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Suppose we would like to keep track of who is related to who in the sense of family relationships in the United States.
We say that person A is related to person B if either
1) A is directly related to B via parent/child or marriage 2) A and B are really the same person (reflexive) 3) B is related to A (symmetric) 4) A is related to A’ and A’ is related to B (transitive)
We can group all the people in the United States into partitions such that everyone in each partition is related to everyone else, and people from different partitions are not related. Think of each partition as a mega-family. The technical terms for each partition is an equivalence class.
We want a fast way of answering the question of whether two people are related.
We also want a fast way to adjust the relationships when
1) new people are born 2) couples get married
The notion of representative is helpful in these regards. Choose one person to be the head of the mega-family, that is, to be the representative. Then ask everyone in the mega-family to remember who the head is.
Example: start with the following partition into disjoint sets. The representative is marked with a #.
{ #Jane } , { #Larry } , { #Susan }, { #Sam } , { #Linda }
That is, no one is related. Everyone is their own representative.
Linda and Sam get married.
{ #Jane } , { #Larry } , { #Susan }, { #Sam , Linda }
Susan and Larry get married.
{ #Jane } , { Larry , #Susan }, { #Sam , Linda }
They have a son named Adam.
{ #Jane } , { Larry , #Susan, Adam }, { #Sam , Linda }
Jane and Adam get married.
{ Jane , Larry , #Susan, Adam }, { #Sam , Linda }
interface DisjointSets<N> {
void make_set(N x);
N find(N x);
N union(N x, N y);
}
Maintain one linked-list for each partition.
Each element in the list also contains a pointer to the list as a whole.
The first element in the list is the representative
make_set: create a list with one element
find: follow the element’s pointer to the whole list
then take the head element
union: append one list to the other, then update the pointer in the
elements of one of the lists.
time complexity
n: the number of make_set operations (i.e., the number of elements)
m: the number of make_set, union, and find_set operations
make_set: O(1)
find: O(1)
Suppose we have n make_set operations followed by n-1
union operations (so there’s only one partition left).
So the total number of operations is m = 2n - 1.
The make_set operations take a total of O(n) time.
The union operations take time proportional to the length of
the lists, which grows from 1 to n-1. So, n-1 unions take
O(n²) time.
Total: O(n²)
weighted union heuristic and time complexity
During a union, always append the shorter list to the longer
one. That way we’re updating fewer pointers.
Now, during the course of the m operations, each of the n elements is moved from one list to another list only \log(n) times.
Total: O(m + n \log(n))
Maintain a separate tree/forest structure in which each tree contains one partition of elements with the root as the representative.
Each node in the tree has a parent pointer.
Basic implementation
class UnionFind<N> implements DisjointSets<N>
{
Map<N,N> parent;
public UnionFind(Map<N,N> p) {
parent = p;
}
public void make_set(N x) {
parent.put(x, x);
}
public N find(N x) {
if (x == parent.get(x))
return x;
else
return find(parent.get(x));
}
public N union(N x, N y) {
N rx = find(x); N ry = find(y);
parent.put(ry, rx);
return rx;
}
}
path compression
public N find(N x) {
if (x == parent.get(x))
return x;
else {
N rep = find(parent.get(x));
parent.put(x, rep);
return rep;
}
}
union-by-rank, by itself O(log n)
We attach numbers called rank to each node. They would be
the same as height except that we let them go stale by not
updating them when we do path compression during a find.
But that’s OK; using the stale ranks to choose the representative
is good enough to stay balanced.
public void make_set(N x) {
parent.put(x, x);
rank.put(x, 0);
}
public N union(N x, N y) {
N rx = find(x);
N ry = find(y);
if (rank.get(rx) > rank.get(ry)) {
parent.put(ry, rx);
return rx;
} else {
parent.put(rx, ry);
if (rank.get(ry) == rank.get(rx))
rank.put(ry, rank.get(ry) + 1);
return ry;
}
}
With both path compression and union-by-rank, the time complexity is O(m \alpha(n)) where alpha is the inverse Ackerman function (grows really slowly).
use the basic union-find to track relatives in the above example
Example:
F = F
/ \ / \
X1 X2 G G
| |
X2 X3
Can also be written as terms:
F(X1, X2) = F(G(X2), G(X3))
The solution is
X1 = G X2 = G X3 = X3
| |
G X3
|
X3
X1=G(G(X3)), X2=G(X3), X3=X3
So we have
F(G(G(X3)), G(X3)) = F(G(G(X3)), G(X3))
Intuition behind how to solve the equations: propagate the equalities to sub-trees.
From the above equation we get the following two equations:
X1 = G X2 = G
| |
X2 X3
In general, the process needs to keep track of several sets of things that are equal, that is, we need to partition all of the trees into sets containing equal trees.
Unification algorithm:
class Node {
String label;
public Node[] kids;
...
};
class Equation {
Equation(Node l, Node r) { lhs = l; rhs = r; }
Node lhs; Node rhs;
}
static void
unify(Node t1, Node t2, DisjointSets<Node> sets)
{
init(t1, sets);
init(t2, sets);
LinkedList<Equation> equations = new LinkedList<Equation>();
equations.add(new Equation(t1,t2));
while (equations.size() != 0) {
Equation e = equations.pop();
Node u = sets.find(e.lhs);
Node v = sets.find(e.rhs);
if (u != v) {
if (u instanceof Variable
|| v instanceof Variable)
sets.union(u,v);
else if (u.label.equals(v.label)) {
sets.union(u, v);
if (u.kids.length != v.kids.length)
return null;
for (int i = 0; i != u.kids.length; ++i)
equations.add(new Equation(u.kids[i], v.kids[i]));
} else
return null;
}
}
return sets;
}
static void init(Node t, DisjointSets<Node> sets) {
sets.make_set(t);
for (int i = 0; i != t.kids.length; ++i) {
init(t.kids[i], sets);
}
}
Def. A connected component is a maximal subset of vertices C in an undirected graph such that for every u and v in C, u \Rightarrow v.
Suppose the graph keeps changing, with additional edges, and we need to continuously keep track of the connected components.
The algorithm for connected components is straightforward to expressing using disjoint sets.
static <V> void connected_components(Graph<V> G,
DisjointSets<V> sets) {
// put every vertex in a partition by itself
for (V v : G.vertices())
sets.make_set(v);
// union partitions that have an edge between them
for (V u : G.vertices())
for (V v : G.adjacent(u))
if (sets.find(u) != sets.find(v))
sets.union(u,v);
}
When another edge (u,v) is added to the graph,
simply call sets.union(u,v).
You can determine which component a vertex v is in by
calling sets.find_set(v).
Example:
a--b e--f h j
| /| | |
|/ | | |
c--d g i
initial partitions {a} {b} {c} {d} {e} {f} {g} {h} {i} {j}
edge processed
(b,d) {a} {b,d} {c} {e} {f} {g} {h} {i} {j}
(e,g) {a} {b,d} {c} {e,g} {f} {h} {i} {j}
(a,c) {a,c} {b,d} {e,g} {f} {h} {i} {j}
(h,i) {a,c} {b,d} {e,g} {f} {h,i} {j}
(a,b) {a,c,b,d} {e,g} {f} {h,i} {j}
(e,f) {a,c,b,d} {e,g,f} {h,i} {j}
(b,c) {a,c,b,d} {e,g,f} {h,i} {j}
Time complexity: (n vertices, m edges) All the work is in the union-find operations, so let’s count those.
make_setunionfindThe time complexity for (3n + m) union-find operations is O((3n+m) \alpha(n)) = O((n + m) \alpha(n)).
Compute the connected components of the following graph.
V = {a,b,c,d,e,f,g,h,i,j,k}
E = { (d,i),(f,k),(g,i),(b,g),(a,h),(i,j),(d,k),(b,j),(d,f),(g,j),(a,e) }
solution:
{a,e,h}
{c}
{b,d,g,i,j,f,k }