FAILURE MAP
← Case archive

FA-13106 / Numerical aggregation / Open access

Frequency left quantile: The rank target uses the number of compressed rows. · case 01

The reduction disagrees with its explicit aggregation oracle.

Verified by executionVariant 1 · 12 checks per implementationDownload source bundle ↓JSON ↗

ROOT CAUSE

The rank target uses the number of compressed rows.

THE FAILURE

The rank target uses the number of compressed rows.

Unsuccessful approach: Distinct support count still ignores frequencies.

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)*len(rows)
    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 fixtureActualExpectedOutcome
regression 1-2-2Passed
regression 218Failed
regression 318Failed
regression 411Passed
regression 522Passed
regression 6NoneNonePassed
regression 7NoneNonePassed
regression 844Passed
regression 999Passed
regression 1033Passed
regression 1122Passed
variable rank11Passed

SHA-256 / caf94784270ca2182b71d5517af4c557f27bf1a9a75a1023724a622e71f61028

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)*len(set(x for x,w in rows))
    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 fixtureActualExpectedOutcome
regression 1-2-2Passed
regression 218Failed
regression 318Failed
regression 411Passed
regression 522Passed
regression 6NoneNonePassed
regression 7NoneNonePassed
regression 844Passed
regression 999Passed
regression 1013Failed
regression 1122Passed
variable rank11Passed

SHA-256 / 78f815d6537e4366dfa71f6775f8bd4788ccf436587ef4f19beee115d2a7f958

HELD IN THE MEMBER ARCHIVE

The verified repair and its recorded checks are member-only.

This mechanism has 12 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.

Member access is invitation-based. Sign in with your invited account to inspect the repair.

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 2026-09-29T14:39:03.883435+00:00.

Case digest / 1a36aa47108a58c9474d5cb4920f871eaf9834836150ca2edeb5b3cb34fd5033