Maybe I don't understand your proof correctly or it has a flaw.

In the three comparisons you propose between the solutions for C and C' you assume there must be a comparison between min and max. This is rarely the case in the implementations regarded as optimal. Thus your proof excludes the relevant algorithms.

Disproving the existence of a certain algorithm is indeed hard. You must not make any assumptions about its function, otherwise you may exclude some algorithms and the proof becomes worthless.

First of all, there is a strong need do define what that algorithm shall be. In this specific case I'd define a solving algorithm as follows:

• It is an nondeterministic state machine.
• The initial input is the number of items.
• The state machine then outputs pairs (i, k) of indices and expects the result of m(i) <=> m(k) as the next input.
• The machine stops after S steps and produces a directed graph G = (V, E) as output with these properties:
• The vertices V are the indices 1 .. N.
• There is an edge i -> k in E if and only if m(i) < m(k) has been directly proven in the running phase of the machine.
• The graph G has one source in min = min m and one sink in max = max m.

There should be a proof that there is no algorithm that produces a valid output with S < 2 N/3 - 2 comparisons. Such a proof is impossible, as the defined algorithms include those that are cheating. An algorithm may just call for N - 1 comparisons producing a straight sequence of subsequent numbers producing the trivial graph min -> p(2) -> ... ->p(n-1) -> max.

Therefore we need to modify the setup. We do not preallocate the numbers, but arrange them in a consistent way while requested from the algorithm. E.g. if we are first asked for comparing a < b, then for c < d and afterwards for b < c, we are free to arrange these such that b > c.

Then we need to prove that an algorithm under such dynamic conditions is not capable of:

• Finding the solution with S < 3 N/2 - 2 comparisons in every attempt (strict). Or
• Finding the solution with an average number of S < 3 N/2 - 2 comparisons (relaxed).

Maybe my expectations are too high, but that is the kind of proof I had expected and I don't feel capable to provide.

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