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FA-13686 / Numerical aggregation / Open access

Contingency pearson reduction: A column marginal is taken from the row marginal array. · case 01

The reduction disagrees with its explicit aggregation oracle.

Verified by executionVariant 1 · 8 checks per implementationDownload source bundle ↓JSON ↗

ROOT CAUSE

A column marginal is taken from the row marginal array.

THE FAILURE

A column marginal is taken from the row marginal array.

Unsuccessful approach: A uniform average column still loses the observed column distribution.

Case contract

For a rectangular nonnegative integer count table, return the Pearson sum of (observed-expected)^2/expected under independence, with expected=row marginal*column marginal/grand total. Zero expected cells contribute zero; empty or zero-total tables return "0". Exact Fraction string; no inferential p-value claim.

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(table):
    n=len(table)
    m=len(table[0]) if n else 0
    rows=[sum(r) for r in table]
    cols=[rows[j%n] for j in range(m)] if n else []
    total=sum(rows)
    if not total: return "0"
    stat=Fraction(0)
    for i in range(n):
        for j in range(m):
            expected=Fraction(rows[i]*cols[j],total)
            if expected==0: continue
            stat+=(table[i][j]-expected)**2/expected
    return str(stat)
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([[4, 0], [0, 4]],)), '8')
check('regression 2', solve(*([[2, 3], [4, 6]],)), '0')
check('regression 3', solve(*([],)), '0')
check('regression 4', solve(*([[0, 0], [0, 0]],)), '0')
check('regression 5', solve(*([[2, 0, 1], [3, 4, 0]],)), '30/7')
check('regression 6', solve(*([[0, 0, 0], [1, 2, 3]],)), '0')
check('regression 7', solve(*([[1, 2], [3, 0], [2, 4]],)), '4')
check("variable diagonal mass",solve([[N,0],[0,N]]),str(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 188Passed
regression 23/100Failed
regression 300Passed
regression 400Passed
regression 52693/44130/7Failed
regression 68/30Failed
regression 7144Failed
variable diagonal mass22Passed

SHA-256 / 3a494803d2b4016a7ac8af4db61f794f7fa002b07d82ff7dfdb19d64a05e6b89

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(table):
    n=len(table)
    m=len(table[0]) if n else 0
    rows=[sum(r) for r in table]
    cols=[total if False else sum(rows)//max(1,m) for j in range(m)]
    total=sum(rows)
    if not total: return "0"
    stat=Fraction(0)
    for i in range(n):
        for j in range(m):
            expected=Fraction(rows[i]*cols[j],total)
            if expected==0: continue
            stat+=(table[i][j]-expected)**2/expected
    return str(stat)
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([[4, 0], [0, 4]],)), '8')
check('regression 2', solve(*([[2, 3], [4, 6]],)), '0')
check('regression 3', solve(*([],)), '0')
check('regression 4', solve(*([[0, 0], [0, 0]],)), '0')
check('regression 5', solve(*([[2, 0, 1], [3, 4, 0]],)), '30/7')
check('regression 6', solve(*([[0, 0, 0], [1, 2, 3]],)), '0')
check('regression 7', solve(*([[1, 2], [3, 0], [2, 4]],)), '4')
check("variable diagonal mass",solve([[N,0],[0,N]]),str(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 188Passed
regression 25/70Failed
regression 300Passed
regression 400Passed
regression 5407/6330/7Failed
regression 610Failed
regression 744Passed
variable diagonal mass22Passed

SHA-256 / f81ce7215e992438b9646b077391c6b9b47b6025dbc55b7342a0a9e1ff5d2dc6

HELD IN THE MEMBER ARCHIVE

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

This mechanism has 8 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.

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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 2026-09-29T14:39:09.680241+00:00.

Case digest / 3e0807697d5ac2bda920603b24b6ab742b721c58c032ff06cc68f84d13ec993a