使用 Haskell 查找网格上两点之间的最短路径

发布于 2024-08-24 21:04:21 字数 306 浏览 5 评论 0原文

这是一个我可以很容易地以非功能性方式解决的问题。

但用 Haskell 解决它给我带来了大问题。我在函数式编程方面缺乏经验肯定是一个原因。

问题:

我有一个二维字段,分为大小相等的矩形。一个简单的网格。有些矩形是空的(并且可以通过),而另一些则无法通过。给定一个起始矩形 A 和一个目标矩形 B,我如何计算两者之间的最短路径?只能垂直和水平移动,步长为单个矩形。

我将如何在 Haskell 中完成这个任务?代码片段当然受欢迎,但也肯定不是必需的。也非常欢迎链接到更多资源!

谢谢!

This is a problem that I can easily enough solve in a non-functional manner.

But solving it in Haskell is giving me big problems. Me being inexperienced when it comes to functional programming is surely a reason.

The problem:

I have a 2D field divided into rectangles of equal size. A simple grid. Some rectangles are empty space (and can be passed through) while others are impassable. Given a starting rectangle A and a destination rectangle B, how would I calculate the shortest path between the two? Movement is possible only vertically and horizontally, in steps a single rectangle large.

How would I go about accomplishing this in Haskell? Code snippets certainly welcome, but also certainly not neccessary. And links to further resources also very welcome!

Thanks!

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可爱咩 2024-08-31 21:04:21

我将网格表示为列表的列表,输入 [[Bool]]。我会定义一个函数来了解网格元素是否已满:

type Grid = [[Bool]]
isFullAt :: Grid -> (Int, Int) -> Bool  -- returns True for anything off-grid

然后我会定义一个函数来查找邻居:

neighbors :: (Int, Int) -> [(Int, Int)]

要查找 point 的非完整邻居,您可以使用 filter 进行过滤(不是 .isFullAt)$ 邻居点

此时,我将定义两个数据结构:

  • 将每个点映射到Maybe Cost
  • 将所有具有已知成本的点存储在堆中

仅使用堆中的起始方A进行初始化,成本为零。

然后按如下方式循环:

  • 从堆中删除一个最小成本方格。
  • 如果它不在有限映射中,请将其及其成本 c 添加到堆中,并使用成本 c+1 将所有非完整邻居添加到堆中。

当堆为空时,您将获得所有可到达点的成本,并且可以在有限映射中查找B。 (这个算法可能被称为“Dijkstra 算法”;我忘了。)

您可以在 Data.Map 中找到有限映射。我假设在巨大的库中的某个地方有一个堆(也称为优先级队列),但我不知道在哪里。

我希望这足以让您开始。

I'd represent the grid as a list of lists, type [[Bool]]. And I'd define a function to know if a grid element is full:

type Grid = [[Bool]]
isFullAt :: Grid -> (Int, Int) -> Bool  -- returns True for anything off-grid

Then I'd define a function to find neighbors:

neighbors :: (Int, Int) -> [(Int, Int)]

To find non-full neighbors of point you can filter with filter (not . isFullAt) $ neighbors point.

At this point I'd define two data structures:

  • Map each point to Maybe Cost
  • Store all points with known cost in a heap

Initialize with only the start square A in the heap, with cost zero.

Then loop as follows:

  • Remove a min-cost square from the heap.
  • If it's not already in the finite map, add it and its cost c, and add all the non-full neighbors to the heap with cost c+1.

When the heap is empty, you will have the costs of all reachable points and can look up B in the finite map. (This algorithm may be called "Dijkstra's algorithm"; I've forgotten.)

You can find finite maps in Data.Map. I assume there's a heap (aka priority queue) somewhere in the vast library, but I don't know where.

I hope this is enough to get you started.

暖树树初阳… 2024-08-31 21:04:21

嗯,你的类型将决定你的算法。

您想使用什么数据类型来表示网格?二维数组?列表的列表?一棵树?图表?

如果您只想有向图中的最短路径,那么使用 FGL(函数图包)中的内容将是最好的。

Well, your types will determine your algorithms.

What data type do you want to use to represent the grid? A two-dimensional array? A list of lists? A tree? A graph?

If you just want shortest path in a directed graph, using something from the FGL (functional graph package) would be best.

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