FA-12781 / Voting rule computation / Open access
A plurality of pairwise victories is mistaken for beating every opponent · case 01
A plurality of pairwise victories is mistaken for beating every opponent.
ROOT CAUSE
The candidate with most head-to-head victories is returned even when a cycle prevents a strict winner.
THE FAILURE
The candidate with most head-to-head victories is returned even when a cycle prevents a strict winner.
Unsuccessful approach: Allowing tied head-to-head contests admits a weak winner where the contract requires a strict winner.
Case contract
Input is a nonempty square matrix of nonnegative head-to-head support counts with zero diagonal. Return the index strictly beating every other index, or None if none exists. A single candidate wins vacuously.
Why this case matters
A deterministic toy ballot model makes the stated counting convention executable.
1 / The failure
Exit 1"""Failure Map reference implementation. Python standard library only."""
import json
N = 1
observations = []
def solve(matrix):
return max(range(len(matrix)),key=lambda i:sum(matrix[i][j]>matrix[j][i] for j in range(len(matrix)) if i!=j))
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('three candidate cycle has no winner', solve([[0,N+1,0],[0,0,N+1],[N+1,0,0]]), None)
check('head to head tie disqualifies', solve([[0,N,N+1],[N,0,N+1],[0,0,0]]), None)
check('strict winner need not be first', solve([[0,0,N],[N+1,0,N+1],[0,0,0]]), 1)
check('single candidate', solve([[0]]), 0)
check('all contests tied', solve([[0,N,N],[N,0,N],[N,N,0]]), None)
check('two candidate strict contest', solve([[0,N],[N+1,0]]), 1)
print(json.dumps({"observations": observations, "passed": all(x["passed"] for x in observations)}, ensure_ascii=False))
raise SystemExit(0 if all(x["passed"] for x in observations) else 1)
| Boundary fixture | Actual | Expected | Outcome |
|---|---|---|---|
| three candidate cycle has no winner | 0 | None | Failed |
| head to head tie disqualifies | 0 | None | Failed |
| strict winner need not be first | 1 | 1 | Passed |
| single candidate | 0 | 0 | Passed |
| all contests tied | 0 | None | Failed |
| two candidate strict contest | 1 | 1 | Passed |
SHA-256 / 474de7714c1250ac0efda05ba94f1238e2bb768aecd9d622d61f3ffc027bf6da
2 / The unsuccessful fix
Exit 1"""Failure Map reference implementation. Python standard library only."""
import json
N = 1
observations = []
def solve(matrix):
return next((i for i in range(len(matrix)) if all(i==j or matrix[i][j]>=matrix[j][i] for j in range(len(matrix)))),None)
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('three candidate cycle has no winner', solve([[0,N+1,0],[0,0,N+1],[N+1,0,0]]), None)
check('head to head tie disqualifies', solve([[0,N,N+1],[N,0,N+1],[0,0,0]]), None)
check('strict winner need not be first', solve([[0,0,N],[N+1,0,N+1],[0,0,0]]), 1)
check('single candidate', solve([[0]]), 0)
check('all contests tied', solve([[0,N,N],[N,0,N],[N,N,0]]), None)
check('two candidate strict contest', solve([[0,N],[N+1,0]]), 1)
print(json.dumps({"observations": observations, "passed": all(x["passed"] for x in observations)}, ensure_ascii=False))
raise SystemExit(0 if all(x["passed"] for x in observations) else 1)
| Boundary fixture | Actual | Expected | Outcome |
|---|---|---|---|
| three candidate cycle has no winner | None | None | Passed |
| head to head tie disqualifies | 0 | None | Failed |
| strict winner need not be first | 1 | 1 | Passed |
| single candidate | 0 | 0 | Passed |
| all contests tied | 0 | None | Failed |
| two candidate strict contest | 1 | 1 | Passed |
SHA-256 / 58e40f9b104d678a2978725a8f16f37aa80ac0fe33e9cf294ff57327c23abd3d
HELD IN THE MEMBER ARCHIVE
The verified repair and its recorded checks are member-only.
This mechanism has 6 recorded checks per implementation. The open-access tier publishes the failure and the unsuccessful fix; the repaired source that passes every check, and the observations that prove it, are available to members.
Every case sharing this mechanism uses the same contract and the same repair, so this one record is held back for all of them.
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Sign in to the archive ↗Verification & scope
Abstract counting rules only; excludes jurisdictional law, ballot authentication and election operations. This reproducer isolates one failure mechanism. Results cover the supplied fixtures. Variants within a family share a test contract and should remain grouped when constructing evaluation splits. Related mechanisms with a shared evaluation_group must also remain together; these controlled models are not independent production incidents.
Observations recorded using Python 3.12.14 at 2026-09-29T14:39:00.110755+00:00.
Case digest / aebb6e0684eeb191761d99289ba13e507e3871c745a2f5249892e97cfb19e5ee