FA-13451 / Numerical aggregation / Open access
Empirical transport distance: Support gaps are forced to at least one even after normalization. · case 01
The reduction disagrees with its explicit aggregation oracle.
ROOT CAUSE
Support gaps are forced to at least one even after normalization.
VERIFIED REPAIR
Preserve the empirical transport distance contract at the identified reduction decision.
Unsuccessful approach: Subtracting a unit treats continuous gap length as interior lattice-point count.
Case contract
For nonempty integer samples with equal total probability after separate normalization, return integral of absolute CDF difference over the real line 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)
points=sorted(set(ca)|set(cb))
pa=pb=Fraction(0)
area=Fraction(0)
for i,x in enumerate(points[:-1]):
pa+=Fraction(ca[x],len(a))
pb+=Fraction(cb[x],len(b))
area+=abs(pa-pb)*(points[i+1]-x+1)
return str(area)
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([0, 0, 9], [1, 4])), '17/6')
check('regression 2', solve(*([2], [7])), '5')
check('regression 3', solve(*([1, 2], [1, 2])), '0')
check('regression 4', solve(*([], [1])), None)
check('regression 5', solve(*([-8, -1, 4], [-3, 9, 9])), '20/3')
check('regression 6', solve(*([0, 2, 8], [0, 8])), '4/3')
check('regression 7', solve(*([0, 10], [4, 6])), '4')
check("variable transport span",solve([0],[N]),str(N))
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 | 4 | 17/6 | Failed |
| regression 2 | 6 | 5 | Failed |
| regression 3 | 0 | 0 | Passed |
| regression 4 | None | None | Passed |
| regression 5 | 8 | 20/3 | Failed |
| regression 6 | 5/3 | 4/3 | Failed |
| regression 7 | 5 | 4 | Failed |
| variable transport span | 2 | 1 | Failed |
SHA-256 / 7f831ec05fb198223e8db985e425f8149ced9f71566a6e6395858416f23ef524
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)
points=sorted(set(ca)|set(cb))
pa=pb=Fraction(0)
area=Fraction(0)
for i,x in enumerate(points[:-1]):
pa+=Fraction(ca[x],len(a))
pb+=Fraction(cb[x],len(b))
area+=abs(pa-pb)*max(0,points[i+1]-x-1)
return str(area)
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([0, 0, 9], [1, 4])), '17/6')
check('regression 2', solve(*([2], [7])), '5')
check('regression 3', solve(*([1, 2], [1, 2])), '0')
check('regression 4', solve(*([], [1])), None)
check('regression 5', solve(*([-8, -1, 4], [-3, 9, 9])), '20/3')
check('regression 6', solve(*([0, 2, 8], [0, 8])), '4/3')
check('regression 7', solve(*([0, 10], [4, 6])), '4')
check("variable transport span",solve([0],[N]),str(N))
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 | 5/3 | 17/6 | Failed |
| regression 2 | 4 | 5 | Failed |
| regression 3 | 0 | 0 | Passed |
| regression 4 | None | None | Passed |
| regression 5 | 16/3 | 20/3 | Failed |
| regression 6 | 1 | 4/3 | Failed |
| regression 7 | 3 | 4 | Failed |
| variable transport span | 0 | 1 | Failed |
SHA-256 / b470f3b1bd1496c5c8a2539407a8175f0bb1b24668f41c591ee7dad60dd81654
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(a, b):
if not a or not b: return None
ca,cb=Counter(a),Counter(b)
points=sorted(set(ca)|set(cb))
pa=pb=Fraction(0)
area=Fraction(0)
for i,x in enumerate(points[:-1]):
pa+=Fraction(ca[x],len(a))
pb+=Fraction(cb[x],len(b))
area+=abs(pa-pb)*(points[i+1]-x)
return str(area)
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([0, 0, 9], [1, 4])), '17/6')
check('regression 2', solve(*([2], [7])), '5')
check('regression 3', solve(*([1, 2], [1, 2])), '0')
check('regression 4', solve(*([], [1])), None)
check('regression 5', solve(*([-8, -1, 4], [-3, 9, 9])), '20/3')
check('regression 6', solve(*([0, 2, 8], [0, 8])), '4/3')
check('regression 7', solve(*([0, 10], [4, 6])), '4')
check("variable transport span",solve([0],[N]),str(N))
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 | 17/6 | 17/6 | Passed |
| regression 2 | 5 | 5 | Passed |
| regression 3 | 0 | 0 | Passed |
| regression 4 | None | None | Passed |
| regression 5 | 20/3 | 20/3 | Passed |
| regression 6 | 4/3 | 4/3 | Passed |
| regression 7 | 4 | 4 | Passed |
| variable transport span | 1 | 1 | Passed |
SHA-256 / aedf34f7a9c69f0b5617f3aefe986b52030289190148a5521303189b768a7b17
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 2026-09-29T14:39:07.193044+00:00.
Case digest / d4277243dc71a7bbf6336b7a52ccb6071305713ecac9172f792de9632a294b74