How could I add more than start point and goal point in A star algorithm?

Question

Below is a Python code for the A* algorithm A* which finds path from the starting point (0,0) to the goal point (4,5) in a 2D simple environment. Can I add another start point and goal point in the same environment(i.e. 2 starting points and 2 goal points on the same environment) and see their paths together and see how the algorithm behaves to more than start and goal points?. Could I get any assistance please to my code or idea?

The figure below shows the operation

This is below the Python code for A* algorithm

from __future__ import print_function
import random

grid = [[0, 1, 0, 0, 0, 0],
        [0, 1, 0, 0, 0, 0],#0 are free path whereas 1's are obstacles
        [0, 1, 0, 0, 0, 0],
        [0, 1, 0, 0, 1, 0],
        [0, 0, 0, 0, 1, 0]]


init = [0, 0]
goal = [len(grid)-1, len(grid[0])-1] #all coordinates are given in format [y,x] 
cost = 1


#the cost map which pushes the path closer to the goal
heuristic = [[0 for row in range(len(grid[0]))] for col in range(len(grid))]
for i in range(len(grid)):    
    for j in range(len(grid[0])):            
        heuristic[i][j] = abs(i - goal[0]) + abs(j - goal[1])




#the actions we can take
delta = [[-1, 0 ], # go up
         [ 0, -1], # go left
         [ 1, 0 ], # go down
         [ 0, 1 ]] # go right


#function to search the path
def search(grid,init,goal,cost,heuristic):

    closed = [[0 for col in range(len(grid[0]))] for row in range(len(grid))]# the referrence grid
    closed[init[0]][init[1]] = 1
    action = [[0 for col in range(len(grid[0]))] for row in range(len(grid))]#the action grid

    x = init[0]
    y = init[1]
    g = 0

    f = g + heuristic[init[0]][init[0]]
    cell = [[f, g, x, y]]

    found = False  # flag that is set when search is complete
    resign = False # flag set if we can't find expand

    while not found and not resign:
        if len(cell) == 0:
            resign = True
            return "FAIL"
        else:
            cell.sort()#to choose the least costliest action so as to move closer to the goal
            cell.reverse()
            next = cell.pop()
            x = next[2]
            y = next[3]
            g = next[1]
            f = next[0]


            if x == goal[0] and y == goal[1]:
                found = True
            else:
                for i in range(len(delta)):#to try out different valid actions
                    x2 = x + delta[i][0]
                    y2 = y + delta[i][1]
                    if x2 >= 0 and x2 < len(grid) and y2 >=0 and y2 < len(grid[0]):
                        if closed[x2][y2] == 0 and grid[x2][y2] == 0:
                            g2 = g + cost
                            f2 = g2 + heuristic[x2][y2]
                            cell.append([f2, g2, x2, y2])
                            closed[x2][y2] = 1
                            action[x2][y2] = i
    invpath = []
    x = goal[0]
    y = goal[1]
    invpath.append([x, y])#we get the reverse path from here
    while x != init[0] or y != init[1]:
        x2 = x - delta[action[x][y]][0]
        y2 = y - delta[action[x][y]][1]
        x = x2
        y = y2
        invpath.append([x, y])

    path = []
    for i in range(len(invpath)):
        path.append(invpath[len(invpath) - 1 - i])
    print("ACTION MAP")
    for i in range(len(action)):
        print(action[i])

    return path

a = search(grid,init,goal,cost,heuristic)
for i in range(len(a)):
    print(a[i]) 

Answer 1

You can have tuples of (init,goal) coordinate pairs in a list that you can iterate through, using the same grid.

init_goal_pairs = [(init_0,goal_0),(init_1,goal_1), (init_2,goal_2), (init_3,goal_3)]

Then, you can do

for init, goal in init_goal_pairs
    path = search(grid,init,goal,cost,heuristic)
    print(path)