FA-14971 / Numerics / Open access

Egyptian greedy fraction: residual numerator · case 01

The exact egyptian greedy fraction result violates the stated contract at residual numerator.

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

ROOT CAUSE

The residual numerator step uses q-pd instead of pd-q.

THE FAILURE

The residual numerator step uses q-p*d instead of p*d-q.

Unsuccessful approach: The partial repair p-q still violates the residual numerator invariant.

Case contract

Input [p,q], 0<p<q<=100; return greedy unit fraction denominators whose sum is p/q. Bounds: q<=31.

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):
    p,q=x;out=[]
    for _ in range(20):
     if p==0:break
     if abs(q).bit_length()>512:return None
     d=(q+p-1)//p
     if d<=0:return None
     out.append(d)
     p,q=q-p*d,q*d
     g=math.gcd(p,q)
     if not g:return None
     p,q=p//g,q//g
    return out
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 3], [2, 6]), ([1, 2], [2]), ([30, 31], [2, 3, 8, 107, 15922, 633759288]), ([1, 3], [3]), ([1, 4], [4]), ([2, 4], [2]), ([3, 4], [2, 4]), ([1, 5], [5])], [([3, 4], [2, 4]), ([1, 4], [4]), ([1, 2], [2]), ([30, 31], [2, 3, 8, 107, 15922, 633759288]), ([3, 7], [3, 11, 231]), ([4, 7], [2, 14]), ([5, 7], [2, 5, 70]), ([6, 7], [2, 3, 42])], [([2, 5], [3, 15]), ([1, 5], [5]), ([1, 2], [2]), ([30, 31], [2, 3, 8, 107, 15922, 633759288]), ([7, 9], [2, 4, 36]), ([8, 9], [2, 3, 18]), ([1, 10], [10]), ([2, 10], [5])], [([3, 5], [2, 10]), ([4, 5], [2, 4, 20]), ([1, 2], [2]), ([30, 31], [2, 3, 8, 107, 15922, 633759288]), ([7, 11], [2, 8, 88]), ([8, 11], [2, 5, 37, 4070]), ([9, 11], [2, 4, 15, 660]), ([10, 11], [2, 3, 14, 231])], [([4, 5], [2, 4, 20]), ([3, 6], [2]), ([1, 2], [2]), ([30, 31], [2, 3, 8, 107, 15922, 633759288]), ([3, 13], [5, 33, 2145]), ([4, 13], [4, 18, 468]), ([5, 13], [3, 20, 780]), ([6, 13], [3, 8, 312])]]
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, 6]`	Failed
explicit oracle 1	`[2]`	`[2]`	Passed
explicit oracle 2	`None`	`[2, 3, 8, 107, 15922, 633759288]`	Failed
explicit oracle 3	`[3]`	`[3]`	Passed
explicit oracle 4	`[4]`	`[4]`	Passed
explicit oracle 5	`[2]`	`[2]`	Passed
explicit oracle 6	`None`	`[2, 4]`	Failed
explicit oracle 7	`[5]`	`[5]`	Passed

SHA-256 / a6171b00d21aa78bd2345eac82704bf989afe427d1206c856b8da53365732b9c

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):
    p,q=x;out=[]
    for _ in range(20):
     if p==0:break
     if abs(q).bit_length()>512:return None
     d=(q+p-1)//p
     if d<=0:return None
     out.append(d)
     p,q=p-q,q*d
     g=math.gcd(p,q)
     if not g:return None
     p,q=p//g,q//g
    return out
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
fixtures = [[([2, 3], [2, 6]), ([1, 2], [2]), ([30, 31], [2, 3, 8, 107, 15922, 633759288]), ([1, 3], [3]), ([1, 4], [4]), ([2, 4], [2]), ([3, 4], [2, 4]), ([1, 5], [5])], [([3, 4], [2, 4]), ([1, 4], [4]), ([1, 2], [2]), ([30, 31], [2, 3, 8, 107, 15922, 633759288]), ([3, 7], [3, 11, 231]), ([4, 7], [2, 14]), ([5, 7], [2, 5, 70]), ([6, 7], [2, 3, 42])], [([2, 5], [3, 15]), ([1, 5], [5]), ([1, 2], [2]), ([30, 31], [2, 3, 8, 107, 15922, 633759288]), ([7, 9], [2, 4, 36]), ([8, 9], [2, 3, 18]), ([1, 10], [10]), ([2, 10], [5])], [([3, 5], [2, 10]), ([4, 5], [2, 4, 20]), ([1, 2], [2]), ([30, 31], [2, 3, 8, 107, 15922, 633759288]), ([7, 11], [2, 8, 88]), ([8, 11], [2, 5, 37, 4070]), ([9, 11], [2, 4, 15, 660]), ([10, 11], [2, 3, 14, 231])], [([4, 5], [2, 4, 20]), ([3, 6], [2]), ([1, 2], [2]), ([30, 31], [2, 3, 8, 107, 15922, 633759288]), ([3, 13], [5, 33, 2145]), ([4, 13], [4, 18, 468]), ([5, 13], [3, 20, 780]), ([6, 13], [3, 8, 312])]]
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, 6]`	Failed
explicit oracle 1	`None`	`[2]`	Failed
explicit oracle 2	`None`	`[2, 3, 8, 107, 15922, 633759288]`	Failed
explicit oracle 3	`None`	`[3]`	Failed
explicit oracle 4	`None`	`[4]`	Failed
explicit oracle 5	`None`	`[2]`	Failed
explicit oracle 6	`None`	`[2, 4]`	Failed
explicit oracle 7	`None`	`[5]`	Failed

SHA-256 / bb5b687eeb002e119bd020e00f4f2ff1c8f0b3877e1d932e68b17396650c0485

HELD IN THE MEMBER ARCHIVE

The verified repair and its recorded checks are member-only.

This mechanism has 8 recorded checks per implementation. The open-access tier publishes the failure and the unsuccessful fix; the repaired source that passes every check, and the observations that prove it, are available to members.

Every case sharing this mechanism uses the same contract and the same repair, so this one record is held back for all of them.

Member access is invitation-based. Sign in with your invited account to inspect the repair.

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 2026-09-29T14:39:22.331795+00:00.

Case digest / 70f3cffab2219daa1bdbb10644488209d123b6c24e9d8debac3933dd17b31d19