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Tournament solutions, counted — five defensible winners from three ballots

The tabulatable evidence behind Tournament solutions — the theory of the win-loss graph. A tournament solution is a rule that reads only the win-loss graph — who beat whom, margins thrown away — which is Fishburn's C1 tier. Ranked Robin is one member of that family (it is Copeland); the academic literature holds a dozen more, and they disagree.

The first two cases are the textbook's own figures turned back into ballots — Brandt, Brill & Harrenstein, "Tournament Solutions," ch. 3 of the Handbook of Computational Social Choice (2016). The chapter draws graphs; McGarvey's theorem guarantees some electorate produces each one, and these three-voter profiles are the smallest we found that do. Candidate labels stay bare A–E on purpose, matching the figures, so the book can be read beside the tabulation.

All three files are LH-only, but only two of them for a reason. The copeland_vs_clones case needs no live tally at all; its point is a disagreement between two published rules. The star_elects_a_covered_candidate case has an engine-independent STAR result, but its Ranked Robin comparison line depends on which tiebreak ladder you run.

five_answers is the exception, and its old rationale was wrong (corrected 2026-08-05). It used to say the case couldn't go to BetterVoting because BV breaks Ranked Robin ties at random — but BV's random rung is never reached here: exactly two candidates tie, they played each other, and A beat B, so BV's head-to-head rung settles it and elects A with tieBreakType: "none". The result is fully derivable from the ballots. In fact this election is already live — BV2270 8h4bvh is this exact profile under tree names (Alder = A, Birch = B, Cedar = C, Dogwood = D), minted for an unrelated display bug and only later recognized as the same graph. So don't mint another one: BV → A, LH → B, pref_voting → {A, B} and declines, all on these three ballots.

Case (source) Ballots What it shows
page · five_answers_one_election_c4_b3.yaml 3 Five different defensible winners from one election. Top cycle {A,B,C,D} · uncovered = Banks = bipartisan {A,B,D} · Copeland {A,B} · Slater = Markov {A}. Ranked Robin lands on {A,B} and breaks it by margin, electing B — where Slater and Markov both elect A. (Chapter Figure 3.3.)
star_elects_a_covered_candidate_c4_b5.yaml 5 STAR lands outside the uncovered set — the weakest structural filter there is. Chicago beats Denver and beats Austin, the only city Denver beats, so Denver is covered; the uncovered set is {Austin, Boston, Chicago} and STAR elects the fourth. Strict ballots throughout, so no result is a tie-breaking artifact. Both halves stated: no graph-only rule would elect Denver, but Denver is not Pareto-dominated and outscores two of the three. Ranked Robin stays inside, as it must
copeland_vs_clones_c5_b3.yaml 3 Copeland vs composition-consistency. {A,B,C} is a component inside a bigger cycle, so "choose the best from the best components" forces a solution to return all five; uncovered, Banks and bipartisan do. Copeland returns {D} alone and Ranked Robin elects D outright — the published failure of composition-consistency, and the arithmetic behind RR's teaming weakness. (Chapter Figures 3.1–3.2.)

Running them

The LH engine tabulates the Copeland / Ranked Robin column and writes the _tabulated mirror:

python STARVote_LH_tabulation_engine/starvote_larry_hastings.py method_comparisons/tournament_solutions/cases/five_answers_one_election_c4_b3.yaml

The other six solutions have no LH implementation. This repo tool prints them all at once — plus the tournament M(T) itself, the outdegrees, and the chapter's axiom table — via pref_voting:

uv run STARVote_LH_tabulation_engine/tools_adam/pref_voting_tabulation_engine/tournament_solutions_report.py method_comparisons/tournament_solutions/cases/five_answers_one_election_c4_b3.yaml

Its closing line is the one that matters for this library:

The tournament does NOT decide this election: Copeland ties {A, B}.
LH's Ranked Robin breaks that tie by TOTAL MARGIN — A +1, B +3 — electing B.
Margins are not in the tournament. The moment Ranked Robin reaches for
them it has stepped out of C1 and is reading C2 information.

Two caveats the tool prints for you

  • A choice set with several names has not failed. Irresoluteness is the normal state of a tournament solution — narrowing to a single winner always takes information from outside the graph, or a lot.
  • A tournament requires no pairwise ties. Real ballots tie, and then the object is a weak tournament (chapter §3.5) where these rules are generalizations with no canonical extension. Both cases here are tie-free by construction; the tool warns when a file isn't.