At the bathhouse, 27 travellers arrive. According to a book they have brought, each one follows a different one of the following rules:

Traveller Yes Unknown No
Sheep Y Y Y
Knightly Sheep Y Y R
Sheepish Knight Y Y N
Confused Sheep Y R Y
Knightly Joker Y R R
Jestful Knight Y R N
Unconfident Sheep Y N Y
Muddled Knight Y N R
Bureaucratic Knight Y N N
Knavish Sheep R Y Y
Credulous Joker R Y R
Mixed Knight R Y N
Unknightly Joker R R Y
Joker R R R
Unknavish Joker R R N
Mixed Knave R N Y
Incredulous Joker R N R
Knightly Bureaucrat R N N
Sheepish Knave N Y Y
Muddled Knave N Y R
Unconfident Bureaucrat N Y N
Jestful Knave N R Y
Knavish Joker N R R
Confused Bureaucrat N R N
Bureaucratic Knave N N Y
Knavish Bureaucrat N N R
Bureaucrat N N N

(The title box is the true answer to the question. Y = definite yes, R = random yes/no, N = definite no)

Unfortunately, the book doesn't say who is who, so you'll have to work this out. You can ask the same question to multiple travellers at once. The travellers seem knowledgeable on many subjects, logic included, but don't know anything like what you had for breakfast or other such personal trivia. What is the lowest number of questions (on average) required to identify all of the travellers?

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    $\begingroup$ These variations on liar problems really don't add anything - anyone could come up with any sort of arbitrarily complicated list of people to ask these questions about. Unless you have some reason to believe that this variant is particularly interesting (for instance, you know it has a nice, "elegant" solution), it is not likely to be a good puzzle. $\endgroup$
    – Deusovi
    Aug 10 '21 at 22:29
  • $\begingroup$ I do not believe it is the right idea to ask for an average number of questions, seeing as there are people who chose randomly, which just adds unnecesarily painful math with no change in logic. Most puzzles like this ask for lowest number of questions required to guarentee identification of all characters. Perhaps you meant to do the same? $\endgroup$
    – Ankit
    Aug 13 '21 at 1:23
  • $\begingroup$ @Deusovi Tbh I don't this it is fair to call this an arbitrarily complicated list. It just has one of each possible character. $\endgroup$
    – Ankit
    Aug 13 '21 at 1:24

I'm checking to see how much this does but I am pretty sure that it will clarify quite a lot

Step 1:

Ask everyone "Will your answer to the next question be the truth?"

Step 2:

Ask everyone "Was your previous answer the truth?"


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