Minimal tilted cycle — 5 voters, margins 3–1–1 (Ranked Robin)¶

Generated from tilted_cycle_c3_b5_rr.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: Ada

Official tie-break (lot) order: Ada > Ben > Cara — consulted only if every deterministic tiebreaker stays tied (how the ladder works).

Scenario¶

The Fig. 1 profile from Brandt, Dong & Peters, "Condorcet-Consistent Choice Among Three Candidates" (arXiv:2411.19857) — the SMALLEST electorate that can produce a tilted (asymmetric) Condorcet cycle:

Ada beats Ben   4–1   (margin +3)
Ben beats Cara  3–2   (margin +1)
Cara beats Ada  3–2   (margin +1)

Every voter here is perfectly transitive; the electorate is not. Copeland (Ranked Robin) ties all three at 1–1 — the "three-way dead rung" shape — and falls through to LH's margins tiebreak, which is ALIVE here (+2 / 0 / −2), so Ada wins deterministically without a lot.

The tilt is what makes this profile useful: on the SYMMETRIC 6-voter cycle (reinforcement_paradox/cases/reinf_north_c3_b6_rr.yaml) every Condorcet method returns all three candidates and even the margins rung is dead. Tilt it by one voter and Copeland STILL ties all three, while the maximin family (Minimax / Ranked Pairs / Schulze / Kemeny — one and the same rule at three candidates) drops Ben and returns {Ada, Cara}. That is the minimal proof that Ranked Robin's Copeland count is NOT in the maximin family.

Why 5 voters and why 3–1–1: see the minimality proof in the folder README — the three cyclic margins always sum to at most n and share n's parity, so n = 3 forces the symmetric (1,1,1) cycle, n = 4 admits NO cycle at all, and n = 5 leaves (3,1,1) as the only tilted shape in existence.

LH-only by design: the Copeland three-way tie is exactly the case where BetterVoting breaks ties at RANDOM, so a BV result here could not be frozen.

Companion: tilted_cycle_c3_b5_irv.yaml (same ballots, RCV-IRV → Cara).

Ballots¶

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

2:Ada>Ben>Cara
1:Ben>Cara>Ada
2:Cara>Ada>Ben

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 5 ballots (ranked ballots).

Ballots:
     2 × Ada > Ben > Cara
     1 × Ben > Cara > Ada
     2 × Cara > Ada > Ben

Round-Robin — every pair, head-to-head (For – Against):
   Ada   beats Ben    4 – 1
   Cara  beats Ada    3 – 2
   Ben   beats Cara   3 – 2

--- Pairwise (Round-Robin) Matrix ---
Head-to-head / pairwise comparison — the Ranked Robin tally
Legend: For - Equal Support - Against   (row vs column)
         |    Ada    |   Ben    |  Cara    |
--------------------------------------------
   Ada > |    ---    |4 - 0 - 1 |2 - 0 - 3 |
   Ben > | 1 - 0 - 4 |   ---    |3 - 0 - 2 |
  Cara > | 3 - 0 - 2 |2 - 0 - 3 |   ---    |

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  Ada        1–1–0         1      +2  Ben
    2  Cara       1–1–0         1      +0  Ada
    3  Ben        1–1–0         1      -2  Cara

Winner — Ranked Robin (RCV-RR): Ada
   *** 3 candidates tie for the most wins (Ada, Ben, Cara) — 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 3): Ada, Ben, Cara
   Outside (0):        —
   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 Ada 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/minimal_tilted_cycle/cases/tilted_cycle_c3_b5_rr.yaml

Minimal tilted cycle — 5 voters, margins 3–1–1 (Ranked Robin)¶

Scenario¶

Ballots¶

What the engine says¶

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

See also¶