FA-14346 / Numerics / Open access

Generalized crt: reduced congruence multiplier · case 01

The exact generalized crt result violates the stated contract at reduced congruence multiplier.

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

ROOT CAUSE

The reduced congruence multiplier step uses ((b-a)pow(m//g,-1,q))%q if q>1 else 0 instead of ((b-a)//gpow(m//g,-1,q))%q if q>1 else 0.

VERIFIED REPAIR

Use ((b-a)//g*pow(m//g,-1,q))%q if q>1 else 0 at the reduced congruence multiplier step.

Unsuccessful approach: The partial repair ((b-a)//g)%q if q>1 else 0 still violates the reduced congruence multiplier invariant.

Case contract

Input [a,m,b,n], positive moduli; return least nonnegative common residue and lcm modulus or None.

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,m,b,n=x
    g=math.gcd(m,n)
    if (b-a)%g != 0: return None
    q=n//g
    k=((b-a)*pow(m//g,-1,q))%q if q>1 else 0
    period=m*(n//g)
    r=a+m*k
    return [r%period,period]
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([-3, 2, -1, 4], [3, 4]), ([-3, 2, -2, 3], [1, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([-3, 1, -2, 1], [0, 1]), ([-3, 1, -1, 1], [0, 1]), ([-3, 1, 0, 1], [0, 1]), ([-3, 1, 1, 1], [0, 1])], [([-3, 2, 3, 4], [3, 4]), ([-3, 2, 2, 3], [5, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([-1, 1, 0, 1], [0, 1]), ([-1, 1, 1, 1], [0, 1]), ([-1, 1, 2, 1], [0, 1]), ([-1, 1, 3, 1], [0, 1])], [([-2, 2, 0, 4], [0, 4]), ([-2, 2, 0, 3], [0, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([1, 1, 3, 1], [0, 1]), ([2, 1, -3, 1], [0, 1]), ([2, 1, -2, 1], [0, 1]), ([2, 1, -1, 1], [0, 1])], [([-1, 2, -3, 4], [1, 4]), ([-1, 2, -3, 3], [3, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([-3, 1, -1, 2], [1, 2]), ([-3, 1, 0, 2], [0, 2]), ([-3, 1, 1, 2], [1, 2]), ([-3, 1, 2, 2], [0, 2])], [([-1, 2, 1, 4], [1, 4]), ([-1, 2, 1, 3], [1, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([-1, 1, 2, 2], [0, 2]), ([-1, 1, 3, 2], [1, 2]), ([0, 1, -3, 2], [1, 2]), ([0, 1, -2, 2], [0, 2])]]
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, 4]`	`[3, 4]`	Failed
explicit oracle 1	`[1, 6]`	`[1, 6]`	Passed
explicit oracle 2	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 3	`[3, 8]`	`[3, 8]`	Passed
explicit oracle 4	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 5	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 6	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 7	`[0, 1]`	`[0, 1]`	Passed

SHA-256 / 80b29ecabbcc33d7fb7e6fd76f6f9e7532c10fd8913a3ad172db42829986b622

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,m,b,n=x
    g=math.gcd(m,n)
    if (b-a)%g != 0: return None
    q=n//g
    k=((b-a)//g)%q if q>1 else 0
    period=m*(n//g)
    r=a+m*k
    return [r%period,period]
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([-3, 2, -1, 4], [3, 4]), ([-3, 2, -2, 3], [1, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([-3, 1, -2, 1], [0, 1]), ([-3, 1, -1, 1], [0, 1]), ([-3, 1, 0, 1], [0, 1]), ([-3, 1, 1, 1], [0, 1])], [([-3, 2, 3, 4], [3, 4]), ([-3, 2, 2, 3], [5, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([-1, 1, 0, 1], [0, 1]), ([-1, 1, 1, 1], [0, 1]), ([-1, 1, 2, 1], [0, 1]), ([-1, 1, 3, 1], [0, 1])], [([-2, 2, 0, 4], [0, 4]), ([-2, 2, 0, 3], [0, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([1, 1, 3, 1], [0, 1]), ([2, 1, -3, 1], [0, 1]), ([2, 1, -2, 1], [0, 1]), ([2, 1, -1, 1], [0, 1])], [([-1, 2, -3, 4], [1, 4]), ([-1, 2, -3, 3], [3, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([-3, 1, -1, 2], [1, 2]), ([-3, 1, 0, 2], [0, 2]), ([-3, 1, 1, 2], [1, 2]), ([-3, 1, 2, 2], [0, 2])], [([-1, 2, 1, 4], [1, 4]), ([-1, 2, 1, 3], [1, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([-1, 1, 2, 2], [0, 2]), ([-1, 1, 3, 2], [1, 2]), ([0, 1, -3, 2], [1, 2]), ([0, 1, -2, 2], [0, 2])]]
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	`[3, 4]`	`[3, 4]`	Passed
explicit oracle 1	`[5, 6]`	`[1, 6]`	Failed
explicit oracle 2	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 3	`[3, 8]`	`[3, 8]`	Passed
explicit oracle 4	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 5	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 6	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 7	`[0, 1]`	`[0, 1]`	Passed

SHA-256 / ca17202ef4362c4f6bedef5cf0294fac605bcb84fb79566a237b578a0450c93d

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,m,b,n=x
    g=math.gcd(m,n)
    if (b-a)%g != 0: return None
    q=n//g
    k=((b-a)//g*pow(m//g,-1,q))%q if q>1 else 0
    period=m*(n//g)
    r=a+m*k
    return [r%period,period]
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([-3, 2, -1, 4], [3, 4]), ([-3, 2, -2, 3], [1, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([-3, 1, -2, 1], [0, 1]), ([-3, 1, -1, 1], [0, 1]), ([-3, 1, 0, 1], [0, 1]), ([-3, 1, 1, 1], [0, 1])], [([-3, 2, 3, 4], [3, 4]), ([-3, 2, 2, 3], [5, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([-1, 1, 0, 1], [0, 1]), ([-1, 1, 1, 1], [0, 1]), ([-1, 1, 2, 1], [0, 1]), ([-1, 1, 3, 1], [0, 1])], [([-2, 2, 0, 4], [0, 4]), ([-2, 2, 0, 3], [0, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([1, 1, 3, 1], [0, 1]), ([2, 1, -3, 1], [0, 1]), ([2, 1, -2, 1], [0, 1]), ([2, 1, -1, 1], [0, 1])], [([-1, 2, -3, 4], [1, 4]), ([-1, 2, -3, 3], [3, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([-3, 1, -1, 2], [1, 2]), ([-3, 1, 0, 2], [0, 2]), ([-3, 1, 1, 2], [1, 2]), ([-3, 1, 2, 2], [0, 2])], [([-1, 2, 1, 4], [1, 4]), ([-1, 2, 1, 3], [1, 6]), ([-3, 1, -3, 1], [0, 1]), ([3, 8, 3, 8], [3, 8]), ([-1, 1, 2, 2], [0, 2]), ([-1, 1, 3, 2], [1, 2]), ([0, 1, -3, 2], [1, 2]), ([0, 1, -2, 2], [0, 2])]]
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	`[3, 4]`	`[3, 4]`	Passed
explicit oracle 1	`[1, 6]`	`[1, 6]`	Passed
explicit oracle 2	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 3	`[3, 8]`	`[3, 8]`	Passed
explicit oracle 4	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 5	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 6	`[0, 1]`	`[0, 1]`	Passed
explicit oracle 7	`[0, 1]`	`[0, 1]`	Passed

SHA-256 / c54e04acf6519dd612e726b9496cfb1702f7b1703e43611dd66e8972f12330a7

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 2026-09-29T14:39:15.908107+00:00.

Case digest / 9a475dc2c2f5887c9512dc0b4fdcc79dc22c7aae9c9dcb25ac2d2bc18b8c0763