FA-13926 / Numerical aggregation / Open access

Adjacent product block reduction: An empty block clears the carried neighbor. · case 01

The reduction disagrees with its explicit aggregation oracle.

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

ROOT CAUSE

An empty block clears the carried neighbor.

THE FAILURE

An empty block clears the carried neighbor.

Unsuccessful approach: Replacing the carried endpoint by zero also breaks adjacency across empty blocks.

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:
            last=None
            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	`-15`	`-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	`2`	`2`	Passed

SHA-256 / 91ed023df952f4ea6cb535b23a2b09c0f0907affd351181bf5b796ddf1a0ecd1

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:
            last=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	`-15`	`-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	`2`	`2`	Passed

SHA-256 / 71a740971e5da68295c0c18b072ae4d6e016bd39ff792e184f2c085d742110ca

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.

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

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 2026-09-29T14:39:12.219008+00:00.

Case digest / 8b5a5c67e9e5df54e0e4a7d318a318b09f1fe64e5f9164b77b3adfd12ea91b36