FA-15121 / Numerics / Open access

Factoradic permutation unrank: permutation block size · case 01

The exact factoradic permutation unrank result violates the stated contract at permutation block size.

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

ROOT CAUSE

The permutation block size step uses math.factorial(i+1) instead of math.factorial(i).

VERIFIED REPAIR

Use math.factorial(i) at the permutation block size step.

Unsuccessful approach: The partial repair max(1,i) still violates the permutation block size invariant.

Case contract

Input [n,r], n<=7 and 0<=r<n!; return lexicographic permutation of 0..n-1 at rank r.

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):
    n,r=x;pool=list(range(n));out=[]
    for i in range(n-1,-1,-1):
     block=math.factorial(i+1)
     digit=r//block
     if digit<0 or digit>=len(pool):return None
     out.append(pool.pop(digit))
     r=r%block
    return out
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 1], [1, 0]), ([4, 3], [0, 2, 3, 1]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([2, 0], [0, 1]), ([3, 0], [0, 1, 2]), ([3, 1], [0, 2, 1]), ([3, 2], [1, 0, 2])], [([3, 1], [0, 2, 1]), ([4, 6], [1, 0, 2, 3]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([4, 8], [1, 2, 0, 3]), ([4, 9], [1, 2, 3, 0]), ([4, 10], [1, 3, 0, 2]), ([4, 11], [1, 3, 2, 0])], [([3, 2], [1, 0, 2]), ([4, 9], [1, 2, 3, 0]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([5, 1], [0, 1, 2, 4, 3]), ([5, 2], [0, 1, 3, 2, 4]), ([5, 3], [0, 1, 3, 4, 2]), ([5, 4], [0, 1, 4, 2, 3])], [([3, 3], [1, 2, 0]), ([4, 12], [2, 0, 1, 3]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([5, 18], [0, 4, 1, 2, 3]), ([5, 19], [0, 4, 1, 3, 2]), ([5, 20], [0, 4, 2, 1, 3]), ([5, 21], [0, 4, 2, 3, 1])], [([3, 4], [2, 0, 1]), ([4, 15], [2, 1, 3, 0]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([5, 35], [1, 2, 4, 3, 0]), ([5, 36], [1, 3, 0, 2, 4]), ([5, 37], [1, 3, 0, 4, 2]), ([5, 38], [1, 3, 2, 0, 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	`None`	`[1, 0]`	Failed
explicit oracle 1	`None`	`[0, 2, 3, 1]`	Failed
explicit oracle 2	`[0]`	`[0]`	Passed
explicit oracle 3	`None`	`[5, 4, 3, 2, 1, 0]`	Failed
explicit oracle 4	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 5	`[0, 1, 2]`	`[0, 1, 2]`	Passed
explicit oracle 6	`None`	`[0, 2, 1]`	Failed
explicit oracle 7	`[0, 2, 1]`	`[1, 0, 2]`	Failed

SHA-256 / 7dc9a329fb9aa09191c18eeb833c8087bbe87aae3a255ff1b82b6dde7c40600e

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):
    n,r=x;pool=list(range(n));out=[]
    for i in range(n-1,-1,-1):
     block=max(1,i)
     digit=r//block
     if digit<0 or digit>=len(pool):return None
     out.append(pool.pop(digit))
     r=r%block
    return out
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 1], [1, 0]), ([4, 3], [0, 2, 3, 1]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([2, 0], [0, 1]), ([3, 0], [0, 1, 2]), ([3, 1], [0, 2, 1]), ([3, 2], [1, 0, 2])], [([3, 1], [0, 2, 1]), ([4, 6], [1, 0, 2, 3]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([4, 8], [1, 2, 0, 3]), ([4, 9], [1, 2, 3, 0]), ([4, 10], [1, 3, 0, 2]), ([4, 11], [1, 3, 2, 0])], [([3, 2], [1, 0, 2]), ([4, 9], [1, 2, 3, 0]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([5, 1], [0, 1, 2, 4, 3]), ([5, 2], [0, 1, 3, 2, 4]), ([5, 3], [0, 1, 3, 4, 2]), ([5, 4], [0, 1, 4, 2, 3])], [([3, 3], [1, 2, 0]), ([4, 12], [2, 0, 1, 3]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([5, 18], [0, 4, 1, 2, 3]), ([5, 19], [0, 4, 1, 3, 2]), ([5, 20], [0, 4, 2, 1, 3]), ([5, 21], [0, 4, 2, 3, 1])], [([3, 4], [2, 0, 1]), ([4, 15], [2, 1, 3, 0]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([5, 35], [1, 2, 4, 3, 0]), ([5, 36], [1, 3, 0, 2, 4]), ([5, 37], [1, 3, 0, 4, 2]), ([5, 38], [1, 3, 2, 0, 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	`[1, 0]`	`[1, 0]`	Passed
explicit oracle 1	`[1, 0, 2, 3]`	`[0, 2, 3, 1]`	Failed
explicit oracle 2	`[0]`	`[0]`	Passed
explicit oracle 3	`None`	`[5, 4, 3, 2, 1, 0]`	Failed
explicit oracle 4	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 5	`[0, 1, 2]`	`[0, 1, 2]`	Passed
explicit oracle 6	`[0, 2, 1]`	`[0, 2, 1]`	Passed
explicit oracle 7	`[1, 0, 2]`	`[1, 0, 2]`	Passed

SHA-256 / 8420e10055d993423ff24134dda9c677c908ba4468bfec409310ff04e85ab7e0

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):
    n,r=x;pool=list(range(n));out=[]
    for i in range(n-1,-1,-1):
     block=math.factorial(i)
     digit=r//block
     if digit<0 or digit>=len(pool):return None
     out.append(pool.pop(digit))
     r=r%block
    return out
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 1], [1, 0]), ([4, 3], [0, 2, 3, 1]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([2, 0], [0, 1]), ([3, 0], [0, 1, 2]), ([3, 1], [0, 2, 1]), ([3, 2], [1, 0, 2])], [([3, 1], [0, 2, 1]), ([4, 6], [1, 0, 2, 3]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([4, 8], [1, 2, 0, 3]), ([4, 9], [1, 2, 3, 0]), ([4, 10], [1, 3, 0, 2]), ([4, 11], [1, 3, 2, 0])], [([3, 2], [1, 0, 2]), ([4, 9], [1, 2, 3, 0]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([5, 1], [0, 1, 2, 4, 3]), ([5, 2], [0, 1, 3, 2, 4]), ([5, 3], [0, 1, 3, 4, 2]), ([5, 4], [0, 1, 4, 2, 3])], [([3, 3], [1, 2, 0]), ([4, 12], [2, 0, 1, 3]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([5, 18], [0, 4, 1, 2, 3]), ([5, 19], [0, 4, 1, 3, 2]), ([5, 20], [0, 4, 2, 1, 3]), ([5, 21], [0, 4, 2, 3, 1])], [([3, 4], [2, 0, 1]), ([4, 15], [2, 1, 3, 0]), ([1, 0], [0]), ([6, 719], [5, 4, 3, 2, 1, 0]), ([5, 35], [1, 2, 4, 3, 0]), ([5, 36], [1, 3, 0, 2, 4]), ([5, 37], [1, 3, 0, 4, 2]), ([5, 38], [1, 3, 2, 0, 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	`[1, 0]`	`[1, 0]`	Passed
explicit oracle 1	`[0, 2, 3, 1]`	`[0, 2, 3, 1]`	Passed
explicit oracle 2	`[0]`	`[0]`	Passed
explicit oracle 3	`[5, 4, 3, 2, 1, 0]`	`[5, 4, 3, 2, 1, 0]`	Passed
explicit oracle 4	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 5	`[0, 1, 2]`	`[0, 1, 2]`	Passed
explicit oracle 6	`[0, 2, 1]`	`[0, 2, 1]`	Passed
explicit oracle 7	`[1, 0, 2]`	`[1, 0, 2]`	Passed

SHA-256 / d62c4d019ace38b470adc9c36870c69e79a9a6653e4c4cf07b1ba57290a2150d

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 2026-09-29T14:39:23.593414+00:00.

Case digest / bb438727a3fd79fc1ec5c7cecbc74b6038373ae81b4be11e2c68f930ec602fb2