FA-14651 / Numerics / Open access
Polynomial taylor shift: coefficient source · case 01
The exact polynomial taylor shift result violates the stated contract at coefficient source.
ROOT CAUSE
The coefficient source step uses A[j] instead of A[i].
VERIFIED REPAIR
Use A[i] at the coefficient source step.
Unsuccessful approach: The partial repair A[n-1-i] still violates the coefficient source invariant.
Case contract
Input [A,h] ascending integer polynomial, integer h; return coefficients of A(t+h), preserving array length.
Why this case matters
Exact discrete arithmetic with observable algorithmic state; no floating point approximation is used.
1 / The failure
Exit 1"""Failure Map reference implementation. Python standard library only."""
import json
import math
import itertools
from fractions import Fraction
N = 1
observations = []
def solve(x):
A,h=x
n=len(A);r=[0]*n
for i in range(n):
for j in range(i+1):
r[j]+=(A[j]) * (math.comb(i,j) if j<=i else 0) * (h**(i-j) if j<=i else 0)
return r
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([[-1, -1, -1, 0], -2], [-3, 3, -1, 0]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, -1, -1], -1], [0, -2, 2, -1]), ([[-1, -1, -1, -1], 0], [-1, -1, -1, -1]), ([[-1, -1, -1, -1], 1], [-4, -6, -4, -1]), ([[-1, -1, -1, -1], 2], [-15, -17, -7, -1]), ([[-1, -1, -1, 0], -1], [-1, 1, -1, 0])], [([[-1, -1, -1, 0], -1], [-1, 1, -1, 0]), ([[-1, -1, -1, 0], 1], [-3, -3, -1, 0]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, -1, 2], 0], [-1, -1, -1, 2]), ([[-1, -1, -1, 2], 1], [-1, 3, 5, 2]), ([[-1, -1, -1, 2], 2], [9, 19, 11, 2]), ([[-1, -1, 0, -1], -2], [9, -13, 6, -1])], [([[-1, -1, -1, 0], 1], [-3, -3, -1, 0]), ([[-1, -1, -1, 1], -1], [-2, 4, -4, 1]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, 0, 1], 2], [5, 11, 6, 1]), ([[-1, -1, 0, 2], -2], [-15, 23, -12, 2]), ([[-1, -1, 0, 2], -1], [-2, 5, -6, 2]), ([[-1, -1, 0, 2], 0], [-1, -1, 0, 2])], [([[-1, -1, -1, 0], 2], [-7, -5, -1, 0]), ([[-1, -1, -1, 1], 2], [1, 7, 5, 1]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, 1, 1], -1], [0, 0, -2, 1]), ([[-1, -1, 1, 1], 0], [-1, -1, 1, 1]), ([[-1, -1, 1, 1], 1], [0, 4, 4, 1]), ([[-1, -1, 1, 1], 2], [9, 15, 7, 1])], [([[-1, -1, -1, 1], -2], [-11, 15, -7, 1]), ([[-1, -1, -1, 2], 0], [-1, -1, -1, 2]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, 2, 0], 1], [0, 3, 2, 0]), ([[-1, -1, 2, 0], 2], [5, 7, 2, 0]), ([[-1, -1, 2, 1], -2], [1, 3, -4, 1]), ([[-1, -1, 2, 1], -1], [1, -2, -1, 1])]]
for i, (args, expected) in enumerate(fixtures[N-1]):
check("explicit oracle %d" % i, solve(args), expected)
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 |
|---|---|---|---|
| explicit oracle 0 | [5, -9, 5, 0] | [-3, 3, -1, 0] | Failed |
| explicit oracle 1 | [5, -9, 5, -1] | [5, -9, 5, -1] | Passed |
| explicit oracle 2 | [30, 34, 14, 2] | [30, 34, 14, 2] | Passed |
| explicit oracle 3 | [0, -2, 2, -1] | [0, -2, 2, -1] | Passed |
| explicit oracle 4 | [-1, -1, -1, -1] | [-1, -1, -1, -1] | Passed |
| explicit oracle 5 | [-4, -6, -4, -1] | [-4, -6, -4, -1] | Passed |
| explicit oracle 6 | [-15, -17, -7, -1] | [-15, -17, -7, -1] | Passed |
| explicit oracle 7 | [0, -2, 2, 0] | [-1, 1, -1, 0] | Failed |
SHA-256 / bb9c3aea771b343f1f3cc713d4263c37adf0741404edf8f9176a12dea84d3602
2 / The unsuccessful fix
Exit 1"""Failure Map reference implementation. Python standard library only."""
import json
import math
import itertools
from fractions import Fraction
N = 1
observations = []
def solve(x):
A,h=x
n=len(A);r=[0]*n
for i in range(n):
for j in range(i+1):
r[j]+=(A[n-1-i]) * (math.comb(i,j) if j<=i else 0) * (h**(i-j) if j<=i else 0)
return r
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([[-1, -1, -1, 0], -2], [-3, 3, -1, 0]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, -1, -1], -1], [0, -2, 2, -1]), ([[-1, -1, -1, -1], 0], [-1, -1, -1, -1]), ([[-1, -1, -1, -1], 1], [-4, -6, -4, -1]), ([[-1, -1, -1, -1], 2], [-15, -17, -7, -1]), ([[-1, -1, -1, 0], -1], [-1, 1, -1, 0])], [([[-1, -1, -1, 0], -1], [-1, 1, -1, 0]), ([[-1, -1, -1, 0], 1], [-3, -3, -1, 0]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, -1, 2], 0], [-1, -1, -1, 2]), ([[-1, -1, -1, 2], 1], [-1, 3, 5, 2]), ([[-1, -1, -1, 2], 2], [9, 19, 11, 2]), ([[-1, -1, 0, -1], -2], [9, -13, 6, -1])], [([[-1, -1, -1, 0], 1], [-3, -3, -1, 0]), ([[-1, -1, -1, 1], -1], [-2, 4, -4, 1]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, 0, 1], 2], [5, 11, 6, 1]), ([[-1, -1, 0, 2], -2], [-15, 23, -12, 2]), ([[-1, -1, 0, 2], -1], [-2, 5, -6, 2]), ([[-1, -1, 0, 2], 0], [-1, -1, 0, 2])], [([[-1, -1, -1, 0], 2], [-7, -5, -1, 0]), ([[-1, -1, -1, 1], 2], [1, 7, 5, 1]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, 1, 1], -1], [0, 0, -2, 1]), ([[-1, -1, 1, 1], 0], [-1, -1, 1, 1]), ([[-1, -1, 1, 1], 1], [0, 4, 4, 1]), ([[-1, -1, 1, 1], 2], [9, 15, 7, 1])], [([[-1, -1, -1, 1], -2], [-11, 15, -7, 1]), ([[-1, -1, -1, 2], 0], [-1, -1, -1, 2]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, 2, 0], 1], [0, 3, 2, 0]), ([[-1, -1, 2, 0], 2], [5, 7, 2, 0]), ([[-1, -1, 2, 1], -2], [1, 3, -4, 1]), ([[-1, -1, 2, 1], -1], [1, -2, -1, 1])]]
for i, (args, expected) in enumerate(fixtures[N-1]):
check("explicit oracle %d" % i, solve(args), expected)
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 |
|---|---|---|---|
| explicit oracle 0 | [6, -9, 5, -1] | [-3, 3, -1, 0] | Failed |
| explicit oracle 1 | [5, -9, 5, -1] | [5, -9, 5, -1] | Passed |
| explicit oracle 2 | [30, 34, 14, 2] | [30, 34, 14, 2] | Passed |
| explicit oracle 3 | [0, -2, 2, -1] | [0, -2, 2, -1] | Passed |
| explicit oracle 4 | [-1, -1, -1, -1] | [-1, -1, -1, -1] | Passed |
| explicit oracle 5 | [-4, -6, -4, -1] | [-4, -6, -4, -1] | Passed |
| explicit oracle 6 | [-15, -17, -7, -1] | [-15, -17, -7, -1] | Passed |
| explicit oracle 7 | [1, -2, 2, -1] | [-1, 1, -1, 0] | Failed |
SHA-256 / 7c48208ae7a8b5594855c02e3436c07ca53ee574169fca87de714000f80b6b9f
3 / The verified repair
Exit 0"""Failure Map reference implementation. Python standard library only."""
import json
import math
import itertools
from fractions import Fraction
N = 1
observations = []
def solve(x):
A,h=x
n=len(A);r=[0]*n
for i in range(n):
for j in range(i+1):
r[j]+=(A[i]) * (math.comb(i,j) if j<=i else 0) * (h**(i-j) if j<=i else 0)
return r
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([[-1, -1, -1, 0], -2], [-3, 3, -1, 0]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, -1, -1], -1], [0, -2, 2, -1]), ([[-1, -1, -1, -1], 0], [-1, -1, -1, -1]), ([[-1, -1, -1, -1], 1], [-4, -6, -4, -1]), ([[-1, -1, -1, -1], 2], [-15, -17, -7, -1]), ([[-1, -1, -1, 0], -1], [-1, 1, -1, 0])], [([[-1, -1, -1, 0], -1], [-1, 1, -1, 0]), ([[-1, -1, -1, 0], 1], [-3, -3, -1, 0]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, -1, 2], 0], [-1, -1, -1, 2]), ([[-1, -1, -1, 2], 1], [-1, 3, 5, 2]), ([[-1, -1, -1, 2], 2], [9, 19, 11, 2]), ([[-1, -1, 0, -1], -2], [9, -13, 6, -1])], [([[-1, -1, -1, 0], 1], [-3, -3, -1, 0]), ([[-1, -1, -1, 1], -1], [-2, 4, -4, 1]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, 0, 1], 2], [5, 11, 6, 1]), ([[-1, -1, 0, 2], -2], [-15, 23, -12, 2]), ([[-1, -1, 0, 2], -1], [-2, 5, -6, 2]), ([[-1, -1, 0, 2], 0], [-1, -1, 0, 2])], [([[-1, -1, -1, 0], 2], [-7, -5, -1, 0]), ([[-1, -1, -1, 1], 2], [1, 7, 5, 1]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, 1, 1], -1], [0, 0, -2, 1]), ([[-1, -1, 1, 1], 0], [-1, -1, 1, 1]), ([[-1, -1, 1, 1], 1], [0, 4, 4, 1]), ([[-1, -1, 1, 1], 2], [9, 15, 7, 1])], [([[-1, -1, -1, 1], -2], [-11, 15, -7, 1]), ([[-1, -1, -1, 2], 0], [-1, -1, -1, 2]), ([[-1, -1, -1, -1], -2], [5, -9, 5, -1]), ([[2, 2, 2, 2], 2], [30, 34, 14, 2]), ([[-1, -1, 2, 0], 1], [0, 3, 2, 0]), ([[-1, -1, 2, 0], 2], [5, 7, 2, 0]), ([[-1, -1, 2, 1], -2], [1, 3, -4, 1]), ([[-1, -1, 2, 1], -1], [1, -2, -1, 1])]]
for i, (args, expected) in enumerate(fixtures[N-1]):
check("explicit oracle %d" % i, solve(args), expected)
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 |
|---|---|---|---|
| explicit oracle 0 | [-3, 3, -1, 0] | [-3, 3, -1, 0] | Passed |
| explicit oracle 1 | [5, -9, 5, -1] | [5, -9, 5, -1] | Passed |
| explicit oracle 2 | [30, 34, 14, 2] | [30, 34, 14, 2] | Passed |
| explicit oracle 3 | [0, -2, 2, -1] | [0, -2, 2, -1] | Passed |
| explicit oracle 4 | [-1, -1, -1, -1] | [-1, -1, -1, -1] | Passed |
| explicit oracle 5 | [-4, -6, -4, -1] | [-4, -6, -4, -1] | Passed |
| explicit oracle 6 | [-15, -17, -7, -1] | [-15, -17, -7, -1] | Passed |
| explicit oracle 7 | [-1, 1, -1, 0] | [-1, 1, -1, 0] | Passed |
SHA-256 / b2b2344e5bd2db4efba9d1add499a7f0b4632e887f9062f273f831e5f0258a4e
Verification & scope
A deterministic bounded teaching model. Inputs are restricted to the explicit contract; this is not a production algebra library. 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 2026-09-29T14:39:19.106785+00:00.
Case digest / 426ca19fe86c2bfd655e50ef8c3a02f2580dcebc32932acb85323560bb253f20