planning_utils.py

from enum import Enum
from queue import PriorityQueue
import numpy as np
import re
from math import sqrt

def create_grid(data, drone_altitude, safety_distance):
    """
    Returns a grid representation of a 2D configuration space
    based on given obstacle data, drone altitude and safety distance
    arguments.
    """

    # minimum and maximum north coordinates
    north_min = np.floor(np.min(data[:, 0] - data[:, 3]))
    north_max = np.ceil(np.max(data[:, 0] + data[:, 3]))

    # minimum and maximum east coordinates
    east_min = np.floor(np.min(data[:, 1] - data[:, 4]))
    east_max = np.ceil(np.max(data[:, 1] + data[:, 4]))

    # given the minimum and maximum coordinates we can
    # calculate the size of the grid.
    north_size = int(np.ceil((north_max - north_min + 1)))
    east_size = int(np.ceil((east_max - east_min + 1)))

    # Initialize an empty grid
    grid = np.zeros((north_size, east_size))

    # Populate the grid with obstacles
    for i in range(data.shape[0]):
        north, east, alt, d_north, d_east, d_alt = data[i, :]
        if alt + d_alt + safety_distance > drone_altitude:
            obstacle = [
                int(np.clip(north - d_north - safety_distance - north_min, 0, north_size-1)),
                int(np.clip(north + d_north + safety_distance - north_min, 0, north_size-1)),
                int(np.clip(east - d_east - safety_distance - east_min, 0, east_size-1)),
                int(np.clip(east + d_east + safety_distance - east_min, 0, east_size-1)),
            ]
            grid[obstacle[0]:obstacle[1]+1, obstacle[2]:obstacle[3]+1] = 1

    return grid, int(north_min), int(east_min)


# Assume all actions cost the same.
class Action(Enum):
    """
    An action is represented by a 3 element tuple.
    The first 2 values are the delta of the action relative
    to the current grid position. The third and final value
    is the cost of performing the action.
    """

    WEST = (0, -1, 1)
    EAST = (0, 1, 1)
    NORTH = (-1, 0, 1)
    SOUTH = (1, 0, 1)
    SOUTH_EAST = (1, 1, sqrt(2))
    NORTH_EAST = (-1, 1, sqrt(2))
    SOUTH_WEST = (1, -1, sqrt(2))
    NORTH_WEST = (-1, -1, sqrt(2))

    @property
    def cost(self):
        return self.value[2]

    @property
    def delta(self):
        return (self.value[0], self.value[1])


def valid_actions(grid, current_node):
    """
    Returns a list of valid actions given a grid and current node.
    """
    valid_actions = list(Action)
    n, m = grid.shape[0] - 1, grid.shape[1] - 1
    x, y = current_node

    # check if the node is off the grid or
    # it's an obstacle

    if x - 1 < 0 or grid[x - 1, y] == 1:
        valid_actions.remove(Action.NORTH)
    if x + 1 > n or grid[x + 1, y] == 1:
        valid_actions.remove(Action.SOUTH)
    if y - 1 < 0 or grid[x, y - 1] == 1:
        valid_actions.remove(Action.WEST)
    if y + 1 > m or grid[x, y + 1] == 1:
        valid_actions.remove(Action.EAST)
    if x + 1 > n or y + 1 > m or grid[x + 1, y + 1] == 1:
        valid_actions.remove(Action.SOUTH_EAST)
    if x - 1 < 0 or y + 1 > m or grid[x - 1, y + 1] == 1:
        valid_actions.remove(Action.NORTH_EAST)
    if x + 1 > n or y - 1 < 0 or grid[x + 1, y - 1] == 1:
        valid_actions.remove(Action.SOUTH_WEST)
    if x - 1 < 0 or y - 1 < 0 or grid[x - 1, y - 1] == 1:
        valid_actions.remove(Action.NORTH_WEST)

    return valid_actions


def a_star(grid, h, start, goal):
    """
    Given a grid and heuristic function returns
    the lowest cost path from start to goal.
    """

    path = []
    path_cost = 0
    queue = PriorityQueue()
    queue.put((0, start))
    visited = set(start)

    branch = {}
    found = False

    while not queue.empty():
        item = queue.get()
        current_cost = item[0]
        current_node = item[1]

        if current_node == goal:
            print('Found a path.')
            found = True
            break
        else:
            # Get the new vertexes connected to the current vertex
            for a in valid_actions(grid, current_node):
                next_node = (current_node[0] + a.delta[0], current_node[1] + a.delta[1])
                new_cost = current_cost + a.cost + h(next_node, goal)

                if next_node not in visited:
                    visited.add(next_node)
                    queue.put((new_cost, next_node))

                    branch[next_node] = (new_cost, current_node, a)

    if found:
        # retrace steps
        n = goal
        path_cost = branch[n][0]
        path.append(goal)
        while branch[n][1] != start:
            path.append(branch[n][1])
            n = branch[n][1]
        path.append(branch[n][1])
    else:
        print('**********************')
        print('Failed to find a path!')
        print('**********************')
    return path[::-1], path_cost

def heuristic(position, goal_position):
    return np.linalg.norm(np.array(position) - np.array(goal_position))

def read_home(filename):
    """
    Reads home (lat, lon) from the first line of the `file`.
    """
    with open(filename) as f:
        first_line = f.readline()
    match = re.match(r'^lat0 (.*), lon0 (.*)$', first_line)
    if match:
        lat = match.group(1)
        lon = match.group(2)
    return np.fromstring(f'{lat},{lon}', dtype='Float64', sep=',')

def collinearity_prune(path, epsilon=1e-5):
    """
    Prune path points from `path` using collinearity.
    """
    def point(p):
        return np.array([p[0], p[1], 1.]).reshape(1, -1)

    def collinearity_check(p1, p2, p3):
        m = np.concatenate((p1, p2, p3), 0)
        det = np.linalg.det(m)
        return abs(det) < epsilon

    pruned_path = [p for p in path]
    i = 0
    while i < len(pruned_path) - 2:
        p1 = point(pruned_path[i])
        p2 = point(pruned_path[i+1])
        p3 = point(pruned_path[i+2])

        # If the 3 points are in a line remove
        # the 2nd point.
        # The 3rd point now becomes and 2nd point
        # and the check is redone with a new third point
        # on the next iteration.
        if collinearity_check(p1, p2, p3):
            # Something subtle here but we can mutate
            # `pruned_path` freely because the length
            # of the list is check on every iteration.
            pruned_path.remove(pruned_path[i+1])
        else:
            i += 1
    return pruned_path