election_title: "A cycle Copeland can't break — three trails tie 1-1, and the refined rules all rescue Alder"

scenario_description: |-
  The first worked profile from "Cycle Resolution — why Minimax, Ranked Pairs,
  and Schulze exist" (05_Ranked_Robin/01_Learn/cycle_resolution.md), now
  runnable. 21 hikers rank four trails; Dogwood is everyone's last choice, and
  the other three form a majority CYCLE:

      Alder beats Birch by 9 · Birch beats Cedar by 11 · Cedar beats Alder by 1

  So there is no Condorcet winner, and Ranked Robin's underlying Copeland count
  (pairwise wins minus losses) ties Alder, Birch and Cedar at +1 each — the
  simple count cannot pick. That tie-proneness is exactly why the refined
  cycle-resolution rules exist.

  Every one of them then agrees on Alder, whose only defeat (to Cedar, by 1) is
  the mildest loss in the cycle: Minimax picks Alder ("least badly beaten"),
  Ranked Pairs locks Birch>Cedar and Alder>Birch and skips the cycle-closing
  Cedar>Alder, Schulze's strongest beat-paths run Alder's way, and Split Cycle
  discards the weakest defeat in the cycle — Cedar>Alder — leaving Alder
  undefeated. Four different philosophies, one winner.

  Verified with pref_voting:
    uv run STARVote_LH_tabulation_engine/tools_adam/pref_voting_tabulation_engine/cycle_resolution_report.py \
      method_comparisons/cycle_resolution/cases/cycle_copeland_ties_c4_b21.yaml

  LH-only (no BetterVoting election): LH and BV break a Copeland tie differently
  — LH by margin then lot, BV at random — so a tie-deciding case cannot be frozen
  on BV. The companion case where the refined rules DISAGREE:
  cycle_schulze_vs_ranked_pairs_c4_b40.yaml

paradoxes: [condorcet-cycle]

voting_method: RankedRobin
num_winners: 1

lot_numbers: [Alder, Birch, Cedar, Dogwood]

ballots: |-
  10:Alder>Birch>Cedar>Dogwood
  6:Birch>Cedar>Alder>Dogwood
  5:Cedar>Alder>Birch>Dogwood

expected_winners:
  - Alder