Connected Component Counting - c++

Question

In the standard algorithm for connected component counting, a disjoint-set data structure called union-find is used.
Why is this data structure used? I've written code to just search the image linearly, maintaining two linear buffers to store the current and next component counts for each connected pixels by just examining four neighbors (E, SE, S, SW), and in case of a connection, update the connection map to join the higher component with the lower component.
Once done, search for all non joined components and report the count.
I just can't see why this approach is less efficient than using union-find.
Here's my code. The input file has been reduced to 0s and 1s. The program outputs the number of connected components formed from 0s.
def CompCount(fname):
fin = open(fname)
b,l = fin.readline().split()
b,l = int(b),int(l)+1
inbuf = '1'*l + fin.read()
prev = curr = [sys.maxint]*l
nextComp = 0
tree = dict()
for i in xrange(1, b+1):
curr = [sys.maxint]*l
for j in xrange(0, l-1):
curr[j] = sys.maxint
if inbuf[i*l+j] == '0':
p = [prev[j+n] for m,n in [(-l+1,1),(-l,0),(-l-1,-1)] if inbuf[i*l + j+m] == '0']
curr[j] = min([curr[j]] + p + [curr[j-1]])
if curr[j] == sys.maxint:
nextComp += 1
curr[j] = nextComp
tree[curr[j]] = 0
else:
if curr[j] < prev[j+1]: tree[prev[j+1]] = curr[j]
if curr[j] < prev[j]: tree[prev[j]] = curr[j]
if curr[j] < prev[j-1]: tree[prev[j-1]] = curr[j]
if curr[j] < curr[j-1]: tree[curr[j-1]] = curr[j]
prev = curr
return len([x for x in tree if tree[x]==0])

Answer 1

I didn't completely understand your question, you'd really gain for yourself in writing this up clearly and structuring your question.
What I understand is that you want to do a connected component labeling in a 0-1 image by using the 8 neighborhood. If this is so your assumption that the resulting neighborhood graph would be planar is wrong. You have crossings at the "diagonals". It should be easy to construct a K_{3,3} or K_{5} in such an image.

Answer 2

Your algorithm is flawed. Consider this example:
11110
01010
10010
11101
Your algorithm says 2 components whereas it has only 1.
To test, I used this slightly-modified version of your code.
import sys
def CompCount(image):
l = len(image[0])
b = len(image)
prev = curr = [sys.maxint]*(l+1)
nextComp = 0
tree = dict()
for i in xrange(b):
curr = [sys.maxint]*(l+1)
for j in xrange(l):
curr[j] = sys.maxint
if image[i][j] == '0':
p = [prev[j+n] for m,n in [(1,1),(-1,0),(-1,-1)] if 0<=i+m<b and 0<=j+n<l and image[i+m][j+n] == '0']
curr[j] = min([curr[j]] + p + [curr[j-1]])
if curr[j] == sys.maxint:
nextComp += 1
curr[j] = nextComp
tree[curr[j]] = 0
else:
if curr[j] < prev[j+1]: tree[prev[j+1]] = curr[j]
if curr[j] < prev[j]: tree[prev[j]] = curr[j]
if curr[j] < prev[j-1]: tree[prev[j-1]] = curr[j]
if curr[j] < curr[j-1]: tree[curr[j-1]] = curr[j]
prev = curr
return len([x for x in tree if tree[x]==0])
print CompCount(['11110', '01010', '10010', '11101'])
Let me try to explain your algorithm in words (in terms of a graph rather than a grid).
Set 'roots' be an empty set.
Iterate over the nodes in the graph.
For a node, n, look at all its neighbours already processed. Call this set A.
If A is empty, pick a new value k, set v[node] to be k, and add k to roots.
Otherwise, let k be the min of v[node] for node in A. Remove v[x] from roots for each x in A with v[x] != k.
The number of components is the number of elements of roots.
(Your tree is the same as my roots: note that you never use the value of tree[] elements, only whether they are 0 or not... this is just implementing a set)
It's like union-find, except that it assumes that when you merge two components, the one with the higher v[] value has never been previously merged with another component. In the counterexample this is exploited because the two 0s in the center column have been merged with the 0s to their left.

Answer 3

My variant:
Split your entire graph into edges. Add each edge to a set.
On next iteration, draw edges between the 2 outer nodes of the edge you made in step 2. This means adding new nodes (with their corresponding sets) to the set the original edge was from. (basically set merging)
Repeat 2 until the 2 nodes you're looking for are in the same set. You will also need to do a check after step 1 (just in case the 2 nodes are adjacent).
At first your nodes will be each in its set,
o-o o-o o1-o3 o2 o3-o4
\ / |
o-o-o-o o2 o1-o3-o4
As the algorithm progresses and merges the sets, it relatively halves the input.
In the example I am checking for components in some graph. After merging all edges to their maximum possible set, I am left with 3 sets giving 3 disconnected components.
(The number of components is the number of sets you are able to get when the algorithm finishes.)
A possible graph (for the tree above):
o-o-o o4 o2
| |
o o3
|
o1