FA-13411 / Numerical aggregation / Open access
Empirical cdf supremum: Each distinct support value receives one count regardless of duplicates. · case 01
The reduction disagrees with its explicit aggregation oracle.
ROOT CAUSE
Each distinct support value receives one count regardless of duplicates.
THE FAILURE
Each distinct support value receives one count regardless of duplicates.
Unsuccessful approach: Normalizing by distinct support count also ignores multiplicity.
Case contract
Return the maximum absolute difference between the right-continuous empirical cumulative distributions of two nonempty integer samples as a Fraction string. Empty either side returns None.
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(a, b):
if not a or not b: return None
ca,cb=Counter(a),Counter(b)
pa=pb=Fraction(0)
best=Fraction(0)
for x in sorted(set(ca)|set(cb)):
pa+=Fraction(int(ca[x]>0),len(a))
pb+=Fraction(cb[x],len(b))
best=max(best,abs(pa-pb))
return str(best)
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([3], [0])), '1')
check('regression 2', solve(*([-5, 3], [-2, 1])), '1/2')
check('regression 3', solve(*([1, 1, 4], [2, 3])), '2/3')
check('regression 4', solve(*([0, 0, 0], [0])), '0')
check('regression 5', solve(*([], [1])), None)
check('regression 6', solve(*([1, 2], [1, 2])), '0')
check('regression 7', solve(*([-5, 0, 8], [-2, 8, 8, 8])), '5/12')
check('regression 8', solve(*([0, 4], [0, 1, 2, 3, 4])), '3/10')
check('regression 9', solve(*([2, 2], [1, 1, 3, 3])), '1/2')
check("variable support",solve([0,N],[N,N]),"1/2")
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 | 1 | 1 | Passed |
| regression 2 | 1/2 | 1/2 | Passed |
| regression 3 | 2/3 | 2/3 | Passed |
| regression 4 | 2/3 | 0 | Failed |
| regression 5 | None | None | Passed |
| regression 6 | 0 | 0 | Passed |
| regression 7 | 5/12 | 5/12 | Passed |
| regression 8 | 3/10 | 3/10 | Passed |
| regression 9 | 1/2 | 1/2 | Passed |
| variable support | 1/2 | 1/2 | Passed |
SHA-256 / e7c7998cd55f66cfb1cb7e1df6bb55abb2be7738d8ac91d3bbdb83da8f8d86bf
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(a, b):
if not a or not b: return None
ca,cb=Counter(a),Counter(b)
pa=pb=Fraction(0)
best=Fraction(0)
for x in sorted(set(ca)|set(cb)):
pa+=Fraction(ca[x],max(1,len(ca)))
pb+=Fraction(cb[x],len(b))
best=max(best,abs(pa-pb))
return str(best)
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([3], [0])), '1')
check('regression 2', solve(*([-5, 3], [-2, 1])), '1/2')
check('regression 3', solve(*([1, 1, 4], [2, 3])), '2/3')
check('regression 4', solve(*([0, 0, 0], [0])), '0')
check('regression 5', solve(*([], [1])), None)
check('regression 6', solve(*([1, 2], [1, 2])), '0')
check('regression 7', solve(*([-5, 0, 8], [-2, 8, 8, 8])), '5/12')
check('regression 8', solve(*([0, 4], [0, 1, 2, 3, 4])), '3/10')
check('regression 9', solve(*([2, 2], [1, 1, 3, 3])), '1/2')
check("variable support",solve([0,N],[N,N]),"1/2")
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 | 1 | 1 | Passed |
| regression 2 | 1/2 | 1/2 | Passed |
| regression 3 | 1 | 2/3 | Failed |
| regression 4 | 2 | 0 | Failed |
| regression 5 | None | None | Passed |
| regression 6 | 0 | 0 | Passed |
| regression 7 | 5/12 | 5/12 | Passed |
| regression 8 | 3/10 | 3/10 | Passed |
| regression 9 | 3/2 | 1/2 | Failed |
| variable support | 1/2 | 1/2 | Passed |
SHA-256 / bc6dc0e92455d1b1073d4772ccbfc6c670d68192cef290c909dd7c14d21e9e81
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This mechanism has 10 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.
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Sign in to the archive ↗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 2026-09-29T14:39:06.766723+00:00.
Case digest / cd1a553154b109cc047cf9b6e08b611caa723f1da034493e163e0406dc1d787f