Question

I’m terribly sorry for the title, but describing my problem with a couple of words is a tad hard. I figured the rest of the post would explain it better! 😉

Description

I basically have a 2D array of tiles/objects/symbols and I want to split it up into two (or more) new 2D arrays whenever a group of tiles is separated by special tiles.

For example, if I have:

[x][x][0][0]

[0][0][x][0]

[0][0][x][0]

[0][0][0][x]

Where the symbol x is not wanted, then that should give me two new arrays:

[x][x][0][0]

[x][x][x][0]

[x][x][x][0]

[x][x][x][x]

and

[x][x][x][x]

[0][0][x][x]

[0][0][x][x]

[0][0][0][x]

One array for each group of interconnecting tiles.

In my specific case, I have null objects as x, and an arbitrary object for the rest. Basically, if I can’t reach tile B from tile A without crossing a null, then those two are two different groups.

I’ve played with it in my mind for a while, and the best I could come up with is surely far worse than O(n^2), given that they even worked in the first place. Flood fill springs to mind as something that could be used to find a group, but other than that, I’m not sure I can come up with any other similar problem to use in this instance.

The question

So what I’m asking is if you happen to know in which direction to take my problem and/or how to solve it. Computational complexity is not that very important since I plan not to execute this often nor on large arrays. Still, I hope i haven’t hit upon a NP-hard problem! :3

Thanks!

Answer 1

I hope i haven’t hit upon a NP-hard problem!

This is far from being a NP problem.

I will explain two different approaches to solve the problem. One will use a Flood Fill as you expected and the other a Disjoint-set data structure.

Flood Fill

Let’s say you have a matrix N x M where a position (row, column) is null when is not used, it contains a value otherwise.

You need to go through each column element 1..M per row 1..N. That’s very simple:

for row in range(1, N + 1):
  for column in range(1, M + 1):
    if matrix[row][column] is not null:
      floodfill(matrix, row, column)

You will need to call the Flood Fill algorithm each time you find a non-null value, the reason will become clearer after I will define the Flood Fill method below.

def floodfill(matrix, row, column):
  # I will use a queue to keep record of the positions we are gonna traverse.
  # Each element in the queue is a coordinate position (row,column) of an element
  # of the matrix.
  Q = Queue()

  # A container for the up, down, left and right directions.
  dirs = { (-1, 0), (1, 0), (0, -1), (0, 1) }

  # Now we will add our initial position to the queue.
  Q.push( (row, column) )

  # And we will mark the element as null. You will definitely need to
  # use a boolean matrix to mark visited elements. In this case I will simply
  # mark them as null.
  matrix[row][column] = null

  # Go through each element in the queue, while there are still elements to visit.
  while Q is not empty:

    # Pop the next element to visit from the queue.
    # Remember this is a (row, column) position.
    (r, c) = Q.pop()

    # Add the element to the output region.
    region.add( (r, c) )

    # Check for non-visited position adjacent to this (r,c) position.
    # These are:
    #   (r + 1, c): down
    #   (r - 1, c): up
    #   (r, c - 1): left
    #   (r, c + 1): right
    for (dr, dc) in dirs:

      # Check if this adjacent position is not null and keep it between
      # the matrix size.
      if matrix[r + dr][c + dc] is not null
         and r + dr <= rows(matrix)
         and c + dc <= colums(matrix):

        # Then add the position to the queue to be visited later
        Q.push(r + dr, c + dc)

        # And mark this position as visited.
        matrix[r + dr][c + dc] = null

  # When there are no more positions to visit. You can return the
  # region visited.
  return region

You can modify this algorithm to mark each region with an specified number in a different array if you keep track of the number of regions recognized. You will notice I am using a queue instead of a recursive function, this will keep you away from hitting a maximum recursion bound limit.

Union-Find Algorithm

Another solution that I considered more expensive is using a Disjoint-set data structure to accomplish the same purpose. I will merely show the change to the floodfill method.

def floodfill(matrix):
  disjoint_set = DisjointSet()

  # Go through each row in the matrix
  for row in range(1, N + 1):

    # Go through each column in the matrix
    for column in range(1, M + 1):

      # Create a set for the current position
      disjoint_set.makeSet(row, column)

      if matrix[row - 1][column] is not null:
        # If the position north of it it is not null then merge them
        disjoint_set.merge((row, column), (row - 1, column))

      if matrix[row][column - 1] is not null:
        # If the position left of it it is not null then merge them
        disjoint_set.merge((row, column), (row, column - 1))

  # You can go through each position identifying its set and do something with it
  for row in range(1, N + 1):
    for column in range(1, M + 1):
      regions[ disjoint_set.find(row, column) ] = (row, column)

  return regions

I hope this help.

I didn’t bother about showing the complexity since you didn’t care about it.