Kruskal component merging

A disjoint-set data structure stores the current partition of a set of vertices into connected components of a graph. The operation $\mathrm{find}(v)$ returns a representative of the block containing $v$. Two vertices are in the same block exactly when their representatives agree. For each candidate edge $uv$, compare $\mathrm{find}(u)$ and $\mathrm{find}(v)$. Equal representatives mean reject; different representatives mean accept and perform $\mathrm{union}(u,v)$, merging the two blocks. The representatives are bookkeeping choices, not preferred vertices of the minimum spanning tree.

Every acceptance reduces the connected component of a graph count by one and increases the accepted edge count by one, preserving $|F|=|V|-c$. Thus the ledger reaches $c=1$ exactly when $|F|=4$. Only ten representative queries and four merges are needed for the five examined edges, counting two queries per examination and using the already found representatives for each merge. Neither breadth-first search nor reconstruction of a graph path is needed for these connectivity queries.

For $n=|V|$ and $m=|E|$, comparison sorting algorithm takes $O(m\log m)$ time for $m\ge 2$, using Big O notation. One simple disjoint-set data structure implementation stores each block as a rooted tree and attaches the smaller block's root below the larger block's root. Whenever a vertex's depth increases, its block size at least doubles, so its depth is at most $\lfloor {\log }_{2}n\rfloor$. Consequently each representative query or merge takes $O(\log n)$ time. Initialization plus the scan costs $O(n+m\log n)$; including sorting algorithm gives $O(n+m\log m+m\log n)$, hence $O(m\log m)$ for a connected input with $n\ge 2$. These are operation bounds, not measured timings.^[ref]