FA-13101 / Numerical aggregation / Open access
Frequency left quantile: An exact cumulative rank boundary advances to the next support value. · case 01
The reduction disagrees with its explicit aggregation oracle.
ROOT CAUSE
An exact cumulative rank boundary advances to the next support value.
VERIFIED REPAIR
Preserve the frequency left quantile contract at the identified reduction decision.
Unsuccessful approach: Adding one count also shifts nonintegral rank thresholds.
Case contract
Rows are integer [value, nonnegative frequency]; 0<=p<=q, q>0. Return the smallest supported value whose cumulative positive frequency reaches p/q of total. At p=0 return minimum positive support. No positive mass 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(rows, p, q):
rows=sorted((x,w) for x,w in rows if w>0)
if not rows: return None
total=sum(w for x,w in rows)
threshold=Fraction(p,q)*total
running=0
for x,w in rows:
running+=w
if running>threshold: return x
return rows[-1][0]
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([(-8, 1), (-2, 1), (3, 1)], 2, 5)), -2)
check('regression 2', solve(*([(1, 2), (8, 5)], 1, 2)), 8)
check('regression 3', solve(*([(1, 1), (8, 2)], 2, 5)), 8)
check('regression 4', solve(*([(9, 1), (1, 3), (5, 2)], 1, 2)), 1)
check('regression 5', solve(*([(8, 0), (2, 1)], 0, 1)), 2)
check('regression 6', solve(*([], 1, 2)), None)
check('regression 7', solve(*([(1, 0), (5, 0)], 1, 2)), None)
check('regression 8', solve(*([(1, 1), (4, 2), (8, 1)], 3, 4)), 4)
check('regression 9', solve(*([(9, 4), (2, 1)], 1, 1)), 9)
check('regression 10', solve(*([(3, 2), (3, 1), (1, 1)], 1, 2)), 3)
check('regression 11', solve(*([(2, 2), (7, 2)], 1, 2)), 2)
check("variable rank",solve([(N,2),(N+4,1)],1,2),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 | -2 | -2 | Passed |
| regression 2 | 8 | 8 | Passed |
| regression 3 | 8 | 8 | Passed |
| regression 4 | 5 | 1 | Failed |
| regression 5 | 2 | 2 | Passed |
| regression 6 | None | None | Passed |
| regression 7 | None | None | Passed |
| regression 8 | 8 | 4 | Failed |
| regression 9 | 9 | 9 | Passed |
| regression 10 | 3 | 3 | Passed |
| regression 11 | 7 | 2 | Failed |
| variable rank | 1 | 1 | Passed |
SHA-256 / f4441b9a1f3250e6e21b8a8027c75308821dcf7fe79b8125b62e28ba39ae7dfc
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, p, q):
rows=sorted((x,w) for x,w in rows if w>0)
if not rows: return None
total=sum(w for x,w in rows)
threshold=Fraction(p,q)*total
running=0
for x,w in rows:
running+=w
if running>=threshold+1: return x
return rows[-1][0]
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([(-8, 1), (-2, 1), (3, 1)], 2, 5)), -2)
check('regression 2', solve(*([(1, 2), (8, 5)], 1, 2)), 8)
check('regression 3', solve(*([(1, 1), (8, 2)], 2, 5)), 8)
check('regression 4', solve(*([(9, 1), (1, 3), (5, 2)], 1, 2)), 1)
check('regression 5', solve(*([(8, 0), (2, 1)], 0, 1)), 2)
check('regression 6', solve(*([], 1, 2)), None)
check('regression 7', solve(*([(1, 0), (5, 0)], 1, 2)), None)
check('regression 8', solve(*([(1, 1), (4, 2), (8, 1)], 3, 4)), 4)
check('regression 9', solve(*([(9, 4), (2, 1)], 1, 1)), 9)
check('regression 10', solve(*([(3, 2), (3, 1), (1, 1)], 1, 2)), 3)
check('regression 11', solve(*([(2, 2), (7, 2)], 1, 2)), 2)
check("variable rank",solve([(N,2),(N+4,1)],1,2),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 | 3 | -2 | Failed |
| regression 2 | 8 | 8 | Passed |
| regression 3 | 8 | 8 | Passed |
| regression 4 | 5 | 1 | Failed |
| regression 5 | 2 | 2 | Passed |
| regression 6 | None | None | Passed |
| regression 7 | None | None | Passed |
| regression 8 | 8 | 4 | Failed |
| regression 9 | 9 | 9 | Passed |
| regression 10 | 3 | 3 | Passed |
| regression 11 | 7 | 2 | Failed |
| variable rank | 5 | 1 | Failed |
SHA-256 / 8c97b0d8991794a8d23a7c28801ec3e74a1d99e66f2fd6137a42ffa9531eefde
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, p, q):
rows=sorted((x,w) for x,w in rows if w>0)
if not rows: return None
total=sum(w for x,w in rows)
threshold=Fraction(p,q)*total
running=0
for x,w in rows:
running+=w
if running>=threshold: return x
return rows[-1][0]
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([(-8, 1), (-2, 1), (3, 1)], 2, 5)), -2)
check('regression 2', solve(*([(1, 2), (8, 5)], 1, 2)), 8)
check('regression 3', solve(*([(1, 1), (8, 2)], 2, 5)), 8)
check('regression 4', solve(*([(9, 1), (1, 3), (5, 2)], 1, 2)), 1)
check('regression 5', solve(*([(8, 0), (2, 1)], 0, 1)), 2)
check('regression 6', solve(*([], 1, 2)), None)
check('regression 7', solve(*([(1, 0), (5, 0)], 1, 2)), None)
check('regression 8', solve(*([(1, 1), (4, 2), (8, 1)], 3, 4)), 4)
check('regression 9', solve(*([(9, 4), (2, 1)], 1, 1)), 9)
check('regression 10', solve(*([(3, 2), (3, 1), (1, 1)], 1, 2)), 3)
check('regression 11', solve(*([(2, 2), (7, 2)], 1, 2)), 2)
check("variable rank",solve([(N,2),(N+4,1)],1,2),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 | -2 | -2 | Passed |
| regression 2 | 8 | 8 | Passed |
| regression 3 | 8 | 8 | Passed |
| regression 4 | 1 | 1 | Passed |
| regression 5 | 2 | 2 | Passed |
| regression 6 | None | None | Passed |
| regression 7 | None | None | Passed |
| regression 8 | 4 | 4 | Passed |
| regression 9 | 9 | 9 | Passed |
| regression 10 | 3 | 3 | Passed |
| regression 11 | 2 | 2 | Passed |
| variable rank | 1 | 1 | Passed |
SHA-256 / 551bcdb223f6329978a6042ae9bafef33f637bb02ab4034980f21b6095458761
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 2026-09-29T14:39:03.583429+00:00.
Case digest / 1a57bcb0ffe7b240f37d01de3ad99dbf24eb2c467e7510be233db2a3aaeed109