FA-15816 / Numerics / Open access
Tonelli shanks square root: nonresidue subgroup generator · case 01
The exact tonelli shanks square root result violates the stated contract at nonresidue subgroup generator.
ROOT CAUSE
The nonresidue subgroup generator step uses z%p instead of pow(z,q,p).
VERIFIED REPAIR
Use pow(z,q,p) at the nonresidue subgroup generator step.
Unsuccessful approach: The partial repair pow(z,s,p) still violates the nonresidue subgroup generator invariant.
Case contract
Input [a,p], odd prime p and nonzero quadratic residue a; return smaller of two square roots modulo p. Bounds: p<=97.
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,p=x
q=p-1;s=0
while q%2==0:q//=2;s+=1
z=2
while pow(z,(p-1)//2,p)!=p-1:z+=1
c=z%p;r=pow(a,(q+1)//2,p);t=pow(a,q,p);m=s
for _ in range(20):
if t==1:break
i=1;v=t*t%p
while i<m and v!=1:v=v*v%p;i+=1
if i>=m:return None
b=pow(c,1<<(m-i-1),p)
r=r*b%p;t=t*b*b%p;c=b*b%p;m=i
return min(r,p-r)
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([4, 13], 2), ([4, 5], 2), ([1, 5], 1), ([96, 97], 22), ([1, 7], 1), ([2, 7], 3), ([4, 7], 2), ([1, 11], 1)], [([10, 13], 6), ([12, 13], 5), ([1, 5], 1), ([96, 97], 22), ([2, 17], 6), ([4, 17], 2), ([8, 17], 5), ([9, 17], 3)], [([12, 13], 5), ([8, 17], 5), ([1, 5], 1), ([96, 97], 22), ([2, 23], 5), ([3, 23], 7), ([4, 23], 2), ([6, 23], 11)], [([4, 29], 2), ([15, 17], 7), ([1, 5], 1), ([96, 97], 22), ([16, 29], 4), ([20, 29], 7), ([22, 29], 14), ([23, 29], 9)], [([5, 29], 11), ([1, 5], 1), ([96, 97], 22), ([18, 31], 7), ([19, 31], 9), ([20, 31], 12), ([25, 31], 5), ([28, 31], 11)]]
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 | 2 | Failed |
| explicit oracle 1 | 2 | 2 | Passed |
| explicit oracle 2 | 1 | 1 | Passed |
| explicit oracle 3 | None | 22 | Failed |
| explicit oracle 4 | 1 | 1 | Passed |
| explicit oracle 5 | 3 | 3 | Passed |
| explicit oracle 6 | 2 | 2 | Passed |
| explicit oracle 7 | 1 | 1 | Passed |
SHA-256 / 4277f78f30c482517593e0b535d11952ab8fb2c6ee53ee9c47a95bb28fb88c9d
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,p=x
q=p-1;s=0
while q%2==0:q//=2;s+=1
z=2
while pow(z,(p-1)//2,p)!=p-1:z+=1
c=pow(z,s,p);r=pow(a,(q+1)//2,p);t=pow(a,q,p);m=s
for _ in range(20):
if t==1:break
i=1;v=t*t%p
while i<m and v!=1:v=v*v%p;i+=1
if i>=m:return None
b=pow(c,1<<(m-i-1),p)
r=r*b%p;t=t*b*b%p;c=b*b%p;m=i
return min(r,p-r)
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([4, 13], 2), ([4, 5], 2), ([1, 5], 1), ([96, 97], 22), ([1, 7], 1), ([2, 7], 3), ([4, 7], 2), ([1, 11], 1)], [([10, 13], 6), ([12, 13], 5), ([1, 5], 1), ([96, 97], 22), ([2, 17], 6), ([4, 17], 2), ([8, 17], 5), ([9, 17], 3)], [([12, 13], 5), ([8, 17], 5), ([1, 5], 1), ([96, 97], 22), ([2, 23], 5), ([3, 23], 7), ([4, 23], 2), ([6, 23], 11)], [([4, 29], 2), ([15, 17], 7), ([1, 5], 1), ([96, 97], 22), ([16, 29], 4), ([20, 29], 7), ([22, 29], 14), ([23, 29], 9)], [([5, 29], 11), ([1, 5], 1), ([96, 97], 22), ([18, 31], 7), ([19, 31], 9), ([20, 31], 12), ([25, 31], 5), ([28, 31], 11)]]
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 | 2 | Failed |
| explicit oracle 1 | None | 2 | Failed |
| explicit oracle 2 | 1 | 1 | Passed |
| explicit oracle 3 | None | 22 | Failed |
| explicit oracle 4 | 1 | 1 | Passed |
| explicit oracle 5 | 3 | 3 | Passed |
| explicit oracle 6 | 2 | 2 | Passed |
| explicit oracle 7 | 1 | 1 | Passed |
SHA-256 / 46e3085791024f4d59572a04c12bbc7643a8d662fd84a015b4d700108f06d46d
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,p=x
q=p-1;s=0
while q%2==0:q//=2;s+=1
z=2
while pow(z,(p-1)//2,p)!=p-1:z+=1
c=pow(z,q,p);r=pow(a,(q+1)//2,p);t=pow(a,q,p);m=s
for _ in range(20):
if t==1:break
i=1;v=t*t%p
while i<m and v!=1:v=v*v%p;i+=1
if i>=m:return None
b=pow(c,1<<(m-i-1),p)
r=r*b%p;t=t*b*b%p;c=b*b%p;m=i
return min(r,p-r)
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([4, 13], 2), ([4, 5], 2), ([1, 5], 1), ([96, 97], 22), ([1, 7], 1), ([2, 7], 3), ([4, 7], 2), ([1, 11], 1)], [([10, 13], 6), ([12, 13], 5), ([1, 5], 1), ([96, 97], 22), ([2, 17], 6), ([4, 17], 2), ([8, 17], 5), ([9, 17], 3)], [([12, 13], 5), ([8, 17], 5), ([1, 5], 1), ([96, 97], 22), ([2, 23], 5), ([3, 23], 7), ([4, 23], 2), ([6, 23], 11)], [([4, 29], 2), ([15, 17], 7), ([1, 5], 1), ([96, 97], 22), ([16, 29], 4), ([20, 29], 7), ([22, 29], 14), ([23, 29], 9)], [([5, 29], 11), ([1, 5], 1), ([96, 97], 22), ([18, 31], 7), ([19, 31], 9), ([20, 31], 12), ([25, 31], 5), ([28, 31], 11)]]
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 | 2 | 2 | Passed |
| explicit oracle 1 | 2 | 2 | Passed |
| explicit oracle 2 | 1 | 1 | Passed |
| explicit oracle 3 | 22 | 22 | Passed |
| explicit oracle 4 | 1 | 1 | Passed |
| explicit oracle 5 | 3 | 3 | Passed |
| explicit oracle 6 | 2 | 2 | Passed |
| explicit oracle 7 | 1 | 1 | Passed |
SHA-256 / 382e58420955eb0267f01926bfab1583d621a031e3b65cf7fe0869916c11d250
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 2026-09-29T14:39:30.492600+00:00.
Case digest / fff3d703272e9aa5c6be708aacd2a284497239ac852ba734ec957d04e7d37613