Skip to content

The Smith Set — the Smallest Club That Beats Everyone Outside It

What "the strongest candidate" means when no single candidate beats everyone. When a cycle erases the Condorcet winner, the principled fallback isn't a person — it's a set: the smallest group whose every member beats every outsider head-to-head. Once you have that club, "any decent method should at least pick from inside it" becomes a testable criterion — and it neatly sorts the Condorcet family.

→ Level: 301 · deep dive — Curriculum 301.5 (limits & theory) · Glossary: Smith set · the tournament math: The math behind Condorcet · the other generalized Condorcet winner: the uncovered set (always a subset of the Smith set, and the Pareto line)

The problem: "beats everyone head-to-head" can come up empty

A Condorcet winner beats every rival one-on-one, and when one exists it's a natural "correct answer" for majority rule. But majority preference isn't guaranteed to be transitive: a majority can prefer A over B, B over C, and C over A — a rock-paper-scissors cycle, and suddenly "the candidate who beats everyone" doesn't exist. That's the Condorcet paradox, and every Condorcet method needs an answer to it.

The Smith set is the principled part of that answer. Before asking "who wins the cycle?", it asks a more modest question with a provably clean answer: who is even in contention?

The definition: the smallest club that beats everyone outside

Call a group of candidates a dominating set if every member beats every non-member head-to-head. The whole field is trivially dominating (there's no one outside), so dominating sets always exist — and a neat theorem says they're nested: of any two, one contains the other. That means there is always a unique smallest one:

The Smith set is the smallest non-empty group of candidates such that every member beats every candidate outside the group in a head-to-head matchup.

Three consequences fall straight out of the definition:

  • If a Condorcet winner exists, the Smith set is just {that winner} — a one-member club. The set only grows when a cycle appears, which is why it's called the generalized Condorcet winner.
  • In a cycle, the whole cyclic clump at the top is in the club. You can't admit "just the strongest one" — whoever you pick, someone else in the cycle beats them.
  • A Condorcet loser can never make the club — everyone beats them, so the club works fine without them.

One set, several names — the literature is messy: Smith set (after mathematician John H. Smith, 1973 — no relation to Warren D. Smith of range voting), top cycle, GETCHA ("Generalized Top-Choice Assumption," Schwartz's older coinage), and textbook framings like "generalized Condorcet candidates" (Börgers, Mathematics of Social Choice).

See it: a cycle plus an outsider (runnable)

Take the Ada/Ben/Cara cycle and change one thing: add a fourth candidate, Dave, whom every voter ranks last (04_smith_set_c4_b7, counted by Ranked Robin):

--- Ranked Robin (RCV-RR / Copeland) Method (single winner) ---
 Tabulating 7 ballots (ranked ballots).

Ballots:
     3 × Ada > Ben > Cara > Dave
     2 × Ben > Cara > Ada > Dave
     2 × Cara > Ada > Ben > Dave

Round-Robin — every pair, head-to-head (For – Against):
   Ada   beats Ben    5 – 2
   Cara  beats Ada    4 – 3
   Ada   beats Dave   7 – 0
   Ben   beats Cara   5 – 2
   Ben   beats Dave   7 – 0
   Cara  beats Dave   7 – 0

--- Pairwise (Round-Robin) Matrix ---
Head-to-head / pairwise comparison — the Ranked Robin tally
Legend: For - Equal Support - Against   (row vs column)
         |    Ada    |   Ben    |  Cara    |  Dave    |
-------------------------------------------------------
   Ada > |    ---    |5 - 0 - 2 |3 - 0 - 4 |7 - 0 - 0 |
   Ben > | 2 - 0 - 5 |   ---    |5 - 0 - 2 |7 - 0 - 0 |
  Cara > | 4 - 0 - 3 |2 - 0 - 5 |   ---    |7 - 0 - 0 |
  Dave > | 0 - 0 - 7 |0 - 0 - 7 |0 - 0 - 7 |   ---    |

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        2–1–0         2      +9  Ben, Dave
    2  Ben        2–1–0         2      +7  Cara, Dave
    3  Cara       2–1–0         2      +5  Ada, Dave
    4  Dave       0–3–0         0     -21  —

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.)

Now run the club-membership check, smallest group first:

  1. {Ada}? Not dominating — Cara beats Ada (4–3). A club member can't be losing to an outsider.
  2. {Ada, Cara}? Not dominating — Ben beats Cara. Same problem.
  3. {Ada, Ben, Cara}? Dominating — the only outsider is Dave, and all three beat Dave 7–0. ✓

No smaller group works, so the Smith set is {Ada, Ben, Cara} — and Dave, despite being on every ballot, is provably out of contention: to get into the club, Dave would have to stop losing to someone inside it. Notice the shortcut the win–loss table hands you: the club is exactly the top block of the Copeland standings (records 2–1, 2–1, 2–1 vs Dave's 0–3) — that's not a coincidence, it's a theorem (dominating sets are nested by Copeland score).

The full pairwise grid, and this exact election as a runnable YAML, are in 04_smith_set_c4_b7. (It's an LH-only case: BetterVoting's Ranked Robin breaks a Copeland tie randomly, so a deliberate three-way tie isn't freezable there — see the tiebreak divergence.) For a Smith set of four in a real library election, see the no-Condorcet-winner comparison set.

The criterion: Smith-efficient methods never leave the club

A method satisfies the Smith criterion (is Smith-efficient) if its winner always comes from the Smith set. Since a lone Condorcet winner is the Smith set, Smith-efficiency implies Condorcet-efficiency — it's the strictly stronger promise: even in a cycle, I won't wander off into the dominated candidates.

Method In the club? Why
Ranked Robin / Copeland ✅ the best win–loss records are the top of the club — however the tie among them is then broken (margins, lot, or BV's random draw), the winner stays inside
Ranked Pairs, Schulze ✅ the "serious" cycle resolvers are Smith-efficient by construction
Minimax ❌ Condorcet-efficient, but in a 4+ candidate cycle its "least bad worst loss" pick can fall outside the Smith set — the classic fine-print failure
RCV-IRV (Hare) ❌ not even Condorcet-efficient (center squeeze), so Smith is out of reach
Borda, Score, Approval, Plurality ❌ point totals can crown a candidate the club collectively beats
STAR ❌ not Condorcet-compliant by design — it trades the guarantee for counting preference strength; see three notions of "winner"

Two footnotes to the table. First, the criterion has spawned a whole construction kit: Smith//X methods first eliminate everyone outside the Smith set, then run method X on the survivors — Smith//IRV (a.k.a. Tideman's Alternative) and Smith//Minimax are the common ones, bolting Smith-efficiency onto methods that lack it. Equal Vote's own tie-breaking protocol reaches for the same idea ("Smith-Minimax" sits in Ranked Robin's tiebreak hierarchy). Second, there's a related independence criterion, ISDA ("Independence of Smith-Dominated Alternatives"): candidates outside the Smith set should be ignorable — delete them from every ballot and the winner shouldn't change. Our demo passes the spot-check: delete Dave and you get exactly the 02_cycle election, where Ranked Robin still elects Ada. (electowiki has the formal statement — advocacy-adjacent wiki, fine for definitions.)

Fine print: Schwartz, ties, and "Smith's method"

  • The Schwartz set (a.k.a. GOCHA) is the Smith set's slightly tighter cousin: always Schwartz ⊆ Smith, and the two differ only when pairwise ties are involved (a tie is enough to keep you in Smith, not in Schwartz). With an odd number of voters and full rankings — like our 7 ballots — no pairwise ties are possible and the two sets coincide.
  • "Smith's method" — electing the Smith set itself — is set-valued: fine for shortlists, but a single-winner election still needs a rule for inside the club. That's exactly the cycle-resolution split, and why the ranked-ballot method zoo files Smith/Schwartz under "set-valued, usually used as a filter."
  • Computing it is graph theory, and cheap: the Smith set is the top strongly connected component of the pairwise-results tournament — or, equivalently, walk down the Copeland standings until the block above the line beats everyone below it. Standard algorithms, quadratic work; the math behind Condorcet maps this corner (tournaments, SCCs, and friends).
  • The Copeland standings can understate the club. The leaders of the win–loss table are always inside the Smith set, but they need not be it. Five candidates where E loses to everyone and, among the rest, A beats B and C, B beats C and D, C beats D, D beats A: the Copeland leaders are just A and B at 3 wins each, while the Smith set is all of {A, B, C, D}. So "who tied at the top?" is a narrower question than "who is still in contention?" — which is why the engine prints the set explicitly rather than leaving you to read it off the standings.

The engine reports it

Every _tabulated mirror for a Ranked Robin or RCV-IRV election now ends with a Smith-set block — the set, whether it is a lone Condorcet winner or a top cycle, and whether the method's winner landed inside it. It is off in the default on-screen report (house rule: lean screen, full mirror); the --full flag puts it on screen, and a file can still opt in with options: { show_smith_set: true }.

The two methods put the block to opposite uses. For Ranked Robin it is descriptive — RR passes by construction, so the line states the guarantee. For RCV-IRV it is a real pass/fail, and the repo's most basic ranked example (RCV_ballot_example) turns out to fail it:

Winner(s) — RCV / Instant-Runoff Voting (single winner)
  A

--- 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 (1 of 3): C
   Outside (2):        A, B
   One member ⇒ C is the Condorcet winner, beating every rival head-to-head.
   RCV-IRV winner A is OUTSIDE the Smith set. ✗
      Every member of the set (C) beats A head-to-head, yet
      RCV-IRV elected A anyway. RCV-IRV is not Smith-efficient (nor
      Condorcet-efficient) — this is the shape a center squeeze leaves behind.

Forty voters rank A>C>B, thirty-five B>C>A, twenty-five C>A>B. C beats A 60–40 and B 65–35 — the Condorcet winner — but holds only 25 first preferences, so C is eliminated in round one and A takes it 65–35. A textbook center squeeze, and the Smith line is what makes it visible without hand-computing the matrix.

How big is the club in practice?

Almost always: one member. Real electorates that spread along a spatial spectrum produce Condorcet winners overwhelmingly often — in one dimension the median-voter theorem guarantees one — so the Smith set is a singleton and every Condorcet method just elects it. The set only becomes interesting when preferences genuinely cycle: rare, likeliest in small, sharply three-way-divided electorates (how often?). The deep-theory footnote is that in 2+ spatial dimensions the guarantee collapses spectacularly — the McKelvey–Schofield chaos theorem says the top cycle can then wander essentially anywhere, which is why theorists wanted a disciplined "still in contention" set in the first place.

So the honest summary matches the repo's house caveat: the Smith set is the right standard for judging cycle behavior, and knowing it costs nothing — but no voter needs it on election day, and a method's Smith-efficiency is fine print that matters only on the rare day the matrix eats its own tail.

Cross-references