Find how many connected groups of nodes in a given adjacency matrix

If the input matrix is guaranteed to describe transitive connectivity, it has a peculiar form that allows for an algorithm probing only a subset of the matrix elements. Here is an example implementation in Python:

def count_connected_groups(adj):
    n = len(adj)
    nodes_to_check = set([i for i in range(n)]) # [] not needed in python 3
    count = 0
    while nodes_to_check:
        count += 1
        node = nodes_to_check.pop()
        adjacent = adj[node]
        other_group_members = set()
        for i in nodes_to_check:
            if adjacent[i]:
                other_group_members.add(i)
        nodes_to_check -= other_group_members
    return count


# your example:
adj_0 = [[1, 1, 0], [1, 1, 0], [0, 0, 1]]
# same with tuples and booleans:
adj_1 = ((True, True, False), (True, True, False), (False, False, True))
# another connectivity matrix:
adj_2 = ((1, 1, 1, 0, 0),
         (1, 1, 1, 0, 0),
         (1, 1, 1, 0, 0),
         (0, 0, 0, 1, 1),
         (0, 0, 0, 1, 1))
# and yet another:
adj_3 = ((1, 0, 1, 0, 0),
         (0, 1, 0, 1, 0),
         (1, 0, 1, 0, 0),
         (0, 1, 0, 1, 0),
         (0, 0, 0, 0, 1))
for a in adj_0, adj_1, adj_2, adj_3:
    print(a)
    print(count_connected_groups(a))


# [[1, 1, 0], [1, 1, 0], [0, 0, 1]]
# 2
# ((True, True, False), (True, True, False), (False, False, True))
# 2
# ((1, 1, 1, 0, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0), (0, 0, 0, 1, 1), (0, 0, 0, 1, 1))
# 2
# ((1, 0, 1, 0, 0), (0, 1, 0, 1, 0), (1, 0, 1, 0, 0), (0, 1, 0, 1, 0), (0, 0, 0, 0, 1))
# 3

An optimized version of the same algorithm (less readable, but faster and more easily translatable into other languages) is the following:

def count_connected_groups(adj):
    n = len(adj)
    nodes_to_check = [i for i in range(n)]  # [0, 1, ..., n-1]
    count = 0
    while n:
        count += 1
        n -= 1; node = nodes_to_check[n]
        adjacent = adj[node]
        i = 0
        while i < n:
            other_node = nodes_to_check[i]
            if adjacent[other_node]:
                n -= 1; nodes_to_check[i] = nodes_to_check[n]
            else:
                i += 1
    return count