井字游戏赢逻辑N×N个板 [英] TicTacToe win logic for NxN board

问题描述

当前逻辑正常工作与3x3的董事会，因为它是静态的。 我怎样才能将其转换成N×N的逻辑是什么？

Current logic works fine with 3x3 board because it is static. How can I convert it into NxN logic?

运逻辑的工作通过将行和列的正方形。

Win logic works by adding the row and column squares.

/*
 * To determine a win condition, each square is "tagged" from left
 * to right, top to bottom, with successive powers of 2.  Each cell
 * thus represents an individual bit in a 9-bit string, and a
 * player's squares at any given time can be represented as a
 * unique 9-bit value. A winner can thus be easily determined by
 * checking whether the player's current 9 bits have covered any
 * of the eight "three-in-a-row" combinations.
 *
 *     273                 84
 *        \               /
 *          1 |   2 |   4  = 7
 *       -----+-----+-----
 *          8 |  16 |  32  = 56
 *       -----+-----+-----
 *         64 | 128 | 256  = 448
 *       =================
 *         73   146   292
 *
 */
wins = [7, 56, 448, 73, 146, 292, 273, 84],

/*
 * Returns whether the given score is a winning score.
 */
win = function (score) {
    for (var i = 0; i < wins.length; i += 1) {
        if ((wins[i] & score) === wins[i]) {
            return true;
        }
    }
    return false;
},

我的小提琴是这里

推荐答案

因此​​，以编程方式做到这一点，你可以使用类来跟踪哪些设置每个单元是，即ROW1或COL1：

So, to do it programatically, you can use classes to keep track of what "set" each cell is in, i.e. "row1" or "col1":

在我 / Ĵ创建循环：

cell.addClass('col' + j); // The cell is in column j
cell.addClass('row' + i); // The cell is in row i
if (i == j) {
    cell.addClass('dia0'); // The cell is in the down/right diagonal
}
if (j == SIZE - i - 1) {
    cell.addClass('dia1'); // The cell is in the up/right diagonal
}

然后，在赢（），传中的最后一个单元格点击。对于每个类的小区属于，检查是否细胞与含有类数 X （或 0 ）是等于表大小：

Then, in win(), pass in the last cell clicked. For each class the cell belongs to, check if the number of cells with that class containing X (or O) is equal to the table size:

win = function (clicked) {
    // Get all of the classes this cell belongs to
    var memberOf = clicked[0].className.split(/\s+/);

    // Check elements with the same class, and see if they contain "turn", i.e. X or O
    for (var i=0; i<memberOf.length; i++) {
        var testClass = '.'+memberOf[i];
        // If the number of elements containing "turn" == SIZE,
        // we have a winning condition
        if( $('#tictactoe').find(testClass+':contains('+turn+')').length == SIZE) {
             return true;   
        }
    }

    return false;
},

的jsfiddle演示

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