FA-14656 / Numerics / Open access

Polynomial taylor shift: binomial translation weight · case 01

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

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

ROOT CAUSE

The binomial translation weight step uses 1 instead of math.comb(i,j) if j<=i else 0.

VERIFIED REPAIR

Use math.comb(i,j) if j<=i else 0 at the binomial translation weight step.

Unsuccessful approach: The partial repair math.comb(i,j) if 0<j<i else 0 still violates the binomial translation weight 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]) * (1) * (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], 1], [-4, -6, -4, -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], 1], [-4, -6, -4, -1]), ([[-1, -1, -1, 0], -1], [-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, 0], 2], [-7, -5, -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, 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, 1], 0], [-1, -1, -1, 1]), ([[-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, -3, 1, -1]`	`[5, -9, 5, -1]`	Failed
explicit oracle 1	`[30, 14, 6, 2]`	`[30, 34, 14, 2]`	Failed
explicit oracle 2	`[0, -1, 0, -1]`	`[0, -2, 2, -1]`	Failed
explicit oracle 3	`[-1, -1, -1, -1]`	`[-1, -1, -1, -1]`	Passed
explicit oracle 4	`[-4, -3, -2, -1]`	`[-4, -6, -4, -1]`	Failed
explicit oracle 5	`[-15, -7, -3, -1]`	`[-15, -17, -7, -1]`	Failed
explicit oracle 6	`[-3, 1, -1, 0]`	`[-3, 3, -1, 0]`	Failed
explicit oracle 7	`[-1, 0, -1, 0]`	`[-1, 1, -1, 0]`	Failed

SHA-256 / 0cd0c54dc2eb5e731ed573281a6c8f57437cbc8394276b7e1587a6f5f7e0592d

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 0<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], 1], [-4, -6, -4, -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], 1], [-4, -6, -4, -1]), ([[-1, -1, -1, 0], -1], [-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, 0], 2], [-7, -5, -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, 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, 1], 0], [-1, -1, -1, 1]), ([[-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	`[0, -8, 6, 0]`	`[5, -9, 5, -1]`	Failed
explicit oracle 1	`[0, 32, 12, 0]`	`[30, 34, 14, 2]`	Failed
explicit oracle 2	`[0, -1, 3, 0]`	`[0, -2, 2, -1]`	Failed
explicit oracle 3	`[0, 0, 0, 0]`	`[-1, -1, -1, -1]`	Failed
explicit oracle 4	`[0, -5, -3, 0]`	`[-4, -6, -4, -1]`	Failed
explicit oracle 5	`[0, -16, -6, 0]`	`[-15, -17, -7, -1]`	Failed
explicit oracle 6	`[0, 4, 0, 0]`	`[-3, 3, -1, 0]`	Failed
explicit oracle 7	`[0, 2, 0, 0]`	`[-1, 1, -1, 0]`	Failed

SHA-256 / 6ba43d2ea9e9014bcf7319d78abff4a648f1384aaf732b52821839c1fdf8dd37

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], 1], [-4, -6, -4, -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], 1], [-4, -6, -4, -1]), ([[-1, -1, -1, 0], -1], [-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, 0], 2], [-7, -5, -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, 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, 1], 0], [-1, -1, -1, 1]), ([[-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 / e9e62bf3b6ece460ab5efb7e2703b3b112fc4d419e730d21b2d19633e911cb98

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 2026-09-29T14:39:19.108723+00:00.

Case digest / f9a189e2478f463aabefef5b9c59f624796817580441f64d119661bb3a87e947