Implementing Disjoint Set System In Python
Solution 1:
Have you looked at any other existingimplementations?
Solution 2:
I don't have the latest edition of the book, but this doesn't look quite like a disjoint-set forest.
I think your mistake is to think that the forest has to be stored in a collection and that you have to traverse this collection to do the operations on the nodes. Remove set
from Merge()
and Find()
and implement Find()
as
def Find(n):
if n != n.parent:
n.parent = Find(n.parent)
return n.parent
just like in the book. Then add a MakeSet()
that returns a single correctly initialized node, and maybe a SameSet()
function too:
defSameSet(n1, n2):
return Find(n1) == Find(n2)
You now have a working disjoint set implementation.
Solution 3:
Presuming that each node is initialised to be its own parent:
defFind(node):
while node isnot node.parent:
node = node.parent
return node
Solution 4:
Using this implementation as a starting point, I've created a new python class to handle disjoint sets, which also supports persistence using a MongoDb.
With my class you should be able, for example, to:
- Create an instance of
UnionFind()
, which uses python build-in dictionaries by default, and decide later toconsolidate()
your results in a MongoDb collection - Create an instance of UnionFind from a MongoDb collection and directly use it.
You might want to check the code on github.
Cheers, Simone
Solution 5:
The Wikipedia page provides pseudocode for the basic operations on disjoint-set data structure. Here's a direct port to Python (employs path compression and union by rank):
classNode:
"""Represents an element of a set."""def__init__(self, id):
self.id = id
self.parent = self
self.rank = 0
self.size = 1def__repr__(self):
return'Node({!r})'.format(self.id)
defFind(x):
"""Returns the representative object of the set containing x."""if x.parent isnot x:
x.parent = Find(x.parent)
return x.parent
defUnion(x, y):
"""Combines the sets x and y belong to."""
xroot = Find(x)
yroot = Find(y)
# x and y are already in the same setif xroot is yroot:
return# x and y are not in same set, so we merge themif xroot.rank < yroot.rank:
xroot, yroot = yroot, xroot # swap xroot and yroot# merge yroot into xroot
yroot.parent = xroot
xroot.size += yroot.size
if xroot.rank == yroot.rank:
xroot.rank = xroot.rank + 1
Demo:
>>>a, b, c = map(Node, 'abc')>>>Find(a), Find(b), Find(c)
(Node('a'), Node('b'), Node('c'))
>>>Find(a).size
1
>>>Union(a, b)>>>Union(b, c)>>>Find(a), Find(b), Find(c)
(Node('a'), Node('a'), Node('a'))
>>>Find(a).size
3
Post a Comment for "Implementing Disjoint Set System In Python"