FA-14661 / Numerics / Open access

Polynomial taylor shift: translation exponent · case 01

The exact polynomial taylor shift result violates the stated contract at translation exponent.

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

ROOT CAUSE

The translation exponent step uses hj instead of h(i-j) if j<=i else 0.

VERIFIED REPAIR

Use h**(i-j) if j<=i else 0 at the translation exponent step.

Unsuccessful approach: The partial repair h**i still violates the translation exponent 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[i]) * (math.comb(i,j) if j<=i else 0) * (h**j)
    return r
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([[-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], -2], [-3, 3, -1, 0]), ([[-1, -1, -1, 0], -1], [-1, 1, -1, 0])], [([[-1, -1, -1, -1], -1], [0, -2, 2, -1]), ([[-1, -1, -1, -1], 2], [-15, -17, -7, -1]), ([[-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, -1], 0], [-1, -1, -1, -1]), ([[-1, -1, -1, 0], 0], [-1, -1, -1, 0]), ([[-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, -1], 2], [-15, -17, -7, -1]), ([[-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, 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, 0], -2], [-3, 3, -1, 0]), ([[-1, -1, -1, 2], -2], [-19, 27, -13, 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	`[-4, 12, -16, 8]`	`[5, -9, 5, -1]`	Failed
explicit oracle 1	`[8, 24, 32, 16]`	`[30, 34, 14, 2]`	Failed
explicit oracle 2	`[-4, 6, -4, 1]`	`[0, -2, 2, -1]`	Failed
explicit oracle 3	`[-4, 0, 0, 0]`	`[-1, -1, -1, -1]`	Failed
explicit oracle 4	`[-4, -6, -4, -1]`	`[-4, -6, -4, -1]`	Passed
explicit oracle 5	`[-4, -12, -16, -8]`	`[-15, -17, -7, -1]`	Failed
explicit oracle 6	`[-3, 6, -4, 0]`	`[-3, 3, -1, 0]`	Failed
explicit oracle 7	`[-3, 3, -1, 0]`	`[-1, 1, -1, 0]`	Failed

SHA-256 / 805f43b78d38fea43411229a0d6a15fe112fafcea49801407e077da6bef6c088

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[i]) * (math.comb(i,j) if j<=i else 0) * (h**i)
    return r
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([[-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], -2], [-3, 3, -1, 0]), ([[-1, -1, -1, 0], -1], [-1, 1, -1, 0])], [([[-1, -1, -1, -1], -1], [0, -2, 2, -1]), ([[-1, -1, -1, -1], 2], [-15, -17, -7, -1]), ([[-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, -1], 0], [-1, -1, -1, -1]), ([[-1, -1, -1, 0], 0], [-1, -1, -1, 0]), ([[-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, -1], 2], [-15, -17, -7, -1]), ([[-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, 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, 0], -2], [-3, 3, -1, 0]), ([[-1, -1, -1, 2], -2], [-19, 27, -13, 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, 18, 20, 8]`	`[5, -9, 5, -1]`	Failed
explicit oracle 1	`[30, 68, 56, 16]`	`[30, 34, 14, 2]`	Failed
explicit oracle 2	`[0, 2, 2, 1]`	`[0, -2, 2, -1]`	Failed
explicit oracle 3	`[-1, 0, 0, 0]`	`[-1, -1, -1, -1]`	Failed
explicit oracle 4	`[-4, -6, -4, -1]`	`[-4, -6, -4, -1]`	Passed
explicit oracle 5	`[-15, -34, -28, -8]`	`[-15, -17, -7, -1]`	Failed
explicit oracle 6	`[-3, -6, -4, 0]`	`[-3, 3, -1, 0]`	Failed
explicit oracle 7	`[-1, -1, -1, 0]`	`[-1, 1, -1, 0]`	Failed

SHA-256 / 404e3731179cb6c8ff1c08828c33fa70ca5254655076b661d166a12cf6133d1e

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, -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], -2], [-3, 3, -1, 0]), ([[-1, -1, -1, 0], -1], [-1, 1, -1, 0])], [([[-1, -1, -1, -1], -1], [0, -2, 2, -1]), ([[-1, -1, -1, -1], 2], [-15, -17, -7, -1]), ([[-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, -1], 0], [-1, -1, -1, -1]), ([[-1, -1, -1, 0], 0], [-1, -1, -1, 0]), ([[-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, -1], 2], [-15, -17, -7, -1]), ([[-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, 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, 0], -2], [-3, 3, -1, 0]), ([[-1, -1, -1, 2], -2], [-19, 27, -13, 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, -1]`	`[5, -9, 5, -1]`	Passed
explicit oracle 1	`[30, 34, 14, 2]`	`[30, 34, 14, 2]`	Passed
explicit oracle 2	`[0, -2, 2, -1]`	`[0, -2, 2, -1]`	Passed
explicit oracle 3	`[-1, -1, -1, -1]`	`[-1, -1, -1, -1]`	Passed
explicit oracle 4	`[-4, -6, -4, -1]`	`[-4, -6, -4, -1]`	Passed
explicit oracle 5	`[-15, -17, -7, -1]`	`[-15, -17, -7, -1]`	Passed
explicit oracle 6	`[-3, 3, -1, 0]`	`[-3, 3, -1, 0]`	Passed
explicit oracle 7	`[-1, 1, -1, 0]`	`[-1, 1, -1, 0]`	Passed

SHA-256 / 81397b21497e7f4aaa838941a2cc1801ffb76f2a88e6b96238efcd94849cd784

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 2026-09-29T14:39:19.161184+00:00.

Case digest / c710328d16f157ddd71075f6ff605dcbdf63767232b0752d32ca892a8cb34032