Data Structures for Disjoint Set Union-Find Data Structure Disjoint - PowerPoint PPT Presentation

Data Structures for Disjoint Set

Union-Find Data Structure

Disjoint Set Data Structure Disjoint Set Data Structure � � ◮ Storing a family of sets S = S 1 , S 2 , . . . , S k with i � j → S i ∩ S j = ∅ . ◮ Each set S i is identified by a representative s i ∈ S i . ◮ Three operations: Make-Set , Union , Find-Set Make-Set ( x ) ◮ Creates a new set { x } . (Clearly, x is representative of the set.) ◮ x cannot be in any other set already. Union ( x , y ) ◮ Merges the sets containing x and y into one set. Find-Set ( x ) ◮ Finds the representative of the set containing x . 3 / 17

Implementation Idea ◮ Represent each set S i as rooted tree (i. e., S is a forest). ◮ Root of tree is representative � � Example: S = { b , c , e , h } , { d , f , д } c f h e d g b 4 / 17

Implementation — Find-Set Find-Set ◮ Follow pointers to root. Example: Find ( b ) = c c f h e d g b 5 / 17

Implementation — Find-Set 1 Procedure Find-Set ( x ) While par ( x ) � x 2 Let x : = par ( x ) . 3 Return x 4 6 / 17

Implementation — Union Union ( x , y ) ◮ Find the representatives r x and r y of x and y (i. e., find roots of trees). ◮ Make r x parent of r y Example: Union( b , д ) c f h e d g b 7 / 17

Implementation — Union 1 Procedure Union ( x , y ) Set par ( Find-Set ( x )) : = Find-Set ( y ) 2 8 / 17

Implementation Questions ◮ What is the worst-case runtime for these operations? ◮ Can we improve the runtime? 9 / 17

Implementation Questions ◮ What is the worst-case runtime for these operations? ◮ Can we improve the runtime? Example ◮ Assume that we perform Union ( 1 , 2 ) , Union ( 1 , 3 ) , Union ( 1 , 4 ) , . . . , Union ( 1 , n ) . ◮ Then, the runtime is in O ( n 2 ) . 9 / 17

Improving Find-Set Observation ◮ If we use Find-Set multiple times on the same element, we have to search for the root each time again. Idea ◮ Update the parent pointer when calling Find-Set such that it points on the root. 1 Procedure Find-Set( x ) If par ( x ) � x Then 2 Set par ( x ) : = Find-Set ( par ( x )) 3 Return par ( x ) 4 10 / 17

Improving Union Idea: Union by Rank ◮ Keep track of height of a tree. Number is denoted as rank of a vertex. ◮ Make root of smaller tree child of root of larger tree. 1 Procedure Make-Set( x ) par ( x ) : = x 2 rank ( x ) : = 0 3 Observation ◮ We only need to keep track of the rank of the root. 11 / 17

Improving Union 1 Procedure Union ( x , y ) Let x : = Find-Set ( x ) . 2 Let y : = Find-Set ( y ) . 3 If rank ( x ) > rank ( y ) Then 4 Set par ( y ) : = x . 5 6 Else Set par ( x ) : = y . 7 If rank ( x ) = rank ( y ) Then 8 Set rank ( y ) : = rank ( y ) + 1 . 9 12 / 17

Runtime Assume our sets contain n elements in total. Runtime ◮ Worst case for single operation: O ( log n ) (Why?) ◮ Worst case for m operations: O ( m · α ( n )) Thus, O ( α ( n )) amortised runtime per operation. α -Function ◮ Inverse Ackermann function ◮ α ( atoms in the universe ) ≤ 4 ◮ Grows extremely slow. However, it is strictly speaking not constant. 13 / 17

Partition Refinement

Partition Re fi nement Union-Find ◮ Start with a partition P = { S 1 , S 2 , . . . , S k } of a set S ◮ Step by step join two sets S i and S j together. � � ◮ Union( i , j ) : P : = P \ { S i , S j } ∪ { S i ∪ S j } Partition Re fi nement ◮ Start with a partition P = { S 1 , S 2 , . . . , S k } of a set S (often P = {S} ) ◮ Step by step, based on a set X ⊆ S , split subsets S i into S i \ X and S i ∩ X . ◮ Refine( X ) : P : = { S \ X , S ∩ X | S ∈ P } 15 / 17

Implementation – Data Structure Data Structure ◮ Set S is an array storing all its elements. ◮ Partition P is a (doubly-linked) list of subsets S i ◮ Subset S i is represented by two integers which describe the interval (i. e., the first and last index) of S i in the array S S 1 S 2 S 3 0 | 2 3 | 5 6 | 7 P s 1 s 2 s 3 s 4 s 5 s 6 s 7 s 8 S 0 1 2 3 4 5 6 7 16 / 17

Implementation – Re fi nement Re fi nement ◮ Flag all elements s i ∈ X . S 1 S 2 S 3 0 | 2 3 | 5 6 | 7 P s 1 s 2 s 3 s 4 s 5 s 6 s 7 s 8 S 0 1 2 3 4 5 6 7 17 / 17

Implementation – Re fi nement Re fi nement ◮ Flag all elements s i ∈ X . ◮ For each subset S i ◮ Reorder such that fl agged elements are in front. S 1 S 2 S 3 0 | 2 3 | 5 6 | 7 P s 1 s 3 s 2 s 4 s 5 s 4 s 7 s 8 S 0 1 2 3 4 5 6 7 17 / 17

Implementation – Re fi nement Re fi nement ◮ Flag all elements s i ∈ X . ◮ For each subset S i ◮ Reorder such that fl agged elements are in front. ◮ Split S i into two sets containing only fl agged or non- fl agged elements. S 1 S 2 S 3 S 4 S 5 0 | 1 2 | 2 3 | 4 5 | 5 6 | 7 P s 1 s 3 s 2 s 4 s 5 s 4 s 7 s 8 S 0 1 2 3 4 5 6 7 17 / 17