Complexity
Where E is the number of edges in the graph and V is the number of vertices, Kruskal's algorithm can be shown to run in O(E log E) time, or equivalently, O(E log V) time, all with simple data structures. These running times are equivalent because:
- E is at most V2 and is O(log V).
- If we ignore isolated vertices, which will each be their own component of the minimum spanning forest, V ≤ E+1, so log V is O(log E).
We can achieve this bound as follows: first sort the edges by weight using a comparison sort in O(E log E) time; this allows the step "remove an edge with minimum weight from S" to operate in constant time. Next, we use a disjoint-set data structure (Union&Find) to keep track of which vertices are in which components. We need to perform O(E) operations, two 'find' operations and possibly one union for each edge. Even a simple disjoint-set data structure such as disjoint-set forests with union by rank can perform O(E) operations in O(E log V) time. Thus the total time is O(E log E) = O(E log V).
Provided that the edges are either already sorted or can be sorted in linear time (for example with counting sort or radix sort), the algorithm can use more sophisticated disjoint-set data structure to run in O(E α(V)) time, where α is the extremely slowly growing inverse of the single-valued Ackermann function.
Read more about this topic: Kruskal's Algorithm
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