[Another logic/lateral thinking problem]

On your travels you come to a fork in the road; unsure which road leads to your destination, you can ask bystanders for assistance. (For the purposes of this question, one road leads to heaven, the other to hell; and you wish to be directed to heaven)

There are three bystanders (all of whom know where each road leads); one bystander always tells the truth, another always gives a false answer and the third gives random answers (randomly answering a question truthfully or with a lie) - you do not know which of the bystanders is which.

You can choose which of the three bystanders is to answer each of your questions.
Your questions must be binary in nature (can be answered yes/no or by one of two alternate responses).

What is the minimum number of questions you need ask to guarantee that you know which road leads to heaven; and what are the questions?

I can solve this conundrum with three questions, can it be done with only two?

 

All I know, is when you come to a fork in the road, take it!

 

[Another logic/lateral thinking problem]


I can solve this conundrum with three questions, can it be done with only two?

Yes.

 

Do the bystanders know each other, and how truthful they are?

 

Do the bystanders know each other, and how truthful they are?

Yes and yes

 

What if you ask all three "will the answer you give to this question be a lie?"

 

What if you ask all three "will the answer you give to this question be a lie?"

The bystander who always tells the truth - will answer No.
The bystander who always tells a lie - will answer No.
The bystander who gives random answers will answer Yes or No.

How will the above responses help you (especially if all answer No)?

 

The bystander who always tells the truth - will answer No.
The bystander who always tells a lie - will answer No.
The bystander who gives random answers will answer Yes or No.

How will the above responses help you (especially if all answer No)?

In the example above, you are counting three questions as being asked - correct?

 

The bystander who always tells the truth - will answer No.
The bystander who always tells a lie - will answer No.
The bystander who gives random answers will answer Yes or No.

How will the above responses help you (especially if all answer No)?

The one who lies cannot answer "no". If he does, that would be a true statement. He lies and is saying that he does, which is truthful.

Edit: I think I might have messed that up. I definitely did not get enough sleep.

 

In the example above, you are counting three questions as being asked - correct?

Yes, that would be three questions.

 

The way I remember the original puzzle, there were two brothers at the fork, one honest and the other a compulsive liar.
The solution went like this. You point to one direction and ask "If I ask your brother if this is the way to town, would he say yes?"
If the answer is "yes", then the other path is the way to town.

I haven't figured out how you would deal with three of them.

 

Edit: This NOT correct as mentioned later. I think I was too enamored with my *thinking out of the box* question and I didn't bother to follow through.

I can do it in three questions but I do not believe it can be done in less.

Spoiler

Roads are 1 and 2.

Bystanders are B1, B2, and B3 and also L (liar) T (truth) and R (random)

Question 1

Ask B1 if both 1 and 2 go to Heaven?

If answer is Y, B1 is L or R
If answer is N, B1 is T or R

Question 2

Ask B2 if both 1 and 2 go to Heaven?

If answer=Y, B2 is L or R and B3 must be T- Q3- ask B3 if 1 goes, if Y, take 1, if N, take 2.
If answer=N, B2 is T or R and B3 must be L- Q3-ask B3 if 1 goes, if Y, take 2, if N take 1.


I just don't see how asking "or" can get it done in two questions, but I am willing to be bested. Edit and I also had to fix some naming.

 

There are four unknowns, so there has to be four equations (questions). The presence of the "random" guy throws a wrench in the works.

 

There are four unknowns, so there has to be four equations (questions). The presence of the "random" guy throws a wrench in the works.

What I often notice about these kinds of problems (and I think this is a good one) is that I end up realizing that there is more information than I think there is - especially at different "steps". I like them because I think that aspect of problem solving generalizes to many other situations. Of course, when I can't/don't solve them, they are obviously bad puzzles

 

I too enjoy these types of puzzles and have posted a few of my favourites on this site. But when I am unable to solve a puzzle, that does not make it a bad puzzle.

What makes a good puzzle is the thought processes needed to reach the correct solution.
The more elegant the solution, the better the puzzle.

 

I too enjoy these types of puzzles and have posted a few of my favourites on this site. But when I am unable to solve a puzzle, that does not make it a bad puzzle.

What makes a good puzzle is the thought processes needed to reach the correct solution.
The more elegant the solution, the better the puzzle.

Please know that I was kidding about the "bad" puzzle part. They are frustrating if they're good (I may have cursed you a time or two on this one ) - what fun would they be if they took only a minute to solve?

 

[Another logic/lateral thinking problem]

On your travels you come to a fork in the road; unsure which road leads to your destination, you can ask bystanders for assistance. ...
...
...

I can solve this conundrum with three questions, can it be done with only two?

i am a man, always sure and don't need to ask for directions. solved with zero questions

 

I can do it in three questions but I do not believe it can be done in less.

There is a flaw in your analysis.

You are asking a binary question to which you know the correct answer is No.

If B1 & B2 both answer Yes to this question, then one must be the liar and the other random.
If B1 & B2 both answer No to this question, then one must be the truth teller and the other random.

But if there is a difference of opinion, then the bystander answering Yes could be the liar or random, the bystander answering No could be the truth teller or the random. So bystander three could be any (truth, liar, random).

 

I can do it in three questions but I do not believe it can be done in less.

There is a flaw in your analysis.

You are asking a binary question to which you know the correct answer is No.

If B1 & B2 both answer Yes to this question, then one must be the liar and the other random.
If B1 & B2 both answer No to this question, then one must be the truth teller and the other random.

But if there is a difference of opinion, then the bystander answering Yes could be the liar or random, the bystander answering No could be the truth teller or the random. So bystander three could be any (truth, liar, random).

You are right, my "solution" only works for YY or NN to my first to questions. Back to the drawing board.

Edited to add: Although at a 50% chance of getting into heaven sounds pretty good at the moment

 

This is an old worn out thought experiment from the 1950s or earlier.

However, I don't know what the right answer is.