FA-5906 / Planar geometry / Open access
Barycentric affine point · case 01
Homogeneous weights are not normalized.
ROOT CAUSE
Homogeneous weights are not normalized.
VERIFIED REPAIR
Apply the specified mathematical contract directly, preserving all terms and boundary cases: return [sum(p[i]*w for p,w in zip(points,weights))/sum(weights) for i in range(2)]
Unsuccessful approach: An unweighted centroid ignores unequal barycentric weights.
Case contract
Integer coordinates and lengths, with exact arithmetic except for indicated division results. Lengths are nonnegative. Rectangles are (xmin,ymin,xmax,ymax) with ordered bounds; coordinate vectors have matching dimensions. points are planar integer vectors; weights are matching nonnegative integers with positive total. Return a two-coordinate weighted affine point. Exact operational definition: [sum(p[i]*w for p,w in zip(points,weights))/sum(weights) for i in range(2)]
Why this case matters
Small exact fixtures expose this error without platform timing, external services, or probabilistic observations. Planar geometry results depend on the stated convention.
1 / The failure
Exit 1"""Failure Map reference implementation. Python standard library only."""
import json
import math
import calendar
import statistics
import itertools
from fractions import Fraction
from datetime import date, datetime, timedelta, timezone
from decimal import Decimal, ROUND_HALF_UP, ROUND_DOWN, ROUND_CEILING, ROUND_FLOOR
N = 1
observations = []
def solve(points, weights):
return [sum(p[i]*w for p,w in zip(points,weights)) for i in range(2)]
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('fixture 1: ([(0, 0), (6, 0), (0, 6)], [1, 1, 1])', solve(*([(0, 0), (6, 0), (0, 6)], [1, 1, 1])), [2, 2])
check('fixture 2: ([(0, 0), (6, 0), (0, 6)], [2, 1, 1])', solve(*([(0, 0), (6, 0), (0, 6)], [2, 1, 1])), [1.5, 1.5])
check('fixture 3: ([(1, 2)], [5])', solve(*([(1, 2)], [5])), [1, 2])
check('fixture 4: ([(0, 0), (4, 4)], [1, 3])', solve(*([(0, 0), (4, 4)], [1, 3])), [3, 3])
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 |
|---|---|---|---|
| fixture 1: ([(0, 0), (6, 0), (0, 6)], [1, 1, 1]) | [6, 6] | [2, 2] | Failed |
| fixture 2: ([(0, 0), (6, 0), (0, 6)], [2, 1, 1]) | [6, 6] | [1.5, 1.5] | Failed |
| fixture 3: ([(1, 2)], [5]) | [5, 10] | [1, 2] | Failed |
| fixture 4: ([(0, 0), (4, 4)], [1, 3]) | [12, 12] | [3, 3] | Failed |
SHA-256 / 80bf2ae94f53e8f65ec07839d2829eaeae0e2e909213178a1c8cb3887bdc523c
2 / The unsuccessful fix
Exit 1"""Failure Map reference implementation. Python standard library only."""
import json
import math
import calendar
import statistics
import itertools
from fractions import Fraction
from datetime import date, datetime, timedelta, timezone
from decimal import Decimal, ROUND_HALF_UP, ROUND_DOWN, ROUND_CEILING, ROUND_FLOOR
N = 1
observations = []
def solve(points, weights):
return [sum(p[i] for p in points)/len(points) for i in range(2)]
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('fixture 1: ([(0, 0), (6, 0), (0, 6)], [1, 1, 1])', solve(*([(0, 0), (6, 0), (0, 6)], [1, 1, 1])), [2, 2])
check('fixture 2: ([(0, 0), (6, 0), (0, 6)], [2, 1, 1])', solve(*([(0, 0), (6, 0), (0, 6)], [2, 1, 1])), [1.5, 1.5])
check('fixture 3: ([(1, 2)], [5])', solve(*([(1, 2)], [5])), [1, 2])
check('fixture 4: ([(0, 0), (4, 4)], [1, 3])', solve(*([(0, 0), (4, 4)], [1, 3])), [3, 3])
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 |
|---|---|---|---|
| fixture 1: ([(0, 0), (6, 0), (0, 6)], [1, 1, 1]) | [2.0, 2.0] | [2, 2] | Passed |
| fixture 2: ([(0, 0), (6, 0), (0, 6)], [2, 1, 1]) | [2.0, 2.0] | [1.5, 1.5] | Failed |
| fixture 3: ([(1, 2)], [5]) | [1.0, 2.0] | [1, 2] | Passed |
| fixture 4: ([(0, 0), (4, 4)], [1, 3]) | [2.0, 2.0] | [3, 3] | Failed |
SHA-256 / d2027b3bc3eb955d1386a83098abbd6464a5ef6f59d7e82c9b4ab697fb0ddf90
3 / The verified repair
Exit 0"""Failure Map reference implementation. Python standard library only."""
import json
import math
import calendar
import statistics
import itertools
from fractions import Fraction
from datetime import date, datetime, timedelta, timezone
from decimal import Decimal, ROUND_HALF_UP, ROUND_DOWN, ROUND_CEILING, ROUND_FLOOR
N = 1
observations = []
def solve(points, weights):
return [sum(p[i]*w for p,w in zip(points,weights))/sum(weights) for i in range(2)]
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('fixture 1: ([(0, 0), (6, 0), (0, 6)], [1, 1, 1])', solve(*([(0, 0), (6, 0), (0, 6)], [1, 1, 1])), [2, 2])
check('fixture 2: ([(0, 0), (6, 0), (0, 6)], [2, 1, 1])', solve(*([(0, 0), (6, 0), (0, 6)], [2, 1, 1])), [1.5, 1.5])
check('fixture 3: ([(1, 2)], [5])', solve(*([(1, 2)], [5])), [1, 2])
check('fixture 4: ([(0, 0), (4, 4)], [1, 3])', solve(*([(0, 0), (4, 4)], [1, 3])), [3, 3])
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 |
|---|---|---|---|
| fixture 1: ([(0, 0), (6, 0), (0, 6)], [1, 1, 1]) | [2.0, 2.0] | [2, 2] | Passed |
| fixture 2: ([(0, 0), (6, 0), (0, 6)], [2, 1, 1]) | [1.5, 1.5] | [1.5, 1.5] | Passed |
| fixture 3: ([(1, 2)], [5]) | [1.0, 2.0] | [1, 2] | Passed |
| fixture 4: ([(0, 0), (4, 4)], [1, 3]) | [3.0, 3.0] | [3, 3] | Passed |
SHA-256 / 7a8c30fd6e1607af5c733907e9daf8983eb730eb2d2c2efe7afe7003d0572381
Verification & scope
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 2026-09-29T14:37:55.330138+00:00.
Case digest / 57b32c538a146d0be3bfa647acd286c807c3a278d4748a49789c3565a05530c1