如何判断游戏结束，在井字棋？ [英] How to determine game end, in tic-tac-toe?

问题描述

我开发井字棋游戏，我需要算法来检查时，游戏结束（谁赢）。 在3×3的比赛，我会检查每一个可能的双赢的局面（有8功能）。但是，在7×7（4所需的迹象在一排或collumn，或对角线）是一个很大的可能取胜的模式。

I'm developing tic-tac-toe game, and I need algorithm to check when game ends(and who win). In 3x3 game I would check each possible win-situation(there is 8 capabilities). But in 7x7(needed 4 signs in a row or collumn, or diagonal)is a lot of possible win patterns.

推荐答案

在一个非常基本的方法是看运行在所有方向的每一个细胞，这里有一种方法则永远只能检查单个单元格行一次。 A线是一个行，列或对角线都不可能赢，就像在拉斯维加斯的老虎机：）

While a very basic approach is to look at runs in all the directions from every single cell, here are an approach then only ever checks a cell in a single "line" once. A "line" is a row, column, or diagonal that can possibly win, like in a Vegas slot machine :)

1. 对于每个线，移动的启动的是行的和;
2. 设置计数器为0。
3. 对于每个单元中的行（穿越，才能行）：
   • 如果该细胞是P1和计数器> = 0，添加一个反击
     • 如果计数器= 4，则P1胜。

1. For each "line", move to start of that "line" and;
2. Set counter to 0.
3. For each cell in the "line" (traversing the line in order):
   • If the cell is P1 and counter is >= 0, add one to counter
     • If counter = 4 then P1 wins.

• 如果计数器= -4然后P2胜

重要编辑：：如果单元格中包含既不P1或P2，重置计数器为0（DOH！）。我忽略这个简单，但需要一步。否则，11-11将被算作一场胜利。

Important If the cell contains neither P1 or P2, reset counter to 0 (doh!). I omitted this trivial but required step. Otherwise "11-11" would be counted as a win.

的线可以被穿越给定起点和行/列每次迭代偏移量（例如开始于（0,0）和提前（1,1），用于最长对角线从NW至SE）。对角线长度与小于4可避免被检查，当然完全。

The "lines" can be traversed given a starting point and row/column offset per iteration (e.g. start at (0,0) and advance (1,1) for longest diagonal from NW to SE). Diagonals with lengths less than 4 can avoid being checked entirely, of course.

快乐编码。

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