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Here is an easy and straightforward liar puzzle.

Four people are in a room, stack reader, boboquack, Rand al'Thor and Brent Hackers.
stack reader states that at least 2 of the other 3 are liars.
boboquack states that Rand al'Thor and Brent Hackers are the only 2 liars or are both truth tellers.
Rand al'Thor states that stack reader is the only liar.
Brent Hackers states that Rand al'Thor is the only liar or the only truth teller.

There is more than 1 liar!
Can you find who are the liars?
I can assure you that no one will find the answer!

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    $\begingroup$ Of course IF stack reader is one of the liars, how can we trust that the above account of who accused who is faithful? :) $\endgroup$ – anything Dec 14 '17 at 6:53
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    $\begingroup$ I am definitely not a liar... o.o $\endgroup$ – Brent Hackers Dec 14 '17 at 7:47
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    $\begingroup$ @BrentHackers how are we too know "I am not a liar" is something both a liar and a truth teller would say :P $\endgroup$ – stack reader Dec 14 '17 at 7:50
  • $\begingroup$ @stackreader and what would happen if I said "I am a liar"? I guess I can't be either. So Wen1now has this one? :) $\endgroup$ – Brent Hackers Dec 14 '17 at 8:27
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    $\begingroup$ Does a liar always lie? $\endgroup$ – einpoklum Dec 14 '17 at 17:14

10 Answers 10

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I think the answer is:

Stack Reader is the only liar

I mean,

Here is an easy and straightforward liar puzzle is a lie - look at how many differing opinions there are

Four people are in a room, stackreader, boboquack, Rand al'Thor and Brent Hackers. is a lie

stackreader states that at least 2 of the other 3 are liars. is a lie (he says at least one of the others is lying, below)

boboquack states that Rand al'Thor and Brent Hackers are the only 2 liars or are both truth tellers. <- lie, bbq never said that. Same for "Rand al'Thor states that stackreader is the only liar." and "Brent Hackers states that Rand al'Thor is the only liar or the only truth teller."

There is more than 1 liar! Lie - stackreader is the only liar

I can assure you that no one will find the answer! lie, somebody will probably find the answer

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    $\begingroup$ This belongs on the skeptic page, not here. $\endgroup$ – insidesin Dec 14 '17 at 7:31
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    $\begingroup$ @insidesin I can't tell if you're being serious... $\endgroup$ – Wen1now Dec 14 '17 at 7:32
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    $\begingroup$ @stackreader I've been trained by the other SE sites to ignore who asks the question and focus on the question itself. Little did I know the "who" was important this time :) $\endgroup$ – Brian J Dec 14 '17 at 15:31
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    $\begingroup$ The problem with this answer being correct is that "stack reader" posted the question and "stackreader" was in the room. "Jonny" and "Jon Ny" are not the same person! A bit pedantic perhaps, but when playing with logic puzzles like this precision is pretty essential. $\endgroup$ – Jayfang Dec 14 '17 at 17:36
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    $\begingroup$ @BrianJ Normally that would be ok, but that is why the "lateral thinking" tag is important :P $\endgroup$ – stack reader Dec 14 '17 at 23:46
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The liars are :

Rand al'Thor and Brent Hackers.

From statement "There is more than 1 liar!", show that Rand al'Thor is a Liar.
This also imply that Brent Hackers is a Liar.

boboquack can be a liar also can be a Truth teller.
boboquack Statement is True for (Rand al'Thor and Brent Hackers are the only 2 liars)
boboquack Statement is False for (Rand al'Thor and Brent Hackers are not the only 2 liars)(Thanks for Sid for your comment)

stackreader is also tell the truth. (At least there are 2 liars)

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  • $\begingroup$ Why can't boboquack be lying? If they were lying, the statement is, "Rand and Brent are NOT the only liars or truth tellers which should fit with the criteria as well $\endgroup$ – Sid Dec 14 '17 at 6:55
  • $\begingroup$ @Sid yes, you are right. $\endgroup$ – Jamal Senjaya Dec 14 '17 at 6:56
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If "There is more than 1 liar!"

Rand al'Thor simply can't be a truth teller because it contradicts this, which makes me a liar (Hmph) because if Rand al'Thor can't be the truth teller that only leaves the "Rand al'Thor is the only liar" part of my statement.

stackreader states that at least 2 of the other 3 are liars.

But this adds nothing beyond "There is more than 1 liar!" and must be true.

boboquack states that Rand al'Thor and Brent Hackers are the only 2 liars or are both truth tellers.

This (" Rand al'Thor and Brent Hackers are the only 2 liars") allows boboquack's statement to be true, so stackreader and boboquack are truth tellers, and Rand al'Thor and Brent Hackers are liars. (hence my other answer)

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    $\begingroup$ "...and Brent Hackers are liars" ACTIVATING PARADOX PREVENTION SYSTEM... $\endgroup$ – Wen1now Dec 14 '17 at 8:15
  • $\begingroup$ I'm secretly >! a joker! $\endgroup$ – Brent Hackers Dec 14 '17 at 8:16
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    $\begingroup$ How can we trust your answer if you openly declare yourself a liar!? :P $\endgroup$ – stack reader Dec 14 '17 at 8:21
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I am mostly asking this because I don't see how this is not a self consistent and therefore plausible explanation:


stackreader - tells the truth - denote them as S

boboquack - impossible to tell from the information given but both options are self consistent - denote them as Q

Rand al'Thor - is a liar - denote them as R

Brent Hackers - is a liar - denote them as H

If someone (say boboquackers) is a truthteller then Q is true. If they are a liar, then Q is false and !Q is true.

Statements in an informal notation

& means and

| means or

! means not

t means true

f means false

  1. $(!Q \& !R)|(!Q \& !H)|(!H \& !R)$ as stated by S
  2. $(!R \& !H \& Q \& S) | (R \& H)$ as stated by Q
  3. $!S \& Q \& R \& H$ as stated by R
  4. $(!R\& S \& Q \& H)|(R\& !S \& !Q \& !H)$ as stated by H
  5. There is more than 1 liar as stated by S
  6. "Can you find who are the liars?" which is a question so we can ignore it
  7. "no one will find the answer" as stated by S

Solution assuming S

  • 1 implies 5 and both are true as we are assuming S.
  • 3 requires !S so is false. As R said 3, then R is false.
  • 4 has two options. The $(!R\& S \& Q \& H)$ contradicts 1 and 5 so is false. The second $(R\& !S \& !Q \& !H)$ is false for several reasons... but here we will just point out that $R$ is false. As both are false, 4 is false and $H$ is a liar (is false) as he said it.
  • As $!R \& !H$, statements 1 and 5 are satified whether Q is true or false.
  • 2 can be simplified to Q implies Q as follows:

$$(!R \& !H \& Q \& S) | (R \& H) \rightarrow Q$$ $$(!f \& !f \& Q \& t) | (f \& f) \rightarrow Q$$ $$(t \& t \& Q \& t) | (f) \rightarrow Q$$ $$(Q \& t) | (f) \rightarrow Q$$ $$Q | f \rightarrow Q$$ $$Q \rightarrow Q$$

  • Basically Q's only statement narrows down to "I am a truth teller" and no one else's statements rely on Q's statement being true of false.
  • We know R (Rand al'Thor) and B (Brent Hackers) are liars and S (stackreader) is a truthteller. We cannot know whether Q (boboquack) is a liar or a truthteller.
  • As there are two indistinguishable self consistent solutions to statements 1-5, we cannot find the full answer so 7 is true (which was required for S to be true).

As all statements 1-7 are true (or a question), $S$'s story is self consistent. As this seemed easy for me at first, statement 0 (Here is an easy and straightforward liar puzzle.) seems true too. While there may be a solutions where S is a liar (as stated by Wen1now), we can find this self consistent solution when we assume S is a truthteller.

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  • $\begingroup$ If this is wrong, could someone please tell me why? I like the core idea for this puzzle with Wen1now's answer.... but hate the fact that stack reader's story is self consistent even including all of his statements. $\endgroup$ – kaine Dec 14 '17 at 15:35
  • $\begingroup$ +1 For usefulness. I'd like so see the logic discounting this neatly spelled out... $\endgroup$ – Brent Hackers Dec 15 '17 at 8:19
  • $\begingroup$ @BrentHackers does this help? I don't know the formal notation so I applogize if I got any wrong. Again, I want this to be wrong as I like the answer where OP is a liar better but OP's story is self consistent. $\endgroup$ – kaine Dec 15 '17 at 19:14
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A meta-answer:

I can assure you that no one will find the answer! is a lie, therefore all the statements in this puzzle are lies. Therefore "stack reader did NOT state that..." etc. Therefore, no valid answer can be found, because there is no puzzle.

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Three liars: Rand al’Thor, Brent Hackers and boboquack.

Rand al'Thor states that stackreader is the only liar:

We know this is wrong so he’s lying.

Thus: Brent Hackers states that Rand al'Thor is the only liar or the only truth teller.

Is also a lie.

Stackreader states that at least 2 of the other 3 are liars:

True by the above.

Thus boboquack states that Rand al'Thor and Brent Hackers are the only 2 liars or are both truth tellers.

Is a lie. Edit: This last line doesn't follow sorry.

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So we've got: Stackreader (P1), Boboquack (P2), Rand al'Thor (P3) and Brent Hackers (P4) and they state as follows:

  • P1.1: [P2,P3,P4] contains at least 2 F(alse)
  • P2.1: [P3,P4] are the only F -> [P1,P2] are thus T(rue)
  • P2.2: [P3,P4] are both T
  • P3.1: [P1] is the only F -> [P1,P2,P4] are thus T
  • P4.1: [P3] is the only F -> [P1,P2,P4] are thus F
  • P4.2: [P3] is the only T -> paradoxal statement

Let's assume that statement P4.1 is T, thus [P1,P2,P4] are T, let's check:

  • P1 = T -> Should be at least two F, thus contradicts P4.1
  • P2 = T -> P4 is not F, so P3 should be T too, thus contradicts P4.1

Let's assume that statement P3.1 is T, thus [P2,P3,P4] are T, let's check:

  • P2 = T -> P3 & P4 are T, thus consistent
  • P4 = T -> either a paradox or a contradiction

Let's assume that statement P2.2 is T, thus [P3.P4] are T, let's check:

  • P3 = T -> P1 is F, so far consistent
  • P4 = T -> either a paradox or contradiction

Let's assume that statement P2.1 is T, thus [P1,P2] are T and [P3,P4] are F:

  • P1 = T -> [P3,P4] are F, so at least 2 are F, thus consistent

Taking all that together means that P1 and P2 are T, whilst P3 and P4 are F.

In other words:

Stackreader & Boboquack are telling the truth.

Rand al'Thor & Brent Hackers are lieing.

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    $\begingroup$ How do you know Boboquack isn't a liar? $\endgroup$ – kaine Dec 14 '17 at 18:25
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Four people are in a room, stack reader, boboquack, Rand al'Thor and Brent Hackers.

This is a lie, so stackreader and boboquack are liars, and Rand al'Thor and Brent Hackers are truth tellers

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My answer:

Rand al'Thor and Brent Hackers are both liars, and boboquack may or may not be a liar. It is impossible to tell whether boboquack is a liar or not, hence why you can assure us no one will get it. Though I suppose people could just guess, and eventually someone would guess right. Though maybe I'm reading too much into that line.

My reasoning:

We know there is more than one liar, so Rand al'Thor must be lying. Again, since there is more than one liar, we know Rand al'Thor cannot be the only liar, and since we know he is lying, he can't be a truth-teller, let alone the only truth teller. So Brent Hackers is lying. stackreader must be telling the truth, since we have have determined that two of the others are liars. We know that Brent Hackers and Rand al'Thor are both liars, so they are not both truth-tellers. So, in order for boboquack to be telling the truth, they must be the only liars, which would be consistent if boboquack is telling the truth. However, if boboquack is a liar, then Rand and Brent are not the only liars, and boboquack is lying, which is consistent. So it is impossible to tell whether boboquack is lying or not, since both possibilities are consistent with the information given to us.

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0
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Stack reader(A), Boboquack(B), Rand al'Thor(C) and Brent Hackers(D)

  • Here Stack reader and Boboquack are the only tuth teller.
  • And Rand al'Thor and Brent Hackers are the liars.

Brief Explanation:

  1. A will be always true.
  2. A cannot be false.
    • If A's statement is false then either B, C and D all are false or one of them is false. And from here you will get the contradiction.
    • If A is false then C is telling truth, according to D, C is the only liar or only truth teller. According to B, C & D are both liars or both truth tellers, so B is liar.
    • So, from here it is deductible that here two are liars, so A is True.
  3. D is always liar. If you assume D is true then all statements will contradict.
  4. From A it is easily deductible that C is a liar.
  5. For B it comes at last that it will be truth teller.
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