A candidate nobody prefers still flips the winner — Schulze's spoiler, and Split Cycle's immunity¶
Generated from split_cycle_schulze_spoiler_c5_b40.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: Cascade
Official tie-break (lot) order: Arches > Bryce > Cascade > Denali > Everglade — consulted only if every deterministic tiebreaker stays tied (how the ladder works).
Scenario¶
A runnable counterexample for Holliday & Pacuit's central claim in "Split Cycle" (arXiv:2004.02350; Public Choice 197, 2023): that Schulze/beat-path is NOT immune to spoilers, and Split Cycle is. Reproduced independently — this is not the paper's own profile — with pref_voting, on the smallest electorate a search turned up. 40 hikers rank five national parks.
Cascade beats Bryce 40-0: NOT ONE VOTER prefers Bryce to Cascade. Bryce wins under no method, in any field. And yet:
Schulze WITHOUT Bryce on the ballot -> Cascade
Schulze WITH Bryce on the ballot -> Everglade
Bryce's mere presence takes the win away from a park she is unanimously behind, and hands it to a third party. That is exactly the paper's Definition 4.1: Bryce spoils the election for Cascade — Cascade wins without Bryce, a majority (here, everyone) prefers Cascade to Bryce, and with Bryce present neither of them wins.
Split Cycle does not do this. It returns {Cascade, Everglade} with Bryce present and {Cascade} without — Cascade is never dropped, so no spoiler effect occurs. That is the paper's "stability for winners" in action, and the reason the authors argue the criterion is worth the cost (Split Cycle sometimes declines to break a tie that Schulze and Ranked Pairs break by convention).
Read the honest fine print on the lesson page: this is a five-candidate profile with no Condorcet winner and a Smith set of ALL FIVE parks — a genuinely knotted election, not an everyday one. Ranked Pairs happens to elect Cascade here too, so this profile does not exhibit its (separately proven) spoiler failure. Nothing here is evidence about STAR or Ranked Robin, neither of which is a C2 method.
Verified with pref_voting, both fields: uv run STARVote_LH_tabulation_engine/tools_adam/pref_voting_tabulation_engine/cycle_resolution_report.py \ method_comparisons/split_cycle/cases/split_cycle_schulze_spoiler_c5_b40.yaml uv run …/cycle_resolution_report.py …/split_cycle_schulze_spoiler_c5_b40.yaml --drop Bryce
LH-only (no BetterVoting election): neither BV nor the LH engine implements Schulze or Split Cycle, and the LH Copeland result here is a tie. Lesson page: 07_Concepts/topics/condorcet/split_cycle.md
Ballots¶
Each row is one voter's ranking, most-preferred first (N: prefix = N identical ballots).
11:Everglade>Denali>Cascade>Bryce>Arches
11:Arches>Denali>Cascade>Everglade>Bryce
8:Everglade>Cascade>Bryce>Arches>Denali
10:Cascade>Bryce>Arches>Denali>Everglade
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 40 ballots (ranked ballots).
Ballots:
11 × Everglade > Denali > Cascade > Bryce > Arches
11 × Arches > Denali > Cascade > Everglade > Bryce
8 × Everglade > Cascade > Bryce > Arches > Denali
10 × Cascade > Bryce > Arches > Denali > Everglade
Round-Robin — every pair, head-to-head (For – Against):
Denali beats Everglade 21 – 19
Cascade beats Everglade 21 – 19
Everglade beats Bryce 30 – 10
Arches beats Everglade 21 – 19
Denali beats Cascade 22 – 18
Denali beats Bryce 22 – 18
Arches beats Denali 29 – 11
Cascade beats Bryce 40 – 0
Cascade beats Arches 29 – 11
Bryce beats Arches 29 – 11
--- Pairwise (Round-Robin) Matrix ---
Head-to-head / pairwise comparison — the Ranked Robin tally
Legend: For - Equal Support - Against (row vs column)
| Everglade | Denali | Cascade | Bryce | Arches |
--------------------------------------------------------------------------------------
Everglade > | --- |19 - 0 - 21 |19 - 0 - 21 |30 - 0 - 10 |19 - 0 - 21 |
Denali > | 21 - 0 - 19 | --- |22 - 0 - 18 |22 - 0 - 18 |11 - 0 - 29 |
Cascade > | 21 - 0 - 19 |18 - 0 - 22 | --- |40 - 0 - 0 |29 - 0 - 11 |
Bryce > | 10 - 0 - 30 |18 - 0 - 22 | 0 - 0 - 40 | --- |29 - 0 - 11 |
Arches > | 21 - 0 - 19 |29 - 0 - 11 |11 - 0 - 29 |11 - 0 - 29 | --- |
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 Cascade 3–1–0 3 +56 Arches, Everglade, Bryce
2 Denali 3–1–0 3 -8 Cascade, Everglade, Bryce
3 Arches 2–2–0 2 -16 Denali, Everglade
4 Everglade 1–3–0 1 +14 Bryce
5 Bryce 1–3–0 1 -46 Arches
Winner — Ranked Robin (RCV-RR): Cascade
*** 2 candidates tie for the most wins (Denali, Cascade) — tied on the tally, not a cycle (some of them beat others head-to-head, but no loop closes). Resolved by total margin, then lot order.
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 (5 of 5): Denali, Cascade, Arches, Everglade, Bryce
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.
Note: the Copeland leaders (Denali, Cascade) are only part of the set — the
win–loss table's top block understates how wide the contention is.
Ranked Robin (RCV-RR) winner Cascade 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/split_cycle/cases/split_cycle_schulze_spoiler_c5_b40.yaml