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Codex/find and fix a bug (#12782)

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* Fix enumeration order in FFT string representation

* updating DIRECTORY.md

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* Update radix2_fft.py

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---------

Co-authored-by: alejandroaldas <alejandroaldas@users.noreply.github.com>
Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
Co-authored-by: Maxim Smolskiy <mithridatus@mail.ru>
This commit is contained in:
alejandroaldas
2025-08-24 17:16:05 -05:00
committed by GitHub
parent b0be304433
commit 8c1c6c1763
1 changed files with 7 additions and 7 deletions
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14
maths/radix2_fft.py
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@@ -39,14 +39,14 @@ class FFT:
>>> x = FFT(A, B) >>> x = FFT(A, B)
Print product Print product
>>> x.product # 2x + 3x^2 + 8x^3 + 4x^4 + 6x^5 >>> x.product # 2x + 3x^2 + 8x^3 + 6x^4 + 8x^5
[(-0-0j), (2+0j), (3-0j), (8-0j), (6+0j), (8+0j)] [(-0-0j), (2+0j), (3-0j), (8-0j), (6+0j), (8+0j)]
__str__ test __str__ test
>>> print(x) >>> print(x)
A = 0*x^0 + 1*x^1 + 2*x^0 + 3*x^2 A = 0*x^0 + 1*x^1 + 0*x^2 + 2*x^3
B = 0*x^2 + 1*x^3 + 2*x^4 B = 2*x^0 + 3*x^1 + 4*x^2
A*B = 0*x^(-0-0j) + 1*x^(2+0j) + 2*x^(3-0j) + 3*x^(8-0j) + 4*x^(6+0j) + 5*x^(8+0j) A*B = (-0-0j)*x^0 + (2+0j)*x^1 + (3-0j)*x^2 + (8-0j)*x^3 + (6+0j)*x^4 + (8+0j)*x^5
""" """
def __init__(self, poly_a=None, poly_b=None): def __init__(self, poly_a=None, poly_b=None):
@@ -159,13 +159,13 @@ class FFT:
# Overwrite __str__ for print(); Shows A, B and A*B # Overwrite __str__ for print(); Shows A, B and A*B
def __str__(self): def __str__(self):
a = "A = " + " + ".join( a = "A = " + " + ".join(
f"{coef}*x^{i}" for coef, i in enumerate(self.polyA[: self.len_A]) f"{coef}*x^{i}" for i, coef in enumerate(self.polyA[: self.len_A])
) )
b = "B = " + " + ".join( b = "B = " + " + ".join(
f"{coef}*x^{i}" for coef, i in enumerate(self.polyB[: self.len_B]) f"{coef}*x^{i}" for i, coef in enumerate(self.polyB[: self.len_B])
) )
c = "A*B = " + " + ".join( c = "A*B = " + " + ".join(
f"{coef}*x^{i}" for coef, i in enumerate(self.product) f"{coef}*x^{i}" for i, coef in enumerate(self.product)
) )
return f"{a}\n{b}\n{c}" return f"{a}\n{b}\n{c}"