Tic Tac脚趾与Minimax:计算机有时会失去球员第一;另外工作
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12-12-2019 - |
题
我正在研究一种用于无与伦比的TIC TAC脚趾的最小新的算法。当计算机首先以及播放器首先时,我需要它来工作。使用当前版本,首先将在移动时丢失。但是,如果播放器首先,似乎最低似乎从未发现最佳举动(总是返回-1作为分数)。
如果玩家首次移动,则导致计算机返回的最小值得分为-1?
示例:
board.addMark( 1, Mark2.PLAYER ); // add a 'PLAYER' mark to an arbitrary location
Minimax.minimax( board, Mark2.COMPUTER ); // will always return -1
.
这是'minimax'类:
public class Minimax {
public static int minimax( Board board, Mark2 currentMark ) {
int score = (currentMark == Mark2.COMPUTER) ? -1 : 1;
int[] availableSpaces = board.getAvailableSpaces();
if ( board.hasWinningSolution() )
score = (board.getWinningMark() == Mark2.COMPUTER) ? 1 : -1;
else if ( availableSpaces.length != 0 ) {
Mark2 nextMark = (currentMark == Mark2.COMPUTER) ? Mark2.PLAYER : Mark2.COMPUTER;
for ( int availableIndex = 0; availableIndex < availableSpaces.length; availableIndex++ ) {
board.addMark( availableSpaces[availableIndex], currentMark );
int nextScore = minimax( board, nextMark );
board.eraseMark( availableSpaces[availableIndex] );
if ( currentMark == Mark2.COMPUTER && nextScore > score
|| currentMark == Mark2.PLAYER && nextScore < score )
score = nextScore;
}
}
return score;
}
}
.
这是'板'类:
public class Board {
private Mark2 gameBoard[];
private int blankSpaces;
private boolean solutionFound;
private Mark2 winningMark;
public final static int winSets[][] = {
{ 0, 1, 2 }, { 3, 4, 5 },
{ 6, 7, 8 }, { 0, 3, 6 },
{ 1, 4, 7 }, { 2, 5, 8 },
{ 0, 4, 8 }, { 2, 4, 6 }
};
public Board() {
gameBoard = new Mark2[9];
blankSpaces = 9;
for ( int boardIndex = 0; boardIndex < gameBoard.length; boardIndex++ ) {
gameBoard[boardIndex] = Mark2.BLANK;
}
}
public void addMark( int spaceIndex, Mark2 mark ) {
if ( gameBoard[spaceIndex] != mark ) {
gameBoard[spaceIndex] = mark;
if ( mark == Mark2.BLANK ) {
blankSpaces++;
} else {
blankSpaces--;
}
}
}
public void eraseMark( int spaceIndex ) {
if ( gameBoard[spaceIndex] != Mark2.BLANK ) {
gameBoard[spaceIndex] = Mark2.BLANK;
blankSpaces++;
}
}
public int[] getAvailableSpaces() {
int spaces[] = new int[blankSpaces];
int spacesIndex = 0;
for ( int boardIndex = 0; boardIndex < gameBoard.length; boardIndex++ )
if ( gameBoard[boardIndex] == Mark2.BLANK )
spaces[spacesIndex++] = boardIndex;
return spaces;
}
public Mark2 getMarkAtIndex( int spaceIndex ) {
return gameBoard[spaceIndex];
}
public boolean hasWinningSolution() {
this.solutionFound = false;
this.winningMark = Mark2.BLANK;
for ( int setIndex = 0; setIndex < winSets.length && !solutionFound; setIndex++ )
checkSpacesForWinningSolution( winSets[setIndex][0], winSets[setIndex][1], winSets[setIndex][2] );
return solutionFound;
}
public Mark2 getWinningMark() {
return this.winningMark;
}
private void checkSpacesForWinningSolution( int first, int second, int third ) {
if ( gameBoard[first] == gameBoard[second] && gameBoard[third] == gameBoard[first]
&& gameBoard[first] != Mark2.BLANK ) {
solutionFound = true;
winningMark = gameBoard[first];
}
}
public void printBoard() {
System.out.printf( " %c | %c | %c\n", getMarkCharacter( gameBoard[0] ), getMarkCharacter( gameBoard[1] ),
getMarkCharacter( gameBoard[2] ) );
System.out.println( "------------" );
System.out.printf( " %c | %c | %c\n", getMarkCharacter( gameBoard[3] ), getMarkCharacter( gameBoard[4] ),
getMarkCharacter( gameBoard[5] ) );
System.out.println( "------------" );
System.out.printf( " %c | %c | %c\n", getMarkCharacter( gameBoard[6] ), getMarkCharacter( gameBoard[7] ),
getMarkCharacter( gameBoard[8] ) );
}
public char getMarkCharacter( Mark2 mark ) {
char result = (mark == Mark2.PLAYER) ? 'O' : ' ';
result = (mark == Mark2.COMPUTER) ? 'X' : result;
return result;
}
}
.
在这里是'mark2'类,如果有什么混淆:
public enum Mark2 {
BLANK,
PLAYER,
COMPUTER
}
. 解决方案 3
延长休息后,借助@Guyadini的回复,我有一个epiphany。我写了一个测试来计算Minimax()返回的三种可能分数的发生。结果,它不会产生0,这使我提出了我需要0所需要的0算法作为可能的分数。
我最初改变了“得分”的初始化,是最低/最高的结果(-1/1)并与它们进行比较。但是,这禁止导致从返回的分数集中获得最低/最高值,而是还包括初始化值。这宠坏了结果。
我添加了与“得分”的条件更改的以下比较:
|| availableIndex == 0
.
这迫使所有剩余的比较,以根据属于返回分数集的值进行。
public class Minimax {
public static int minimax( Board board, Mark2 currentMark ) {
int score = 0;
int[] availableSpaces = board.getAvailableSpaces();
if ( board.hasWinningSolution() )
score = (board.getWinningMark() == Mark2.COMPUTER) ? 1 : -1;
else if ( availableSpaces.length != 0 ) {
Mark2 nextMark = (currentMark == Mark2.COMPUTER) ? Mark2.PLAYER : Mark2.COMPUTER;
for ( int availableIndex = 0; availableIndex < availableSpaces.length; availableIndex++ ) {
board.addMark( availableSpaces[availableIndex], currentMark );
int nextScore = minimax( board, nextMark );
board.eraseMark( availableSpaces[availableIndex] );
if ( currentMark == Mark2.COMPUTER && nextScore > score
|| currentMark == Mark2.PLAYER && nextScore < score
|| availableIndex == 0 )
score = nextScore;
}
}
return score;
}
}
. 其他提示
让我们看看一些更简单的东西 - 电路板是1x1,而第一个播放器在那里赢得标记。 现在计算机启动,得分= -1。 没有获胜解决方案(检查一个获胜集,它不是一个连续的), 还有一个可用的空间。 所以我们通过回溯来尝试一个可用的位置。
现在标记=播放器,电路板有一个获胜的解决方案。所以电脑获胜,得分= -1。
返回第一个呼叫,行“int nextscore= minimax(板,Nextmark);”再次返回-1,最终得分为-1。
在较大的问题上发生了同样的事情。
在一个tic tac toe游戏中,有3个可能性而不是2:玩家1胜,玩家2胜,没有人赢了。
你应该替换这样的行:
int score = (currentMark == Mark2.COMPUTER) ? -1 : 1;
.
这样的东西:
int score = (currentMark == Mark2.COMPUTER) ? -1 : ((currentMark == Mark2.PLAYER) ? 1 : 0);
.