FAILURE MAP
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FA-15626 / Numerics / Open access

Binary polynomial division: quotient monomial accumulation · case 01

The exact binary polynomial division result violates the stated contract at quotient monomial accumulation.

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

ROOT CAUSE

The quotient monomial accumulation step uses q^shift instead of q^(1<<shift).

VERIFIED REPAIR

Use q^(1<<shift) at the quotient monomial accumulation step.

Unsuccessful approach: The partial repair q|(1<<max(0,shift-1)) still violates the quotient monomial accumulation 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^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 = [[([1, 1], [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, 1], [2, 0]), ([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])], [([2, 2], [1, 0]), ([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])], [([2, 3], [1, 1]), ([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])], [([3, 1], [3, 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[0, 0][1, 0]Failed
explicit oracle 1[1, 0][2, 0]Failed
explicit oracle 2[0, 0][0, 0]Passed
explicit oracle 3[2, 3][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 / bf1ee162c310881e27c31aa019d34f113b53c934e2b8c7b8a2ed99afec7e427d

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<<max(0,shift-1))
     a=a^(b<<shift)
    return [q,a]
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([1, 1], [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, 1], [2, 0]), ([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])], [([2, 2], [1, 0]), ([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])], [([2, 3], [1, 1]), ([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])], [([3, 1], [3, 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[1, 0][2, 0]Failed
explicit oracle 2[0, 0][0, 0]Passed
explicit oracle 3[2, 3][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 / 48a9f64aa0b7a580ecc5f08fe35615562ab5cf26759b06b0ea5ed82889ff65a9

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 = [[([1, 1], [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, 1], [2, 0]), ([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])], [([2, 2], [1, 0]), ([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])], [([2, 3], [1, 1]), ([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])], [([3, 1], [3, 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 / 043b13da8ee9c3741a9cb6f2918b4ee5ad4e51993d27954434eca5431d3e1603

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 2026-09-29T14:39:28.733924+00:00.

Case digest / 389fa997c899ea38a3c6d570fdb80b4de620e764a57bba6c509b46b81822e758