FA-15736 / Numerics / Open access

Finite field discrete transform: transform normalization convention · case 01

The exact finite field discrete transform result violates the stated contract at transform normalization convention.

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

ROOT CAUSE

The transform normalization convention step uses total*pow(n,-1,p)%p instead of total.

VERIFIED REPAIR

Use total at the transform normalization convention step.

Unsuccessful approach: The partial repair total*n%p still violates the transform normalization convention invariant.

Case contract

Input [values,p,w], len(values)=n and w has exact multiplicative order n modulo prime p; forward transform sum values[j]*w^(j*k) modulo p.

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):
    values,p,w=x;n=len(values);out=[]
    for k in range(n):
     total=0
     for j in range(n):
      factor=pow(w,j*k,p)
      total=(total+values[j]*factor)%p
     out.append(total*pow(n,-1,p)%p)
    return out
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[2, 1], 5, 4], [3, 1]), ([[2, 1], 5, 4], [3, 1]), ([[3, 0], 5, 4], [3, 3]), ([[0, 2], 5, 4], [2, 3]), ([[1, 3], 5, 4], [4, 3]), ([[0, 4], 5, 4], [4, 1])], [([[2, 1], 5, 4], [3, 1]), ([[3, 0], 5, 4], [3, 3]), ([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[0, 2], 5, 4], [2, 3]), ([[1, 1], 5, 4], [2, 0]), ([[4, 3], 5, 4], [2, 1]), ([[2, 0], 5, 4], [2, 2])], [([[2, 1], 5, 4], [3, 1]), ([[0, 4], 5, 4], [4, 1]), ([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[4, 0], 5, 4], [4, 4]), ([[1, 4], 5, 4], [0, 2]), ([[1, 2], 5, 4], [3, 4]), ([[0, 3], 5, 4], [3, 2])], [([[3, 0], 5, 4], [3, 3]), ([[3, 1], 5, 4], [4, 2]), ([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[4, 3], 5, 4], [2, 1]), ([[4, 1], 5, 4], [0, 3]), ([[1, 0], 5, 4], [1, 1]), ([[3, 3], 5, 4], [1, 0])], [([[0, 2], 5, 4], [2, 3]), ([[4, 0], 5, 4], [4, 4]), ([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[2, 5, 0], 7, 2], [0, 5, 1]), ([[0, 0, 4], 7, 2], [4, 2, 1]), ([[5, 5, 2], 7, 2], [5, 2, 1]), ([[1, 6, 1], 7, 2], [1, 3, 6])]]
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	`[4, 1]`	`[3, 2]`	Failed
explicit oracle 1	`[4, 6, 16, 6, 9, 6, 0, 14]`	`[15, 14, 9, 14, 4, 14, 0, 10]`	Failed
explicit oracle 2	`[4, 3]`	`[3, 1]`	Failed
explicit oracle 3	`[4, 3]`	`[3, 1]`	Failed
explicit oracle 4	`[4, 4]`	`[3, 3]`	Failed
explicit oracle 5	`[1, 4]`	`[2, 3]`	Failed
explicit oracle 6	`[2, 4]`	`[4, 3]`	Failed
explicit oracle 7	`[2, 3]`	`[4, 1]`	Failed

SHA-256 / de50158390fe78f29fc4fd07be0abfedcd6672edf8faafa704d745011a159acc

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):
    values,p,w=x;n=len(values);out=[]
    for k in range(n):
     total=0
     for j in range(n):
      factor=pow(w,j*k,p)
      total=(total+values[j]*factor)%p
     out.append(total*n%p)
    return out
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[2, 1], 5, 4], [3, 1]), ([[2, 1], 5, 4], [3, 1]), ([[3, 0], 5, 4], [3, 3]), ([[0, 2], 5, 4], [2, 3]), ([[1, 3], 5, 4], [4, 3]), ([[0, 4], 5, 4], [4, 1])], [([[2, 1], 5, 4], [3, 1]), ([[3, 0], 5, 4], [3, 3]), ([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[0, 2], 5, 4], [2, 3]), ([[1, 1], 5, 4], [2, 0]), ([[4, 3], 5, 4], [2, 1]), ([[2, 0], 5, 4], [2, 2])], [([[2, 1], 5, 4], [3, 1]), ([[0, 4], 5, 4], [4, 1]), ([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[4, 0], 5, 4], [4, 4]), ([[1, 4], 5, 4], [0, 2]), ([[1, 2], 5, 4], [3, 4]), ([[0, 3], 5, 4], [3, 2])], [([[3, 0], 5, 4], [3, 3]), ([[3, 1], 5, 4], [4, 2]), ([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[4, 3], 5, 4], [2, 1]), ([[4, 1], 5, 4], [0, 3]), ([[1, 0], 5, 4], [1, 1]), ([[3, 3], 5, 4], [1, 0])], [([[0, 2], 5, 4], [2, 3]), ([[4, 0], 5, 4], [4, 4]), ([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[2, 5, 0], 7, 2], [0, 5, 1]), ([[0, 0, 4], 7, 2], [4, 2, 1]), ([[5, 5, 2], 7, 2], [5, 2, 1]), ([[1, 6, 1], 7, 2], [1, 3, 6])]]
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, 2]`	Failed
explicit oracle 1	`[1, 10, 4, 10, 15, 10, 0, 12]`	`[15, 14, 9, 14, 4, 14, 0, 10]`	Failed
explicit oracle 2	`[1, 2]`	`[3, 1]`	Failed
explicit oracle 3	`[1, 2]`	`[3, 1]`	Failed
explicit oracle 4	`[1, 1]`	`[3, 3]`	Failed
explicit oracle 5	`[4, 1]`	`[2, 3]`	Failed
explicit oracle 6	`[3, 1]`	`[4, 3]`	Failed
explicit oracle 7	`[3, 2]`	`[4, 1]`	Failed

SHA-256 / bddb1a7a99f2244edb2871e248077b968d3889011a6d867baec595239843eacf

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):
    values,p,w=x;n=len(values);out=[]
    for k in range(n):
     total=0
     for j in range(n):
      factor=pow(w,j*k,p)
      total=(total+values[j]*factor)%p
     out.append(total)
    return out
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[2, 1], 5, 4], [3, 1]), ([[2, 1], 5, 4], [3, 1]), ([[3, 0], 5, 4], [3, 3]), ([[0, 2], 5, 4], [2, 3]), ([[1, 3], 5, 4], [4, 3]), ([[0, 4], 5, 4], [4, 1])], [([[2, 1], 5, 4], [3, 1]), ([[3, 0], 5, 4], [3, 3]), ([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[0, 2], 5, 4], [2, 3]), ([[1, 1], 5, 4], [2, 0]), ([[4, 3], 5, 4], [2, 1]), ([[2, 0], 5, 4], [2, 2])], [([[2, 1], 5, 4], [3, 1]), ([[0, 4], 5, 4], [4, 1]), ([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[4, 0], 5, 4], [4, 4]), ([[1, 4], 5, 4], [0, 2]), ([[1, 2], 5, 4], [3, 4]), ([[0, 3], 5, 4], [3, 2])], [([[3, 0], 5, 4], [3, 3]), ([[3, 1], 5, 4], [4, 2]), ([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[4, 3], 5, 4], [2, 1]), ([[4, 1], 5, 4], [0, 3]), ([[1, 0], 5, 4], [1, 1]), ([[3, 3], 5, 4], [1, 0])], [([[0, 2], 5, 4], [2, 3]), ([[4, 0], 5, 4], [4, 4]), ([[0, 3], 5, 4], [3, 2]), ([[10, 15, 12, 4, 14, 0, 16, 12], 17, 2], [15, 14, 9, 14, 4, 14, 0, 10]), ([[2, 5, 0], 7, 2], [0, 5, 1]), ([[0, 0, 4], 7, 2], [4, 2, 1]), ([[5, 5, 2], 7, 2], [5, 2, 1]), ([[1, 6, 1], 7, 2], [1, 3, 6])]]
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, 2]`	`[3, 2]`	Passed
explicit oracle 1	`[15, 14, 9, 14, 4, 14, 0, 10]`	`[15, 14, 9, 14, 4, 14, 0, 10]`	Passed
explicit oracle 2	`[3, 1]`	`[3, 1]`	Passed
explicit oracle 3	`[3, 1]`	`[3, 1]`	Passed
explicit oracle 4	`[3, 3]`	`[3, 3]`	Passed
explicit oracle 5	`[2, 3]`	`[2, 3]`	Passed
explicit oracle 6	`[4, 3]`	`[4, 3]`	Passed
explicit oracle 7	`[4, 1]`	`[4, 1]`	Passed

SHA-256 / 5884d77836ac915b12824d2e7ca49f84f2f810cffeb214f5c7103cba6b49ed27

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 2026-09-29T14:39:29.758791+00:00.

Case digest / edaaa3b19b8279321982435172d8d427c872209e5fa81a3c9fba42f32ae74773