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I have a task at hand where I need to detect angle of an image like the following sample (part of microchip photograph). The image does contain orthogonal features, but they could have different size, with different resolution/sharpness. Image will be slightly imperfect due to some optical distortion and aberrations. Sub-pixel angle detection accuracy is required (i.e. it should be well under <0.1° error, something like 0.01° would be tolerable). For reference, for this image optimal angle is around 32.19°.

enter image description here Currently I've tried 2 approaches: Both do a brute-force search for a local minimum with 2° step, then gradient descend down to 0.0001° step size.

  1. Merit function is sum(pow(img(x+1)-img(x-1), 2) + pow(img(y+1)-img(y-1)) calculated across the image. When horizontal/vertical lines are aligned - there is less change in horizontal/vertical directions. Precision was about 0.2°.
  2. Merit function is (max-min) over some stripe width/height of the image. This stripe is also looped across the image, and merit function is accumulated. This approach also focuses on smaller change of brightness when horizontal/vertical lines are aligned, but it can detect smaller changes over larger base (stripe width - which could be around 100 pixels wide). This gives better precision, up to 0.01° - but has a lot of parameters to tweak (stripe width/height for example is quite sensitive) which could be unreliable in real world.

Edge detection filter did not helped much.

My concern is very small change in merit function in both cases between worst and best angles (<2x difference).

Do you have any better suggestions on writing merit function for angle detection?

Update: Full size sample image is uploaded here (51 MiB)

After all processing it will end up looking like this.

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    $\begingroup$ It is very sad that it was transitioned from stackoverflow to dsp. I do not see a DSP-like solution here, and chances are now much reduced. 99.9% of DSP algorithms and tricks are useless for this task. It seems like custom algorithm or approach is needed here, not an FFT. $\endgroup$ – BarsMonster May 10 at 8:50
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    $\begingroup$ I'm super happy to tell you that it's totally wrong to be sad; DSP.SE is the absolute right place to ask this! (not so much stackoverflow. It's not a programming question. You know your programming. You don't know how to process this image.) Images are signals, and DSP.SE concerns itself very much with image processing! Also, a great deal of general DSP tricks (even as known for e.g. communication signals) are very applicable to your problem :) $\endgroup$ – Marcus Müller May 10 at 16:47
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    $\begingroup$ How important is efficiency? $\endgroup$ – Cedron Dawg May 10 at 19:11
  • $\begingroup$ by the way, even when running with a 0.04° resolution, I'm pretty sure the rotation is exactly 32°, not 32.19° – what are the resolutions of your original photography? Because at 800 px width, an uncorrected rotation of 0.01° is but 0.14 px height difference, and that would even under sinc interpolation be barely noticeable. $\endgroup$ – Marcus Müller May 10 at 20:42
  • $\begingroup$ @CedronDawg Definitely no real-time requirements, I can tolerate some 10-60 seconds of computation on some 8-12 cores. $\endgroup$ – BarsMonster May 10 at 23:43
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If I understand your method 1 correctly, with it, if you used a circularly symmetrical region and did the rotation about the center of the region, you would eliminate the region's dependency on the rotation angle and get a more fair comparison by the merit function between different rotation angles. I will suggest a method that is essentially equivalent to that, but uses the full image and does not require repeated image rotation, and will include low-pass filtering to remove pixel grid anisotropy and for denoising.

Gradient of isotropically low-pass filtered image

First, let's calculate a local gradient vector at each pixel for the green color channel in the full size sample image.

I derived horizontal and vertical differentiation kernels by differentiating the continuous-space impulse response of an ideal low-pass filter with a flat circular frequency response that removes the effect of the choice of image axes by ensuring that there is no different level of detail diagonally compared to horizontally or vertically, by sampling the resulting function, and by applying a rotated cosine window:

$$\begin{gather}h_x[x, y] = \begin{cases}0&\text{if }x = y = 0,\\-\displaystyle\frac{\omega_c^2\,x\,J_2\left(\omega_c\sqrt{x^2 + y^2}\right)}{2 \pi\,(x^2 + y^2)}&\text{otherwise,}\end{cases}\\ h_y[x, y] = \begin{cases}0&\text{if }x = y = 0,\\-\displaystyle\frac{\omega_c^2\,y\,J_2\left(\omega_c\sqrt{x^2 + y^2}\right)}{2 \pi\,(x^2 + y^2)}&\text{otherwise,}\end{cases}\end{gather}\tag{1}$$

where $J_2$ is a 2nd order Bessel function of the first kind, and $\omega_c$ is the cut-off frequency in radians. Python source (does not have the minus signs of Eq. 1):

import matplotlib.pyplot as plt
import scipy
import scipy.special
import numpy as np

def rotatedCosineWindow(N):  # N = horizontal size of the targeted kernel, also its vertical size, must be odd.
  return np.fromfunction(lambda y, x: np.maximum(np.cos(np.pi/2*np.sqrt(((x - (N - 1)/2)/((N - 1)/2 + 1))**2 + ((y - (N - 1)/2)/((N - 1)/2 + 1))**2)), 0), [N, N])

def circularLowpassKernelX(omega_c, N):  # omega = cutoff frequency in radians (pi is max), N = horizontal size of the kernel, also its vertical size, must be odd.
  kernel = np.fromfunction(lambda y, x: omega_c**2*(x - (N - 1)/2)*scipy.special.jv(2, omega_c*np.sqrt((x - (N - 1)/2)**2 + (y - (N - 1)/2)**2))/(2*np.pi*((x - (N - 1)/2)**2 + (y - (N - 1)/2)**2)), [N, N])
  kernel[(N - 1)//2, (N - 1)//2] = 0
  return kernel

def circularLowpassKernelY(omega_c, N):  # omega = cutoff frequency in radians (pi is max), N = horizontal size of the kernel, also its vertical size, must be odd.
  kernel = np.fromfunction(lambda y, x: omega_c**2*(y - (N - 1)/2)*scipy.special.jv(2, omega_c*np.sqrt((x - (N - 1)/2)**2 + (y - (N - 1)/2)**2))/(2*np.pi*((x - (N - 1)/2)**2 + (y - (N - 1)/2)**2)), [N, N])
  kernel[(N - 1)//2, (N - 1)//2] = 0
  return kernel

N = 41  # Horizontal size of the kernel, also its vertical size. Must be odd.
window = rotatedCosineWindow(N)

# Optional window function plot
#plt.imshow(window, vmin=-np.max(window), vmax=np.max(window), cmap='bwr')
#plt.colorbar()
#plt.show()

omega_c = np.pi/4  # Cutoff frequency in radians <= pi
kernelX = circularLowpassKernelX(omega_c, N)*window
kernelY = circularLowpassKernelY(omega_c, N)*window

# Optional kernel plot
#plt.imshow(kernelX, vmin=-np.max(kernelX), vmax=np.max(kernelX), cmap='bwr')
#plt.colorbar()
#plt.show()

enter image description here
Figure 1. 2-d rotated cosine window.

enter image description here
enter image description here
enter image description here
Figure 2. Windowed horizontal isotropic-low-pass differentiation kernels, for different cut-off frequency $\omega_c$ settings. Top: omega_c = np.pi, middle: omega_c = np.pi/4, bottom: omega_c = np.pi/16. The minus sign of Eq. 1 was left out. Vertical kernels look the same but have been rotated 90 degrees. A weighted sum of the horizontal and the vertical kernels, with weights $\cos(\phi)$ and $\sin(\phi)$, respectively, gives an analysis kernel of the same type for gradient angle $\phi$.

Differentiation of the impulse response does not affect the bandwidth, as can be seen by its 2-d fast Fourier transform (FFT), in Python:

# Optional FFT plot
absF = np.abs(np.fft.fftshift(np.fft.fft2(circularLowpassKernelX(np.pi, N)*window)))
plt.imshow(absF, vmin=0, vmax=np.max(absF), cmap='Greys', extent=[-np.pi, np.pi, -np.pi, np.pi])
plt.colorbar()
plt.show()

enter image description here
Figure 3. Magnitude of the 2-d FFT of $h_x$. In frequency domain, differentiation appears as multiplication of the flat circular pass band by $\omega_x$, and by a 90 degree phase shift which is not visible in the magnitude.

To do the convolution for the green channel and to collect a 2-d gradient vector histogram, for visual inspection, in Python:

import scipy.ndimage

img = plt.imread('sample.tif').astype(float)
X = scipy.ndimage.convolve(img[:,:,1], kernelX)[(N - 1)//2:-(N - 1)//2, (N - 1)//2:-(N - 1)//2]  # Green channel only
Y = scipy.ndimage.convolve(img[:,:,1], kernelY)[(N - 1)//2:-(N - 1)//2, (N - 1)//2:-(N - 1)//2]  # ...

# Optional 2-d histogram
#hist2d, xEdges, yEdges = np.histogram2d(X.flatten(), Y.flatten(), bins=199)
#plt.imshow(hist2d**(1/2.2), vmin=0, cmap='Greys')
#plt.show()
#plt.imsave('hist2d.png', plt.cm.Greys(plt.Normalize(vmin=0, vmax=hist2d.max()**(1/2.2))(hist2d**(1/2.2))))  # To save the histogram image
#plt.imsave('histkey.png', plt.cm.Greys(np.repeat([(np.arange(200)/199)**(1/2.2)], 16, 0)))

This also crops the data, discarding (N - 1)//2 pixels from each edge that were contaminated by the rectangular image boundary, before the histogram analysis.

enter image description here$\pi$ enter image description here$\frac{\pi}{2}$ enter image description here$\frac{\pi}{4}$
enter image description here$\frac{\pi}{8}$ enter image description here$\frac{\pi}{16}$ enter image description here$\frac{\pi}{32}$ enter image description here$\frac{\pi}{64}$ enter image description here$0$
Figure 4. 2-d histograms of gradient vectors, for different low-pass filter cutoff frequency $\omega_c$ settings. In order: first with N=41: omega_c = np.pi, omega_c = np.pi/2, omega_c = np.pi/4 (same as in the Python listing), omega_c = np.pi/8, omega_c = np.pi/16, then: N=81: omega_c = np.pi/32, N=161: omega_c = np.pi/64 . Denoising by low-pass filtering sharpens the circuit trace edge gradient orientations in the histogram.

Vector length weighted circular mean direction

There is the Yamartino method of finding the "average" wind direction from multiple wind vector samples in one pass through the samples. It is based on the mean of circular quantities, which is calculated as the shift of a cosine that is a sum of cosines each shifted by a circular quantity of period $2\pi$. We can use a vector length weighted version of the same method, but first we need to bunch together all the directions that are equal modulo $\pi/2$. We can do this by multiplying the angle of each gradient vector $[X_k,Y_k]$ by 4, using a complex number representation:

$$Z_k = \frac{(X_k + Y_k i)^4}{\sqrt{X_k^2 + Y_k^2}^3} = \frac{X_k^4 - 6X_k^2Y_k^2 + Y_k^4 + (4X_k^3Y_k - 4X_kY_k^3)i}{\sqrt{X_k^2 + Y_k^2}^3},\tag{2}$$

satisfying $|Z_k| = \sqrt{X_k^2 + Y_k^2}$ and by later interpreting that the phases of $Z_k$ from $-\pi$ to $\pi$ represent angles from $-\pi/4$ to $\pi/4$, by dividing the calculated circular mean phase by 4:

$$\phi = \frac{1}{4}\operatorname{atan2}\left(\sum_k\operatorname{Im}(Z_k), \sum_k\operatorname{Re}(Z_k)\right)\tag{3}$$

where $\phi$ is the estimated image orientation.

The quality of the estimate can be assessed by doing another pass through the data and by calculating the mean weighted square circular distance, $\text{MSCD}$, between phases of the complex numbers $Z_k$ and the estimated circular mean phase $4\phi$, with $|Z_k|$ as the weight:

$$\begin{gather}\text{MSCD} = \frac{\sum_k|Z_k|\bigg(1 - \cos\Big(4\phi - \operatorname{atan2}\big(\operatorname{Im}(Z_k), \operatorname{Re}(Z_k)\big)\Big)\bigg)}{\sum_k|Z_k|}\\ = \frac{\sum_k\frac{|Z_k|}{2}\left(\left(\cos(4\phi) - \frac{\operatorname{Re}(Z_k)}{|Z_k|}\right)^2 + \left(\sin(4\phi) - \frac{\operatorname{Im}(Z_k)}{|Z_k|}\right)^2\right)}{\sum_k|Z_k|}\\ = \frac{\sum_k\big(|Z_k| - \operatorname{Re}(Z_k)\cos(4\phi) - \operatorname{Im}(Z_k)\sin(4\phi)\big)}{\sum_k|Z_k|},\end{gather}\tag{4}$$

which was minimized by $\phi$ calculated per Eq. 3. In Python:

absZ = np.sqrt(X**2 + Y**2)
reZ = (X**4 - 6*X**2*Y**2 + Y**4)/absZ**3
imZ = (4*X**3*Y - 4*X*Y**3)/absZ**3
phi = np.arctan2(np.sum(imZ), np.sum(reZ))/4

sumWeighted = np.sum(absZ - reZ*np.cos(4*phi) - imZ*np.sin(4*phi))
sumAbsZ = np.sum(absZ)
mscd = sumWeighted/sumAbsZ

print("rotate", -phi*180/np.pi, "deg, RMSCD =", np.arccos(1 - mscd)/4*180/np.pi, "deg equivalent (weight = length)")

Based on my mpmath experiments (not shown), I think we won't run out of numerical precison even for very large images. For different filter settings (annotated) the outputs are, as reported between -45 and 45 degrees:

rotate 32.29809399495655 deg, RMSCD = 17.057059965741338 deg equivalent (omega_c = np.pi)
rotate 32.07672617150525 deg, RMSCD = 16.699056648843566 deg equivalent (omega_c = np.pi/2)
rotate 32.13115293914797 deg, RMSCD = 15.217534399922902 deg equivalent (omega_c = np.pi/4, same as in the Python listing)
rotate 32.18444156018288 deg, RMSCD = 14.239347706786056 deg equivalent (omega_c = np.pi/8)
rotate 32.23705383489169 deg, RMSCD = 13.63694582160468 deg equivalent (omega_c = np.pi/16)

Strong low-pass filtering appear useful, reducing the root mean square circular distance (RMSCD) equivalent angle calculated as $\operatorname{acos}(1 - \text{MSCD})$. Without the 2-d rotated cosine window, some of the results would be off by a degree or so (not shown), which means that it is important to do proper windowing of the analysis filters. The RMSCD equivalent angle is not directly an estimate of the error in the angle estimate, which should be much less.

Alternative square-length weight function

Let's try square of the vector length as an alternative weight function, by:

$$Z_k = \frac{(X_k + Y_k i)^4}{\sqrt{X_k^2 + Y_k^2}^2} = \frac{X_k^4 - 6X_k^2Y_k^2 + Y_k^4 + (4X_k^3Y_k - 4X_kY_k^3)i}{X_k^2 + Y_k^2},\tag{5}$$

In Python:

absZ_alt = X**2 + Y**2
reZ_alt = (X**4 - 6*X**2*Y**2 + Y**4)/absZ_alt
imZ_alt = (4*X**3*Y - 4*X*Y**3)/absZ_alt
phi_alt = np.arctan2(np.sum(imZ_alt), np.sum(reZ_alt))/4

sumWeighted_alt = np.sum(absZ_alt - reZ_alt*np.cos(4*phi_alt) - imZ_alt*np.sin(4*phi_alt))
sumAbsZ_alt = np.sum(absZ_alt)
mscd_alt = sumWeighted_alt/sumAbsZ_alt

print("rotate", -phi_alt*180/np.pi, "deg, RMSCD =", np.arccos(1 - mscd_alt)/4*180/np.pi, "deg equivalent (weight = length^2)")

The square length weight reduces the RMSCD equivalent angle by about a degree:

rotate 32.264713568426764 deg, RMSCD = 16.06582418749094 deg equivalent (weight = length^2, omega_c = np.pi, N = 41)
rotate 32.03693157762725 deg, RMSCD = 15.839593856962486 deg equivalent (weight = length^2, omega_c = np.pi/2, N = 41)
rotate 32.11471435914187 deg, RMSCD = 14.315371970649874 deg equivalent (weight = length^2, omega_c = np.pi/4, N = 41)
rotate 32.16968341455537 deg, RMSCD = 13.624896827482049 deg equivalent (weight = length^2, omega_c = np.pi/8, N = 41)
rotate 32.22062839958777 deg, RMSCD = 12.495324176281466 deg equivalent (weight = length^2, omega_c = np.pi/16, N = 41)
rotate 32.22385477783647 deg, RMSCD = 13.629915935941973 deg equivalent (weight = length^2, omega_c = np.pi/32, N = 81)
rotate 32.284350817263906 deg, RMSCD = 12.308297934977746 deg equivalent (weight = length^2, omega_c = np.pi/64, N = 161)

This seems a slightly better weight function. I added also cutoffs $\omega_c = \pi/32$ and $\omega_c = \pi/64$. They use larger N resulting in a different cropping of the image and not strictly comparable MSCD values.

1-d histogram

The benefit of the square-length weight function is more apparent with a 1-d weighted histogram of $Z_k$ phases. Python script:

# Optional histogram
hist_plain, bin_edges = np.histogram(np.arctan2(imZ, reZ), weights=np.ones(absZ.shape)/absZ.size, bins=900)
hist, bin_edges = np.histogram(np.arctan2(imZ, reZ), weights=absZ/np.sum(absZ), bins=900)
hist_alt, bin_edges = np.histogram(np.arctan2(imZ_alt, reZ_alt), weights=absZ_alt/np.sum(absZ_alt), bins=900)
plt.plot((bin_edges[:-1]+(bin_edges[1]-bin_edges[0]))*45/np.pi, hist_plain, "black")
plt.plot((bin_edges[:-1]+(bin_edges[1]-bin_edges[0]))*45/np.pi, hist, "red")
plt.plot((bin_edges[:-1]+(bin_edges[1]-bin_edges[0]))*45/np.pi, hist_alt, "blue")
plt.xlabel("angle (degrees)")
plt.show()

enter image description here enter image description here
Figure 5. Linearly interpolated weighted histogram of gradient vector angles, wrapped to $-\pi/4\ldots\pi/4$ and weighted by (in order from bottom to top at the peak): no weighting (black), gradient vector length (red), square of gradient vector length (blue). The bin width is 0.1 degrees. Filter cutoff was omega_c = np.pi/4, same as in the Python listing. The bottom figure is zoomed at the peaks.

Steerable filter math

We have seen that the approach works, but it would be good to have a better mathematical understanding. The $x$ and $y$ differentiation filter impulse responses given by Eq. 1 can be understood as the basis functions for forming the impulse response of a steerable differentiation filter that is sampled from a rotation of the right side of the equation for $h_x[x, y]$ (Eq. 1). This is more easily seen by converting Eq. 1 to polar coordinates:

$$\begin{align}h_x(r, \theta) = h_x[r\cos(\theta), r\sin(\theta)] &= \begin{cases}0&\text{if }r = 0,\\-\displaystyle\frac{\omega_c^2\,r\cos(\theta)\,J_2\left(\omega_c r\right)}{2 \pi\,r^2}&\text{otherwise}\end{cases}\\ &= \cos(\theta)f(r),\\ h_y(r, \theta) = h_y[r\cos(\theta), r\sin(\theta)] &= \begin{cases}0&\text{if }r = 0,\\-\displaystyle\frac{\omega_c^2\,r\sin(\theta)\,J_2\left(\omega_c r\right)}{2 \pi\,r^2}&\text{otherwise}\end{cases}\\ &= \sin(\theta)f(r),\\ f(r) &= \begin{cases}0&\text{if }r = 0,\\-\displaystyle\frac{\omega_c^2\,r\,J_2\left(\omega_c r\right)}{2 \pi\,r^2}&\text{otherwise,}\end{cases}\end{align}\tag{6}$$

where both the horizontal and the vertical differentiation filter impulse responses have the same radial factor function $f(r)$. Any rotated version $h(r, \theta, \phi)$ of $h_x(r, \theta)$ by steering angle $\phi$ is obtained by:

$$h(r, \theta, \phi) = h_x(r, \theta - \phi) = \cos(\theta - \phi)f(r)\tag{7}$$

The idea was that the steered kernel $h(r, \theta, \phi)$ can be constructed as a weighted sum of $h_x(r, \theta)$ and $h_x(r, \theta)$, with $\cos(\phi)$ and $\sin(\phi)$ as the weights, and that is indeed the case:

$$\cos(\phi) h_x(r, \theta) + \sin(\phi) h_y(r, \theta) = \cos(\phi) \cos(\theta) f(r) + \sin(\phi) \sin(\theta) f(r) = \cos(\theta - \phi) f(r) = h(r, \theta, \phi).\tag{8}$$

We will arrive at an equivalent conclusion if we think of the isotropically low-pass filtered signal as the input signal and construct a partial derivative operator with respect to the first of rotated coordinates $x_\phi$, $y_\phi$ rotated by angle $\phi$ from coordinates $x$, $y$. (Derivation can be considered a linear-time-invariant system.) We have:

$$\begin{gather}x = \cos(\phi)x_\phi - \sin(\phi)y_\phi,\\ y = \sin(\phi)x_\phi + \cos(\phi)y_\phi\end{gather}\tag{9}$$

Using the chain rule for partial derivatives, the partial derivative operator with respect to $x_\phi$ can be expressed as a cosine and sine weighted sum of partial derivatives with respect to $x$ and $y$:

$$\begin{gather}\frac{\partial}{\partial x_\phi} = \frac{\partial x}{\partial x_\phi}\frac{\partial}{\partial x} + \frac{\partial y}{\partial x_\phi}\frac{\partial}{\partial y} = \frac{\partial \big(\cos(\phi)x_\phi - \sin(\phi)y_\phi\big)}{\partial x_\phi}\frac{\partial}{\partial x} + \frac{\partial \big(\sin(\phi)x_\phi + \cos(\phi)y_\phi\big)}{\partial x_\phi}\frac{\partial}{\partial y} = \cos(\phi)\frac{\partial}{\partial x} + \sin(\phi)\frac{\partial}{\partial y}\end{gather}\tag{10}$$

A question that remains to be explored is how a suitably weighted circular mean of gradient vector angles is related to the angle $\phi$ of in some way the "most activated" steered differentiation filter.

Possible improvements

To possibly improve results further, the gradient can be calculated also for the red and blue color channels, to be included as additional data in the "average" calculation.

I have in mind possible extensions of this method:

1) Use a larger set of analysis filter kernels and detect edges rather than detecting gradients. This needs to be carefully crafted so that edges in all directions are treated equally, that is, an edge detector for any angle should be obtainable by a weighted sum of orthogonal kernels. A set of suitable kernels can (I think) be obtained by applying the differential operators of Eq. 11, Fig. 6 (see also my Mathematics Stack Exchange post) on the continuous-space impulse response of a circularly symmetric low-pass filter.

$$\begin{gather}\lim_{h\to 0}\frac{\sum_{N=0}^{4N + 1} (-1)^n f\bigg(x + h\cos\left(\frac{2\pi n}{4N + 2}\right), y + h\sin\left(\frac{2\pi n}{4N + 2}\right)\bigg)}{h^{2N + 1}},\\ \lim_{h\to 0}\frac{\sum_{N=0}^{4N + 1} (-1)^n f\bigg(x + h\sin\left(\frac{2\pi n}{4N + 2}\right), y + h\cos\left(\frac{2\pi n}{4N + 2}\right)\bigg)}{h^{2N + 1}}\end{gather}\tag{11}$$

enter image description here
Figure 6. Dirac delta relative locations in differential operators for construction of higher-order edge detectors.

2) The calculation of a (weighted) mean of circular quantities can be understood as summing of cosines of the same frequency shifted by samples of the quantity (and scaled by the weight), and finding the peak of the resulting function. If similarly shifted and scaled harmonics of the shifted cosine, with carefully chosen relative amplitudes, are added to the mix, forming a sharper smoothing kernel, then multiple peaks may appear in the total sum and the peak with the largest value can be reported. With a suitable mixture of harmonics, that would give a kind of local average that largely ignores outliers away from the main peak of the distribution.

Alternative approaches

It would also be possible to convolve the image by angle $\phi$ and angle $\phi + \pi/2$ rotated "long edge" kernels, and to calculate the mean square of the pixels of the two convolved images. The angle $\phi$ that maximizes the mean square would be reported. This approach might give a good final refinement for the image orientation finding, because it is risky to search the complete angle $\phi$ space at large steps.

Another approach is non-local methods, like cross-correlating distant similar regions, applicable if you know that there are long horizontal or vertical traces, or features that repeat many times horizontally or vertically.

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  • $\begingroup$ How accurate the result you got? $\endgroup$ – Royi May 15 at 3:48
  • $\begingroup$ @Royi Maybe around 0.1 deg. $\endgroup$ – Olli Niemitalo May 16 at 10:25
  • $\begingroup$ @OlliNiemitalo which is pretty impressive, given the limited resolution! $\endgroup$ – Marcus Müller May 21 at 11:37
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    $\begingroup$ @OlliNiemitalo speaking of impressive: this. answer. is. that. word's. very. definition. $\endgroup$ – Marcus Müller May 21 at 11:37
  • $\begingroup$ @MarcusMüller Thanks Marcus, I anticipate the first extension to be very interesting too. $\endgroup$ – Olli Niemitalo May 21 at 11:50
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There is a similar DSP trick here, but I don't remember the details exactly.

I read about it somewhere, some while ago. It has to do with figuring out fabric pattern matches regardless of the orientation. So you may want to research on that.

Grab a circle sample. Do sums along spokes of the circle to get a circumference profile. Then they did a DFT on that (it is inherently circular after all). Toss the phase information (make it orientation independent) and make a comparison.

Then they could tell whether two fabrics had the same pattern.

Your problem is similar.

It seems to me, without trying it first, that the characteristics of the pre DFT profile should reveal the orientation. Doing standard deviations along the spokes instead of sums should work better, maybe both.

Now, if you had an oriented reference image, you could use their technique.

Ced


Your precision requirements are rather strict.

I gave this a whack. Taking the sum of the absolute values of the differences between two subsequent points along the spoke for each color.

Here is a graph of around the circumference. Your value is plotted with the white markers.

enter image description here

You can sort of see it, but I don't think this is going to work for you. Sorry.


Progress Report: Some

I've decided on a three step process.

1) Find evaluation spot.

2) Coarse Measurement

3) Fine Measurement

Currently, the first step is user intevention. It should be automatible, but I'm not bothering. I have a rough draft of the second step. There's some tweaking I want to try. Finally, I have a few candidates for the third step that is going to take testing to see which works best.

The good news is it is lighting fast. If your only purposed is to make an image look level on a web page, then your tolerances are way too strict and the coarse measurement ought to be accurate enough.

This is the coarse measurement. Each pixel is about 0.6 degrees. (Edit, actually 0.3)

enter image description here


Progress Report: Able to get good results

enter image description here

Most aren't this good, but they are cheap (and fairly local) and finding spots to get good reads is easy..... for a human. Brute force should work fine for a program.

The results can be much improved on, this is a simple baseline test. I'm not ready to do any explaining yet, nor post the code, but this screen shot ain't photoshopped.


Progress Report: The code is posted, I'm done with this for a while.

This screenshot is the program working on Marcus' 45 degree shot.

enter image description here

The color channels are processed independently.

A point is selected as the sweep center.

A diameter is swept through 180 degrees at discrete angles

At each angle, "volatility" is measuring across the diameter. A trace is made for each channel gathering samples. The sample value is a linear interpolation of the four corner values of whichever grid square the sample spot lands on.

For each channel trace

The samples are multiplied by a VonHann window function

A Smooth/Differ pass is made on the samples

The RMS of the Differ is used as a volatility measure

The lower row graphs are:

First is the sweep of 0 to 180 degrees, each pixel is 0.5 degrees. Second is the sweep around the selected angle, each pixel is 0.1 degrees. Third is the sweep around the selected angle, each pixel is 0.01 degrees. Fourth is the trace Differ curve

The initial selection is the minimal average volatility of the three channels. This will be close, but usually not on, the best angle. The symmetry at the trough is a better indicator than the minimum. A best fit parabola in that neighborhood should yield a very good answer.

The source code (in Gambas, PPA gambas-team/gambas3) can be found at:

https://forum.gambas.one/viewtopic.php?f=4&t=707

It is an ordinary zip file, so you don't have to install Gambas to look at the source. The files are in the ".src" subdirectory.

Removing the VonHann window yields higher accuracy because it effectively lengthens the trace, but adds wobbles. Perhaps a double VonHann would be better as the center is unimportant and a quicker onset of "when the teeter-totter hits the ground" will be detected. Accuracy can easily be improved my increasing the trace length as far as the image allows (Yes, that's automatible). A better window function, sinc?

The measures I have taken at the current settings confirm the 3.19 value +/-.03 ish.

This is just the measuring tool. There are several strategies I can think of to apply it to the image. That, as they say, is an exercise for the reader. Or in this case, the OP. I'll be trying my own later.

There's head room for improvement in both the algorithm and the program, but already they are really useful.

Here is how the linear interpolation works

'---- Whole Number Portion

        x = Floor(rx)
        y = Floor(ry)

'---- Fractional Portions

        fx = rx - x
        fy = ry - y

        gx = 1.0 - fx
        gy = 1.0 - fy

'---- Weighted Average

        vtl = ArgValues[x, y] * gx * gy         ' Top Left
        vtr = ArgValues[x + 1, y] * fx * gy     ' Top Right
        vbl = ArgValues[x, y + 1] * gx * fy     ' Bottom Left
        vbr = ArgValues[x + 1, y + 1] * fx * fy ' Bottom Rigth

        v = vtl + vtr + vbl + vbr

Anybody know the conventional name for that?

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  • 1
    $\begingroup$ hey, you don't need to be sorry for something that was a very clever approach, and might be super helpful for someone with a similar problem who'll come here later! +1 $\endgroup$ – Marcus Müller May 10 at 16:44
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    $\begingroup$ @BarsMonster, I am making good progess. You will want to install Gambas (PPA: gambas-team/gambas3) on your Linux box. (Likely, you too Marcus and Olli, if you can.) I'm working on a program that will not only tackle this problem, but will also serve as a good base for other image processing tasks. $\endgroup$ – Cedron Dawg May 14 at 2:38
  • $\begingroup$ looking forward! $\endgroup$ – Marcus Müller May 14 at 6:16
  • $\begingroup$ @CedronDawg that's called bilinear interpolation, here's why, indicating also to an alternative implementation. $\endgroup$ – Olli Niemitalo May 16 at 11:57
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    $\begingroup$ @OlliNiemitalo,Thanks Olli. In this situation, I don't think going bicubic would improve results over bilinear, in fact, it may even be detrimental. Later, I will play around with different volatility metrics along the diameter, and different shaped window function. At this point I am thinking of using a VonHann at the ends of the diameter like paddles or "teeter-totter seats hitting the mud". The flat bottom in the curve is where the teeter-totter hasn't his the ground (edge) yet. Half way between the two corners is a good read. The current settings are good to less than 0.1 degrees, $\endgroup$ – Cedron Dawg May 16 at 14:01
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I've went ahead and basically adjusted the Hough transform example of opencv to your use case. The idea is nice, but since your image already has plenty of edges due to its edgy nature, the edge detection shouldn't have much benefit.

So, what I did above said example was

  • Omit the edge detection
  • decompose your input image into color channels and process them separately
  • count the occurrences of lines in a specific angle (after quantizing the angles and taking them modulo 90°, since you have plenty right angles)
  • combine the counters of the color channels
  • correct these rotations

What you could do to further improve the quality of estimation (as you'll see below, the top guess wasn't right – the second was) would probably amount to converting of the image to a grayscale image that represents the actual differences between different materials best – clearly, the RGB channels aren't the best. You're the semiconductor expert, so find a way to combine the color channels in a way that maximizes the difference between e.g. metallization and silicon.

My jupyter notebook is here. See the results below.

To increase the angular resolution, increase the QUANT_STEP variable, and the angular precision in the hough_transform call. I didn't, because I wanted this code to be written in < 20 min, and thus didn't want to invest a minute in computation.

import cv2
import numpy
from matplotlib import pyplot
import collections

QUANT_STEPS = 360*2
def quantized_angle(line, quant = QUANT_STEPS):
    theta = line[0][1]
    return numpy.round(theta / numpy.pi / 2 * QUANT_STEPS) / QUANT_STEPS * 360 % 90

def detect_rotation(monochromatic_img):
    # edges = cv2.Canny(monochromatic_img, 50, 150, apertureSize = 3) #play with these parameters
    lines = cv2.HoughLines(monochromatic_img, #input
                           1, # rho resolution [px]
                           numpy.pi/180, # angular resolution [radian]
                           200) # accumulator threshold – higher = fewer candidates
    counter = collections.Counter(quantized_angle(line) for line in lines)
    return counter
img = cv2.imread("/tmp/HIKRe.jpg") #Image directly as grabbed from imgur.com
total_count = collections.Counter()
for channel in range(img.shape[-1]):
    total_count.update(detect_rotation(img[:,:,channel]))

most_common = total_count.most_common(5)
for angle,_ in most_common:
    pyplot.figure(figsize=(8,6), dpi=100)
    pyplot.title(f"{angle:.3f}°")
    rotation = cv2.getRotationMatrix2D((img.shape[0]/2, img.shape[1]/2), -angle, 1)
    pyplot.imshow(cv2.warpAffine(img, rotation, img.shape[:2]))

output_4_0

output_4_1

output_4_2

output_4_3

output_4_4

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Rather performance intensive, but should get you accuracy as wanted:

  • Edge detect the image
  • Hough transform to a space where you have enough pixels for the wanted accuracy.
  • Because there are enough orthogonal lines; the image in the hough space will contain maxima lying on two lines. These are easily detectable and give you the desired angle.
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  • $\begingroup$ Nice, exactly my approach: I'm kind of sad that I didn't see it before I went on my train ride and thus didn't incorporate it in my answer. A clear +1! $\endgroup$ – Marcus Müller May 10 at 16:45
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This is a go at the first suggested extension of my previous answer.

Ideal circularly symmetric band-limiting filters

We construct an orthogonal bank of four filters bandlimited to inside a circle of radius $\omega_c$ on the frequency plane. The impulse responses of these filters can be linearly combined to form directional edge detection kernels. An arbitrarily normalized set of orthogonal filter impulse responses are obtained by applying the first two pairs of "beach-ball like" differential operators to the continuous-space impulse response of the circularly symmetric ideal band-limiting filter impulse response $h(x,y)$:

$$h(x,y) = \frac{\omega_c}{2\pi \sqrt{x^2 + y^2} } J_1 \big( \omega_c \sqrt{x^2 + y^2} \big)\tag{1}$$

$$\begin{align}h_{0x}(x, y) &\propto \frac{d}{dx}h(x, y),\\ h_{0y}(x, y) &\propto \frac{d}{dy}h(x, y),\\ h_{1x}(x, y) &\propto \left(\left(\frac{d}{dx}\right)^3-3\frac{d}{dx}\left(\frac{d}{dy}\right)^2\right)h(x, y),\\ h_{1y}(x, y) &\propto \left(\left(\frac{d}{dy}\right)^3-3\frac{d}{dy}\left(\frac{d}{dx}\right)^2\right)h(x, y)\end{align}\tag{2}$$

$$\begin{align}h_{0x}(x, y) &= \begin{cases}0&\text{if }x = y = 0,\\-\displaystyle\frac{\omega_c^2\,x\,J_2\left(\omega_c\sqrt{x^2 + y^2}\right)}{2 \pi\,(x^2 + y^2)}&\text{otherwise,}\end{cases}\\ h_{0y}(x, y) &= h_{0x}[y, x],\\ h_{1x}(x, y) &= \begin{cases}0&\text{if }x = y = 0,\\\frac{\begin{array}{l}\Big(ω_cx(3y^2 - x^2)\big(J_0\left(ω_c\sqrt{x^2 + y^2}\right)ω_c\sqrt{x^2 + y^2}(ω_c^2x^2 + ω_c^2y^2 - 24)\\ - 8J_1\left(ω_c\sqrt{x^2 + y^2}\right)(ω_c^2x^2 + ω_c^2y^2 - 6)\big)\Big)\end{array}}{2π(x^2 + y^2)^{7/2}}&\text{otherwise,}\end{cases}\\ h_{1y}(x, y) &= h_{1x}[y, x],\end{align}\tag{3}$$

where $J_\alpha$ is a Bessel function of the first kind of order $\alpha$ and $\propto$ means "is proportional to". I used Wolfram Alpha queries ((ᵈ/dx)³; ᵈ/dx; ᵈ/dx(ᵈ/dy)²) to carry out differentiation, and simplified the result.

Truncated kernels in Python:

import matplotlib.pyplot as plt
import scipy
import scipy.special
import numpy as np

def h0x(x, y, omega_c):
  if x == 0 and y == 0:
    return 0
  return -omega_c**2*x*scipy.special.jv(2, omega_c*np.sqrt(x**2 + y**2))/(2*np.pi*(x**2 + y**2))

def h1x(x, y, omega_c):
  if x == 0 and y == 0:
    return 0
  return omega_c*x*(3*y**2 - x**2)*(scipy.special.j0(omega_c*np.sqrt(x**2 + y**2))*omega_c*np.sqrt(x**2 + y**2)*(omega_c**2*x**2 + omega_c**2*y**2 - 24) - 8*scipy.special.j1(omega_c*np.sqrt(x**2 + y**2))*(omega_c**2*x**2 + omega_c**2*y**2 - 6))/(2*np.pi*(x**2 + y**2)**(7/2))

def rotatedCosineWindow(N):  # N = horizontal size of the targeted kernel, also its vertical size, must be odd.
  return np.fromfunction(lambda y, x: np.maximum(np.cos(np.pi/2*np.sqrt(((x - (N - 1)/2)/((N - 1)/2 + 1))**2 + ((y - (N - 1)/2)/((N - 1)/2 + 1))**2)), 0), [N, N])

def circularLowpassKernel(omega_c, N):  # omega = cutoff frequency in radians (pi is max), N = horizontal size of the kernel, also its vertical size, must be odd.
  kernel = np.fromfunction(lambda x, y: omega_c*scipy.special.j1(omega_c*np.sqrt((x - (N - 1)/2)**2 + (y - (N - 1)/2)**2))/(2*np.pi*np.sqrt((x - (N - 1)/2)**2 + (y - (N - 1)/2)**2)), [N, N])
  kernel[(N - 1)//2, (N - 1)//2] = omega_c**2/(4*np.pi)
  return kernel

def prototype0x(omega_c, N):  # omega = cutoff frequency in radians (pi is max), N = horizontal size of the kernel, also its vertical size, must be odd.
  kernel = np.zeros([N, N])
  for y in range(N):
    for x in range(N):
      kernel[y, x] = h0x(x - (N - 1)/2, y - (N - 1)/2, omega_c)
  return kernel

def prototype0y(omega_c, N):  # omega = cutoff frequency in radians (pi is max), N = horizontal size of the kernel, also its vertical size, must be odd.
  return prototype0x(omega_c, N).transpose()

def prototype1x(omega_c, N):  # omega = cutoff frequency in radians (pi is max), N = horizontal size of the kernel, also its vertical size, must be odd.
  kernel = np.zeros([N, N])
  for y in range(N):
    for x in range(N):
      kernel[y, x] = h1x(x - (N - 1)/2, y - (N - 1)/2, omega_c)
  return kernel

def prototype1y(omega_c, N):  # omega = cutoff frequency in radians (pi is max), N = horizontal size of the kernel, also its vertical size, must be odd.
  return prototype1x(omega_c, N).transpose()

N = 321  # Horizontal size of the kernel, also its vertical size. Must be odd.
window = rotatedCosineWindow(N)

# Optional window function plot
#plt.imshow(window, vmin=-np.max(window), vmax=np.max(window), cmap='bwr')
#plt.colorbar()
#plt.show()

omega_c = np.pi/8  # Cutoff frequency in radians <= pi
lowpass = circularLowpassKernel(omega_c, N)
kernel0x = prototype0x(omega_c, N)
kernel0y = prototype0y(omega_c, N)
kernel1x = prototype1x(omega_c, N)
kernel1y = prototype1y(omega_c, N)

# Optional kernel image save
plt.imsave('lowpass.png', plt.cm.bwr(plt.Normalize(vmin=-lowpass.max(), vmax=lowpass.max())(lowpass)))
plt.imsave('kernel0x.png', plt.cm.bwr(plt.Normalize(vmin=-kernel0x.max(), vmax=kernel0x.max())(kernel0x)))
plt.imsave('kernel0y.png', plt.cm.bwr(plt.Normalize(vmin=-kernel0y.max(), vmax=kernel0y.max())(kernel0y)))
plt.imsave('kernel1x.png', plt.cm.bwr(plt.Normalize(vmin=-kernel1x.max(), vmax=kernel1x.max())(kernel1x)))
plt.imsave('kernel1y.png', plt.cm.bwr(plt.Normalize(vmin=-kernel1y.max(), vmax=kernel1y.max())(kernel1y)))
plt.imsave('kernelkey.png', plt.cm.bwr(np.repeat([(np.arange(321)/320)], 16, 0)))

enter image description here
enter image description here
Figure 1. Color-mapped 1:1 scale plot of circularly symmetric band-limiting filter impulse response, with cut-off frequency $\omega_c = \pi/8$. Color key: blue: negative, white: zero, red: maximum.

enter image description hereenter image description here
enter image description hereenter image description here
enter image description here
Figure 2. Color-mapped 1:1 scale plots of sampled impulse responses of filters in the filter bank, with cut-off frequency $\omega_c = \pi/8$, in order: $h_{0x}$, $h_{0y}$, $h_{1x}$, $h_{0y}$. Color key: blue: minimum, white: zero, red: maximum.

Directional edge detectors can be constructed as weighted sums of these. In Python (continued):

composite = kernel0x-4*kernel1x
plt.imsave('composite0.png', plt.cm.bwr(plt.Normalize(vmin=-composite.max(), vmax=composite.max())(composite)))
plt.imshow(composite, vmin=-np.max(composite), vmax=np.max(composite), cmap='bwr')
plt.colorbar()
plt.show()

composite = (kernel0x+kernel0y) + 4*(kernel1x+kernel1y)
plt.imsave('composite45.png', plt.cm.bwr(plt.Normalize(vmin=-composite.max(), vmax=composite.max())(composite)))
plt.imshow(composite, vmin=-np.max(composite), vmax=np.max(composite), cmap='bwr')
plt.colorbar()
plt.show()

enter image description hereenter image description here
enter image description here
Figure 3. Directional edge detection kernels constructed as weighted sums of kernels of Fig. 2. Color key: blue: minimum, white: zero, red: maximum.

The filters of Fig. 3 should be better tuned for continuous edges, compared to gradient filters (first two filters of Fig. 2).

Gaussian filters

The filters of Fig. 2 have a lot of oscillation due to strict band limiting. Perhaps a better staring point would be a Gaussian function, as in Gaussian derivative filters. Relatively, they are much easier to handle mathematically. Let's try that instead. We start with the impulse response definition of a Gaussian "low-pass" filter:

$$h(x, y, \sigma) = \frac{e^{-\displaystyle\frac{x^2 + y^2}{2 \sigma^2}}}{2\pi \sigma^2}.\tag{4}$$

We apply the operators of Eq. 2 to $h(x, y, \sigma)$ and normalize each filter $h_{..}$ by:

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}h_{..}(x, y, \sigma)^2\,dx\,dy = 1.\tag{5}$$

$$\begin{align}h_{0x}(x, y, \sigma) &= 2\sqrt{2\pi}σ^2 \frac{d}{dx}h(x, y, \sigma) = - \frac{\sqrt{2}}{\sqrt{\pi}σ^2} x e^{-\displaystyle\frac{x^2 + y^2}{2σ^2}},\\ h_{0y}(x, y, \sigma) &= h_{0x}(y, x, \sigma),\\ h_{1x}(x, y, \sigma) &= \frac{2\sqrt{3\pi}σ^4}{3}\left(\left(\frac{d}{dx}\right)^3-3\frac{d}{dx}\left(\frac{d}{dy}\right)^2\right)h(x, y, \sigma) = - \frac{\sqrt{3}}{3\sqrt{\pi}σ^4} (x^3 - 3xy^2) e^{-\displaystyle\frac{x^2 + y^2}{2σ^2}},\\ h_{1y}(x, y, \sigma) &= h_{1x}(y, x, \sigma).\end{align}\tag{6}$$

We would like to construct from these, as their weighted sum, the impulse response of a vertical edge detector filter that maximizes specificity $S$ which is the mean sensitivity to a vertical edge over the possible edge shifts $s$ relative to the mean sensitivity over the possible edge rotation angles $\beta$ and possible edge shifts $s$:

$$S = \frac{2\pi\displaystyle\int_{-\infty}^{\infty}\Bigg(\int_{-\infty}^{\infty}\bigg(\int_{-\infty}^{s}h_x(x, y, \sigma)dx - \int_{s}^{\infty}h_x(x, y, \sigma)dx\bigg)dy\Bigg)^2ds} {\Bigg(\displaystyle\int_{-\pi}^{\pi}\int_{-\infty}^{\infty}\bigg(\int_{-\infty}^{\infty}\Big(\int_{-\infty}^{s}h_x\big(\cos(\beta)x- \sin(\beta)y, \sin(\beta)x + \cos(\beta)y\big)dx \\- \displaystyle\int_{s}^{\infty}h_x\big(\cos(\beta)x - \sin(\beta)y, \sin(\beta)x + \cos(\beta)y\big)dx\Big)dy\bigg)^2ds\,d\beta\Bigg)}.\tag{7}$$

We only need a weighted sum of $h_{0x}$ with variance $\sigma^2$ and $h_{1x}$ with optimal variance. It turns out that $S$ is maximized by an impulse response:

$$\begin{align}h_x(x, y, \sigma) &= \frac{\sqrt{7625 - 2440\sqrt{5}}}{61} h_{0x}(x, y, \sigma) - \frac{2\sqrt{610\sqrt{5} - 976}}{61} h_{1x}(x, y, \sqrt{5}\sigma)\\ &= - \frac{\sqrt{(15250 - 4880\sqrt{5}}}{61\sqrt{\pi}σ^2}xe^{-\displaystyle\frac{x^2 + y^2}{2σ^2}} + \frac{\sqrt{1830\sqrt{5} - 2928}}{4575 \sqrt{\pi} σ^4}(2x^3 - 6xy^2)e^{-\displaystyle\frac{x^2 + y^2}{10 σ^2}}\\ &= \frac{2\sqrt{\pi}σ^2\sqrt{15250 - 4880\sqrt{5}}}{61}\frac{d}{dx}h(x, y, \sigma) - \frac{100\sqrt{\pi}σ^4\sqrt{1830\sqrt{5} - 2928}}{183}\left(\left(\frac{d}{dx}\right)^3-3\frac{d}{dx}\left(\frac{d}{dy}\right)^2\right)h(x, y,\sqrt{5}\sigma)\\ &\approx 3.8275359956049814\,\sigma^2\frac{d}{dx}h(x, y, \sigma) - 33.044650082417731\,\sigma^4\left(\left(\frac{d}{dx}\right)^3-3\frac{d}{dx}\left(\frac{d}{dy}\right)^2\right)h(x, y,\sqrt{5}\sigma),\end{align}\tag{8}$$

also normalized by Eq. 5. To vertical edges, this filter has a specificity of $S = \frac{10\times5^{1/4}}{9}$ $+$ $2$ $\approx$ $3.661498645$, in contrast to the specificity $S = 2$ of a first-order Gaussian derivative filter with respect to $x$. The last part of Eq. 8 has normalization compatible with separable 2-d Gaussian derivative filters from Python's scipy.ndimage.gaussian_filter:

import matplotlib.pyplot as plt
import numpy as np
import scipy.ndimage

sig = 8;
N = 161
x = np.zeros([N, N])
x[N//2, N//2] = 1
ddx = scipy.ndimage.gaussian_filter(x, sigma=[sig, sig], order=[0, 1], truncate=(N//2)/sig)
ddx3 = scipy.ndimage.gaussian_filter(x, sigma=[np.sqrt(5)*sig, np.sqrt(5)*sig], order=[0, 3], truncate=(N//2)/(np.sqrt(5)*sig))
ddxddy2 = scipy.ndimage.gaussian_filter(x, sigma=[np.sqrt(5)*sig, np.sqrt(5)*sig], order=[2, 1], truncate=(N//2)/(np.sqrt(5)*sig))

hx = 3.8275359956049814*sig**2*ddx - 33.044650082417731*sig**4*(ddx3 - 3*ddxddy2)
plt.imsave('hx.png', plt.cm.bwr(plt.Normalize(vmin=-hx.max(), vmax=hx.max())(hx)))

h = scipy.ndimage.gaussian_filter(x, sigma=[sig, sig], order=[0, 0], truncate=(N//2)/sig)
plt.imsave('h.png', plt.cm.bwr(plt.Normalize(vmin=-h.max(), vmax=h.max())(h)))
h1x = scipy.ndimage.gaussian_filter(x, sigma=[sig, sig], order=[0, 3], truncate=(N//2)/sig) - 3*scipy.ndimage.gaussian_filter(x, sigma=[sig, sig], order=[2, 1], truncate=(N//2)/sig)
plt.imsave('ddx.png', plt.cm.bwr(plt.Normalize(vmin=-ddx.max(), vmax=ddx.max())(ddx)))
plt.imsave('h1x.png', plt.cm.bwr(plt.Normalize(vmin=-h1x.max(), vmax=h1x.max())(h1x)))
plt.imsave('gaussiankey.png', plt.cm.bwr(np.repeat([(np.arange(161)/160)], 16, 0)))

enter image description hereenter image description hereenter image description hereenter image description hereenter image description here
Figure 4. Color-mapped 1:1 scale plots of, in order: A 2-d Gaussian function, derivative of the Gaussian function with respect to $x$, a differential operator $\big(\frac{d}{dx}\big)^3-3\frac{d}{dx}\big(\frac{d}{dy}\big)^2$ applied to the Gaussian function, the optimal two-component Gaussian-derived vertical edge detection filter $h_x(x, y, \sigma)$ of Eq. 8. The standard deviation of each Gaussian was $\sigma = 8$ except for the hexagonal component in the last plot which had standard deviation $\sqrt{5}\times8$. Color key: blue: minimum, white: zero, red: maximum.

TO BE CONTINUED...

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