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I have a grayscale image consisting of MxN 1 byte pixels. The image shows K vertical lines. Each vertical line has the same thickness.

Due to a systematic error each line is imaged in a different horizontal position than the actual position. The horizontal shift differs between lines.

Consider a scenario where there are K=2 lines. This figure shows the position of the imaged lines as red and the correct position as blue:

enter image description here

We see that the left line is shifted 1 to the right of its actual position and the right line is shifted 2 to the right of its actual position.

In this case I may use $$RMSE=\sqrt{1^2+2^2}$$ as the error of the imaging.

I am looking for an algorithm that automatically calculates such an error/penalty function. Using this algorithm I will then tune two parameters (using gradient descent or similar) until the error is minimized.

Consider a part of a row in such an image. This figure shows the pixel values and one vertical line as a red rectangle:

enter image description here

One first step may be to find the center of the vertical line as a fractional horizontal pixel coordinate.

How could I do this?

Having found this I could for each row find a set of K fractional horizontal coordinates representing the K imaged lines: $$\{x_{imaged_i}\}$$.

I could then perhaps for each imaged line find the distance to the closest actual line: $$d_i=|x_{imaged_i}-x_{actual_i}|$$ and sum up the square of these distances and take the square root?

$$RMSE=\sqrt{\Sigma d_i^2}$$

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The location could be found using a simple weighting of the pixel values:

Pixel location 0 1 2 3 4 5
Pixel value 0 50 255 200 0 0
Product 0 1 $\times$ 50 = 50 2 $\times$ 255 = 510 3 $\times$ 200 = 600 4 $\times$ 0 = 0 5 $\times$ 0 = 0

And then sum the products up and divide by the summed pixel values.

$$ \bar{x} = \frac{0 + 50 + 510 + 600 + 0 + 0}{0 + 50 + 255 + 200 + 0 + 0} = \frac{1160}{505} = 2.297 $$

The trick will be to decide when one line stops and the other starts.

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  • $\begingroup$ Is this a weighted version of the centroid formulae in 1 dimension? $\endgroup$
    – Andy
    Sep 5, 2023 at 20:03
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    $\begingroup$ @Andy Yes, it's just finding the weighted centroid. $\endgroup$
    – Peter K.
    Sep 5, 2023 at 20:10

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