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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, and its Ranked Robin comparison line rests on a two-way Copeland tie (Boston and Chicago at 2–1) that the 1st Degree settles for Boston, still inside the uncovered set.

five_answers is the exception, and its rationale has now been wrong twice. It first said 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" (corrected 2026-08-05). The second correction is bigger: this page then reported LH → B as an equally defensible reading of the same tie, and it was not. Head-to-head between two finalists is Ranked Robin's 1st Degree; the LH engine simply had no such rung and jumped to the 2nd Degree, total margin over the whole field. Since 2026-08-19 it elects A as well. 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 → A, 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 Four defensible answers to "who should win," from three ballots. 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 the 1st Degree — A beat B head-to-head — electing A, agreeing with Slater and Markov. (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

What matters for this library is what happens after the tournament runs out. The LH engine prints both rungs side by side, and they point opposite ways:

Abridged for the lesson — the win–loss block of the full report
Win–loss record — Copeland score = wins + ½·ties (highest score wins; ties broken by the Ranked Robin degrees, then lot order):
    #  Candidate  W–L–T  Copeland  Margin  vs finalists  Beats
    1  A          2–1–0         2      +1            +1  B, C
    2  B          2–1–0         2      +3            -1  C, D
    3  C          1–2–0         1      -3             —  D
    4  D          1–2–0         1      -1             —  A

Winner — Ranked Robin (RCV-RR): A
   *** 2 candidates tie for the most wins (A, B) — tied on the tally, not a cycle (some of them beat others head-to-head, but no loop closes). Resolved by the 1st Degree tiebreaker: A has the greatest sum of win margins over the other finalists (+1).

The vs finalists column is the 1st Degree, and with exactly two finalists it holds one number: their own match, which A wins 2–1. That is who-beat-whom — still C1 information, so on this profile the tiebreak never leaves the tournament. The Margin column is the 2nd Degree, margins over the whole field, and those are not in the tournament at all: they run the other way (B +3 to A's +1), which is exactly the answer this engine printed until 2026-08-19, when it was applying the 2nd Degree in place of the 1st. So Ranked Robin does step out of C1 to break a tie — one rung further down than this page used to claim, and only when the finalists' own matches leave them level.

(The standalone tournament_solutions_report.py still closes with the pre-fix line about total margin electing B; its summary has not been brought forward yet.)

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.