FA-15296 / Numerics / Open access
Josephus elimination order: starting position · case 01
The exact josephus elimination order result violates the stated contract at starting position.
ROOT CAUSE
The starting position step uses 0 instead of start.
VERIFIED REPAIR
Use start at the starting position step.
Unsuccessful approach: The partial repair (start+1)%n still violates the starting position invariant.
Case contract
Input [n,k,start], n>0 k>0 0<=start<n; return removal order counting current position as one.
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,k,start=x
ring=list(range(n));index=0;out=[]
for _ in range(n):
if not ring:break
index=(index+k-1)%len(ring)
out.append(ring.pop(index))
index=index
return out
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 1, 1], [1, 0]), ([2, 1, 0], [0, 1]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([1, 2, 0], [0]), ([1, 3, 0], [0]), ([1, 4, 0], [0]), ([1, 5, 0], [0])], [([2, 2, 1], [0, 1]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([2, 5, 1], [1, 0]), ([2, 6, 0], [1, 0]), ([2, 6, 1], [0, 1]), ([2, 7, 0], [0, 1]), ([2, 7, 1], [1, 0])], [([2, 3, 1], [1, 0]), ([2, 4, 0], [1, 0]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([3, 4, 1], [1, 0, 2]), ([3, 4, 2], [2, 1, 0]), ([3, 5, 0], [1, 2, 0]), ([3, 5, 1], [2, 0, 1])], [([2, 4, 1], [0, 1]), ([2, 5, 1], [1, 0]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([4, 1, 3], [3, 0, 1, 2]), ([4, 2, 0], [1, 3, 2, 0]), ([4, 2, 1], [2, 0, 3, 1]), ([4, 2, 2], [3, 1, 0, 2])], [([2, 5, 1], [1, 0]), ([2, 7, 0], [0, 1]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([4, 6, 0], [1, 0, 3, 2]), ([4, 6, 1], [2, 1, 0, 3]), ([4, 6, 2], [3, 2, 1, 0]), ([4, 6, 3], [0, 3, 2, 1])]]
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, 1] | [1, 0] | Failed |
| explicit oracle 1 | [0, 1] | [0, 1] | Passed |
| explicit oracle 2 | [0] | [0] | Passed |
| explicit oracle 3 | [7, 4, 2, 1, 3, 6, 10, 9, 5, 0, 8] | [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7] | Failed |
| explicit oracle 4 | [0] | [0] | Passed |
| explicit oracle 5 | [0] | [0] | Passed |
| explicit oracle 6 | [0] | [0] | Passed |
| explicit oracle 7 | [0] | [0] | Passed |
SHA-256 / 2671e22350f61e25e45fe8af0e1261f8e74af49d415ae970e32f13559cd20dbf
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,k,start=x
ring=list(range(n));index=(start+1)%n;out=[]
for _ in range(n):
if not ring:break
index=(index+k-1)%len(ring)
out.append(ring.pop(index))
index=index
return out
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 1, 1], [1, 0]), ([2, 1, 0], [0, 1]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([1, 2, 0], [0]), ([1, 3, 0], [0]), ([1, 4, 0], [0]), ([1, 5, 0], [0])], [([2, 2, 1], [0, 1]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([2, 5, 1], [1, 0]), ([2, 6, 0], [1, 0]), ([2, 6, 1], [0, 1]), ([2, 7, 0], [0, 1]), ([2, 7, 1], [1, 0])], [([2, 3, 1], [1, 0]), ([2, 4, 0], [1, 0]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([3, 4, 1], [1, 0, 2]), ([3, 4, 2], [2, 1, 0]), ([3, 5, 0], [1, 2, 0]), ([3, 5, 1], [2, 0, 1])], [([2, 4, 1], [0, 1]), ([2, 5, 1], [1, 0]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([4, 1, 3], [3, 0, 1, 2]), ([4, 2, 0], [1, 3, 2, 0]), ([4, 2, 1], [2, 0, 3, 1]), ([4, 2, 2], [3, 1, 0, 2])], [([2, 5, 1], [1, 0]), ([2, 7, 0], [0, 1]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([4, 6, 0], [1, 0, 3, 2]), ([4, 6, 1], [2, 1, 0, 3]), ([4, 6, 2], [3, 2, 1, 0]), ([4, 6, 3], [0, 3, 2, 1])]]
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, 1] | [1, 0] | Failed |
| explicit oracle 1 | [1, 0] | [0, 1] | Failed |
| explicit oracle 2 | [0] | [0] | Passed |
| explicit oracle 3 | [7, 4, 2, 1, 3, 6, 10, 9, 5, 0, 8] | [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7] | Failed |
| explicit oracle 4 | [0] | [0] | Passed |
| explicit oracle 5 | [0] | [0] | Passed |
| explicit oracle 6 | [0] | [0] | Passed |
| explicit oracle 7 | [0] | [0] | Passed |
SHA-256 / edbcf609111dfce67702a2ffb90355188582f605d7ea5fbc34e902ffb72ba2dd
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,k,start=x
ring=list(range(n));index=start;out=[]
for _ in range(n):
if not ring:break
index=(index+k-1)%len(ring)
out.append(ring.pop(index))
index=index
return out
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 1, 1], [1, 0]), ([2, 1, 0], [0, 1]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([1, 2, 0], [0]), ([1, 3, 0], [0]), ([1, 4, 0], [0]), ([1, 5, 0], [0])], [([2, 2, 1], [0, 1]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([2, 5, 1], [1, 0]), ([2, 6, 0], [1, 0]), ([2, 6, 1], [0, 1]), ([2, 7, 0], [0, 1]), ([2, 7, 1], [1, 0])], [([2, 3, 1], [1, 0]), ([2, 4, 0], [1, 0]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([3, 4, 1], [1, 0, 2]), ([3, 4, 2], [2, 1, 0]), ([3, 5, 0], [1, 2, 0]), ([3, 5, 1], [2, 0, 1])], [([2, 4, 1], [0, 1]), ([2, 5, 1], [1, 0]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([4, 1, 3], [3, 0, 1, 2]), ([4, 2, 0], [1, 3, 2, 0]), ([4, 2, 1], [2, 0, 3, 1]), ([4, 2, 2], [3, 1, 0, 2])], [([2, 5, 1], [1, 0]), ([2, 7, 0], [0, 1]), ([1, 1, 0], [0]), ([11, 8, 10], [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7]), ([4, 6, 0], [1, 0, 3, 2]), ([4, 6, 1], [2, 1, 0, 3]), ([4, 6, 2], [3, 2, 1, 0]), ([4, 6, 3], [0, 3, 2, 1])]]
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, 1] | [0, 1] | Passed |
| explicit oracle 2 | [0] | [0] | Passed |
| explicit oracle 3 | [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7] | [6, 3, 1, 0, 2, 5, 9, 8, 4, 10, 7] | Passed |
| explicit oracle 4 | [0] | [0] | Passed |
| explicit oracle 5 | [0] | [0] | Passed |
| explicit oracle 6 | [0] | [0] | Passed |
| explicit oracle 7 | [0] | [0] | Passed |
SHA-256 / 0dc344327ed66d6f5dc67b1c26ce69be7e5c44c482f60b37bffc8bfa70162e44
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 2026-09-29T14:39:25.331513+00:00.
Case digest / 2e8e27bbe5fb3b9f75ce2cc875b29463eb9732a1a5c9cdf3b97dc243f29e8558