FA-13931 / Numerical aggregation / Open access
Adjacent product block reduction: A singleton block is ignored because it has no local neighboring pair. · case 01
The reduction disagrees with its explicit aggregation oracle.
ROOT CAUSE
A singleton block is ignored because it has no local neighboring pair.
VERIFIED REPAIR
Preserve the adjacent product block reduction contract at the identified reduction decision.
Unsuccessful approach: Retaining the singleton endpoint still omits its product with the previous nonempty block.
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 len(block)<2: 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 | 0 | -21 | Failed |
| 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 | 0 | 2 | Failed |
SHA-256 / 1c9a3e5e43c7f39a5e2bbb13c9699edffb35045a5d3aef5b0d002e9385679c61
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
if len(block)==1:
last=block[0]
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 | 0 | -21 | Failed |
| 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 | 0 | 2 | Failed |
SHA-256 / e962e99947bb1e42ea8db71d02168651c3671282661bc00cfc88ae4a14da99f8
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 2026-09-29T14:39:11.959091+00:00.
Case digest / 83e6f38df38a5fd26ca65ccc3c4220c5869f6cf1e58bc439e1a7410cb2e86b33