36

One day after having his morning Bat-coffee, Batman received the following message on his Bat-computer:

Hello there Dark Knight. I, the Riddler, along with Joker, Two-Face, and Penguin, are preparing to pull off the heist of the century. But before we go, we're going to play a little game. If you win, we'll surrender to the police. If we win, the heist goes off as planned. Here's how the game works, in 1 hour you'll call the number at the bottom of this message. One of us will pick up the phone, and you'll ask 1 yes/no question. Try to ask any other kind of question, and we hang up and you lose. After you ask and we answer, we pass the phone to another of our members. We know the order we'll pass in, but you won't. The following facts will hold:

  1. I, the Riddler, will always answer truthfully.

  2. Penguin will always lie.

  3. Two-face will flip his coin, if it comes up heads he will tell the truth, tails he lies.

  4. Joker will lie or tell the truth depending on what he feels will most disrupt you.

  5. Each of us will answer before anyone repeats, and the cycle will not change while we talk.(i.e. The order we answer will be the same each round)

  6. If we do not know the answer to a question, we will say "Pass". Note: Joker will never pass.

The task is simple, figure out our order. Good luck.

What is the smallest number of questions Batman will have to ask in order to figure out the villain's order, regardless of what it is?

Bonus: Does it matter if a villain decides to hang-up if they hear a repeated question?

Note: I just came up with this puzzle, so I do not know the answer at this point in time.

Clarification of Joker's behavior: If a) Joker knows the answer to a question AND b) There is a way to answer the question while remaining logically consistent, then the Joker answers the question (lying or truthing depending on how to most screw-up Batman)

If a) Joker doesn't know the answer OR b) There is no way to remain logically consistent while answering, then Joker will pick yes/no based on what he believes would most screw-up Batman.

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Saladani
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  • If the Joker is deliberately trying to disrupt you, wouldn't he always tell the truth, thus making himself indistinguishable from the Riddler, and making it impossible for Batman to win? Alternately the Joker would seem to be able to disrupt you by flipping a coin like two-face. It is unclear what you mean the Joker will do when you word the question like this. Please specify what the Joker does. – Braydon Sep 18 '17 at 23:20
  • I meant that the Joker decides whether or not to lie on a given question based on what makes it most difficult for Batman to determine the order. If that means mimicing Two-face, he'll mimic Two-face. If that means always lying, he'll always lie. – Saladani Sep 18 '17 at 23:24
  • For example: If asked "Are you the Joker?" Joker would lie and say "No", unless he feels that answering truthfully would give Batman a harder time. – Saladani Sep 18 '17 at 23:25
  • But that makes this puzzle completely impossible. If the behavior of the Joker is the same as two face then there is no way to distinguish between the two. – Braydon Sep 18 '17 at 23:25
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    Nothing stops you from asking something like "Is the xth person after you Joker?" since that is a yes/no question. – Saladani Sep 18 '17 at 23:26
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    @Braydon The yes/no questions could be "are you the Riddler" or things of that nature, so if the Joker always answers truthfully, like the Riddler, then he would be less effective than "always answer truthfully, except for this question when it makes it too easy for Bats." – Mister B Sep 18 '17 at 23:27
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    How do you know that they are telling you the truth in their message? – boboquack Sep 18 '17 at 23:28
  • Riddler sent it, and if you can't trust him, the puzzle falls apart. He hates when that happens, so he's telling the truth. – Saladani Sep 18 '17 at 23:28
  • Are you allowed to ask questions you already know the answer to, such as "Is one plus one equal to two?"? I feel that that would make it too easy though... – DqwertyC Sep 18 '17 at 23:33
  • As long as it is a yes/no question, Batman can ask. If that means "Is 1+1=2?" so be it. – Saladani Sep 18 '17 at 23:34
  • @Saladani Well then it would still appear there is no answer to how many rounds, even though the answer can be found. The reason for this is because of Two-Faces erratic behavior. First in order to identify if someone is the Joker you must first positively identify either the Riddler or Penguin. The Joker can mimic one of there behaviors but not both, so one can always be found. The only issue is Two-face can tell repeated truths or lies. You could rule out Two-face in the second round if he contradicts himself, as he is then ruled out. However it is equally possible he will repeat. . . cont. – Braydon Sep 18 '17 at 23:35
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    @Braydon If you believe there is no solution, write it in an answer – Saladani Sep 18 '17 at 23:36
  • @Saladani It is equally possible Two-face will repeat his initial lie or truth in the second round, thus making it remain impossible to tell in a situation where the Joker imitates one consistent villain and Two-face randomly mimics the other. You would have to keep repeating questions until Two-face slips up, so the minimum rounds is decided by Two-face's randomness. – Braydon Sep 18 '17 at 23:37
  • If they can't respond, will they just pass the phone on? – boboquack Sep 18 '17 at 23:39
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    @boboquack Added a 6th rule to cover that case – Saladani Sep 18 '17 at 23:42
  • @Saladani Joker may be forced to pass though – boboquack Sep 18 '17 at 23:42
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    @boboquack Since Joker is trying to disrupt Batman as much as possible, he will always answer, even if he doesn't know the answer – Saladani Sep 18 '17 at 23:44
  • Just to make sure I understand rule 5 correctly, will the order they are asking questions stay the same each round, or will that change between rounds (so long as they all answer a question before the next round)? – DqwertyC Sep 18 '17 at 23:44
  • @DqwertyC The order will stay the same between rounds – Saladani Sep 18 '17 at 23:45
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    Let the heist go as planned. Batman will be there to save the day anyway. – Noldor130884 Sep 19 '17 at 06:56
  • Bat-coffee?? Really? Bat-coffee? WT....? – Marius Sep 19 '17 at 07:28
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    What happens if you catch a character in a logical fallacy? For example, if you ask The Penguin "Is the next word out of your mouth going to be 'Yes'?", he can't answer "Yes" because he would be telling the truth, he can't answer "No" because he would be telling the truth, and he can't pass because he technically knows what the answer will be. – oobug Sep 19 '17 at 08:06
  • Thinking out loud, if the Batman asks a question to which the answer is only only known by Penguin and Batman, then two villains will 'pass', and two will answer 'no' (or 'yes' - but both answers will be the same because this is the only way for the Joker to be the most disruptive) – Strawberry Sep 19 '17 at 11:33
  • You say that the Joker always answers what will make it hardest for Batman, therefore he'll never pass. This doesn't seem to follow to me; like, that would allow Bats to identify the Joker by asking a yes-or-no question that none of them know the answer to ('did the coin I just flipped here fall heads?'). In that circumstance, the Joker passing is what would make things hardest for Bats. So is "the Joker will never pass" inaccurate, or should it be considered an additional rule (not necessarily a logical consequence of "Joker answers in a way that makes things hardest")? – Oosaka Sep 19 '17 at 13:28
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    @Rozenn Keribin It is an additional rule, Joker likes the sound of his own voice too much to pass – Saladani Sep 19 '17 at 13:37
  • On this question: "Does it matter if a villain decides to hang-up if they hear a repeated question?" -> Is this repeated to that villain, or repeated overall ? It sounds from the setup that the villains don't hear your questions to the others so I assume the first, but the second is a stronger constraint. – Oosaka Sep 19 '17 at 14:05
  • @Rozenn Keribin The intent was the former, but if you can find a way to do the latter go for it – Saladani Sep 19 '17 at 14:09
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    Side channel attack, hook up the phone to the batcomputer and make it run voice analysis one the voice samples... – ratchet freak Sep 19 '17 at 15:26
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    Wait, you WROTE a puzzle and posted it without solving it yourself?? – feelinferrety Sep 19 '17 at 19:24
  • @Marius Not a fan of the 60's Batman TV show, I take it? – Kyle Strand Sep 19 '17 at 21:45
  • So, Saladani, can you please confirm the behaviour of Joker if both lying and telling the truth result in a logical contradiction? – boboquack Sep 19 '17 at 23:50
  • @boboquack Joker selects an answer based on what will most disrupt Batman. He doesn't care if he is logically consistent or not. (If you have a specific question in mind, I can provide better guidance). – Saladani Sep 19 '17 at 23:56
  • @Saladani For example, a question like the following: "Will you lie and say yes or tell the truth and say no?" - if the Joker lies and says yes, they are telling the truth, if they lie and say no, they are telling the truth, if they tell the truth and say yes they are lying, and if they tell the truth and say no they are lying. So the Joker can neither tell the truth nor lie. – boboquack Sep 20 '17 at 00:02
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    @boboquack He'd pick one. His behavior is such that if he knows the answer to a question and can answer consistently he will do so (lying or truthing as he sees fit). Otherwise, he selects yes/no to make Batman's life most difficult if a) Joker doesn't know the answer or b) The question cannot be answered consistently. Anyone else would pass such a question. – Saladani Sep 20 '17 at 00:07
  • @Saladani then you should probably say that – boboquack Sep 20 '17 at 00:11
  • Remembering how yes/no questions work, a perfectly valid and truthful answer to the question "Will you lie and say yes or tell the truth and say no?" is "No, I will not do either of those things." – oobug Sep 20 '17 at 07:56
  • @Braydon It's possible to distinguish between Two-Face and the Riddler, or Two-Face and Penguin. For instance, let A= "will you flip a coin?" B= "Would you answer 'no' to A?" Riddler will say 'no' to A, therefore 'yes' to B. If Two-Face flips a coin and it says to tell the truth, then he'll say 'yes' to A, and therefore 'yes' to B. If the coin says to lie, then he'll say 'no' to A, therefore 'yes' to B. – Acccumulation Sep 20 '17 at 18:37
  • @oobug But if "no" is a truthful answer, then they told the truth and said "no", so the answer to the question it "yes", which is a contradiction. And if you're insisting that "No, I will not do either of those things" is different from just "no", that contradicts the idea that these are yes/no question, and besides, the question can be reworded as "Will you lie and say yes or tell the truth and answer in the negative?" – Acccumulation Sep 20 '17 at 18:41
  • @Acccumulation The problem is that the question as asked is not a yes or no question. It's asking the answerer to select a value from a list.

    Take, for example, the question, "Is your favorite word 'Yes', 'No', or 'Potato'?" If the answerer's favorite word happens to be Potato, how would they answer truthfully? Simple! They would say "Yes," as in "Yes, my favorite word is one of 'Yes', 'No', or 'Potato'."

    If their favorite word is "Proscuitto", they would truthfully answer "No." The problem is with a misinterpretation of what a "Yes or No question" is.

     – oobug Sep 20 '17 at 19:17
  • I need clarification: If I were to ask Two-Face "If I were to ask you 'How many fingers am I holding up?' would you pass?" How would he answer? Would he pass because he isn't allowed to answer Yes/No unless by random chance, or would he answer Yes/No randomly since he knows for sure that the answer will be pass? – Ethan The Brave Sep 21 '17 at 16:02
  • Adding a new comment since I for some reason can't edit mine: Or would he simply flip the coin since he knows the answer, he just can't say it? – Ethan The Brave Sep 21 '17 at 16:04
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    @Ethan the Brave Since Two-Face knows the true answer is "Yes", he'd flip the coin and truth or lie based on what it came up – Saladani Sep 21 '17 at 16:12

12 Answers12

37

This answer is kind of cheap and relies on some technicalities, but here goes:

First round, ask everyone:

"Do I know how many more questions I'll ask?"
As per rule 6, all should say "PASS", except Joker. Now you know where he is.

Next, ask everyone:

"Does 1+1=2?"
Riddler says yes, Penguin says no, Two-Face is random, and we ignore Joker. If there are 2 "yes"es (ignoring Joker), then we know where Penguin is, and 2 "no"s mean we know where Riddler is.

Now, we have one person we know is either always lying or always telling the truth. Now, we ask everyone:

"Does Penguin answer before Two-Face?" (If we know Riddler's Position) OR
"Does Riddler answer before Two-Face?" (If we know Penguin's Position)
Riddler's answer to the first question, or Penguin's answer to the second, is enough to find everyone's order. (We already knew Joker's and one other's position, and their answer gives us the order of the other two)

So, at most, it would take:

12 questions, regardless of order.

I think I'll put a bit more work into this though, because the first question still seems cheap.

DqwertyC
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    You could reduce the number of questions by at least one. If the Joker is the last, he is not required for the last round of questions. If he is not last, you can start the second round of questions at least one question earlier. – Apep Sep 19 '17 at 00:30
  • +1 for the work so far. I may add a second bonus where a villain hangs-up if he is asked a question he cannot answer, but this answer is valid for the problem as stated – Saladani Sep 19 '17 at 00:38
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    @Saladani How is the Joker answering to mess with you if he allows himself to be found by refusing to pass? If he wants to mess up Batman he should pass. If this is an acceptable answer the question was again unclear. – Braydon Sep 19 '17 at 02:30
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    @Braydon the puzzle clearly states that the Joker will never pass. – oobug Sep 19 '17 at 05:52
  • @DqwertyC I don't think your first question is cheap. Batman could ask them "Is my right index finger itchy?" and it would be an equally valid question. – oobug Sep 19 '17 at 05:53
  • @oobug Oh yes, new rules were added after I had posted my answer, hurray for changing puzzle rules. – Braydon Sep 19 '17 at 06:32
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    The problem is that if they "no" to your first question, it probably becomes true. I'm not certain about this but I think it makes sense. – Samthere Sep 19 '17 at 12:18
  • @dqwertyC This answer can actually be much faster if the Joker goes first, as he will identify himself and there is no need to ask the next 3 pass questions. – Braydon Sep 19 '17 at 14:20
  • @Apep (and Braydon): Does your comment work under the assumption that Batman can ask to skip to the next round? As it was described, it's a sequence of ask-answer-pass-repeat (and any deviation from this will end in losing the game), so it may be impossible to "skip ahead". – Flater Sep 20 '17 at 08:15
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    @Flater It works under the assumption that Batman does not have to ask the same question 4 times. The 'rounds' was those set up in the answer (asking the same question 4 times), not those in the question (the villains answering in order). Once the Joker is found, Batman can stop asking the first question and ask the second question 4 times. In the case that the Joker is not the 4th, this saves at least one question. In the case that the Joker is the 4th, Batman does not need to ask him the last question, so that also saves one question. – Apep Sep 20 '17 at 11:59
  • @Apep: true, but that does have some implications for questions such as "Does Penguin answer before Two-Face?", where Batman must observe the original round order as opposed to his changed round order. – Flater Sep 20 '17 at 12:07
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The answer is

3

The reason is

Because he's Batman. The conversation is happening over the phone, which means he can hear their voices. He knows what each of them sound like so he just needs to ask three BS questions and he'll know who's answering simply by the sound of their voice. Once he knows the first three, the fourth is deduced from the process of elimination.

This will also work if

They attempt to disguise their voice. Batman is in his BatCave and has access to his highly-sophisticated audio analysis equipment. No doubt he would be recording the conversation to potentially use as evidence later. If the voices are masked, he'll be able to run them through his computer to remove that mask and determine who the speaker is.

And just for kicks,

Once he's identified the order, he won't actually reveal that he knows. Instead, he'll start asking questions designed to cause tension between the group. For example, he might ask the Riddler "Do you think TwoFace will get a bigger cut of this heist than the Riddler?" or maybe he'll ask Penguin "Did Joker tell you about that nuke he's planning on setting off in Gotham later today?". This is just to mess with them, possibly incite some in-fighting and more importantly, keep them occupied until he can show up and catch them. Because you see, they gave him their phone number and he'll have easily traced it to its location. He's asking these questions while en route and just needs to make sure they don't escape before he arrives.

However,

This won't work if they don't actually respond directly with their voices. Perhaps they have some machine that does the speaking for them and all they do is press the 'yes', 'no' or 'pass' buttons. But the question makes no mention of such a device so I think it's safe to say they won't be using one. I do realize this answer is probably going against the spirit of the question but this is Batman we're talking about. He's always one step ahead those bumbling buffoons.

aleppke
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    HOLY out of box answer, bat-aleppke +1 – Mindwin Remember Monica Sep 19 '17 at 21:55
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    This was my first thought as an answer. Your "however" area can still be worked around, even if they are disguising their voice. Batman can probably still tell from their speech patterns. For example, Riddler will probably use lots of big words, Joker will use lots of terrible puns and manic laughter, Two-Face will probably make references to binary dichotomies, and Penguin will talk like Danny DeVito. Also make bird jokes. You know those 4 could never be content with a simple yes/no answer, they must taunt Batman whenever they can. – Cody Sep 19 '17 at 22:48
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    ... Because he's Batman. –  Sep 20 '17 at 15:59
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At least half the time, Batman can get the order correct in only 4 questions. For certain sequences, it will take 5 questions.

Batman asks the first three villains in the sequence the same question:

If I were to ask you the question, "Are you The Riddler?", would your answer be "Yes"?
  • The Riddler's future answer would be "Yes", and since he answers truthfully, his answer is "Yes".
  • The Penguin's future answer would be "Yes", and since he answers untruthfully, his answer is "No".
  • Two-Face doesn't know what his future answer will be, so he answers "Pass".
  • The Joker doesn't know what his future answer will be, but since he never passes, he answers either "Yes" or "No".

At this point, Batman has asked 3 questions. Now for the fourth (and possibly the fifth).

What Batman does next depends on the answers that he's received so far.

  1. If the answers are either 2 "Yes" answers and 1 "No" answer, or 1 "Yes" answer and 2 "No" answers, the last villain in the sequence is Two-Face, since no one has passed yet. Batman will ask this question to Two-Face: (replacing [order number] with the order of one of the duplicate answerers)
    If I were to ask the villain in position [order number] if they are The Riddler, would they answer "Yes"?
    If Two-Face gives an answer, Batman knows that the character in [order number] is The Riddler or The Penguin, depending on whether the duplicate was "Yes" or "No". If Two-Face passes, that character is The Joker. With this information, Batman can figure out the order.
  2. If the answers are a "Pass" and either 2 "Yes" answers or 2 "No" answers, the "Pass" reveals Two-Face's location and Batman knows the last character is either Riddler (for 2 "No" answers) or Penguin (for 2 "Yes" answers). Batman can straight-out ask Riddler/Penguin about one of the duplicates (e.g. "Is The Joker the villain in position [order number]?"), and their Yes/No answer will reveal the entire order. This scenario is unlikely, as The Joker would have had to mess up by giving a duplicate answer.
  3. If the answers are "Yes", "No", and "Pass", the last character is either The Riddler, The Penguin, or The Joker. Batman will ask the last villain the following question:
    I am thinking of a number between 1 and 10. Is it the number 7?
    If the villain gives a Yes/No answer, the villain must be The Joker, since Riddler or Penguin would have to pass.

In any of the above scenarios, Batman solves the puzzle in only 4 questions.

In scenario 3 above, if the last villain does not give a Yes/No answer but passes instead, Batman knows that the last character is not The Joker, but doesn't know whether they are The Riddler or The Penguin. What Batman asks as his 5th question depends on who is first in the order.

  • If the first person is Two-Face (that is, the original answer was a Pass), Batman will ask him:
    If I ask the next villain the question "Are you The Riddler?", will they say "Yes"?
    Two-Face does not know what The Joker's answer would be, so if Two-Face passes, the next character is The Joker. If he answers, the asked-about character is The Riddler or Penguin (depending on a whether the first-round answer was Yes or No). Either way, Batman knows the order.
  • If the first person had a Yes/No answer, Batman will ask them the question from scenario 3. If they answer, the character is The Joker. If they pass, the character is Riddler or Penguin, depending on their original Yes/No answer.

In that scenario, Batman solves the puzzle in 5 questions.

Using this method, Batman is guaranteed to solve the puzzle in 4 questions if Two-Face or The Joker are the last villain in the order. Otherwise, Batman will solve it in at most 5 questions.

BONUS: With this technique, Batman will never repeat the same question to a single villain.

oobug
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  • Why is 8 the minimum? – boboquack Sep 19 '17 at 07:54
  • @boboquack You're right, I could actually refine it a lot. If I started with question #2 I could find Two-Face and The Joker faster. But I need to go to sleep. I'll take a look in the morning. – oobug Sep 19 '17 at 08:10
  • @boboquack Evidently my math was bad. But that makes my answer better! – oobug Sep 19 '17 at 08:18
  • I just completely rewrote my answer. I've been thinking about this puzzle way too much. – oobug Sep 19 '17 at 22:12
  • Re-edited my answer; still using basically the same technique. The minimum number of questions is now 4. – oobug Sep 20 '17 at 06:38
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    I don't think the qualifier 'on the next round of questions' is needed. I'd also hand-wave 2 away as Joker can't answer that way as it gives away too much information and lets Batman find him early. – Danikov Sep 20 '17 at 15:09
  • @Danikov Thanks for the suggestions! I've cleaned up the original question. I don't want to entirely remove scenario 2 since it proves this technique will work for any set of answers, but I have edited the wording to clarify the scenario's unlikelihood. – oobug Sep 20 '17 at 16:53
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    I think you're right, the more I think of it, Joker should be treated as a random selection of yes/no. He cannot predict Batman's strategy so, while sometimes he might answer later and there's an obvious answer that gives away information to avoid, sometimes he might go first and he cannot predict Batman will ask the same question again later on. – Danikov Sep 21 '17 at 10:34
  • I have a nagging feeling it can be done in 4 reliably, but it's tricksy. – Danikov Sep 21 '17 at 13:15
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    Ah, maybe I'm wrong; the Joker's misinformation multiplies out the 24 possibilities into 48 possibilities, which is less than the 3^3 = 27 bits of information obtainable from the other three villains. – Danikov Sep 21 '17 at 13:42
  • Very, very nice try, however there is a problem with scenario 1: "If Two-Face passes, that character is The Joker." You would have to prove that The Joker will answer randomly. I don't believe that is the case, because if that character is The Joker, he will apply the same logic as you do, and he will answer Yes or No to mislead you. Now assuming Two-Face is on the same thinking level, he will also know which answer is the misleading answer that The Joker will give, and therefore he will flip a coin and give an answer himself. – Bogdan Alexandru Sep 22 '17 at 09:25
  • @BogdanAlexandru you're half-right. Two-Face's answers are all logically consistent: if he doesn't pass, he enables the Joker to be malicious, making the Joker's answer predictable, but if he does pass, the Joker doesn't know what to choose and his action does become unknowable. The rules suggest Two-Face will flip a coin and attempt to answer truthfully or not before considering passing, though, so I agree with your gotcha, but I don't think it undermines that it is possible in 4 or 5 questions. – Danikov Sep 22 '17 at 12:57
  • @BogdanAlexandru The problem with that logic is that, if The Joker is in position 2, an answer of either yes or no will be equally misleading. Two-Face truly doesn't know what The Joker's answer will be, and since he must either answer truthfully, untruthfully, or pass, he is guaranteed to pass. – oobug Sep 22 '17 at 22:08
  • There's a flaw in this answer. The first time you asked The Joker whether he is The Riddler, he answered either yes or no. Since he tries to be consistent, he knows what he'd answer if you were to actually ask him that question. So do the other villains. Your answer relies on them on knowing that, so it isn't going to work... – Jasper Sep 23 '17 at 08:19
  • @Jasper Where in the original puzzle does it say that the Joker tries to be consistent? – oobug Sep 23 '17 at 16:35
  • @oobug "AND b) There is a way to answer the question while remaining logically consistent" – Jasper Sep 23 '17 at 21:33
  • @Jasper Oh I see. That (and the following) paragraph was added later as a provision to ensure Joker will never pass, even if the question is logically unanswerable. The Joker's behavior should really be described more like, "Joker will answer yes or no to every question, choosing his answer each time based on what will most mislead Batman." Also, the puzzle does not indicate that the villains will know each other's answer, and even if they do, The Joker is an inherently unpredictable character, so the other villains will never know and will always have to pass when asked about his answers. – oobug Sep 23 '17 at 22:43
  • @Jasper Example of a logically unanswerable question: "Is 'No' the answer to this question?" – oobug Sep 23 '17 at 22:46
  • That is logically answerable - "No" (lying) – boboquack Sep 24 '17 at 06:20
  • @boboquack Correct, I should have clarified that the question is logically impossible to answer truthfully. The equivalent of the logically unanswerable question for characters forced to lie is "Is 'Yes' the answer to this question?" – oobug Sep 24 '17 at 07:07
  • @oobug Yes, that's why it was added. However, I'd say it also logically flows from his core tenet of "misleading as much as possible". If you pin him down as either the Joker or The Penguin (by eliminating Two-Face) his best strategy to mislead you is to keep impersonating The Penguin. It is not mentioned whether the others can hear the questions and answers of each other, but the assumption that they don't is a pretty big one (as is that they do). It makes sense to assume good logicians who know the rules of the game and calling someone "inherently unreliable" doesn't change that. – Jasper Sep 24 '17 at 13:04
  • Sorry for Necroing; took a bit to get the points, but I've wanted to point this out since I first read this after it was posted: "If I were to ask the villain in position [order number] if they are The Riddler, what would they answer?" is not a Yes/No question. While it's true that its possible answers are "Yes" and "No", this is the equivalent of writing a program that outputs true or false vs a program that outputs the strings "true" or "false". As such, the villains would hang up and Gotham is doomed. – Suthek Jan 19 '18 at 12:10
  • @Suthek Whoops! Good catch! I have rephrased the question so that it still possesses the question's original logic (thereby not changing the answer), but has now been converted to proper boolean yes/no logic. Thank you for spotting that! – oobug Jan 19 '18 at 18:39
9

I can do (and this is minimal):

3 questions (for both variations, this is minimal because there are only 3 possible responses and 32<4!)

Ask the following questions to three people:

If I took the four letters J, P, R and T, put them in the order corresponding to the order you are in where J=Joker, P=Penguin, R=Riddler and T=Two-face, indexed it into the following dictionary (see dictionary below, Batman will have to recite this but I won't add it here), and took the first/second/third change this for each question - each corresponding to the question number asked letter of the identifier, will either the letter be Y and you are telling the truth, the letter be N and you are lying, or the letter be P and either you will lie and say yes or tell the truth and say no?

Then:

Yes means Y, no means N and pass means P since the option for P is a head-exploding statement. Then you can reconstruct the three letter identifier, back-index into dictionary and find the order!

Dictionary:

JPRT: YNY
JPTR: YNN
JRPT: YNP
JRTP: YPY
JTPR: YPN
JTRP: YPP
PJRT: NYY
PJTR: NYN
PRJT: NNY
PRTJ: PPP
PTJR: PNY
PTRJ: PNP
RJPT: NYP
RJTP: PYY
RPJT: NPY
RPTJ: NPP
RTJP: PPY
RTPJ: NNP
TJPR: PYN
TJRP: PYP
TPJR: NNN
TPRJ: YYP
TRJP: PNN
TRPJ: YYN

Examples:

Riddler, letter=Y: "The letter is Y, and I am telling the truth. So true, and I say yes."
Riddler, letter=2: "The letter is not Y. The letter is N, but I am telling the truth, so false. The digit is not 0. So false, and I say no."
Riddler, letter=P: "The letter is not Y or N. The letter is P, and I am telling the truth. If I say no, then it's true, because I say no. If I say yes, then it's false, because I don't say no. So pass."
Penguin, letter=Y: "The letter is Y, but I am lying. So false, but I say yes."
Penguin, letter=N: "The letter is not Y. The letter is N, and I am lying. So true, but I say no."
Penguin, letter=P: "The letter is not Y or N. The letter is P, and I am lying. If I say yes, then it's true, and I should have said no. If I say no, then it's false, and I should have said yes. So pass."
Two-face/Joker: refer to Riddler if telling the truth, Penguin if lying. Note - the Joker will never be given a head-exploding question

boboquack
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    Could you please provide an example, I'm having trouble tracking the logic all the way through – Saladani Sep 19 '17 at 02:37
  • @Saladani done! Also, the Joker has to pass since there is no way out. – boboquack Sep 19 '17 at 03:19
  • Thanks, I believe this is as good as it can get (assuming I'm not overlooking some flaw that means this actually would not work). I think I'll wait ~ a day to be sure. – Saladani Sep 19 '17 at 04:42
  • Based on the original puzzle description, Batman cannot ask a question seeking a True/False answer, he has to ask a question seeking a Yes/No answer. Answering "Yes" to a question in the form "Is the following statement true or false?" means that the statement is either true or false; answering "No" to the question means that the statement is neither true nor false. This would continue to be true even if the question were phrased, "Is the answer to the following question yes or no?" – oobug Sep 19 '17 at 05:48
  • @oobug True and false is virtually the same as yes or no, you can easily convert between them. And don't take my thing so literally - I'm saying 'is the following thing true or false' to elicit a true or false answer. – boboquack Sep 19 '17 at 06:49
  • @boboquack Well, it's a pretty literal puzzle. There's no way to ask an "Is this statement true/false/yes/no/red/blue" question if the asker is asking a yes or no question. The answer will always be "Yes, the statement is true or false" or "No, the statement is neither true nor false".

    And yes, there can be statements that are neither true nor false. For example: "Is the following statement true or false: 'Count the gumdrops!'".

     – oobug Sep 19 '17 at 07:35
  • @oobug Well, does this satisfy your constraints? – boboquack Sep 19 '17 at 07:50
  • @boboquack Almost! But I think you need to change the last part of the statement from a head-exploding logical impossibility (since then the correct answer to "is the statement true" will simply be "no") into something that it wouldn't be possible for the characters to know, like "I have exactly 70,452 hairs on my body". – oobug Sep 19 '17 at 07:57
  • @boboquack The other problem being, of course, that in an actual Yes/No scenario, you can't force The Joker to pass. – oobug Sep 19 '17 at 07:59
  • @oobug then the Joker could answer without problems. The beauty of the head-exploding statement is that if they answer yes, no matter whether they are telling the truth or lying, the only answer is no. And if they answer no, no matter whether they are telling the truth or lying, the only answer is yes. And the Joker has to pass since no matter whether they are lying or not, neither answer is logically consistent. – boboquack Sep 19 '17 at 08:00
  • Let us continue this discussion in chat. – oobug Sep 19 '17 at 08:25
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    This raises the question of whether the "Joker never passes" rule is consistent with the "Joker lies or tells the truth depending on which messes you worse" rule (which isn't the same as "Joker answers whatever messes you worse", in which case it would indeed be impossible to force him to pass). – Oosaka Sep 19 '17 at 15:06
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    Also, I had to really struggle to understand this answer - maybe it's just me, but I feel it isn't completely clear that you aren't asking three times the same question, and "ones/threes/nines" represent a variable part where in the first question you plug in "ones", the second you plug in "threes" and the third you plug in "nines". – Oosaka Sep 19 '17 at 15:31
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    @oobug batman always uses oxford commas – Mindwin Remember Monica Sep 19 '17 at 21:54
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    Correct me if I'm wrong, but I just realized the scenario JRPT (which, assuming correct interpretation indexes to 010) won't work because Joker cannot pass. He will either answer yes or no, both of which screw up the scheme. – Saladani Sep 19 '17 at 21:54
  • Yeah, this answer unfortunately doesn't hold up if you enforce the "Joker doesn't pass" rule. – oobug Sep 19 '17 at 22:13
  • @Saladani but how can he answer yes or no? – boboquack Sep 19 '17 at 22:59
  • I'll reiterate what I said in chat: if the question you're actually asking is "Is the following statement true?" and then the following statement is a logical impossibility, the truthful answer is "No, the statement is not true, it is nonsensical." – oobug Sep 20 '17 at 00:07
  • @Saladani now it's fixed. The joker never has to pass – boboquack Sep 20 '17 at 00:41
  • @oobug now it's fixed for your quibble as well – boboquack Sep 20 '17 at 00:43
  • @boboquack Again, anybody's answer to the head-exploding statement will be "No, that is not true, that doesn't make sense." So no one will ever pass. It's an ingenious solution, but it's hampered by the limitations of spoken language. – oobug Sep 20 '17 at 00:58
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    @boboquack: It's been made explicit now that the Joker doesn't necessarily lie or say the truth, he says yes/no based on what will screw with Batman best. So it is impossible to force him to give any given answer, head-exploding as the question may be. And if the Joker doesn't pass then the strategy doesn't work, as it relies on there being three possible answers. It's too bad because it was an impressively minimal and parsimonious answer. – Oosaka Sep 20 '17 at 11:14
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    @RozennKeribin That is not the intention. If Joker knows the answer, he will lie or truth based on what messes with Batman the most. However, if truthing and lying give the same answer, then that is his answer. He only "randomly" (but not really) picks yes/no if the question is either head-exploding or he doesn't know the answer. – Saladani Sep 20 '17 at 19:10
  • Copying comment from above... the problem with this solution is that the question as asked is not a yes or no question. It's asking the answerer to select a value from a list. Take, for example, the question, "Is your favorite word 'Yes', 'No', or 'Potato'?" If the answerer's favorite word happens to be Potato, how would they answer truthfully? Simple! They would say "Yes," as in "Yes, my favorite word is one of 'Yes', 'No', or 'Potato'." If their favorite word is "Proscuitto", they would truthfully answer "No." This solution misinterprets what a "Yes or No question" is. – oobug Sep 20 '17 at 19:19
  • @oobug The question is not: "Is X, Y or Z?" The question is: "Will X, Y, or Z occur?". If one of X, Y and Z occur, then the only correct answer is yes (though they might lie and say no). If none of X, Y nor Z occur, then the only correct answer is no (though they might lie and say yes). If they are telling the truth and one of X, Y and Z occur when they say no but none occur if they say yes, there is no logical answer. If they are lying and one of X, Y and Z occur when they say yes but none occur if they say no, there is no logical answer. I stress to you, read the question again! – boboquack Sep 20 '17 at 22:25
  • @Saladani: yes, I later saw my comment was wrong the way I phrased it, thank you for clarifying. But the Joker being able to answer whatever to a head-exploding question is enough to break this solution. – Oosaka Sep 21 '17 at 08:46
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    @RozennKeribin This solution has changed, now the Joker will never have to answer a head-exploding question. – boboquack Sep 21 '17 at 09:06
  • @boboquack: Oh, I see! I saw the solution had changed but hadn't realized how this keeps the Joker from head-exploding questions; I see now that you've build the YNP order <=> villain order dictionary so that the Joker is never in a position where the letter is "P". Well done ! I didn't see how you could compact even more information into this solution but you pulled it off. – Oosaka Sep 21 '17 at 10:13
  • Yeah, given the Joker now gives a binary answer, your possibility space for 3 questions mightly only be 3 x 3 x 2 = 18, < 24, so 3 questions will never be enough unambiguously determine the order. – Danikov Sep 21 '17 at 10:13
  • @Danikov No, if we account for the Joker giving a Yes/No answer in our dictionary, it allows us to take advantage of the fact that there are still 3^3 outcomes, even if not all 3^3 can be realised in a specific case. – boboquack Sep 22 '17 at 00:03
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    How do you go from "If we do not know the answer to a question, we will say 'Pass'. Note: Joker will never pass." to "If neither yes or no is a correct answer we will say 'Pass'"? – Taemyr Sep 22 '17 at 08:53
  • @Taemyr There are correct answers, such as 'I don't understand that', but they don't 'know' them since they are confined to saying yes, no or pass – boboquack Sep 22 '17 at 10:03
  • @boboquack They are confined to saying yes, no, pass, or hang up. The issue depends on how you interpret 'yes/no question'. An interpretation that would cause problems would be 'A yes/no question is one that could, given perfect information, be answered truthfully with either 'yes' or 'no'.' – Taemyr Sep 22 '17 at 10:31
  • @Taemyr I would define it as a question that expects an answer of yes or no, even if that's not necessarily what's given. For example, "Are you happy?" is a yes/no question, but "What is 8-3?" is a number question and "What colour is a rose?" is a colour question. But I could answer the first with, "Very much so," the second with, "Depends what number system you use," and the third with, "What species of rose?" – boboquack Sep 22 '17 at 10:56
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    @boboquack I am not saying that there are no definitions that causes your question to work. I am saying that there are definitions that causes your questions not to work. – Taemyr Sep 22 '17 at 11:45
  • @Taemyr Point taken, then. I think my answer is fine as it is, if it is not what the OP is looking for, I can change it. – boboquack Sep 22 '17 at 11:48
  • @boboquack The joker deliberately lies and/or is random, so at least one bit of information is unreliable (and you don't know which it is). So you must expend a second bit to correct the unreliable bit. This means the Joker's unreliable bit of information actually costs us a bit; the 3 x 3 x 3 = 27 from the other villains must be halved and we fall short of the required dictionary size. – Danikov Sep 22 '17 at 12:45
  • @Danikov The joker will answer any single question in a reliably true or false manner, so you can account for his lies by indirection. – Taemyr Sep 22 '17 at 13:40
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    Tbh, I'm not sure whether or not to accept this as the answer. Upon looking at it further, the question Batman would ask seems to be stretching the definition of a yes/no question. – Saladani Sep 22 '17 at 17:58
  • @Saladini IMO this is a yes/no question, even if it is very convoluted - it's no different to saying, "Depending on the temperature, do you wear long sleeves when it's cold and short sleeves when it's hot?" or "If I made you wash up the dishes for a week, would you really learn your lesson and stop misbehaving?" – boboquack Sep 22 '17 at 19:51
  • I think that Joker listening to the question would pick the "screw him the most" and add to that "screw the game" and hang up somewhere between word "indexed" and "dictionary". – quetzalcoatl Sep 23 '17 at 18:49
  • @quetzalcoatl If the Joker has that short an attention span, screw him :P – boboquack Sep 23 '17 at 19:47
  • @boboquack In your "Depending on the temperature" example, remember that valid "No" answers include "No, I wear long sleeves all the time" and "No, I wear short sleeves all the time." The problem with your solution is that nobody will pass, because the question is never unanswerable—it can always be truthfully answered "Yes" or "No". – oobug Sep 23 '17 at 22:52
  • @oobug how is that related? – boboquack Sep 23 '17 at 23:05
  • @boboquack I'm just saying you're not thinking creatively enough about the possible answers to your question. Imagine you ask The Penguin your "would either the letter be..." question. Regardless of the letter in the dictionary, The Penguin could answer dishonestly and say "No, the letter is 'H'." H isn't an option, but that's fine because The Penguin knows the answer—he's just lying. – oobug Sep 24 '17 at 01:18
  • @oobug no means that none of the things are true xor the character is lying. From that we can conclude what we want, logically, it doesn't matter what the penguin is saying no for. – boboquack Sep 24 '17 at 05:07
  • @boboquack But it means that The Penguin never has to pass. If the letter is "P", he can simply say "No." Your solution implies that The Penguin and Two-Face will be forced to pass. But if they're lying, that's not actually true. If the letter is "P", the answer can be "No, the letter is Q." Except they only give a one-word answer, which is "No." So you will never know if the answer is P. – oobug Sep 24 '17 at 05:15
  • @oobug It does not matter what they would like to trick you into thinking the letter is, the truthiness or falsity of the statement is not changed – boboquack Sep 24 '17 at 06:17
  • @Danikov We don't need to determine whether the Joker is lying or telling the truth. 27 possible sets of answers is more than enough to cover for 24 permutations, provided you use a working bijection. – boboquack Sep 24 '17 at 06:22
  • @boboquack but you're not guaranteed 27 answers when the Joker is in the mix (a 3 in 4 chance of happening) and you haven't convinced me how the Joker is compelled to answer cooperatively with your dictionary. What's to stop him from lying about his answer to your question and saying no when you expect a yes? – Danikov Sep 25 '17 at 08:51
  • @Danikov Have you looked at my examples section? It will show how we can extract the right answer regardless of his disposition – boboquack Sep 25 '17 at 10:21
  • @boboquack You say refer to Riddler/Penguin depending on his disposition, but if Joker is doing his best to screw with Batman, lying Yes when you figure he can only say No is the best way to do so. I don't see why you're treating his answers as reliable. – Danikov Sep 26 '17 at 08:57
  • @Danikov in the question statement, it says the Joker will tell the truth or lie if possible, and iff it is not possible, they will lie. But since they are always fed a question where it is possible to consistently answer, they will answer consistently, and give the answer we want. – boboquack Sep 26 '17 at 09:13
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ETA: Just saw this is DqwertyC's answer. Sooo... I counted the questions better?

This answer relies on the fact that the Joker never passes in a way that makes things easier; so it really depends on whether it's true the Joker never passes (the question makes it look like the Joker makes things hardest for Batman and therefore never passes but that looks wrong to me).

[Looks like he does: "It is an additional rule, Joker likes the sound of his own voice too much to pass"]

Step 1

First ask a question that none of the villains know the answer to, forcing them to pass. Such as: "Did the coin I've just flipped fall heads?". The first person that doesn't pass is the Joker. Say it's the Nth person.

This takes 1 to 4 questions.

Step 2

Next, ask the next three people: "Is the Nth person the Joker?" (or any question both you and they know that the answer to is "yes"). The Riddler and the Penguin will answer differently, while Two-face answers like either of them. So you'll get two "yes"s and one "no", or one "yes" and two "no"s. Either way, you know that the answer you didn't get two of is the Penguin (if it was "no") or the Riddler (if it was "yes"). Actually if you're lucky and you get two "yes"s or two "no"s right away you know who the third person is without asking them. Say the newly-identified person is in position M, and one of the other two is in position L.

This takes 2 to 3 questions.

Step 3 (only necessary if Step 2 took 3 questions)

At this point you're at the Joker again; ask a dummy question and ignore the answer ("Are you that bad at makeup or is it on purpose?").

This takes 1 question.

Step 4

Now you just need to get to the Riddler or Penguin you identified, and ask them "Is the Lth person Two-face"? They'll tell the truth or lie, and since you know which they're doing that allows you to tell the last two people apart. If your first two answers in Step 2 were identical then you just need to ask this of the fourth guy; otherwise you need to do the round again after having passed the Joker, thus getting to the second or third.

This takes 1 to 2 questions.

Total questions : 10 at most, 4 at least.

BONUS QUESTION:

This already doesn't repeat any question to a single villain. It can also be made so that no question is repeated at all:

make three questions they don't know the answer to ("Did the coin I just flipped fall tails?", "Did I just flip a coin?", "Do I have a coin on me?") and two answers that you all know the answer to is "yes" ("Is 1+1=2?", "Am I Batman?"), and different dummy questions ("Did you really think this stupid plan would work?").

Oosaka
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These types of questions assume that the villains are perfect logicians and either tell the truth completely, or lie completely. No partial truths allowed.

Therefore, a simple way to get them to answer consistently is:

To the question "Are you X?", is your answer "yes"?

If the villain is not speaking the truth, then if you were to plainly ask them if they are X, they would lie. But in this question, they would lie about lying, and therefore end up giving the same answer as someone who speaks the truth.

Given that, it's trivial to simply

find out from everyone if they are the Joker, then if they are the Riddler, etc.

The worst case number of questions for this is

12, since if you know three villains, you can deduce the fourth

The best case number of questions for this is

3, if you get lucky and guess right the first three times

hvd
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    I see how this allows you to get the identity of The Riddler and The Penguin, but both Two-Face and The Joker answer randomly. How would you extrapolate this logic to identify them? – oobug Sep 19 '17 at 06:01
  • @oobug They do not randomly decide to give a yes or no answer, they randomly decide to tell the truth or lie. Therefore, this should work just as well for them, if I'm understanding correctly. – hvd Sep 19 '17 at 06:11
  • My understanding is that they can randomly choose each time they are asked a question whether to tell the truth or lie. Two-Face and The Joker's answers will not be consistent. – oobug Sep 19 '17 at 07:38
  • @oobug I can sort of see that the wording I picked allowed it to be treated as two separate questions, so they could tell the truth for one and lie for the other. Re-worded so that it's no longer a hypothetical separate question. – hvd Sep 19 '17 at 07:55
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    The Joker could still answer "Yes" or "No" because he can still choose each time the question is asked to tell the truth or lie. But the specific wording of your question could be interpreted as always forcing Two-Face to pass, which is interesting. – oobug Sep 19 '17 at 08:03
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It can be done in 5 questions, depending on if Two-Face is first or not.

Round 1:

"Next round, will you say yes if I ask if you are the Joker or the Penguin?" Two-face immediately identifies himself by passing, since he doesn't know. The Riddler will truthfully answer no. The penguin will say yes. At this point, the Joker needs to decide whether he will tell the truth or lie next turn ahead of time. If his truthiness is the same next turn as the current turn, he will say yes, and the Riddler will be identified. Otherwise he will say no, and the Penguin will be identified.

Round 2:

This depends on the order now. If Two-Face is not first, and, if we have someone we've identified we just ask them if the next non-Two-Face is the Joker, and it's solved. Otherwise, ask "Are you the joker and did you tell the truth last time I asked you a question?" If you didn't identify the Riddler yet, the Riddler will always say no, and the Joker will always say yes. If you didn't identify the Penguin yet, the Penguin will say yes and the Joker will say no. If Two-Face is the first answerer, ask "If the coin flip were reversed, would you say [the next person I haven't identified] is the Joker?" Two-Face is guaranteed to lie in response, so you invert his answer and the problem is solved.

M.Herzkamp
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    The Joker cannot be forced to give a single answer. For one thing, nothing's keeping the Joker from changing his mind about lying or telling the truth from one round to the next. But it's also been updated in the rules that the Joker can answer yes/no regardless of truth value if he doesn't know an answer or can neither lie or tell the truth (which in this case can be the case with the first question) – Oosaka Sep 20 '17 at 11:24
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    "Next round, what will be you answer if I ask if you are the Joker or the Penguin." might be considered not to be a "yes/no" question... You are asking for their answer after all! Better ask "... would your answer to the question ... be 'yes'?" – M.Herzkamp Sep 20 '17 at 15:39
  • @M.Herzkamp Thanks, I cleaned that up. – Ethan Sep 20 '17 at 17:18
  • @RozennKeribin The clarification right now states that if the Joker can remain logically consistent, he will. In this situation, it is possible for the Joker to remain logically consistent with the behavior specified in the answer. The thing keeping him from changing his mind is that he is required to remain logically consistent. – Ethan Sep 20 '17 at 17:20
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    @Ethan: Nothing requires the Joker to decide in advance what he will answer on a later questions, meaning you can't guarantee that he will know the answer to the question "How will you answer next round", and if he doesn't know the answer he can say whatever. Also I don't see how changing one's mind is logically inconsistent. Truth or lies depend on your state of knowledge; if you make a statement you think is true, and later new information shows it to be false, it doesn't retroactively turn it into a lie. And changing your mind is new information that wasn't available before the change. – Oosaka Sep 21 '17 at 08:42
  • @RozennKeribin The Joker MUST remain logically consistent, and he CAN decide ahead of time what his truthiness will be. The ONLY way he can remain logically consistent in this case is by deciding ahead of time. If he adopts a strategy of changing his mind, his answer is no longer logically consistent, which as per the clarification he CAN NOT do. – Ethan Sep 21 '17 at 16:01
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    I didn't talk about a strategy of changing his mind; that would come down to him not deciding in advance what to answer. I talked about just... changing his mind from one round to the next. The kind you don't predict in advance because that's what "changing one's mind" means. Also, the rules don't say "if he CAN know the answer", they say "if he knows the answer", on the same priority level as "if he can be logically consistent". No suggestion that being logically consistent is a higher constraint that forces him to decide things just so he can know them when he wouldn't know them otherwise. – Oosaka Sep 21 '17 at 22:53
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The minimum questions is 8, but this is only possible with a specific order and behavior from Two-face.

Explanation:

The Joker will likely imitate the behavior of another villain in order to make it hard to distinguish between the two of them. In order for him to be the most problematic he will either mimic the behavior of the Riddler or Penguin. In order to find the joker we must first identify either Riddler or Penguin, so we begin by asking all 4 questions with known true of false answers.

After one round we will not have verified anyone's identity, so we ask again questions with known answers. In this second round we hope to get either two repeated lies and one truth, or two repeated truths and a lie, plus a behavior change from Two-face. (However if Two-face repeats his answer from chance, you must repeat this round of questioning until Two-face repeats.)

Assuming however that Two-Face, Joker, and the villain he is mimicking are the first three, we will then be able to identify the villain Joker is not mimicking after the 7th question. Facing two repeating answers and one non-repeating you know the non-repeating is Two-Face and the other two are the Joker and the villain he mimics, this gives us the identity of the fourth based on if he lies or tells the truth. You must then ask the predictable villain with the known identity if one of the two identical villains is the Joker, and then interpret his answer to deduce which is the Joker, and then which must be the last remaining unknown villain.

Alpha
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  • Why is 8 the minimum? – boboquack Sep 19 '17 at 07:54
  • Umm. . . See the explanation? If you have a more specific question about it then maybe I can explain. – Braydon Sep 19 '17 at 13:38
  • I see why it is the minimum in this answer, but why would a different set of questions not be able to reveal the identities sooner? – boboquack Sep 19 '17 at 23:02
  • @boboquack Well at the time of this answer it was not yet clarified that the Joker always answers, since it simply said he would answer in a way intended to mess you up. (This would seem to indicate he would not reveal himself by refusing to pass.) Without the "Joker never passes" rule I had to find him last, which makes it longer. – Braydon Sep 19 '17 at 23:20
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Best Case Scenario: 3 questions
Average: 4.5 questions
Worst Case Scenario: 6 questions
The Questions:

"How many fingers am I holding up" until you find the Joker "Would you say yes if I asked you next round if the penguin was ahead of the riddler?"
"Would you say yes if I asked you next round if the riddler was ahead of the penguin"

The Logic:

First you have to figure out who the Joker is, so ask some arbitrary question that the villains can't possibly know the answer to. The Joker is the only one who says he doesn't know. Worst case scenario is that the Joker is last so that's 4 questions.

The second part relies on the fact that a double positive and a double negative is a positive. By asking if they would say yes to an embedded question, it doesn't matter if they're lying or not. Consider the example "If I asked you if 1 plus 1 is two, would you say yes?" The Riddler would say yes to 1 plus 1 being two, so he says yes. The Penguin would say yes because he would say no to 1 plus being 2, so he lies and says yes. Now consider if I asked "If I asked you if 1 plus 1 is three, would you say yes?" The Riddler would say no to 1 plus 1 being 3, so he would honestly say no. The Penguin would say yes to 1 plus 1 being 3, so he lies and says no. Since we've proven we can ask questions without worries if they're lying, we just need to find the order of the three. Since Two-Face doesn't know what he's gonna say next round, he has to pass. So with 2 questions, you can find out if the penguin is ahead of the riddler, and where two-face is.

Qwepoiwo
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I can do it reliably in 6 Questions:

Ask the following question four times:

If I ask the next person if they are the Riddler, will they say 'Yes'?

If the Riddler answers the question and the next person is the Penquin, he will say 'Yes' otherwise he will 'Pass'.

If the Penquin answers the question and the next person is the Riddler, he will say 'No', otherwise he will 'Pass'

If TwoFace answers the question, he will 'Pass' if the next person is the Joker, otherwise he will answer 'Yes' or 'No'

If Joker answers the question, he will just answer 'Yes' or 'No'

This will tell us some thing about the order and might actually tell us what the order is:

If the Joker follows TwoFace, then we will have two Passes together, the first from either the Riddler or the Penguin, and the second from TwoFace.

The previous answer, before the first 'Pass' will tell us whether the Penguin is before or after the Riddler, e.g.

Yes,Pass,Pass,No - would be Riddler, Penguin, TwoFace, Joker
and ...
No,Pass,Pass,No - would be Penguin, Riddler, TwoFace, Joker

alternatively:

If TwoFace follows the Joker, then there will be only 1 'Pass' (the Penguin or the Riddler) which will identify the Joker, but TwoFace will be the next person, so:

Yes,Pass,No,Yes - would be Riddler, Penguin, Joker, TwoFace
Likewise,
No,Yes,Pass,No - would be TwoFace, Riddler, Penguin, Joker

So anything ordered similarly to the above will only need 4 questions.

Other orders are dealt with as below:

Otherwise, the Riddler and the Penguin will be alternating with TwoFace and Joker, and there will be two Passes, one from the Riddler, and one from the Penguin. So in this case it is simply a matter of telling the Joker from TwoFace, after the second 'Pass' answer is given - and then telling the Riddler from the Penguin

If you ask TwoFace/Joker, 'Will the Joker tell me the truth?', TwoFace must always 'Pass', the Joker must answer 'Yes' or 'No'

If you ask the Riddler/Penguin, 'Will the Penguin tell me the truth?', the Riddler will always answer 'No', the Penguin will always answer 'Yes'.

Bonus: No one will be asked the same question twice.

So, Batman can work the order out in a maximum of 6 questions.

Lee Leon
  • 910
  • 4
  • 9
0

0 Because it is possible to nullify some character's abilities to answer correctly, making logical deduction uncertain.

Ask each person:

"Are you going to pass? "There is no way for the Riddler to tell the truth, so there is no one correct answer that is always true. The puzzlemaker has to eliminate some question(s) beforehand. In other words, if you ask the Riddler, and he says yes, then he cannot pass because he already said yes. In fact, none of them can be relied on, since they can only answer yes or no.

johnny
  • 101
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0

Well, I would say 0. Bats got a phone number to dial in 1 hour. In that time, he finishes his Bat-coffee, suits up, enters the Bat wing and flies over Gotham, searches the number in the Gotham Communication Services Database and knows the exact address of the holder of said phone number. He will fly to that address, jump out of the wing and tadaaa... after some punches, there is a Batman in the middle of 4 knocked out villains. Should it be a mobile number, Bats will use his technology to trace back the belonging cell phone by the providers mobile network. He is Batman, he has access to everything.

Alfred
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