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"""

This script implements the Dijkstra algorithm on a binary grid.

The grid consists of 0s and 1s, where 1 represents

a walkable node and 0 represents an obstacle.

The algorithm finds the shortest path from a start node to a destination node.

Diagonal movement can be allowed or disallowed.

"""

from heapq import heappop, heappush

import numpy as np

def dijkstra(

grid: np.ndarray,

source: tuple[int, int],

destination: tuple[int, int],

allow_diagonal: bool,

) -> tuple[float | int, list[tuple[int, int]]]:

"""

Implements Dijkstra's algorithm on a binary grid.

Args:

grid (np.ndarray): A 2D numpy array representing the grid.

1 represents a walkable node and 0 represents an obstacle.

source (Tuple[int, int]): A tuple representing the start node.

destination (Tuple[int, int]): A tuple representing the

destination node.

allow_diagonal (bool): A boolean determining whether

diagonal movements are allowed.

Returns:

Tuple[Union[float, int], List[Tuple[int, int]]]:

The shortest distance from the start node to the destination node

and the shortest path as a list of nodes.

>>> dijkstra(np.array([[1, 1, 1], [0, 1, 0], [0, 1, 1]]), (0, 0), (2, 2), False)

(4.0, [(0, 0), (0, 1), (1, 1), (2, 1), (2, 2)])

>>> dijkstra(np.array([[1, 1, 1], [0, 1, 0], [0, 1, 1]]), (0, 0), (2, 2), True)

(2.0, [(0, 0), (1, 1), (2, 2)])

>>> dijkstra(np.array([[1, 1, 1], [0, 0, 1], [0, 1, 1]]), (0, 0), (2, 2), False)

(4.0, [(0, 0), (0, 1), (0, 2), (1, 2), (2, 2)])

"""

rows, cols = grid.shape

dx = [-1, 1, 0, 0]

dy = [0, 0, -1, 1]

if allow_diagonal:

dx += [-1, -1, 1, 1]

dy += [-1, 1, -1, 1]

queue, visited = [(0, source)], set()

matrix = np.full((rows, cols), np.inf)

matrix[source] = 0

predecessors = np.empty((rows, cols), dtype=object)

predecessors[source] = None

while queue:

(dist, (x, y)) = heappop(queue)

if (x, y) in visited:

continue

visited.add((x, y))

if (x, y) == destination:

path = []

while (x, y) != source:

path.append((x, y))

x, y = predecessors[x, y]

path.append(source) # add the source manually

path.reverse()

return float(matrix[destination]), path

for i in range(len(dx)):

nx, ny = x + dx[i], y + dy[i]

if 0 <= nx < rows and 0 <= ny < cols:

next_node = grid[nx][ny]

if next_node == 1 and matrix[nx, ny] > dist + 1:

heappush(queue, (dist + 1, (nx, ny)))

matrix[nx, ny] = dist + 1

predecessors[nx, ny] = (x, y)

return np.inf, []

if __name__ == "__main__":

import doctest

doctest.testmod()

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