Now let’s make it a little harder. You have three guards: one tells the truth, one lies, one answers randomly. The guards understand you, but only answer either “da” or “ja”. One means yes, one means no, but you don’t know which is which. You get to ask each guard one question.
When I was a substitute teacher I would give the kids logic puzzles of varying difficulty. I would offer $100 if anyone could provide me with the answer to this one. If they looked it up on Wikipedia and could then explain it to me, I’d give them a king size candy bar.
Give them a paradox by encoding the other two’s potential responses into the question (similarly to the two guard solution, but this time the random response is included). If they are able to answer, then you asked the random one, because the liar and truth teller have no idea what the random one would answer so can’t answer only yes or no without potentially violating their truthiness rule.
This isn’t to solve the puzzle but to see what the other two would do in that situation. If I figured out the random one with the first question, I’d use the 2nd to ask the same thing of one of the others. Then, if it’s still 2 doors, the two guard solution will work on the last one to figure it out.
But if the first guard asked explodes or something when asked, I think that there wouldn’t be enough questions left to find both the random guard (which I believe you have to do first) and the door. Though if you change the question to only ask about one other’s answer instead of both, you’ll be able to find both the random guard and the safe door.
Though hopefully the whole setup isn’t a lie and everyone present is a strategic liar that wants you dead. Imagine doing one of those riddles and when you step through the door you notice both doors lead into the same room whose walls now seem to be closing in and the last thing you hear is one of the guards asking another why riddles seem to get people to let their guard down anyways.
It’s still trivial, assuming the three guards guard three doors: just ask each guard: “Would the guard that always lies say this door is safe?” The random guard will give a random answer while the other two give the inverted answer. Even better if don’t ask the random guard first, then you can repeat the question about the other doors to the same guard and only need two questions
Now let’s make it a little harder. You have three guards: one tells the truth, one lies, one answers randomly. The guards understand you, but only answer either “da” or “ja”. One means yes, one means no, but you don’t know which is which. You get to ask each guard one question.
When I was a substitute teacher I would give the kids logic puzzles of varying difficulty. I would offer $100 if anyone could provide me with the answer to this one. If they looked it up on Wikipedia and could then explain it to me, I’d give them a king size candy bar.
I never had to pay out.
https://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever
Give them a paradox by encoding the other two’s potential responses into the question (similarly to the two guard solution, but this time the random response is included). If they are able to answer, then you asked the random one, because the liar and truth teller have no idea what the random one would answer so can’t answer only yes or no without potentially violating their truthiness rule.
This isn’t to solve the puzzle but to see what the other two would do in that situation. If I figured out the random one with the first question, I’d use the 2nd to ask the same thing of one of the others. Then, if it’s still 2 doors, the two guard solution will work on the last one to figure it out.
But if the first guard asked explodes or something when asked, I think that there wouldn’t be enough questions left to find both the random guard (which I believe you have to do first) and the door. Though if you change the question to only ask about one other’s answer instead of both, you’ll be able to find both the random guard and the safe door.
Though hopefully the whole setup isn’t a lie and everyone present is a strategic liar that wants you dead. Imagine doing one of those riddles and when you step through the door you notice both doors lead into the same room whose walls now seem to be closing in and the last thing you hear is one of the guards asking another why riddles seem to get people to let their guard down anyways.
It’s still trivial, assuming the three guards guard three doors: just ask each guard: “Would the guard that always lies say this door is safe?” The random guard will give a random answer while the other two give the inverted answer. Even better if don’t ask the random guard first, then you can repeat the question about the other doors to the same guard and only need two questions