Hired by the Council of Magic - Language Barrier at the Circle:
After your two successful visits to call upon the aid of the Council of Magic, the Council has called in one of the favors the King of Puzzlington owes them for their help. As your luck would have it, the favor they have requested is for you to go and get them help from the Circle of Magic in Answerland, across the Great Sea.

There are three types of spellcasters in Answerland: Wizards, Priests and Warlocks. They work the same as in previous missions, but I'll repeat them here.

Wizards follow one of two paths: the path of Fire or the path of Water. Wizards on the Path of Fire always tell the truth when asked a question. Wizards on the Path of Water always lie when asked a question.

Priests worship one of two Gods: Yes, god of life or No, god of death. When asked a question, instead of answering, priests just say their God's name. That is, a Priest of Yes will always answer "Yes" to any question.

Warlocks are unpredictable tricksters. When asked a question, Warlocks will tell the truth or lie, as they wish. Warlocks can also choose not to answer and remain silent.

Language Barrier:
There's just one major issue. They don't speak English in Answerland. Instead, they speak Rewsna. You've studied Rewsna in the past, but you don't know it that well.

You only know Rewsna well enough to ask simple questions. You can't ask hypothetical questions (such as "what would X say if I asked him Y?") or nested questions (such as "Out of statements X and Y, is exactly one true?").

You do know that in Rewsna, Bek is "Yes" and Dok is "No"... or was it the other way around? You're not entirely sure, but it's one or the other.

The names of the Gods are translated into Rewsna as well... into Bek and Dok. They retain their meaning in the process; that is Yes is translated into the word meaning Yes.

The Circle:
When your boat arrives, you make your way to the great hall of the Circle of Magic. There are four spellcasters in the Circle: Abigail, Brian, Camilla and Dylan.

The Council knows a great deal about the Circle and have kindly shared their knowledge with you. The Council knows the Circle consists of two Wizards (one Fire, one Water), a Priest (of an unknown god) and a Warlock. The Council does not know which member of the Circle is which; figuring that out is your job. The members of the Circle know each other's types and subtypes.

You arrive at the hall at noon. You can, once every 12 hours, ask any one member of the Circle any one question that can be answered Yes or No. If a spellcaster is asked a question they can't answer (because they don't know the answer), they remain silent. Priests are an exception to this rule; to them, their god is always the answer, no matter what the question.

More like your normal luck, the Warlock here hates you as well; when asked a question, the Warlock will choose truth, lies or silence, whichever they think will disrupt your strategy the most.

Your task is the same as it has always been: Learn the type and sub-type of all Circle members as fast (in as few questions) as you can.

I can do it in 11 questions, but I'm sure it can be done faster.

  • $\begingroup$ I can do it in 11 if I have to ask yes or no questions. However, I can do it much easier if a can ask a simple non yes/no question, like what color is the sky. Is my language skills enough to manage a question like that? $\endgroup$
    – dsollen
    Commented May 19, 2017 at 14:34
  • 1
    $\begingroup$ @dsollen Your language skills are irrelevant in that case. The yes-no restriction is a silly rule spellcasters follow; you aren't allowed to break it. $\endgroup$
    – qwertyu63
    Commented May 19, 2017 at 14:36

3 Answers 3


The hard part is going to be distinguishing the warlock. My approach is to identify a wizard, identify their school, and then ask them to resolve any remaining uncertainty.

Fortunately the ban on recursive questions doesn't rule out the classic

"Are you a liar?"

First I ask all four people the same question.

"Did I have coffee for breakfast 17 years ago last Tuesday?" Or any other question which I'm certain they can't answer.

There are essentially two cases:

Either three remain silent and one answers, in which case that one is the priest and I know their god's name in Rewsna, or two remain silent and two answer, in which case the two silent ones are the wizards and the others are the priest and the warlock.

Taking the second case first, because it's easier:

I ask one of the wizards "Are you a wizard of water?" (fifth question). The answer must be "No", so I now know the words for "Yes" and "No". I ask any question which is definitely true (e.g. "Are you a wizard?" (sixth question)) to know which wizard is which, and finally I ask "Is {one of the two non-wizards} a priest?" (seventh question). Solved.

The first case requires more follow-up questions. I've identified the priest, but I don't know which of the other three is the warlock. I ask each of those three the same question:

For the sake of argument, "Are you a warlock?". They must answer this question, and there are only two possible answers, so two of them will give one answer and the third will give the other answer. The odd one out is a wizard.

Now the procedure is fairly similar:

"Are you a wizard of water?" (eighth question). I know how they answered "Are you a warlock?", so putting the two together I know which type of wizard they are. Finally I ask "Is {one of the unknown ones} a warlock?" (ninth question).

So in the worst case it takes nine questions.

  • $\begingroup$ I applaud your thinking of using the non-answer option to simplify this, most would forget that option. $\endgroup$
    – dsollen
    Commented May 19, 2017 at 14:41
  • $\begingroup$ @dsollen, it took me a long time to think of a way that it could be useful. $\endgroup$ Commented May 19, 2017 at 16:27

I pick 3, ask each one if they are a wizard of fire, then of water.

This helps because:

The priest will answer the same to both. The wizards will answer yes to fire and no to water. The warlock will masquerade as wizard or priest. So three either answer YN YN YN - fourth is priest, YN YY/NN YY/NN - fourth is wizrd, YN YN YY/NN

Then, we have a few paths:

YN YY/NN YY/NN = warlock answers like priest - fourth is a wizard, and what first answered for "fire" is yes. essentially I have 2 wizards and 2 priests. I ask a "priest" a question none there know (what my middle name is?). The priest will answer, the warlock cannot. So I determine priest and warlock. I ask 1 wizard if he is a wizard. If he answers yes, he's fire, if no, he's water. I can easily figure out the other wizard. Since I know yes and no, I also know the priests' god.

That was total of 8 questions.

Next path:

YN YN YN = warlock answers like wizard - what they all answered for "fire" is yes. Ask priest (fourth person) anything - you know the clan. Ask two wizards if they are wizards. Both yes - one of them is warlock, the other fire, third is water. Both no - one of them is warlock, the other water, third is fire. If they answer different, ask the third too. Out of two unknown ask one a personal detail about his presumable self ("is the fire/water wizard...") - no answer, it's the warlock, any answer, it's the wizard.

That was total of 10 or 11 questions.

Next path:

YN YN YY/NN = don't know what warlock is doing yet - what the two presumable wizards answered for "fire" is yes. Ask the fourth the two questions. then proceed as fitting option above.

This can result in 11 or 12.

  • $\begingroup$ That looks like it checks out to me (other than your example unanswerable question breaking the yes/no only rule). Well played; you've got it if no one beats your question count. $\endgroup$
    – qwertyu63
    Commented May 18, 2017 at 4:24
  • $\begingroup$ It can be a yes/no question $\endgroup$
    – JNF
    Commented May 18, 2017 at 5:16
  • $\begingroup$ I think your analysis of the first six answers is wrong. The warlock could answer NY to mess with your strategy. $\endgroup$ Commented May 19, 2017 at 7:37
  • $\begingroup$ He wouldn't do that because that's giving himself away $\endgroup$
    – JNF
    Commented May 19, 2017 at 7:38
  • $\begingroup$ @JNF, no, because you don't know which of Bek-Dok is NY and which is YN. $\endgroup$ Commented May 19, 2017 at 7:56

Can be done in...

10 questions.


First, try asking each if Bek means yes/is an affirmative answer (4 questions). The spellcasters say Bek if they're telling the truth and Dok if they're lying. One of the spellcasters telling the truth in the first question is the truth-telling wizard and one of the lying ones is the lying wizard.

- If only one of the answers/reactions is unique, the one giving it is a truth-telling or lying wizard depending on the answer. Ask him/her a trivial question to find out the meaning of both words. Then ask if a certain other spellcaster is a wizard. At least 2 such questions after learning the meaning of the words, you'll find out both wizards. Then ask any of the wizards if a certain one of the non-wizards is a priest, and you'll find out about the type and sub-type of everyone (8 questions in total).

- If one of them stays silent, it must be the warlock, who can then be eliminated. The only person giving a unique non-silent answer is a wizard. Ask one of the remaining two if Bek is a negative answer. If that person contradicts him/herself, he/she is the priest. Otherwise he/she is a wizard. Don't forget to ask a predictable wizard your trivial question though (6 questions in total).

- If 2 of the very first 4 answers are true, ask each if Bek is a negative answer (3 more questions - no need for the fourth). The spellcasters will say Bek if they're lying and Dok if they're telling the truth. One of the spellcasters telling the truth in the first question is the truth-telling wizard and one of the lying ones is the lying wizard, so the two lie/tell the truth both times. The priest tells the truth only once. That means at least one of the four will contradict themself or would have. After the first contradiction, the non-contradicting spellcaster who has made a statement before with the same truth value as the contradicting one's will be revealed as one of the wizards. Ask the wizard a trivial question. After that, ask if a given one telling a statement of the opposite truth value of that of his/hers in one of the first 4 questions is a wizard (if the wizard is a TT, ask about one of the two who lied in the first stage or the other way around) so that both wizards are found. Lastly, ask a wizard if a given non-wizard is a priest.


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