How to find all possible unique paths in a grid? - algorithm

Question

I have a 3 x 3 grid with randomly placed obstacles in which there is a random starting point but no endpoint. The endpoint is created when there are no more cells to occupy. Movement can occur up, down, left or right.
How can I see all possible unique paths within the grid?
Example:
Once a cell is used when looking for a path, it cannot be used again (1 becomes a 0).
If there are no more neighbouring cells to move into the path has ended. Regardless of weather all the cells have been visited.
# bottom left is (0, 0) and top right is (2, 2)...
# start is (1, 1) and obstacle is (2, 1)
[1] [1] [1]
[1] [S] [0]
[1] [1] [1]
S = starting point
0 = obstacle
1 = open cell
With the above example there would be 6 unique paths.
path_1 = (1, 2), (2, 2) # up, right
path_2 = (1, 0), (2, 0) # down, right
path_3 = (0, 1), (0, 2), (1, 2), (2, 2) # left, up, right, right
path_4 = (0, 1), (0, 0), (1, 0), (2, 0) # left, down, right, right
path_5 = (1, 2), (0, 2), (0, 1), (0, 0), (1, 0), (2, 0) # up, left, down, down, right, right
path_6 = (1, 0), (0, 0), (0, 1), (0, 2) (1, 2), (2, 2) # down, left, up, up, right, right

Answer 1

To get all the paths, you can use DFS or BFS but each path needs to have a unique visited set to keep track that you:
do not go back to the same coordinate twice in a single path
allow different paths to go on the same coordinate
I wrote a DFS implementation for a grid here and the solution will rely on this example.
Solution
To do graph search one would need to define the states, which are the coordinates in this case, but for this problem we will keep track of two parameters, for convenience:
The path taken will be documented via Crack code (right=0, down=1, left=2, up=3) which is a form of chain code
the visited set for each path will de documented for the reasons noted above
The implementation in Python is as follows (in my case the top left matches coordinate (0,0) and lower right matches (n-1,n-1) for nXn grid)
import collections
def find_paths(grid, start_coord):
paths = set() # paths will be added here
state_queue = collections.deque([]) # Pending states which have not been explored yet
# State is a tuple consists of:
# 1. current coordinate
# 2. crack code path
# 3. set of visited coordinates in that path
state = [start_coord, [], {start_coord}] # Starting state
while True:
# Getting all possible neighboring states
# Crack code (right=0, down=1, left=2, up=3)
state_right = [(state[0][0],state[0][1]+1), state[1] + [0], state[2].copy()] if state[0][1]+1 < len(grid[state[0][0]]) else None
state_down = [(state[0][0]+1,state[0][1]), state[1] + [1], state[2].copy()] if state[0][0]+1 < len(grid) else None
state_left = [(state[0][0],state[0][1]-1), state[1] + [2], state[2].copy()] if state[0][1]-1 >= 0 else None
state_up = [(state[0][0]-1,state[0][1]), state[1] + [3], state[2].copy()] if state[0][0]-1 >= 0 else None
# Adding to the queue all the unvisited states, as well as to the visited to avoid returning to states
blocked_counter = 0
for next_state in [state_right, state_down, state_left, state_up]:
if next_state is None:
blocked_counter += 1
elif next_state[0] in state[2] or grid[next_state[0][0]][next_state[0][1]] == 0:
blocked_counter += 1
else:
next_state[2].add(next_state[0])
state_queue.append(next_state)
# After checking all directions' if reached a 'dead end', adding this path to the path set
if blocked_counter == 4:
paths.add(tuple(state[1]))
# Popping next state from the queue and updating the path if needed
try:
state = state_queue.pop()
except IndexError:
break
return paths
Explanation
At the beginning we create the initial state, as well as the initial path which is an empty list and the visited set, containing only the start coordinate
For each state we are in we do the following:
2.1. create the four neighboring states (advancing right, up, left or down)
2.2. checking if we can advance in each direction by checking if the path we are in already visited this coordinate and if the coordinate is valid
2.3. if we can advance in the said direction, add this next_state to the state_queue as well as to the visited set of the path
2.4. if we can not advance in any of the four directions, we reached a 'dead end' and we add the path we are in to the paths set
2.5. pop the next state from state_queue which is kind of a bad name since it is a stack (due to the appending and popping from the same side of the deque()
When the state_queue is empty, we finished the search and we can return all the found paths
When running this with the given example, i.e.
start_coord = (1,1)
grid = [[1, 1, 1],
[1, 1, 0],
[1, 1, 1]]
We get:
find_paths(grid, start_coord)
# {(2, 3, 0, 0), (3, 2, 1, 1, 0, 0), (3, 0), (2, 1, 0, 0), (1, 0), (1, 2, 3, 3, 0, 0)}

Answer 2

This is actually not as difficult as it seems. Since it seems to me that you are learning, I avoid giving you code and will focus on the ideas.
The solution
Since you need to find all solutions, this is a classical backtracking problem. The idea of backtracking is to try out every possibility and store all the solutions. Particularities about your problem:
grid
movement
obstacles
unique solutions
How to check everything?
You need to loop your grid and for each point and for each point, starting from a depth of 0, you repeat the following:
if depth < available points map the possible moves (no obstacles, not visited) (1)
for each point:
increase depth with 1
mark the current point as visited
check whether there was a solution and handle it if so
with the current status, jump to (1)
unmark the current point from being visited
decrease depth by 1
Handling the solution
You need to have a unique, predictable signature for your solution, like the points you have moved to in their order and store all solutions you had so far. Before you enter your new solution, check whether it's already among the solutions and only append it if it's not there.
Pruning
You can prune your findings based on earlier findings, if the problem-space is big-enough to make this a performance optimization, but you should only consider that when you already have a solution.

Answer 3

I was looking to solve this with dynamic programming to avoid the exponential time complexity but failed to do so. However, here is a dense version in Javascript which uses DFS (which given this problem is the same as brute-forcing since we are interested in all possible paths) and recursion in a functional style with es6 features. We also make use of a single-dimension array to represent the board with the directions method accounting for which positions are reachable from a given position. The single-dimension array contains the rows of the board laid out sequentially from top to bottom.
[ 0 1 2 ]
[ 3 4 5 ] => [0, 1, 2, 3, 4, 5, 6, 7, 8]
[ 6 7 8 ]
The algorithm works for any m by n grid and so we must specify the width as an input parameter.
const directions = (pos, board, width) => {
const [n, s, w, e] = [pos - width, pos + width, pos - 1, pos + 1]
return [
n >= 0 && board[n] ? n : null,
s < board.length && board[s] ? s : null,
pos % width !== 0 && board[w] ? w : null,
e % width !== 0 && board[e] ? e : null
]
.filter(pos => pos !== null)
}
const solve = (pos, board, width, path) => {
const next = directions(pos, board, width)
return next.length === 0 ?
[[...path, pos]] // have a complete path
:
next
.map(nextPos => solve(nextPos, [...board.slice(0, pos), 0, ...board.slice(pos + 1)], width, [...path, pos]))
.flat()
}
const oneDim2TwoDim = (oneDimCoord, width) =>
[Math.floor(oneDimCoord / width), oneDimCoord % width]
const res = solve(4, [1, 1, 1, 1, 1, 0, 1, 1, 1], 3, [])
console.log(res.map(path => path.map(step => oneDim2TwoDim(step, 3))))
At each step, we check which directions are possible to go in. If no direction is available, we return the current path, otherwise we make a recursive call in each of the directions.
Use it like
solve(4, [1, 1, 1, 1, 1, 0, 1, 1, 1], 3, [])
/*
[
[ 4, 1, 0, 3, 6, 7, 8],
[ 4, 1, 2 ],
[ 4, 7, 6, 3, 0, 1, 2],
[ 4, 7, 8 ],
[ 4, 3, 0, 1, 2 ],
[ 4, 3, 6, 7, 8 ]
]
*/
Each path starts with the starting position and you can easily convert it back to (x, y) coordinate style with
const oneDim2TwoDim = (oneDimCoord, width) =>
[Math.floor(oneDimCoord / width), oneDimCoord % width]
const res = solve(4, [1, 1, 1, 1, 1, 0, 1, 1, 1], 3, [])
.map(path => path.map(step => oneDim2TwoDim(step, 3)))
/*
[
[ [ 1, 1 ], [ 0, 1 ], [ 0, 0 ], [ 1, 0 ], [ 2, 0 ], [ 2, 1 ], [ 2, 2 ] ],
[ [ 1, 1 ], [ 0, 1 ], [ 0, 2 ] ],
[ [ 1, 1 ], [ 2, 1 ], [ 2, 0 ], [ 1, 0 ], [ 0, 0 ], [ 0, 1 ], [ 0, 2 ] ],
[ [ 1, 1 ], [ 2, 1 ], [ 2, 2 ] ],
[ [ 1, 1 ], [ 1, 0 ], [ 0, 0 ], [ 0, 1 ], [ 0, 2 ] ],
[ [ 1, 1 ], [ 1, 0 ], [ 2, 0 ], [ 2, 1 ], [ 2, 2 ] ]
]
*/