A Puzzle in Tribute to Raymond Smullyan

[Galendir]

A list of eighteen three-letter words is written on a chalkboard. The logic professor hands each of his three ace students a small piece of paper and announces: “I have selected one of the words written on the board and have given each of you a different letter of that word. Without revealing your letter to your fellows, can you deduce the selected word?”
(Each of the students has displayed flawless reasoning to date and each expects that the same will be true for this challenge.)
The professor waits several moments for the students to consider the list and then begins to ask each one in turn if they have deduced the selected word pausing for several moments between each query.
Each student in turn answers “No”.
The professor waits a few minutes more and in the same order as before asks each student a second time (again pausing after each response) and again each student answers “No”.
The professor waits for several more minutes and for the third time asks each student as before if they have deduced the selected word. This time each student in turn answers “Yes”.


What is the selected word, and which letter does each student have?


ANT APE ASP AUK BAT BOA CAT DOG EMU
FLY FOX GNU JAY OWL PIG RAT RAY YAK




- In tribute to Raymond Smullyan -

[CliveStaples]

I've got a bit of a start.

First, everyone looks at their letter. They each generate a set S containing every word that contains that letter. If S had only one word in it, that person would immediately know which word the Professor had chosen. So each person's S has to contain at least two words. Moreover, after round 1, everyone can eliminate from S words that contain a unique letter (i.e. that are the only word that contains that letter). If the Professor's word contained a unique letter, then someone would have been handed that letter (since the word is only 3 letters long and each person got a different letter), and would have picked out the Professor's word in the first round. This didn't happen, so the Professor's word doesn't contain a unique letter. Ergo, for each character C in the Professor's word, there must exist at least one other word that contains C.

I'm not sure where to go from there.

[Galendir]

Version 2
A list of twenty three-letter words is written on a chalkboard. The logic professor hands each of his three ace students a small piece of paper and announces: “I have selected one of the words written on the board and have given each of you a different letter of that word. Without revealing your letter to your fellows, can you deduce the selected word?”
(Each of the students has displayed flawless reasoning to date and each expects that the same will be true for this challenge.)
The professor waits several moments for the students to consider the list and then begins to ask each one in turn if they have deduced the selected word pausing for several moments between each query.
Each student in turn answers “No”.
The professor waits a few minutes more and in the same order as before asks each student a second time (again pausing after each response) and again each student answers “No”.
The professor waits for several more minutes and for the third time asks each student as before if they have deduced the selected word. This time each student in turn answers “Yes”.


What is the selected word, and which letter does each student have?


APE ASP AUK BAT BOA BUG CAT COW DOG ELK
EMU FLY FOX GNU HOG JAY OWL PIG RAT YAK

[MindTrap028]

Answer is YAK It's a long process of elimination of course simply repeat the list of unique and it whittles down. But my logic isn't perfect, and I may have screwed up. First guy has a "K", second guy has a "Y" and last guy got screwed with the "A"

P.S. I'm not doing the second version Its not about smarts, it's about processing power. (or time for us slow people)

As a side note, I don't consider it an interesting puzzle.

[Galendir]

Mindtrap, it looks like you were close to the correct answer, but you erred somewhere.
Since you don't find this puzzle interesting, thanks for giving it a go anyway. It may in fact be more interesting (and challenging) to create than to solve.

[CowboyX]

Jay or Yak

[MindTrap028]

Mindtrap, it looks like you were close to the correct answer, but you erred somewhere.
Since you don't find this puzzle interesting, thanks for giving it a go anyway. It may in fact be more interesting (and challenging) to create than to solve.

Thanks. If you created the puzzle, then I'm more impressed with that then the process of answering it.. very impressed

I caught one error I made in the process and figured there was bound to be another, just couldn't muster the will to double check. For me it was more challenge to my dyslexia then logic *J*

[Galendir]

CowboyX , there is only one solution to a given puzzle (but each version of the puzzle offers a different solution) and you must indicate which letter each student had.
Thanks for the attempt.

---------- Post added at 06:55 PM ---------- Previous post was at 06:46 PM ----------

Clive, that is a good start. Hint:

~ "Moreover" Iterate ~

[CowboyX]

Ok, so after the first round of questioning all three students should know neither of them have the word where one letter only appears once - D, H, I, J, M, N, R, S, X So that leaves APE AUK BAT BOA BUG CAT COW ELK FLY OWL YAK Round two removes F, G, P, U leaving BAT BOA CAT COW ELK OWL YAK But all three would know simultaneously it was YAK and I can't figure how we know which letter each student had...hmmm

[Galendir]

Ok, so after the first round of questioning all three students should know neither of them have the word where one letter only appears once - D, H, I, J, M, N, R, S, X So that leaves APE AUK BAT BOA BUG CAT COW ELK FLY OWL YAK Round two removes F, G, P, U leaving BAT BOA CAT COW ELK OWL YAK But all three would know simultaneously it was YAK and I can't figure how we know which letter each student had...hmmm

CowboyX,
Look again.

~ You've removed one too many words in round two. There should be one more word in your choice list for round three leaving three words each with a unique letter. ~

[CowboyX]

Ok, so after the first round of questioning all three students should know neither of them have the word where one letter only appears once - D, H, I, J, M, N, R, S, X So that leaves APE AUK BAT BOA BUG CAT COW ELK FLY OWL YAK Round two removes F, G, P leaving AUK BAT BOA CAT COW ELK OWL YAK So at the beginning of round three we have AUK ELK YAK and since the first student knows the word when asked they must have the letter U E L or Y

[Galendir]

So close, CowboyX

~ You've listed one too many letter possibilities for the first student to have in round three. But you've almost got it. ~

[CowboyX]

Ok, so after the first round of questioning all three students should know neither of them have the word where one letter only appears once - D, H, I, J, M, N, R, S, X So that leaves APE AUK BAT BOA BUG CAT COW ELK FLY OWL YAK Round two removes F, G, P leaving AUK BAT BOA CAT COW ELK OWL YAK So at the beginning of round three we have AUK ELK YAK and since the first student knows the word when asked they must have the letter U (E or L) or Y The second student can have an A or an E or L (whatever the first student doesn't have) but it can't be A because they wouldn't know if it was AUK or YAK so it's E or L. The last student has a K

I'm surprised I made it this far.

[Galendir]

That's good, CowboyX, but you didn't make the correction from my last hint.

... So at the beginning of round three we have AUK ELK YAK and since the first student knows the word when asked they must have the letter U (E or L) or Y. ...

~ There are eight words left after round two, of which only three have a letter unique to the set (and they each have only ONE such letter). The first student must have one of those THREE unique letters to be able to identify the selected word. ~

[CowboyX]

I'm surprised I made it this far.

Ok, so after the first round of questioning all three students should know neither of them have the word where one letter only appears once - D, H, I, J, M, N, R, S, X So that leaves APE AUK BAT BOA BUG CAT COW ELK FLY OWL YAK Round two removes F, G, P leaving AUK BAT BOA CAT COW ELK OWL YAK So at the beginning of round three we have AUK ELK YAK and since the first student knows the word when asked they must have the letter U E or Y (L isn't unique since it is also in OWL) The second student has the L because it can't be the U E or Y (he knows the first student has one of those), but he knows the word so it can't be A or K. The last student has a K The word is ELK, the first student has an E

[Galendir]

Yes, CowboyX. That is the correct solution to the second version of the puzzle.

[Galendir]

SOLUTION KEY

~ Raymond Smullyan Tribute Puzzle Version 1 ANT APE ASP AUK BAT BOA CAT DOG EMU FLY FOX GNU JAY OWL PIG RAT RAY YAK Since no student can immediately identify the word, it follows that no student has a letter unique to a single word of the original set eliminating the possibility of any words with a unique letter from being the selected word. Round 1 Eliminate: ASP CAT DOG EMU FOX JAY OWL PIG Leaving: ANT APE AUK BAT BOA FLY GNU RAT RAY YAK Since no student can identify the word after the first round of elimination, it follows that no student has a letter unique to a single word in the remaining set eliminating the possibility of any words with a unique letter of the current set from being the selected word. Round 2 Eliminate: APE BOA FLY GNU Leaving: ANT AUK BAT RAT RAY YAK Round 3 Student 1 knows the word so must have a letter unique to the current set (N or U or B) thus eliminating RAT RAY YAK. Student 2 also knows the word so must have a second letter unique to the last three possibilities (ANT AUK BAT) - the only such letter being K. The word is thus AUK. 1 has U, 2 has K, 3 has A. Raymond Smullyan Tribute Puzzle Version 2 APE ASP AUK BAT BOA BUG CAT COW DOG ELK EMU FLY FOX GNU HOG JAY OWL PIG RAT YAK Since no student can immediately identify the word, it follows that no student has a letter unique to a single word of the original set eliminating the possibility of any words with a unique letter from being the selected word. Round 1 Eliminate: ASP DOG EMU FOX GNU HOG JAY PIG RAT Leaving: APE AUK BAT BOA BUG CAT COW ELK FLY OWL YAK Since no student can identify the word after the first round of elimination, it follows that no student has a letter unique to a single word in the remaining set eliminating the possibility of any words with a unique letter of the current set from being the selected word. Round 2 Eliminate: APE BUG FLY Leaving: AUK BAT BOA CAT COW ELK OWL YAK Round 3 Student 1 knows the word so must have a letter unique to the current set (U or E or Y) thus eliminating BAT BOA CAT COW OWL. Student 2 also knows the word so must have a second letter unique to the last three possibilities (AUK ELK YAK) - the only such letter being L. The word is thus ELK. 1 has E, 2 has L, 3 has K. ~