python - 二维网格中的快速寻路和距离矩阵

标签 python numpy path-finding

上下文: 我正在研究支持不同楼层设计并模拟一个或多个负责订单拣选的代理的仓库模拟。一个订单可以包含多个产品。拣选产品的路由被解决为一个有能力的 vehicle routing problem (CVRP)。这需要产品位置之间的距离矩阵,为此 A* algorithm用来。目前,距离矩阵是按订单生成的,就在该订单开始拣货之前。准确测量需要多次模拟运行,因此计算效率非常重要。为了完整起见,我在下面附上了模拟的屏幕截图,其中包含产品位置(深色)、代理(白色)、当前订单中的产品(蓝色)以及当前 route 正在挑选的产品(绿色/红色)。请注意,白线代表当前的拣选优先级,而不是确切的路径。

Simulation screenshot

问题: 距离矩阵的大小随着每个订单的产品数量呈二次方增长。因此,用 A* 计算它的时间很快变得无法接受。

问题: 我需要一种使距离矩阵的计算更加高效的方法。这可以是精确方法或启发式方法,只要不牺牲太多准确性即可。我不是在寻找实现或特定的代码片段,而是在寻找用于我可以自己实现的类似问题的想法和/或方法。

尝试过的方法/思路: 以下是我考虑过或尝试实现但未成功的一些方法:

  • 整个仓库的距离矩阵:不可行,因为产品位置的数量实在是太多了。
  • 使用欧氏距离:不够好。这假设一排仓库相对两侧的产品靠得很近,而实际上代理人必须在两者之间绕远路。
  • 使用聚类算法来识别彼此靠近的区域,并将距离矩阵基于聚类而不是单个位置:这将减少矩阵的总大小,从而可以完全预先计算它。但是,这会大大降低准确性,而且我还没有找到一种聚类算法可以可靠地解决不同楼层布局的这个问题。

布局示例: 白色像素表示地板单元,黑色像素表示产品位置。订单中的产品是从所有可能的位置随机选择的。所选方法应支持更多楼层布局(任何楼层布局!)。

Layout 1

Layout 2

如果有人想弄乱它,这里有一个布局 1 的可粘贴数组:

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[0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0],
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[0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0],
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最佳答案

我知道通常不鼓励仅链接的答案,但“哪些算法可以使 A* 更快”是一个非常复杂的主题,在过去 50 年里一直是一个活跃的研究领域。因此,除了在 Stackoverflow 答案中给出一个模糊的总结之外,实际上不可能给出任何其他内容。

对于像您自己的 2D 网格,有两种常用技术可以提供巨大的加速:

另外,既然你提到它不需要是最优的,

最后,如果您的网格随时间变化,则整个字段为 incremental algorithms表现优于 A*。

关于python - 二维网格中的快速寻路和距离矩阵,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/68918811/

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