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fft division for fast polynomial division

I'm trying to implement fast polynomial division using Fast Fourier Transform (fft).

Here is what I have got so far:

from numpy.fft import fft, ifft
def fft_div(C1, C2):
    # fft expects right-most for significant coefficients
    C1 = C1[::-1]
    C2 = C2[::-1]
    d = len(C1)+len(C2)-1
    c1 = fft(list(C1) + [0] * (d-len(C1)))
    c2 = fft(list(C2) + [0] * (d-len(C2)))
    res = list(ifft(c1-c2)[:d].real)
    # Reorder back to left-most and round to integer
    return [int(round(x)) for x in res[::-1]]

This works well for polynomials of same length, but if length is different then the result is wrong (I benchmark against RosettaCode's extended_synthetic_division() function):

# Most signficant coefficient is left
N = [1, -11, 0, -22, 1]
D = [1, -3, 0, 1, 2]
# OK case, same length for both polynomials
fft_div(N, D)
>> [0, 0, 0, 0, 0, -8, 0, -23, -1]
extended_synthetic_division(N, D)
>> ([1], [-8, 0, -23, -1])

# NOT OK case, D is longer than N (also happens if shorter)
D = [1, -3, 0, 1, 2, 20]
fft_div(N, D)
>> [0, 0, 0, 0, -1, 4, -11, -1, -24, -19]
extended_synthetic_division(N, D)
>> ([], [1, -11, 0, -22, 1])

What is weird is that it seems it's very close, but still a bit off. What did I do wrong? In other words: how to generalize fast polynomial division (using FFT) to vectors of different sizes.

Also bonus if you can tell me how to compute the division quotient (currently I only have the remainder).

like image 821
gaborous Avatar asked Jun 27 '17 00:06

gaborous


1 Answers

Here's a direct implementation of a fast polynomial division algorithm found in these lecture notes.

The division is based on the fast/FFT multiplication of dividend with the divisor's reciprocal. My implementation below strictly follows the algorithm proven to have O(n*log(n)) time complexity (for polynomials with degrees of the same order of magnitude), but it's written with emphasis on readability, not efficiency.

from math import ceil, log
from numpy.fft import fft, ifft

def poly_deg(p):
    return len(p) - 1


def poly_scale(p, n):
    """Multiply polynomial ``p(x)`` with ``x^n``.
    If n is negative, poly ``p(x)`` is divided with ``x^n``, and remainder is
    discarded (truncated division).
    """
    if n >= 0:
        return list(p) + [0] * n
    else:
        return list(p)[:n]


def poly_scalar_mul(a, p):
    """Multiply polynomial ``p(x)`` with scalar (constant) ``a``."""
    return [a*pi for pi in p]


def poly_extend(p, d):
    """Extend list ``p`` representing a polynomial ``p(x)`` to
    match polynomials of degree ``d-1``.
    """
    return [0] * (d-len(p)) + list(p)


def poly_norm(p):
    """Normalize the polynomial ``p(x)`` to have a non-zero most significant
    coefficient.
    """
    for i,a in enumerate(p):
        if a != 0:
            return p[i:]
    return []


def poly_add(u, v):
    """Add polynomials ``u(x)`` and ``v(x)``."""
    d = max(len(u), len(v))
    return [a+b for a,b in zip(poly_extend(u, d), poly_extend(v, d))]


def poly_sub(u, v):
    """Subtract polynomials ``u(x)`` and ``v(x)``."""
    d = max(len(u), len(v))
    return poly_norm([a-b for a,b in zip(poly_extend(u, d), poly_extend(v, d))])


def poly_mul(u, v):
    """Multiply polynomials ``u(x)`` and ``v(x)`` with FFT."""
    if not u or not v:
        return []
    d = poly_deg(u) + poly_deg(v) + 1
    U = fft(poly_extend(u, d)[::-1])
    V = fft(poly_extend(v, d)[::-1])
    res = list(ifft(U*V).real)
    return [int(round(x)) for x in res[::-1]]


def poly_recip(p):
    """Calculate the reciprocal of polynomial ``p(x)`` with degree ``k-1``,
    defined as: ``x^(2k-2) / p(x)``, where ``k`` is a power of 2.
    """
    k = poly_deg(p) + 1
    assert k>0 and p[0] != 0 and 2**round(log(k,2)) == k

    if k == 1:
        return [1 / p[0]]

    q = poly_recip(p[:k/2])
    r = poly_sub(poly_scale(poly_scalar_mul(2, q), 3*k/2-2),
                 poly_mul(poly_mul(q, q), p))

    return poly_scale(r, -k+2)


def poly_divmod(u, v):
    """Fast polynomial division ``u(x)`` / ``v(x)`` of polynomials with degrees
    m and n. Time complexity is ``O(n*log(n))`` if ``m`` is of the same order
    as ``n``.
    """
    if not u or not v:
        return []
    m = poly_deg(u)
    n = poly_deg(v)

    # ensure deg(v) is one less than some power of 2
    # by extending v -> ve, u -> ue (mult by x^nd)
    nd = int(2**ceil(log(n+1, 2))) - 1 - n
    ue = poly_scale(u, nd)
    ve = poly_scale(v, nd)
    me = m + nd
    ne = n + nd

    s = poly_recip(ve)
    q = poly_scale(poly_mul(ue, s), -2*ne)

    # handle the case when m>2n
    if me > 2*ne:
        # t = x^2n - s*v
        t = poly_sub(poly_scale([1], 2*ne), poly_mul(s, ve))
        q2, r2 = poly_divmod(poly_scale(poly_mul(ue, t), -2*ne), ve)
        q = poly_add(q, q2)

    # remainder, r = u - v*q
    r = poly_sub(u, poly_mul(v, q))

    return q, r

The poly_divmod(u, v) function returns a (quotient, remainder) tuple for polynomials u and v (like Python's standard divmod for numbers).

For example:

>>> print poly_divmod([1,0,-1], [1,-1])
([1, 1], [])
>>> print poly_divmod([3,-5,10,8], [1,2,-3])
([3, -11], [41, -25])
>>> print poly_divmod([1, -11, 0, -22, 1], [1, -3, 0, 1, 2])
([1], [-8, 0, -23, -1])
>>> print poly_divmod([1, -11, 0, -22, 1], [1, -3, 0, 1, 2, 20])
([], [1, -11, 0, -22, 1])

I.e:

  • (x^2 - 1) / (x - 1) == x + 1
  • (2x^3 - 5x^2 + 10x + 8) / (x^2 + 2x -3) == 3x - 11, with remainder 41x - 25
  • etc. (Last two examples are yours.)
like image 67
randomir Avatar answered Sep 18 '22 00:09

randomir