题
我已经写了一个游戏的井字游戏,我的前的方法确定比赛结束帐户为以下可能的方案,为游戏结束:
- 委员会是完整的,并没有赢家尚未宣布:游戏是一个平局。
- 跨已经赢了。
- 圈子已经赢了。
不幸的是,这样做时,它会读取通过一个预先设定的这些方案从一个表中。这不一定是坏的考虑,只有9位在一个董事会,因而表是有点小,但是有一个更好的计算方式的确定如果游戏就结束了?确定是否有人已经赢了还不是肉类的问题,因为检查,如果9空间完全是微不足道的。
该表述方法可能的解决方案,但如果没有,是什么?此外,如果委员会被不小 n=9
?如果它是一个更大的板说 n=16
, n=25
, 等等,导致数的连续放项目,以赢得可以 x=4
, x=5
, 等?一般算法用于所有 n = { 9, 16, 25, 36 ... }
?
解决方案
你知道获胜的移动可能只发生之后X或O已经做了他们最近的移动,所以可以只搜索行列可选diag中包含的这一行动限制检索的空间,当试图确定一个奖委员会。此外,由于有一个固定数量的动画井字游戏,一旦最后一步是,如果它不是一个赢得移动它的默认一个绘画比赛。
编辑:这个代码是一个n的n局与正在连续赢得(3x3板requries3行,等等)
编辑:加入代码来检查防diag,我不想出一个不循环的方法来确定如果点是在防diag所以这就是为什么这一步骤是失踪
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
}
}
}
其他提示
你可以使用一个魔方 http://mathworld.wolfram.com/MagicSquare.html 如果任何行列,或者diag增加了15然后玩家获胜。
这是类似于 乌萨马*本*ALASSIRY的答案, 但它的交易恒空间和线性时间线性空间和恒的时间。也就是说,没有循环后的初始化。
初始化的一对 (0,0)
对于每一行,每个柱,这两条对角线(对角线和反对角线).这些对代表的积累了 (sum,sum)
在件在相应的行列,或是对角,这里
A piece from player A has value (1,0) A piece from player B has value (0,1)
当玩家的地方一片,更新相应的行对,列对,并对角线对(如果在对角线).如果任何新更新的行列,或是对角的对等于或者 (n,0)
或 (0,n)
然后A或B会,分别。
渐近的分析:
O(1) time (per move) O(n) space (overall)
对于存储使用中,你使用 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:你可以看到如何测试一个画在O(1)时间每移动?如果这样,就能结束比赛早就一个平局。
这个怎么样的伪:
后播放下一块位置(x,y):
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
我会用一系列char[n,n]O、X和空间空。
- 简单的。
- 一个循环。
- 五个简单变量:4整数和一个布尔。
- 扩展到任何尺寸的正。
- 仅检查当前件。
- 没有魔法。:)
这是我的解决方案,我写了一个项目,我正在在javascript。如果你不介意的存储费用的一些阵列,这可能是最快和最简单的解决方案,你会找到。假设你知道位置的最后一步。
/*
* 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;
}
我只写了这个我的C程序的类。
我是张贴的,因为没有其他的例子在这里工作的任何尺寸 矩 电网,以及任何数量 N-在一个排,连续标志赢。
你会找到我的算法,如是,在这 checkWinner()
功能。它不使用魔法的数字或任何幻想检查一个胜利者,它只是使用四个循环的代码是很好的注释所以我会让它为自己说话,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.
如果委员会是 n × n 然后还有 n 行, n 列和2条对角线上。检查每一所有-X或全O's找到一个胜利者。
如果只需要 x < n 方块连续赢了,那么它就更复杂一点。最明显的解决办法是检查每 x × x 广场上的赢家。这里的一些代码的说明。
(我并没有实际测试这种*咳嗽*,但它 有没有 汇编的第一次尝试,耶我!)
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;
}
}
}
我不知道Java,好吧,但我不知道C,所以我试着 adk的幻方的想法 (沿用 Hardwareguy的搜索限制).
// 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;
}
它编制和测试。
% 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,思考它,你不需要一个神奇的方,只是一个最为每个行列/对角线。这是一个小更易于推广的一个魔方 n
× n
矩阵,因为你只需数到 n
.
我被要求同样的问题在我的访谈。我的想法:初始化的矩阵,用0.保持3阵列 1)sum_row(大小n) 2)sum_column(大小n) 3)对角线(2)
每个移动通过(X)减少框值通过1并为每个移动通过(0)增加它由1.在任何一点,如果该行列/角,这已经被修改当前的行动已总结的任-3或+3意味着有人已经赢得了游戏.对于提请我们可以使用上述方法保持moveCount变量。
你觉得我漏了什么吗?
编辑:同样可以用于为矩阵。总和应该甚至+3或-3.
非循环的方法来确定如果点是在防diag:
`if (x + y == n - 1)`
我做了一些优化在该行,col,对角线检查。它的主要决定在第一套循环,如果我们需要检查一个特定列或是对角线。因此,我们避免检查的列或斜杆节省时间。这使得大的影响,当板的尺寸是更多的和重大数量的细胞是没有填补。
这里是java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;
}
我迟到了派对,但我想指出一个好处,我发现利用一个 魔方, ,即它可以被用来获取基准的方会导致赢得或损失在下一回合,而不是仅仅被用于计算当一个游戏结束。
采取这种神奇广场:
4 9 2
3 5 7
8 1 6
第一,设置了一个 scores
阵列,是增加每次移动。看看 这个答案 对于细节。现在,如果我们非法发挥X连续两次在[0,0]和[0,1],然后 scores
阵列是这样的:
[7, 0, 0, 4, 3, 0, 4, 0];
和董事会看起来像这样:
X . .
X . .
. . .
然后,我们所要做的只为了得到一个参考哪一方赢/块是:
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
}
}
}
在现实中,执行需要一些额外的技巧,像处理编号的钥匙(JavaScript),但是我发现它挺简单的,并享有娱乐的数学。
我喜欢这个算法的,因为它使用一个1x9vs3x3代表性的委员会。
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;
}
另一种选择:产生您的表用的代码。起来对称,只有三种方式来赢得:边行,中间行,或是对角线。采取这三个旋和他们周围每个可能的方式:
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))))
这些对称性可以有更多的使用在你玩游戏码:如果你得到一个董事会,你已经看到了一个转版本,你可以把缓存的价值或缓存的最好的行动从那一个(以及unrotate它回)。这通常是远远快于评价的游戏子树。
(翻转左右可以帮同样的方式;它不是在这里需要是因为设定的轮换的获胜的模式是镜子-对称的。)
这里是一个解决方案,我想到,这个商店的符号字符和使用的炭的int值图,如果X或O赢得(来看看裁判的代码)
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;
}
}
}
还在这里是我的单元的测试,以验证它的实际工作
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());
}
}
全面解决方案: https://github.com/nashjain/tictactoe/tree/master/java
如何关于以下方法为9插槽?宣布9整数的变量3x3矩阵(a1、a2。...a9)在a1、a2、a3代表行-1和a1、a4、a7会形式列-1(你得到的想法)。使用'1'表示Player-1和'2'表示Player-2.
有8个可能赢的组合:赢-1:a1+a2+a3(答案可能是3或6个基于哪个球员会) 赢-2:a4+a5+a6 赢-3:a7+a8+a9 赢-4:a1+a4+a7 ....赢的-7:a1+a5+a9 赢-8:a3+a5+a7
现在我们知道,如果放一个跨越a1然后我们需要重新评估总结的3个变量:赢-1,赢-4和赢-7.取'赢-?'变量达到3或6的第一次赢得了比赛。如果赢-1变量达到6中的第一个然后播放-2的胜利。
我不明白,这个方案是不可扩展很容易。
这是一个非常简单的方式检查。
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);
}
}
如果你有的边境领域5*5举个例子来说,我用下一个方法的检查:
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,这是更清楚,但可能不是最优化的方式。
这里是我的溶液使用2维阵列:
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;
}
恒定时间O(8),平均4短。播放器=短号码。需要额外的检查,确保移是有效的。
// 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));
我发明了一种算法为这一部分作为一个科学项目一次。
你基本上递归的划分委员会成堆的重叠2x2的矩形,测试不同的组合可能赢得在2×2平方。
它是缓慢的,但它具有的优势作任何规模板,有相当线性存储器的要求。
我希望我能找到我的执行情况