3
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Evelyn, Skylar, Alayna, Madelyn, Jonathon, Lara, Alyin, Felix, Iris, Bob and Roslina live on an island where some people always tell the truth, and the rest always lie.

  • Roslina says "Alyin and Evelyn lie"

  • Roslina says "Skylar would say 'Either Alayna lies, or At least one of Felix or Lara is truthful'"

  • Jonathon says "Evelyn and Skylar lie, and also Alayna is truthful"

  • Roslina says "Either at least one of Iris or Lara lies, or Bob is truthful"

  • Felix says "At least one of Bob or Alayna lies"

  • Iris says "Alayna and Felix lie, and also Evelyn is truthful"

  • Felix says "Jonathon is truthful"

  • Madelyn says "Jonathon, Bob and Roslina lie"

Can you figure out who are the liars and who are the truth-tellers?

[Adapted from this crazy generator I found.]

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  • $\begingroup$ Can you clarify what is meant by "always lie" in this context? Does it mean that every clause is a lie or just that the statement as a whole is? In other words does "True is True and False is True" count as "sometimes telling the truth" since the first clause is true or "always lying" since the statement as a whole is false? $\endgroup$ – Barker Apr 17 at 18:44
  • $\begingroup$ @Barker The complete statement made by a liar would be false. $\endgroup$ – Flog Edoc Apr 19 at 0:09
5
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Let's call everyone by their initial, except that since there are two As we'll use N for Alyin and A for Alayna. Then the answer is

A,B,L,R,S tell the truth; E,F,I,J,M,N lie

Derivation:

Begin

R = ~N & ~E
R = (S = (~A | (F|L)))
J = ~E & ~S & A
R = ~I | ~L | B
F = ~A | ~B
I = ~A & ~F & E
F = J
M = ~B & ~J & ~R
Eliminate J (=F)
R = ~N & ~E
R = (S = (~A | (F|L)))
F = ~E & ~S & A
R = ~I | ~L | B
F = ~A | ~B
I = ~A & ~F & E
M = ~B & ~F & ~R
Eliminate F (=~A|~B)
R = ~N & ~E
R = (S = (~A|~B|L))
~A|~B = ~E & ~S & A
R = ~I | ~L | B
I = ~A & A & B & E
M = ~B & A & B & ~R
Eliminate I (=false) and M (=false, not actually used elsewhere)
R = ~N & ~E
R = (S = (~A|~B|L))
~A|~B = ~E & ~S & A
R = true | ~L | B
Eliminate R (=true)
~N & ~E
S = (~A|~B|L)
~A|~B = ~E & ~S & A
Eliminate N (=false) and E (=false)
S = (~A|~B|L)
~A|~B = ~S & A
Note that if ~A then LHS of second of those is true, hence RHS is, hence A, contradiction
Eliminate A (=true)
S = (~B|L)
~B = ~S
Eliminate B (=S)
S = (~S|L)
Note that if ~S then RHS is true so LHS is so S is, contradiction
Eliminate S (=true)
false|L
Conclude that L is true

If you want to check my work, put the following into a Python interpreter and verify that you get a bunch of Trues out (I do):

a,b,l,r,s = True,True,True,True,True; e,f,i,j,m,n = False,False,False,False,False,False
r == ((not n) and (not e))
r == (s == ((not a) or f or l))
j == ((not e) and (not s) and a)
r == ((not i) or (not l) or b)
f == ((not b) or (not a))
i == ((not a) and (not f) and e)
f == j
m == ((not j) and (not b) and (not r))

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  • $\begingroup$ Well that would make sense, considering it's a story about Liars ;) $\endgroup$ – Mr Pie Apr 16 at 20:56
  • $\begingroup$ Well, maybe it would but in fact on checking it appears to be very wrong so I goofed somewhere. $\endgroup$ – Gareth McCaughan Apr 16 at 20:59
  • $\begingroup$ oh, wait, I think maybe I just mistranscribed something $\endgroup$ – Gareth McCaughan Apr 16 at 21:00
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    $\begingroup$ Bleh. Being a moderator is no guarantee of anything. You could trust it because I'm a mathematician and that would make more sense but still be crazy. I make a lot of mistakes. $\endgroup$ – Gareth McCaughan Apr 16 at 21:03
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    $\begingroup$ Welp, there is a reason why you are a moderator; and there might be users who look up to you as one, too :) ....but let's not get off-topic here, because as they say, comments aren't for discussions. Nice answer, I will upvote once my voting limit is over :P $\endgroup$ – Mr Pie Apr 16 at 21:06
0
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Typed this up on my phone, hoping I didn't end up being too slow haha.

My answer is:

Truth tellers are Rosalina, Bob, Alayna, Skylar, & Lara
Liars are Alyin, Evelyn, Iris, Madelyn, Jonathon, & Felix

My reasoning is below. Names are shortened to first letters, except Alyin is AN and Alayna is AA. Truth tellers are labeled with a T and liars with an X.

Assume R is a T. Then Clue #1 states that AN and E are both X.
Thus Clue #6 is untrue, given E is not a T. Thus I is an X.
Thus Clue #8 is untrue, given R is not an X. Thus M is an X.
Consider Clue #2. R is a T, thus if S is an X it would require AA to be a T, and for F and L to both be X. But then Clue #7 would require J to be an X. However, this would contradict Clue #3 which would be true in this scenario. Thus we have a contradiction, and S must be a T.
Thus Clue #3 is untrue, given S is a T. Thus J is an X.
Thus Clue #7 is untrue, given J is an X. Thus F is an X.
Thus Clue #5 is untrue, given F is an X. Thus B and AA are T.
Finally, given S is a T, one element of Clue #2 must be true. To avoid contradictions, the sole option is that L is a T.

The above satisfies all 8 clues. Also, if we took our initial assumption to be

R is an X then upon inspection Clue #4 would make B an X & I and L both T, and then Clue #6 would make E a T & AA and F both X. However Clue #5 would be a true statement, thus F cannot be an X, and so we have a contradiction.

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-1
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If we assume Felix = T then Jonathon = T
Evelyn = L
Skylar = L
Alayna = T
Iris = L
Madelyn = L
Bob = T
Roslina = T and if Bob = T then Iris and Lara both are T, but Iris = L
Therefore Felix is a LIAR.

Therefore:

Felix = Liar
Jonathon = Liar
Evelyn = Truth
Skylar = Truth
Alayna = Liar
Iris = Truth (claims Alayna is a liar)
Madelyn = Truth (claims Jonathan is a liar)
Bob = Liar
Roslina = Liar
Lara = Liar
Which leaves Alyin, whom I missed before, and = Truth

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  • $\begingroup$ Updated to include a person I missed before $\endgroup$ – mkinson Apr 16 at 22:13
  • $\begingroup$ Undeleted this post as it at least gives an answer for each person based on "logic". I do, however, believe that logic based on the parameters creates an impossibility. $\endgroup$ – mkinson Apr 17 at 18:25
-1
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Okay, I found an issue in my original post and instead of changing everything I'm reposting..

First, some logic:

Let's assume Felix is Truthful (T)

If Felix = T then
Jonathon = T which means
Evelyn = L and
Skylar = L and
Alayna = T, and since Iris says Alayna = L
Iris = L, which makes..
Evelyn = L, and since Madelyn claims Jonathon is L..
Madelyn = L, which makes..
Bob = T and

BUT...

Felix says at least one of Bob or Alayna is L, but in the above they are both T proving Felix a LIAR!!

This is the same result I keep getting time and again.. but this time:

Felix = L
Jonathon = L
Evelyn = T
Skylar = T
Alayna = L
Iris = T
Evelyn = T
Madelyn = T
Bob = L ... But wait!! Felix says at least one of Bob or Alayna are Liars, which is now TRUE.. but he's a proven Liar! Something is up..

What if this riddle isn't about what it says it's about?


Bob (B), Lara (L), Alayna (A), Madelyn (M), Evelyn (E), and Skyler (S).. Blames? Has kind of a negative connotation to it. Could be these be the Liars?
Felix (F), Alyin (A), Iris (I), Roslina (R). Wait... Fair? Sounds positive..
Could these be the truthful ones?


I do realize that Alayna and Alyin can be interchanged.. but the point of the puzzle remains the same that their names spell out Blames and Fair.

It's out of the box thinking, but it seems to me that the logic above proves that no combination can fit 100%.

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  • $\begingroup$ When you start by supposing F is a liar, how do you infer that E tells the truth? My guess: you reckon that if a liar says "X and Y and Z" you're allowed to infer that X, Y, and Z are all false, but I don't think that's what's intended: all you know is that at least one of X,Y,Z is false. $\endgroup$ – Gareth McCaughan Apr 17 at 15:28
  • $\begingroup$ Because the beginning states: some people always tell the truth, and the rest always lie. So if A is a liar and they say B is also a liar, B has to be truthful. $\endgroup$ – mkinson Apr 17 at 15:32
  • $\begingroup$ But if they say B and C are both also liars, then it suffices for either B or C to be truthful, because either way A will have lied. $\endgroup$ – Gareth McCaughan Apr 17 at 16:47
  • $\begingroup$ (The usual convention in these puzzles is that each complete statement a "liar" makes is false.) $\endgroup$ – Gareth McCaughan Apr 17 at 16:47
  • $\begingroup$ @GarethMcCaughan "Roslina says "Alyin and Evelyn lie" & Roslina says "Either at least one of Iris or Lara lies, or Bob is truthful" I take these as two separate statements, both of which would need to be reversed due to Roslina being a Liar. Are you suggesting only one of the two statements needs to be false, or are you saying that only one side of the "Either at least one of Iris or Lara lies, or Bob is truthful" needs to be false? $\endgroup$ – mkinson Apr 19 at 11:28
-1
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So let’s start with Alayna, she’s a

Lie, this is because, according to the rule given in the link in the question, states that “If A says “A would say X” you know that X is true whether or not A is a liar” as Skylar say she’s a lie,

We can say Skylar & Evelyn

are a truth-tellers

Roslina & Jonathon

are liars, Jonathon mention Evelyn & Skylar are a lie, Roslina mention a lie too with saying Evelyn is a lie

Iris

is a truth-teller, 2/3 person that she mentioned which is Alayna & Evelyn are correct, so we pick

Felix

as a lie, we may use Felix’s statement on Jonathon for this too

Madelyn

also saying the truth by 2/3 person mentioned by her makes

Bob

a lie

Iris & Lara

truth-tellers by using Skylar only statement we can say Roslina lie about her statement on Iris & Lara

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  • $\begingroup$ Your first statement is incorrect. The important fact about the "A says 'A would say X'" is that the person making the statement is the same person is being talked about. In other words the statement is about something the person talking would say. In this case Roslina is talking about what Skylar would say so the statement is "A says 'B would say X'". For the rule to apply it would need to say "Roslina says 'Roslina would say X'". $\endgroup$ – Barker Apr 17 at 19:57
  • $\begingroup$ No it’s like Alayna says “Alyin would say ’Felix is...’” $\endgroup$ – Casablanca Kookie Apr 17 at 23:15
  • $\begingroup$ Look at the explanation on the site again. The logic only works because the speaker and the person who "would say" something are the same person. That is why it says "A says 'A would say X'", A refers to the same person both times. That is why the site says it is rare to see a clue like that. In your example Alayna and Alyin are two different people so they would be A and B with the 'Felix is...' being X. $\endgroup$ – Barker Apr 17 at 23:26

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