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There are two types of logicians: knights and knaves. Knights will always give a statement that is true whereas knaves will always give a statement that is false.

After listening in on a conversation between 10 logicians, you gather the following information:

  • Logician 1 claims that (Logician 10 is a knight) and (Logician 1 is a knight)
  • Logician 2 claims that (Logician 9 is a knave) or (Logician 2 is a knave)
  • Logician 3 claims that (Logician 8 is a knight) and (Logician 3 is a knight)
  • Logician 4 claims that (Logician 7 is a knave) or (Logician 4 is a knave)
  • Logician 5 claims that (Logician 6 is a knight) and (Logician 5 is a knight)
  • Logician 6 claims that (Logician 5 is a knave) or (Logician 6 is a knave)
  • Logician 7 claims that (Logician 4 is a knight) and (Logician 7 is a knight)
  • Logician 8 claims that (Logician 3 is a knave) or (Logician 8 is a knave)
  • Logician 9 claims that (logician 2 is a knight) and (Logician 9 is a knight)
  • Logician 10 claims that (logician 1 is a knave) or (Logician 10 is a knave)

Just by using the information gathered, is it possible to determine who the truthtellers and liars are? Explain why or why not.

Note: Do not confuse 'or' with 'xor'.

Bonus: What would the answer become if all the or's became and's?

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    $\begingroup$ I'll have to give the check to @d'alar'cop since he came up with the correct explanation first (which I have to add was very well written). $\endgroup$
    – Allan
    Commented Apr 17, 2015 at 4:41
  • $\begingroup$ Are the parts in brackets considered separate statements or is the whole claim a single statement? For example the claim he is a knight and I am a knight must that be completely true or false, or can it be part truth part falsehood? $\endgroup$
    – Bob
    Commented Apr 17, 2015 at 5:01
  • $\begingroup$ The entire claim is one statement. Each statement is a combination of two statements using a bitwise and/or. $\endgroup$
    – Allan
    Commented Apr 17, 2015 at 5:02
  • $\begingroup$ So you mean the latter ie that a single negative renders the whole negative regardless of a positive?! $\endgroup$
    – Bob
    Commented Apr 17, 2015 at 5:12
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    $\begingroup$ Like a bad apple that ruins the barrel. $\endgroup$
    – Bob
    Commented Apr 17, 2015 at 5:26

5 Answers 5

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No Logician could possibly claim that they are Knave via the rules. Thus half of these statements are impossible to be made. However, I will just assume that an impossible statement is FALSE.

I notice that there is a natural grouping of the Logicians and the Logicians they mention.

Using this information, we have:

  • Logician 1 claims that (Logician 10 is a knight) and (Logician 1 is a knight) $\land$ Logician 10 claims that (logician 1 is a knave)

  • Logician 2 claims that (Logician 9 is a knave) $\land$ Logician 9 claims that (logician 2 is a knight) and (Logician 9 is a knight)

  • Logician 3 claims that (Logician 8 is a knight) and (Logician 3 is a knight) $\land$ Logician 8 claims that (Logician 3 is a knave)

  • Logician 4 claims that (Logician 7 is a knave) $\land$ Logician 7 claims that (Logician 4 is a knight) and (Logician 7 is a knight)

  • Logician 5 claims that (Logician 6 is a knight) and (Logician 5 is a knight) $\land$ Logician 6 claims that (Logician 5 is a knave)

Clearly this is a case of 5 instances of the same pattern, when cast into formal logic formulae:
$LX = (LY \land LX)$
$LY = \lnot LX$

Suppose that $LY$ is a Knight. This means that $LX$ is a Knave. Therefore $(LY \land LX) = (TRUE \land FALSE) = FALSE$ therefore $LX=false$ i.e. $LX$ is a knave - this is consistent. Suppose that $LY$ is a Knave. This means that $LX$ is a Knight and would need to be (correctly) claiming that $LY$ is a Knight - this is a contradiction (inconsistent).

This implies that:
Logicians: 2,4,6,8,10 are KNIGHTS.
Logicians: 1,3,5,7,9 are KNAVES.

For the bonus, the 5 instances of the same pattern, when cast into formal logic formulae are now:
$LX = (LY \land LX)$
$LY = (\lnot LX \land \lnot LY)$

Clearly $LY$ cannot be a Knight - otherwise he'd be correctly saying that his own statement is false. So we deduce that $LY$ is a Knave.

$LX$ is trying to claim that $LY$ is a Knight - but we just deduced that he is a Knave. Thus $LX$ is also a Knave. But if $LY$ is a Knave then it implies that one of $LX$, $LY$ must be a Knight (in order to make his statement false). This is a paradox. Since $LY$ is a Knave, and $LX$ cannot be both Knight and Knave. It is not possible to determine anything other than the fact that these Logicians are in fact $TROLLS$.

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  • $\begingroup$ You are very close for the bonus. What would LX and LY both being a knave imply about LY (look back to your second statement)? $\endgroup$
    – Allan
    Commented Apr 17, 2015 at 4:07
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    $\begingroup$ Hahaha, trolls indeed :) $\endgroup$
    – Allan
    Commented Apr 17, 2015 at 4:21
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    $\begingroup$ That what all those little pyramids were about! :) If it was written !(A&&B)=!A||!B I might have got it. $\endgroup$
    – Bob
    Commented Apr 17, 2015 at 5:22
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    $\begingroup$ Haha, yes. It's nice to use the symbols as a short-hand (especially when you're writing it out). $\endgroup$
    – Allan
    Commented Apr 17, 2015 at 5:24
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    $\begingroup$ Logician 2 claims that (Logician 9 is a knave) or (Logician 2 is a knave) - this is not an impossible statement. Logician 2 claims that Logician 2 is a knave would be impossible, as would Logician 2 claims that (Logician 9 is a knave) and (Logician 2 is a knave) $\endgroup$
    – user20574
    Commented Apr 17, 2015 at 11:20
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For the first question...

It is possible. If 1 were a knight, then his statement would imply 10 was a knight. But then 10's true statement would mean 1 was a knave, a contradiction. Thus, 1 is a knave. This means 10's statement is true, so 10 is a knight. By identical reasoning, odd numbered logicians are liars and evens tell the truth.

For the bonus...

The conversation could not have happened. 10 can't be a knight since he is calling himself a knave. Thus, 10's statement implies NOT[(1 is a knave) AND (10 is a knave)]=(1 is a knight) OR (10 is a knight). The latter condition doesn't hold, so we now know 1 is a knight. However, 1 is calling 10 a knight, which is false, so we know 1 is not a knight. We have deduced a contradiction from this conversation, so someone broke character.

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  • $\begingroup$ You're very close on the bonus. What does 1 being a knight imply about 10? $\endgroup$
    – Allan
    Commented Apr 17, 2015 at 3:58
  • $\begingroup$ I'll refer back to this statement: Logician 1 claims that (Logician 10 is a knight) and (Logician 1 is a knight). $\endgroup$
    – Allan
    Commented Apr 17, 2015 at 4:11
  • $\begingroup$ Ahh, I was confused, I somehow read the bonus as "all the or's became and all the and's became or's" I shall fix. $\endgroup$
    – Mike Earnest
    Commented Apr 17, 2015 at 4:14
  • $\begingroup$ Oh ok. You would be right then if that was the case. $\endgroup$
    – Allan
    Commented Apr 17, 2015 at 4:15
  • $\begingroup$ Correct! Nicely done! $\endgroup$
    – Allan
    Commented Apr 17, 2015 at 4:30
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The odd numbered logicians are all Knaves and the even are all Knights.

$(L_1\iff(L_{10}\land L_1))\land (L_2\iff(\lnot L_9\lor\lnot L_2))\land (L_3\iff(L_8\land L_3))\land (L_4\iff(\lnot L_7\lor\lnot L_4))\land (L_5\iff(L_6\land L_5))\land (L_6\iff(\lnot L_5\lor\lnot L_6))\land (L_7\iff(L_4\land L_7))\land (L_8\iff(\lnot L_3\lor\lnot L_8))\land (L_9\iff(L_2\land L_9))\land (L_{10}\iff(\lnot L_1\lor\lnot L_{10}))$

In the second case, if all the or's became and's the list of claim's would be contradictory, similar to the liar's paradox.

If L1 is a Knight then L10 must also be a Knight because of claim 1. But they can't both be Knights because if L10 is a Knight they must be a both be Knaves by claim 10 (remember, or->and).

If L1 is a Knave then maybe L10 is also a Knave...but wait, L10 can't be a Knave, because he would be telling the truth, so he must be a Knight...no that doesn't work either because, paradoxically, he would be lying about being a Knave.

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  • $\begingroup$ Well done! You got both the question and the bonus correct. Could you explain how you got the first statement (not the bonus)? $\endgroup$
    – Allan
    Commented Apr 17, 2015 at 4:20
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    $\begingroup$ I cheated a little and used a truth table >_> $\endgroup$
    – Kyle G
    Commented Apr 17, 2015 at 4:21
  • $\begingroup$ Fair enough. Could you maybe post a picture of it (if you still have it) so that I can look it over? $\endgroup$
    – Allan
    Commented Apr 17, 2015 at 4:23
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    $\begingroup$ Copy the text here into the link above. I don't see an easy way to link the table directly unfortunately. $\endgroup$
    – Kyle G
    Commented Apr 17, 2015 at 4:27
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There's no need for 10 logicians in this puzzle. There's 5 identical pairs of logicians, so you can get the answer for all 10 of them just by considering two of them. In each pair, A claims that (B is a knight) and (A is a knight), and B claims that (A is a knave) or (B is a knave).

The key insight is that

Knaves can never say "X, or I am a knave", and knights can only say it if X is true.

So the answer is that

the odd-numbered logicians (corresponding to A) are knaves and the even-numbered logicians (corresponding to B) are knights.

Bonus:

Knights can never say "X, and I am a knave", and knaves can only say it if X is false.

So it seems like

B would be a knave and A would be a knight.

But

A's statement would then be impossible to make, since knights can't claim that knaves are fellow knights. So the bonus setup would be impossible/paradoxical.

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If we represent each logician as a boolean variable, True for knight and False for knave, then a logician L making a claim C is equivalent to (L == C). So the first statement is

$L1 = (L10 \land L1)$

The tenth statement is

$L10 = ( (\neg L1) \lor (\neg L10) )$

By substitution,

$$L10 = ( (\neg (L10 \land L1)) \lor (\neg L10) )$$ Distribute the negation: $$L10 = ( (\neg L10) \lor (\neg L1)) \lor (\neg L10) )$$ $\lor$ is associative and communitive, so $$L10 = ( (\neg L10) \lor (\neg L10)) \lor (\neg L1) )$$ $$L10 = ( \neg L10)) \lor (\neg L1) )$$ $L10$ can't be equal to its own negation, so $L10 = \neg L1$
Substituting this back into the first statement, we get $L1 = (\neg L1 \land L1)$, so clearly $L1 = False$, and $L10 = True$, i.e. L1 is a knave and L10 is a knight.

If we replace all the ors with ands, we get

$$L10 = ( (\neg (L10 \land L1)) \land (\neg L10) )$$ $$L10 = ( (\neg L10) \lor (\neg L1)) \land (\neg L10) )$$

But this reduces to $L10 =\neg L10$, so this is impossible.

xor is the same as $\neq$, so if we replace all the ors with xors, we get

$$L10 = ( (\neg (L10 \land L1)) \neq (\neg L10) )$$ $$L10 = ( (\neg L10) \lor (\neg L1)) \neq (\neg L10) )$$

The only way that or-ing a boolean variable with another variable can change that variable is if you started with False and or-ed it with True. So $\neg L10$ is False and $\neg L1$ is True, so $L10$ is true and $L1$ is False, i.e. L10 is a knight and a L1 is a knave.

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