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## Disjoint Sets

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- A relation R is defined on a set S if for every pair of elements (a,b), a,bS, a R b is either true or false. If a R b is true, then we say that a is related to b.
- An equivalence relation is a relation R that satisfy three properties:
- (reflexive) a R a, for all a S.
- (symmetric) a R b if and only if b R a.
- (transitive) a R b and b R c implies that a R c.

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The Dynamic Equivalence Problem

- The equivalence class of an element aS is the subset of S that contains all the elements that are related to a.
- Equivalence classes form a partition of S: every member of S appears in exactly one equivalence class.
- To decide if two member are related, only need to check whether the two are in the same equivalence class.

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- Make the input a collection of N sets, each with one element.
- All relations (except reflexive) are false;
- Each set has a different element: SiSj= => it makes the sets disjoint.
- Find operation: returns the name of the set containing a given element.
- Add/Union operation (e.g., add relation a~b)
- Check if a and b are already related: if they are in the same equivalence class.
- If not, apply “union”: merge the two equivalence classes containing a and b into a new equivalence class.

5

On a set of subsets, the three operations amount to:

- Create a set of n disjoint subsets with every node in its own subset
- Test whether A and B are in the same subset
- If A and B are in the same subset, then do nothing, else unify the subsets to which A and B belong

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The most efficient way to implement the operation is:

- For every A it is possible to ask for the index of the subset to which A belongs
- This will be denoted as find(A). And we have A ~ B <--> find(A) == find(B). The operation of adding a relation between A and B will be denoted by union(A, B) (even though nothing needs to happen). Thus, for union(A, B) one performs

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- All the elements in S are numbered sequentially from 0 to N-1—the numbering can be determined by hashing —we have Si={i} for i=0 through N-1.
- Maintain, in an array, the name of the equivalence class for each element.
- “find” is just a simple O(1) lookup.
- “union(a,b)”: suppose that a is in equivalence class i and b is in equivalence class j. Scan through the array, changing each i to j. It takes O(N) for one operation and O(N2) for N number of union

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- Keep all the elements that are in the same equivalence class in a linked list.
- By tracking the size of each equivalence class, we, when “union”, change the name of the smaller equivalence class to the larger. Thus the total time spent for N “union” is O(NlogN). Using this strategy, any sequence of M finds and up to N-1unions takes at most O(M+NlogN) time.

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Class DisjSets

{

pubic:

explicit DisjSets(int numElements);

int find(int x) const;

int find(int x);

void unionSets(int root1,int root2);

private:

vector<int> s;

}

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DisjSets::DisjSets(int numElements):s(numElements)

{

for(int j=0; j<s.size(); j++)

s[j]=-1;

}

Void DisjSets::unionSets(root1, root2)

{

s[root2]=root1;

}

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- Union-by-size: make the smaller tree a subtree of the larger.
- If union-by-size, the depth of any node is never more than logN: a find operation is O(logN), and O(MlogN) for a sequence of M. The worse-case trees are binomial trees.

17

18

Path Compression for faster find

- Path compression for find(x): every node on the path from x to the root has its parent changed to the root: make the tree shallow.

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/* Union-by-height: make the shallow tree a subtree of the deeper. */

void DisjSets::unionSets(root1, root2)

{

if(s[root2]<s[root1]) //root2 is deeper

s[root1]=root2;

else { //update height if same

if(s[root1]==s[root2])

s[root1]--;

s[root2]=root1;

}

}

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The Union-Find problem starts with n elements (numbered 1 to n), each one representing a singleton set. We allow two operations to be performed:

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Find (i) returns the ID number of the set that i

currently is in. We assume the ID number of

each set is the number of one of the elements

in it.

- Union (i; j) combines the elements in the sets

with ID numbers i and j into a single set.

The ID number of the new set will be either

i or j . This is a destructive operation in the

sense that the original sets i and j are lost

when they are combined to form the new set.

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We can implement Union-Find by maintaining

an array A[1 : : : n] of integers. If i is the ID

number of a set, then A[i] will be the negative

of the number of elements in this set. Otherwise, A[i] will the number of another element in this set.

- We begin with each A[i] equal to -1The

Union operation is easily implemented as follows:

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{ if (A [i] < A [j])

{ A [i] += A [j];

A [j] = i; }

else

{ A [j] += A [i];

A [i] = j; }

return;

}

This simply makes the \leader" of the smaller

set point to the leader of the larger set, and

adjusts the size of the leader. The time complexity

is O(1)

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To implement Find (i) we just follow the pointers

to the leader:

int Find (int i)

{

while (A [i] > 0)

i = A [i];

return i;

}

This yields O(lg n) time. We can do better, though, with path compression. After we find the leader, we make all the nodes we've passed through point directly to it:

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{

int j = i, k;

if (A [i] < 0)

return i;

while (A [j] > 0)

j = A [j];

while (A [i] != j)

{

k = A [i];

A [i] = j;

i = k;

}

return j;

}

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Using Union by size (rank) and Find with path compression, we get the following result:

Starting with n singleton sets, any sequence of M Union and/or Find operations takes O(M logN) time.

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