FA-13231 / Numerical aggregation / Open access

Positive run block summary: An entirely positive incoming block replaces the old suffix. · case 01

The reduction disagrees with its explicit aggregation oracle.

Verified by executionVariant 1 · 10 checks per implementationDownload source bundle ↓JSON ↗

ROOT CAUSE

An entirely positive incoming block replaces the old suffix.

VERIFIED REPAIR

Preserve the positive run block summary contract at the identified reduction decision.

Unsuccessful approach: Always extending also crosses a break inside the incoming block.

Case contract

Return [item count, positive prefix length, positive suffix length, longest strictly positive contiguous run] over ordered integer blocks. Empty blocks are identity, zero breaks a run.

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
    previous=None
    prefix=suffix=best=0
    all_positive=True
    for block in blocks:
        k=len(block)
        p=next((i for i,x in enumerate(block) if x<=0),k)
        s=next((i for i,x in enumerate(reversed(block)) if x<=0),k)
        local=run=0
        for x in block:
            run=run+1 if x>0 else 0
            local=max(local,run)
        best=max(best,local,suffix+p)
        if all_positive: prefix+=p
        suffix=s
        all_positive=all_positive and p==k
        total+=k
    return [total,prefix,suffix,best]
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([[1, 1, 0, 1, 1, 0]],)), [6, 2, 0, 2])
check('regression 2', solve(*([[1, 0, 1, 0, 1, 1, 0]],)), [7, 1, 0, 2])
check('regression 3', solve(*([[1, 2], [3, 0, 4]],)), [5, 3, 1, 3])
check('regression 4', solve(*([[1, 0, 2], [3, 4, -1, 5]],)), [7, 1, 1, 3])
check('regression 5', solve(*([],)), [0, 0, 0, 0])
check('regression 6', solve(*([[], [0, 0], []],)), [2, 0, 0, 0])
check('regression 7', solve(*([[1, 2], [], [3, 4]],)), [4, 4, 4, 4])
check('regression 8', solve(*([[-1, 2, 3], [4, -1], [-2, 1]],)), [7, 0, 1, 3])
check('regression 9', solve(*([[0], [1, 1, 1], [1, 0]],)), [6, 0, 0, 4])
check("variable positive run",solve([[1]*N,[2]*N+[0]]),[2*N+1,2*N,0,2*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 fixture	Actual	Expected	Outcome
regression 1	`[6, 2, 0, 2]`	`[6, 2, 0, 2]`	Passed
regression 2	`[7, 1, 0, 2]`	`[7, 1, 0, 2]`	Passed
regression 3	`[5, 3, 1, 3]`	`[5, 3, 1, 3]`	Passed
regression 4	`[7, 1, 1, 3]`	`[7, 1, 1, 3]`	Passed
regression 5	`[0, 0, 0, 0]`	`[0, 0, 0, 0]`	Passed
regression 6	`[2, 0, 0, 0]`	`[2, 0, 0, 0]`	Passed
regression 7	`[4, 4, 2, 2]`	`[4, 4, 4, 4]`	Failed
regression 8	`[7, 0, 1, 3]`	`[7, 0, 1, 3]`	Passed
regression 9	`[6, 0, 0, 4]`	`[6, 0, 0, 4]`	Passed
variable positive run	`[3, 2, 0, 2]`	`[3, 2, 0, 2]`	Passed

SHA-256 / 8430a9efa80f52754799998c6a4bc14628b8c5c8d53a2c9430de7cf139aa8cb0

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
    previous=None
    prefix=suffix=best=0
    all_positive=True
    for block in blocks:
        k=len(block)
        p=next((i for i,x in enumerate(block) if x<=0),k)
        s=next((i for i,x in enumerate(reversed(block)) if x<=0),k)
        local=run=0
        for x in block:
            run=run+1 if x>0 else 0
            local=max(local,run)
        best=max(best,local,suffix+p)
        if all_positive: prefix+=p
        suffix=suffix+s
        all_positive=all_positive and p==k
        total+=k
    return [total,prefix,suffix,best]
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([[1, 1, 0, 1, 1, 0]],)), [6, 2, 0, 2])
check('regression 2', solve(*([[1, 0, 1, 0, 1, 1, 0]],)), [7, 1, 0, 2])
check('regression 3', solve(*([[1, 2], [3, 0, 4]],)), [5, 3, 1, 3])
check('regression 4', solve(*([[1, 0, 2], [3, 4, -1, 5]],)), [7, 1, 1, 3])
check('regression 5', solve(*([],)), [0, 0, 0, 0])
check('regression 6', solve(*([[], [0, 0], []],)), [2, 0, 0, 0])
check('regression 7', solve(*([[1, 2], [], [3, 4]],)), [4, 4, 4, 4])
check('regression 8', solve(*([[-1, 2, 3], [4, -1], [-2, 1]],)), [7, 0, 1, 3])
check('regression 9', solve(*([[0], [1, 1, 1], [1, 0]],)), [6, 0, 0, 4])
check("variable positive run",solve([[1]*N,[2]*N+[0]]),[2*N+1,2*N,0,2*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 fixture	Actual	Expected	Outcome
regression 1	`[6, 2, 0, 2]`	`[6, 2, 0, 2]`	Passed
regression 2	`[7, 1, 0, 2]`	`[7, 1, 0, 2]`	Passed
regression 3	`[5, 3, 3, 3]`	`[5, 3, 1, 3]`	Failed
regression 4	`[7, 1, 2, 3]`	`[7, 1, 1, 3]`	Failed
regression 5	`[0, 0, 0, 0]`	`[0, 0, 0, 0]`	Passed
regression 6	`[2, 0, 0, 0]`	`[2, 0, 0, 0]`	Passed
regression 7	`[4, 4, 4, 4]`	`[4, 4, 4, 4]`	Passed
regression 8	`[7, 0, 3, 3]`	`[7, 0, 1, 3]`	Failed
regression 9	`[6, 0, 3, 4]`	`[6, 0, 0, 4]`	Failed
variable positive run	`[3, 2, 1, 2]`	`[3, 2, 0, 2]`	Failed

SHA-256 / 46812cfe3c27da5ca3dcfcb8bf16f7c402a3c17834bcd5bca3212f13b9df9a9e

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
    previous=None
    prefix=suffix=best=0
    all_positive=True
    for block in blocks:
        k=len(block)
        p=next((i for i,x in enumerate(block) if x<=0),k)
        s=next((i for i,x in enumerate(reversed(block)) if x<=0),k)
        local=run=0
        for x in block:
            run=run+1 if x>0 else 0
            local=max(local,run)
        best=max(best,local,suffix+p)
        if all_positive: prefix+=p
        suffix=suffix+k if s==k else s
        all_positive=all_positive and p==k
        total+=k
    return [total,prefix,suffix,best]
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('regression 1', solve(*([[1, 1, 0, 1, 1, 0]],)), [6, 2, 0, 2])
check('regression 2', solve(*([[1, 0, 1, 0, 1, 1, 0]],)), [7, 1, 0, 2])
check('regression 3', solve(*([[1, 2], [3, 0, 4]],)), [5, 3, 1, 3])
check('regression 4', solve(*([[1, 0, 2], [3, 4, -1, 5]],)), [7, 1, 1, 3])
check('regression 5', solve(*([],)), [0, 0, 0, 0])
check('regression 6', solve(*([[], [0, 0], []],)), [2, 0, 0, 0])
check('regression 7', solve(*([[1, 2], [], [3, 4]],)), [4, 4, 4, 4])
check('regression 8', solve(*([[-1, 2, 3], [4, -1], [-2, 1]],)), [7, 0, 1, 3])
check('regression 9', solve(*([[0], [1, 1, 1], [1, 0]],)), [6, 0, 0, 4])
check("variable positive run",solve([[1]*N,[2]*N+[0]]),[2*N+1,2*N,0,2*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 fixture	Actual	Expected	Outcome
regression 1	`[6, 2, 0, 2]`	`[6, 2, 0, 2]`	Passed
regression 2	`[7, 1, 0, 2]`	`[7, 1, 0, 2]`	Passed
regression 3	`[5, 3, 1, 3]`	`[5, 3, 1, 3]`	Passed
regression 4	`[7, 1, 1, 3]`	`[7, 1, 1, 3]`	Passed
regression 5	`[0, 0, 0, 0]`	`[0, 0, 0, 0]`	Passed
regression 6	`[2, 0, 0, 0]`	`[2, 0, 0, 0]`	Passed
regression 7	`[4, 4, 4, 4]`	`[4, 4, 4, 4]`	Passed
regression 8	`[7, 0, 1, 3]`	`[7, 0, 1, 3]`	Passed
regression 9	`[6, 0, 0, 4]`	`[6, 0, 0, 4]`	Passed
variable positive run	`[3, 2, 0, 2]`	`[3, 2, 0, 2]`	Passed

SHA-256 / dd6d50270aaa436369bc50b1b751995ab08b42b558a8b268f92e74559b1aba29

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 2026-09-29T14:39:04.941054+00:00.

Case digest / b779d35fe095f6e1a7564f836cbe763d1671d9a55948d68dc864c4d4106e6993