FA-15711 / Numerics / Open access

Binary field minimal polynomial: monic linear term contribution · case 01

The exact binary field minimal polynomial result violates the stated contract at monic linear term contribution.

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

ROOT CAUSE

The monic linear term contribution step uses d[j+1]^root instead of d[j+1]^b.

VERIFIED REPAIR

Use d[j+1]^b at the monic linear term contribution step.

Unsuccessful approach: The partial repair d[j+1]^mul(b,root) still violates the monic linear term contribution invariant.

Case contract

Input [a,irreducible bitmask]; return ascending 0/1 coefficients of the minimal polynomial of a over F2, through its distinct Frobenius orbit.

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,poly=x;n=poly.bit_length()-1
    def mul(a,b):
     z=0
     while b:
      if b&1:z^=a
      b>>=1;a<<=1
      if a&(1<<n):a^=poly
     return z
    orbit=[];v=a
    for _ in range(n+1):
     if v in orbit:break
     orbit.append(v);v=mul(v,v)
    c=[1]
    for root in orbit:
     d=[0]*(len(c)+1)
     for j,b in enumerate(c):
      d[j]=d[j]^mul(b,root)
      d[j+1]=d[j+1]^root
     c=d
    return c
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([1, 7], [1, 1]), ([2, 7], [1, 1, 1]), ([3, 7], [1, 1, 1]), ([0, 11], [0, 1]), ([1, 11], [1, 1]), ([2, 11], [1, 1, 0, 1])], [([2, 7], [1, 1, 1]), ([0, 11], [0, 1]), ([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([5, 19], [1, 1, 0, 0, 1]), ([6, 19], [1, 1, 1]), ([7, 19], [1, 1, 1]), ([8, 19], [1, 1, 1, 1, 1])], [([3, 7], [1, 1, 1]), ([4, 11], [1, 1, 0, 1]), ([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([6, 25], [1, 1, 0, 0, 1]), ([7, 25], [1, 1, 0, 0, 1]), ([8, 25], [1, 1, 1, 1, 1]), ([9, 25], [1, 0, 0, 1, 1])], [([0, 11], [0, 1]), ([7, 11], [1, 0, 1, 1]), ([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([7, 31], [1, 1, 0, 0, 1]), ([8, 31], [1, 1, 1, 1, 1]), ([9, 31], [1, 0, 0, 1, 1]), ([10, 31], [1, 1, 0, 0, 1])], [([2, 11], [1, 1, 0, 1]), ([3, 19], [1, 1, 0, 0, 1]), ([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([8, 37], [1, 0, 1, 1, 1, 1]), ([9, 37], [1, 0, 0, 1, 0, 1]), ([10, 37], [1, 0, 1, 1, 1, 1]), ([11, 37], [1, 0, 0, 1, 0, 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, 0]`	`[0, 1]`	Failed
explicit oracle 1	`[1, 0, 1, 26, 9, 15]`	`[1, 0, 0, 1, 0, 1]`	Failed
explicit oracle 2	`[1, 1]`	`[1, 1]`	Passed
explicit oracle 3	`[1, 2, 3]`	`[1, 1, 1]`	Failed
explicit oracle 4	`[1, 3, 2]`	`[1, 1, 1]`	Failed
explicit oracle 5	`[0, 0]`	`[0, 1]`	Failed
explicit oracle 6	`[1, 1]`	`[1, 1]`	Passed
explicit oracle 7	`[1, 2, 3, 6]`	`[1, 1, 0, 1]`	Failed

SHA-256 / 4d013d4d28b06a311c131ac3b81d0f3252cf85eeec392960b7c14aa501be8bc1

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,poly=x;n=poly.bit_length()-1
    def mul(a,b):
     z=0
     while b:
      if b&1:z^=a
      b>>=1;a<<=1
      if a&(1<<n):a^=poly
     return z
    orbit=[];v=a
    for _ in range(n+1):
     if v in orbit:break
     orbit.append(v);v=mul(v,v)
    c=[1]
    for root in orbit:
     d=[0]*(len(c)+1)
     for j,b in enumerate(c):
      d[j]=d[j]^mul(b,root)
      d[j+1]=d[j+1]^mul(b,root)
     c=d
    return c
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([1, 7], [1, 1]), ([2, 7], [1, 1, 1]), ([3, 7], [1, 1, 1]), ([0, 11], [0, 1]), ([1, 11], [1, 1]), ([2, 11], [1, 1, 0, 1])], [([2, 7], [1, 1, 1]), ([0, 11], [0, 1]), ([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([5, 19], [1, 1, 0, 0, 1]), ([6, 19], [1, 1, 1]), ([7, 19], [1, 1, 1]), ([8, 19], [1, 1, 1, 1, 1])], [([3, 7], [1, 1, 1]), ([4, 11], [1, 1, 0, 1]), ([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([6, 25], [1, 1, 0, 0, 1]), ([7, 25], [1, 1, 0, 0, 1]), ([8, 25], [1, 1, 1, 1, 1]), ([9, 25], [1, 0, 0, 1, 1])], [([0, 11], [0, 1]), ([7, 11], [1, 0, 1, 1]), ([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([7, 31], [1, 1, 0, 0, 1]), ([8, 31], [1, 1, 1, 1, 1]), ([9, 31], [1, 0, 0, 1, 1]), ([10, 31], [1, 1, 0, 0, 1])], [([2, 11], [1, 1, 0, 1]), ([3, 19], [1, 1, 0, 0, 1]), ([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([8, 37], [1, 0, 1, 1, 1, 1]), ([9, 37], [1, 0, 0, 1, 0, 1]), ([10, 37], [1, 0, 1, 1, 1, 1]), ([11, 37], [1, 0, 0, 1, 0, 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, 0]`	`[0, 1]`	Failed
explicit oracle 1	`[1, 1, 0, 0, 1, 1]`	`[1, 0, 0, 1, 0, 1]`	Failed
explicit oracle 2	`[1, 1]`	`[1, 1]`	Passed
explicit oracle 3	`[1, 0, 1]`	`[1, 1, 1]`	Failed
explicit oracle 4	`[1, 0, 1]`	`[1, 1, 1]`	Failed
explicit oracle 5	`[0, 0]`	`[0, 1]`	Failed
explicit oracle 6	`[1, 1]`	`[1, 1]`	Passed
explicit oracle 7	`[1, 1, 1, 1]`	`[1, 1, 0, 1]`	Failed

SHA-256 / ebbf755c7af7069d3b3d559aad1cebf6fca3a8b9dfe45d0bca010da7977f64c4

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,poly=x;n=poly.bit_length()-1
    def mul(a,b):
     z=0
     while b:
      if b&1:z^=a
      b>>=1;a<<=1
      if a&(1<<n):a^=poly
     return z
    orbit=[];v=a
    for _ in range(n+1):
     if v in orbit:break
     orbit.append(v);v=mul(v,v)
    c=[1]
    for root in orbit:
     d=[0]*(len(c)+1)
     for j,b in enumerate(c):
      d[j]=d[j]^mul(b,root)
      d[j+1]=d[j+1]^b
     c=d
    return c
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([1, 7], [1, 1]), ([2, 7], [1, 1, 1]), ([3, 7], [1, 1, 1]), ([0, 11], [0, 1]), ([1, 11], [1, 1]), ([2, 11], [1, 1, 0, 1])], [([2, 7], [1, 1, 1]), ([0, 11], [0, 1]), ([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([5, 19], [1, 1, 0, 0, 1]), ([6, 19], [1, 1, 1]), ([7, 19], [1, 1, 1]), ([8, 19], [1, 1, 1, 1, 1])], [([3, 7], [1, 1, 1]), ([4, 11], [1, 1, 0, 1]), ([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([6, 25], [1, 1, 0, 0, 1]), ([7, 25], [1, 1, 0, 0, 1]), ([8, 25], [1, 1, 1, 1, 1]), ([9, 25], [1, 0, 0, 1, 1])], [([0, 11], [0, 1]), ([7, 11], [1, 0, 1, 1]), ([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([7, 31], [1, 1, 0, 0, 1]), ([8, 31], [1, 1, 1, 1, 1]), ([9, 31], [1, 0, 0, 1, 1]), ([10, 31], [1, 1, 0, 0, 1])], [([2, 11], [1, 1, 0, 1]), ([3, 19], [1, 1, 0, 0, 1]), ([0, 7], [0, 1]), ([31, 37], [1, 0, 0, 1, 0, 1]), ([8, 37], [1, 0, 1, 1, 1, 1]), ([9, 37], [1, 0, 0, 1, 0, 1]), ([10, 37], [1, 0, 1, 1, 1, 1]), ([11, 37], [1, 0, 0, 1, 0, 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, 1]`	`[0, 1]`	Passed
explicit oracle 1	`[1, 0, 0, 1, 0, 1]`	`[1, 0, 0, 1, 0, 1]`	Passed
explicit oracle 2	`[1, 1]`	`[1, 1]`	Passed
explicit oracle 3	`[1, 1, 1]`	`[1, 1, 1]`	Passed
explicit oracle 4	`[1, 1, 1]`	`[1, 1, 1]`	Passed
explicit oracle 5	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 6	`[1, 1]`	`[1, 1]`	Passed
explicit oracle 7	`[1, 1, 0, 1]`	`[1, 1, 0, 1]`	Passed

SHA-256 / cdd9ced0b3eb3d05ad8b1d9fafc5bbcb3f13c3a36f44a64d006eb6a55c7d2330

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 2026-09-29T14:39:29.474246+00:00.

Case digest / db9c91751a902e10be0492fcc92838ab4b072726e07b86218650fc1ead029292