R23-T055: Add extended_euclid trace and non-coprime doctests to chinese_remainder_theorem - #65

Description

Traces extended_euclid(10, 6) and chinese_remainder_theorem(5, 1, 7, 3) step by step (by hand and with pdb) and turns the observations into doctests: intermediate results of the recursion, the gcd identity, and the documented behaviour for non-coprime moduli.

Related Issue

Fixes: #55

Type of Change

• Bug Fix
• New Feature
• UI/UX Improvement
• Performance Improvement
• Tests
• Documentation
• Refactoring / Maintenance
• CI/CD

Changes Made

• Added doctests to extended_euclid: intermediate results (2,0)->(1,0), (4,2)->(0,1), (6,4)->(1,-1), the identity ax + by == gcd (2 for (10, 6)), and the a < b swap case (5, 7).
• Added doctests to chinese_remainder_theorem: the result satisfies both congruences for coprime moduli, and a non-coprime input (6, 1, 4, 3) documented as unsupported.
• Existing 8 doctests unchanged (21 total).
• Added work/R23-T055/TRACE.md (tables of both traces, pdb steps, the surprise) and trace_driver.py.

Screenshots / Recordings

Testing

• python -m pytest --doctest-modules maths/chinese_remainder_theorem.py -> passed (21 doctests)
• ruff check maths/chinese_remainder_theorem.py -> all checks passed
• pdb traces recorded in work/R23-T055/TRACE.md and checked by hand

Checklist

• I have tested my changes locally.
• I have added or updated tests, if needed.
• I have updated documentation, if needed.
• I have followed the project's coding guidelines.
• I have run the relevant tests and checks.
• I have addressed previous review feedback, if applicable.

Additional Notes

Finding: the existing doctest chinese_remainder_theorem(6, 1, 4, 3) -> 14 passes but 14 is not a valid solution (14 % 6 = 2, 14 % 4 = 2). 6 and 4 are not coprime, so extended_euclid(6, 4) gives ax + by = 2 instead of 1 and the formula breaks silently. A valid answer exists (7, modulo 12). Not fixed here (doctests only); documented in the new doctests and in work/R23-T055/TRACE.md. A follow-up could raise ValueError when gcd(n1, n2) != 1.