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You are located outside of two rooms which have a door between them. Each room has one or more computers whose total power is unknown, so they cannot be reliably used as proof-of-work to compare times, however they are not strong enough to break cryptographic secrets. You cannot see who is in the rooms.
There are only two options:
there are two different people in the rooms: in each room there is a different, honest person who stays in their room, or
there is one dishonest person who pretends to be two different people, by moving between the rooms.
You and the person in each room can verbally communicate with each other only through a computer network. You can also send stuff back and forth between you and the rooms, e.g. cards, balls — in order to find out if these are two different people or one person who pretends to be two. However there is a random delay of up to one minute in all methods of communication due to limits of the communication infrastructure.
The people in the rooms can communicate with each other through you, assuming they are two different people.
However, you cannot do anything that would reveal the actual identity of a person, e.g. check their DNA, take their fingerprints, take their photo, collect a sample of their handwriting, ask to see their ID card, listen to their voice, etc.
What strategy would you devise in order to make the dishonest person reveal their pretension, if indeed there is only one person in the rooms?
P.S. It's always possible to send to each room a different secret number, and if a person can tell me both numbers, then they have proven that they have access to both rooms. However to achieve this, I will probably have to reward the dishonest person e.g. by paying them, or by threatening to punish honest people who cannot prove having access to both rooms. Which is exactly the opposite of what a good solution would achieve. A good solution would make a dishonest person expose their pretension without punishing (or punishing too severely) honest people.
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$\\begingroup$ Is there a reason why you call these person(s) honest or dishonest? I ask to make sure this is not another one liar and one truth-teller riddle. $\\endgroup$ – Abbas 19 hours ago
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$\\begingroup$ @Abbas In each of the two cases, the occupants of the rooms would claim that they are two different people. In the 2nd case it would be a lie. $\\endgroup$ – rapt 18 hours ago
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$\\begingroup$ same principle applies I think. $\\endgroup$ – Abbas 18 hours ago
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$\\begingroup$ what about rot13(Jbhyq lbh fnl lrf vs V nfxrq lbh, ner gurer gjb crbcyr va gur ebbz?) $\\endgroup$ – Abbas 18 hours ago
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1$\\begingroup$ @Adam This tag covers several other concepts. The dishonest person in my question is a spy, if I understand the term correctly. They may lie if they think it would help hide their secret. $\\endgroup$ – rapt 18 hours ago
8 Answers
How about
All you have to do is preoccupy both sides using a method which cannot be abused by the dishonest person. For example any of these will do:
1. Hand both sides a woodwind instrument and tell them both to use it at the same time.
2. Send them both a locked container which contains a code word which they can only unlock by keeping all of their limbs on the container for an extended amount of time.
3. Send a device to each side that contains a code word and will only open when a button on it is held for an extended amount of time. The devices are set so that they will (for example) explode if they come within a certain proximity of each other.
4. You can give both sides a bottle which can only be drained if drunk. They must drink from it and give it back. You can check for tampering or if the bottle wasn't drained after you are handed it back. One of the bottles contain water; the other contains strong sleeping medication. Simply speak to both rooms afterwards, it will be very straightforward to figure out!
5. Give a typical and unique construction task to both rooms and time them. If they take longer than the average person then you should start to suspect it is the same person.
6. Have both sides read a long (never been released) book in a short amount of time and then ask about specific details of the books. Maybe the books chosen are such that someone who reads both of them fast will confuse the details which is another technicality on top of just reading the content.
7. Give both rooms a large cipher to decrypt.
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$\\begingroup$ Your several variations of proof of work are interesting but possible to circumvent. Idea #4 (poison) is good, but it can be a sleep medication instead. How would you adjust it to my updated question, i.e. computer network communication, where people are not supposed to be knocked out? $\\endgroup$ – rapt 19 hours ago
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1$\\begingroup$ @rapt if I understand the scenario correctly, a delay in communication - with the communication through computer makes this question unsolvable without resorting to psychology. P.s. the methods I listed are all way more complicated to pull off in practice however I summarised them to be sufficient examples. $\\endgroup$ – Adam 18 hours ago
I may:
Preparing a rope with enough length, and put each end to each room. Then I'll also send each room a clock, and same note written "Pull the rope at XX:XX". If the rope could be pulled at same time, 2 individuals in separate room is proved. If only 1 end is pulled or nothing happened, maybe just 1 dishonest guy in the connected rooms.
Notes: This solution stands only when someone is impossible to stay in the middle door and could reach both rope ends with both hands, due to the distance between the entries to put/send stuffs is not mentioned.
Improved:
Maybe more simple.
Just send in same clocks and same notes written: "Show me your left(or right) hand(or foot) at XX:XX". If I could see same hands(or foots) at same time via the put/send stuff entries, then it's true that 2 individuals in separate rooms, else only 1 dishonest guy in connected rooms.
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$\\begingroup$ What if there is no way to communicate between me and the person other than sending notes/stuff back and forth, and there is some random delay due to the communication infrastructure (in other words, imagine we communicate through a computer network)? $\\endgroup$ – rapt 20 hours ago
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$\\begingroup$ Hmm, that should be a totally different question compared to the original one. If that, something authentication concept in IT may help, like public/private key or digital signature, to identify whom is the real guy to contact or just someone pretended. $\\endgroup$ – Conifers 20 hours ago
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$\\begingroup$ A dishonest person may get two different sets of keys. Your task is through a clever communication find out if they are one person or two. $\\endgroup$ – rapt 20 hours ago
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$\\begingroup$ Then you might to have modify your question or ask a new one, due to all of the answers should works, through common sense, simple logical deduction or physic property. Anyway, time to think more deeper :D $\\endgroup$ – Conifers 19 hours ago
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$\\begingroup$ I have updated the question. It's hard to keep it simple and accurate at the same time. I hope this time around I did not miss anything. $\\endgroup$ – rapt 19 hours ago
This may be cruel yet very effective.
ask them to donate 1.5L of blood each. If it's a single person, they will have to donate 3L of blood, which will be fatal. Either person will confess to you, or you'll realise, once they stop responding.
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$\\begingroup$ This would identify the person. $\\endgroup$ – Veedrac 5 hours ago
You could:
send one side a leaky propane tank and, after a little bit of messaging, send the other side a lit match.
Or more humanely:
Send one side a strong smelling compound. Ask the other side to send back a sample of clothing. Use a trained dog to detect if the compound is on the clothing.
I would:
Get 2 clocks, and make sure they are set to the same time. Hand one clock over to the person in each room.
Ask them to pass a note (or anything) to me at exactly a specific time on the clock.
If there is only one person, that person would be unable to pass a note from both rooms at exactly the same time
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1$\\begingroup$ What if there is no way to communicate between me and the person other than sending notes/stuff back and forth, and there is some random delay due to the communication infrastructure (in other words, imagine we communicate through a computer network)? $\\endgroup$ – rapt 20 hours ago
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$\\begingroup$ @rapt Nice extra challenge, will give it some thought! $\\endgroup$ – MKBakker 20 hours ago
I think the best way to come up with an answer to this question is to look at this backwards - instead of thinking of ways to reveal if there is one person or two, think of how you would try to keep your secret if you were the one in the rooms.
To start with, I would make sure that all my interactions were carefully recorded so that I wouldn't unintentionally use information I'd received from the wrong room. If the computer doesn't support reviewing messages received, I'd have a notepad in each room on which I'd write down all the information I was given while in that room. I'd also have a box in which to put any objects I'd been given, so that I wouldn't unintentionally give back an object to the wrong room. Being a well-trained spy, you can be certain that I would never slip up and reveal myself through information or objects received from the room that I wasn't currently in.
The random delay in communication certainly helps with this - I could consistently wait longer than necessary to reply to messages and to pick up or return objects. This would give me time to review whatever notes I needed in order to maintain the deception.
Now then, how would I uncover a spy like this? If I sent them anything to eat or drink, the spy would wisely say "I'm sorry, I'm not allowed to eat or drink anything that's been sent to me" and return the item to me. I could try to trip them up on the communications delay by sending them an item that can detect when it has been handled - after it has been returned to me I would be able to see when they first touched it, and when they last touched it, and check if it matches the communication delays that they have been claiming. However, if the spy maintains proper discipline and waits a while to even touch objects, this method still won't work.
I can only think of one way to guarantee that you can determine if there is one person or two (two if you're allowed to kill someone):
Determine what information the spy is trying to get, and send it to them encrypted with a nonce, and send that nonce to the other room. When a nonce is generated in a crytographically secure manner and only used once, it is a perfect form of encryption that cannot be broken by any amount of computational power. The only trick here is that you need to obtain some information that is important enough to the spy that they will be willing to reveal themself.
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$\\begingroup$ 1. "only used once" - you mean only once for encryption (not necessarily for decryption)? 2. the simplest info is such that would stop something that would make life too difficult for a person who claims to be honest (i.e. who claims to stay in only one room), e.g. some code to prevent shutting the power off in the room of each honest person; however that goes against the principle of not punishing honest people. In that case instead of fancy encryption, just send each room a number, if a person knows both #s, I won't shut their power off. Any way to get the truth without punishing good guys? $\\endgroup$ – rapt 8 hours ago
What do you think?
Tell one end to speak continuously and upon receiving speech on your end (without interrupting the ongoing speech), tell the other end to do the same. Then abruptly tell the first end to pronounce a secret code. If and only if the secret code is pronounced without the second speech interrupting, they are two honest people.
Since I can't yet comment ("reputation" issues) I'm posting this as an answer, and would just like to point out that all "Ask the two rooms to perform a one-shot action on the same time" solutions (and by "one-shot" I mean it's not continuous work, you click a button, for example, and are done, as opposed, say, to typing in the entire text of "Hamlet" in one sitting) are wrong.
Since there is a RANDOM delay of up to a minute, and assuming (as it is not explicitly stated) that a single lying person can move between the rooms at near-instantaneous speed, it is easily possible that they perform the one-shot operation in both rooms themselves, relying on the randomness of the delay to not give their secret up: the fact that the key was REPORTED as being pressed in room A first does NOT actually mean it WAS the first pressed!
In contrast, typing a large body of text at on sitting defeats said randomness (I'm assuming, for this claim, that once an input stream from one of the rooms started it is continuous, that is, if the person on the room started typing and I started getting its response once the random delay finished, I'll keep getting that user's typing in REAL-TIME, AS LONG AS THE USER KEEPS TYPING. Actually, a pause would then be the ultimate indicator).
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$\\begingroup$ The random DELAY was added TO THE problem later to manage TIME based solutions, so THAT'S why it's in THERE. As for continuous TASKS, like typing A long SET OF characters, couldn't an HONEST person just need a BREAK every once in a while, KIND OF like how it's hard to CAPITALIZE ARBITRARY PARTS of a comment WHILE typing at the SAME SPEED all the TIME? $\\endgroup$ – hdsdv 1 hour ago
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$\\begingroup$ Didn't know that about the delay, I don't usually look at the edit log. As to the "need a break"... well, of course, I'm making some assumptions about the person(s) abilities. Even so, if we assume the input stream is consistent once the delay has worn off and until a meaningful pause in the typing is detected, I stand by my solution. Even this comment, if you type it by hand without pause should do the trick of proving. Basically, any operation taking longer than 1 minute to complete would do. And BTW, I didn't capitalize ARBITRARY parts of my answer. Look better, you'll see the logic... :D $\\endgroup$ – O.F. 45 mins ago
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$\\begingroup$ I claim it's arbitrary because if you read your answer, with all of the capitalized words uncapitalized, it doesn't change the meaning or anyone's ability to understand it in any meaningful way. People like to italicize (or capitalize, when they don't know the markup) things to make them seem important. But that doesn't make them important. $\\endgroup$ – hdsdv 33 mins ago
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$\\begingroup$ I can't upvote (thank you stupid reputation limitations), so consider your comment upvoted. :) I got your point about arbitrariness, I just, as I'm sure you know, disagree. Thanks for this cheerful discussion (really, not being sarcastic), but it feels like we're having a private chat that, on top of all, is off-topic. Upvoting you again, and have a great day! $\\endgroup$ – O.F. 21 mins ago
