====================================================================== SOURCE FILE: copeland_vs_clones_c5_b3.yaml TABULATED FILE: copeland_vs_clones_c5_b3_tabulated.txt ====================================================================== election_title: "Copeland picks one, composition-consistency demands all five (tournament solutions)" scenario_description: >- The companion case: where Ranked Robin's simple count and the literature's favourite structural axiom flatly disagree. Three voters, five candidates: A>B>C>E>D, D>C>A>B>E, E>D>B>C>A. Again every head-to-head is decided, so this is a genuine tournament, and again there is no Condorcet winner. The graph has a shape worth seeing: A, B and C form a rock-paper-scissors 3-cycle; D beats all three of them; E beats D; and A, B, C all beat E. So the whole thing is one big cycle at a higher level — {A,B,C} -> E -> D -> {A,B,C}. Because {A,B,C} is a COMPONENT (all three stand in the same relation to D and to E), the tournament decomposes into {A,B,C}, {D}, {E} with a 3-cycle summary. Composition-consistency — "choose the best from the best components" — then forces a solution to select ALL FIVE candidates, since nonemptiness and neutrality make a 3-cycle unsplittable. The uncovered, Banks and bipartisan sets duly return all five. Copeland does not. D has the most wins (3, over A, B and C) so the Copeland set is {D} alone, and Ranked Robin elects D outright with no tiebreak. That is not a bug in the engine — it is Copeland failing composition-consistency, a known and published limit of the rule (chapter Section 3.3.1), and the same arithmetic behind Ranked Robin's one clone-independence weakness, teaming (see 05_Ranked_Robin/01_Learn/rr_clone_independence.md). This is Figure 3.1 / 3.2 of Brandt, Brill & Harrenstein, "Tournament Solutions" (Handbook of Computational Social Choice, 2016, ch. 3) — their own three-voter profile, run through this repo's engines. Bare A-E labels match the figure on purpose. Verified two ways: LH's Ranked Robin below, and pref_voting's C1 module via tournament_solutions_report.py. LH-only: nothing here needs a live BetterVoting election, and the teaching point is the disagreement between two published rules, not a live tally. voting_method: RankedRobin num_winners: 1 ballots: |- A>B>C>E>D D>C>A>B>E E>D>B>C>A expected_winners: - D # file: copeland_vs_clones_c5_b3.yaml ====================================================================== TABULATION RESULTS ====================================================================== --- Ranked Robin (RCV-RR / Copeland) Method (single winner) --- Tabulating 3 ballots (ranked ballots). Ballots: 1 × A > B > C > E > D 1 × D > C > A > B > E 1 × E > D > B > C > A Round-Robin — every pair, head-to-head (For – Against): A beats B 2 – 1 C beats A 2 – 1 A beats E 2 – 1 D beats A 2 – 1 B beats C 2 – 1 B beats E 2 – 1 D beats B 2 – 1 C beats E 2 – 1 D beats C 2 – 1 E beats D 2 – 1 --- Pairwise (Round-Robin) Matrix --- Head-to-head / pairwise comparison — the Ranked Robin tally Legend: For - Equal Support - Against (row vs column) | A | B | C | E | D | --------------------------------------------------------------- A > | --- |2 - 0 - 1 |1 - 0 - 2 |2 - 0 - 1 |1 - 0 - 2 | B > | 1 - 0 - 2 | --- |2 - 0 - 1 |2 - 0 - 1 |1 - 0 - 2 | C > | 2 - 0 - 1 |1 - 0 - 2 | --- |2 - 0 - 1 |1 - 0 - 2 | E > | 1 - 0 - 2 |1 - 0 - 2 |1 - 0 - 2 | --- |2 - 0 - 1 | D > | 2 - 0 - 1 |2 - 0 - 1 |2 - 0 - 1 |1 - 0 - 2 | --- | 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 D 3–1–0 3 +2 A, B, C 2 A 2–2–0 2 +0 B, E 3 B 2–2–0 2 +0 C, E 4 C 2–2–0 2 +0 A, E 5 E 1–3–0 1 -2 D Winner — Ranked Robin (RCV-RR): D the most head-to-head wins (3). --- 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): D, A, B, C, E 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 (D) are only part of the set — the win–loss table's top block understates how wide the contention is. Ranked Robin (RCV-RR) winner D 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