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   Author  Topic: SUPER-VILLAIN TRANSPARENCIES  (Read 18094 times)
Khal
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Re: SUPER-VILLAIN TRANSPARENCIES  
« Reply #25 on: Oct 16th, 2008, 12:49pm »
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That almost works. Sine the transparencies are a subtractive color process, you need pens that are cyan, magenta, and yellow (rather than red, green, and blue).  
 
With RGB pens you would get a black pixel on the screen where any two of them overlapped and you only want black where all three overlap.
« Last Edit: Oct 16th, 2008, 12:51pm by Khal » IP Logged
impartial_james
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Re: SUPER-VILLAIN TRANSPARENCIES  
« Reply #26 on: Oct 2nd, 2013, 2:37pm »
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Assuming darkness is additive (2 black pixels on top of each other is twice as dark as just 1), I think that S. Owen's method surprisingly does not allow you to see the image when the transparencies are layered.
 
His idea, in order to show image P, was to have A and B be slides made of random pixels, and C = A xor B xor P. Then the darkness A, B and C layed on top of each other will be 1 or 3 when P has a black pixel, and 0 or 2 when P does not. There are 4 equally likely ways for A+B+C to be odd: letting 1 be black and 0 clear, these are
 
100, 010, 001, 111
 
In the first 3 cases, we have a darkness of 1, and the last a darkness of 3, so the average darkness in the region where P is back will be (3/4)*1 + (1/4)*3 = 1.5. This means that when the mind blurs together the pixels, it will see this region as one uniform color, of darkness halfway between 1 and 2.
 
Similarly, the average darkness in the region where P is clear will be (1/4)*0 + (3/4)*2 = 1.5. But this means that when the transparencies are layered, we will see the same color everywhere on the slide, so we can't recover the information P had!
 
My argument depends on these two assumptions.
 
1. Darkness is additive (maybe two black pixels on top of each other would only make a pixel which we would perceive as 1.5 times as dark).
 
2. Our perceived darkness of a bunch of pixels is there average darkness (maybe its the darkness that occurs most, or something else weird.).
 
Neither of these, I think, are intuitively true or false; they can only be verified by studying how the human eye/brain works. If this is only supposed to be a CS puzzle, it shouldn't require this kind of knowledge to come up with a solution, so hopefully there is a another answer. I've tried generalizing this method to n slides, where A1, A2, . . . An-1 are all random, and the last slide is the xor of all of these with with desired slide, but at least for n up to 5, the same thing happens! Tongue
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dantruong
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Re: SUPER-VILLAIN TRANSPARENCIES  
« Reply #27 on: Apr 22nd, 2014, 1:47am »
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I also contributed some comments with you.
Follow me because the problem said "several" slides and not a specific amount, I would use 5. Then I would mimic a RAID 5 array onto the slides. If I remember correctly, in a 5 disk RAID 5 array, you can loose 2 disks, and still keep the array.
 However, if you loose 3, meaning only 2 are still good, your data is lost.
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dudiobugtron
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Re: SUPER-VILLAIN TRANSPARENCIES  
« Reply #28 on: Jun 19th, 2014, 7:29pm »
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Here is a(nother?) solution to this puzzle.
 
Note that the CMY solution won't work, because you will know that a pixel should be non-black whenever two slides share the same colour for a particular pixel.
 
My solution is to use 5 slides, each with either a grey or empty pixel (as for previous solutions).  The black pixels in the original image will be grey on exactly 3 slides, and the white pixels on the original will be grey on exactly 2 slides.  This way the black pixels will show up as darker (3 greys vs 2) when the slides are overlapped.  But, if you have any two slides, the fact that they have a grey or non-grey pixel in a particular location will tell you nothing.
 
You need 5 slides, since with 3 or 4 slides, you would be able to tell whether some pixels were black or white on the original. (If, say, you got two slides which shared a non-grey pixel, then you know the original must be white (or two grey pixels would mean the answer must be black).)
 
This extends to a general solution pretty obviously (n = 2k + 1), providing a nice upper bound.  Note that you may need to use a different number instead of 1 if you wish the shade-difference to be more discernable.
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