Union-find operation bounds ($O(n+m\alpha (n))$)

Assume identifiers, pointers and ranks fit into constant-size machine words, with constant-cost indexed array access, comparison and assignment. Initializing $n\ge 1$ members costs $O(n)$ time and machine words. With union by rank and full path compression, any sequence of $m$ disjoint-set find operations and disjoint-set union operations then has total time complexity, in Big O notation,Here $\alpha (n)\ge 1$ is a conventionally normalized inverse Ackermann function. The documented path-halving implementation has the same amortized analysis bound.^[ref]

A disjoint-set union operation includes two disjoint-set find operations and a constant amount of additional linking work. Including $n$ in the bound accounts for initialization even if no queries follow.

This is a bound on an entire sequence. It is not a worst-case constant bound for each call: a disjoint-set find operation can still encounter a previously uncompressed route of logarithmic length. The inverse Ackermann function grows extraordinarily slowly but is unbounded, so it cannot be replaced by a mathematical constant in this asymptotic statement. The general distinction between a per-operation maximum and a whole-sequence guarantee is explained in amortized analysis.

Robert Tarjan's 1975 paper established an inverse Ackermann function bound for set union. Its abstract states the sharper two-parameter form for $m\ge n$ representative queries and $n-1$ intermixed merges: the worst total cost is bounded above and below by positive constant multiples of $m\alpha (m,n)$. Those variables count different operations from the convenient $n+m\alpha (n)$ formulation above; the abstract's matching lower bound is about its analyzed method, rather than a claim that every conceivable representation obeys the same lower bound.^[ref]

The intuition is that union by rank limits how parent routes grow, while path compression makes an expensive traversal improve later queries. The rank argument alone proves only a logarithmic bound. Establishing the inverse Ackermann function bound requires a finer accounting of traversals across rank ranges; it does not follow merely from drawing a flatter rooted tree. The useful result is that repeated merging and membership queries can share the work of maintaining a changing partition of a set.