This is a hard riddle – it took me a couple of days (wall-time) to solve, but it was definitely worth it. Be warned – a certain mathematical maturity is needed…

There are three prisoners and a guard. The guard has an infinite sequence that is either all 1′s (e.g. 1111111111…) or it starts with a finite number of 1′s and eventually turns into 2′s (e.g. 1112222222…). The guard creates 3 bit sequences (i.e. composed entirely of 0′s and 1′s) such that the sum of the i’th bit in the 3 sequences is equal to the i’th element in his sequence (1 or 2).

He then gives each of the 3 prisoners one of the 3 bit sequences. Each prisoner sees only his own sequence and has to guess whether the guard’s sequence consists only of 1′s or whether it turns into 2′s eventually. Devise a method that will make sure the majority of the prisoners (i.e. at least 2) guess correctly.

Thanks Haran for giving me this cool riddle!

A cute and easy algorithmic riddle.

You have an array of N bit strings each of length M. You know that there is at least one element that appears more than N/8 times in the array. Using O(M+log(N)) memory and O(NM) time, find such an element.

An Easier Version

Well, its actually almost exactly the same, but solve the above riddle in case there is an element that appears more than N/2 times in the array. I managed to find 3 distinct solutions to this easier variation, but only one of which generalizes easily.

Thanks Nemo for giving me this riddle!

The game of Reverse & Clean is a one player game, played on an N by M board, filled with Reversi pieces. Each Reversi piece has a black side and a white side. The game starts with all pieces showing their white side, except for the piece in the lower right corner, which shows its black side. See the figure for a 3 by 3 example.
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There are 10 cells in a line. A transparent rabbit is in one of them. You have a shotgun, and obviously you want to shoot the rabbit.

If you hit the cell with the rabbit, you kill him (and win). Otherwise, if you shoot an empty cell, the rabbit hears the shot, gets scared of the noise and jumps one cell to the right or one cell to the left. In case the rabbit is in the right-most cell, it can only jump to the left (and similarly, if the rabbit is in the left-most cell, it jumps to the right).

Can you kill the rabbit? If so, what is the minimum number of shots needed to guarantee a kill?

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Two aunts are living each in her own (0-dimensional) house. There are two non-intersecting (1-dimensional) roads between the houses.

Last year, both aunts were doing a lot of exercise, and so they were slim (0-dimensional). They managed to walk together from House 1 to House 2, taking different roads, while each was holding one end of a rope of length less than L.

This year, they gained weight, and each became a sphere of radius L/2. One aunt is in House 1 and the other is in House 2. Can they exchange houses without bumping into each other (their centers must always remain on the roads)?

You have a set of 2N+1 natural numbers, with the following property: if you remove any one element, you can partition the remaining 2N elements into two sets A and B, each of size N, such that the sum of the N numbers in set A equals the sum of the N numbers in set B. Prove that all the numbers in the original set are equal.

Try to solve the riddle in the more general case, where the numbers are not necessarily natural, but arbitrary reals (some knowledge of algebra is helpful here).

For inspirational music, click here.

Il Buono (nice and easy)

Prove that in a subset of size n+1 of the set {1,2,…,2n} there are two numbers such that one divides the other.

Il Cattivo (beautiful and hard)

Prove that in a sequence of r*s+1 distinct numbers, there is either a monotonically increasing sub-sequence of length r+1 or a monotonically decreasing sub-sequence of length s+1.

Il Brutto (just easy)

Prove that in a subset of size n+1 of the set {1,2,…,2n} there are two relatively prime numbers (i.e. numbers whose gcd is 1).

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This riddle is my take on the Monty Hall problem. If you know the original version, this one should be very easy for you.

You are watching the TV series “Lets Make a Deal”.

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This riddle is a very cool extension of a well known (and easy) riddle, involving people with hats waiting in a line.

So, lets begin with the original:

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I haven’t written new posts for a while now, as I have been very busy lately.

I think this is a very interesting post and I hope it will make up for the lack of updates. I also want to take this opportunity to thank all of you for posting lots of interesting comments and for sending me many ideas and riddles, thank you!

In this post (and its sequel) I will describe Hilbert’s 3rd problem and show how it is solved. I based the posts mainly on a lecture by Prof. David Gilat.

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