FA-13911 / Numerical aggregation / Open access
Adjacent product block reduction: Cross-block neighboring products are omitted. · case 01
The reduction disagrees with its explicit aggregation oracle.
ROOT CAUSE
Cross-block neighboring products are omitted.
VERIFIED REPAIR
Preserve the adjacent product block reduction contract at the identified reduction decision.
Unsuccessful approach: Adding boundary observations does not form their neighboring product.
Case contract
Concatenate ordered integer blocks logically and sum products of neighboring observations. Empty blocks are identity; there is exactly one cross-boundary product between consecutive nonempty blocks.
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(blocks):
total=0
last=None
for block in blocks:
if not block: continue
total+=sum(a*b for a,b in zip(block,block[1:]))
if False: total+=last*block[0]
last=block[-1]
return total
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([[1, 2], [3, 4]],)), 20)
check('regression 2', solve(*([[2], [], [-3], [5]],)), -21)
check('regression 3', solve(*([],)), 0)
check('regression 4', solve(*([[0, 4], [3, 0]],)), 12)
check('regression 5', solve(*([[-2, -1], [3, -4, 5]],)), -33)
check('regression 6', solve(*([[], [7], []],)), 0)
check('regression 7', solve(*([[2, 3, 4]],)), 18)
check("variable interblock product",solve([[N],[N+1]]),N*(N+1))
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 | 14 | 20 | Failed |
| regression 2 | 0 | -21 | Failed |
| regression 3 | 0 | 0 | Passed |
| regression 4 | 0 | 12 | Failed |
| regression 5 | -30 | -33 | Failed |
| regression 6 | 0 | 0 | Passed |
| regression 7 | 18 | 18 | Passed |
| variable interblock product | 0 | 2 | Failed |
SHA-256 / de33e4b94d13a8705309866247d09ef8f4ca12c0f98a58f08030c195cebed50b
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(blocks):
total=0
last=None
for block in blocks:
if not block: continue
total+=sum(a*b for a,b in zip(block,block[1:]))
if last is not None: total+=last+block[0]
last=block[-1]
return total
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([[1, 2], [3, 4]],)), 20)
check('regression 2', solve(*([[2], [], [-3], [5]],)), -21)
check('regression 3', solve(*([],)), 0)
check('regression 4', solve(*([[0, 4], [3, 0]],)), 12)
check('regression 5', solve(*([[-2, -1], [3, -4, 5]],)), -33)
check('regression 6', solve(*([[], [7], []],)), 0)
check('regression 7', solve(*([[2, 3, 4]],)), 18)
check("variable interblock product",solve([[N],[N+1]]),N*(N+1))
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 | 19 | 20 | Failed |
| regression 2 | 1 | -21 | Failed |
| regression 3 | 0 | 0 | Passed |
| regression 4 | 7 | 12 | Failed |
| regression 5 | -28 | -33 | Failed |
| regression 6 | 0 | 0 | Passed |
| regression 7 | 18 | 18 | Passed |
| variable interblock product | 3 | 2 | Failed |
SHA-256 / f7603abf32e78203fd090471e3ecc746a081bdeb6578fafa520ba4aaf179f3fa
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(blocks):
total=0
last=None
for block in blocks:
if not block: continue
total+=sum(a*b for a,b in zip(block,block[1:]))
if last is not None: total+=last*block[0]
last=block[-1]
return total
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([[1, 2], [3, 4]],)), 20)
check('regression 2', solve(*([[2], [], [-3], [5]],)), -21)
check('regression 3', solve(*([],)), 0)
check('regression 4', solve(*([[0, 4], [3, 0]],)), 12)
check('regression 5', solve(*([[-2, -1], [3, -4, 5]],)), -33)
check('regression 6', solve(*([[], [7], []],)), 0)
check('regression 7', solve(*([[2, 3, 4]],)), 18)
check("variable interblock product",solve([[N],[N+1]]),N*(N+1))
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 | 20 | 20 | Passed |
| regression 2 | -21 | -21 | Passed |
| regression 3 | 0 | 0 | Passed |
| regression 4 | 12 | 12 | Passed |
| regression 5 | -33 | -33 | Passed |
| regression 6 | 0 | 0 | Passed |
| regression 7 | 18 | 18 | Passed |
| variable interblock product | 2 | 2 | Passed |
SHA-256 / 57613ef4fdd7ea759e8e79c19b3f78b846fe3c80a8c64f8eca1cbd31f91df0ea
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 2026-09-29T14:39:11.685748+00:00.
Case digest / 6bfaad874c4a4b690f49180ca01015c345fe4c2c3c292065d06347deda33b5ac