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After a lifetime of searching, you have discovered the legendary city of El Dorado. You are approached by three inhabitants. You know from your research that these are the three highest ranking members of their society: the High Priest, who always tells the truth, the Chieftain, who always lies, and the Royal Advisor, who answers yes to every question. They know each other's identities, but you do not.

Legend has it that the El Dorado people will only answer yes/no questions which are directed at a single person, and will always respond with the single word for "yes" or "no" in their language. The problem is, you don't speak the El Dorado language. From overhearing the citizens chattering, you've surmised that the only syllables are in their language are "el," "do," and "ra." Therefore, the words for yes and no are something like "ra-el" and "do-el-do-do," but you have no idea precisely what.

With just three questions, determine which person is which.

Remark: The language constraint in this puzzle is similar to, but more difficult than, the variant where you know the words for yes and no are "ja" and "da" in some order.

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  • $\begingroup$ another variant of en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever ? $\endgroup$ – Jan Ivan Mar 28 '17 at 5:25
  • $\begingroup$ @JanIvan, not quite... But if you want a copy of that question on site, go here. $\endgroup$ – boboquack Mar 28 '17 at 5:26
  • $\begingroup$ …but you can assume that "Yes teller" is random one and solution is same… or what is the difference? $\endgroup$ – Jan Ivan Mar 28 '17 at 5:36
  • $\begingroup$ @JanIvan That you don't know what the words are $\endgroup$ – boboquack Mar 28 '17 at 5:44
  • $\begingroup$ Doesn't that basically mean you have 2 bits of information to determine one of the 6 possible arrangements? Because no matter what answers you get, your method should still give the same output if all the answers were replaced with their counterparts. $\endgroup$ – BaSzAt Mar 28 '17 at 8:34
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If my answer is right, the first two questions (addressed to the same person) are surprisingly simple:

Is your word for yes alphabetically before your word for no?

Is your word for no alphabetically before your word for yes?

We then have that

The Advisor is the only one that can respond the same to both questions. If "ra-el" is yes and "do-el-do-do" is no, then the High Priest will say do-el-do-do to the first and ra-el to the second. The Chieftain will reverse these responses. If, instead, "el-ra" is yes and "ra-ra-ra" is no, then the High Priest will respond el-ra and ra-ra-ra respectively. The High Priest will always respond with the alphabetically first response, the Chieftain the reverse, so you can always identify the first person.

After you know the identity of the first, the rest is pretty standard if you're familiar with the original "ja" and "da" puzzle, since there are only two possibilities left for the remaining identities, you've identified someone who isn't he adivisor, and you know at least one of the words for yes and no. For example, if the first person is the Advisor and you know that "ra-ra-ra" is a word for yes or no, ask the second person "If I asked you if 1+1 = 2, would you say ra-ra-ra?" A "ra-ra-ra" means they are the High Priest, any other response they are the Chieftain. If the first person isn't the Advisor, then "If I asked you if the second person was the Advisor, would you answer ra-ra-ra?" does the trick.

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  • $\begingroup$ This must be it. Amazing idea! $\endgroup$ – BaSzAt Mar 29 '17 at 5:47
  • $\begingroup$ The second part confuses me a a little. Any chance of a diagram or something? $\endgroup$ – Weckar E. Mar 29 '17 at 8:27
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    $\begingroup$ That's the trick, nicely done! (I deleted an earlier comment which said something incorrect). $\endgroup$ – Mike Earnest Mar 29 '17 at 18:36
  • $\begingroup$ Mike -- hadn't checked puzzling in months, and randomly decided to yesterday. I was so happy to see a Mike Earnest puzzle on the front page! $\endgroup$ – Tyler Seacrest Mar 30 '17 at 4:54
  • $\begingroup$ @WeckarE. I left out a lot of details on the second part. If I get a chance I can try to add more detail if you think that would be helpful. But ultimately it's the exact same logic behind this puzzle originally linked to by Jan Ivan above. $\endgroup$ – Tyler Seacrest Mar 30 '17 at 5:00
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Done!

Initial thoughts:

We need to have identified a 'not-advisor' after one question. Assuming we have a vague idea of what yes and no are, it seems unlikely that both would contain all 3 syllables of the language.

Question 1:

'ra' seems to be a unique syllable, so:
'Would your answer include 'ra' if I asked you if person 2 was the advisor?', asked of person 1.
The priest would answer the second clause truthfully. Let's assume person 2 is the advisor. Thus if 'ra' is in the word for 'no', their answer would not include it, and so they'll answer 'no' overall (containing 'ra'). If 'ra' is in the word for 'yes', their answer will include it, and so they'll answer 'yes' overall, containing 'ra'. If person 2 is not the advisor, and 'ra' is in 'no', their answer overall would be yes, which does not contain 'ra', and so on.
Essentially, this forces an answer containing 'ra' for a positive response, and not containing 'ra' for a negative.

Outcomes:

A: If the answer contains 'ra', either person 2 is the advisor or 'ra' is in the word for 'yes' and person 1 is the advisor. Either way, person 3 is not the advisor
B: If the answer does not contain 'ra', either person 3 is the advisor or 'ra' is not in the word for yes, and person 1 is the advisor. Either way, person 2 is clear.

Question 2,3 :

Case A: Ask person 3: 'Would your answer include 'ra' if I asked you if person 2 was the advisor?'
and:
'Would your answer include 'ra' if I asked you if you were the priest?'

Case B: Ask person 2:
'Would your answer include 'ra' if I asked you if person 3 was the advisor?'
and:
'Would your answer include 'ra' if I asked you if you were the priest?'

Outcomes:

Case A:
'ra','ra' - chieftain, advisor, priest
'ra', other - priest, chieftain, advisor
other, 'ra' - advisor, chieftain, priest
other, other - advisor, priest, chieftain
case B:
'ra','ra' - chieftain, priest, advisor
'ra', other - priest, chieftain, advisor
other, 'ra' - advisor, priest, chieftain
other, other - advisor, chieftain, priest

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  • 3
    $\begingroup$ You are assuming knowledge of "ra" being unique, which makes the puzzle just another variation of: puzzling.stackexchange.com/questions/2665/… $\endgroup$ – Zizy Archer Mar 28 '17 at 13:49
  • $\begingroup$ @ZizyArcher "Would your answer be closer to 'ra-el' than to 'do-el-do-do' if..." for the first one, and then you could use the actual word subsequently? Still feels too close to the linked puzzle, though $\endgroup$ – LogicianWithAHat Mar 28 '17 at 14:00
  • $\begingroup$ Some basic PHP - you can't divide any set with $\ge3$ elements into two disjoints sets such that any pair of elements from their union do not both come from the same set. $\endgroup$ – boboquack Mar 29 '17 at 7:41
  • $\begingroup$ What if both "yes" and "no" contain the syllable "ra", e.g., "yes" is "ra-el" and "no" is "ra-do"? I don't see anything in the question that precludes this - my interpretation is that the "ra-el" and "do-el-do-do" were just arbitrary examples of words in the language, actual words for "yes" and "no" could even be "ra" and "ra-ra". $\endgroup$ – Arkku Mar 29 '17 at 15:17
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As the details provided in the question,

My first question would be truth based or that require yes for answer
Q1-

Can you tell me which word is used for yes (or any affirmative question)

Outcome

If it would be High Priest, he would say "Yes" and for Royal Advisor too. But for Chieftain, it would be no.
Here I learned what sounds like yes and also identified who is chieftain.

Q2-

Are you a chieftain?

Outcome

Now its a negative question, for this High Priest have to tell truth and Royal advisor will again go with "yes". So, in two questions there identity is revealed.

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  • $\begingroup$ Except a question can only be directed at one person, not all of them. Otherwise not a bad approach. $\endgroup$ – Weckar E. Mar 29 '17 at 7:39
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Let's arbitrarily call them A, B and C. You can ask these questions:

1. Ask A if he always tells the truth. You'll get the answer "yes" no matter who A is.

2. Now that you know what means yes, ask A if B will tell the truth if you ask him a question at the moment/a certain point. If A hesitates, B is the advisor. Otherwise, if A is the priest or chieftain, he'll say no, and he'll say yes if he's the advisor. If anyone says something other than the yes word, it can only mean no.

3. Now there are 3 possibilities:

- If A has just hesitated, B is the advisor. Then just ask A or C an easy yes/no question like "Does 2+2 make 4?" to tell who each of them are.

- If A says yes, he's the advisor. Just do the same with B or C this time.

- If A says no, he's the priest or chieftain and B isn't the advisor, so C is. Just do the same with A or B this time.

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  • $\begingroup$ Hold on, once you use your first question to get a word, you've only got 4 possibilities of answers but 3!=6 orderings of people? Also, what does hesitation mean in this case? There isn't any 'random' person in the setup. $\endgroup$ – boboquack Mar 29 '17 at 8:18
  • $\begingroup$ @boboquack There are more possibilities - the non-advisors can't predict if the advisor would tell the truth or lie when asked any question, so they can't answer at all, which is what I meant by "hesitation". $\endgroup$ – Nautilus Mar 29 '17 at 8:24
  • $\begingroup$ No, the advisor will just say yes in any case, so they can predict. $\endgroup$ – boboquack Mar 29 '17 at 8:29
  • $\begingroup$ Yes they can, but only if they know the question you'll ask the advisor. You can read the 2nd question as something like "If I ask B a question at 12:00 that I'd randomly choose from a pot/wheel of fortune etc, will he tell the truth or lie?" if you like. $\endgroup$ – Nautilus Mar 29 '17 at 8:46
  • $\begingroup$ I would argue that "will B answer truthfully" is not a yes/no question, because there is the third option "depends on the question", so A would not answer the question at all. Alternatively, one could make the case that this is the same question as "will B answer any question truthfully", in which the answer for the advisor is "no" (it may coincidentally be the truthful answer, but B did not intentionally answer truthfully). $\endgroup$ – Arkku Mar 29 '17 at 15:08
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Q1 (Ask the first person): Do you always say 'yes'?

Anyone you will ask will answer 'yes', by this you will know what corresponding word means 'yes'.

Q2: (Ask the second and third people) Are you lying?

If the person will hesitate or won't answer, then that person is the Chieftain. (Thanks to the Liar paradox)

If the person answers 'yes', then that person is the Advisor.

If the person answers 'no', then that person is the High Priest.

P.S. What kind of society has a chieftain who always lies? Does this make the Chieftain a bad leader?

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