllCellsAreFilledTheGameIsADraw() {
game.move(0, 2);
game.move(1, 1);
game.move(1, 0);
game.move(2, 1);
game.move(2, 2);
game.move(0, 0);
game.move(0, 1);
game.move(1, 2);
assertEquals("Its a Draw!", game.move(2, 0));
}
private void assertBoardIs(char[][] expectedBoard) {
assertArrayEquals(expectedBoard, game.displayBoard());
}
}
完整解决方案：https://github.com/nashjain/tictactoe/tree/master/java
答案 22 :(得分：0)
另一种选择：使用代码生成表格。达到对称性，只有三种获胜方式：边缘行，中间行或对角线。拿这三个并尽可能地旋转它们：
def spin(g): return set([g, turn(g), turn(turn(g)), turn(turn(turn(g)))])
def turn(g): return tuple(tuple(g[y][x] for y in (0,1,2)) for x in (2,1,0))
X,s = 'X.'
XXX = X, X, X
sss = s, s, s
ways_to_win = ( spin((XXX, sss, sss))
| spin((sss, XXX, sss))
| spin(((X,s,s),
(s,X,s),
(s,s,X))))
这些对称性可以在你的游戏代码中有更多的用途：如果你已经看到了旋转版本的板子，你可以从缓存中获取缓存值或缓存最佳移动（并取消旋转）背部）。这通常比评估游戏子树要快得多。
（向左和向右翻转可以帮助相同的方式;这里不需要它，因为获胜模式的旋转集是镜像对称的。）
答案 23 :(得分：-1)
恒定时间O（8），平均4个短AND。玩家=短号码。需要额外的检查以确保移动有效。
// O(8)
boolean isWinner(short X) {
for (int i = 0; i < 8; i++)
if ((X & winCombinations[i]) == winCombinations[i])
return true;
return false;
}
short[] winCombinations = new short[]{
7, 7 << 3, 7 << 6, // horizontal
73, 73 << 1, 73 << 2, // vertical
273, // diagonal
84 // anti-diagonal
};
for (short X = 0; X < 511; X++)
System.out.println(isWinner(X));
答案 24 :(得分：-2)
我为此开发了一种算法，作为科学项目的一部分。
你基本上递归地将棋盘划分为一堆重叠的2x2 rects，测试不同的可能组合，以便在2x2的方格上获胜。
它很慢，但它具有在任何尺寸的电路板上工作的优势，具有相当线性的存储器要求。
我希望我能找到我的实施
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