10 Prisoners, 10 Keys, 2 Weeks

One day, the warden of a prison is, like most wardens in puzzles, feeling a little capricious and decides that he wants to get rid of his prisoners, one way or another. He gathers all the prisoners in the yard and explains to them – “Tonight, I will go to each of you, hand you a key, and tell you who has your key. Each day after that, while the others are out of the cells and no one is watching, I will allow each of you to place your key in someone else’s cell – and each night, you may collect the keys in your own cell. If at any point, you are certain that everyone has the key to their own cell, you may summon me, at which point each of you will open your own cell and walk free. If anyone has the wrong key, everyone will be executed then and there. You may discuss your strategy before tonight, but afterward, no communication will be allowed regarding my game. – Oh, and by the way, if you are still here two weeks from today, I will execute everyone – it’ll be a big birthday for me and I want to retire!”

That night, just as promised, the warden went to each cell and gave each prisoner a key. As he handed each prisoner the key, he whispered to them the name of the person possessing the key to their cell. The keys were entirely indistinguishable from one another, but that was okay, because the prisoners had not counted on being able to tell anything about them. Indeed, the prisoners all seemed confident.

What was their strategy? How could they beat the warden’s game?

We assume the cells in the prison are arranged in a circle. The prisoners agree every day each of them to pass the key they receive to their left neighbor, except for their own key which they keep. It is easy to figure out which key is their own since they can easily calculate when they will receive it. For example, if prisoner 8 knows that his key is at prisoner 3 in the beginning, then he will get it on the 5th day. Therefore, within 10 days, all prisoners will have their own keys.

Source:

Puzzling StackExchange

Impossible!

Is it possible the following chess position to occur in a game?

No, it is impossible. The White’s pawn from e2 should have captured the Black’s bishop from c8. In order for the bishop to get there, the pawn on c6 should have captured one of White’s rooks. It couldn’t be the rook from h1, so it should have been the rook from a1. But in order for the rook from a1 to get to c6, the pawns from b2 and c2 should have been moved to b3 and c4 respectively. However, in that case, the bishop from f1 couldn’t get to a4, since it has been blocked before the capture e2xf3.

Larger or Smaller

Alice secretly picks two different integers by an unknown process and puts them in two envelopes. Bob chooses one of the two envelopes randomly (with a fair coin toss) and shows you the number in that envelope. Now you must guess whether the number in the other, closed envelope is larger or smaller than the one you have seen.

Is there a strategy which gives you a better than 50% chance of guessing correctly, no matter what procedure Alice used to pick her numbers?

Choose any strictly decreasing function F on the set of all integers which takes values between 0 and 1. Now, if you see the number X in Bob’s envelope, guess with probability F(X) that this number is smaller. If the two numbers in the envelopes are A and B, then your probability of guessing correctly is equal to:

F(A) * 0.5 + (1 – F(B)) * 0.5 = 0.5 + 0.5 * (F(A) – F(B)) > 50%.

High Tide

A boat has a ladder with six rungs on it. The rungs are spaced one foot from each other, the lowest one starting a foot above the water. The tide rises with 10 inches every 15 minutes.

How many rungs will be still above the water 2 hours later?

Six rings – the ship and the ladder will be rising with the tide.

Seven Dwarfs

In one house deep in the forest, seven dwarfs are living alone. The first dwarf is reading a book, the second dwarf is cooking, the third dwarf is playing chess, the fourth dwarf is tidying up the house, the fifth dwarf is washing the clothes, and the sixth dwarf is gardening. What is the seventh dwarf doing?

The seventh dwarf is also playing chess; 2 people are needed for this.

Wizards with Hats

There are 2 wizards and each of them has infinitely many hats on his head. Every hat has 50-50 chance to be white or black, and the wizards can see the hats of the other person, but not their own. Each wizard is asked to identify a black hat on his head without looking, and they win if both succeed to guess correctly. If the wizards are allowed to devise a strategy in advance, can they increase their chance of winning to more than 25%?

Each wizard guesses the position of the lowest black hat on the head of the other wizard. Then the chance of winning becomes 1/4 + 1/16 + 1/64 + … = 1/3. It can be shown that this is an optimal strategy as well.

Rubik’s Chess

Last week we found out that Puzzle Pranks Co. have invented a new type of puzzle – Rubik’s Chess. The goal is simple – you get a scrambled cube with various chess pieces on its sides, and you must unscramble it so that on each side there is one mated King, assuming the kings cannot capture the neighboring pieces (Queens, Rooks, Bishops, kNights).

We are usually good with this type of puzzles, but we spent our entire weekend trying to solve this one without any success. We even started wondering if it can be actually solved, so decided to share it with you and see if you can help us figure that out.

Below you can see the way the cube looks when seen from 8 different angles:

Remark: The orientations of the pieces are irrelevant to the final solution, i.e. they don’t need to be consistent on each side.

The Rubik’s Chess puzzle cannot be solved. You can see a detailed solution HERE.