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I try to find the rotation matrix to align two vectors.
I have a vector A = [ax, ay, az] and I would like to align it on the vector B = [1, 0, 0] (x-axis unit vector).
I found the following explanation and tried to implement it: https://math.stackexchange.com/questions/180418/calculate-rotation-matrix-to-align-vector-a-to-vector-b-in-3d/897677#897677
def align_vectors(a, b):
v = np.cross(a, b)
s = np.linalg.norm(v)
c = np.dot(a, b)
v1, v2, v3 = v
h = 1 / (1 + c)
Vmat = np.array([[0, -v3, v2],
[v3, 0, -v1],
[-v2, v1, 0]])
R = np.eye(3, dtype=np.float64) + Vmat + (Vmat.dot(Vmat) * h)
return R
When I apply it to find the rotation of a point, this is what I have :
x_axis = np.array([1, 0, 0], dtype=np.float64)
direction = np.array([-0.02, 1.004, -0.02], dtype=np.float64)
Ralign = align_vectors(x_axis, direction)
point = 1000 * np.array([-0.02, 1.004, -0.02], dtype=np.float64) # Point in the direction of the unit vector
result = Ralign.dot(point)
The resulting point is not aligned with the unit vector.
If you only want to rotate ONE vector a to align with b, not the entire coordinate contain that vector, use simple vector projection and the length of a:
a_norm = np.linalg.norm(a)
b_norm = np.linalg.norm(b)
result = b * a_norm / b_norm
The following fixes the issue in the question that input are not unit vector by vector normalization.
def align_vectors(a, b):
b = b / np.linalg.norm(b) # normalize a
a = a / np.linalg.norm(a) # normalize b
v = np.cross(a, b)
# s = np.linalg.norm(v)
c = np.dot(a, b)
v1, v2, v3 = v
h = 1 / (1 + c)
Vmat = np.array([[0, -v3, v2],
[v3, 0, -v1],
[-v2, v1, 0]])
R = np.eye(3, dtype=np.float64) + Vmat + (Vmat.dot(Vmat) * h)
return R
testing:
def angle(a, b):
"""Angle between vectors"""
a = a / np.linalg.norm(a)
b = b / np.linalg.norm(b)
return np.arccos(a.dot(b))
point = np.array([-0.02, 1.004, -0.02])
direction = np.array([1., 0., 0.])
rotation = align_vectors(point, direction)
# Rotate point in align with direction. The result vector is aligned with direction
result = rotation.dot(point)
print(result)
print('Angle:', angle(direction, point)) # 0.0
print('Length:', np.isclose(np.linalg.norm(point), np.linalg.norm(result))) # True
# Rotate direction by the matrix, result does not align with direction but the angle between the original vector (direction) and the result2 are the same.
result2 = rotation.dot(direction)
print(result2)
print('Same Angle:', np.isclose(angle(point,result), angle(direction,result2))) # True
print('Length:', np.isclose(np.linalg.norm(direction), np.linalg.norm(result2))) # True
So here is what I have tried:
def bezier(a):
# find order of curve from number of control points
n = np.shape(a)[0]-1
# initialise arrays
B = np.zeros([101, 2])
terms = np.zeros([n+1, 2])
# create an array of values for t from 0 to 1 in 101 steps
t = np.linspace(0, 1, 101)
# loop through all t values
for i in range(0, 101):
#calculate terms inside sum in equation 13
for j in range(0, n + 1):
terms[j,:] = (1 - t[i])**2 * a[0,:] + 2 * t[i] * (1 - t[i]) * a[1,:] + t[i]**2 * a[2,:]
#sum terms to find Bezier curve
B[i, :] = sum(terms, 0)
# plot Bezier
pl.plot(B[:, 0], B[:, 1])
# plot control points
pl.plot(a[:, 0], a[:, 1],'ko')
# plot control polygon
pl.plot(a[:, 0], a[:, 1],'k')
return B
with:
a = np.array([[0, 0], [0.5, 1], [1, 0]])
B = bezier(a)
and it is returning:
this graph
which as you can see does not correspond to my control points
Any help appreciated, thanks.
The sum over j is redundant. What happens is that you create your Bezier curve but sum it three times, thus obtaining something that is three times as big as it should.
import numpy as np
import matplotlib.pyplot as pl
def bezier(a):
# find order of curve from number of control points
n = np.shape(a)[0]-1
# initialise arrays
B = np.zeros([101, 2])
terms = np.zeros([n+1, 2])
# create an array of values for t from 0 to 1 in 101 steps
t = np.linspace(0, 1, 101)
# loop through all t values
for i in range(0, 101):
#calculate terms inside sum in equation 13
B[i, :] = (1 - t[i])**2 * a[0,:] + 2 * t[i] * (1 - t[i]) * a[1,:] + t[i]**2 * a[2,:]
# plot Bezier
pl.plot(B[:, 0], B[:, 1])
# plot control points
pl.plot(a[:, 0], a[:, 1],'ko')
# plot control polygon
pl.plot(a[:, 0], a[:, 1],'k')
return B
a = np.array([[0, 0], [0.5, 1], [1, 0]])
B = bezier(a)
pl.show()
I would also recommend renaming a to something more descriptive such as controlPts.
Is there something like Matlab's procrustes function in NumPy/SciPy or related libraries?
For reference. Procrustes analysis aims to align 2 sets of points (in other words, 2 shapes) to minimize square distance between them by removing scale, translation and rotation warp components.
Example in Matlab:
X = [0 1; 2 3; 4 5; 6 7; 8 9]; % first shape
R = [1 2; 2 1]; % rotation matrix
t = [3 5]; % translation vector
Y = X * R + repmat(t, 5, 1); % warped shape, no scale and no distortion
[d Z] = procrustes(X, Y); % Z is Y aligned back to X
Z
Z =
0.0000 1.0000
2.0000 3.0000
4.0000 5.0000
6.0000 7.0000
8.0000 9.0000
Same task in NumPy:
X = arange(10).reshape((5, 2))
R = array([[1, 2], [2, 1]])
t = array([3, 5])
Y = dot(X, R) + t
Z = ???
Note: I'm only interested in aligned shape, since square error (variable d in Matlab code) is easily computed from 2 shapes.
I'm not aware of any pre-existing implementation in Python, but it's easy to take a look at the MATLAB code using edit procrustes.m and port it to Numpy:
def procrustes(X, Y, scaling=True, reflection='best'):
"""
A port of MATLAB's `procrustes` function to Numpy.
Procrustes analysis determines a linear transformation (translation,
reflection, orthogonal rotation and scaling) of the points in Y to best
conform them to the points in matrix X, using the sum of squared errors
as the goodness of fit criterion.
d, Z, [tform] = procrustes(X, Y)
Inputs:
------------
X, Y
matrices of target and input coordinates. they must have equal
numbers of points (rows), but Y may have fewer dimensions
(columns) than X.
scaling
if False, the scaling component of the transformation is forced
to 1
reflection
if 'best' (default), the transformation solution may or may not
include a reflection component, depending on which fits the data
best. setting reflection to True or False forces a solution with
reflection or no reflection respectively.
Outputs
------------
d
the residual sum of squared errors, normalized according to a
measure of the scale of X, ((X - X.mean(0))**2).sum()
Z
the matrix of transformed Y-values
tform
a dict specifying the rotation, translation and scaling that
maps X --> Y
"""
n,m = X.shape
ny,my = Y.shape
muX = X.mean(0)
muY = Y.mean(0)
X0 = X - muX
Y0 = Y - muY
ssX = (X0**2.).sum()
ssY = (Y0**2.).sum()
# centred Frobenius norm
normX = np.sqrt(ssX)
normY = np.sqrt(ssY)
# scale to equal (unit) norm
X0 /= normX
Y0 /= normY
if my < m:
Y0 = np.concatenate((Y0, np.zeros(n, m-my)),0)
# optimum rotation matrix of Y
A = np.dot(X0.T, Y0)
U,s,Vt = np.linalg.svd(A,full_matrices=False)
V = Vt.T
T = np.dot(V, U.T)
if reflection != 'best':
# does the current solution use a reflection?
have_reflection = np.linalg.det(T) < 0
# if that's not what was specified, force another reflection
if reflection != have_reflection:
V[:,-1] *= -1
s[-1] *= -1
T = np.dot(V, U.T)
traceTA = s.sum()
if scaling:
# optimum scaling of Y
b = traceTA * normX / normY
# standarised distance between X and b*Y*T + c
d = 1 - traceTA**2
# transformed coords
Z = normX*traceTA*np.dot(Y0, T) + muX
else:
b = 1
d = 1 + ssY/ssX - 2 * traceTA * normY / normX
Z = normY*np.dot(Y0, T) + muX
# transformation matrix
if my < m:
T = T[:my,:]
c = muX - b*np.dot(muY, T)
#transformation values
tform = {'rotation':T, 'scale':b, 'translation':c}
return d, Z, tform
There is a Scipy function for it: scipy.spatial.procrustes
I'm just posting its example here:
>>> import numpy as np
>>> from scipy.spatial import procrustes
>>> a = np.array([[1, 3], [1, 2], [1, 1], [2, 1]], 'd')
>>> b = np.array([[4, -2], [4, -4], [4, -6], [2, -6]], 'd')
>>> mtx1, mtx2, disparity = procrustes(a, b)
>>> round(disparity)
0.0
You can have both Ordinary Procrustes Analysis and Generalized Procrustes Analysis in python with something like this:
import numpy as np
def opa(a, b):
aT = a.mean(0)
bT = b.mean(0)
A = a - aT
B = b - bT
aS = np.sum(A * A)**.5
bS = np.sum(B * B)**.5
A /= aS
B /= bS
U, _, V = np.linalg.svd(np.dot(B.T, A))
aR = np.dot(U, V)
if np.linalg.det(aR) < 0:
V[1] *= -1
aR = np.dot(U, V)
aS = aS / bS
aT-= (bT.dot(aR) * aS)
aD = (np.sum((A - B.dot(aR))**2) / len(a))**.5
return aR, aS, aT, aD
def gpa(v, n=-1):
if n < 0:
p = avg(v)
else:
p = v[n]
l = len(v)
r, s, t, d = np.ndarray((4, l), object)
for i in range(l):
r[i], s[i], t[i], d[i] = opa(p, v[i])
return r, s, t, d
def avg(v):
v_= np.copy(v)
l = len(v_)
R, S, T = [list(np.zeros(l)) for _ in range(3)]
for i, j in np.ndindex(l, l):
r, s, t, _ = opa(v_[i], v_[j])
R[j] += np.arccos(min(1, max(-1, np.trace(r[:1])))) * np.sign(r[1][0])
S[j] += s
T[j] += t
for i in range(l):
a = R[i] / l
r = [np.cos(a), -np.sin(a)], [np.sin(a), np.cos(a)]
v_[i] = v_[i].dot(r) * (S[i] / l) + (T[i] / l)
return v_.mean(0)
For testing purposes, the output of each algorithm can be visualized as follows:
import matplotlib.pyplot as p; p.rcParams['toolbar'] = 'None';
def plt(o, e, b):
p.figure(figsize=(10, 10), dpi=72, facecolor='w').add_axes([0.05, 0.05, 0.9, 0.9], aspect='equal')
p.plot(0, 0, marker='x', mew=1, ms=10, c='g', zorder=2, clip_on=False)
p.gcf().canvas.set_window_title('%f' % e)
x = np.ravel(o[0].T[0])
y = np.ravel(o[0].T[1])
p.xlim(min(x), max(x))
p.ylim(min(y), max(y))
a = []
for i, j in np.ndindex(len(o), 2):
a.append(o[i].T[j])
O = p.plot(*a, marker='x', mew=1, ms=10, lw=.25, c='b', zorder=0, clip_on=False)
O[0].set(c='r', zorder=1)
if not b:
O[2].set_color('b')
O[2].set_alpha(0.4)
p.axis('off')
p.show()
# Fly wings example (Klingenberg, 2015 | https://en.wikipedia.org/wiki/Procrustes_analysis)
arr1 = np.array([[588.0, 443.0], [178.0, 443.0], [56.0, 436.0], [50.0, 376.0], [129.0, 360.0], [15.0, 342.0], [92.0, 293.0], [79.0, 269.0], [276.0, 295.0], [281.0, 331.0], [785.0, 260.0], [754.0, 174.0], [405.0, 233.0], [386.0, 167.0], [466.0, 59.0]])
arr2 = np.array([[477.0, 557.0], [130.129, 374.307], [52.0, 334.0], [67.662, 306.953], [111.916, 323.0], [55.119, 275.854], [107.935, 277.723], [101.899, 259.73], [175.0, 329.0], [171.0, 345.0], [589.0, 527.0], [591.0, 468.0], [299.0, 363.0], [306.0, 317.0], [406.0, 288.0]])
def opa_out(a):
r, s, t, d = opa(a[0], a[1])
a[1] = a[1].dot(r) * s + t
return a, d, False
plt(*opa_out([arr1, arr2, np.matrix.copy(arr2)]))
def gpa_out(a):
g = gpa(a, -1)
D = [avg(a)]
for i in range(len(a)):
D.append(a[i].dot(g[0][i]) * g[1][i] + g[2][i])
return D, sum(g[3])/len(a), True
plt(*gpa_out([arr1, arr2]))
Probably you want to try this package with various flavors of different Procrustes methods, https://github.com/theochem/procrustes.
I am trying to solve a complicated problem.
For example, I have a batch of 2D predicted images (softmax output, value between 0 and 1) with size: Batch x H x W and ground truth Batch x H x W
The light gray color pixels are the background with value 0, and the dark gray color pixels are the foreground with value 1. I try to compute the mass center coordinates using scipy.ndimage.center_of_mass on each ground truth image. Then I get the center location point C (red color) for each ground truth. The C points set is Batch x 1.
Now, for each pixel A (yellow color) in the predicted images, I want to get three pixels B1, B2, B3 (blue color) which are the closest to A on the line AC (here C is corresponding location of mass center in ground truth).
I used following code to get the three closest points B1, B2, B3.
def connect(ends, m=3):
d0, d1 = np.abs(np.diff(ends, axis=0))[0]
if d0 > d1:
return np.c_[np.linspace(ends[0, 0], ends[1, 0], m + 1, dtype=np.int32),
np.round(np.linspace(ends[0, 1], ends[1, 1], m + 1))
.astype(np.int32)]
else:
return np.c_[np.round(np.linspace(ends[0, 0], ends[1, 0], m + 1))
.astype(np.int32),
np.linspace(ends[0, 1], ends[1, 1], m + 1, dtype=np.int32)]
So the B points set is Batch x 3 x H x W.
Then, I want to compute like this: |Value(A)-Value(B1)|+|Value(A)-Value(B2)|+|Value(A)-Value(B3)|. The size of the result should be Batch x H x W.
Is there any numpy vectorization tricks that can be used to update the value of each pixel in predicted images? Or can this be solved using pytorch functions? I need to find a method to update the whole image. The predicted image is the softmax output. I cannot use for loop to compute each single value since it will become non-differentiable. Thanks a lot.
As suggested by #Matin, you could consider Bresenham's algorithm to get your points on the AC line.
A simplistic PyTorch implementation could be as follows (directly adapted from the pseudo-code here ; could be optimized):
import torch
def get_points_from_low(x0, y0, x1, y1, num_points=3):
dx = x1 - x0
dy = y1 - y0
xi = torch.sign(dx)
yi = torch.sign(dy)
dy = dy * yi
D = 2 * dy - dx
y = y0
x = x0
points = []
for n in range(num_points):
x = x + xi
is_D_gt_0 = (D > 0).long()
y = y + is_D_gt_0 * yi
D = D + 2 * dy - is_D_gt_0 * 2 * dx
points.append(torch.stack((x, y), dim=-1))
return torch.stack(points, dim=len(x0.shape))
def get_points_from_high(x0, y0, x1, y1, num_points=3):
dx = x1 - x0
dy = y1 - y0
xi = torch.sign(dx)
yi = torch.sign(dy)
dx = dx * xi
D = 2 * dx - dy
y = y0
x = x0
points = []
for n in range(num_points):
y = y + yi
is_D_gt_0 = (D > 0).long()
x = x + is_D_gt_0 * xi
D = D + 2 * dx - is_D_gt_0 * 2 * dy
points.append(torch.stack((x, y), dim=-1))
return torch.stack(points, dim=len(x0.shape))
def get_points_from(x0, y0, x1, y1, num_points=3):
is_dy_lt_dx = (torch.abs(y1 - y0) < torch.abs(x1 - x0)).long()
is_x0_gt_x1 = (x0 > x1).long()
is_y0_gt_y1 = (y0 > y1).long()
sign = 1 - 2 * is_x0_gt_x1
x0_comp, x1_comp, y0_comp, y1_comp = x0 * sign, x1 * sign, y0 * sign, y1 * sign
points_low = get_points_from_low(x0_comp, y0_comp, x1_comp, y1_comp, num_points=num_points)
points_low *= sign.view(-1, 1, 1).expand_as(points_low)
sign = 1 - 2 * is_y0_gt_y1
x0_comp, x1_comp, y0_comp, y1_comp = x0 * sign, x1 * sign, y0 * sign, y1 * sign
points_high = get_points_from_high(x0_comp, y0_comp, x1_comp, y1_comp, num_points=num_points) * sign
points_high *= sign.view(-1, 1, 1).expand_as(points_high)
is_dy_lt_dx = is_dy_lt_dx.view(-1, 1, 1).expand(-1, num_points, 2)
points = points_low * is_dy_lt_dx + points_high * (1 - is_dy_lt_dx)
return points
# Inputs:
# (#todo: extend A to cover all points in maps):
A = torch.LongTensor([[0, 1], [8, 6]])
C = torch.LongTensor([[6, 4], [2, 3]])
num_points = 3
# Getting points between A and C:
# (#todo: what if there's less than `num_points` between A-C?)
Bs = get_points_from(A[:, 0], A[:, 1], C[:, 0], C[:, 1], num_points=num_points)
print(Bs)
# tensor([[[1, 1],
# [2, 2],
# [3, 2]],
# [[7, 6],
# [6, 5],
# [5, 5]]])
Once you have your points, you could retrieve their "values" (Value(A), Value(B1), etc.) using torch.index_select() (note that as of now, this method only accept 1D indices, so you need to unravel your data). All things put together, this would look like something such as the following (extending A from shape (Batch, 2) to (Batch, H, W, 2) is left for exercise...)
# Inputs:
# (#todo: extend A to cover all points in maps):
A = torch.LongTensor([[0, 1], [8, 6]])
C = torch.LongTensor([[6, 4], [2, 3]])
batch_size = A.shape[0]
num_points = 3
map_size = (9, 9)
map_num_elements = map_size[0] * map_size[1]
map_values = torch.stack((torch.arange(0, map_num_elements).view(*map_size),
torch.arange(0, -map_num_elements, -1).view(*map_size)))
# Getting points between A and C:
# (#todo: what if there's less than `num_points` between A-C?)
Bs = get_points_from(A[:, 0], A[:, 1], C[:, 0], C[:, 1], num_points=num_points)
# Get map values in positions A:
A_unravel = torch.arange(0, batch_size) * map_num_elements
A_unravel = A_unravel + A[:, 0] * map_size[1] + A[:, 1]
values_A = torch.index_select(map_values.view(-1), dim=0, index=A_unravel)
print(values_A)
# tensor([ 1, -4])
# Get map values in positions A:
A_unravel = torch.arange(0, batch_size) * map_num_elements
A_unravel = A_unravel + A[:, 0] * map_size[1] + A[:, 1]
values_A = torch.index_select(map_values.view(-1), dim=0, index=A_unravel)
print(values_A)
# tensor([ 1, -78])
# Get map values in positions B:
Bs_flatten = Bs.view(-1, 2)
Bs_unravel = (torch.arange(0, batch_size)
.unsqueeze(1)
.repeat(1, num_points)
.view(num_points * batch_size) * map_num_elements)
Bs_unravel = Bs_unravel + Bs_flatten[:, 0] * map_size[1] + Bs_flatten[:, 1]
values_B = torch.index_select(map_values.view(-1), dim=0, index=Bs_unravel)
values_B = values_B.view(batch_size, num_points)
print(values_B)
# tensor([[ 10, 20, 29],
# [-69, -59, -50]])
# Compute result:
res = torch.abs(values_A.unsqueeze(-1).expand_as(values_B) - values_B)
print(res)
# tensor([[ 9, 19, 28],
# [ 9, 19, 28]])
res = torch.sum(res, dim=1)
print(res)
# tensor([56, 56])
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).
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.)