《吃豆人》中的幽灵遵循或多或少可预测的模式，即尝试先匹配X或Y，直到实现目标。我一直认为这对于眼睛寻找回去的路是完全一样的。

我建议幽灵存储他从洞到吃豆人的路径。所以一旦鬼魂死了，他就可以沿着这条存储路径向相反的方向移动。

我认为你的解决方案是正确的，比这更简单，就是制作一个更“现实”的新版本，鬼魂的眼睛可以穿过墙壁=)

最初的《吃豆人》并没有使用寻径或花哨的AI。它只是让玩家觉得游戏比实际更有深度，但实际上它是随机的。正如Ian Millington和John Funge在《ai Intelligence for Games》中所述。

Not sure if it's true or not, but it makes a lot of sense to me. Honestly, I don't see these behaviors that people are talking about. Red/Blinky for ex is not following the player at all times, as they say. Nobody seems to be consistently following the player, on purpose. The chance that they will follow you looks random to me. And it's just very tempting to see behavior in randomness, especially when the chances of getting chased are very high, with 4 enemies and very limited turning options, in a small space. At least in its initial implementation, the game was extremely simple. Check out the book, it's in one of the first chapters.

我不太清楚你是如何执行游戏的，但你可以这么做:

Determine the eyes location relative position to the gate. i.e. Is it left above? Right below? Then move the eyes opposite one of the two directions (such as make it move left if it is right of the gate, and below the gate) and check if there are and walls preventing you from doing so. If there are walls preventing you from doing so then make it move opposite the other direction (for example, if the coordinates of the eyes relative to the pin is right north and it was currently moving left but there is a wall in the way make it move south. Remember to keep checking each time to move to keep checking where the eyes are in relative to the gate and check to see when there is no latitudinal coordinate. i.e. it is only above the gate. In the case it is only above the gate move down if there is a wall, move either left or right and keep doing this number 1 - 4 until the eyes are in the den. I've never seen a dead end in Pacman this code will not account for dead ends. Also, I have included a solution to when the eyes would "wobble" between a wall that spans across the origin in my pseudocode.

一些伪代码:

   x = getRelativeOppositeLatitudinalCoord()
   y
   origX = x
    while(eyesNotInPen())
       x = getRelativeOppositeLatitudinalCoordofGate()
       y = getRelativeOppositeLongitudinalCoordofGate()
       if (getRelativeOppositeLatitudinalCoordofGate() == 0 && move(y) == false/*assume zero is neither left or right of the the gate and false means wall is in the way */)
            while (move(y) == false)
                 move(origX)
                 x = getRelativeOppositeLatitudinalCoordofGate()
        else if (move(x) == false) {
            move(y)
    endWhile

假设你已经有了追逐吃豆人所需的逻辑，为什么不重用它呢?只要改变目标。这似乎比尝试使用完全相同的逻辑创建一个全新的例程要少得多。