Removing checkered background baked onto image

Question

I have an image as follows:

The checkered pattern is not the transparency, it is impressed directly on the image, and runs across the whole graphic.

I would like to know if there is a way to remove the checkered background, extracting the original graphic without (much) damage. I suppose it is impossible, but maybe someone knows a trick.

Answer 1

Alpha blending is:

C_out = α * C_FG+(1-α) * C_BG

Now there are two alphas on the image: a full transparency and the object transparency. The first order of business is to calculate the full transparency. Because we know the color and structure of the checker, we get a system of equations out of two pixels that have the same color.

C = α * C_FGdark+(1-α) * C_BGdark

C = α * C_FGDLight+(1-α) * C_BGLight

In these 2 equations there are 2 unknowns, so first let's solve for alpha. Since C is same in both equations we get:

α * C_FGdark+(1-α) * C_BGdark = α * C_FGDLight+(1-α) * C_BGLight

And then:

α * C_FGdark+ C_BGdark - α* C_BGdark = α * C_FGDLight+C_BGLight -α * C_BGLight

Collect by terms:

α * C_FGdark - α* C_BGdark - α * C_FGDLight + α * C_BGLight= C_BGLight - C_BGdark

break out alpha and divide

α = (C_BGLight - C_BGdark) / (C_FGdark - C_BGdark - C_FGDLight + C_BGLight)

Whew, now that we have calculated the alpha we can go use the first formula and solve it in terms of C_BG:

C_out = α * C_FG+(1-α) * C_BG

collect by terms and divide:

C_FG = -((1-α) * C_BG - C_out)/ α

OK let's test this! Take the images:

bg.png and 0.5.png (click to zoom)

Where the second is a synthetic check with 50% opacity stroke on the background. We can use ImageMagick to do the calculation. Run the command

magick 05.png bg.png -fx "-((0.5)*u[1]-u[0])/0.5" out.png 

you get out.png

Seems to work rather well. You can now do a difference alpha for an alpha mask.

Now the original image does not seem to be alpha blended. Maybe the mode is multiply or screen? Let us test those next.

Multiplication mode is:

C_out = C_FG * C_BG

Thus

C_FG = C_out/C_BG

Testing

Results in:

looks ok. But thats not it either. Maybe multiply with a blend?

Screen mode is:

C_out = 1 - (1 - C_FG) * (1 - C_BG)

So:

C_FG = (C_BG - C_out) / (1 - C_BG)

But that wouldnt be recoverable.

OK so not easy. But theres still stuff to go through. Maybe they have multiple alphas? What if the multiplication or blend was done color correctly and not naively like photoshop software tend to do? (Need to do actual work also sorry)