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!

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!

In each turn of the game, you remove one of the black pieces and flip all of its (remaining) neighbours (the pieces up, down, left and right of it). The figures below demonstrate 2 possible first moves (note that the very first move of the game is always to remove the bottom right piece, which is the only black piece).

The goal of the game is to clean the board (i.e. remove all pieces). Note that if the board is 1 by N, you can always win, while if the board is 2 by 2, winning is impossible (make sure you see why).

For what numbers, N and M, is a win possible?

Extra Credit: What about 3-dimensions? D-dimensions?

Thanks to Liron Raz for giving me this one.

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?

Extra Credit

Spoiler Warning – read after solving the riddle above!

Instead of considering the cells in a row, the riddle can be generalized to a graph.

If the graph has cycles, no solution exists (make sure you see why!).

What happens if the graph is a general tree?

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)?

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).

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).

Il Non Collegato (another easy one)

Prove that a degree 7 polynomial with integer coefficients, that receives the values +1 or -1 on 7 integer points is irreducible over the integers.

Unusually, I decided to omit the second page, so post your solution related comments below.

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

You are very excited as you know your friend Heidi is participating today. You watch attentively throughout the show, only to find out that it ends without Heidi ever appearing in it.

You call Heidi, several times during the following day, but you keep getting her answering machine. Maybe something happened to her?

You recall the Rules of the Show:

The host presents to each candidate three closed doors. Behind one of these doors hides a big prize, behind the other two there is an empty bucket. The candidate chooses one of the doors (this choice is random as he has no information whatsoever on the location of the prize). Then the host opens one of the other doors behind which there is no prize. Then the candidate is given the option to remain with his initial choice or to switch doors and choose the other closed door. Then the remaining doors are opened, and the candidate is awarded whatever is behind the door finally chose. The goal of the candidate is (obviously) to choose the door with the prize.

Pondering about the rules, you develop the following theory:

Maybe the host does not know in advance which of the doors contains the prize. In that case, if the host happens to choose to reveal the door with the prize (thus ruining the show) the candidate is killed (finally, some proper riddle scenario) and his scene is cut-out during the editing of the show.

Maybe that is what happened to Heidi?

Assuming you have a lot of episodes (as many as you need), how can your theory be ruled out or strengthened?

So, lets begin with the original:

The Original Riddle

100 men are standing in a line, such that each of them sees all those that are in-front of him (so the last man sees the 99 others, etc.). Each guy has a hat on his head colored either black or white. Each of the men guesses the color of his own hat, out-loud. The guy standing last (the one that sees all the other ones) guesses first, then the guy in-front of him, etc. The men’s task is to make sure no more than one of them is wrong.

Just to make things explicit – the men are allowed to agree upon an algorithm before actually being given the hats, but once the task starts they cannot communicate in any way other than the fact that each of them hears the guess of the guys standing behind him.

Also, to conform to the universal standards of riddles, note that should more than one man be mistaken, they will all be brutally killed.

If you do not know this riddle yet, take a minute to think about its solution before reading on.

The Solution of the Basic Riddle

I give here the solution of the basic riddle so that there are no misunderstandings regarding the much more interesting extension. It goes like this – The last man (the one that guesses first) sacrifices himself (with a probability of 0.5). He regards each black hat as a 1 and each white hat as a 0. He then guesses out-loud that he has a black hat if the sum of all the hats of the other people is odd, and guesses white otherwise. The other people can all guess correctly (make sure that you understand why).

Some Trivial Extensions

Obviously the numbers 100 and 2 (the number of possible hat colors) in the original riddle are completely arbitrary. If m,n are two natural numbers then the riddle with m men and n possible hat colors is a trivial generalization of the previous result.

Now, note that if we let n, the number of possible hat colors, be equal to infinity (well, aleph 0, the cardinality of the set of natural numbers) then again we have a trivial generalization (make sure you see why).

The Real Deal

Now the extension of the riddle we are interested in consists of letting m, the number of people, be equal to aleph 0. For simplicity, lets assume that n equals two (the hats are either black or white).

Is the riddle still solvable? I.e. can the men devise an algorithm such that there will be only one mistake?

As usual, post solution related comments to the second page.

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.

Lets begin by considering a polygon in the Euclidean plane. Here is an example of such a polygon:

We want to devise a method for somehow “measuring” the area of this and similar polygons.

In order to do this, we begin by defining the area of a rectangle that is parallel to the coordinate axes as the product of its horizontal side by its vertical side.

We now define the area of a polygon to be the infimum of the sum of all the areas of sets of rectangles that constitute a covering of the polygon. An example of such a covering is depicted here: 

This is the Lebesgue measure of the polygon. This area happens to have some very desirable properties:

• It is invariant to translations (i.e. a polygon’s area will remain the same if it is moved to a different location in the plane).
• It is invariant to rotations (i.e. a polygon’s area will remain the same if it is rotated).
• It is invariant to reflections (i.e. a  polygon’s area will remain the same if it is reflected over any straight line).
• It has the finite-additivity property (explained below).

I will not prove these properties here as it is very easy and somewhat tedious to do so.

Lets focus on the last property, namely finite-additivity. Finite-additivity means that if we take two polygons that are disjoint (actually, that have disjoint interior) and regard their union as one new polygon, then the area of the new polygon will be equal to the sum of the areas of the two original polygons.

Lets call two polygons, P and Q, congruent in parts, if we can divide P to polygons p1,…,pn and Q to polygons q1,…,qn such that pi can be obtained from qi by a finite number of translations, rotations and reflections.

The properties of area listed above, give the following important (though a little obvious) conclusion:

Theorem 1: Two polygons P and Q that are congruent in parts have the same area.

It suffices, by finite-additivity, to show that the areas of the polygons pi and qi are the same for all i. But by the invariance of the area under rotations, translations and reflections, this is obvious.

Now an interesting question arises: Given two polygons P and Q that have an equal area, are they congruent in parts?

The question can be given an affirmative answer. Lets give an outline of the proof:

1. Every polygon can be divided to finitely many triangles. This is obvious in the case of a convex polygon. In the case of a concave polygon, select any diagonal that lies completely inside the polygon. This diagonal divides the polygon into two polygons each with fewer sides than the original. By induction we can divide the polygon into finitely many triangles. It still remains to be shown that we can always find a diagonal that lies completely inside our polygon, but that is not hard (I will leave this as an exercise to the reader ).
2. Every triangle is congruent in parts to a rectangle (see illustration below).
3. Every rectangle is congruent in parts to a square.
4. Two squares are congruent in parts to a square whose area is the sum of their areas.

An illustration showing that a triangle is congruent in parts to a rectangle follows:

We have shown that every polygon is congruent in parts to a square of the same area. We thus have the inverse of Theorem 1, namely,

Theorem 2: Two polygons that have the same area are congruent in parts.

On to three dimensions. We define a polyhedron’s volume (a polyhedron is the three dimensional analogue of a polygon), it’s Lebesgue measure, in a completely analogous way to the two dimensional Lebesgue measure. Given a box whose facets are parallel to the coordinate plane we define its volume to be the product of three of its perpendicular sides. Given any polyhedron we now define its volume as the infimum of the sum of the volumes of all the boxes in a covering of the polyhedron, where the infimum is taken over all the countable coverings of the polyhedron by boxes with facets parallel to the coordinate planes.

We say that two polyhedrons are congruent in parts if we can divide them into polyhedrons p1,…,pn and q1,…,qn such that the pi’s have disjoint interiors (and similarly for the qi’s) and such that qi can be obtained from pi by translations, rotations and reflections (around planes).

The three dimensional Lebesgue measure has the same properties as the two dimensional Lebesgue measure mentioned above. Thus Theorem 1 can be generalized to:

Theorem 3: Two polyhedrons that are congruent in parts have the same volume.

It was conjectured by Gauss that Theorem 3′s inverse (the generalization of Theorem 2 to three dimensions) does not hold. This question later became known as Hilbert’s Third Problem (see Hilbert’s Problems).

Lets state it again, for clarity:

Hilbert’s Third Problem: Are two polyhedrons that have the same volume necessarily congruent in parts?

One might try to answer the question affirmatively by employing a method similar to the one we used for Theorem 2. I.e. find a “generic” 3D polyhedron (the 3D analogue of the 2D triangle) and divide the polyhedrons to finitely many such “generic” polyhedrons. The problem is that, contrary to the 2D triangle, the existence of such a “generic” polyhedron is not obvious.

The problem was indeed given a negative answer by Hilbert’s student Max Dehn. Dehn managed to construct a function D of polyhedrons that remained invariant under congruence in parts (i.e. two polyhedrons that are congruent in parts produce the same value of D). He managed to show that the values of D on the regular tetrahedron and on the cube are different even if they have the same volume.

Dehn’s invariant, D, is defined as follows:

D(P) = Σl(m)f(q(m))

Where the Σ is taken over all edges m of the polyhedron, l(m) denotes the length of the edge, q(m) denotes the angle between the two planes formed by the facets m connects, and finally f:R->Q is a linear transformation (from the real line into the field of rational numbers, where R is regraded as an infinite dimensional vector space over Q) with the following properties:

1. f(1) = 0
2. f(qx) = qf(x), q in Q, x in R
3. f(x+y) = f(x) + f(y)
4. f(pi) = 0
5. f(x) != 0, if x is not linearly dependent on pi and 1 (again, regarding R as a vector space over Q).

Cautious readers might notice that the existence of such a function depends on the axiom of choice (as it relies on the possibility of constructing a base for R, when regarded as a vector space over Q). Such a construction is not really necessary, i.e. the theorem does not depend upon the axiom of choice. I leave the matter of arranging a finite dimensional vector space over Q that is sufficient for the proof as an exercise to the reader.

In the next installment of the series I will prove that Dehn’s invariant is indeed an invariant (under congruence in parts), and calculate its value for the cube and for the tetrahedron, thus finishing the proof.

To Be Continued…