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=== Optimal algorithm === [[Seth Pettie]] and [[Vijaya Ramachandran]] have found a {{not a typo|provably}} optimal deterministic comparison-based minimum spanning tree algorithm.<ref name=PettieRamachandran2002>{{citation | last1 = Pettie | first1 = Seth | last2 = Ramachandran | first2 = Vijaya | doi = 10.1145/505241.505243 | mr = 2148431 | issue = 1 | journal = [[Journal of the Association for Computing Machinery]] | pages = 16β34 | title = An optimal minimum spanning tree algorithm | url = https://web.eecs.umich.edu/~pettie/papers/jacm-optmsf.pdf | volume = 49 | year = 2002| s2cid = 5362916 }}.</ref> The following is a simplified description of the algorithm. # Let {{math|1=''r'' = log log log ''n''}}, where {{mvar|n}} is the number of vertices. Find all optimal decision trees on {{mvar|r}} vertices. This can be done in time {{math|''O''(''n'')}} (see [[#Decision trees|Decision trees]] above). # Partition the graph to components with at most {{mvar|r}} vertices in each component. This partition uses a [[soft heap]], which "corrupts" a small number of the edges of the graph. # Use the optimal decision trees to find an MST for the uncorrupted subgraph within each component. # Contract each connected component spanned by the MSTs to a single vertex, and apply any algorithm which works on [[#Dense graphs|dense graphs]] in time {{math|''O''(''m'')}} to the contraction of the uncorrupted subgraph # Add back the corrupted edges to the resulting forest to form a subgraph guaranteed to contain the minimum spanning tree, and smaller by a constant factor than the starting graph. Apply the optimal algorithm recursively to this graph. The runtime of all steps in the algorithm is {{math|''O''(''m'')}}, ''except for the step of using the decision trees''. The runtime of this step is unknown, but it has been proved that it is optimal - no algorithm can do better than the optimal decision tree. Thus, this algorithm has the peculiar property that it is ''{{not a typo|provably}} optimal'' although its runtime complexity is ''unknown''.
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