The first of these problems involving counting graphs. A graph is a collection of vertices with a bunch of edges connecting some or all of these vertices to each other.

A tree is an acyclic graph, i.e a graph containing no cycles.

Open question: How many trees are there with n vertices?

If we consider labelled trees then this question has any easy solution: there are labelled trees on n vertices. Counting the total number of *graphs* on n vertices is even easier. However, no closed form solution is known for unlabelled trees and this problem has been open for well over a century.

Another problem involving graphs is that of computing the diagonal Ramsey numbers. Before stating this problem, consider the following puzzle:

How many people must be present at a party to ensure that there are guaranteed to be three people present who either all know each other or are mutual strangers?

For the sake of this puzzle we’ll assume that any pair of people either know each other or they don’t (if I know you then you know me and I can’t kinda-know-you-a-bit). To make this puzzle more maths-y, we’ll represent each party guest as the vertex of a complete graph (i.e. a graph in which every vertex is connected to every other vertex). We colour each edge of the graph red or blue according as the relevant two people do or don’t know each other. Denoting the complete graph on n vertices by , the puzzle now becomes:

What is the least value of n such that however we colour the edges of in red and blue there must exist either a red triangle or a blue triangle?

It’s easy to see that n=5 is too small (just draw and you’ll easily find a suitable colouring). To see that the solution is n=6 we need to prove that whenever you colour the edges of in red and blue there must exist either a red or blue triangle. So let’s suppose that you’re given any colouring of . Pick a vertex v and consider the edges coming out of v. There are five such edges and so either at least three of them are coloured blue or at least three are coloured red. Let’s assume that at least three are coloured blue (otherwise just swap the words “red” and “blue” in what follows!).

Pick three vertices that are connected to v by blue edges and call them a, b and c. If any of the edges a-b, b-c or a-c are coloured blue then this edge, together with the edges from v, form a blue triangle. Otherwise, all three of the edges a-b, b-c and a-c are red. But in this case, these edges form a red triangle.

So we’ve solved the puzzle and the answer is “six people”. Now suppose that instead of asking for a red or blue triangle we asked for four vertices such that all of the edges between them were the same colour. i.e., instead of asking for a monochromatic copy of , we’re after a monochromatic copy of .

This question is a little harder but it’s still not terribly hard to prove that the solution is n=18. i.e., the following two facts are true:

1. Whenever you colour the edges of in red and blue there must exist four vertices such that the edges between them form a red or blue copy of .

2. It’s possible to colour the edges of such that, for any set of four vertices in the graph, there exists at least one blue edge and at least one red edge between them.

We’ll denote the answer to the first puzzle by R(3) and the answer to the second puzzle by R(4), i.e. R(3)=6 and R(4)=18. What if we ask for an n large enough that any colouring of the edges of in two colours must contain a monochromatic copy of or ? i.e., what are the values of R(5) and R(6)? No-one knows! Like the problem of counting trees, this problem has been open for decades.

Erdős asks us to imagine an alien force, vastly more powerful than us, landing on Earth and demanding the value of R(5) or they will destroy our planet. In that case, he claims, we should marshal all our computers and all our mathematicians and attempt to find the value. But suppose, instead, that they ask for R(6). In that case, he believes, we should attempt to destroy the aliens.

– Joel Spencer

Not only are these values unknown, mathematicians aren’t even close to being able to compute them.

Another open problem concerns colouring in the natural numbers (i.e. the numbers 0,1,2,3,…). It’s known that, for any colouring of the naturals in finitely many colours (e.g. colour 1 in blue, 2 in red, 3 in blue, 4 in green,…), and any n, there exists a sequence of numbers of the form a, a+b, a+2b, … , a+nb such that every number in this sequence has the same colour. In fact, it’s known that for any colouring of the naturals in finitely many colours there must exist an *infinite* sequence , , , … such that every finite sum of these numbers gets the same colour. Even stronger results are known, but I’ll not state them here as they’re quite technical. The upshot of these results is that whenever you colour the natural numbers using finitely many colours there must exist large monochromatic substructures with each of many different properties. However, these structures are all defined only in terms of addition. What if we also consider multiplication? Well, in that case we know almost nothing! The following problem is open:

Is it the case that whenever the natural numbers are coloured in finitely many colours there must exist numbers y and z such that y+z and have the same colour?

Well, actually, the answer to that question is “obviously yes, just set y=z=2”! However, if we rule out that case then the answer is unknown.

The inability of mathematicians to answer the previous question is part of a wider trend: mathematicians are generally pretty useless at solving problems that involve relating the additive and multiplicative structures of the naturals. A fundamental property that’s defined in terms of multiplication is that of being prime: a number is prime if it has no divisors other than itself and one. The prime numbers have been studied for centuries and a vast quantity of deep results have been proved about them. However, the following two problems have been unresolved for centuries:

Is every even number greater than two the sum of two primes?

Are there infinitely many numbers n such that n and n+2 are both prime?

For the next open problem, consider the following algorithm:

Given input n, set . For all , set if is even and otherwise.

Call the resulting sequence the *Collatz sequence with seed n*. What do these sequences look like? I’ve listed the Collatz sequences for the first few values of n below.

{1,4,2,1,4,2,1,…}

{2,1,4,2,1,4,2,1,…}

{3,10,5,16,8,4,2,1,4,2,1,…}

{4,2,1,4,2,1,…}

{5,16,8,4,2,1,1,…..}

It looks like all sequences eventually just alternate between the values 4, 2 and 1, and computers have verified this fact for all n less than . However, given the theme of this post, you’ll not be surprised to hear that no-one has been able to prove this fact. In fact, Paul Erdos famously said that “mathematics is not yet ready for such problems.”

The final example of an “embarrassing” open problem is one of the (possibly *the*) most important questions in maths. In fact, it’s one of the Clay Institute’s seven “millennium problems” and there’s a million dollar prize for answering it.

Suppose that you work for a sales company and wish to visit every city in some country to promote your wares. Given your limited budget, you want to know if there exists a route that visits every city (travelling by road, say) with total distance less than some fixed amount. How would you work out if such a route exists? Well, you could try every single sequence of roads that passes through every city, but it’s easy to see that this quickly becomes untenable (if there are n cities in a country, the number of possible routes grows exponentially in n). In fact, there are no known algorithms for efficiently checking whether such a route exists. However, suppose someone claimed to have found a short enough route. In this case you could trivially check their claim: just add up the lengths of the roads in their proposed route.

Returning to colouring graphs, suppose that you’re challenged to colour the *vertices* of a graph using only the colours red, blue and green in such a way that no vertices with an edge between them have the same colour. This again seems like a pretty tricky task: there are possible colourings of the vertices of a graph with n vertices and for large n this number is far too large for you to merely try all colourings. As in the previous example though, you can immediately test the correctness of any proposed solution: just colour the vertices in the proposed colours and check that no adjacent vertices get the same colour.

There are thousands of other problems from many areas in maths and industry that share a key feature with these two examples, namely that finding a solution to them appears to be very tricky but checking a solution is very easy. The P vs NP questions asks whether there are actually algorithms that quickly find solutions to *all* of these problems. In other words, is it generally true that finding solutions to problems is as easy as verifying them? The answer to this question is surely “no, obviously not”, but after decades of research by thousands of mathematicians no-one is even close to being able to prove this.

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**The paradoxes**

*Zeno’s paradox*

The simplest version of Zeno’s paradox involves a woman walking along a chalked line. She starts at one end of the line and walks until she reaches the other end. For the sake of easy numbers we will imagine that the line is one metre long, that the woman is travelling at a constant speed of one metre per second, and that the woman is called Alice. After one second Alice will have walked the length of the line. The paradox comes from dividing the distance to be covered into infinitely many intervals of decreasing length. Label each point on the line by its distance in metres from Alice’s starting point (in particular, she starts at point 0 and must reach point 1). To walk the line, Alice must first get halfway across, i.e. travel from 0 to 1/2. We will call this the first task. In the course of getting the rest of the way along the line, she must travel from 1/2 to 3/4. We call this the second task. After performing the second task, she must travel from 3/4 to 7/8 (the third task), and so on. Continuing in this way we see that Alice must perform infinitely many tasks in order to walk the line: for each natural number , she must walk from to . Zeno claims that it is impossible to complete infinitely many tasks in finite time and so, contrary to our claim above, Alice never does make it to the far end of the line. In fact, we can scale down the intervals being considered to show that Alice cannot cover any positive distance in finite time and so in fact all movement is impossible. There are some obvious objections to this paradox and nowadays it’s generally considered to be resolved. Before discussing the standard resolution in detail, I’ll introduce the other paradoxes.

*The Ross-Littlewood paradox*

The previous paradox claimed to demonstrate something about reality. The remaining paradoxes discuss situations which appear to be clearly `non-physical’ but which none the less manage to produce surprising outcomes. In the Ross-Littlewood paradox, we imagine a man (let’s call him Bob) who possesses an infinite collection of labelled balls: for each natural number , he possesses exactly one ball with the label . He also possesses a bag capable of containing all of these balls. We consider Bob to be capable of putting balls into the bag and removing balls from the bag instantaneously (it suffices for him to be able to perform these tasks arbitrarily quickly, but we’ll assume he can do them instantaneously when stating the paradox). Now imagine Bob performing the following task: at 11pm, Bob puts the ball with label 1 in the bag. After half an hour, he removes the ball labelled 1 and adds the balls labelled 2 to 10. Fifteen minutes later, he removes the ball labelled 2 and adds the balls labelled 11 to 20. After a further seven and a half minutes, he removes the ball labelled 3 and adds balls 21 to 30, and so on. i.e., we divide the time between 11pm and midnight into infinitely many intervals so that the first interval is of length 30 minutes and the st interval is always half the length of the th interval. At the end of each interval Bob removes the ball with smallest label from the bag and from among the balls which have not yet been in the bag adds those with the next ten smallest labels. As the time approaches midnight, the bag gets increasingly full of balls (every time Bob removes a ball he adds ten more). However, when we reach midnight the bag is empty. To see this, consider the ball with label . At the end of the th interval, this ball is removed from the bag and never returned. Thus none of the balls remain in the bag at midnight.

*Thompson’s lamp*

For this paradox we assume that a man (Thompson) possesses a lamp that he is capable of switching on or off instantaneously (as with the Ross-Littlewood paradox, `arbitrarily quickly’ will suffice). At 11pm the lamp is off. At 11.30pm, he turns the lamp on. At 11.45pm he turns the lamp back off and he continues turning the lamp alternately on and off using the same time intervals as Bob did in the previous paradox. When we reach midnight, is the lamp on or off? There appears to be no definite answer to the question. Any time before midnight at which the lamp is turned on is followed by another time before midnight at which the lamp was switch off, and vice versa. Hence our completely deterministic process appears to have produced an indeterminate outcome.

*Zeno’s balls*

The final paradox involves a woman (Zeno) in possession of an infinite set of balls. For each she possesses a ball of radius of mass 1kg (the precise diameter doesn’t matter; we just need the balls to get smaller sufficiently quickly). We suppose Zeno possesses a frictionless table placed in a room with frictionless atmosphere and that when the balls are rolled at each other along the surface of this table they collide elastically. Zeno picks a point on the table (`the origin’) and labels it 0. In some fixed direction, she then puts labels at distance metres from 0 for . For each she places the th ball on the point labelled . Having done all this, she sets her timer to zero and pushes the first ball ball towards the origin, giving it a speed of 1m/s. The first ball then collides with the second ball, coming to a halt and transferring all of its energy. The second ball then rolls until it hits the third ball, which rolls until it hits the fourth ball, and so on. After one second, none of the balls are moving: for any , the th ball collided with the st ball at time strictly less than one second and came to a halt. Thus energy is not conserved in Newtonian mechanics: at time zero the first ball has positive kinetic energy, while at time one all of the balls are stationary. Running this process in reverse, we have a row of balls that spontaneously start moving; now we’ve not only broken energy conservation, we also have an effect without a cause!

**Some `resolutions’**

All of the paradoxes centre on the issue of performing infinitely many tasks in finite time. Zeno’s paradox asserts that this is not possible and claims to deduce that movement is impossible, while the others ask us to consider what happens *after* performing some specified infinite collection of tasks.

Since Weierstrass’s definition of limit, and subsequent basic results on infinite series, Zeno’s paradox has largely been considered to be resolved (although this opinion is not universal). The generally accepted solution is that infinitely many tasks can indeed be completed in finite time if the time taken for each task decreases sufficiently quickly. Completing the first task takes Alice half a second, completing the second takes a quarter of a second and so on: in general, completing the th task takes seconds. Thus Alice will complete the first tasks in seconds. As tends to infinity this sum tends to 1. Alice thus really can complete infinitely many tasks in unit time and she defeats Zeno. However, it’s worth pausing for a moment to consider this argument more carefully. What role does the technical machinery of series and limits play in it? On second glance, not an awful lot. The sum given above doesn’t tell us anything we didn’t already know; you don’t need a lot of maths to work out that when travelling at 1 metre per second it takes seconds to walk metres! And you certainly don’t need to use any technical machinery to calculate that it should take Alice one second to walk the entire line. So why do most people treat the machinary of limits and series as having `resolved’ the paradox? It appears that the crucial role played by Weierstrass’s mathematical innovation is demystifying the notion of infinity. If you’re used to the idea of taking limits then it seems natural to state that Alice’s position at time 1 is the limit of her positions at times and the division of her task into infinitely many diminishing subtasks no longer appears problematic. The resolution to Zeno’s paradox is then just to realise that there doesn’t seem to be any good reason to deny that infinitely many tasks can be performed in finite time. The paradox is merely a linguistic trick. The remaining paradoxes however, appear to be less clear-cut.

Now consider the Ross-Littlewood paradox. The main issue with this paradox is that what happens in the limit (i.e. at time midnight) is not well-defined. We are told the state of Bob’s bag after any finite number of intervals (and so at any time before midnight), but the set-up doesn’t actually tell us what happens after an infinite number of intervals (i.e. at midnight). Let be the characteristic function of Bob’s bag at stage . i.e., equals 1 if the ball labelled is in Bob’s bag at the end of interval and 0 otherwise. Let be the characteristic function of Bob’s bag at midnight. The argument that Bob’s bag eventually contains no balls amounts to saying that should be chosen as the pointwise limit of the , i.e. that for each , . However, there doesn’t seem to be any particular reason to choose to take the limit of the in the norm of pointwise convergence. We could instead decide that the number of balls in the bag at midnight should be the limit of the number of balls in the bag at the end of each interval. If we adopt this approach then we will decide that the bag should contain infinitely many balls at midnight. One or the other of these arguments might appeal to you more. However, in this paradox I think the set-up leaves us with a genuine choice as to which limiting process we want to use to define the state of the bag at midnight. The paradox then amounts to the observation that when describing infinitary processes we should make sure that we carefully specify what happens when we take limits.

The case of Thompson’s lamp is a little different again. In this scenario, there is no sensible way to take a limit of the state of the lamp. Unlike the Ross-Littlewood paradox we don’t have a choice of which limit to take. Instead, we can only reasonably conclude that the information given tells us nothing about the state of the lamp at midnight. I don’t find this particularly troubling, as the ability to toggle a current on or off infinitely often in a finite period of time is clearly nonphysical (contrast this with Alice’s walk; her `tasks’ become increasingly trivial while Thompson’s switching is just as tricky every time he does it). Our intuition that the lamp should have a definite state at midnight and that its state at all earlier stages should tell us this is based on our experiences with normal, finitary processes. In this case though, the talk of a man and his lamp is merely a way of tricking our intuition into treating the scenario as being somehow physical. The claim that not all sequences have limits is neither surprising nor mysterious. As with Zeno’s paradox then, it appears that this paradox dissolves when considered more carefully.

Finally, let’s consider the paradox of Zeno’s balls. An obvious first observation here is that we shouldn’t be unduly disturbed if it turns out that a model designed to predict the motion of finite numbers of sensibly behaved physical bodies gives screwy results when applied to a question involving an infinite number of objects of arbitrarily small diameter and arbitrarily high density. After reading my discussion of the previous paradoxes, you shouldn’t be too surprised to hear that my `resolution’ to this paradox is again that there’s nothing to resolve. Unlike the Ross-Littlewood paradox, we don’t appear to have a choice here as to which metric to take limits in. The velocities (and hence kinetic energies) of the balls tend pointwise to zero, while the kinetic energy of the whole system is constant for all times less than one, so I suppose we could decide that energy should change continously and conclude that the kinetic energy of the system is indeed conserved at time one. However, this invites the obvious question of where that kinetic energy lies! It seems ludicruous to try and claim that a collection of stationary balls somehow possesses non-zero kinetic energy. I think a better solution is just to say that this system is non-physical and so we shouldn’t be surprised that it doesn’t behave according to our physical intuitions. We’re perfectly free to decide that the state of the system at time one is indeed the pointwise limit of the states at earlier times, but we shouldn’t read much of anything into this.

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Parrot sings “Let the bodies hit the floor”

Mandatory minimums increase drug purity

The Alot is better than you at everything

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On first glance, this looks pretty tough for Bob: Alice has infinitely many choices, and isn’t even constrained in the degree of the polynomial she chooeses. However, there’s a really easy winning strategy for him: first ask for p(1) and then ask for p(m) for any m greater than p(1). That’s it!

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The continuum hypothesis (CH) is the claim that , or equivalently that there are no cardinalities intermediate between the cardinality of the naturals and the cardinality of the reals. It’s well known that this is independent of ZFC, and this is often taken to mean that its truthhood is “unknowable”, or that we are free to assume CH or its negation as we please. In a sense this is true. However, it is worth stressing that there is nothing inherently special about the ZFC axioms. We have good reasons for believing that they are consistent and they are powerful enough to formalise most of mathematics in, but just as choice and replacement were added for pratical purposes it is reasonable to suppose that we might produce other useful axioms that we wish to add that appear “plausible”. The study of large cardinal axioms is a well established branch of set theory that considers various strong axioms that could be added to ZFC (“strong” in the sense of consistency strength; all large cardinal axioms prove Con(ZFC)) that are motivated by various heuristics or desired consequences. However, none of the current large cardinal axioms settle the continuum hypothesis.

We now introduce the equivalent statement to not CH. The statement seems to me to be very intuitive, but I’m not sure if this is really just because my intuition of how infinite sets behave is a bit suspect (I am also somewhat drawn to the axiom, and clearly I’m not allowed both!).

Denote the unit interval by . Given a set and , define to be the set of all , i.e. is the “row” of at height . We say that is *thin* if for every , is countable. Given , define the *reflection* of , denoted , to be the set of all such that (i.e. the reflection of along the line ). Intuitively, it seems that taking the union of any thin set with its reflection should never give us all of ; thin sets “don’t fill up very much of “. In fact, it’s easy to show that this claim is equivalent to the continuum hypothesis being false.

**Theorem **

The negation of the continuum hypothesis is equivalent to there being no thin set with .

**Proof**

Suppose that CH is true and let be a bijection. Define a set as follows: for each , define to be the set of all such that . Each is clearly countable (by definition of !) and .

Now suppose that CH is not true and let be any thin subset of . Let be a set of cardinality . Let . so there exists such that . As is countable there exists some . So by our choice of and by our choice of .

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Medic discovers the trapezium rule, gets 75 citations

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