如何找到二維數組中兩個坐標之間的最短路徑？

Question

我試圖找到從二維數組中的一個點（一個坐標，x 和 y 值代表它在數組中的 position）到另一個點的最短方法。

我想 output 一個坐標數組，必須經過這些坐標才能從初始坐標到最終坐標。

像這樣的數組的一個例子可能是

arr = [
          [15, 7, 3],
          [1, 2, 6],
          [7, 4, 67]
      ]

在這種情況下，我們可以說我們將從arr[0][0]開始並在arr[2][2]結束。 因此，坐標將是(0, 0)和(2, 2) 。

預期的 output 將是： [(0, 2), (1, 2), (2, 2), (2, 1)]或相同長度的東西。

---

我試過的

我設法在下面制作了一個半成功的 function，但在較大的情況下效率非常低且耗時。

import math

arr = [
          [0, 1, 2],
          [3, 4, 5],
          [6, 7, 8]
      ]

coor1 = (0, 0) # seen as 2 in the arr array
coor2 = (2, 2) # seen as 7 in the arr array

def pythagoras(a, b):

    # find pythagorean distances between the two
    distance_y = max(a[0], b[0]) - min(a[0], b[0])
    distance_x = max(a[1], b[1]) - min(a[1], b[1])

    # calculate pythagorean distance to 3 d.p.
    pythag_distance = round(math.sqrt(distance_x**2 + distance_y**2), 3)

    return pythag_distance


def find_shortest_path(arr, position, target):
    ''' finds shortest path between two coordinates, can't go diagonally '''
    coordinates_for_distances = []
    distances = []

    for i in range(len(arr)):
        for r in range(len(arr)):
            coordinates_for_distances.append((i, r))
            distances.append(pythagoras((i, r), target))

    route = []

    while position != target:
        acceptable_y_range = [position[1] + 1, position[1] - 1]
        acceptable_x_range = [position[0] + 1, position[0] - 1]

        possibilities = []
        distance_possibilities = []

        for i in range(len(coordinates_for_distances)):
            if coordinates_for_distances[i][0] == position[0] and coordinates_for_distances[i][1] in acceptable_y_range:
                possibilities.append(coordinates_for_distances[i])
                distance_possibilities.append(distances[i])

            elif coordinates_for_distances[i][1] == position[1] and coordinates_for_distances[i][0] in acceptable_x_range:
                possibilities.append(coordinates_for_distances[i])
                distance_possibilities.append(distances[i])

        zipped_lists = zip(distance_possibilities, possibilities)
        minimum = min(zipped_lists)
        position = minimum[1]
        route.append(position)

    return route

Answer 1

為了找到一對坐標之間的最短路徑，我們可以將其轉化為一個圖問題，其中每個坐標都是一個圖節點。 現在在這種設置下，找到兩個節點之間的最短路徑是一個眾所周知的圖論問題，並且使用正確的工具很容易解決。

我們可以使用NetworkX ，它實際上有一個Graph generator ，它返回mxn個節點的 2d 網格圖，每個節點都連接到它最近的鄰居。 這是完美的案例：

import networkx as nx
from matplotlib import pyplot as plt

G = nx.grid_2d_graph(3,3)

plt.figure(figsize=(6,6))
pos = {(x,y):(y,-x) for x,y in G.nodes()}
nx.draw(G, pos=pos, 
        node_color='lightgreen', 
        with_labels=True,
        node_size=600)

現在我們可以使用 networkX 的nx.bidirectional_shortest_path來找到兩個坐標之間的最短路徑：

coor1 = (0, 2) # seen as 2 in the arr array
coor2 = (2, 1) # seen as 7 in the arr array

nx.bidirectional_shortest_path(G, source=coor1, target=coor2)
# [(0, 2), (1, 2), (2, 2), (2, 1)]

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請注意， nx.grid_2d_graph將生成最多具有任意大m和n的網格圖，通過定位標簽，您還可以 plot 坐標網格，就像上面一樣：