election_title: "Five defensible answers, one three-ballot election (tournament solutions)"
scenario_description: >-
  The smallest election that makes the whole tournament-solutions literature
  necessary. Three voters, four candidates, ballots that are just one ranking
  rotated: A>B>C>D, B>C>D>A, D>A>B>C. Every head-to-head is decided (no ties),
  so the pairwise results form a genuine TOURNAMENT — a complete directed graph.
  There is no Condorcet winner: D beats A, so the top of the graph cycles.
  Now the famous C1 rules, which read ONLY that graph, split five ways:
  Top cycle / Schwartz = {A, B, C, D} (everyone); Uncovered set = Banks set =
  Bipartisan set = {A, B, D} (C is COVERED — B beats C and beats everyone C
  beats, so C is strictly redundant); Copeland set = {A, B} (both win 2, C and D
  win 1); Slater set = Markov set = {A}. Five different answers to "who should
  win," each with a published defense, from three ballots.
  Ranked Robin is the Copeland set, so the LH engine lands on {A, B} and must
  break the tie — by TOTAL MARGIN, electing B (+3 vs A's +1). That step is the
  lesson: margins are not in the tournament. The moment Ranked Robin consults
  them it has left C1 and is reading C2 information, and it elects B where
  Slater and Markov both elect A.
  This is Figure 3.3 of Brandt, Brill & Harrenstein, "Tournament Solutions"
  (Handbook of Computational Social Choice, 2016, ch. 3), turned back into
  ballots — the chapter gives the graph, and McGarvey's theorem guarantees some
  profile produces it; this three-voter rotation is one. Candidate labels are
  kept as bare A/B/C/D deliberately, matching the figure, so the book can be
  read side by side with the tabulation. Verified two ways: the LH engine's
  Ranked Robin below, and pref_voting's independent C1 module via
  tournament_solutions_report.py.
  This is also the picture every write-up of the subject opens with. Wikipedia's
  "Tournament solution" leads with A = {1,2,3,4} and the pairs (1,2) (1,4) (2,4)
  (3,1) (3,2) (4,3); read 3=A, 1=B, 2=C, 4=D and that is this matrix, arrow for
  arrow. Not a coincidence: there are exactly four tournaments on four vertices
  up to relabelling and only ONE is strongly connected, so every four-candidate
  election with a cycling top and no pairwise ties draws this graph.
  LH-only as a file, but NOT because BetterVoting can't count it — the earlier
  note claiming that was wrong. Exactly two candidates tie and they played each
  other, so BV's head-to-head rung settles it deterministically (tieBreakType
  "none") and elects A, where LH's margin rung elects B. The same profile is
  live on BV under tree names as BV2270 (8h4bvh): Alder=A, Birch=B, Cedar=C,
  Dogwood=D. Three tabulators, three positions: BV -> A, LH -> B,
  pref_voting -> the leader set {A, B}, declining to choose.
voting_method: RankedRobin
num_winners: 1
ballots: |-
  A>B>C>D
  B>C>D>A
  D>A>B>C
expected_winners:
  - B

# file: five_answers_one_election_c4_b3.yaml