Let's play a little rule game in here. Take this doodle sketch, you can do anything with it (redraw or whatever), but you need to share the process of making it in this comment section. (monyet.cc)

submitted 10 months ago* (last edited 10 months ago) by Otherwise_Direction7@monyet.cc to c/196@lemmy.blahaj.zone

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[-] PM_ME_VINTAGE_30S@lemmy.sdf.org 39 points 10 months ago* (last edited 10 months ago)

I created this on my phone in MATLAB. You can probably do this in Octave with similar or the same code.

Figure 2024-01-14 17_18_24~2

First, I downloaded the image from Lemmy, then uploaded it into my MATLAB app. I renamed the image to image.jpg, then ran the following code:

image=imread("image.jpg") imagesc(log10(abs( fftshift(fft2(image)) )))

fft2 applies a 2D Fast Fourier transform to the image, which creates a complex (as in complex numbers) image. abs takes the magnitude of the complex image elementwise. log10 scales the result for display.

Then I downloaded the image from the MATLAB app, went into the Photos app and (badly) cropped out the white border.

Despite how dramatically different it looks, it actually contains the same [1] information as the original image. Said differently, you can actually go back to the original with the inverse functions, specifically by undoing the logarithm and applying the inverse FFT.

[1] Almost. (1). There will be border problems potentially caused by me sloppily cropping some pixels out of the image. (2). It looks like MATLAB resized the image when rendering the figure. However, if I actually saved the matrix (raw image) rather than the figure, then it would be the correct size. (3) (Thank you to @itslilith@lemmy.blahaj.zone for pointing this out.) You need the phase information to reconstruct the original signal, which I (intentionally) threw out (to get a real image) when I took the absolute value but then completely forgot about it.

[-] asbestos@lemmy.world 20 points 10 months ago* (last edited 10 months ago)

Can guarantee OP didn’t expect this when he was making the post, damn

[-] itslilith@lemmy.blahaj.zone 11 points 10 months ago

don't you lose information by taking the abs(), since to restore the full information you need both complex and imaginary parts of the fft? You could probably get away with encoding Re and Im in different color channels tho

[-] drailin@kbin.social 5 points 10 months ago* (last edited 10 months ago)

Another nice way one could preserve the complex data when visualizing it would be to make a 3d color mesh and display the imaginary components as the height in z and the real component as the color scale (or vice-versa).

Edit* now I am trying to think if there would be a clever way to show the abs, Re and Im values in one 3d plot, but drawing a blank. Maybe tie Im to the alpha value to make the transparency change as the imaginary component goes up and down? It would just require mapping the set of all numbers from -inf:inf to 0:1, which is doable in a 1-1 transformation iirc since they both have cardinality C. I think it would be

alpha = 1 - 1/(1-e^{Im(z)})

Which looks a lot like the equation for Bose-Einstein statistics in Stat. Mech. I was never very good at complex analysis or group theory though, so I don't really know what to make of that.

[-] itslilith@lemmy.blahaj.zone 4 points 10 months ago

Bose-Einstein isn't a great fit, since you'd need to integrate, and it only goes from 0:inf. For mapping the reals to 0:1 you could use arctan and shift it a bit.

Now I'm thinking, instead of color and alpha, you could use two out of hue, saturation and value for Re and Im (or all three, and plot Abs as well)

[-] drailin@kbin.social 4 points 10 months ago

Ah, you're right, I haven't taken Stat. Mech. in almost 5 years so my brain just latched on to the general form. Analysis in frequency space is always fun

[-] PM_ME_VINTAGE_30S@lemmy.sdf.org 4 points 10 months ago

Correct. I updated my comment.

[-] drailin@kbin.social 8 points 10 months ago

Apply a nice gaussian kernel convolution to the fft and smooth that doodle out! Lets get blurry up in this doodle party!

[-] PM_ME_VINTAGE_30S@lemmy.sdf.org 12 points 10 months ago

Apply a nice gaussian kernel convolution to the fft

I applied the following code in MATLAB:

new_image = abs(ifft2(conv2(fft2(image),fspecial('gaussian',69)))); imwrite(new_image,"new_image.png")

And got this:

Figure 2024-01-15 05_40_24

[-] drailin@kbin.social 7 points 10 months ago

Oh fuck yeah

[-] drailin@kbin.social 5 points 10 months ago

Just noticed that you chose the nicest kernel size. Even better.

[-] Otherwise_Direction7@monyet.cc 5 points 10 months ago

Ok, this is really mad cool, I like it

[-] HerbalGamer@sh.itjust.works 4 points 10 months ago

Saving this comment in case I ever need a new phone background.

[-] Jackie_meaiii@lemmy.blahaj.zone 30 points 10 months ago

I used the default photos app on my phone to draw a totally normal cow

[-] PM_ME_VINTAGE_30S@lemmy.sdf.org 11 points 10 months ago

So I took your image and ~~ruined my MATLAB account~~ used the most normal part of your totally normal cow as a 3D [1] ~~cockvolution~~ convolution kernel. So in some sense, I dragged the red and purple part all across your image and added up the results. Here's the result:

Figure 2024-01-15 14_46_27

Here's the MATLAB code:

normal_image = imread("totally_normal_image.png");

feature=normal_image(272:350,205:269,:);

feature_expansion = padarray(feature,[0,ceil((79-65)/2),0],'replicate');

for m = 1:1:3
new_normal_image(:,:,m) = conv2(normal_image(:,:,m),feature_expansion(:,:,m));

new_normal_image(:,:,m) = new_normal_image(:,:,m)/max(max(new_normal_image(:,:,m))); 
end

imshow(new_normal_image)

[1] The original image was practically grayscale, so only a 2D convolution is required, i.e. over 2 spatial dimensions. Since you added color, it adds a extra dimension, one per color channel. Which makes it more annoying to work with in MATLAB. I mean, I could have just dumped everything into grayscale, but I need practice with processing color images anyways.