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SOURCE FILE:     cycle_copeland_ties_c4_b21.yaml
TABULATED FILE:  cycle_copeland_ties_c4_b21_tabulated.txt
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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

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TABULATION RESULTS
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--- Ranked Robin (RCV-RR / Copeland) Method (single winner) ---
 Tabulating 21 ballots (ranked ballots).

Ballots:
    10 × Alder > Birch > Cedar > Dogwood
     6 × Birch > Cedar > Alder > Dogwood
     5 × Cedar > Alder > Birch > Dogwood

Round-Robin — every pair, head-to-head (For – Against):
   Alder    beats Birch     15 –  6
   Cedar    beats Alder     11 – 10
   Alder    beats Dogwood   21 –  0
   Birch    beats Cedar     16 –  5
   Birch    beats Dogwood   21 –  0
   Cedar    beats Dogwood   21 –  0

--- Pairwise (Round-Robin) Matrix ---
Head-to-head / pairwise comparison — the Ranked Robin tally
Legend: For - Equal Support - Against   (row vs column)
            |    Alder     |   Birch     |   Cedar     |  Dogwood    |
----------------------------------------------------------------------
    Alder > |     ---      |15 -  0 -  6 |10 -  0 - 11 |21 -  0 -  0 |
    Birch > |  6 -  0 - 15 |    ---      |16 -  0 -  5 |21 -  0 -  0 |
    Cedar > | 11 -  0 - 10 | 5 -  0 - 16 |    ---      |21 -  0 -  0 |
  Dogwood > |  0 -  0 - 21 | 0 -  0 - 21 | 0 -  0 - 21 |    ---      |

Win–loss record — Copeland score = wins + ½·ties (highest score wins; ties broken by total margin, then lot order):
    #  Candidate  W–L–T  Copeland  Margin  Beats
    1  Alder      2–1–0         2     +29  Birch, Dogwood
    2  Birch      2–1–0         2     +23  Cedar, Dogwood
    3  Cedar      2–1–0         2     +11  Alder, Dogwood
    4  Dogwood    0–3–0         0     -63  —

Winner — Ranked Robin (RCV-RR): Alder
   *** 3 candidates tie for the most wins (Alder, Birch, Cedar) — a Condorcet cycle (no candidate beats all others). Resolved by total margin, then lot order. (This is where Minimax / Ranked Pairs / Schulze differ — see 05_Ranked_Robin/01_Learn/cycle_resolution.md.)

--- Smith Set (the generalized Condorcet winner) ---
The smallest group whose every member beats every candidate outside it —
the honest answer to "who is even in contention?".
   Smith set (3 of 4): Alder, Birch, Cedar
   Outside (1):        Dogwood
   More than one member ⇒ NO Condorcet winner: the top of the tournament is a
   cycle, so the strongest "candidate" is a set, not a person. Which member of
   the set should win is exactly what Minimax / Ranked Pairs / Schulze disagree
   about — see 05_Ranked_Robin/01_Learn/cycle_resolution.md.
   Ranked Robin (RCV-RR) winner Alder is INSIDE the Smith set. ✓
      Guaranteed: Ranked Robin (Copeland) is Smith-efficient — every member of
      the set outscores every outsider, so the top of the win–loss table is
      always inside the set, however the tie among them is then broken.
   More: 07_Concepts/topics/smith_set.md