# Talk:Maximal elements algorithms

As far as I know, Kosaraju's algorithm finds the "strongly connected components" of a directed graph. But the Schwartz set is not identical to the "strongly connected components", but to the "communicating classes" of a directed graph.

A strongly connected component SCC is a set of knots with the following property:

- If knot A is in SCC and knot B is in SCC \ {A}, then there is a directed path from knot A to knot B, that consists only of knots in SCC, and a directed path from knot B to knot A, that consists only of knots in SCC.

On the other side, a communicating class CC is a set of knots with the following properties:

- If knot A is in CC and knot B is in CC \ {A}, then there is a directed path from knot A to knot B and a directed path from knot B to knot A.
- If knot A is in CC and there is a directed path from knot B to knot A, then also knot B is in CC.

Markus Schulze 04:44, 27 October 2006 (PDT)

Isn't the Beats and Beats-or-ties relation for Smith and Schwartz the wrong way around? The maximal element for Beat is the set whose members beat everybody outside of it -- that's Smith. And the maximal element of Beat-or-tie is the set whose members beat or tie everybody outside of it -- that's Schwartz, if I'm not mistaken. Because Beat-or-tie is more lenient than Beat, the set can be smaller (can exclude more candidates)... and the Schwartz set is a subset of the Smith set. Kristomun (talk) 23:27, 25 January 2020 (UTC)

- I don't know for sure, but let me point out that Smith is associated with the beat-or-tie path, and Schwartz with the beatpath. Taking the Wikipedia example of 3 candidates (https://en.wikipedia.org/wiki/Schwartz_set#Smith_set_comparison), with A>B, B>C, A=C, we can see that the only candidate with a beatpath to all others is A (A>B>C), since B doesn't beat anyone who beats A, and C beats nobody. But a beat-or-tie path can be constructed from any of the 3 candidates to the others, since they all beat or tie someone who beats or ties someone else. This lines up with A being the only member of the Schwartz set and all 3 being in the Smith set. Since this page mentions "The Schwartz set is associated with the beatpath order" and "The Smith set is associated with the beat-or-tie order", I'm guessing this explains it. BetterVotingAdvocacy (talk) 20:00, 26 February 2020 (UTC)