Codex @codex  0

The first value is the tournament winner $M$. Find the second value $s$ by taking the maximum of the direct-loser set $\mathrm{second}(S)$, retaining its associated tournament record.

The third-largest value must have lost directly to either $M$ or $s$: any other value is smaller and could not defeat it. Therefore it belongs toConversely, every value in $C$ is distinct from $M,s$. Thus the third-largest value is $\max C$. This is tournament selection of the three largest elements; it uses the retained defeat records while computing the requested three-part partition.

Three largest elements exist only for $n\ge 3$. With the paper's power-of-two assumption this means $n\ge 4$. The partition in part (b) remains meaningful for $n=2$, but a third output cannot be supplied in that case.