FA-15631 / Numerics / Open access

Binary polynomial division: carryless leading cancellation · case 01

The exact binary polynomial division result violates the stated contract at carryless leading cancellation.

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

ROOT CAUSE

The carryless leading cancellation step uses a-(b<<shift) instead of a^(b<<shift).

VERIFIED REPAIR

Use a^(b<<shift) at the carryless leading cancellation step.

Unsuccessful approach: The partial repair a^(b>>shift) still violates the carryless leading cancellation 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()-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, 3], [1, 1]), ([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])], [([4, 3], [3, 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])], [([4, 5], [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])], [([4, 6], [1, 2]), ([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, 7], [1, 3]), ([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 fixture	Actual	Expected	Outcome
explicit oracle 0	`[1, -1]`	`[1, 1]`	Failed
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 / 80796fb245bf84e9add4d7e0637bfc412e1f1e9c75df6898b5ddcb90b4230c3f

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=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, 3], [1, 1]), ([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])], [([4, 3], [3, 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])], [([4, 5], [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])], [([4, 6], [1, 2]), ([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, 7], [1, 3]), ([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 fixture	Actual	Expected	Outcome
explicit oracle 0	`[1, 1]`	`[1, 1]`	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 / dc8be2214c00d2107b2972a7be29b17094c54185b1adead9fd590f496f1a6513

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, 3], [1, 1]), ([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])], [([4, 3], [3, 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])], [([4, 5], [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])], [([4, 6], [1, 2]), ([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, 7], [1, 3]), ([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 fixture	Actual	Expected	Outcome
explicit oracle 0	`[1, 1]`	`[1, 1]`	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 / 060822c7d3e390ad88c6c4dbdde3c108979b99660e92feda0a485d31d37eaa20

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 2026-09-29T14:39:28.733780+00:00.

Case digest / d72eb8af69821239529b6c424d652f676954418073d9105e1fe6ad089d07f1a6