FA-13921 / Numerical aggregation / Open access
Adjacent product block reduction: The carried endpoint is the first observation of the old block. · case 01
The reduction disagrees with its explicit aggregation oracle.
ROOT CAUSE
The carried endpoint is the first observation of the old block.
VERIFIED REPAIR
Preserve the adjacent product block reduction contract at the identified reduction decision.
Unsuccessful approach: Carrying block total loses the identity of its final observation.
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 last is not None: total+=last*block[0]
last=block[0]
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 | 17 | 20 | Failed |
| regression 2 | -21 | -21 | Passed |
| regression 3 | 0 | 0 | Passed |
| regression 4 | 0 | 12 | Failed |
| regression 5 | -36 | -33 | Failed |
| regression 6 | 0 | 0 | Passed |
| regression 7 | 18 | 18 | Passed |
| variable interblock product | 2 | 2 | Passed |
SHA-256 / 827d0a0917cb4b5a96fe5371980e27e6cf696b184f787fbeac51e40746b93cb8
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=sum(block)
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 | 23 | 20 | Failed |
| regression 2 | -21 | -21 | Passed |
| regression 3 | 0 | 0 | Passed |
| regression 4 | 12 | 12 | Passed |
| regression 5 | -39 | -33 | Failed |
| regression 6 | 0 | 0 | Passed |
| regression 7 | 18 | 18 | Passed |
| variable interblock product | 2 | 2 | Passed |
SHA-256 / e0c238aed9ac1e1c4ef584e4ae6227c202c88412616c3f895a6425ec176587ee
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 2026-09-29T14:39:11.685760+00:00.
Case digest / 711ca562945ce7e882147d2b272ba279c75834af0020d390c71009fcb1015716