FA-13316 / Numerical aggregation / Open access
Frequency symmetric trim mean: A negative zero slice end removes all untrimmed observations. · case 01
The reduction disagrees with its explicit aggregation oracle.
ROOT CAUSE
A negative zero slice end removes all untrimmed observations.
VERIFIED REPAIR
Preserve the frequency symmetric trim mean contract at the identified reduction decision.
Unsuccessful approach: Forcing a nonzero end offset discards an observation for k=0.
Case contract
Expand nonnegative integer frequencies. Remove exactly k lowest and k highest observations by multiplicity. Return the exact mean as a Fraction string, or None when no observations survive.
Why this case matters
Exact bounded examples isolate a reduction defect without floating-point or external-service effects.
1 / The failure
Exit 1"""Failure Map reference implementation. Python standard library only."""
import json
from fractions import Fraction
from collections import Counter, defaultdict
import math
import itertools
N = 1
observations = []
def solve(rows, k):
xs=sorted(x for x,w in rows for _ in range(w))
if not xs or 2*k>=len(xs): return None
return str(Fraction(sum(xs[k:-k]),len(xs)-2*k))
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([(9, 1), (1, 4), (5, 2)], 2)), '7/3')
check('regression 2', solve(*([(2, 4)], 1)), '2')
check('regression 3', solve(*([], 0)), None)
check('regression 4', solve(*([(1, 2), (4, 2)], 2)), None)
check('regression 5', solve(*([(0, 1), (10, 3), (-5, 2)], 1)), '15/4')
check('regression 6', solve(*([(3, 0), (8, 1)], 0)), '8')
check('regression 7', solve(*([(1, 1), (8, 4), (9, 1)], 1)), '8')
check("variable weighted trim",solve([(N,3),(N+6,2)],1),str(Fraction(3*N+6,3)))
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 |
|---|---|---|---|
| regression 1 | 7/3 | 7/3 | Passed |
| regression 2 | 2 | 2 | Passed |
| regression 3 | None | None | Passed |
| regression 4 | None | None | Passed |
| regression 5 | 15/4 | 15/4 | Passed |
| regression 6 | 0 | 8 | Failed |
| regression 7 | 8 | 8 | Passed |
| variable weighted trim | 3 | 3 | Passed |
SHA-256 / 7d953eaf9b9a1436ad6197fa5f6ba4d77bad56e62056cafa9759e0afc61b5d12
2 / The unsuccessful fix
Exit 1"""Failure Map reference implementation. Python standard library only."""
import json
from fractions import Fraction
from collections import Counter, defaultdict
import math
import itertools
N = 1
observations = []
def solve(rows, k):
xs=sorted(x for x,w in rows for _ in range(w))
if not xs or 2*k>=len(xs): return None
return str(Fraction(sum(xs[k:-max(1,k)]),len(xs)-2*k))
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([(9, 1), (1, 4), (5, 2)], 2)), '7/3')
check('regression 2', solve(*([(2, 4)], 1)), '2')
check('regression 3', solve(*([], 0)), None)
check('regression 4', solve(*([(1, 2), (4, 2)], 2)), None)
check('regression 5', solve(*([(0, 1), (10, 3), (-5, 2)], 1)), '15/4')
check('regression 6', solve(*([(3, 0), (8, 1)], 0)), '8')
check('regression 7', solve(*([(1, 1), (8, 4), (9, 1)], 1)), '8')
check("variable weighted trim",solve([(N,3),(N+6,2)],1),str(Fraction(3*N+6,3)))
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 |
|---|---|---|---|
| regression 1 | 7/3 | 7/3 | Passed |
| regression 2 | 2 | 2 | Passed |
| regression 3 | None | None | Passed |
| regression 4 | None | None | Passed |
| regression 5 | 15/4 | 15/4 | Passed |
| regression 6 | 0 | 8 | Failed |
| regression 7 | 8 | 8 | Passed |
| variable weighted trim | 3 | 3 | Passed |
SHA-256 / a4b16c314bc62790d7634ddee6a604caa736ae1dd539b8bfa53b7e51113ca83a
3 / The verified repair
Exit 0"""Failure Map reference implementation. Python standard library only."""
import json
from fractions import Fraction
from collections import Counter, defaultdict
import math
import itertools
N = 1
observations = []
def solve(rows, k):
xs=sorted(x for x,w in rows for _ in range(w))
if not xs or 2*k>=len(xs): return None
return str(Fraction(sum(xs[k:len(xs)-k]),len(xs)-2*k))
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([(9, 1), (1, 4), (5, 2)], 2)), '7/3')
check('regression 2', solve(*([(2, 4)], 1)), '2')
check('regression 3', solve(*([], 0)), None)
check('regression 4', solve(*([(1, 2), (4, 2)], 2)), None)
check('regression 5', solve(*([(0, 1), (10, 3), (-5, 2)], 1)), '15/4')
check('regression 6', solve(*([(3, 0), (8, 1)], 0)), '8')
check('regression 7', solve(*([(1, 1), (8, 4), (9, 1)], 1)), '8')
check("variable weighted trim",solve([(N,3),(N+6,2)],1),str(Fraction(3*N+6,3)))
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 |
|---|---|---|---|
| regression 1 | 7/3 | 7/3 | Passed |
| regression 2 | 2 | 2 | Passed |
| regression 3 | None | None | Passed |
| regression 4 | None | None | Passed |
| regression 5 | 15/4 | 15/4 | Passed |
| regression 6 | 8 | 8 | Passed |
| regression 7 | 8 | 8 | Passed |
| variable weighted trim | 3 | 3 | Passed |
SHA-256 / 9ee402af70a2f625a20c7f5ad40164924bd05deaf5e177e5b46ad614402e637d
Verification & scope
Small offline integer/rational inputs only; no performance, statistical inference, or production-library conformance claim. 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 2026-09-29T14:39:05.776771+00:00.
Case digest / 12092fa424847f96d1b1908e76e5afa1f7c6bcf940f18c5f66d17c5279f23a8c