feat(Function): Add __eq__ method and improve quadratic_solve stability #23

Merged
jono merged 1 commits from v0.6.2-dev into main 2025-11-02 16:53:38 +00:00
3 changed files with 76 additions and 3 deletions

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@@ -5,7 +5,7 @@ build-backend = "setuptools.build_meta"
[project]
# --- Core Metadata ---
name = "polysolve"
version = "0.6.1"
version = "0.6.2"
authors = [
{ name="Jonathan Rampersad", email="jonathan@jono-rams.work" },
]

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@@ -746,6 +746,21 @@ class Function:
return NotImplemented
return self
def __eq__(self, other: object) -> bool:
"""
Checks if two Function objects are equal by comparing
their coefficients.
"""
# Check if the 'other' object is even a Function
if not isinstance(other, Function):
return NotImplemented
# Ensure both are initialized before trying to access .coefficients
if not self._initialized or not other._initialized:
return False
return np.array_equal(self.coefficients, other.coefficients)
def quadratic_solve(self) -> Optional[List[float]]:
@@ -770,9 +785,35 @@ class Function:
return None # No real roots
sqrt_discriminant = math.sqrt(discriminant)
root1 = (-b + sqrt_discriminant) / (2 * a)
root2 = (-b - sqrt_discriminant) / (2 * a)
# 1. Calculate the first root.
# We use math.copysign(val, sign) to get the sign of b.
# This ensures (-b - sign*sqrt) is always an *addition*
# (or subtraction of a smaller from a larger number),
# avoiding catastrophic cancellation.
root1 = (-b - math.copysign(sqrt_discriminant, b)) / (2 * a)
# 2. Calculate the second root using Vieta's formulas.
# We know that root1 * root2 = c / a.
# This is just a division, which is numerically stable.
# Handle the edge case where c=0.
# If c=0, then root1 is 0.0, and root2 is -b/a
# We can't divide by root1=0, so we check.
if root1 == 0.0:
# If c is also 0, the other root is -b/a
if c == 0.0:
root2 = -b / a
else:
# This case (root1=0 but c!=0) shouldn't happen
# with real numbers, but it's safe to just
# return the one root we found.
return [0.0]
else:
# Standard case: Use Vieta's formula
root2 = (c / a) / root1
# Return roots in a consistent order
return [root1, root2]
# Example Usage

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@@ -37,6 +37,12 @@ def m_func_2() -> Function:
f.set_coeffs([5, -4])
return f
@pytest.fixture
def base_func():
f = Function(2)
f.set_coeffs([1, 2, 3])
return f
# --- Core Functionality Tests ---
def test_solve_y(quadratic_func):
@@ -95,6 +101,32 @@ def test_function_multiplication(m_func_1, m_func_2):
assert result.largest_exponent == 3
assert np.array_equal(result.coefficients, [10, 7, -7, -4])
def test_equality(base_func):
"""Tests the __eq__ method for the Function class."""
# 1. Test for equality with a new, identical object
f_identical = Function(2)
f_identical.set_coeffs([1, 2, 3])
assert base_func == f_identical
# 2. Test for inequality (different coefficients)
f_different = Function(2)
f_different.set_coeffs([1, 9, 3])
assert base_func != f_different
# 3. Test for inequality (different degree)
f_diff_degree = Function(1)
f_diff_degree.set_coeffs([1, 2])
assert base_func != f_diff_degree
# 4. Test against a different type
assert base_func != "some_string"
assert base_func != 123
# 5. Test against an uninitialized Function
f_uninitialized = Function(2)
assert base_func != f_uninitialized
# --- Genetic Algorithm Root-Finding Tests ---
def test_get_real_roots_numpy(quadratic_func):