Disjointset/Unionfind Forest  

Type  multiway tree  
Invented  1964  
Invented by  Bernard A. Galler and Michael J. Fischer  

In computer science, a disjointset data structure, also called a union–find data structure or merge–find set, is a data structure that stores a collection of disjoint (nonoverlapping) sets. Equivalently, it stores a partition of a set into disjoint subsets. It provides operations for adding new sets, merging sets (replacing them by their union), and finding a representative member of a set. The last operation makes it possible to find out efficiently if any two elements are in the same or different sets.
While there are several ways of implementing disjointset data structures, in practice they are often identified with a particular implementation called a disjointset forest. This is a specialized type of forest which performs unions and finds in nearconstant amortized time. To perform a sequence of m addition, union, or find operations on a disjointset forest with n nodes requires total time O(mα(n)), where α(n) is the extremely slowgrowing inverse Ackermann function. Disjointset forests do not guarantee this performance on a peroperation basis. Individual union and find operations can take longer than a constant times α(n) time, but each operation causes the disjointset forest to adjust itself so that successive operations are faster. Disjointset forests are both asymptotically optimal and practically efficient.
Disjointset data structures play a key role in Kruskal's algorithm for finding the minimum spanning tree of a graph. The importance of minimum spanning trees means that disjointset data structures underlie a wide variety of algorithms. In addition, disjointset data structures also have applications to symbolic computation, as well as in compilers, especially for register allocation problems.
Disjointset forests were first described by Bernard A. Galler and Michael J. Fischer in 1964.^{[2]} In 1973, their time complexity was bounded to , the iterated logarithm of , by Hopcroft and Ullman.^{[3]} In 1975, Robert Tarjan was the first to prove the (inverse Ackermann function) upper bound on the algorithm's time complexity,.^{[4]} He also proved it to be tight. In 1979, he showed that this was the lower bound for a certain class of algorithms, that include the GallerFischer structure.^{[5]} In 1989, Fredman and Saks showed that (amortized) words of bits must be accessed by any disjointset data structure per operation,^{[6]} thereby proving the optimality of the data structure in this model.
In 1991, Galil and Italiano published a survey of data structures for disjointsets.^{[7]}
In 1994, Richard J. Anderson and Heather Woll described a parallelized version of Union–Find that never needs to block.^{[8]}
In 2007, Sylvain Conchon and JeanChristophe Filliâtre developed a semipersistent version of the disjointset forest data structure and formalized its correctness using the proof assistant Coq.^{[9]} "Semipersistent" means that previous versions of the structure are efficiently retained, but accessing previous versions of the data structure invalidates later ones. Their fastest implementation achieves performance almost as efficient as the nonpersistent algorithm. They do not perform a complexity analysis.
Variants of disjointset data structures with better performance on a restricted class of problems have also been considered. Gabow and Tarjan showed that if the possible unions are restricted in certain ways, then a truly linear time algorithm is possible.^{[10]}
Each node in a disjointset forest consists of a pointer and some auxiliary information, either a size or a rank (but not both). The pointers are used to make parent pointer trees, where each node that is not the root of a tree points to its parent. To distinguish root nodes from others, their parent pointers have invalid values, such as a circular reference to the node or a sentinel value. Each tree represents a set stored in the forest, with the members of the set being the nodes in the tree. Root nodes provide set representatives: Two nodes are in the same set if and only if the roots of the trees containing the nodes are equal.
Nodes in the forest can be stored in any way convenient to the application, but a common technique is to store them in an array. In this case, parents can be indicated by their array index. Every array entry requires Θ(log n) bits of storage for the parent pointer. A comparable or lesser amount of storage is required for the rest of the entry, so the number of bits required to store the forest is Θ(n log n). If an implementation uses fixed size nodes (thereby limiting the maximum size of the forest that can be stored), then the necessary storage is linear in n.
Disjointset data structures support three operations: Making a new set containing a new element; Finding the representative of the set containing a given element; and Merging two sets.
The MakeSet
operation adds a new element into a new set containing only the new element, and the new set is added to the data structure. If the data structure is instead viewed as a partition of a set, then the MakeSet
operation enlarges the set by adding the new element, and it extends the existing partition by putting the new element into a new subset containing only the new element.
In a disjointset forest, MakeSet
initializes the node's parent pointer and the node's size or rank. If a root is represented by a node that points to itself, then adding an element can be described using the following pseudocode:
function MakeSet(x) is if x is not already in the forest then x.parent := x x.size := 1 // if nodes store size x.rank := 0 // if nodes store rank end if end function
This operation has constant time complexity. In particular, initializing a disjointset forest with n nodes requires O(n) time.
Lack of a parent assigned to the node implies that the node is not present in the forest.
In practice, MakeSet
must be preceded by an operation that allocates memory to hold x. As long as memory allocation is an amortized constanttime operation, as it is for a good dynamic array implementation, it does not change the asymptotic performance of the randomset forest.
The Find
operation follows the chain of parent pointers from a specified query node x until it reaches a root element. This root element represents the set to which x belongs and may be x itself. Find
returns the root element it reaches.
Performing a Find
operation presents an important opportunity for improving the forest. The time in a Find
operation is spent chasing parent pointers, so a flatter tree leads to faster Find
operations. When a Find
is executed, there is no faster way to reach the root than by following each parent pointer in succession. However, the parent pointers visited during this search can be updated to point closer to the root. Because every element visited on the way to a root is part of the same set, this does not change the sets stored in the forest. But it makes future Find
operations faster, not only for the nodes between the query node and the root, but also for their descendants. This updating is an important part of the disjointset forest's amortized performance guarantee.
There are several algorithms for Find
that achieve the asymptotically optimal time complexity. One family of algorithms, known as path compression, makes every node between the query node and the root point to the root. Path compression can be implemented using a simple recursion as follows:
function Find(x) is if x.parent ≠ x then x.parent := Find(x.parent) return x.parent else return x end if end function
This implementation makes two passes, one up the tree and one back down. It requires enough scratch memory to store the path from the query node to the root (in the above pseudocode, the path is implicitly represented using the call stack). This can be decreased to a constant amount of memory by performing both passes in the same direction. The constant memory implementation walks from the query node to the root twice, once to find the root and once to update pointers:
function Find(x) is root := x while root.parent ≠ root do root := root.parent end while while x.parent ≠ root do parent := x.parent x.parent := root x := parent end while return root end function
Tarjan and Van Leeuwen also developed onepass Find
algorithms that retain the same worstcase complexity but are more efficient in practice.^{[4]} These are called path splitting and path halving. Both of these update the parent pointers of nodes on the path between the query node and the root. Path splitting replaces every parent pointer on that path by a pointer to the node's grandparent:
function Find(x) is while x.parent ≠ x do (x, x.parent) := (x.parent, x.parent.parent) end while return x end function
Path halving works similarly but replaces only every other parent pointer:
function Find(x) is while x.parent ≠ x do x.parent := x.parent.parent x := x.parent end while return x end function
The operation Union(x, y)
replaces the set containing x and the set containing y with their union. Union
first uses Find
to determine the roots of the trees containing x and y. If the roots are the same, there is nothing more to do. Otherwise, the two trees must be merged. This is done by either setting the parent pointer of x's root to y's, or setting the parent pointer of y's root to x's.
The choice of which node becomes the parent has consequences for the complexity of future operations on the tree. If it is done carelessly, trees can become excessively tall. For example, suppose that Union
always made the tree containing x a subtree of the tree containing y. Begin with a forest that has just been initialized with elements and execute Union(1, 2)
, Union(2, 3)
, ..., Union(n  1, n)
. The resulting forest contains a single tree whose root is n, and the path from 1 to n passes through every node in the tree. For this forest, the time to run Find(1)
is O(n).
In an efficient implementation, tree height is controlled using union by size or union by rank. Both of these require a node to store information besides just its parent pointer. This information is used to decide which root becomes the new parent. Both strategies ensure that trees do not become too deep.
In the case of union by size, a node stores its size, which is simply its number of descendants (including the node itself). When the trees with roots x and y are merged, the node with more descendants becomes the parent. If the two nodes have the same number of descendants, then either one can become the parent. In both cases, the size of the new parent node is set to its new total number of descendants.
function Union(x, y) is // Replace nodes by roots x := Find(x) y := Find(y) if x = y then return // x and y are already in the same set end if // If necessary, swap variables to ensure that // x has at least as many descendants as y if x.size < y.size then (x, y) := (y, x) end if // Make x the new root y.parent := x // Update the size of x x.size := x.size + y.size end function
The number of bits necessary to store the size is clearly the number of bits necessary to store n. This adds a constant factor to the forest's required storage.
For union by rank, a node stores its rank, which is an upper bound for its height. When a node is initialized, its rank is set to zero. To merge trees with roots x and y, first compare their ranks. If the ranks are different, then the larger rank tree becomes the parent, and the ranks of x and y do not change. If the ranks are the same, then either one can become the parent, but the new parent's rank is incremented by one. While the rank of a node is clearly related to its height, storing ranks is more efficient than storing heights. The height of a node can change during a Find
operation, so storing ranks avoids the extra effort of keeping the height correct. In pseudocode, union by rank is:
function Union(x, y) is // Replace nodes by roots x := Find(x) y := Find(y) if x = y then return // x and y are already in the same set end if // If necessary, rename variables to ensure that // x has rank at least as large as that of y if x.rank < y.rank then (x, y) := (y, x) end if // Make x the new root y.parent := x // If necessary, increment the rank of x if x.rank = y.rank then x.rank := x.rank + 1 end if end function
It can be shown that every node has rank or less.^{[11]} Consequently each rank can be stored in O(log log n) bits and all the ranks can be stored in O(n log log n) bits. This makes the ranks an asymptotically negligible portion of the forest's size.
It is clear from the above implementations that the size and rank of a node do not matter unless a node is the root of a tree. Once a node becomes a child, its size and rank are never accessed again.
A disjointset forest implementation in which Find
does not update parent pointers, and in which Union
does not attempt to control tree heights, can have trees with height O(n). In such a situation, the Find
and Union
operations require O(n) time.
If an implementation uses path compression alone, then a sequence of n MakeSet
operations, followed by up to n − 1 Union
operations and f Find
operations, has a worstcase running time of .^{[11]}
Using union by rank, but without updating parent pointers during Find
, gives a running time of for m operations of any type, up to n of which are MakeSet
operations.^{[11]}
The combination of path compression, splitting, or halving, with union by size or by rank, reduces the running time for m operations of any type, up to n of which are MakeSet
operations, to .^{[4]}^{[5]} This makes the amortized running time of each operation . This is asymptotically optimal, meaning that every disjoint set data structure must use amortized time per operation.^{[6]} Here, the function is the inverse Ackermann function. The inverse Ackermann function grows extraordinarily slowly, so this factor is 4 or less for any n that can actually be written in the physical universe. This makes disjointset operations practically amortized constant time.
The precise analysis of the performance of a disjointset forest is somewhat intricate. However, there is a much simpler analysis that proves that the amortized time for any m Find
or Union
operations on a disjointset forest containing n objects is O(m log^{*} n), where log^{*} denotes the iterated logarithm.^{[12]}^{[13]}^{[14]}^{[15]}
Lemma 1: As the find function follows the path along to the root, the rank of node it encounters is increasing.
claim that as Find and Union operations are applied to the data set, this fact remains true over time. Initially when each node is the root of its own tree, it's trivially true. The only case when the rank of a node might be changed is when the Union by Rank operation is applied. In this case, a tree with smaller rank will be attached to a tree with greater rank, rather than vice versa. And during the find operation, all nodes visited along the path will be attached to the root, which has larger rank than its children, so this operation won't change this fact either.
Lemma 2: A node u which is root of a subtree with rank r has at least nodes.
Initially when each node is the root of its own tree, it's trivially true. Assume that a node u with rank r has at least 2^{r} nodes. Then when two trees with rank r are merged using the operation Union by Rank, a tree with rank r + 1 results, the root of which has at least nodes.
Lemma 3: The maximum number of nodes of rank r is at most
From lemma 2, we know that a node u which is root of a subtree with rank r has at least nodes. We will get the maximum number of nodes of rank r when each node with rank r is the root of a tree that has exactly nodes. In this case, the number of nodes of rank r is
For convenience, we define "bucket" here: a bucket is a set that contains vertices with particular ranks.
We create some buckets and put vertices into the buckets according to their ranks inductively. That is, vertices with rank 0 go into the zeroth bucket, vertices with rank 1 go into the first bucket, vertices with ranks 2 and 3 go into the second bucket. If the Bth bucket contains vertices with ranks from interval then the (B+1)st bucket will contain vertices with ranks from interval
We can make two observations about the buckets.
Let F represent the list of "find" operations performed, and let
Then the total cost of m finds is
Since each find operation makes exactly one traversal that leads to a root, we have T_{1} = O(m).
Also, from the bound above on the number of buckets, we have T_{2} = O(mlog^{*}n).
For T_{3}, suppose we are traversing an edge from u to v, where u and v have rank in the bucket [B, 2^{B} − 1] and v is not the root (at the time of this traversing, otherwise the traversal would be accounted for in T_{1}). Fix u and consider the sequence that take the role of v in different find operations. Because of path compression and not accounting for the edge to a root, this sequence contains only different nodes and because of Lemma 1 we know that the ranks of the nodes in this sequence are strictly increasing. By both of the nodes being in the bucket we can conclude that the length k of the sequence (the number of times node u is attached to a different root in the same bucket) is at most the number of ranks in the buckets B, that is, at most
Therefore,
From Observations 1 and 2, we can conclude that
Therefore,
The worstcase time of the Find
operation in trees with Union by rank or Union by weight is (i.e., it is and this bound is tight).
In 1985, N. Blum gave an implementation of the operations that does not use path compression, but compresses trees during . His implementation runs in time per operation,^{[16]} and thus in comparison with Galler and Fischer's structure it has a better worstcase time per operation, but inferior amortized time. In 1999, Alstrup et al. gave a structure that has optimal worstcase
time together with inverseAckermann amortized time.^{[17]}
The regular implementation as disjointset forests does not react favorably to the deletion of elements,
in the sense that the time for Find
will not improve as a result of the decrease in the number of elements. However, there exist modern implementations that allow for constanttime deletion and where the timebound for Find
depends on the current number of elements^{[18]}^{[19]}
Disjointset data structures model the partitioning of a set, for example to keep track of the connected components of an undirected graph. This model can then be used to determine whether two vertices belong to the same component, or whether adding an edge between them would result in a cycle. The Union–Find algorithm is used in highperformance implementations of unification.^{[20]}
This data structure is used by the Boost Graph Library to implement its Incremental Connected Components functionality. It is also a key component in implementing Kruskal's algorithm to find the minimum spanning tree of a graph.
The HoshenKopelman algorithm uses a UnionFind in the algorithm.