Python/data_structures/disjoint_set/disjoint_set.py
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Python

"""
Disjoint set.
Reference: https://en.wikipedia.org/wiki/Disjoint-set_data_structure
"""
class Node:
def __init__(self, data: int) -> None:
self.data = data
self.rank: int
self.parent: Node
def make_set(x: Node) -> None:
"""
Make x as a set.
"""
# rank is the distance from x to its' parent
# root's rank is 0
x.rank = 0
x.parent = x
def union_set(x: Node, y: Node) -> None:
"""
Union of two sets.
set with bigger rank should be parent, so that the
disjoint set tree will be more flat.
"""
x, y = find_set(x), find_set(y)
if x == y:
return
elif x.rank > y.rank:
y.parent = x
else:
x.parent = y
if x.rank == y.rank:
y.rank += 1
def find_set(x: Node) -> Node:
"""
Return the parent of x
"""
if x != x.parent:
x.parent = find_set(x.parent)
return x.parent
def find_python_set(node: Node) -> set:
"""
Return a Python Standard Library set that contains i.
"""
sets = ({0, 1, 2}, {3, 4, 5})
for s in sets:
if node.data in s:
return s
msg = f"{node.data} is not in {sets}"
raise ValueError(msg)
def test_disjoint_set() -> None:
"""
>>> test_disjoint_set()
"""
vertex = [Node(i) for i in range(6)]
for v in vertex:
make_set(v)
union_set(vertex[0], vertex[1])
union_set(vertex[1], vertex[2])
union_set(vertex[3], vertex[4])
union_set(vertex[3], vertex[5])
for node0 in vertex:
for node1 in vertex:
if find_python_set(node0).isdisjoint(find_python_set(node1)):
assert find_set(node0) != find_set(node1)
else:
assert find_set(node0) == find_set(node1)
if __name__ == "__main__":
test_disjoint_set()