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Well put. The title is indeed completely factual! Including '23 lines' was just to let people know at first glance that the solution is simple. > If you had to

Well put. The title is indeed completely factual! Including '23 lines' was just to let people know at first glance that the solution is simple. > If you had to write the un-swirling, gray-scaling, Sobel filtering and summation code in Python you'd be looking at a much larger pile-o-code I would also like to point out these are also all very simple operations, which would only take a few lines of Python/C/whatever to do. Python just happens to have a pre-written libraries to save reinventing the wheel.

robomartin 14y ago
> I would also like to point out these are also all very simple operations, which would only take a few lines of Python/C/whatever to do. Yikes! They are not. That's a lot of code. I've done a ton of image processing work in both hardware (FPGA) and software (various languages). Swirling, gray-scaling, Sobel filtering and summation collectively are far from "a few lines of Python/C/whatever to do" > Python just happens to have a pre-written libraries to save reinventing the wheel That's not true. OpenCV is NOT a Python library, it's nearly 100% C++ as far as I can tell: https://github.com/Itseez/opencv https://github.com/Itseez/opencv That was, in many ways, my point. This is not a solution in 23 lines of Python. Go install Python from Python.org --and NOTHING ELSE-- and solve the problem. Let's see how many lines of Python it takes. There's nothing wrong with the intent of the author in terms of showing how one can break these CAPTCHA's using edge-length evaluation. I never put any thought into this myself so, yes, I learned something of value from his post. I just wish the title was more genuine, that's all.
- cidquick 14y ago
  My turn to nitpick! RGB to grayscale: V = r x g x b/3 (one line). The Sobel operation is at most 4 nested loops (really only 3) for a total of ~7 lines of code, depending on how you like your white space. Fold the summation into your Sobel function without requiring another line. The most complicated thing used from the OpenCV library (which I'm well aware is C++!) is JPEG decoding. OK, I'm avoiding swirling, but then it isn't needed to solve the CAPTCHA anyway.
  - robomartin 14y ago
    Show me working code and then we can talk. :) Oh, BTW, "V = r x g x b/3" is incorrect. This is not you convert a color image to a grayscale image.
    - deleted 14y ago
      [deleted]
    - chmod775 14y ago
      Ok, I'll do. I will assume that we already have the image in an 1-dimensional array, each element containing an array of the RGB values. Then we can convert it to grayscale with one line of python. gray = map(lambda p:sum(p)/3,IMG) If you want to test it, you can use this 3x3 sample image (or just load one): IMG = [[5, 5, 5], [6, 8, 9], [94, 123, 4], [54, 5, 32], [44, 3, 3], [34, 234, 33], [5, 5, 5], [6, 8, 9], [94, 123, 4]] The result will look like this: [5, 7, 73, 30, 16, 100, 5, 7, 73] If your image happens to be loaded into a 2-dimensional array, use this: gray = map(lambda row:map(lambda p:sum(p)/3,row), IMG) Sobel is slightly more complicated, but can be written like this (you'll need a larger sample image though): width, height #width and height of our image IMG #2D array of our RGB values sobel #result image # The actual filter starts here: for x in range(1,width-1): for y in range(1,height-1): sx = IMG[x-1][y-1]+IMG[x][y-1]*2+IMG[x+1][y-1]-IMG[x-1][y+1]-IMG[x][y+1]*2-IMG[x+1][y+1] sy = IMG[x+1][y-1]+IMG[x+1][y]*2+IMG[x+1][y+1]-IMG[x-1][y-1]-IMG[x-1][y]*2-IMG[x-1][y+1] sobel[x][y] = Math.sqrt(sx*sx+sy*sy) A total of 5 lines. And I have to thank you, because I finally just learned what convolutions are while doing this.
      - robomartin 14y ago
        Excellent! Now we are starting down the right path. I don't have time to fully look at your code right now. I will later. Just one quick observation. You can't convert from RGB to grayscale by simply averaging the three values. Each color channel influences the perceived luminance (grayscale) differently, with green being, by far, the largest component and blue the least significant. Rather than give you the answer I'd suggest you research "converting RGB to grayscale" or "converting RGB to luminance" as this is an important subject to understand if you are dealing with images. I'll take a look at the rest of it later and comment.

robomartin 14y ago

> I would also like to point out these are also all very simple operations, which would only take a few lines of Python/C/whatever to do. Yikes! They are not. That's a lot of code. I've done a ton of image processing work in both hardware (FPGA) and software (various languages). Swirling, gray-scaling, Sobel filtering and summation collectively are far from "a few lines of Python/C/whatever to do" > Python just happens to have a pre-written libraries to save reinventing the wheel That's not true. OpenCV is NOT a Python library, it's nearly 100% C++ as far as I can tell: https://github.com/Itseez/opencv https://github.com/Itseez/opencv That was, in many ways, my point. This is not a solution in 23 lines of Python. Go install Python from Python.org --and NOTHING ELSE-- and solve the problem. Let's see how many lines of Python it takes. There's nothing wrong with the intent of the author in terms of showing how one can break these CAPTCHA's using edge-length evaluation. I never put any thought into this myself so, yes, I learned something of value from his post. I just wish the title was more genuine, that's all.

cidquick 14y ago
My turn to nitpick! RGB to grayscale: V = r x g x b/3 (one line). The Sobel operation is at most 4 nested loops (really only 3) for a total of ~7 lines of code, depending on how you like your white space. Fold the summation into your Sobel function without requiring another line. The most complicated thing used from the OpenCV library (which I'm well aware is C++!) is JPEG decoding. OK, I'm avoiding swirling, but then it isn't needed to solve the CAPTCHA anyway.
- robomartin 14y ago
  Show me working code and then we can talk. :) Oh, BTW, "V = r x g x b/3" is incorrect. This is not you convert a color image to a grayscale image.
  - deleted 14y ago
    [deleted]
  - chmod775 14y ago
    Ok, I'll do. I will assume that we already have the image in an 1-dimensional array, each element containing an array of the RGB values. Then we can convert it to grayscale with one line of python. gray = map(lambda p:sum(p)/3,IMG) If you want to test it, you can use this 3x3 sample image (or just load one): IMG = [[5, 5, 5], [6, 8, 9], [94, 123, 4], [54, 5, 32], [44, 3, 3], [34, 234, 33], [5, 5, 5], [6, 8, 9], [94, 123, 4]] The result will look like this: [5, 7, 73, 30, 16, 100, 5, 7, 73] If your image happens to be loaded into a 2-dimensional array, use this: gray = map(lambda row:map(lambda p:sum(p)/3,row), IMG) Sobel is slightly more complicated, but can be written like this (you'll need a larger sample image though): width, height #width and height of our image IMG #2D array of our RGB values sobel #result image # The actual filter starts here: for x in range(1,width-1): for y in range(1,height-1): sx = IMG[x-1][y-1]+IMG[x][y-1]*2+IMG[x+1][y-1]-IMG[x-1][y+1]-IMG[x][y+1]*2-IMG[x+1][y+1] sy = IMG[x+1][y-1]+IMG[x+1][y]*2+IMG[x+1][y+1]-IMG[x-1][y-1]-IMG[x-1][y]*2-IMG[x-1][y+1] sobel[x][y] = Math.sqrt(sx*sx+sy*sy) A total of 5 lines. And I have to thank you, because I finally just learned what convolutions are while doing this.
    - robomartin 14y ago
      Excellent! Now we are starting down the right path. I don't have time to fully look at your code right now. I will later. Just one quick observation. You can't convert from RGB to grayscale by simply averaging the three values. Each color channel influences the perceived luminance (grayscale) differently, with green being, by far, the largest component and blue the least significant. Rather than give you the answer I'd suggest you research "converting RGB to grayscale" or "converting RGB to luminance" as this is an important subject to understand if you are dealing with images. I'll take a look at the rest of it later and comment.