FA-15091 / Numerics / Open access
Lehmer permutation rank: unused alphabet initialization · case 01
The exact lehmer permutation rank result violates the stated contract at unused alphabet initialization.
ROOT CAUSE
The unused alphabet initialization step uses list(range(1,n+1)) instead of list(range(n)).
VERIFIED REPAIR
Use list(range(n)) at the unused alphabet initialization step.
Unsuccessful approach: The partial repair list(range(n))[::-1] still violates the unused alphabet initialization invariant.
Case contract
Input permutation of 0..n-1 with n<=7; return zero-based lexicographic rank.
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):
p=x;n=len(p)
remaining=list(range(1,n+1));rank=0
for i,v in enumerate(p):
if v not in remaining:return None
digit=remaining.index(v)
rank=rank+digit*math.factorial(n-i-1)
remaining.pop(remaining.index(v))
return rank
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([0], 0), ([0, 1], 0), ([5, 4, 3, 2, 1, 0], 719), ([1, 0], 1), ([0, 1, 2], 0), ([0, 2, 1], 1), ([1, 0, 2], 2), ([1, 2, 0], 3)], [([0, 1], 0), ([0, 2, 1], 1), ([0], 0), ([5, 4, 3, 2, 1, 0], 719), ([1, 2, 0, 3], 8), ([1, 2, 3, 0], 9), ([1, 3, 0, 2], 10), ([1, 3, 2, 0], 11)], [([1, 0], 1), ([2, 0, 1], 4), ([0], 0), ([5, 4, 3, 2, 1, 0], 719), ([0, 1, 2, 4, 3], 1), ([0, 1, 3, 2, 4], 2), ([0, 1, 3, 4, 2], 3), ([0, 1, 4, 2, 3], 4)], [([0, 1, 2], 0), ([0, 1, 3, 2], 1), ([0], 0), ([5, 4, 3, 2, 1, 0], 719), ([0, 4, 1, 2, 3], 18), ([0, 4, 1, 3, 2], 19), ([0, 4, 2, 1, 3], 20), ([0, 4, 2, 3, 1], 21)], [([0, 2, 1], 1), ([0, 3, 1, 2], 4), ([0], 0), ([5, 4, 3, 2, 1, 0], 719), ([1, 2, 4, 3, 0], 35), ([1, 3, 0, 2, 4], 36), ([1, 3, 0, 4, 2], 37), ([1, 3, 2, 0, 4], 38)]]
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 | None | 0 | Failed |
| explicit oracle 1 | None | 0 | Failed |
| explicit oracle 2 | None | 719 | Failed |
| explicit oracle 3 | None | 1 | Failed |
| explicit oracle 4 | None | 0 | Failed |
| explicit oracle 5 | None | 1 | Failed |
| explicit oracle 6 | None | 2 | Failed |
| explicit oracle 7 | None | 3 | Failed |
SHA-256 / 70dde553f1d27e1adb889d823462393a541b39e8a4c9cf1a1515209794ff14c8
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):
p=x;n=len(p)
remaining=list(range(n))[::-1];rank=0
for i,v in enumerate(p):
if v not in remaining:return None
digit=remaining.index(v)
rank=rank+digit*math.factorial(n-i-1)
remaining.pop(remaining.index(v))
return rank
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([0], 0), ([0, 1], 0), ([5, 4, 3, 2, 1, 0], 719), ([1, 0], 1), ([0, 1, 2], 0), ([0, 2, 1], 1), ([1, 0, 2], 2), ([1, 2, 0], 3)], [([0, 1], 0), ([0, 2, 1], 1), ([0], 0), ([5, 4, 3, 2, 1, 0], 719), ([1, 2, 0, 3], 8), ([1, 2, 3, 0], 9), ([1, 3, 0, 2], 10), ([1, 3, 2, 0], 11)], [([1, 0], 1), ([2, 0, 1], 4), ([0], 0), ([5, 4, 3, 2, 1, 0], 719), ([0, 1, 2, 4, 3], 1), ([0, 1, 3, 2, 4], 2), ([0, 1, 3, 4, 2], 3), ([0, 1, 4, 2, 3], 4)], [([0, 1, 2], 0), ([0, 1, 3, 2], 1), ([0], 0), ([5, 4, 3, 2, 1, 0], 719), ([0, 4, 1, 2, 3], 18), ([0, 4, 1, 3, 2], 19), ([0, 4, 2, 1, 3], 20), ([0, 4, 2, 3, 1], 21)], [([0, 2, 1], 1), ([0, 3, 1, 2], 4), ([0], 0), ([5, 4, 3, 2, 1, 0], 719), ([1, 2, 4, 3, 0], 35), ([1, 3, 0, 2, 4], 36), ([1, 3, 0, 4, 2], 37), ([1, 3, 2, 0, 4], 38)]]
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 | Passed |
| explicit oracle 1 | 1 | 0 | Failed |
| explicit oracle 2 | 0 | 719 | Failed |
| explicit oracle 3 | 0 | 1 | Failed |
| explicit oracle 4 | 5 | 0 | Failed |
| explicit oracle 5 | 4 | 1 | Failed |
| explicit oracle 6 | 3 | 2 | Failed |
| explicit oracle 7 | 2 | 3 | Failed |
SHA-256 / f46aee9706981f75d9450279add10785223f454e9f3315bfb507eb270380c392
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):
p=x;n=len(p)
remaining=list(range(n));rank=0
for i,v in enumerate(p):
if v not in remaining:return None
digit=remaining.index(v)
rank=rank+digit*math.factorial(n-i-1)
remaining.pop(remaining.index(v))
return rank
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([0], 0), ([0, 1], 0), ([5, 4, 3, 2, 1, 0], 719), ([1, 0], 1), ([0, 1, 2], 0), ([0, 2, 1], 1), ([1, 0, 2], 2), ([1, 2, 0], 3)], [([0, 1], 0), ([0, 2, 1], 1), ([0], 0), ([5, 4, 3, 2, 1, 0], 719), ([1, 2, 0, 3], 8), ([1, 2, 3, 0], 9), ([1, 3, 0, 2], 10), ([1, 3, 2, 0], 11)], [([1, 0], 1), ([2, 0, 1], 4), ([0], 0), ([5, 4, 3, 2, 1, 0], 719), ([0, 1, 2, 4, 3], 1), ([0, 1, 3, 2, 4], 2), ([0, 1, 3, 4, 2], 3), ([0, 1, 4, 2, 3], 4)], [([0, 1, 2], 0), ([0, 1, 3, 2], 1), ([0], 0), ([5, 4, 3, 2, 1, 0], 719), ([0, 4, 1, 2, 3], 18), ([0, 4, 1, 3, 2], 19), ([0, 4, 2, 1, 3], 20), ([0, 4, 2, 3, 1], 21)], [([0, 2, 1], 1), ([0, 3, 1, 2], 4), ([0], 0), ([5, 4, 3, 2, 1, 0], 719), ([1, 2, 4, 3, 0], 35), ([1, 3, 0, 2, 4], 36), ([1, 3, 0, 4, 2], 37), ([1, 3, 2, 0, 4], 38)]]
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 | Passed |
| explicit oracle 1 | 0 | 0 | Passed |
| explicit oracle 2 | 719 | 719 | Passed |
| explicit oracle 3 | 1 | 1 | Passed |
| explicit oracle 4 | 0 | 0 | Passed |
| explicit oracle 5 | 1 | 1 | Passed |
| explicit oracle 6 | 2 | 2 | Passed |
| explicit oracle 7 | 3 | 3 | Passed |
SHA-256 / 24c7bf7aa0c8fd4364656ecce46a9ecfca8113d0dd87e1fdd9ce7aec8ea1540e
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 2026-09-29T14:39:23.276007+00:00.
Case digest / ce28885fe7022784702277e4b55567c0579b93a332084d2c43741fc862fe412c