King Calvin's Conundrum
The cruel, devious King Calvin has captured sixty hapless Brilliant users! He puts all of them in a prison cell, ties each of them down to a chair, and seals their mouths up with duct tape so they can never communicate with each other.
Fortunately, their prison guard is the compliant, yet rigorously virtuous Andrew. He takes a bunch of sticky notes, writes a 0 or a 1 on each of them, and places them at random on the forehead of each of the prisoners. Each prisoner does not know the number on their own forehead, but they can see the numbers on the others’ foreheads. Through this, the guard offers a way out: if someone can guess the number on their forehead correctly, he will let the prisoner out during the night while King Calvin is sleeping. But there is one condition: each prisoner has only one chance to guess, and if they choose the wrong number, then they will remain in the prison, forever! (This means that prisoners will only guess if they know that they are absolutely correct.)
Guard Andrew scoffs at the prisoners’ begs for mercy, flippantly providing a seemingly trivial hint: at least one of the prisoners has a 0 on their forehead. However, Brilliant users tend to be rather...well, brilliant! After only six nights, King Calvin checks the prison himself and is furious to find that all of his prisoners have successfully escaped!
Find the largest possible number of prisoners with 0s on their foreheads.
All of these prisoners are level 5 in logic. (In other words, assume that they are perfectly rational and have infinite intelligence in this situation.)
Hint:
There might be days when the prisoners are unable to make a definite guess without a night passing.
Details and Assumptions:
- A prisoner will only know if their guess was correct in the night when Andrew sets, or does not set, them free. Assuming that all of the prisoners sleep in the night, all the other prisoners will only know if someone has guessed correctly the next morning, if that person is no longer in the prison.
This problem was inspired by a quite tricky puzzle which my friend Josh Y. told me.
Image Credit: King Harkinian by AuronKaizer, Zelda Wikia .
The answer is 5.
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1 solution
Here is a general solution.Let no. of days after which all prisoners have been released be 'n'.Let us take cases:-
(1). 1 zero: This man sees all others with '1' as label and escapes in a single day.
(2). 2 zeroes: These men see a man except himself with a zero.They do not leave that night but wait.The next day,they see none of them have left.Thus they both deduce that there are exactly two zero and leave in the second night.The others then ensure that there were exactly two zeroes and all leave on the third night.
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N zeroes: Take a person with a zero.He can see (n-1) zeroes.He waits for (n-1) days and see none of them have left because he also had a zero which others were able to see.He deduces this fact and escapes along with all prisoners with zero on the (n)th day.The others labelled as '1' leave on the next day,i.e (n+1)th day which is the total no. of days of escape.
Here (n+1)=6.Therefore n=5.
THE NO. OF ZEROS IS FIVE.