The Busy Beaver Game

21 May, 2016

This month, a bunch of ‘logic hackers’ have been seeking to determine the precise boundary between the knowable and the unknowable. The challenge has been around for a long time. But only now have people taken it up with the kind of world-wide teamwork that the internet enables.

A Turing machine is a simple model of a computer. Imagine a machine that has some finite number of states, say N states. It’s attached to a tape, an infinitely long tape with lots of squares, with either a 0 or 1 written on each square. At each step the machine reads the number where it is. Then, based on its state and what it reads, it either halts, or it writes a number, changes to a new state, and moves either left or right.

The tape starts out with only 0’s on it. The machine starts in a particular ‘start’ state. It halts if it winds up in a special ‘halt’ state.

The Busy Beaver Game is to find the Turing machine with N states that runs as long as possible and then halts.

The number BB(N) is the number of steps that the winning machine takes before it halts.

In 1961, Tibor Radó introduced the Busy Beaver Game and proved that the sequence BB(N) is uncomputable. It grows faster than any computable function!

A few values of BB(N) can be computed, but there’s no way to figure out all of them.

As we increase N, the number of Turing machines we need to check increases faster than exponentially: it’s

(4(n+1))^{2n}

Of course, many could be ruled out as potential winners by simple arguments. But the real problem is this: it becomes ever more complicated to determine which Turing machines with N states never halt, and which merely take a huge time to halt.

Indeed, matter what axiom system you use for math, as long as it has finitely many axioms and is consistent, you can never use it to correctly determine BB(N) for more than some finite number of cases.

So what do people know about BB(N)?

For starters, BB(0) = 0. At this point I should admit that people don’t count the halt state as one of our N states. This is just a convention. So, when we consider BB(0), we’re considering machines that only have a halt state. They instantly halt.

Next, BB(1) = 1.

Next, BB(2) = 6.

Next, BB(3) = 21. This was proved in 1965 by Tibor Radó and Shen Lin.

Next, BB(4) = 107. This was proved in 1983 by Allan Brady.

Next, BB(5). Nobody knows what BB(5) equals!

The current 5-state busy beaver champion was discovered by Heiner Marxen and Jürgen Buntrock in 1989. It takes 47,176,870 steps before it halts. So, we know

BB(5) ≥ 47,176,870.

People have looked at all the other 5-state Turing machines to see if any does better. But there are 43 machines that do very complicated things that nobody understands. It’s believed they never halt, but nobody has been able to prove this yet.

We may have hit the wall of ignorance here… but we don’t know.

That’s the spooky thing: the precise boundary between the knowable and the unknowable is unknown. It may even be unknowable… but I’m not sure we know that.

Next, BB(6). In 1996, Marxen and Buntrock showed it’s at least 8,690,333,381,690,951. In June 2010, Pavel Kropitz proved that

\displaystyle{ \mathrm{BB}(6) \ge 7.412 \cdot 10^{36,534} }

You may wonder how he proved this. Simple! He found a 6-state machine that runs for

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13052411058118309311147944260478608

steps and then halts!

Of course, I’m just kidding when I say this was simple. The machine is easy enough to describe, but proving it takes exactly this long to run takes real work! You can read about such proofs here:

• Pascal Michel, The Busy Beaver Competition: a historical survey.

I don’t understand them very well. All I can say at this point is that many of the record-holding machines known so far are similar to the famous Collatz conjecture. The idea there is that you can start with any positive integer and keep doing two things:

• if it’s even, divide it by 2;

• if it’s odd, triple it and add 1.

The conjecture is that this process will always eventually reach the number 1. Here’s a graph of how many steps it takes, as a function of the number you start with:

Nice pattern! But this image shows how it works for numbers up to 10 million, and you’ll see it doesn’t usually take very long for them to reach 1. Usually less than 600 steps is enough!

So, to get a Turing machine that takes a long time to halt, you have to take this kind of behavior and make it much more long and drawn-out. Conversely, to analyze one of the potential winners of the Busy Beaver Game, people must take that long and drawn-out behavior and figure out a way to predict much more quickly when it will halt.

Next, BB(7). In 2014, someone who goes by the name Wythagoras showed that

\displaystyle{ \textrm{BB}(7) > 10^{10^{10^{10^{10^7}}}} }

It’s fun to prove lower bounds on BB(N). For example, in 1964 Milton Green constructed a sequence of Turing machines that implies

\textrm{BB}(2N) \ge 3 \uparrow^{N-2} 3

Here I’m using Knuth’s up-arrow notation, which is a recursively defined generalization of exponentiation, so for example

\textrm{BB}(10) \ge 3 \uparrow^{3} 3 = 3 \uparrow^2 3^{3^3} = 3^{3^{3^{3^{\cdot^{\cdot^\cdot}}}}}

where there are 3^{3^3} threes in that tower.

But it’s also fun to seek the smallest N for which we can prove BB(N) is unknowable! And that’s what people are making lots of progress on right now.

Sometime in April 2016, Adam Yedidia and Scott Aaronson showed that BB(7910) cannot be determined using the widely accepted axioms for math called ZFC: that is, Zermelo—Fraenkel set theory together with the axiom of choice. It’s a great story, and you can read it here:

• Scott Aaronson, The 8000th Busy Beaver number eludes ZF set theory: new paper by Adam Yedidia and me, Shtetl-Optimized, 3 May 2016.

• Adam Yedidia and Scott Aaronson, A relatively small Turing machine whose behavior is independent of set theory, 13 May 2016.

Briefly, Yedidia created a new programming language, called Laconic, which lets you write programs that compile down to small Turing machines. They took an arithmetic statement created by Harvey Friedman that’s equivalent to the consistency of the usual axioms of ZFC together with a large cardinal axiom called the ‘stationary Ramsey property’, or SRP. And they created a Turing machine with 7910 states that seeks a proof of this arithmetic statement using the axioms of ZFC.

Since ZFC can’t prove its own consistency, much less its consistency when supplemented with SRP, their machine will only halt if ZFC+SRP is inconsistent.

Since most set theorists believe ZFC+SRP is consistent, this machine probably doesn’t halt. But we can’t prove this using ZFC.

In short: if the usual axioms of set theory are consistent, we can never use them to determine the value of BB(7910).

The basic idea is nothing new: what’s new is the explicit and rather low value of the number 7910. Poetically speaking, we know the unknowable starts here… if not sooner.

However, this discovery set off a wave of improvements! On the Metamath newsgroup, Mario Carneiro and others started ‘logic hacking’, looking for smaller and smaller Turing machines that would only halt if ZF—that is, Zermelo–Fraenkel set theory, without the axiom of choice—is inconsistent.

By just May 15th, Stefan O’Rear seems to have brought the number down to 1919. He found a Turing machine with just 1919 states that searches for an inconsistency in the ZF axioms. Interestingly, this turned out to work better than using Harvey Friedman’s clever trick.

Thus, if O’Rear’s work is correct, we can only determine BB(1919) if we can determine whether ZF set theory is consistent. However, we cannot do this using ZF set theory—unless we find an inconsistency in ZF set theory.

For details, see:

• Stefan O’Rear, A Turing machine Metamath verifier, 15 May 2016.

I haven’t checked his work, but it’s available on GitHub.

What’s the point of all this? At present, it’s mainly just a game. However, it should have some interesting implications. It should, for example, help us better locate the ‘complexity barrier’.

I explained that idea here:

• John Baez, The complexity barrier, Azimuth, 28 October 2011.

Briefly, while there’s no limit on how much information a string of bits—or any finite structure—can have, there’s a limit on how much information we can prove it has!

This amount of information is pretty low, perhaps a few kilobytes. And I believe the new work on logic hacking can be used to estimate it more accurately!


Shelves and the Infinite

6 May, 2016

Infinity is a very strange concept. Like alien spores floating down from the sky, large infinities can come down and contaminate the study of questions about ordinary finite numbers! Here’s an example.

A shelf is a set with a binary operation \rhd that distributes over itself:

a \rhd (b \rhd c) = (a \rhd b) \rhd (a \rhd c)

There are lots of examples, the simplest being any group, where we define

g \rhd h = g h g^{-1}

They have a nice connection to knot theory, which you can see here if you think hard:

My former student Alissa Crans, who invented the term ‘shelf’, has written a lot about them, starting here:

• Alissa Crans, Lie 2-Algebras, Chapter 3.1: Shelves, Racks, Spindles and Quandles, Ph.D. thesis, U.C. Riverside, 2004.

I could tell you a long and entertaining story about this, including the tale of how shelves got their name. But instead I want to talk about something far more peculiar, which I understand much less well. There’s a strange connection between shelves, extremely large infinities, and extremely large finite numbers! It was first noticed by a logician named Richard Laver in the late 1980s, and it’s been developed further by Randall Dougherty.

It goes like this. For each n, there’s a unique shelf structure on the numbers \{1,2, \dots ,2^n\} such that

a \rhd 1 = a + 1 \bmod 2^n

So, the elements of our shelf are

1

1 \rhd 1 = 2

2 \rhd 1 = 3

and so on, until we get to

2^n \rhd 1 = 1

However, we can now calculate

1 \rhd 1

1 \rhd 2

1 \rhd 3

and so on. You should try it yourself for a simple example! You’ll need to use the self-distributive law. It’s quite an experience.

You’ll get a list of 2^n numbers, but this list will not contain all the numbers \{1, 2, \dots, 2^n\}. Instead, it will repeat with some period P(n).

Here is where things get weird. The numbers P(n) form this sequence:

1, 1, 2, 4, 4, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, …

It may not look like it, but the numbers in this sequence approach infinity!

if we assume an extra axiom, which goes beyond the usual axioms of set theory, but so far seems consistent!

This axiom asserts the existence of an absurdly large cardinal, called an I3 rank-into-rank cardinal.

I’ll say more about this kind of cardinal later. But, this is not the only case where a ‘large cardinal axiom’ has consequences for down-to-earth math, like the behavior of some sequence that you can define using simple rules.

On the other hand, Randall Dougherty has proved a lower bound on how far you have to go out in this sequence to reach the number 32.

And, it’s an incomprehensibly large number!

The third Ackermann function A_3(n) is roughly 2 to the nth power. The fourth Ackermann function A_4(n) is roughly 2 raised to itself n times:

2^{2^{2^{2^{\cdot^{\cdot^\cdot}}}}}

And so on: each Ackermann function is defined by iterating the previous one.

Dougherty showed that for the sequence P(n) to reach 32, you have to go at least

n = A(9,A(8,A(8,255)))

This is an insanely large number!

I should emphasize that if we use just the ordinary axioms of set theory, the ZFC axioms, nobody has proved that the sequence P(n) ever reaches 32. Neither is it known that this is unprovable if we only use ZFC.

So, what we’ve got here is a very slowly growing sequence… which is easy to define but grows so slowly that (so far) mathematicians need new axioms of set theory to prove it goes to infinity, or even reaches 32.

I should admit that my definition of the Ackermann function is rough. In reality it’s defined like this:

A(m, n) = \begin{cases} n+1 & \mbox{if } m = 0 \\ A(m-1, 1) & \mbox{if } m > 0 \mbox{ and } n = 0 \\ A(m-1, A(m, n-1)) & \mbox{if } m > 0 \mbox{ and } n > 0. \end{cases}

And if you work this out, you’ll find it’s a bit annoying. Somehow the number 3 sneaks in:

A(2,n) = 2 + (n+3) - 3

A(3,n) = 2 \cdot (n+3) - 3

A(4,n) = 2^{n+3} - 3

A(5,n) = 2\uparrow\uparrow(n+3) - 3

where a \uparrow\uparrow b means a raised to itself b times,

A(6,n) = 2 \uparrow\uparrow\uparrow(n+3) - 3

where a \uparrow\uparrow\uparrow b means a \uparrow\uparrow (a \uparrow\uparrow (a \uparrow\uparrow \cdots )) with the number a repeated b times, and so on.

However, these irritating 3’s scarcely matter, since Dougherty’s number is so large… and I believe he could have gotten an even larger upper bound if he wanted.

Perhaps I’ll wrap up by saying very roughly what an I3 rank-into-rank cardinal is.

In set theory the universe of all sets is built up in stages. These stages are called the von Neumann hierarchy. The lowest stage has nothing in it:

V_0 = \emptyset

Each successive stage is defined like this:

V_{\lambda + 1} = P(V_\lambda)

where P(S) is the the power set of S, that is, the set of all subsets of S. For ‘limit ordinals’, that is, ordinals that aren’t of the form \lambda + 1, we define

\displaystyle{ V_\lambda = \bigcup_{\alpha < \lambda} V_\alpha }

An I3 rank-into-rank cardinal is an ordinal \lambda such that V_\lambda admits a nontrivial elementary embedding into itself.

Very roughly, this means the infinity \lambda is so huge that the collection of sets that can be built by this stage can mapped into itself, in a one-to-one but not onto way, into a smaller collection that’s indistinguishable from the original one when it comes to the validity of anything you can say about sets!

More precisely, a nontrivial elementary embedding of V_\lambda into itself is a one-to-one but not onto function

f: V_\lambda \to V_\lambda

that preserves and reflects the validity of all statements in the language of set theory. That is: for any sentence \phi(a_1, \dots, a_n) in the language of set theory, this statement holds for sets a_1, \dots, a_n \in V_\lambda if and only if \phi(f(a_1), \dots, f(a_n)) holds.

I don’t know why, but an I3 rank-into-rank cardinal, if it’s even consistent to assume one exists, is known to be extraordinarily big. What I mean by this is that it automatically has a lot of other properties known to characterize large cardinals. It’s inaccessible (which is big) and ineffable (which is bigger), and measurable (which is bigger), and huge (which is even bigger), and so on.

How in the world is this related to shelves?

The point is that if

f, g : V_\lambda \to V_\lambda

are elementary embeddings, we can apply f to any set in V_\lambda. But in set theory, functions are sets too: sets of ordered pairs So, g is a set. It’s not an element of V_\lambda, but all its subsets g \cap V_\alpha are, where \alpha < \lambda. So, we can define

f \rhd g = \bigcup_{\alpha < \lambda} f (g \cap V_\alpha)

Laver showed that this operation distributes over itself:

f \rhd (g \rhd h) = (f \rhd g) \rhd (f \rhd h)

And, he showed that if we take one elementary embedding and let it generate a shelf by this this operation, we get the free shelf on one generator!

The shelf I started out describing, the numbers \{1, \dots, 2^n \} with

a \rhd 1 = a + 1 \bmod 2^n

also has one generator namely the number 1. So, it’s a quotient of the free shelf on one generator by one relation, namely the above equation.

That’s about all I understand. I don’t understand how the existence of a nontrivial elementary embedding of V_\lambda into itself implies that the function P(n) goes to infinity, and I don’t understand Randall Dougherty’s lower bound on how far you need to go to reach P(n) = 32. For more, read these:

• Richard Laver, The left distributive law and the freeness of an algebra of elementary embeddings, Adv. Math. 91 (1992), 209–231.

• Richard Laver, On the algebra of elementary embeddings of a rank into itself, Adv. Math. 110 (1995), 334–346.

• Randall Dougherty and Thomas Jech, Finite left distributive algebras and embedding algebras, Adv. Math. 130 (1997), 201–241.

• Randall Dougherty, Critical points in an algebra of elementary embeddings, Ann. Pure Appl. Logic 65 (1993), 211–241.

• Randall Dougherty, Critical points in an algebra of elementary embeddings, II.


Statistical Laws of Darwinian Evolution

18 April, 2016

guest post by Matteo Smerlak

Biologists like Steven J. Gould like to emphasize that evolution is unpredictable. They have a point: there is absolutely no way an alien visiting the Earth 400 million years ago could have said:

Hey, I know what’s gonna happen here. Some descendants of those ugly fish will grow wings and start flying in the air. Others will walk the surface of the Earth for a few million years, but they’ll get bored and they’ll eventually go back to the oceans; when they do, they’ll be able to chat across thousands of kilometers using ultrasound. Yet others will grow arms, legs, fur, they’ll climb trees and invent BBQ, and, sooner or later, they’ll start wondering “why all this?”.

Nor can we tell if, a week from now, the flu virus will mutate, become highly pathogenic and forever remove the furry creatures from the surface of the Earth.

Evolution isn’t gravity—we can’t tell in which directions things will fall down.

One reason we can’t predict the outcomes of evolution is that genomes evolve in a super-high dimensional combinatorial space, which a ginormous number of possible turns at every step. Another is that living organisms interact with one another in a massively non-linear way, with, feedback loops, tipping points and all that jazz.

Life’s a mess, if you want my physicist’s opinion.

But that doesn’t mean that nothing can be predicted. Think of statistics. Nobody can predict who I’ll vote for in the next election, but it’s easy to tell what the distribution of votes in the country will be like. Thus, for continuous variables which arise as sums of large numbers of independent components, the central limit theorem tells us that the distribution will always be approximately normal. Or take extreme events: the max of N independent random variables is distributed according to a member of a one-parameter family of so-called “extreme value distributions”: this is the content of the famous Fisher–Tippett–Gnedenko theorem.

So this is the problem I want to think about in this blog post: is evolution ruled by statistical laws? Or, in physics terms: does it exhibit some form of universality?

Fitness distributions are the thing

One lesson from statistical physics is that, to uncover universality, you need to focus on relevant variables. In the case of evolution, it was Darwin’s main contribution to figure out the main relevant variable: the average number of viable offspring, aka fitness, of an organism. Other features—physical strength, metabolic efficiency, you name it—matter only insofar as they are correlated with fitness. If we further assume that fitness is (approximately) heritable, meaning that descendants have the same fitness as their ancestors, we get a simple yet powerful dynamical principle called natural selection: in a given population, the lineage with the highest fitness eventually dominates, i.e. its fraction goes to one over time. This principle is very general: it applies to genes and species, but also to non-living entities such as algorithms, firms or language. The general relevance of natural selection as a evolutionary force is sometimes referred to as “Universal Darwinism”.

The general idea of natural selection is pictured below (reproduced from this paper):

It’s not hard to write down an equation which expresses natural selection in general terms. Consider an infinite population in which each lineage grows with some rate x. (This rate is called the log-fitness or Malthusian fitness to contrast it with the number of viable offspring w=e^{x\Delta t} with \Delta t the lifetime of a generation. It’s more convenient to use x than w in what follows, so we’ll just call x “fitness”). Then the distribution of fitness at time t satisfies the equation

\displaystyle{ \frac{\partial p_t(x)}{\partial t} =\left(x-\int d y\, y\, p_t(y)\right)p_t(x) }

whose explicit solution in terms of the initial fitness distribution p_0(x):

\displaystyle{ p_t(x)=\frac{e^{x t}p_0(x)}{\int d y\, e^{y t}p_0(y)} }

is called the Cramér transform of p_0(x) in large deviations theory. That is, viewed as a flow in the space of probability distributions, natural selection is nothing but a time-dependent exponential tilt. (These equations and the results below can be generalized to include the effect of mutations, which are critical to maintain variation in the population, but we’ll skip this here to focus on pure natural selection. See my paper referenced below for more information.)

An immediate consequence of these equations is that the mean fitness \mu_t=\int dx\, x\, p_t(x) grows monotonically in time, with a rate of growth given by the variance \sigma_t^2=\int dx\, (x-\mu_t)^2\, p_t(x):

\displaystyle{ \frac{d\mu_t}{dt}=\sigma_t^2\geq 0 }

The great geneticist Ronald Fisher (yes, the one in the extreme value theorem!) was very impressed with this relationship. He thought it amounted to an biological version of the second law of thermodynamics, writing in his 1930 monograph

Professor Eddington has recently remarked that “The law that entropy always increases—the second law of thermodynamics—holds, I think, the supreme position among the laws of nature”. It is not a little instructive that so similar a law should hold the supreme position among the biological sciences.

Unfortunately, this excitement hasn’t been shared by the biological community, notably because this Fisher “fundamental theorem of natural selection” isn’t predictive: the mean fitness \mu_t grows according to the fitness variance \sigma_t^2, but what determines the evolution of \sigma_t^2? I can’t use the identity above to predict the speed of evolution in any sense. Geneticists say it’s “dynamically insufficient”.

Two limit theorems

But the situation isn’t as bad as it looks. The evolution of p_t(x) may be decomposed into the evolution of its mean \mu_t, of its variance \sigma_t^2, and of its shape or type

\overline{p}_t(x)=\sigma_t p_t(\sigma_t x+\mu_t).

(We also call \overline{p}_t(x) the “standardized fitness distribution”.) With Ahmed Youssef we showed that:

• If p_0(x) is supported on the whole real line and decays at infinity as

-\ln\int_x^{\infty}p_0(y)d y\underset{x\to\infty}{\sim} x^{\alpha}

for some \alpha > 1, then \mu_t\sim t^{\overline{\alpha}-1}, \sigma_t^2\sim t^{\overline{\alpha}-2} and \overline{p}_t(x) converges to the standard normal distribution as t\to\infty. Here \overline{\alpha} is the conjugate exponent to \alpha, i.e. 1/\overline{\alpha}+1/\alpha=1.

• If p_0(x) has a finite right-end point x_+ with

p(x)\underset{x\to x_+}{\sim} (x_+-x)^\beta

for some \beta\geq0, then x_+-\mu_t\sim t^{-1}, \sigma_t^2\sim t^{-2} and \overline{p}_t(x) converges to the flipped gamma distribution

\displaystyle{ p^*_\beta(x)= \frac{(1+\beta)^{(1+\beta)/2}}{\Gamma(1+\beta)} \Theta[x-(1+\beta)^{1/2}] }

\displaystyle { e^{-(1+\beta)^{1/2}[(1+\beta)^{1/2}-x]}\Big[(1+\beta)^{1/2}-x\Big]^\beta }

Here and below the symbol \sim means “asymptotically equivalent up to a positive multiplicative constant”; \Theta(x) is the Heaviside step function. Note that p^*_\beta(x) becomes Gaussian in the limit \beta\to\infty, i.e. the attractors of cases 1 and 2 form a continuous line in the space of probability distributions; the other extreme case, \beta\to0, corresponds to a flipped exponential distribution.

The one-parameter family of attractors p_\beta^*(x) is plotted below:

These results achieve two things. First, they resolve the dynamical insufficiency of Fisher’s fundamental theorem by giving estimates of the speed of evolution in terms of the tail behavior of the initial fitness distribution. Second, they show that natural selection is indeed subject to a form of universality, whereby the relevant statistical structure turns out to be finite dimensional, with only a handful of “conserved quantities” (the \alpha and \beta exponents) controlling the late-time behavior of natural selection. This amounts to a large reduction in complexity and, concomitantly, an enhancement of predictive power.

(For the mathematically-oriented reader, the proof of the theorems above involves two steps: first, translate the selection equation into a equation for (cumulant) generating functions; second, use a suitable Tauberian theorem—the Kasahara theorem—to relate the behavior of generating functions at large values of their arguments to the tail behavior of p_0(x). Details in our paper.)

It’s useful to consider the convergence of fitness distributions to the attractors p_\beta^*(x) for 0\leq\beta\leq \infty in the skewness-kurtosis plane, i.e. in terms of the third and fourth cumulants of p_t(x).

The red curve is the family of attractors, with the normal at the bottom right and the flipped exponential at the top left, and the dots correspond to numerical simulations performed with the classical Wright–Fisher model and with a simple genetic algorithm solving a linear programming problem. The attractors attract!

Conclusion and a question

Statistics is useful because limit theorems (the central limit theorem, the extreme value theorem) exist. Without them, we wouldn’t be able to make any population-level prediction. Same with statistical physics: it only because matter consists of large numbers of atoms, and limit theorems hold (the H-theorem, the second law), that macroscopic physics is possible in the first place. I believe the same perspective is useful in evolutionary dynamics: it’s true that we can’t predict how many wings birds will have in ten million years, but we can tell what shape fitness distributions should have if natural selection is true.

I’ll close with an open question for you, the reader. In the central limit theorem as well as in the second law of thermodynamics, convergence is driven by a Lyapunov function, namely entropy. (In the case of the central limit theorem, it’s a relatively recent result by Arstein et al.: the entropy of the normalized sum of n i.i.d. random variables, when it’s finite, is a monotonically increasing function of n.) In the case of natural selection for unbounded fitness, it’s clear that entropy will also be eventually monotonically increasing—the normal is the distribution with largest entropy at fixed variance and mean.

Yet it turns out that, in our case, entropy isn’t monotonic at all times; in fact, the closer the initial distribution p_0(x) is to the normal distribution, the later the entropy of the standardized fitness distribution starts to increase. Or, equivalently, the closer the initial distribution p_0(x) to the normal, the later its relative entropy with respect to the normal. Why is this? And what’s the actual Lyapunov function for this process (i.e., what functional of the standardized fitness distribution is monotonic at all times under natural selection)?

In the plots above the blue, orange and green lines correspond respectively to

\displaystyle{ p_0(x)\propto e^{-x^2/2-x^4}, \quad p_0(x)\propto e^{-x^2/2-.01x^4}, \quad p_0(x)\propto e^{-x^2/2-.001x^4} }

References

• S. J. Gould, Wonderful Life: The Burgess Shale and the Nature of History, W. W. Norton & Co., New York, 1989.

• M. Smerlak and A. Youssef, Limiting fitness distributions in evolutionary dynamics, 2015.

• R. A. Fisher, The Genetical Theory of Natural Selection, Oxford University Press, Oxford, 1930.

• S. Artstein, K. Ball, F. Barthe and A. Naor, Solution of Shannon’s problem on the monotonicity of entropy, J. Am. Math. Soc. 17 (2004), 975–982.


Diamonds and Triamonds

11 April, 2016

The structure of a diamond crystal is fascinating. But there’s an equally fascinating form of carbon, called the triamond, that’s theoretically possible but never yet seen in nature. Here it is:


k4_crystal

In the triamond, each carbon atom is bonded to three others at 120° angles, with one double bond and two single bonds. Its bonds lie in a plane, so we get a plane for each atom.

But here’s the tricky part: for any two neighboring atoms, these planes are different. In fact, if we draw the bond planes for all the atoms in the triamond, they come in four kinds, parallel to the faces of a regular tetrahedron!

If we discount the difference between single and double bonds, the triamond is highly symmetrical. There’s a symmetry carrying any atom and any of its bonds to any other atom and any of its bonds. However, the triamond has an inherent handedness, or chirality. It comes in two mirror-image forms.

A rather surprising thing about the triamond is that the smallest rings of atoms are 10-sided. Each atom lies in 15 of these 10-sided rings.

Some chemists have argued that the triamond should be ‘metastable’ at room temperature and pressure: that is, it should last for a while but eventually turn to graphite. Diamonds are also considered metastable, though I’ve never seen anyone pull an old diamond ring from their jewelry cabinet and discover to their shock that it’s turned to graphite. The big difference is that diamonds are formed naturally under high pressure—while triamonds, it seems, are not.

Nonetheless, the mathematics behind the triamond does find its way into nature. A while back I told you about a minimal surface called the ‘gyroid’, which is found in many places:

The physics of butterfly wings.

It turns out that the pattern of a gyroid is closely connected to the triamond! So, if you’re looking for a triamond-like pattern in nature, certain butterfly wings are your best bet:

• Matthias Weber, The gyroids: algorithmic geometry III, The Inner Frame, 23 October 2015.

Instead of trying to explain it here, I’ll refer you to the wonderful pictures at Weber’s blog.

Building the triamond

I want to tell you a way to build the triamond. I saw it here:

• Toshikazu Sunada, Crystals that nature might miss creating, Notices of the American Mathematical Society 55 (2008), 208–215.

This is the paper that got people excited about the triamond, though it was discovered much earlier by the crystallographer Fritz Laves back in 1932, and Coxeter named it the Laves graph.

To build the triamond, we can start with this graph:


k4_graph_colored_sunada

It’s called \mathrm{K}_4, since it’s the complete graph on four vertices, meaning there’s one edge between each pair of vertices. The vertices correspond to four different kinds of atoms in the triamond: let’s call them red, green, yellow and blue. The edges of this graph have arrows on them, labelled with certain vectors

e_1, e_2, e_3, f_1, f_2, f_3 \in \mathbb{R}^3

Let’s not worry yet about what these vectors are. What really matters is this: to move from any atom in the triamond to any of its neighbors, you move along the vector labeling the edge between them… or its negative, if you’re moving against the arrow.

For example, suppose you’re at any red atom. It has 3 nearest neighbors, which are blue, green and yellow. To move to the blue neighbor you add f_1 to your position. To move to the green one you subtract e_2, since you’re moving against the arrow on the edge connecting blue and green. Similarly, to go to the yellow neighbor you subtract the vector f_3 from your position.

Thus, any path along the bonds of the triamond determines a path in the graph \mathrm{K}_4.

Conversely, if you pick an atom of some color in the triamond, any path in \mathrm{K}_4 starting from the vertex of that color determines a path in the triamond! However, going around a loop in \mathrm{K}_4 may not get you back to the atom you started with in the triamond.

Mathematicians summarize these facts by saying the triamond is a ‘covering space’ of the graph \mathrm{K}_4.

Now let’s see if you can figure out those vectors.

Puzzle 1. Find vectors e_1, e_2, e_3, f_1, f_2, f_3 \in \mathbb{R}^3 such that:

A) All these vectors have the same length.

B) The three vectors coming out of any vertex lie in a plane at 120° angles to each other:


k4_graph_colored_sunada

For example, f_1, -e_2 and -f_3 lie in a plane at 120° angles to each other. We put in two minus signs because two arrows are pointing into the red vertex.

C) The four planes we get this way, one for each vertex, are parallel to the faces of a regular tetrahedron.

If you want, you can even add another constraint:

D) All the components of the vectors e_1, e_2, e_3, f_1, f_2, f_3 are integers.

Diamonds and hyperdiamonds

That’s the triamond. Compare the diamond:

Here each atom of carbon is connected to four others. This pattern is found not just in carbon but also other elements in the same column of the periodic table: silicon, germanium, and tin. They all like to hook up with four neighbors.

The pattern of atoms in a diamond is called the diamond cubic. It’s elegant but a bit tricky. Look at it carefully!

To build it, we start by putting an atom at each corner of a cube. Then we put an atom in the middle of each face of the cube. If we stopped there, we would have a face-centered cubic. But there are also four more carbons inside the cube—one at the center of each tetrahedron we’ve created.

If you look really carefully, you can see that the full pattern consists of two interpenetrating face-centered cubic lattices, one offset relative to the other along the cube’s main diagonal.

The face-centered cubic is the 3-dimensional version of a pattern that exists in any dimension: the Dn lattice. To build this, take an n-dimensional checkerboard and alternately color the hypercubes red and black. Then, put a point in the center of each black hypercube!

You can also get the Dn lattice by taking all n-tuples of integers that sum to an even integer. Requiring that they sum to something even is a way to pick out the black hypercubes.

The diamond is also an example of a pattern that exists in any dimension! I’ll call this the hyperdiamond, but mathematicians call it Dn+, because it’s the union of two copies of the Dn lattice. To build it, first take all n-tuples of integers that sum to an even integer. Then take all those points shifted by the vector (1/2, …, 1/2).

In any dimension, the volume of the unit cell of the hyperdiamond is 1, so mathematicians say it’s unimodular. But only in even dimensions is the sum or difference of any two points in the hyperdiamond again a point in the hyperdiamond. Mathematicians call a discrete set of points with this property a lattice.

If even dimensions are better than odd ones, how about dimensions that are multiples of 4? Then the hyperdiamond is better still: it’s an integral lattice, meaning that the dot product of any two vectors in the lattice is again an integer.

And in dimensions that are multiples of 8, the hyperdiamond is even better. It’s even, meaning that the dot product of any vector with itself is even.

In fact, even unimodular lattices are only possible in Euclidean space when the dimension is a multiple of 8. In 8 dimensions, the only even unimodular lattice is the 8-dimensional hyperdiamond, which is usually called the E8 lattice. The E8 lattice is one of my favorite entities, and I’ve written a lot about it in this series:

Integral octonions.

To me, the glittering beauty of diamonds is just a tiny hint of the overwhelming beauty of E8.

But let’s go back down to 3 dimensions. I’d like to describe the diamond rather explicitly, so we can see how a slight change produces the triamond.

It will be less stressful if we double the size of our diamond. So, let’s start with a face-centered cubic consisting of points whose coordinates are even integers summing to a multiple of 4. That consists of these points:

(0,0,0)   (2,2,0)   (2,0,2)   (0,2,2)

and all points obtained from these by adding multiples of 4 to any of the coordinates. To get the diamond, we take all these together with another face-centered cubic that’s been shifted by (1,1,1). That consists of these points:

(1,1,1)   (3,3,1)   (3,1,3)   (1,3,3)

and all points obtained by adding multiples of 4 to any of the coordinates.

The triamond is similar! Now we start with these points

(0,0,0)   (1,2,3)   (2,3,1)   (3,1,2)

and all the points obtain from these by adding multiples of 4 to any of the coordinates. To get the triamond, we take all these together with another copy of these points that’s been shifted by (2,2,2). That other copy consists of these points:

(2,2,2)   (3,0,1)   (0,1,3)   (1,3,0)

and all points obtained by adding multiples of 4 to any of the coordinates.

Unlike the diamond, the triamond has an inherent handedness, or chirality. You’ll note how we used the point (1,2,3) and took cyclic permutations of its coordinates to get more points. If we’d started with (3,2,1) we would have gotten the other, mirror-image version of the triamond.

Covering spaces

I mentioned that the triamond is a ‘covering space’ of the graph \mathrm{K}_4. More precisely, there’s a graph T whose vertices are the atoms of the triamond, and whose edges are the bonds of the triamond. There’s a map of graphs

p: T \to \mathrm{K}_4

This automatically means that every path in T is mapped to a path in \mathrm{K}_4. But what makes T a covering space of \mathrm{K}_4 is that any path in T comes from a path in \mathrm{K}_4, which is unique after we choose its starting point.

If you’re a high-powered mathematician you might wonder if T is the universal covering space of \mathrm{K}_4. It’s not, but it’s the universal abelian covering space.

What does this mean? Any path in \mathrm{K}_4 gives a sequence of vectors e_1, e_2, e_3, f_1, f_2, f_3 and their negatives. If we pick a starting point in the triamond, this sequence describes a unique path in the triamond. When does this path get you back where you started? The answer, I believe, is this: if and only if you can take your sequence, rewrite it using the commutative law, and cancel like terms to get zero. This is related to how adding vectors in \mathbb{R}^3 is a commutative operation.


k4_graph_colored_sunada

For example, there’s a loop in \mathrm{K}_4 that goes “red, blue, green, red”. This gives the sequence of vectors

f_1, -e_3, e_2

We can turn this into an expression

f_1 - e_3 + e_2

However, we can’t simplify this to zero using just the commutative law and cancelling like terms. So, if we start at some red atom in the triamond and take the unique path that goes “red, blue, green, red”, we do not get back where we started!

Note that in this simplification process, we’re not allowed to use what the vectors “really are”. It’s a purely formal manipulation.

Puzzle 2. Describe a loop of length 10 in the triamond using this method. Check that you can simplify the corresponding expression to zero using the rules I described.

A similar story works for the diamond, but starting with a different graph:


diamond_graph_sunada

The graph formed by a diamond’s atoms and the edges between them is the universal abelian cover of this little graph! This graph has 2 vertices because there are 2 kinds of atom in the diamond. It has 4 edges because each atom has 4 nearest neighbors.

Puzzle 3. What vectors should we use to label the edges of this graph, so that the vectors coming out of any vertex describe how to move from that kind of atom in the diamond to its 4 nearest neighbors?

There’s also a similar story for graphene, which is hexagonal array of carbon atoms in a plane:


graphene

Puzzle 4. What graph with edges labelled by vectors in \mathbb{R}^2 should we use to describe graphene?

I don’t know much about how this universal abelian cover trick generalizes to higher dimensions, though it’s easy to handle the case of a cubical lattice in any dimension.

Puzzle 5. I described higher-dimensional analogues of diamonds: are there higher-dimensional triamonds?

References

The Wikipedia article is good:

• Wikipedia, Laves graph.

They say this graph has many names: the K4 crystal, the (10,3)-a network, the srs net, the diamond twin, and of course the triamond. The name triamond is not very logical: while each carbon has 3 neighbors in the triamond, each carbon has not 2 but 4 neighbors in the diamond. So, perhaps the diamond should be called the ‘quadriamond’. In fact, the word ‘diamond’ has nothing to do with the prefix ‘di-‘ meaning ‘two’. It’s more closely related to the word ‘adamant’. Still, I like the word ‘triamond’.

This paper describes various attempts to find the Laves graph in chemistry:

• Stephen T. Hyde, Michael O’Keeffe, and Davide M. Proserpio, A short history of an elusive yet ubiquitous structure in chemistry, materials, and mathematics, Angew. Chem. Int. Ed. 47 (2008), 7996–8000.

This paper does some calculations arguing that the triamond is a metastable form of carbon:

• Masahiro Itoh et al, New metallic carbon crystal, Phys. Rev. Lett. 102 (2009), 055703.

Abstract. Recently, mathematical analysis clarified that sp2 hybridized carbon should have a three-dimensional crystal structure (\mathrm{K}_4) which can be regarded as a twin of the sp3 diamond crystal. In this study, various physical properties of the \mathrm{K}_4 carbon crystal, especially for the electronic properties, were evaluated by first principles calculations. Although the \mathrm{K}_4 crystal is in a metastable state, a possible pressure induced structural phase transition from graphite to \mathrm{K}_4 was suggested. Twisted π states across the Fermi level result in metallic properties in a new carbon crystal.

The picture of the \mathrm{K}_4 crystal was placed on Wikicommons by someone named ‘Workbit’, under a Creative Commons Attribution-Share Alike 4.0 International license. The picture of the tetrahedron was made using Robert Webb’s Stella software and placed on Wikicommons. The pictures of graphs come from Sunada’s paper, though I modified the picture of \mathrm{K}_4. The moving image of the diamond cubic was created by H.K.D.H. Bhadeshia and put into the public domain on Wikicommons. The picture of graphene was drawn by Dr. Thomas Szkopek and put into the public domain on Wikicommons.


Computing the Uncomputable

2 April, 2016

I love the more mind-blowing results of mathematical logic:

Surprises in logic.

Here’s a new one:

• Joel David Hamkins, Any function can be computable.

Let me try to explain it without assuming you’re an expert on mathematical logic. That may be hard, but I’ll give it a try. We need to start with some background.

First, you need to know that there are many different ‘models’ of arithmetic. If you write down the usual axioms for the natural numbers, the Peano axioms (or ‘PA’ for short), you can then look around for different structures that obey these axioms. These are called ‘models’ of PA.

One of them is what you think the natural numbers are. For you, the natural numbers are just 0, 1, 2, 3, …, with the usual way of adding and multiplying them. This is usually called the ‘standard model’ of PA. The numbers 0, 1, 2, 3, … are called the ‘standard’ natural numbers.

But there are also nonstandard models of arithmetic. These models contain extra numbers beside the standard ones! These are called ‘nonstandard’ natural numbers.

This takes a while to get used to. There are several layers of understanding to pass through.

For starters, you should think of these extra ‘nonstandard’ natural numbers as bigger than all the standard ones. So, imagine a whole bunch of extra numbers tacked on after the standard natural numbers, with the operations of addition and multiplication cleverly defined in such a way that all the usual axioms still hold.

You can’t just tack on finitely many extra numbers and get this to work. But there can be countably many, or uncountably many. There are infinitely many different ways to do this. They are all rather hard to describe.

To get a handle on them, it helps to realize this. Suppose you have a statement S in arithmetic that is neither provable nor disprovable from PA. Then S will hold in some models of arithmetic, while its negation not(S) will hold in some other models.

For example, the Gödel sentence G says “this sentence is not provable in PA”. If Peano arithmetic is consistent, neither G nor not(G) is provable in PA. So G holds in some models, while not(G) holds in others.

Thus, you can intuitively think of different models as “possible worlds”. If you have an undecidable statement, meaning one that you can’t prove or disprove in PA, then it holds in some worlds, while its negation holds in other worlds.

In the case of the Gödel sentence G, most mathematicians think G is “true”. Why the quotes? Truth is a slippery concept in logic—there’s no precise definition of what it means for a sentence in arithmetic to be “true”. All we can precisely define is:

1) whether or not a sentence is provable from some axioms

and

2) whether or not a sentence holds in some model.

Nonetheless, mathematicians are human, so they have beliefs about what’s true. Many mathematicians believe that G is true: indeed, in popular accounts one often hears that G is “true but unprovable in Peano arithmetic”. So, these mathematicians are inclined to say that any model where G doesn’t hold is nonstandard.

The result

Anyway, what is Joel David Hamkins’ result? It’s this:

There is a Turing machine T with the following property. For any function f from the natural numbers to the natural numbers, there is a model of PA such that in this model, if we give T any standard natural n as input, it halts and outputs f(n).

So, take f to be your favorite uncomputable function. Then there’s a model of arithmetic such that in this model, the Turing machine computes f, at least when you feed the machine standard numbers as inputs.

So, very very roughly, there’s a possible world in which your uncomputable function becomes computable!

But you have to be very careful about how you interpret this result.

The trick

What’s the trick? The proof is beautiful, but it would take real work to improve on Hamkins’ blog article, so please read that. I’ll just say that he makes extensive use of Rosser sentences, which say:

“For any proof of this sentence in theory T, there is a smaller proof of the negation of this sentence.”

Rosser sentences are already mind-blowing, but Hamkins uses an infinite sequence of such sentences and their negations, chosen in a way that depends on the function f, to cleverly craft a model of arithmetic in which the Turing machine T computes this function on standard inputs.

But what’s really going on? How can using a nonstandard model make an uncomputable function become computable for standard natural numbers? Shouldn’t nonstandard models agree with the standard one on this issue? After all, the only difference is that they have extra nonstandard numbers tacked on after all the standard ones! How can that make a Turing machine succeed in computing f on standard natural numbers?

I’m not 100% sure, but I think I know the answer. I hope some logicians will correct me if I’m wrong.

You have to read the result rather carefully:

There is a Turing machine T with the following property. For any function f from the natural numbers to the natural numbers, there is a model of PA such that in this model, if we give T any standard natural n as input, it halts and computes f(n).

When we say the Turing machine halts, we mean it halts after N steps for some natural number N. But this may not be a standard natural number! It’s a natural number in the model we’re talking about.

So, the Turing machine halts… but perhaps only after a nonstandard number of steps.

In short: you can compute the uncomputable, but only if you’re willing to wait long enough. You may need to wait a nonstandard amount of time.

It’s like that old Navy saying:

the-difficult-we-do-immediately-the-impossible-takes-a-little-longer

But the trick becomes more evident if you notice that one single Turing machine T computes different functions from the natural numbers to the natural numbers… in different models. That’s even weirder than computing an uncomputable function.

The only way to build a machine that computes n+1 in one model and n+2 in another to build a machine that doesn’t halt in a standard amount of time in either model. It only halts after a nonstandard amount of time. In one model, it halts and outputs n+1. In another, it halts and outputs n+2.

A scary possibility

To dig a bit deeper—and this is where it gets a bit scary—we have to admit that the standard model is a somewhat elusive thing. I certainly didn’t define it when I said this:

For you, the natural numbers are just 0, 1, 2, 3, …, with the usual way of adding and multiplying them. This is usually called the standard model of PA. The numbers 0, 1, 2, 3, … are called the ‘standard’ natural numbers.

The point is that “0, 1, 2, 3, …” here is vague. It makes sense if you already know what the standard natural numbers are. But if you don’t already know, those three dots aren’t going to tell you!

You might say the standard natural numbers are those of the form 1 + ··· + 1, where we add 1 to itself some finite number of times. But what does ‘finite number’ mean here? It means a standard natural number! So this is circular.

So, conceivably, the concept of ‘standard’ natural number, and the concept of ‘standard’ model of PA, are more subjective than most mathematicians think. Perhaps some of my ‘standard’ natural numbers are nonstandard for you!

I think most mathematicians would reject this possibility… but not all. Edward Nelson tackled it head-on in his marvelous book Internal Set Theory. He writes:

Perhaps it is fair to say that “finite” does not mean what we have always thought it to mean. What have we always thought it to mean? I used to think that I knew what I had always thought it to mean, but I no longer think so.

If we go down this road, Hamkins’ result takes on a different significance. It says that any subjectivity in the notion of ‘natural number’ may also infect what it means for a Turing machine to halt, and what function a Turing machine computes when it does halt.


Mathematics in Biochemical Kinetics and Engineering

23 March, 2016

Anyone who was interested in the Workshop on Mathematical Trends in Reaction Network Theory last summer in Copenhagen might be interested in this:

Mathematics in (bio)Chemical Kinetics and Engineering (MaCKiE 2017), Budapest, 25–27 May, 2017.

This conference is planned so that it starts right after another one: the 14th Joint European Thermodynamics Conference will be in Budapest from the 21st to the 25th.

Since its first event in 2002, the MaCKiE workshop is organized in every second year. The previous meetings were held in Ghent (Belgium), Chennai (India), Heidelberg (Germany), and Houston (USA). The meeting aims to bring together scientists interested in the application of advanced mathematical methods to describe kinetic phenomena, especially chemists, mathematicians, physicist, biologists, and engineers. The acronym MaCKiE naturally comes from the title of the conference, but is also part of the German name of Mack the Knife in Brecht and Weill’s Threepenny Opera, Mackie Messer, and is phonetically indistinguishable from “makkie” in Dutch, optimistically meaning “a cinch”.

Conference papers will be published in Reaction Kinetics, Mechanisms and Catalysis in early 2018.


The Involute of a Cubical Parabola

22 March, 2016

In his remarkable book The Theory of Singularities and its Applications, Vladimir Arnol’d claims that the symmetry group of the icosahedron is secretly lurking in the problem of finding the shortest path from one point in the plane to another while avoiding some obstacles that have smooth boundaries.

Arnol’d nicely expresses the awe mathematicians feel when they discover a phenomenon like this:

Thus the propagation of waves, on a 2-manifold with boundary, is controlled by an icosahedron hidden at an inflection point at the boundary. This icosahedron is hidden, and it is difficult to find it even if its existence is known.

I would like to understand this!

I think the easiest way for me to make progress is to solve this problem posed by Arnol’d:

Puzzle. Prove that the generic involute of a cubical parabola has a cusp of order 5/2 on the straight line tangent to the parabola at the inflection point.

There’s a lot of jargon here! Let me try to demystify it. (I don’t have the energy now to say how the symmetry group of the icosahedron gets into the picture, but it’s connected to the ‘5’ in the cusp of order 5/2.)

A cubical parabola is just a curve like y = x^3:

cubical_parabola

It’s a silly name. I guess y = x^3 looked at y = x^2 and said “I want to be a parabola too!”

The involute of a curve is what you get by attaching one end of a taut string to that curve and tracing the path of the string’s free end as you wind the string onto that curve. For example:

Here our original curve, in blue, is a catenary: the curve formed by a hanging chain. Its involute is shown in red.

There are a couple of confusing things about this picture if you’re just starting to learn about involutes. First, Sam Derbyshire, who made this picture, cleverly moved the end of the string attached to the catenary at the instant the other end hit the catenary! That allowed him to continue the involute past the moment it hits the catenary. The result is a famous curve called a tractrix.

Second, it seems that the end of the string attached to the catenary is ‘at infinity’, very far up.

But you don’t need to play either of these tricks if you’re trying to draw an involute. Take a point p on a curve C. Take a string of length \ell, nail down one end at p, and wind the string along C. Then the free end of your string traces out a curve D.

D is called an involute of C. It consists of all the points you can get to from p by a path of length \ell that doesn’t cross C.

So, Arnol’d’s puzzle concerns the involute of the curve y = x^3.

He wants you to nail down one end of the string at any ‘generic’ location. So, don’t nail it down at x = 0, y = 0, since that point is different from all the rest. That point is an inflection point, where the curve y = x^3 switches from curving down to curving up!

He wants you to wind the string along the curve y = x^3, forming an involute. And he wants you to see what the involute does when it crosses the line y = 0.

This is a bit tricky, since the region y \le x^3 is not convex. If you nail your string down at x = -1, y = -1, your string will have to start out above the curve y = x^3. But when the free end of your string crosses the line y = 0, the story changes. Now your string will need to go below the curve y = x^3.

It’s a bit hard to explain this both simply and accurately, but if you imagine drawing the involute with a piece of string, I think you’ll encounter the issue I’m talking about. I hope I understand it correctly!

Anyway, suppose you succeed in drawing the involute. What should you see?

Arnol’d says the involute should have a ‘cusp of order 5/2’ somewhere on the line y = 0.

A cusp of order 5/2 is a singularity in an otherwise smooth curve that looks like y^2 = x^5 in some coordinates. In a recent post I described various kinds of cusps, and in a comment I mentioned that the cusp of order 5/2 was called a rhamphoid cusp. Strangely, I wrote all that before knowing that Arnol’d places great significance on the cusp of order 5/2 in the involute of a cubical parabola!

Simon Burton drew some nice cusps of order 5/2. The curve y^2 = x^5 looks like this:

Rhamphoid Cusp

This is a more typical curve with a cusp of order 5/2:

(x-4y^2)^2 - (y+ 2x)^5 = 0

It looks like this:

Rhamphoid Cusp

It’s less symmetrical than the curve y^2 = x^5. Indeed, it looks like a bird’s beak: the word ‘rhamphoid’ means ‘beak-like’.
Arnol’d emphasizes that you should usually expect this sort of shape for a cusp of order 5/2:

It is easy to recognize this curve in experimental data, since after a generic diffeomorphism the curve consists of two branches that have equal curvatures at the common point, and hence are convex from the same side [….]

So, if we draw the involutes of a cubical parabola we should see something like this! And indeed, Marshall Hampton has made a great online program that draws these involutes. Here’s one:

Involute of cubical parabola - Marshall Hampton

The blue curve is the involute. It looks like it has a cusp of order 5/2 where it hits the line y = 0. It also has a less pointy cusp where it hits the red curve y = x^3. Like the cusp in the tractrix, this should be a cusp of order 3/2, also known as an ordinary cusp.

Hints

Regarding the easier puzzle I posed above, Arnol’d gives this hint:

HINT. The curvature centers of both branches of the involute, which meet at the point of the inflectional tangent, lie at the inflection point, hence both branches have the same convexity (they are both concave from the side of the inflection point of the boundary).

That’s not what I’d call crystal clear! However, I now understand what he means by the two ‘branches’ of the involute. They come from how you need to change the rules of the game as the free end of your string crosses the line y = 0. Remember, I wrote:

If you nail your string down at x = -1, y = -1, your string will have to start out above the curve y = x^3. But when the free end of your string crosses the line y = 0, the story changes. Now your string will need to go below the curve y = x^3.

When the rules of the game change, he claims there’s a cusp of order 5/2 in the involute.

I also think I finally understand the picture that Arnol’d uses to explain what’s going on:

It shows the curve y = x^3 in bold, and three involutes of this curve. One involute is not generic: it goes through the special point x = 0, y = 0. The other two are. They each have a cusp of order 5/2 where they hit the line y = 0, but also a cusp of order 3/2 where they hit the curve y = x^3. We can recognize the cusps of order 5/2, if we look carefully, by the fact that both branches are convex on the same side.

But again, the challenge is to prove that these involutes have cusps of order 5/2 where they hit the line y = 0. A cusp of order 7/2 would also have two branches that are convex on the same side!

Here’s one more hint. Wikipedia says that if we have a curve

C :\mathbb{R} \to \mathbb{R}^2

parametrized by arclength, so

|C^\prime(s)|=1

for all s, then its involute is the curve

D :\mathbb{R} \to \mathbb{R}^2

given by

D(s) = C(s)- s C^\prime(s)

Strictly speaking, this must be an involute. And it must somehow handle the funny situations I described, where the involute fails to be smooth. I don’t know it does this.


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