FA-14401 / Numerics / Open access
Bounded diophantine: divisibility filter · case 01
The exact bounded diophantine result violates the stated contract at divisibility filter.
ROOT CAUSE
The divisibility filter step uses t == 0 instead of t%b != 0.
VERIFIED REPAIR
Use t%b != 0 at the divisibility filter step.
Unsuccessful approach: The partial repair t%b == 0 still violates the divisibility filter invariant.
Case contract
Input [a,b,c,L], positive a,b and nonnegative L; lexicographically ordered nonnegative solutions a*u+b*v=c with u,v<=L.
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,c,L=x
out=[]
for u in range(L+1):
t=c-a*u
if t == 0: continue
v=t//b
if 0<=v<=L: out.append([u,v])
return out
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([1, 1, 0, 0], [[0, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, -2, 1], []), ([1, 1, -2, 2], []), ([1, 1, -2, 3], []), ([1, 1, -2, 4], []), ([1, 1, -2, 5], [])], [([1, 1, 0, 1], [[0, 0]]), ([1, 1, 0, 3], [[0, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, 0, 5], [[0, 0]]), ([1, 1, 1, 0], []), ([1, 1, 1, 1], [[0, 1], [1, 0]]), ([1, 1, 1, 2], [[0, 1], [1, 0]])], [([1, 1, 0, 2], [[0, 0]]), ([1, 1, 1, 1], [[0, 1], [1, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, 3, 4], [[0, 3], [1, 2], [2, 1], [3, 0]]), ([1, 1, 3, 5], [[0, 3], [1, 2], [2, 1], [3, 0]]), ([1, 1, 4, 0], []), ([1, 1, 4, 1], [])], [([1, 1, 0, 3], [[0, 0]]), ([1, 1, 1, 4], [[0, 1], [1, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, 6, 3], [[3, 3]]), ([1, 1, 6, 4], [[2, 4], [3, 3], [4, 2]]), ([1, 1, 6, 5], [[1, 5], [2, 4], [3, 3], [4, 2], [5, 1]]), ([1, 1, 7, 0], [])], [([1, 1, 0, 4], [[0, 0]]), ([1, 1, 2, 2], [[0, 2], [1, 1], [2, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, 9, 2], []), ([1, 1, 9, 3], []), ([1, 1, 9, 4], []), ([1, 1, 9, 5], [[4, 5], [5, 4]])]]
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]] | Failed |
| explicit oracle 1 | [] | [] | Passed |
| explicit oracle 2 | [[0, 3], [1, 2], [2, 1], [3, 0]] | [] | Failed |
| explicit oracle 3 | [] | [] | Passed |
| explicit oracle 4 | [] | [] | Passed |
| explicit oracle 5 | [] | [] | Passed |
| explicit oracle 6 | [] | [] | Passed |
| explicit oracle 7 | [] | [] | Passed |
SHA-256 / 56134983f5c434c8d7afff328df6c9cc45ca22af2c5c3bf45116c3943cf20273
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,c,L=x
out=[]
for u in range(L+1):
t=c-a*u
if t%b == 0: continue
v=t//b
if 0<=v<=L: out.append([u,v])
return out
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([1, 1, 0, 0], [[0, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, -2, 1], []), ([1, 1, -2, 2], []), ([1, 1, -2, 3], []), ([1, 1, -2, 4], []), ([1, 1, -2, 5], [])], [([1, 1, 0, 1], [[0, 0]]), ([1, 1, 0, 3], [[0, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, 0, 5], [[0, 0]]), ([1, 1, 1, 0], []), ([1, 1, 1, 1], [[0, 1], [1, 0]]), ([1, 1, 1, 2], [[0, 1], [1, 0]])], [([1, 1, 0, 2], [[0, 0]]), ([1, 1, 1, 1], [[0, 1], [1, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, 3, 4], [[0, 3], [1, 2], [2, 1], [3, 0]]), ([1, 1, 3, 5], [[0, 3], [1, 2], [2, 1], [3, 0]]), ([1, 1, 4, 0], []), ([1, 1, 4, 1], [])], [([1, 1, 0, 3], [[0, 0]]), ([1, 1, 1, 4], [[0, 1], [1, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, 6, 3], [[3, 3]]), ([1, 1, 6, 4], [[2, 4], [3, 3], [4, 2]]), ([1, 1, 6, 5], [[1, 5], [2, 4], [3, 3], [4, 2], [5, 1]]), ([1, 1, 7, 0], [])], [([1, 1, 0, 4], [[0, 0]]), ([1, 1, 2, 2], [[0, 2], [1, 1], [2, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, 9, 2], []), ([1, 1, 9, 3], []), ([1, 1, 9, 4], []), ([1, 1, 9, 5], [[4, 5], [5, 4]])]]
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]] | Failed |
| explicit oracle 1 | [] | [] | Passed |
| explicit oracle 2 | [[0, 3], [1, 2], [2, 1], [3, 0]] | [] | Failed |
| explicit oracle 3 | [] | [] | Passed |
| explicit oracle 4 | [] | [] | Passed |
| explicit oracle 5 | [] | [] | Passed |
| explicit oracle 6 | [] | [] | Passed |
| explicit oracle 7 | [] | [] | Passed |
SHA-256 / 17ba7f8886bfd1522c1f1d71076ea35b6f80efd28d0572baf1f8fad799cd8523
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,c,L=x
out=[]
for u in range(L+1):
t=c-a*u
if t%b != 0: continue
v=t//b
if 0<=v<=L: out.append([u,v])
return out
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([1, 1, 0, 0], [[0, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, -2, 1], []), ([1, 1, -2, 2], []), ([1, 1, -2, 3], []), ([1, 1, -2, 4], []), ([1, 1, -2, 5], [])], [([1, 1, 0, 1], [[0, 0]]), ([1, 1, 0, 3], [[0, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, 0, 5], [[0, 0]]), ([1, 1, 1, 0], []), ([1, 1, 1, 1], [[0, 1], [1, 0]]), ([1, 1, 1, 2], [[0, 1], [1, 0]])], [([1, 1, 0, 2], [[0, 0]]), ([1, 1, 1, 1], [[0, 1], [1, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, 3, 4], [[0, 3], [1, 2], [2, 1], [3, 0]]), ([1, 1, 3, 5], [[0, 3], [1, 2], [2, 1], [3, 0]]), ([1, 1, 4, 0], []), ([1, 1, 4, 1], [])], [([1, 1, 0, 3], [[0, 0]]), ([1, 1, 1, 4], [[0, 1], [1, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, 6, 3], [[3, 3]]), ([1, 1, 6, 4], [[2, 4], [3, 3], [4, 2]]), ([1, 1, 6, 5], [[1, 5], [2, 4], [3, 3], [4, 2], [5, 1]]), ([1, 1, 7, 0], [])], [([1, 1, 0, 4], [[0, 0]]), ([1, 1, 2, 2], [[0, 2], [1, 1], [2, 0]]), ([1, 1, -2, 0], []), ([4, 4, 14, 5], []), ([1, 1, 9, 2], []), ([1, 1, 9, 3], []), ([1, 1, 9, 4], []), ([1, 1, 9, 5], [[4, 5], [5, 4]])]]
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, 0]] | Passed |
| explicit oracle 1 | [] | [] | Passed |
| explicit oracle 2 | [] | [] | Passed |
| explicit oracle 3 | [] | [] | Passed |
| explicit oracle 4 | [] | [] | Passed |
| explicit oracle 5 | [] | [] | Passed |
| explicit oracle 6 | [] | [] | Passed |
| explicit oracle 7 | [] | [] | Passed |
SHA-256 / be36e30cdd0408a8aeb6bd73184a932c29c0c513d1ddaf49d4f244f6164af87e
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 2026-09-29T14:39:16.490041+00:00.
Case digest / 5d8e66e1f4fcfc264cb2271f17af1785099628e8768874b0464997c621ab9392