4
$\begingroup$

(I made up this riddle myself, hope you enjoy it)

A criminal was using magic to flatten a building, while everyone, except 2 people, ran away being super scared. The police of magic came, and with further investigations, took the criminal and the two people to the prison. Here, each person, criminal or innocent, chose a curse to be given to him before calling an expert logician.

The first person chose the INFERFECIUDAX-VEDATUR curse, that meant he told the truth about the person on his right, only if he was a criminal, or else lied. If he was asked a question about anything else he will lie.

The second person thought the INFERFECIUDAX-VEDATUR curse wasn't so bad so he chose it too.

The third person chose the SILENTINPOSUISSET curse. That means he can't talk or answer your questions, but he could have been the criminal.

You can ask any of the persons a question, but you aren't allowed to ask him any question containing the word "criminal", "innocent" etc. You can ask 2 questions total, and then you'd have to chose the two innocents. If you chose the criminal, he will quickly run and flatten the prison, killing you. If you chose the right people, the criminal will get a life sentence.

How do you know who is the criminal? What questions do you ask?

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ Could you clarify whether the effect of the IV curse is that the person so affected tells the truth about the person on his right (1) if and only if the person on his right is a criminal, or (2) if and only if the person under the IV curse is a criminal? Thanks. $\endgroup$ – Gareth McCaughan♦ May 13 '18 at 18:34
  • $\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful. $\endgroup$ – Rubio♦ May 19 '18 at 23:46
8
$\begingroup$

The description of the IV curse at present is a bit ambiguous. So:

If those so affected tell the truth about the person on their right if and only if the person on the right is a criminal:

Ask each of those two people "Is the person on your right human?". If either of them says "yes" then the person on their right is the criminal. If not, then the criminal is whichever of those two isn't on the right.

If those so affected tell the truth about the person on their right if and only if the person asked is a criminal:

Ask each of those people "Is the person on your right human?". (Same question as before.) Now if either of them says "yes" then that person is the criminal. If not, then the criminal is the third person (under the silence curse).

Improve this answer
$\endgroup$
0
$\begingroup$

How about

Asking the first two "Did you flat a building using magic?". If they both say "yes", the third one is the criminal, otherwise the first who says "no" is the criminal.

Improve this answer
$\endgroup$
-1
$\begingroup$

Ended up being very similar to Gareth's solution:

- If an IV-cursed person could only tell the truth about the person on their right if the latter was guilty:

Ask 1 if 2 will tell the truth when asked if 3 has the silence curse. If 1 is guilty, he'll lie about 2, and also 2 will lie about 3, so 1 will say 2 will tell the truth. If 2 is guilty, 1 will tell the truth and 2 will lie about 3, meaning 1 will say no. If 3 is guilty, 1 will lie and 2 will tell the truth, so 1 will say no. Assuming he said no, just ask him a trivial question about 2, and you'll know who is guilty.


- If an IV-cursed person could only tell the truth about the person on their right if he himself was guilty:

Ask 1 if 2 will tell the truth when asked if 3 has the silence curse. If 1 is guilty, he'll tell the truth about 2, and 2 will lie about 3, so 1 will say 2 will lie. If 2 is guilty, 1 will lie and 2 will tell the truth, meaning 1 will say no. If 3 is guilty, both 1 and 2 will lie, so 1 will say yes. Assuming he said yes, just ask him a trivial question about 2, and you'll know who is guilty.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.