FAILURE MAP
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FA-15651 / Numerics / Open access

Binary field inversion: Euclidean polynomial cancellation · case 01

The exact binary field inversion result violates the stated contract at Euclidean polynomial cancellation.

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

ROOT CAUSE

The Euclidean polynomial cancellation step uses u^(v>>j) instead of u^(v<<j).

THE FAILURE

The Euclidean polynomial cancellation step uses u^(v>>j) instead of u^(v<<j).

Unsuccessful approach: The partial repair u-(v<<j) still violates the Euclidean polynomial cancellation invariant.

Case contract

Input [a,irreducible bitmask], a nonzero with degree below modulus; return multiplicative inverse in binary field. Bounds: modulus degree is between two and six.

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,p=x;u=a;v=p;g1=1;g2=0
    for _ in range(60):
     if u==1:break
     if u==0:return None
     j=u.bit_length()-v.bit_length()
     if j<0:
      u,v=v,u;g1,g2=g2,g1;j=-j
     u=u^(v>>j)
     g1=g1^(g2<<j)
    for _ in range(60):
     if g1.bit_length()<p.bit_length():break
     g1=g1^(p<<(g1.bit_length()-p.bit_length()))
    return g1
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 7], 3), ([3, 11], 6), ([1, 7], 1), ([63, 67], 32), ([3, 7], 2), ([1, 11], 1), ([2, 11], 5), ([4, 11], 7)], [([3, 7], 2), ([3, 19], 14), ([1, 7], 1), ([63, 67], 32), ([8, 19], 15), ([9, 19], 2), ([10, 19], 12), ([11, 19], 5)], [([2, 11], 5), ([6, 19], 7), ([1, 7], 1), ([63, 67], 32), ([10, 25], 11), ([11, 25], 10), ([12, 25], 2), ([13, 25], 9)], [([3, 11], 6), ([10, 19], 12), ([1, 7], 1), ([63, 67], 32), ([12, 31], 13), ([13, 31], 12), ([14, 31], 11), ([15, 31], 2)], [([4, 11], 7), ([14, 19], 3), ([1, 7], 1), ([63, 67], 32), ([14, 37], 6), ([15, 37], 13), ([16, 37], 11), ([17, 37], 24)]]
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 003Failed
explicit oracle 106Failed
explicit oracle 211Passed
explicit oracle 3032Failed
explicit oracle 402Failed
explicit oracle 511Passed
explicit oracle 605Failed
explicit oracle 707Failed

SHA-256 / 76ca606a849f83fe9fa5360e8d1df3ff079d23008d5c885a8b8ffdb97a30c03f

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,p=x;u=a;v=p;g1=1;g2=0
    for _ in range(60):
     if u==1:break
     if u==0:return None
     j=u.bit_length()-v.bit_length()
     if j<0:
      u,v=v,u;g1,g2=g2,g1;j=-j
     u=u-(v<<j)
     g1=g1^(g2<<j)
    for _ in range(60):
     if g1.bit_length()<p.bit_length():break
     g1=g1^(p<<(g1.bit_length()-p.bit_length()))
    return g1
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 7], 3), ([3, 11], 6), ([1, 7], 1), ([63, 67], 32), ([3, 7], 2), ([1, 11], 1), ([2, 11], 5), ([4, 11], 7)], [([3, 7], 2), ([3, 19], 14), ([1, 7], 1), ([63, 67], 32), ([8, 19], 15), ([9, 19], 2), ([10, 19], 12), ([11, 19], 5)], [([2, 11], 5), ([6, 19], 7), ([1, 7], 1), ([63, 67], 32), ([10, 25], 11), ([11, 25], 10), ([12, 25], 2), ([13, 25], 9)], [([3, 11], 6), ([10, 19], 12), ([1, 7], 1), ([63, 67], 32), ([12, 31], 13), ([13, 31], 12), ([14, 31], 11), ([15, 31], 2)], [([4, 11], 7), ([14, 19], 3), ([1, 7], 1), ([63, 67], 32), ([14, 37], 6), ([15, 37], 13), ([16, 37], 11), ([17, 37], 24)]]
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 033Passed
explicit oracle 136Failed
explicit oracle 211Passed
explicit oracle 32232Failed
explicit oracle 422Passed
explicit oracle 511Passed
explicit oracle 655Passed
explicit oracle 737Failed

SHA-256 / 2d4e8e424d7d18fc54e7c2f85128ae709faac2dd3542a4d88eff37b32412e4aa

HELD IN THE MEMBER ARCHIVE

The verified repair and its recorded checks are member-only.

This mechanism has 8 recorded checks per implementation. The open-access tier publishes the failure and the unsuccessful fix; the repaired source that passes every check, and the observations that prove it, are available to members.

Every case sharing this mechanism uses the same contract and the same repair, so this one record is held back for all of them.

Member access is invitation-based. Sign in with your invited account to inspect the repair.

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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 2026-09-29T14:39:28.776869+00:00.

Case digest / 3064a13bb463f38ab63631a99be5f268760120d5b8233aa08665106019d03b45