# The 10 Logicians

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?

• 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). – Allan Apr 17 '15 at 4:41
• 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? – Bob Apr 17 '15 at 5:01
• The entire claim is one statement. Each statement is a combination of two statements using a bitwise and/or. – Allan Apr 17 '15 at 5:02
• So you mean the latter ie that a single negative renders the whole negative regardless of a positive?! – Bob Apr 17 '15 at 5:12
• Like a bad apple that ruins the barrel. – Bob Apr 17 '15 at 5:26

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 $TROLLS$.

• 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)? – Allan Apr 17 '15 at 4:07
• Hahaha, trolls indeed :) – Allan Apr 17 '15 at 4:21
• That what all those little pyramids were about! :) If it was written !(A&&B)=!A||!B I might have got it. – Bob Apr 17 '15 at 5:22
• Haha, yes. It's nice to use the symbols as a short-hand (especially when you're writing it out). – Allan Apr 17 '15 at 5:24
• 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) – user253751 Apr 17 '15 at 11:20

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.

• You're very close on the bonus. What does 1 being a knight imply about 10? – Allan Apr 17 '15 at 3:58
• I'll refer back to this statement: Logician 1 claims that (Logician 10 is a knight) and (Logician 1 is a knight). – Allan Apr 17 '15 at 4:11
• 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. – Mike Earnest Apr 17 '15 at 4:14
• Oh ok. You would be right then if that was the case. – Allan Apr 17 '15 at 4:15
• Correct! Nicely done! – Allan Apr 17 '15 at 4:30

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.

• Well done! You got both the question and the bonus correct. Could you explain how you got the first statement (not the bonus)? – Allan Apr 17 '15 at 4:20
• I cheated a little and used a truth table >_> – Kyle G Apr 17 '15 at 4:21
• Fair enough. Could you maybe post a picture of it (if you still have it) so that I can look it over? – Allan Apr 17 '15 at 4:23
• Copy the text here into the link above. I don't see an easy way to link the table directly unfortunately. – Kyle G Apr 17 '15 at 4:27