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

Generated from cycle_copeland_ties_c4_b21.yaml — do not edit by hand. Regenerate: python STARVote_LH_tabulation_engine/tools_adam/scripts/build_yaml_pages.py.

Method: Ranked Robin (RCV-RR / Copeland) · 1 seat · Expected winner: Alder

Official tie-break (lot) order: Alder > Birch > Cedar > Dogwood — consulted only if every deterministic tiebreaker stays tied (how the ladder works).

Scenario¶

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

Ballots¶

Each row is one voter's ranking, most-preferred first (N: prefix = N identical ballots).

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

What the engine says¶

The count, step by step — the rounds and how the winner is reached:

--- 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.)

Full audit — preference matrix, Condorcet, and score distribution¶

--- 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

Everything in one file: the _tabulated mirror (regenerated on every run; every analysis forced on).

Run it yourself:

python STARVote_LH_tabulation_engine/starvote_larry_hastings.py method_comparisons/cycle_resolution/cases/cycle_copeland_ties_c4_b21.yaml

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

Scenario¶

Ballots¶

What the engine says¶

Full audit — preference matrix, Condorcet, and score distribution¶

See also¶