去模拟 - 标记死石 [英] Go Simulation - Mark dead stones

问题描述

Hiya，我正在尝试创建一个 Go （游戏） ）在PHP中模拟，问题是如何标记死石。


每块石头都有一些自由，附近是空旷的地方，当一组石头被完全封闭在另一种颜色的石头中时，已经死了。


这里是我迄今为止所做的例子：
http://www.k4s.ch/test/sca/test.php


p =播放器[1：黑色，2：白色]

x，y = [移动坐标]

& reset to wipe


我们可以在石头附近有4件事，让'' s说黑色（1）：

- 另一种相同颜色的石头（1）

- 另一种颜色的另一种石头（2）

- 没有（在板外，即在边界上）

- 一个空的空间（标记为0）


'这就是我想要做的事情......例如每次我放一块新石头：


[PHP]

功能检查（$ x，$ y）{

// $ array是一个包含所有st和st的地方的数组一个即。 [4] [5] = 2，[4] [6] = 1，..

全球$数组;

// $我是玩家1或2

$ me = $ array [$ x] [$ y];

// $ near是一个包含北，南，东，西石头的阵列（如果有的话） ）

$ near [0] = $ array [$ x] [$ y-1];

$ near [1] = $ array [$ x + 1] [$ y];

$ near [2] = $ array [$ x] [$ y + 1];

$ near [3] = $ array [$ x-1] [$ y];

for（$ a = 0; $ a< = 3; $ a ++）{

//我怎么能重新执行下一块石头的功能相同吗？

if（（$ near [$ a]！== $ me）&&（$ near [$ a]！== 0）&&（isset（$ near [$ a））））{OMG有一块敌人的石头！让我们看看它是否有一些自由......否则它已经死了！ }

}

}

[/ PHP]


就是这样......希望有人能理解我在说什么：P


最后一块石头有多少自由并不重要..重要的是靠近那块石头的石头（因为如果它'自由它死了）O_o


（对不起英语不好意思）

解决方案

x，

y）{

//

数组是一个包含所有地方的数组，即。 [4] [5] = 2，[4] [6] = 1，..

global

Hiya, i''m trying to create a Go (the game) simulation in PHP, the problem is how to mark dead stones.

each stone has some liberties wich are the empty territories nearby, when a group of stones is completly enclosed in stones of another color, is dead.

here the example of what i''ve done so far:
http://www.k4s.ch/test/sca/test.php

p= player [1: black, 2: white]
x,y = [move coordinates]
&reset to wipe

We can have 4 things near a stone, let''s say im black (1):
- another stone of the same color (1)
- another stone of another color (2)
- nothing (out of the board, ie. on the border)
- an empty space (marked as ''0'')

that''s what i was thinking to do.. for instance each time i place a new stone:

[PHP]
function check($x,$y) {
//$array is an array with all the places with some stone ie. [4][5] = 2, [4][6] = 1, ..
global $array;
//$me is the player 1 or 2
$me = $array[$x][$y];
//$near is an array with the north, south, east, west stones (if any)
$near[0] = $array[$x][$y-1];
$near[1] = $array[$x+1][$y];
$near[2] = $array[$x][$y+1];
$near[3] = $array[$x-1][$y];
for ($a=0;$a<=3;$a++) {
//how can i re-exec the same function to the next stone ?
if (($near[$a] !== $me) && ($near[$a] !== 0) && (isset($near[$a]))) { OMG there is an enemy stone ! let''s see if it has some liberty.. otherwise it''s dead! }
}
}
[/PHP]

That''s it.. hopefully someone will understand what im talking about :P

In the end how many liberties a stone has is not important.. the important thing is the stone near that one (because if it ain''t liberties it dies) O_o

(sorry for the bad english lol)

解决方案

x,

y) {
//

array is an array with all the places with some stone ie. [4][5] = 2, [4][6] = 1, ..
global

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