Steven greeted me with a puzzle when I entered the office today. I got it after thinking a bit, though I definitely have seen this a long time ago:

Given 51 numbers between 1 and 100, show that you can find two of them that are relatively prime

Proof:

By pigeonhole, we must get at least two numbers of the form {x, x+1}, where x is odd. These two are relatively prime.

Steven then gave me a strictly harder followup. This took me a bit longer, and afterwards I was very happy to have come up with an unorthodox solution, and hence a new post:

Given 51 numbers between 1 and 100, show that you can find two of them such that one divides the other.

Proof and analysis:

Start by putting every number between 1 and 100 into a bucket. Now, starting with 100 and moving down to 2, move *all* the numbers in the bin of every even number of the form into the bucket where is. Each time we do this, a bucket becomes empty. So at the end we have 50 buckets. By pigeonhole, we must get at least two numbers in the same bucket. The quotient of those two numbers is a power of two, so we’re done.

Note I could have just said something like: “for every odd number , put all numbers of the form into the same equivalence class” and then counted equivalence classes. However, it seems that in this language, counting equivalence classes requires at least a couple of steps of thought. For some unknown reason, by using this handwavy “bucket” talk, the number of equivalence classes is obviously 50, whereas by using the more rigorous language above, I don’t see an immediate way to get the answer at quite the same speed.

People used to competitions would probably recognize the first problem as an old chestnut; this is actually a puzzle Uncle Erdos used to test mathematical ability in young kids (‘epsilons’). I don’t know where the second problem comes from.

Happy Thanksgiving,
-Yan

Today we have a M-2… but it feels more like a M-3 or M-4. I’ll explain after the two problems.

The first problem is fairly classical. It appears in many texts, and was first brought to my attention by Chris Evans:

N people stand in line, all facing right, where each person can only see the people in front. They each get a hat (red or blue), and in order (from left to right), they are asked to guess the color of their own hat. Find a strategy to maximize the number of people who guess their hat colors correctly. Note that the strategy can depend on what people further right hear from the guesses of people further left.

Solution:

The funny thing about this problem is that you can ask it as a stronger statement (“find a strategy so that N-1 people guess their hat color correctly”), which makes the problem a lot easier =) This is why I ask the problem in the above form.

Now that I’ve said that, you may want to take a moment to try to do it again.

Give up? Do this: Think of a red hat as a 0 and a blue hat as a 1. The first person guesses the color corresponding to the sum of everyone else (mod 2). Note that this gives each person after him enough information; the second person can just subtract off what he sees from what the first person says. The third person, after hearing the second person’s guess, knows the sum of all hats after and including the third, so he can just subtract off what he sees to get his own hat. This gets at least N-1 right, and half of the times gets all N right. Note we can’t possibly do anything better since the first person can only ever get his own hat right with probability 1/2.

I heard the second problem recently from Nick Rosenbloom and Dustin Clausen:

Countably infinite people stand in line, all facing right… the rest of the setup is the same. Find a strategy to maximize the cardinality of people who guess their hat colors correctly.

Sketch:

The awesome thing about this problem is there is also a way to ask this problem that hints at the answer. What is slightly different about this formulation of asking the problem is that it sounds like a strictly harder problem has been asked: one can restrict the problem so that the strategy cannot rely on hearing what people left of each person answers.

This looks kind of impossible, which is perfect for things like the Axiom of choice (or equivalently, Zorn’s Lemma). Consider all countably infinite binary strings and divide them up into equivalence classes, where two strings A and B are equivalent if they share a truncation (i.e. the bits after the k_th bit of A are equal to the bits after the l_th bit of B). Note it takes a bit of work to show that this is an equivalence class, which I’ll leave as an exercise. You might want to do the case of the string being periodic after a certain point differently from the case where it is not.

Now, just pick a representative from each equivalence class. I leave it as another exercise to see that you are done =) It basically involves each person to try to fit himself into a string which has the same equivalence class at what he sees. The earliest people may get this wrong, but the later people (once you get to the common truncation) will get them all right. So you get all but finitely many guesses correct.

I think the coolest thing about this set of problems is the way you can rephrase each one and still ask the problem at the same strength. Note this becomes a M-3 if there was a solution to the second problem that breaks the restriction given at the beginning of the proof. So if any of you have ideas, let me know!

-Y

Source: Chris Evans, Nick Rosenbloom, Dustin Clausen

Here is another Matryoshka problem – well not really, since I can’t prove that the second problem is “strictly harder” than the first in any meaningful way, but it does have the property that the most intuitive trick needed to solve the first one fails on the second one:

Timothy Gowers posted a fairly classical “paradox” in probability involving two boxes:

original post

My analysis:

Note that the guy who gives you the box must pick (a, 2a) with equal probability for every a – i.e., he needs to be able to pick a real number uniformly. Unfortunately, this is impossible.

This is actually a common sleight of hand that many probability “paradoxes” use at some level. You can increase your “defense against probability paradoxes” by adding this to the list of things to check against every time you find a strange probability situation, in the same way that recognizing Russell’s Paradox counters many logic “paradoxes.”

—

Now the inner doll. Prof. Gowers attributes this one to Noga Alon:

original post

These two problems are similar in spirit. However, the trick that works for the first one no longer works here: we have a very well-defined distribution! So we have to dig a little deeper. Here is my take:

In the paradox, we considered the random variable of the values of the two envelopes (call them and , and proceeded to divide the probability space into a series of sums x_i (where ), seeing that for all i. Then we did an implicit sum of all of these equations to conclude that .

The problem is in the sum – we cannot do this summation when the expectation of X and Y are infinite (we can easily check, by the way, that and are both infinite), so we cannot conclude that one is larger than the other. Here is a clearer example: consider the list of inequalities , , … each one of these statements is true, but it makes no sense to then argue that , because the sums diverge and cannot be defined. If you track the list of inequalities in the problem, you will notice the exact analogous chicanery going on, just with different numbers.

The cool lesson form this M-2 pair is that I am reminded again to appreciate why math books always mention “boundary cases” like infinity and deal with them very carefully. Usually we see them as just stupid limitations for corner cases, but this is a case where not thinking about the clauses on these border cases can mislead (to be honest, I never remember these boundary cases, like the algebra theorems that hold for every ring except the zero ring).

By the way, I bet there is some way to turn each one of these into a “paradox” in their respective fields. Probability is just riper for these conundrums since the problems offer more intuitive statements.

-Y

Source: Gowers’ Weblog

As I stated, we’ll start with something easy:

We put in a grid. Call two squares adjacent if they touch either on a corner or on a side. Show that there are two adjacent numbers which differ by at least .

Proof and analysis:

Look at 1 and n^2. They differ by at most n-1 steps (where a ’step’ is a chess King-move), which means their average gap is (n^2 – 1) / (n-1) = n+1 – so at least one step must actually take this value or more.

The concepts are simple – the pigeonhole principle and the “strip away irrelevant information” metatechnique. Being “adjacent” is a weird thing to quantify, but an idea of “distance” is not – so we turn things to “distance.”

I saw the first problem a really long time ago (elementary/middle school?) when I was practicing problem-solving, so I don’t remember exactly where it was from. So I was delighted to see the second problem in Bela Bollobas’s The Art of Mathematics:

We put in a grid. Call two squares adjacent if they touch on a side. Show that there are two adjacent numbers which differ by at least .

Proof and analysis:

For each row and column of n elements, there is a maximum element. Find the smallest of these (possibly overlapping) maximum elements, and call it k. Note that for each other row, there is at least one ’small’ number less than k (namely the element in its column) but at least one ‘big’ number greater than k (namely the maximal element in that row). Since every number is one of these two types, there exists at least two adjacent numbers in that row where one is ’small’ and the other is ‘big’.

Since there are n-1 of these pairs, at least one of the pairs must differ by n (this is because at least one number must be less than or equal to (k-n+1), and the smallest ‘big’ number it can pair with is (k+1), which already gives the difference n), so we are done.

This is very elegant, surprisingly nonconstructive, but shows the cool things you can do with seemingly arbitrary combinatorial constructions. Okay, I admit I cheated, since this problem is not “strictly harder” than the first one (it would if we changed “n+1″ to “n” in the first problem, which loosens it and doesn’t really change it much). But whatever, let’s stop being picky mathematicians.

True to the nature with their obsession with sets, combinatorialists like to collect stuff assorted by theme. Thus, we have Richard Stanley’s collection of Catalan-number-enumerated sets, Bridgit Tenner’s collection of permutation-avoidance occurrences, and of course… this. One day I’ll make a collection of these collections just for the sake of having a metacollection, but until then my main collectible hobby will be of Matryoshka problems, a term I just made up today.

Like its namesake, a “matryoshka problem” is not a single problem but a family (a “chain” really – although in practice the length is usually 2) of problems that increase strictly in difficulty. Then, as I am bewildered at discovering a smaller doll inside another, I realize I can push the boundaries of what I can prove.

Of course, this nested structure is not rare – in fact, it occurs very naturally in mathematics. Why? Because mathematics come from theory-building and generalization. Usually, some real-world (I use this term loosely) problem needs to be solved, so some smart person (the other mathematicians) goes ahead and generates just enough math to solve that. Then we want to generalize – say from to Hilbert spaces, and then from Hilbert Spaces to Banach spaces… and we naturally get a nested sequence of problems.

Usually, these families all have solutions which are at heart the same (just a little harder technically at each step) – like nested dolls who look exactly alike. For me, the fun comes when the techniques used to solve each level of the problem are completely different. It is great practice for think ing outside the box, because doing the “easier” versions of the problems tend to trap me into modes of thinking that makes it harder to do later problems.

To start this series, I’ll just do a pair of problems (again, most of my posts will probably end up being pairs, so I’ll call them “M-2 problems“). This one isn’t particularly nice, but I hope it demonstrates the spirit of what I’m trying to do (as a warning, since I give these in tuples, it is tempting to move through the “easy” ones quickly. But I think a lot of the fun and appreciation of these problems come from having thought about and being able to appreciate the “easy” versions of the roblems before going ahead, so keep this in mind).

-Y