FAILURE MAP
← Case archive

FA-15621 / Numerics / Open access

Binary polynomial division: leading term shift · case 01

The exact binary polynomial division result violates the stated contract at leading term shift.

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

ROOT CAUSE

The leading term shift step uses a.bit_length()-1 instead of a.bit_length()-b.bit_length().

VERIFIED REPAIR

Use a.bit_length()-b.bit_length() at the leading term shift step.

Unsuccessful approach: The partial repair max(0,a.bit_length()-b.bit_length()-1) still violates the leading term shift invariant.

Case contract

Input [a,b] integer bitmasks for F2 polynomials with b>0; return [quotient,remainder] of carryless division. Bounds: a,b<2^32.

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,b=x;q=0
    for _ in range(40):
     if a.bit_length()<b.bit_length():break
     shift=a.bit_length()-1
     if shift<0:return None
     q=q^(1<<shift)
     a=a^(b<<shift)
    return [q,a]
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 2], [1, 0]), ([2, 1], [2, 0]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([0, 2], [0, 0]), ([0, 3], [0, 0]), ([0, 4], [0, 0]), ([0, 5], [0, 0])], [([2, 3], [1, 1]), ([4, 2], [2, 0]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([0, 18], [0, 0]), ([0, 19], [0, 0]), ([0, 20], [0, 0]), ([0, 21], [0, 0])], [([3, 2], [1, 1]), ([5, 2], [2, 1]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([1, 4], [0, 1]), ([1, 5], [0, 1]), ([1, 6], [0, 1]), ([1, 7], [0, 1])], [([3, 3], [1, 0]), ([6, 2], [3, 0]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([1, 21], [0, 1]), ([1, 22], [0, 1]), ([1, 23], [0, 1]), ([1, 24], [0, 1])], [([4, 2], [2, 0]), ([7, 2], [3, 1]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([2, 7], [0, 2]), ([2, 8], [0, 2]), ([2, 9], [0, 2]), ([2, 10], [0, 2])]]
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 fixtureActualExpectedOutcome
explicit oracle 0[2199023255550, 4398046511102][1, 0]Failed
explicit oracle 1[2, 0][2, 0]Passed
explicit oracle 2[0, 0][0, 0]Passed
explicit oracle 3[6235740319278519117669055286256140883865312183360, 180836469259077054412402603301428085632094053317567][4, 3]Failed
explicit oracle 4[0, 0][0, 0]Passed
explicit oracle 5[0, 0][0, 0]Passed
explicit oracle 6[0, 0][0, 0]Passed
explicit oracle 7[0, 0][0, 0]Passed

SHA-256 / f0245bc180e77329ae52f9e5e8e201bf3f86942f3da2fc2a004f11071fb93823

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,b=x;q=0
    for _ in range(40):
     if a.bit_length()<b.bit_length():break
     shift=max(0,a.bit_length()-b.bit_length()-1)
     if shift<0:return None
     q=q^(1<<shift)
     a=a^(b<<shift)
    return [q,a]
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 2], [1, 0]), ([2, 1], [2, 0]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([0, 2], [0, 0]), ([0, 3], [0, 0]), ([0, 4], [0, 0]), ([0, 5], [0, 0])], [([2, 3], [1, 1]), ([4, 2], [2, 0]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([0, 18], [0, 0]), ([0, 19], [0, 0]), ([0, 20], [0, 0]), ([0, 21], [0, 0])], [([3, 2], [1, 1]), ([5, 2], [2, 1]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([1, 4], [0, 1]), ([1, 5], [0, 1]), ([1, 6], [0, 1]), ([1, 7], [0, 1])], [([3, 3], [1, 0]), ([6, 2], [3, 0]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([1, 21], [0, 1]), ([1, 22], [0, 1]), ([1, 23], [0, 1]), ([1, 24], [0, 1])], [([4, 2], [2, 0]), ([7, 2], [3, 1]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([2, 7], [0, 2]), ([2, 8], [0, 2]), ([2, 9], [0, 2]), ([2, 10], [0, 2])]]
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 fixtureActualExpectedOutcome
explicit oracle 0[1, 0][1, 0]Passed
explicit oracle 1[0, 2][2, 0]Failed
explicit oracle 2[0, 0][0, 0]Passed
explicit oracle 3[0, 127][4, 3]Failed
explicit oracle 4[0, 0][0, 0]Passed
explicit oracle 5[0, 0][0, 0]Passed
explicit oracle 6[0, 0][0, 0]Passed
explicit oracle 7[0, 0][0, 0]Passed

SHA-256 / 383a36e2945430f784de8ccf34e51fb34fe49474cd3ceb84a3a9a2be1f41b4df

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,b=x;q=0
    for _ in range(40):
     if a.bit_length()<b.bit_length():break
     shift=a.bit_length()-b.bit_length()
     if shift<0:return None
     q=q^(1<<shift)
     a=a^(b<<shift)
    return [q,a]
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 2], [1, 0]), ([2, 1], [2, 0]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([0, 2], [0, 0]), ([0, 3], [0, 0]), ([0, 4], [0, 0]), ([0, 5], [0, 0])], [([2, 3], [1, 1]), ([4, 2], [2, 0]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([0, 18], [0, 0]), ([0, 19], [0, 0]), ([0, 20], [0, 0]), ([0, 21], [0, 0])], [([3, 2], [1, 1]), ([5, 2], [2, 1]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([1, 4], [0, 1]), ([1, 5], [0, 1]), ([1, 6], [0, 1]), ([1, 7], [0, 1])], [([3, 3], [1, 0]), ([6, 2], [3, 0]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([1, 21], [0, 1]), ([1, 22], [0, 1]), ([1, 23], [0, 1]), ([1, 24], [0, 1])], [([4, 2], [2, 0]), ([7, 2], [3, 1]), ([0, 1], [0, 0]), ([127, 31], [4, 3]), ([2, 7], [0, 2]), ([2, 8], [0, 2]), ([2, 9], [0, 2]), ([2, 10], [0, 2])]]
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 fixtureActualExpectedOutcome
explicit oracle 0[1, 0][1, 0]Passed
explicit oracle 1[2, 0][2, 0]Passed
explicit oracle 2[0, 0][0, 0]Passed
explicit oracle 3[4, 3][4, 3]Passed
explicit oracle 4[0, 0][0, 0]Passed
explicit oracle 5[0, 0][0, 0]Passed
explicit oracle 6[0, 0][0, 0]Passed
explicit oracle 7[0, 0][0, 0]Passed

SHA-256 / 0781959e260db8538699427e6b1a6a29cdbce1bde25d71621dbd5fd517da0760

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 2026-09-29T14:39:28.731440+00:00.

Case digest / 1962bdee925432b831ef98ec80574afac59bf4d5a46958cf9318e9ce81ee3577