The "super-villain transparencies" brain-teaser
Here's a good brain-teaser from [wu:riddles] --
You're a super-villain and you want to prepare a transparency (the kind that goes on an overhead projector) with the key points of your plan for world domination so you can present them to the hero/superagent before you attempt to kill him in some ridiculously novel way. You don't want this information to fall into the wrong hands before you're ready. Smart villain that you are, you know you can share the information across several slides so that if the enemy agents capture any 2 of your slides, they won't learn even the tiniest bit of information about your plan. How?Super-villain transparencies
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Perhaps some variation of putting a small amount of information (a single letter, perhaps) per transparency, then stacking them on the projector simultaneously so together they form your dastardly scheme?
You make the diagrams in layers right? Each sheet would have like one line or something, and when you put them all together you get the whole image.
@1: that's the first thing I thought of.
or just scattering random letters on a single transparency so that they all lined up when you stacked them... same idea.
You could get even fancier by "printing" the letters as polarization filters onto the plastic, so they would only show up with just the right orientation.
Needlessly over-the-top, and not necessarily any more secure, in perfect Super Villain Style.
I was thinking off that; putting a few little lines on each sheet, but it seems, I don't know, too simple? I think there may be a more complicated solution, otherwise this a very simple brain teaser.
@ Nosehat
If it's transparencies, you don't even need to use letters on each, really. Just a random sampling of the black parts; each transparency will look like just random noise, if not just slightly dirty.
How is this a brain-teaser? Didn't Robert Downy, Jr. do this exact thing in Iron Man?
That's a silly puzzle. All true super-villains use Power Point. Not only is the boredom from a long presentation enough to kill the hero/superagent on its own, but you can password protect the document to keep it from falling into the wrong hands.
Um, clearly you write it in lemon juice.
Make each slide the left or right half of a random-dot stereogram. When they're combined you can cross your eyes and see the plan spelled out in eerie floating letters. With only one slide you have only random dots.
The random bits on each slide to be overlaid on each other was my first thought too, but in my book that doesn't count as not the "tiniest bit of information". Statistically, you could get quite a bit out of that if it wasn't done very, very carefully - like with really irregular fonts.
Using an m of n cipher would be the safest, most general way to do this, but it wouldn't require transparencies.
The polarization thing mentioned above would be cool if you could do it like LCDs where a region could impose a certain twist in the polarization. That way, each layer could have twists that might add to the others or subtract from them.
Brilliant super-villains will have brilliant nemesis. I would kill him or her first. Then brag.
Divide a slide into pairs of pixels. Randomly make one pixel of each pair black. Make a duplicate of that slide, but where you want black to spell out your evil plan, swap the black and clear pixels of the pair. Alone each slide is still random, but when you overlay the two you can read the letters, black on gray.
Something like this would be perfect:
http://people.csail.mit.edu/rivest/voting/papers/Chaum-SecretBallotReceiptsTrueVoterVerifiableElections.pdf
Makes a great voter-verified reciept too ....
Make gazillions of slides (you're a supervillain, you have the resources), hide your world domination slides in the pile somewhere.
By covering the entire planet in overhead projector transparencies you may have achieved your supervillain objectives anyway.
Shut up and shoot.
Make each slide have parts of the map so that only a combination of all the map will show the entire map.
The villain could make the diagram on a standard letter sized transparency, then cut it up into tiny pieces and neglect to number them. If the hero gets even a few, he'll have no idea.
Seriously though, if the hero manages to get the overheads, how good of super-villain are we dealing with here?
Print the slides using an ink very high in plutonium, such that combining any two slides results in criticality.
Show the completed presentation to the secret agent, solve two problems at once.
(I loved the lemon juice idea :)
Hey, wasn't this taken from an episode of The Prisoner? The one where Number Six gets his mind switched with some military leader? They have this whole subplot about a message hidden in a stack of slides. I think he has to wear polarizing glasses in order to view the message.
At any rate, here's my answer--you make your transparencies seem like they are encoded, but in fact include no information of any value whatsoever. Then, while the good guys are busy messing around with the overhead projector, take over the world.
Most of the solutions seem not to be following the requirement that we do not want even the tiniest bit of information available to someone with 2 of the 3 slides.
I think nosehat in #4 is on to something, but I worry that the method will not work if the amount of "black" on the transparency is not very much. If the final result were mostly black with some white writing then make it so that all of the slides will be white in each pixel where we want the writing, and for each other pixel that needs to be black in the final result make 1 black and the other two white.
Superimposing any two will reveal what appears to be a random sampling of white and black dots, but with all three in place you will see exactly what you want.
Maybe something like this?
Multiple layers was my first thought.
Unfortunately, there doesn't seem to be any difinitive answer on the site http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_cs;action=display;num=1029437951;start=
When the hero wakes up in your deathtrap, do your little slideshow. Fail to mention the fact that your plan has already been put into effect while he was unconscious, and that you already have achieved your goals.
Then shoot him in the head a few times with a shotgun, just in case.
What most people here are missing is that only having two sheets should give no information at all. Many people (#1, #2, #3, #5 and #16, for instance) for instance have suggested some variation of "Print every third letter on different sheets, with just two you wont have the whole thing", but the problem with that is that every sheet gives away WAAAY to much information. You could modify it by increasing the number of sheets and decreasing the amount of information (so if your plan is a hundred characters long, you have a hundred sheets each with one character on them), but I think that's sort-of a violation of the original riddle, it's murderously impractical (and you still get a tiny bit of information from two sheets, sometimes just two characters is enough).
A few other people have missed that it's supposed to be unreadable even if you have two sheets, not just one. #9's clever idea of using stereograms for instance, fails under this condition. If that were the problem, the solution is dead easy, much simpler (and much older) than the random dot stereogram: one page is a page full of completely random characters, and the other is a Cryptographic grille. No need for fancy schmancy computer-generated cross-eyed images, this is cryptography 16th century-style.
The problem with this riddle as stated is that it doesn't specify how many pages you can have, which makes it much less challenging ("I know! You have a gazillion sheets each with one pixel on it!"). The most bad-ass way to do it would be if the villain would only use three sheets, and the hero would be completely unable to get any information if he had any two of the sheets.
The only way I can think of to do it right now is to use the grille method, but with two grilles instead of one. If you have the two grille sheets, you obviously couldn't get any information out of it, and if you had the character-sheet and only one grille, you would get just a string of randomcharacters. But if you had all three, the characters where the two grilles intersect would become a readable sentence. The problem with this is that the two grilles would have to intersect only rarely to throw in enough entropy in it so that it would be unreadable with just one grille. Can't pack much information in there (unless you were projecting it on an IMAX screen and you had microfilm-sized letters). That's a solution that would be relatively simple and work pretty well.
Polarized transparencies.
I heard once that if you look through two pairs of polarized sunglasses, they're opaque... but if you add a THIRD pair, you can see through them again.
Assuming I'm remember correctly, it'd be easy to divide the chart among three slides. Each slide would have 1/3rd of the information, but if you arranged the text right I think you could mask the content.
Use a pen.
Draw as you explain.
Wipe away as necessary.
@24 Goldmineguttd:
Sadly, it's not true. Once the light is blocked out, adding more polarizers can't get it back. Try playing around with them yourself, if you get a chance - they have a lot of interesting properties.
You could use three sheets covered, mostly at random, by 1/3 gray dots. Very often two of them would line up to give 2/3 gray, but only in select places would all three line up to give black.
Done properly the result is black text on a 1/2 gray background, but a computer is necessary to make sure it's not too obvious from pairs of sheets.
The only way this even rates as a riddle is if you are indeed limited to 3 sheets.
Sheet 1: "Hold the world to ransom with death ray"
Sheet 2: "Kidnap the presidents daughter"
Sheet 3: "Start a new super-race"
Mwah hah hah! So you've seen two of my three sheets of plans do you? But you really still have no idea what I'm actually going to do until I actually put it on the projector!
... Actually I had another idea and faffed around trying to get my image proggie to display the message with negative space - but my photoshop masking skills are weak.
I've got it! First you write a hashing function that maps binary numbers (say, 128 bits) into letters, numbers and punctuation. Then you type up the plan, and convert each character to a block of binary that will hash back to that character.
Then you have, say, three transparencies that each contain every third binary digit, so that when they are overlaid you get the full binary sequence and each block hashes to a character and the plan is revealed.
With only two of the three transparencies, each block has a third of its bits missing and could be anything, and the agent has absolutely no information.
Of course then you'd need a computer to convert and display all the text, but I'm sure an evil genius can manage that.
Assuming a three sheet solution...
Sheets 1, 2 and 3 contain all the words in the message... plus random words.
How do you tell which words are relevant?
Each sheet has the words in seemingly random colors. When just two sheets are lined up you see another rainbow of colors. When the third sheet is added the relevant information is easily seen as the words that don't let light through... perfect for an evil genius, no?
@26... No, actually Goldmine is correct. Check out:
http://demonstrations.wolfram.com/LightBeamsThroughMultiplePolarizers/
@26, 31
in between is boring. after is the point:
http://www.scientificamerican.com/article.cfm?id=a-do-it-yourself-quantum-eraser
Throw people off with garbage characters.
Many of them will match on two sheets, but only proper letters will match on all three.
Sample 3 sheets:
1:HFTBGHUIOYFETYIE THYWGHEJDRFTTHYFVEIKJNFBGFL
2:SFFBHGUIOLKEPOIU QAZWSXEEFBHTTHFKVXIASDFJGFL
3:ERTBPPENYLRELKJH TGZYHNEUJYHYGRFEVXIOKNUHYGL
1-2-3 Combined:
###B#######E#### ######E#########V#I#######L
(# signifies a garbled blob
produced by layers of different letters)
1 and 2 combined:
##F#B##UIO##E###I# ###W###E####T#H##V#I##F##GFL
(2 & 3 or 3& 1 would be equally useless)
Notes to nitpickers:
Yes, they would be able to view full characters, but the more garbage you added the less useful this would become. Data and comprehendible information are not the same thing. Would *you* try to tell the police that someone will plant a Bomb, Bog, or Oboe at
123,12,or 15263 Main street in
1,17,or 17987 days?
-Yes, you would need a font that lines up better than this one does.
Afterthought:
To really make this work, the garbage characters that still show up when only 2 sheets are used should spell proper words, so that one can’t just look for proper words amidst the garbage.
1.) COLLECT UNDERPANTS 2.) ? 3.) BIG PROFIT!
#30, I tip my hat to you for scooping me while I was making an overly detailed explanation. I just couldn't resists spending too long delighting in explaining my evil scheme.....
Building on #23 Oskar's Grille idea - you can do this in three sheets. One has the base information sprinkled as random letters through a field of text. The other two sheets are polarizing grilles, where the polarization is changed throughout the slide to make the grille. Where the two grilles have a 90 degree rotation between their polarizations no light goes through. So with the base sheet and one grill you still see everything. With both grilles and no info sheet you see random clear spots. But with the information sheet and both grilles the opacities block out the incorrect data leaving just the presentation - in light letters on a black background.
#35 - I finally created an account to make that comment... somehow this pleases me even more now.
@ 26 and 31,
Once the light beam is blocked (two filters, with the last one at a 90 degree angle) there is no way of getting it back. What you can do is have a third filter in between at a 45 degree angle. With that the light beam will continue, although less brighter.
@38
again, nope. if u don't like my link just look up quantum eraser. from tha article:
"We will show you how to set up an experiment that illustrates what is known as quantum erasure. This effect involves one of the oddest features of quantum mechanics--the ability to take actions that change our basic interpretation of what happened in past events."
I have discovered a solution, so if you don't want to know one, don't read this comment.
I tend to think that the intended solution is something like:
have n slides, for each opaque pixel, assign it randomly to one of the n slides. Also scatter some random opaque pixels throughout the rest of the slides.
Having seen this done in a cryptography course with four transparencies, in real-world terms, this works pretty well. It's very difficult to really make out anything at all, especially if original image is a dithered black and white image. However, as many of the posters have correctly pointed out, this still reveals some information, if a very small amount. Any pixel which is opaque on any of the slides will be black in the final image. So you have gained some information about the final image.
If we were doing something similar in information security with files, we would use XOR for combination instead of OR (which is what laying transparencies on top of each other does). Let's say our villain has a boolean value x, and he XORs it with a random boolean value r (chosen 50/50 as 0 or 1), then if we snoop on him and find out this new value (x XOR r), this gives us no information about what x was originally unless also know r. If we know that x XOR r is 0, then x=0 if r=0 and x=1 if r=1. But since we don't know what r is and it was randomly chosen, it could be either just as likely. Similarly, if x XOR r is 1, then x=0 if r=1 and x=1 if r=0. But again, since we have no idea what r is, all we know is that x is 0 or 1, which is no information at all. This is used as the basis for one-time-pad, which is a cryptographic system which has perfect security (meaning that it is not just difficult, but, in fact, impossible to crack).
We can use this to encode a black-and-white image by dividing it into pixels and using the same technique on each pixel. Let's say our original image is I and we want to divide it into three new images, A, B, and C. For pixel I(x,y), A(x,y) will be a random 0 or 1, B(x,y) will be a random 0 or 1, and C(x,y) will be I(x,y) XOR A(x,y) XOR B(x,y). Obviously, A and B are just full of random pixels so they don't contain any information by themselves or even together. Because of the XORs with random numbers, C won't contain any information by itself. And even if you combine C and A, the best you can get out is I(x,y) XOR B(x,y), which is still no information about I(x,y) since B(x,y) is random. Similarly, if you combined C and B, you could find I(x,y) XOR A(x,y). But only when you combine A, B, and C can you get I back.
What we really wish to do is have something like XOR which works for slides. That's the real tricky part to this. 1 XOR 1 is 0, but you can't combine two opaque pixels to form a transparent one (I'm quite certain that the polarization technique one of the previous posters suggested for doing that would not work).
But there is a way to do it, and that's the real trick to the problem. In order to make this scheme work, we must have at least six different inks. There should be four different color regions (ranges of frequencies of light) such that for each ink, it absorbs two frequency ranges of our choosing and lets the other two pass through. This is pretty feasible given the way lighting gels tend to work, but I'm not going to spend my time actually digging up gels which have the right gel, nevermind verifying that there is an ink which gives the same effect.
We will have three slides: A, B, and C. Each of these slides will be colored by two of the inks. We will call these inks, A0, A1, B0, B1, C0, and C1. Every slide will be divided into pixels and then completely colored. Every pixel on A will be colored either A0 or A1. Likewise for B with B0 and B1 and for C with C0 and C1. The image we're encoding will be black-and-white. In the final result, we'll treat any opaque pixel as black and any pixel through which any color of light shines as white.
So, what we do is have the following pattern for the four color regions (where P means that the color range passes through it, and S means it absorbs it (soaks it up)):
A0 : T,T,S,S
A1 : S,S,T,T
B0 : T,S,T,S
B1 : S,T,S,T
C0 : T,S,S,T
C1 : S,T,T,S
When we layer the sheets on top of each other, any one pixel will have either A0 or A1, B0 or B1, and C0 or C1. If there is some color range which makes it through (T for all three sheets), then the final pixel will be considered "white" (0). If there is no color range which makes it through all of them (at least one S in each region), then it will be black(1).
So, for example, if we do A0, B0, and C0, then that should be 0 (since 0 XOR 0 XOR 0 is 0), and surely enough the first color range makes it through since they all start with T. Likewise A1, B0, C0 will be opaque because there is no column with all Ts. I won't go through all of them, but you can verify that the XOR table for combining the three values works out.
It doesn't work quite the same when just combining two slides, because given any two combinable inks (for example: A0 and B1 (not A0 and A1 since they will never be on two separate slides)) they always overlap at one color range. But specifically, they will overlap at a color range such that the coloring on the final slide will determine whether or not light will make it through. So, if A1, B0, C0 is black, but A1, B0, C1 is white, so if you just have A1 and B0 for some pixel, you don't have any information about whether the final pixel will be black or white. In fact, to make this true, it has to be the case that any two combinable inks overlap in at least one color range. If they didn't then you would know that the resulting pixel would be black. Anyway, even though it's not exactly the same as XOR in the two slide case, it has the same property: if you have two of the slides, you have no information about the original image.
So, in the overall, we split our image using the standard cryptographic technique employing randomness and XOR and then we color it using these three inks, and the problem is solved. World domination, here we come.
Have all the consonants on sheet and all the vowels on another.
Nobody will be able to understand the message without the vowels!
Vry fnny, nnyms. Vry fnny.
@ 39 Teufelsdroch
Quantum erasers do exist, but I couldn't find anything that suggests you can have three polarizers in sequence, and have the third restore light after the first two have blocked it. We're talking about transparency, not interference patterns, right?
@39
There's nothing about polarized filters in that article. (Although 'filter' is a bit of a misnomer.)
Why the polarization solution doesn't work:
Let's assume that Teufelsdroch and Goldmineguttd are correct, and that the effect they describe works. In my Physics II class we never did more than 2 polarized sheets, and I dropped out of quantum because I had a lousy professor, so I don't really know. But let's make that assumption. What I want to show is that this doesn't solve the brainteaser.
Let's further assume that we have a polarizing ink or a laser cutter and we can create arbitrary polarizing gradients on our slides down to the pixel level. In particular, what's been proposed is strict perpendicular polarization so that horizontal + horizontal = transparent, horizontal + vertical = opaque, but horizontal + vertical + horizontal = transparent.
The problem with this is that the plan will be completely revealed when the opponent has captured two sheets, if they are the right two sheets (or the wrong two sheets, depending on your perspective).
The order of the sheets matters because HHV will be opaque, but HVH will be transparent. So we have a distinct first, second, and third sheet. The problem arises when our adversary gets a hold of the first and third sheets. If the first sheet and the third sheet have the same polarization for a pixel, then that pixel will be transparent in the final result because both HVH and HHH are transparent (and likewise VHV and VVV). If they are different, then the final pixel will be opaque since HVV and HHV are both opaque, as are VHH and VVH. So, really, the middle sheet winds up not mattering at all. If you have the first and third sheet, you win.
Even if we expand to four sheets or more, this will still be the case (unless something really strange happens with HVHV or HVHVH). If you have the first sheet the photons pass through and the last sheet they pass through, you will know the answer because that will tell you whether the number of changes in polarization is odd or even.
So it was a good guess, but it doesn't really work as a solution to the problem. I'm still curious to know who's right, though.
This reminds me of one of the initial puzzles in the groundbraking ARG "The Beast". We got a gif file that was a simple green square, which needed to be set to transparent to read.
I thought that was pretty clever. I had no idea what was to follow.
@43
meaningful ambiguity begets solving camps begets argument begets solution. don't like my camp, find you own!
Each of the sheets is covered in different type of deadly poison and also the antidote to one of the poisons on another sheet.
Unless you have all of them, you will be missing one antidote and die before you read them.
nanobots.
the polarising films do work, because a polarising film rotates the the plane of polarisation by 90 degrees
I'm telling you guys, all these fancy-schmanzy quantum mechanical tricks are more trouble than they're worth! This problem was solved 450 years ago! Grille, baby, grille!
You make 3 identical transparencies describing your "plans" and let them fall into enemy hands. Then when the super-agent comes to foil your "plans" he finds out that it was a trap. This is the point where you place the transparencies on the projector and explain it all to him before you dispatch him.
My recall of dealing with polarized filters yielded results given following placement and listed angles:
0 + _ + _ = light
_ + 90 + _ = light
0 + 90 + _ = no light
0 + 90 + 45 = no light
0 + _ + 90 = no light
0 + 45 + 90 = light (!)
The last result is less surprising once you stop thinking of the word 'filter' in the traditional sense. 'Filter' tends to mean to eliminate something. Whereas a 'polarized filter' isn't eliminating so much as twisting the light waves. It just happens that a 90 degree 'twist' also reduces its magnitude to zero - thus 'filtering' out waves at a 90 degree difference.
(Yes, 'twist' isn't technically correct, but it is easier to think of it this way for simplicity of explanation.)
#12 Anonymous and #13 jasonjayr have it right for two sheets. The Chaum paper uses a method from Naor and Shamir, and #12's answer is pretty much the same.
For three sheets here's a way:
Four (or a multiple of 4) pixels per block. Say # = black and _ = clear.
Sheet 1: _#_# or #_#_ at random
Sheet 2: __## or ##__ at random
Sheet 3: _##_ or #__# according to sheets 1 and 2 and the desired color for the block.
Each sheet is 1/2 clear, the overlay of any two sheets is 1/4 clear.
If you randomly choose, for each block, which of the three sheets follows which of the three patterns, you also obscure which sheet is which. Bwah hah haa.
@40
Great idea! FYI you could implement the same crypto with no dyes; just divide each pixel of your message into four subregions (say North, South, East & West) and have the same transmission table you used, but here the columns are the transmission of the subregions.
(from 40, T=transmit, S=block):
N,S,E,W
A0 : T,T,S,S
A1 : S,S,T,T
B0 : T,S,T,S
B1 : S,T,S,T
C0 : T,S,S,T
C1 : S,T,T,S
Do the same thing 40 did: generate strings of different random 0s and 1s for the B sheet and the C sheet, and fill in each B pixel's subregions with a B0 or B1 pixel depending on whether that B pixel was 0 or 1; ditto for C. Now, for the A sheet, generate a bitstream of "message XOR B_random XOR C_random".
Obviously, neither the B sheet nor the C sheet contains the message; both are random. If the good guys capture the A sheet plus the B sheet, they'll be able to decode "message XOR C_random", which is still totally random.
Warning to all mad genii out there: you'll need to use a different B and C sheet for each slide you want to show; otherwise any good guy able to capture two A slides will be able to find the difference between the messages in the two slides captured, and will likely be able to discover large parts of your evil plan. This is the classic rookie mistake of reusing your one-time pad; don't fall for it.
I don't buy anything with polarization. I can't imagine that a super villain could take the time to polarize his overheads. Better things to do and all that. How about this for a 3 slide version:
Slide 1: Your plan written out, but with all kinds of random letters scattered through the message. So, suppose, your whole plan was the word, "JUGGLE," you would write: "7GJYOUGM1OOGRSLY11EPLQ." Easy right, "JUGGLE" is in there.
Slide 2: Lay a fresh transparency over the top of the first one and put a mark (dot, slash, whatever) above all the correct letters. Then go and put dots above a whole random mess of the wrong letters.
Slide 3: Lay a third slide over the top. Put a mark beneath all the correct letters, and then underneath a whole random mess of the wrong letters, making sure NOT to put a mark beneath any of the wrong letters that have a mark on the top.
When you lay all 3 down, you read only the letters with 2 marks. If you have just the two slides with marks, you have a mess of stupid. If you have the one with letters and just the top or bottom mark version, you can't tell which marks are for the right letters and which are for the random wrong ones.
I may be incorrect, of course. But that's how I'd do it.
what kind of riddle gets solved through polarising filters or (n)functions. The simplest answer will be the correct one. I think we're wrong so far. Especially you Keith -six different inks, indeed.
Draw the presentation on the glass of the projector. All the transparency slides are blank.
Ha!
Obvious: put all info on multiple slides, in such a way that you have to stack two or more slides on the overhead simultaneously, arranged in only one way, so that when combined they form the complete information (for example, 1/2 a diagram on one slide, 1/2 on another, and 1/2 of words/numbers on two others-- though you could stack as many as you like within reason: spread the info over 7 or 8 sheets that all have to be stacked, and use a mark in one corner to align them (in fact you could put the mark in different places, and have the sheets align at odd angles). You could even further obscure the info by using red and blue marker, and to view the true info you have to wear either red or blue glasses (like a kids game on a cereal box).
IF the G-men (or X-men) find some of the transparencies, I assume a brilliant supervillain knows enough of his plans that he can just fill in the rest from memory during the presentation.
C'mon guys. You are WAY overthinking.
Simply use up to 15 transparencies, maximum. Then, use an LCD letter font like this: http://www-inst.eecs.berkeley.edu/~ee40/calbot/webpage/lcd-ch1a.gif
Only put one or a few strokes of each letter on each transparency - you could get a bit creative at this point and require that some sheets be rotated or something.
Ta da! An encrypted message.
You can do it with straight black and white transparencies.
Divide each pixel into 4 quarters. Let us call them ABCD.
For the first transparency, pick two quarters at random (A and D, say) and make them dark, leaving the other two quarters (B and C) light.
For the second transparency, pick one dark quarter and one light quarter at random from the first transparency, (A and B, say) leaving the other two (C and D) light.
For the third transparency, if you want a dark pixel, pick the quarters that are the same in both the previous transparencies (A and C) and make them dark. If you want a light pixel, then pick the other two quarters (B and D) and make them dark. This way the one quarter that is clear on both the first two separations will also be clear on the third.
The first two transparencies clearly contain no data. However, the third transparency also contains no data, even when combined with any one of the other two. Pick any pixel on any two transparencies, and the four quarters will be light-light, light-dark, dark-light, and dark-dark in some random arrangement. Without the remaining transparency you have no idea whether the addition of all three will be light or dark.
Your stack of transparencies will pass at best 25% of the light, but your supervillan OHP contains an atomic pile and frickin' huge lasers, so you can live with that.
Mwhahahaa!!
Here's a real life version of the one time pad cryptography with sheets of polaroid film I did a few years ago:
http://www.segerman.org/misc_art.html#otpc
(it's the eye of Horus thing on the right)
The only way to satisfy the conditions of the problem presented is to have totally blank slides. Any slide with anything at all on it, even if encoded, is still information. They must be absolutely blank.
The problem doesn't say you must draw on the slides. It says "you know you can share the information across several slides." It doesn't say you have to actually use any slides.
So drawing on the glass of the projector means the enemy would have to capture the projector. But that is not a part of the problem presented. So you solve the puzzle by making the glass your transparency.
The obvious mathematical way (but one that may result in transparencies that are not readable by humans) is to divide the the message into pixels. On the three transparencies, each pixel is ON or OFF (shaded or unshaded).
On two of the transparencies, randomly shade the pixels. The third sheet is generated by using odd-even parity. The pixels that are ON in the complete message should have an odd number of pixels turned ON across all three sheets. That is, all three pixels are ON across all three sheets; or only one pixel is turned ON across all three sheets. The third transparency will also be random (this can be mathematically proved) and the combination of any two will be random. Only the combination of all three will have the required information.
The algorithm is as follows. Let sheet #1 and #2 be generated randomly. Here, x = ON and y = OFF.
MESSAGE PIXEL ON:
1:oxox
2:ooxx
3:xoox
MESSAGE PIXEL OFF:
1:oxox
2:ooxx
3:oxxo
If the transparencies are all printed in black, the message can be found, but it may be hard to see with the human eye. If we make the transparencies in primary colors (blue, red, and yellow), then the ON parts of the message will appear in primary colors and brown/black, while the OFF parts of the message will appear in secondary colors (orange, green, and purple) and white.
how about just not writing it down until its needed
#64 Alessandro Cima: The point is that random information is useless information. But it has to be really random. A lot of people suggested stuff that would look random but really isn't. Turing and his crew would get 'em.
#40 KeithIrwin: thank you for pointing out it's XOR. Of course! The slide does "and" and your eyes do "or"! (Which means you can implement any logic function that way!)
It looks like you and I (#55) are doing the same thing... but... colored ink?
#56 LeDopore: Yeah, NSEW. You can also use repetitions, like
ABBA
CDDC
CDDC
ABBA So the pixels finally come out centered.
#62 Richard Kirk: Yours is the right and most elegant way. But you need better visuals.
By the way, with all these methods, you must never let the slides touch each other prior to the fateful moment, or else CSI will recover the traces of ink on the backs of slides.
#67,
Absolutely not. The only way to prevent information from getting to the enemy is with totally blank slides. There's no other possible way to keep the information from the enemy if they capture slides. So where do you put the information?
There's no code that can't be cracked. So you have only one option with the slides - if you use slides. The riddle doesn't force you to use the slides. It only really requires a 'transparency.'
I think the rather vague language of the riddle is where the solution lies. The language of the riddle needs to be closely examined, I think.
What does 'share the information across several slides' mean? Share.? Doesn't that imply duplication? If each slide held only a part of the information, wouldn't the language say something like 'spread the information across several slides?' Also, 'share' does not mean 'print.' The riddle does not say that the information is 'printed' across several slides.
Also, what the hell does 'several' mean? It can sometimes actually mean 2 according to Merriam-Webster. It can also mean more than two but not many. Why doesn't the riddle say 3 slides? Or 4? Why 'several?'
Maybe thinking about printing on 3 slides is a mistake. Why print the information at all? Why not use 2 or three pieces of totally black slide material on the projector to simply make shapes or letters as you go? Sort of a live-action approach. If the enemy captures 2 pieces of your material they just have 2 pieces of black plastic.
Heck, 'several' can mean just about any number. You could make twenty pieces of black slide material and arrange them into words on your projector.
Here is Richard Kirk’s solution, with more elegant visuals. Label the three transparencies #1, #2, and #3. Make the message into pixels which are ON or OFF. Let’s use the terminology ON for where light is transmitted and OFF for where light is not transmitted (i.e. the transparency is black). Each pixel on each transparency is translated by a 2 by 2 grid of squares in states A (left) or B (right) as follows:
Transparency #1
◼◼ | ◻◻
◻◻ | ◼◼
Transparency #2
◼◻ | ◻◼
◼◻ | ◻◼
Transparency #3
◻◼ | ◼◻
◼◻ | ◻◼
Transparencies #1 and #2 are randomly generated. Transparency #3 is created so that for pixels that are ON, the number of transparencies in state A is ODD. For pixels that are OFF, the number of transparencies in state A is EVEN. Each ON pixel will have one square that transmits light and three that do not. Each OFF pixel will transmit no light.
All transparencies have no information and the combination of any two is also random. (Combining any two, every pixel will have one square transmitting light.)
I would use the method that the secure voting guy came up with.
You have two "metapixels" composed of four smaller pixels:
*-
-*
and
-*
*-
When the same type of metapixels are overlapped, the result is 2/4 black pixels, or a medium grey. If opposite pixels are overlapped, the result is 4/4 black pixels, or totally grey. Construct a one time pad for the secret transparency by choosing a random set of these 2 pixel metapixels as a one time pad, and then xor your real transparency with the one time pad (if the real transparency pixel should be black, choose the opposite metapixel to the one in the one time pad, and if the pixel should be grey, choose the same metapixel as the one time pad) so that only by overlaying the two transparencies will the secret plan be revealed.
This could be made more impressive by just polarizing each pixel vertically or horizontally, thereby allowing black/white areas to be created instead of only black/grey.
The solution I think the puzzle writer was looking for is as follows.
Use 6 sheets say (more will work even better). Each sheet will be divided up into pixels. Every pixel is either "black" or "white." (It will turn out that they will actually be black or some other dark color.)
White: The white pixels are easier: Pick two random colors out of R,G,B (say R,G for example) then for each sheet, flip a coin and either put a R dot or a G dot.
Black: For the black pixels, pick a majority color and two minority colors. With probability 4/6's put the main color on a slide, and with probability 1/6 use the first minority color and with probability 1/6 use the other minority color.
What does Eve see which she looks at two slides? For a white pixel, she see the same color 1/2 the time and two different colors 1/2 the time. For a black pixel, she sees the same color 1/2 the time and a different color 1/2 the time. So she learns nothing?
But what about Bob the intended audience? Any white pixel will come out something like purple. But any black pixel will have all three colors represented (at least with high probability) and will look perfectly black. So he should be able to read the text with out too much eye strain.
What I don't like about my solution is that sometimes a black gets printed as a white. Oops. I think this could be fixed by carefully using a stack of slides that always had all three colors represented. But I haven't gotten the math to work out yet. So I'll leave this for further discussion.
@69
You are still transmitting information. It is possible that two transparencies will combine to fill in one pixel entirely. One pixel of your plan will be absolutely known.
The standard way of handling this is to use XOR. You fill two files with randomly generated data and fill the third in the only remaining way to produce the correct result when you XOR each of the inputs together. This works because every data point can be reversed when you add another input file.
This doesn't work with transparencies. You are *always* reduced to some scheme by which some number of transparencies must vote for a particular pixel to be filled. The attacker can know some values for certain if he has enough of the transparencies. For example, if two transparencies have to vote yes in order to fill a pixel, the attacker will be able to tell that some pixels will be filled and some will be empty if he has two out of three transparencies. Some will be in doubt, but not all.
Using more transparencies and requiring more votes means that knowing any pixel for certain will require more transparencies. However, you will always get some information: the probability that a pixel is filled is different based on the number of transparencies you have that vote for it.
So there is no solution to this problem.
#72
Check again the solution in #69. Two transparencies cannot combine to transmit any information. The combination of any 2 × 2 element from two DIFFERENT rows (remember that each transparency only contains elements from one row) will NEVER black out all pixels.
Further note. The solution in #69 is based on XOR. Notice the pixel elements for Transparency #3 are derived XOR operation performed on the other two transparencies. By using four pixels, you can guarantee that all four are black or that three are black and one is white when all three transparencies are combined.
Hmmm... Interesting...
My plan would be to create slides with thin, parallel lines on them. The lines would have slight shifts in angle and thickness. When stacked at the proper angle, the text appears in the Moiré effect.
Bwa-hah-hah-hah!!!
#68 Alessandro Cima, you say about my #67, "Absolutely not. The only way to prevent information from getting to the enemy is with totally blank slides. There's no other possible way to keep the information from the enemy if they capture slides. So where do you put the information?
"There's no code that can't be cracked."
This paradox is why the puzzle is cool. Quoting from the puzzle: "Smart villain that you are, you know you can share the information across several slides so that if the enemy agents capture any 2 of your slides, they won't learn even the tiniest bit of information about your plan."
The key words are the last three. There is tons of information on each slide. But in the kinds of one-time-pad situations some of us are talking about, there's zero, exactly zero, information about the plan, until you put all three slides together.
One kind of crypto uses clever tricks and functions that are hard, maybe incredibly hard, but still not theoretically impossible, to decode. For instance, the mainstay of public-key crypto, RSA. People try to use a key length such that deducing the private key from the public key will take more computing than the NSA is likely to have in the next century. They call this "cryptographically secure." But still, all the information you need is there in the public key.
Then there is "information-theoretic" cryptography. This is the kind of crypto where decoding is actually impossible without the right information, because the necessary information just isn't there. It isn't there in a hard-to-decode form, it just isn't there.
This multiple-slide setup is of this stronger, information-theoretic kind. With any two of the slides you get information related to each of the bits of the original message. But when you put that information together to tell you the odds of the original bit being one or zero, you get exactly fifty-fifty, until you have the third slide. In information-theoretic terms, that is zero information.
It's easier to see using the example of a one bit message and a one-bit one-time pad.
Alice tosses an exactly-fair coin. She memorizes whether it was heads or tails, and also tells Bob, who also memorizes it. Edgar, an evesdropper working for the FBI, overhears this. This information is called the pad, because traditionally Alice and Bob would have matching sheets of many random numbers and they would use one sheet for each message and tear it off and destroy it after use. Okay, now Bob travels to Afghanistan.
Later, Alice has a message for Bob that's either one or zero. If the coin toss was heads, she writes down just what her message is. If tails, she writes down the opposite. Now, she calls Bob on the phone.
You are Eve, the evesdropper, in the police wiretapping office in Afghanistan, and you hear Alice say to Bob, "The encoded message is one."
What do you now know about her intended message? Either she flipped heads and intended one, or she flipped tails and intended zero. The odds are 50/50, and no amount of godlike computational or intellectual power will change that, because, in crypto lingo, there's no "toehold". But Bob, knowing the coin flip, knows the intended message.
So to answer your question, "Where do you put the information," it's in the combination only. Knowing just the coin toss (as Edgar does), or just the encoded message (as Eve does), you know zero bits of the message. But knowing both (as Bob does), you know one bit. The three-slide technique is like that except for fun details.
Of course any imperfection, leaked information, or mistake creates a toehold. If Edgar and Eve get together they can share information and learn the message. But with these systems, all other things being the way we're assuming (I mean after all even your "blank" slide can have chemical or physical or optically-induced evidence traces in it), you have zero useful information until you have the info from all three slides.
I don't know a better way to explain how each slide, and even each pair of slides, contains no useful information but the three together contain all the useful information. Maybe by saying, the slides contain information about the relationships between the coin tosses and the message, but it's just all "either", "or else" until you get all three.
#69 posted by KP,
The beauty of Richard Kirk’s solution is that any of the slides can use any of those three pairs of patterns in each block, and his wording makes them automatically, but randomly, use three different pairs, per block. So instead of each slide having a characteristic look (verticals, horizontals, or diagonals), all three have a random mixture.
By the way, how did you do that with the black and white squares!? But hm, "◼◼ | ◻◻," I can cut and paste...
Bwah hah haa, over?
#72 posted by dhasenan
Here is how we have been doing XOR with slides.
A pixel in a stack of three slides is clear if it's clear in the first slide AND the second slide AND the third slide, otherwise it's black.
A group of four neighboring pixels shows some light if the first pixel is clear OR the second is clear OR the third is clear OR the fourth is clear.
Letting clear be 1 and black be 0, you get NOT X -- let's say ~X -- by painting clear if X=0 and black if X=1.
Those are all the tools we need besides the definition of XOR, which is:
A xor B xor C =
( ~A and ~B and C ) or
( ~A and B and ~C ) or
( A and ~B and ~C ) or
( A and B and C ).
Now we can paint the pixels on the slides by reading down the columns and arranging in a clump of four:
Slide 1:
~A ~A
A A
Slide 2:
~B B
~B B
Slide 3:
C ~C
~C C
There will be one clear pixel when these three patterns are overlaid, iff A xor B xor C, otherwise all four will be black.
Notice the resemblance to KP's cool diagrams in #69 (except mine has been more uglified by Wordpress).
Pff.
here you go.
... and they called me mad? Hah!
Claiming "there is no solution
just because you don't know it is more arrogant than the average evil genius - and will lead to your downfall!!!
Taking the purist approach that "the only secret is one that you don't tell anyone" is ridiculous and against the point of the riddle.
We are writing this down to convey the message - in the knowledge that part of it may fall into enemy hands.
If it does not convey the message, it's not worth writing it down, and your method has failed.
I've illustrated a straightforward solution (well, 3x more straightforward than half those long ones put forward here) with animations and source code even!
- All information is on the three slides.
- No useful knowledge on any two slides
- Simple, clear, black-and white text and freehand line drawings on photocopy-able transparent sheets.
- no colors, no polaroids, no out-of-problem-space tricks.
- The process can be repeated automatically for any similar source material.
No hidden key or withheld, outside knowledge. I can give these three slides to my henchman and he can put up the display without any further help.
If you feel this system can't work, feel free to crack the example # 2 on that page.
I'll enjoy being proved wrong - just paste up the text of my encrypted evil plan #2 here. Or admit it works and I'll upload slide 3 for proof.
Guys, it's really more simple if you do it backwords.
Provide two sheets with plans, one completely contradicting the other (thusly negating the information you provide).
Anyone provided with the two sheets will be at a complete loss.
You, on the other hand, know which one to use for the presentation, and which one to ignore.
You need two sheets at most, but just one for your presentation.
Where's my prize?
I like solution 69. It is much better than my suggestion. I didn't think you could encode XOR in black and white--but this is totally cool! Well done!
The solution in 69 is identical to the solutions in 55 and 56 and 62 and 65, just with a more impressive diagram. All of them are correct.
I like the fact that it can be done without the colored inks, but I don't understand why people seem to think that using six different colors of ink is a big deal. I'm all good with subdividing pixels into blocks, but I'd much rather subdivide the spectrum.
And as for #58
Well, physics riddles are solved through polarizing filters and computer science riddles are solved through logical functions and algorithm. This comes from a page of computer science riddles, so hey, look, XOR. This is a cryptographic riddle, so it's no big surprise that it has a cryptographic solution.
The other thing about computer science riddles is that they frequently have more than one solution. This is why I stated that I had a solution, rather than the solution. So anyone who rejects a correct solution because they think it's too complicated, as anonymous is doing, is an idiot.
For anyone who comes back years later, here's a nice visualization of an XOR-based code, although not with the slide method.
Imagine a 2x2x2 cube of little black and white blocks in a checkerboard pattern. So the top layer is
◼ ◻
◻ ◼
and the bottom layer is
◻ ◼
◼ ◻
I can encode one bit with three bits by giving the coordinates of one of the little blocks:
Bit 1: top or bottom
Bit 2: front or back
Bit 3: left or right
Any two of those bits locates a pair of one black and one white block. For instance top left, or back right, or bottom back. Only when you have all three coordinates do you know whether the original message bit was black or white.
There are four ways to encode black and four ways to encode white. If I were to fall into a habit of always (or usually) encoding black one way and white another, my individual bits would give away my message (or give statistical clues). But if I always choose one of the four options completely randomly, the message is obscure until all three bits are known.
This is a visualization of XOR, and there's a further trick to doing XOR with overlapped slides.
The Richard Kirk solution (and the visualizing by KP) are such a wonderful thing, it continues to haunt my waking hours.
Of course, reading the solution I can visualize how it works, but actually working out a sample by hand on graph paper is an interesting experience (sadly I lack the higher functions necessary to write a program to output the required images). I generated the random elements of slide #1 and slide #2 using a fist full of pennies. My brain kept expecting the patterns to have an evenly distributed "staticy picture" quality, but instead it was very clumpy. There were longer strips of "heads" or "tails" with bits of singular elements here and there. Slide #3 was created to embed a simple image of a plus sign in the 3 slide set. It was fascinating to see how the two layers of random pixel slides underneath drove the content of Slide #3, it was like my brain wanted to push a pattern into the 3rd slide, but the lower slides pushed back. They created an equilibrium where NONE of the information existing on any two slides. Any two slides could be the foundation of any answer, but with the third slide there is only one answer.
Quite fun, if a bit time consuming. Thanks!