# Introducing The Humming of the Strings

Hello! I’m sorry that I haven’t been around much in the last few months, which have been rather busy for me. I’m hoping to continue more regular posting now, though, starting with the announcement of a new project: Point at Infinity‘s first spin-off blog, The Humming of the Strings.

Loyal readers will remember a few posts here on Point at Infinity about music. We had a post about the Shepard tone, an aural illusion that seems to perpetually rise in pitch, two posts about the Risset rhythm, the Shepard tone’s rhythmic cousin, and a post about Euclidean rhythms and their surprising connections with nuclear physics, computer graphics, and the Hebrew calendar. These were among the most enjoyable posts for me to write, and some of them were among the favorites of readers as well. Mathematics and music are two passions of mine, and there is a vast field of ideas to explore here, many of which don’t necessarily fit here on Point at Infinity. This is why I’m starting The Humming of the Strings, to provide a place for more in-depth exploration of musical topics that might be out of place here.

There is another, more technical reason for this blog. Over the last couple of years, I’ve sporadically been playing around with SuperCollider, an audio synthesis platform and programming language. It is a wonderful tool for exploring the connections between math and music, and I plan on using it to create music and audio samples for The Humming of the Strings, and on sharing what I learn in this process with the readers. I’m hoping that the details of the computer programming will be of interest to some readers, and I will try to make it as unintrusive as possible for those readers not interested in it. (I’m anticipating that most posts will consist of a non-technical body section followed by a technical appendix with details of the code.)

I hope you will join me over at The Humming of the Strings. I will likely primarily be posting there rather than here in the near future, though there may be some new content on Point at Infinity from time to time. Thanks for reading, and Happy New Year!

# Cantor v. Crank

No one shall expel us from the paradise that Cantor has created for us.

-David Hilbert

To the extent that a mathematical theorem can be considered controversial, Cantor’s Theorem has historically been quite a controversial statement. The theorem, which, we remind you, states that, for any set $A$, the power set $\mathcal{P}(A)$, which consists of all subsets of $A$, is strictly larger than $A$ (or, in a commonly cited special case, the set of all real numbers, $\mathbb{R}$, is strictly larger than the set of all natural numbers, $\mathbb{N}$), was attacked by a number of Cantor’s illustrious contemporaries, among them Kronecker and Poincaré, who objected to Cantor’s manipulation of infinite sets as mathematical objects in their own right.

Cantor’s work also had prominent defenders, though, most notably David Hilbert, whose quote in the header of this post has become among the most iconic proclamations in modern mathematics. As mathematicians settled into the twentieth century, Cantor’s Theorem and the field of set theory that it helped establish became widely accepted in the community. Today, it is safe to say that Cantor’s Theorem is uncontroversial among the vast majority of mathematicians. Resistance to the theorem or to its extrapolations has not died out, though; it remains present in certain pockets of mathematics, such as ultrafinitism, in some philosophy departments, and among amateur mathematicians. Today, we will look at an example from this last category.

Let us turn now to Dilworth’s paper. The impetus for the work appears to be Dilworth’s displeasure with the Banach-Tarski Paradox, which is a theorem stating roughly that one may take a three-dimensional ball, decompose it into finitely many pieces, and move these pieces around in space so that they form two complete balls, each of the same size as the original ball. On its face, this seems absurd. We have seemingly doubled the amount of stuff we have! Of course, closer inspection dispels the aura of paradox here. The pieces needed to perform the construction are infinitely complicated and non-definable and certainly could not be used in some brilliant gold-proliferating get-rich-quick scheme in the real world. And, when we think about it, there are just as many points in two balls as there are in one (continuum-many, in both cases), so maybe this theorem isn’t so surprising after all?

Dilworth was apparently not convinced, though, He found the Banach-Tarski Paradox to be clearly wrong, and since the result depended on the techniques of the then-young field of set theory, and since set theory was born out of Cantor’s Theorem, there must be something wrong with Cantor’s Theorem.

The paper makes for alternately entertaining and maddening reading. His reasoning is often muddled, but Dilworth does sometimes have a way with words and, though he perhaps goes a bit too far in this direction, the writing has character and flair in a way that I sometimes wish more technical mathematical writing did. For example, when describing Cantor’s diagonal argument and what he sees as its undeserved acceptance by the mathematical community, Dilworth writes,

Historically and up to this date, he has won. The horrendous “alephs” of his endless infinities thunder through the evening skies of academe “with hooves of steel”, as the songwriter put it.

There is a decided lack of Johnny Cash references in today’s mathematical writing. (And yes, I know that Cash’s version of this song was released in 1979 and therefore can’t be the version referenced in this 1974 paper…)

The paper also ends memorably, with an assertion of a conspiracy among mathematicians to cover up the fallacies in Cantor’s proof, followed by a final, simple, cryptic directive:

Remember the spheres.

In substance, though, Dilworth’s writing is at first incomprehensible, and when its meaning has been at least partially uncovered, his error is immediately evident. In attempting to show that the size of the set of all real numbers is exactly the same as the size of the set of all natural numbers, he is not in fact considering all real numbers, but just those whose decimal representations only have finitely many digits (or, in a different reading of his argument, only those whose decimal representations end in endlessly repeating patterns; or, in a still different reading, only those which are definable by a finite sentence). And he is correct in the sense that this subset of the real numbers does have the same size as the set of natural numbers. But by restricting his vision to this subset, he is excluding the vast majority of real numbers.

This is a mistake that could have been pointed out to Dilworth by any mathematician or any sufficiently advanced student of mathematics. And it apparently was pointed out to him on multiple occasions, for Dilworth includes accounts of his interactions with mathematics professors in the paper. For example, in one passage, Dilworth acknowledges the obvious objection to his argument, namely that he is not considering all of the real numbers, but dismisses it out of hand without argument. In another passage, when discussing Cantor’s claim that, if one tries to pair the real numbers off in a one-to-one fashion with the integers, the integers will necessarily “become exhausted” before the process is complete, he relates the following incident:

“Yes sir,” the head of the mathematics department of a Univ. of Illinois section said matter-of-factly to my face, “The integers will become exhausted.” Believe it or not, Georg Cantor made these remarkable claims stick with the world’s mathematicians of his time, and they stick unto this day. The effects of the Cantorian grip on the professional mind have to be experienced to be believed.

It is difficult to judge how much of this disconnect between Dilworth and the professional mathematicians is due to Dilworth’s own stubbornness and unwillingness to admit his error and how much of it is due to the mathematicians’ inability to clearly articulate somewhat complex mathematical ideas to a non-mathematician (something that, as I have discovered through writing this blog, is often quite difficult to do). In any event, something went wrong in this case, which is unfortunate. It is encouraging, and all too rare, to see non-mathematicians become enthusiastic about and engaged in mathematics, so it is always a shame to see them go astray.

Let us take a slight detour here to make an observation about education and training. I don’t think it will be particularly controversial to say that the primary aim of mathematics education, particularly at the graduate level, is not the learning of calculation techniques or the statements and proofs of theorems (though this will of course come), but rather a training in the actual practice of doing mathematics, of thinking mathematically, and of being able to communicate mathematical ideas with other mathematicians. In my experience, competence in this practice of mathematics is much harder to come by than mathematical knowledge, perhaps almost impossible to come by without a dedicated mentor, a role typically filled by one’s PhD advisor. It is precisely this conversance with mathematical practice that, at least in modern times, seems to be almost a prerequisite to making any sort of substantial contribution to mathematical knowledge, and it seems to be precisely what Dilworth was lacking.

There is a common and mildly insulting label that often gets applied to people such as William Dilworth: crank. In 1992, Underwood Dudley, then a professor at DePauw University in Indiana, published a comprehensive and encyclopedic compendium of crankery, Mathematical Cranks. In the introduction to the book, Dudley clarifies what he means by the word.

They’re not nuts. Well, a few are, but most aren’t. A lot of them are amateurs — mathematical amateurs who don’t know much mathematics but like to work on mathematical problems. Sometimes, when you can’t convince them that they haven’t done what they thought they’ve done, they turn into cranks, but cranks aren’t nuts, they’re just people who have a blind spot in one direction.

One chapter of Dudley’s book is devoted to cranks attacking Cantor’s Theorem. We meet a person who found five separate mistakes in Cantor’s proof, a person who held that Cantor’s conception of “number” was incorrect, and people who dispute the existence of mathematical infinity. At the end of the chapter, we meet William Dilworth. After running through the problems with Dilworth’s paper, Dudley concludes the chapter with the following sentences (Dudley does not identify the state responsible for publishing Dilworth’s paper, and also refers to Dilworth as W.D.):

His article reads as if it is by someone convinced, whose mind is not going to be changed by anything. It is by, in two words, a crank, and it is no credit to the state of X.

Dilworth was not happy about his inclusion in Mathematical Cranks. As an engineer, an academic outsider, it was hard enough for him to publish his ideas and to be taken seriously by the mathematical establishment. After being publicly labeled a crank, it would become nearly impossible. And so, in 1995, he sued Underwood Dudley for defamation.

The suit was dismissed by a district judge for “failure to state a claim.” More precisely, the judge held that the word “crank” is incapable of being defamatory; it is mere “rhetorical hyperbole.” Dilworth appealed this ruling, and the case ended up in the Seventh Circuit Court of Appeals, before none other than Judge Richard Posner, the “most cited legal scholar of the 20th century” and one of the most prominent modern American judges not to have been appointed to the Supreme Court. And if this whole story gave us nothing else, it would have been worth it solely for Posner’s remarkable decision.

Posner begins by noting that it is crucial for Dilworth’s claim to establish that Dudley acted with “actual malice” in calling Dilworth a crank:

The allegation of actual malice is necessary because the plaintiff is a public figure. Not, it is true, a “public figure” in the lay sense of the term. Dilworth is an obscure engineer. But anyone who publishes becomes a public figure in the world bounded by the readership of the literature to which he has contributed.

(It is good to know that I am, at least legally speaking, a public figure!)

Posner then goes on to consider the district judge’s ruling that “crank” cannot be defamatory because it is mere “rhetorical hyperbole.” He begins by considering past cases that deal with precisely this issue.

Among the terms or epithets that have been held (all in the cases we’ve cited) to be incapable of defaming because they are mere hyperbole rather than falsifiable assertions of discreditable fact are “scab,” “traitor,” “amoral,” “scam,” “fake,” “phony,” “a snake-oil job,” “he’s dealing with half a deck,” and “lazy, stupid, crap-shooting, chicken-stealing idiot.”

These terms, Posner asserts, have both literal and figurative usages. Figurative usages cannot be defamatory; they are mere “rhetorical hyperbole.” Literal usages, on the other hand, that make real factual claims, can be defamatory if false. Decisions on the defamatory nature of these terms, therefore, hinge on a judgment as to whether they are intended literally or figuratively. Posner then goes on to consider whether “crank” falls into this category.

“Crank” might seem the same type of word, but we think not. A crank is a person inexplicably obsessed by an obviously unsound idea — a person with a bee in his bonnet. To call a person a crank is to say that because of some quirk of temperament he is wasting his time pursuing a line of thought that is plainly without merit or promise. An example of a math crank would be someone who spent his time trying to square the circle. To call a person a crank is basically just a colorful and insulting way of expressing disagreement with his master idea, and it therefore belongs to the language of controversy rather than to the language of defamation.

And, before affirming the district judge’s opinion, Posner includes this sentence designed to flatter set theorists everywhere.

As we emphasized in the Underwager case, judges are not well equipped to resolve academic controversies, of which a controversy over Cantor’s diagonal process is a daunting illustration…

Acknowledgments: Our thanks to Peter Smith and his blog, Logic Matters, where we first learned of this story. Cover image from the Bad Postcards Tumblr page.

# L’escalier du Diable

Welcome one, welcome all to the Point at Infinity sideshow, where today we present a tantalizing and diabolical selection of musical and mathematical curiosities. Just watch your step; these stairs can be a bit tricky.

A few months ago, you may recall, we published two posts about the Shepard tone and the Risset rhythm, aural illusions in which a tone or rhythm seems to perpetually rise or fall in pitch or in tempo but is actually repeating the same pattern over and over again, the musical equivalents of Penrose stairs.

To accompany the posts we created some sound samples so the readers could hear the illusions themselves. A couple of weeks ago, one of these samples was used in an internet radio program on audio paradoxes released by Eat This Radio, paired with some work of Jean-Claude Risset. The entire program is really excellent, ranging from a piece by J.S. Bach to mid-twentieth century audio experiments to modern electronic music, and I encourage all of you to listen to it.

One of the pieces in the radio program is a piano étude written by György Ligeti in the late twentieth century. The étude is named L’escalier du diable, or The Devil’s Staircase, and its repeated ascents of the keyboard have a striking resonance with the never-ending ascent of the Shepard tone.

The Devil’s Staircase is also the colloquial name given to a particular mathematical function introduced by Georg Cantor in the 1880s. It is a function defined on the set of real numbers between 0 and 1 and taking values in the same interval, and it has some quite curious properties. Before we discuss it, let’s take a look at (an approximation to) the graph of the function.

To appreciate the strangeness of this function, let us recall some definitions regarding functions of real numbers. Very roughly speaking, a function is called continuous if it has no sudden jumps, or if its graph can be drawn without lifting the pencil from the page. Continuous functions satisfy a number of nice properties, such as the intermediate value theorem.

The derivative of a function at a given point of its domain, if it exists, measures the rate of change of the function at that point. If the x-axis measures time and the y-axis measures the position of an object along some one-dimensional track, then the derivative can be thought of as the velocity of that object. If a function is differentiable at a point (i.e., if its derivative exists there) then it must be continuous at that point, but the converse is not necessarily true. (For example, if the graph of a function has a sharp corner at a point, then the function cannot be differentiable there.)

Let’s think about what it means for a function to have a derivative of 0 at a point. It means that, at that point, the rate of change of the function has vanished. It means that, if we zoom in sufficiently close to that point, the function should look like a constant function. Its graph should look like a horizontal line. What would it mean for a function to have a derivative of 0 almost everywhere? (Here “almost everywhere” is a technical term (which I’m not going to define) and not just me being vague.) One might think that this must imply that the function is a constant function. At almost every point in its domain, the rate of change of the function is 0, so how can the value of the function change?

One will quickly discover that this is not quite right. Consider the function defined on the real numbers whose value is 0 at all negative numbers and 1 at all non-negative numbers.

This function has derivative 0 everywhere except at 0 itself, and yet it increases from 0 to 1. It does this quite easily by being discontinuous at 0, which, in hindsight, seems sort of like cheating. So what if we also require our function to be continuous? Now we need more exotic examples, and this is where the Devil’s Staircase comes in, for the Devil’s Staircase is a continuous function, it is differentiable almost everywhere, it has a  derivative of 0 wherever its derivative is defined, and yet it still manages to increase from 0 to 1. Wild!

What is the Devil’s Staircase exactly? I’ll give two different definitions. The first proceeds via an iterative construction. Start with the function $f_0(x) = x$. Its graph, between 0 and 1, is simply a straight line segment increasing from (0,0) to (1,1). Now, look at the midpoint of this increasing line segment, and draw a horizontal line segment centered there whose length is 1/3 of the horizontal line of the original increasing segment. Now connect the ends of this line segment via straight lines to (0,0) and (1,1). This new curve is the graph of a function that we call $f_1$. It consists of two increasing line segments with one horizontal line segment between them. Now repeat the process that took us from $f_0$ to $f_1$ on each of these increasing line segments, and let $f_2$ be the function whose graph is the result. Continue in this manner, constructing $f_n$ for every natural number $n$.

It turns out that, as $n$ goes to infinity, the sequence of functions $\langle f_n \mid n \in \mathbb{N} \rangle$ converges (uniformly) to a single function. This function is the Devil’s Staircase.

A more direct but also more opaque definition is as follows: Given a real number $x$ between 0 and 1, first express $x$ in base 3 (i.e., using only 0s, 1s, and 2s). If this base 3 representation contains a 1, then replace every digit after the first 1 with a 0. Next, replace all 2s with 1s. The result has only 0s and 1s, so we can interpret it as a binary (i.e., base 2) number, and we let $f(x)$ be this value. Then the function $f$ defined in this manner is the Devil’s Staircase. Play around with this definition, and you might get a feel for what it’s doing.

And now, on our way out, some musical addenda. An encore, if you will. First, after making the Risset rhythms for the aforementioned post, I did some further coding and wrote a little program that can take any short audio snippet and make a Risset rhythm out of it. Here’s an example, first accelerating and then decelerating, using a bit from a Schubert piano trio.

You may recognize the sample from the soundtrack to Barry Lyndon.

Finally, I can’t help but include here one of my favorite pieces by Ligeti, Poema sinfónico para 100 Metrónomos.

Cover image: Devil’s Staircase Wilderness, Oregon, USA

# Hippasus and the Infinite Descent

Today, dear Reader, we bring you a story. A story of Mathematics and Music, of Reason and Passion, of Drama and Irony. It is the story of Hippasus of Metapontum, of his remarkable life and his equally remarkable death. Before we begin, a note of warning. Not everything presented here is true, but all of it is meaningful.

A Greek philosopher and mathematician from the 5th century BCE, Hippasus was a follower of Pythagoras. The Pythagoreans believed in the transmigration of souls, subscribed to the belief that All is Number (where “Number” is, of course, “Whole Number”), made great strides in the study of musical harmony, and eschewed the eating of beans. Our hero was a particularly illustrious Pythagorean. He performed experiments linking the sizes of metal discs to the tones they emit upon being struck, developed a theory of the musical scale and a theory of proportions, and showed how to inscribe a regular dodecahedron in a sphere. The regular dodecahedron is a twelve-sided solid whose faces are regular pentagons, shapes which were dear to the Pythagoreans and central to our story. The pentagram, the five-sided star formed by extending the sides of a regular pentagon, and whose tips themselves form a regular pentagon, was a religious symbol of the Pythagoreans and a mark of recognition amongst themselves.

A corollary to the Pythagorean doctrine that All is Number was a belief, at the time, that any two lengths are commensurable, i.e., that given any two lengths, they are both whole number multiples of some fixed smaller length. In modern language, this amounts to the assertion that, given any two lengths, their ratio is a rational number, i.e., can be expressed as the ratio of whole numbers. It is only fitting that the first evidence to the contrary would come from the pentagram.

It happened when Hippasus was stargazing. He saw five stars forming a perfect regular pentagon, and inside this regular pentagon he formed a pentagram, at the center of which lay another regular pentagon, into which he formed another pentagram, at the center of which lay another regular pentagon, into which he formed another pentagram. An infinite web of similar triangles was woven through his mind and, in a flash of insight, he realized something terrible: the lengths of one side of the regular pentagon and one side of the pentagram found inside it are incommensurable.

The next day, he told his fellow Pythagoreans of his discovery, and they were horrified. They could not have this knowledge, which struck at the core of their belief system, getting out into the wider world. So they took Hippasus far out to sea, and they threw him overboard.

Infinite descent is a proof technique that morally dates back to the ancient Greeks but really came into its own in the work of Pierre de Fermat in the 17th century. The idea behind it is simple and immediately appealing. Suppose we want to prove that there are no positive integers satisfying a particular property. One way to prove this would be to show that, given any positive integer $n$ satisfying this property, we could always find a smaller positive integer $n'$ satisfying the same property. Repeating the argument with $n'$ in place of $n$, we could find a still smaller positive integer $n''$ satisfying the same property. Continuing, we would construct a decreasing sequence of positive integers, $n > n' > n'' > n''' > \ldots$, the aforementioned “infinite descent”. But of course there can be no infinite decreasing sequences of positive integers (if a decreasing sequence starts with $n$, it can of course have at most $n$ elements). One thus reaches a contradiction and concludes that there are no positive integers with the given property.

Fermat made great use of the method of infinite descent in his work on number theory. One particularly striking application came in the proof of a special case of his famous Last Theorem: there are no positive integers $a,b,c$ such that $a^4 + b^4 = c^4$. Fermat showed that, if $a_0,b_0,c_0$ are positive integers such that $a_0^4 + b_0^4 = c_0^4$, then we can construct positive integers $a_1, b_1, c_1$ such that $a_1^4 + b_1^4 = c_1^4$ and $c_1 < c_0$. One can then continue, with $a_1, b_1, c_1$ in place of $a_0, b_0, c_0$, and obtain $a_2, b_2, c_2$ with $a_2^4 + b_2^4 = c_2^4$ and $c_2 < c_1$. This of course leads to the infinite descent $c_0 > c_1 > c_2 > \ldots$ and a contradiction.

We will use the method of infinite descent to prove the result of Hippasus mentioned above. This will not be exactly the way the ancient Greeks would have presented the proof, but it is very similar in spirit, and we make no apologies for the anachronism. Let’s get started.

Suppose, for the sake of an eventual contradiction, that we have a regular pentagon whose side length is commensurable with the side length of the inscribed pentagram. Let $s$ denote the side length of the pentagon and $t$ denote the side length of the pentagram (i.e., the diagonal of the pentagon). In modern language, our assumption is that $\frac{t}{s}$ is rational, i.e., that there are positive integers $p$ and $q$ such that $\frac{t}{s} = \frac{p}{q}$. By scaling the pentagon, we can in fact assume that $t = p$ and $s = q$. So let this be our starting assumption: there is a regular pentagon whose side length $s$ and diagonal length $t$ are both positive integers.

Let us label the vertices of the pentagon by the letters A,B,C,D,E. The center of the pentagram forms another regular pentagon, whose vertices we shall call a,b,c,d,e. This is shown in the diagram below. Note that $s$ is the length of the line segment connecting A and B (we will denote this length by |AB|), and $t$ is |AC|. Let $s'$ denote the side length of the inner pentagon, i.e., |ab|, and let $t'$ denote the length of the diagonal of the inner pentagon, i.e., |ac|.

We now make use of the wealth of congruent triangles in the diagram. We first observe that, whenever a pentagram is inscribed into a regular pentagon, the diagonals that form the pentagram exactly trisect the angles of the pentagon. (Exercise: Prove this!) Therefore, the triangle formed by A, E and D is congruent to that formed by A, e, and D, on account of their sharing a side and having the same angles on either end of that side. In particular, we have |Ae| = |AE| = $s$.

Next, consider the triangle formed by A, b, and d. By our observation at the start of the previous paragraph, the angle at b in this triangle has the same measure as the angle at A. But then this triangle is isosceles, so we have |Ad| = |db| = $t'$. Of course, we clearly have |Ad| = |Bd| = |Be| = |Ce| =…, so all of these lengths are equal to $t'$.

Let’s now see what we have. Consider first |AC|. By definition, we have |AC| = $t$. But also |AC| = |Ae| + |Ce|, and we saw previously that |Ae| = $s$ and |Ce| = $t'$. All together, we have $t$ = |AC| = |Ae| + |Ce| = $s + t'$, or $t' = t-s$.

Next, consider |Ae|. We have already seen that |Ae| = $s$. But we also have |Ae| = |Ad| + |de|. We saw previously that |Ad| = $t'$, and by definition we have |de| = $s'$. All together, we have $s$ = |Ae| = |Ad| + |de| = $t' + s'$, or $s' = s - t'$. Since we already know that $t' = t-s$, this yields $s' = 2s - t$.

This might not seem like much, but we’re actually almost done now! The key observation is that, since $s$ and $t$ are integers, and since $s' = 2s - t$ and $t' = t-s$, it follows that $s'$ and $t'$ are also (positive) integers. We started with a regular pentagon whose side length $s$ and diagonal length $t$ were integers and produced a smaller pentagon whose side length $s'$ and diagonal length $t'$ are also integers. But now we can continue this process, producing smaller and smaller regular pentagons and producing infinite descending sequences of positive integers, $s > s' > s'' > \ldots$ and $t > t' > t'' > \ldots$.

This is a contradiction, and we have thus shown that $\frac{t}{s}$ is irrational. But what is this ratio exactly? Well, it turns out to be a fascinating number, easily #2 on the list of Most Famous Ratios of All Time: $\phi$, a.k.a. the golden ratio! But that is a story for another time…

As a short addendum to today’s story, here’s a little-known fact about the end of Hippasus’ life. It turns out that, after being thrown overboard by his fellow Pythagoreans, he has not and will never in fact reach the seabed! For to get there, he would first have to travel halfway down, and then he would have to travel half of the remaining distance, and then half of the still remaining distance, and so on and so forth, completing an endless sequence of tasks, and thus he remains to this date in the midst of an infinite descent of his own.

Notes:

(1) $\sqrt{2}$ is more commonly cited as being the first irrational number discovered by the Pythagoreans, and it is almost always the first number proven to be irrational in classrooms today. However, the proof of the irrationality of $\sqrt{2}$ possessed by the Greeks is not the simple number-theoretic proof used today and is in fact a rather complex elaboration of the proof of the incommensurability of the side and diagonal lengths of a regular pentagon. Given this fact and the centrality of the pentagram in Pythagorean intellectual life, some scholars have suggested that perhaps this was in fact the first proof of the existence of incommensurability and that the proof of the irrationality of $\sqrt{2}$ came later. We have adopted this hypothesis for the purpose of our story today.

(2) This story is derived from legends that significantly post-date the death of Hippasus. It seems unlikely actually to have happened as presented here.

(3) This post was in part inspired by the episode “Drowned at Sea”, from the excellent podcast, Hi-Phi Nation, by philosopher Barry Lam. Check it out!

Cover image: “Rainstorm Over the Sea” by John Constable

# Beats by Bjorklund

You gotta keep ’em separated.

-The Offspring, “Come Out And Play”

##### (Friendly note to the reader before we begin: If you’re looking for music to listen to while reading this post, just scroll to the bottom…)

The Spallation Neutron Source (SNS), at Oak Ridge National Laboratory in Tennessee, provides the world’s most intense pulsed neutron beams for scientific research. By accelerating proton pulses and directing them towards a mercury target, the laboratory ejects neutrons from the target. By measuring the energies and angles of these scattered neutrons, scientists can uncover information about the magnetic and molecular structure of various materials.

Pulses at the SNS are grouped into super-cycles, each 10 seconds long and consisting of 600 pulses. Commonly, experimenters will want to perform a particular action on a fixed number of pulses ($k$, say) during each super-cycle and will want to have the executions of the actions spaced as evenly as possible across time. If $k$ evenly divides 600, then the solution to this problem is trivial: just perform the action on pulses $0, \frac{600}{k}, \frac{2 \cdot 600}{k}, \frac{3 \cdot 600}{k}, \ldots, \frac{(k-1) \cdot 600}{k}$. If $k$ does not evenly divide 600, though, then the solution is not as obvious.

This is of course a special case of a general problem: Given cycles of length $n$, how can $k$ events be spaced as evenly as possible across a single cycle? (Naturally, we should have $k \leq n$ here.) As seen above, the problem is trivial if $k$ evenly divides $n$, so we shall assume this is not the case. In fact, we shall assume that $k$ and $n$ are relatively prime, i.e., that they have no common positive divisors other than 1. We will also assume that $k > 1$, since it is also trivial to evenly distribute one event across a cycle.

This exact problem was tackled by E. Bjorklund, motivated by timing questions at the SNS. In this paper, he presents an algorithm for generating solutions. For ease of notation, let us denote a cycle by a sequence of 0s and 1s, with a 1 representing an event taking place and a 0 representing an event not taking place. For definiteness, we will always start the sequence with a 1, but, because we are dealing with cyclic phenomena, we will consider two “rotations” of the same base sequence to be the same. For example, we consider 10010 and 10100 to be the same, since the second is just the first “rotated” (and wrapped around) three spaces to the right.

We will not give a formal account of Bjorklund’s algorithm here, but we will illustrate it with an example that we hope will indicate how to carry it out in general. Suppose we are given the specific problem of evenly spacing eight events in a cycle of length thirteen. Our final sequence will thus have eight 1s and five 0s. We start with thirteen individual sequences, each of length one, with all of the 1s first and then all of the 0s:

[1] [1] [1] [1] [1] [1] [1] [1] [0] [0] [0] [0] [0]

You will notice that we have two different sequences in this list ([1] and [0], here). This will be true for every step of the algorithm. In the next step, we take all instances of the second sequence ([0], here) and append them in turn to instances of the first sequence. Since there are five instances of the second sequence, five instances of the first sequence will become longer, and we obtain:

[10] [10] [10] [10] [10] [1] [1] [1]

We again have two different sequences ([10] and [1]). And we again take all instances of the second sequence and append them in turn to instances of the first sequence, obtaining:

[101] [101] [101] [10] [10]

Repeating this process again yields:

[10110][10110][101]

Technically, since there is only one instance of the second sequence in this stage, we could stop here and just append all of the sequences in order to obtain the correct solution, but let us continue for illustration’s sake. Next, we get:

[10110101][10110]

And, finally:

[1011010110110}

(which is just a rotation of what we would have gotten two stages earlier: [1011010110101]). Thus, the optimal way of evenly distributing eight events in a single cycle of length 13 is given by the sequence 1011010110110. A similar process will yield the correct answer in any other specific instance of this problem.

(The astute reader might have noticed that this algorithm bears a resemblance to Euclid’s famous algorithm for calculating the greatest common divisor of two positive integers. This is not a coincidence.)

You might be slightly disturbed by my lack of rigor at this point. Indeed, there are two ways in which I am slightly cheating the reader. First, I have not really said what I mean by having events be spaced “as evenly as possible.” Intuitively, 1011010110110 is more even than 111111110000, or 1101101101100, or, if one thinks about it, 1011011011010, but how do we know it is “as even as possible,” and what do we even mean by this? Second, even if I had given a rigorous definition of “evenness,” I have not proven that Bjorklund’s algorithm does in fact deliver the optimal result (again, this should be intuitively plausible, but should perhaps not be blindly accepted). I will do neither of these things here, but will instead direct the interested reader to the excellent paper, “The distance geometry of music,” by Demaine, Gomez-Martin, Meijer, Rappaport, Taslakian, Toussaint, Winograd, and Wood, from which much of this post was derived.

Unsurprisingly, the problem of evenly distributing $k$ events or items into cycles of length $n$ does not show up only in nuclear physics. Let us briefly look at a couple of other interesting appearances.

Computer graphics: Suppose you want to depict a straight line on a computer screen with a slope of $\frac{k}{n}$. Computer screens consist of grids of pixels, so, unless the line is horizontal or vertical, it cannot be drawn perfectly straight. Here is an example of a segment of a line with slope $-\frac{5}{11}$:

It should be clear that, at least as long as $k < n$, the problem of drawing this line consists of choosing $k$ out of every $n$ horizontal spaces after which to move your line one space vertically. To make your line as straight as possible, this distribution should be as even as possible. Indeed, the situation depicted above can be represented by the sequence 10101001010, where 1 indicates a vertical shift and 0 represents the absence of a vertical shift. This is precisely the result one gets (after a rotation) if one applies Bjorklund’s algorithm to the numbers 5 and 11.

Leap year distribution: The problem of reconciling the fact that one revolution of the earth around the sun does not take an exact integer number of days with the desire to have years that last an exact integer number of days is one that affects every calendar design. It is handled particularly interestingly in the Hebrew calendar. The Hebrew calendar is a lunisolar calendar, in which each month lasts one full cycle of the moon, but in which adjustments are made so that each month occurs at roughly the same stage of the solar year during each cycle. The Hebrew calendar reconciles these two contradictory demands by having twelve months in a typical year but, during certain “leap years,” having one of these months (namely, Adar) repeated to yield a year of thirteen months. It turns out that, to keep the Hebrew calendar very close to being in sync with the solar cycles, one needs to have seven out of every nineteen years be leap years. It is also natural to want these leap years to be spaced as evenly as possible. And, indeed, the distribution of these seven leap years in nineteen-year cycles in the Hebrew calendar is given precisely by a rotation of the result of Bjorklund’s algorithm applied to the numbers seven and nineteen.

Perhaps the most interesting appearances of the problem of evenly distributing events in a cycle, though, are in music. If one thinks of a musical measure consisting of $n$ beats as being a cycle of length $n$, of a 1 as being a note being played, and of a 0 as being the absence of a note being played, then our strings of 1s and 0s can be thought of as musical rhythms. And it turns out that applications of Bjorklund’s algorithm, when interpreted in this way, yield a startling variety of musical rhythms used in practice around the world. We end this post by presenting a few prominent examples.

In what follows, we notationally replace 1s with ‘x’s and 0s with dots. This is more conducive to following along with musical rhythms, as it is visually easier to get lost in strings of 1s and 0s than in strings of ‘x’s and dots. An ‘x’ thus denotes a note being played and a dot denotes the absence of a note being played (or, often, an ‘x’ denotes an accented beat and a dot denotes an unaccented beat).  Given numbers $k < n$, we denote by $E(k,n)$ the result of Bjorklund’s algorithm applied to $k$ and $n$. The sequence that follows is the rotation of this result that is actually heard in the piece of music. As before, we restrict ourselves to the interesting cases in which $k$ and $n$ are relatively prime and $k > 1$. (Of course, the standard waltz rhythm [x .  . ], for example, would be E(1,3), and the standard backbeat rhythm [. x . x ] would be a rotation of E(2,4), but these are sort of boring in this context.)

There’s some good music here. Enjoy!

E(2,5) – (x . . x . ): “Take Five” by The Dave Brubeck Quartet.

E(3,7) – (x . x . x . . ): “Money” by Pink Floyd.

E(3,8) – (x . . x . . x . ): This is the famous tresillo rhythm, which can be heard throughout Latin American and sub-Saharan African music as well as in such songs as “Hound Dog,” “Despacito,” and the song we feature here, “Intro” by The xx.

E(5,8) – (x . x x . x x . ): This is the cinquillo rhythm, an elaboration of the tresillo that is also ubiquitous in Latin American and sub-Saharan African music. It also features prominently in “Come Out and Play” by The Offspring.

E(4,9) – (x . x . x . x . . ): This is our second track from the classic Time Out album, which was inspired by musical styles (especially time signatures) observed during travels in Europe and Asia. Brubeck saw this particular beat being played by Turkish street musicians and then used it for  “Blue Rondo à la Turk” by The Dave Brubeck Quartet.

E(4,11) – (x . . x . . x . . x . ): “Outside Now” by Frank Zappa.

E(5,11) – (x . x . x . x . x . .): “Promenade” from Pictures at an Exhibition by Modest Mussorgsky.

E(7,12) – (x . x . x x . x . x . x): This is one of the most important rhythms in sub-Saharan Africa, often used as a bell pattern. It is also a common beat in the Palo music of the Caribbean. It can be heard played by the claves player in this performance by Los Calderones, which I have been listening to non-stop for the last hour. (Note also that, if one re-interprets this sequence in a harmonic rather than rhythmic setting, in which each step of the cycle represents a half-step in pitch, an ‘x’ represents the inclusion of that note in a scale, and a dot represents the exclusion of that note from a scale, then this sequence (this exact rotation, in fact) precisely corresponds to the major diatonic scale, the most common musical scale in Western music!)

E(4,13) – (x . . x . . x . . x . . . ): This rhythm can be heard in the instrumental sections of “Golden Brown” by The Stranglers.

E(5,16) – (x . . x . . x . . . x . . x . . ): This rhythm forms the basis for much bossa nova music and, when significantly slowed down, “Pyramid Song” by Radiohead.

Cover image: Detail from “Wall Drawing #260” by Sol LeWitt

# Common Knowledge

The phrase “common knowledge,” in its common usage, refers to knowledge that is held by (nearly) everyone, at least within some given community. For example, it is common knowledge in modern society that the Earth is round and orbits the sun, or that 2+2=4. These are facts that I can assume that you, my readers, already know or can easily verify, and it’s common practice not to give citations or acknowledge sources for them.

Today, though, we’re not really going to be discussing this common, informal usage of “common knowledge.” Nor are we going to devote this post to facts about the musician behind the classic hip hop album, Resurrection. (Though did you know that the track “I Used to Love H.E.R.” launched a feud with Ice Cube that was eventually settled at a private meeting organized by Louis Farrakhan at his house??) Rather, we’re going to delve into “common knowledge” as a fascinating technical concept in philosophical logic, epistemology, and social science.

Let’s start with the idea of mutual knowledge. Suppose that there is a group of people, and there is a fact that is known by everyone in the group. Then this fact is mutual knowledge among the members of the group. The classification of a fact as mutual knowledge seems like a pretty strong assertion, but we can (and often do) ask for even more. We can ask, for example, that not only does everyone in the group know the fact, but everyone in the group knows that everyone in the group knows the fact. Or we can go further, and ask that everyone in the group knows that everyone in the group knows that everyone in the group knows the fact. Carrying this process to infinity, we obtain the concept of common knowledge.

To be more precise, let us say that a fact is first-order knowledge among members of a group if everyone knows it, i.e., it is mutual knowledge. Let us say it is second-order knowledge if everyone in the group knows that everyone in the group knows it. Continue in this manner. In general, if n is a natural number and we have already defined the notion of n-th order knowledge, then a fact is (n+1)-st order knowledge among members of a group if it is n-th order knowledge and everyone in the group knows that it is n-th order knowledge. A fact is then common knowledge among members of a group if it is n-th order knowledge for every natural number n.

Let’s look at an example to try to clarify the distinction between mutual knowledge and common knowledge. Suppose that all of the students in a class have arrived to a classroom early and are sitting at their desks, and there is a large beetle crawling on the wall. Every student sees the beetle, so the presence of the beetle in the classroom is mutual knowledge among the students. None of the students particularly wants to deal with the beetle, though, so each student pretends to ignore it. As far as each student is concerned, they might be the only student who has noticed the beetle. The presence of the beetle is thus not even second-order knowledge among the students. Then the professor walks in the room, sees the beetle, points at it, and loudly exclaims, “Ah, I see we have a beetle auditing the class today!” Now, suddenly, the students know that they all know that there is a beetle in the room, and they know that they all know that they all know that there is a beetle in the room, and so on. The presence of the beetle has suddenly become common knowledge.

In the grand scheme of things, the study of common knowledge, and higher order knowledge in general, is a rather recent development in philosophy; David Hume was perhaps the first major philosopher to consider it, in A Treatise of Human Nature. It has become extensively studied since then, though, and, if one knows where to look, it can be found almost anywhere.

Consider the following example, due to philosopher Alan Hájek. You are in a restaurant, and a server walking by you trips on his shoelaces and spills a bit of soup on your shirt. You are, naturally, upset, and the server becomes apologetic and says, “It was my fault.” This is of course something almost everyone would expect him to say, but let’s look at it more closely. Why exactly did he say it, and why is it important? Everyone involved in the incident already knew that the server was at fault, so this utterance would seem to add no new information. However, it actually can be the key to ensuring that the incident does not escalate further, by establishing the server’s fault as common knowledge. Indeed, before the server’s admission, you might have thought that the server considered you at fault, which surely would have made you even angrier. After the server’s statement, some common ground has been established, and, with luck and maybe a free dessert, the situation can calmly be resolved.

The notion of higher order knowledge also shows up all the time in game theory (and, therefore, economics, international relations, etc.). When formulating an ideal strategy for a game, one must consider not only the current state and rules of the game, but also the knowledge and strategies of all of one’s partners and adversaries. But they, in turn, of course, are considering your knowledge and strategies when formulating their own, and one quickly reaches vertiginous levels of higher order knowledge. If you want to see common knowledge repeatedly raise its many confusing heads, just head to a high stakes poker game. (“I know that she would almost never bluff in this spot. But we’ve been playing together for days, so I also know that she knows that I know this, so maybe she’s actually more likely to bluff. But maybe she knows that I know that she knows that I know this, which means…I don’t know. I call.”)

We end today with a puzzle that has common knowledge at its core: There are two prisoners, Louis and Marie, each in their own prison cell. Each looks out on their own little yard. Louis can see 8 trees in his yard; Marie can see 12. One day, the warden brings them together and tells them both, “Between the two of you, you can see a total of either 18 or 20 trees. Every day at 5 pm, I will go to each of you. If one of you tells me the correct total number of trees in your two yards, you will immediately be set free. If you are incorrect, you will immediately be executed. If you say nothing, you will stay in prison for another day.”

Louis and Marie are not able to directly communicate with one another during this meeting with the warden, and no communication is possible between their two cells. Each was fully aware of themselves during the meeting, though; the contents of the warden’s speech have become common knowledge between them.

Q: Assuming Louis and Marie are entirely rational (and that their rationality is common knowledge between them) can they guarantee their release? If so, how? How long will it take?

We’ll talk about the solution beneath this nice painting of trees. Meanwhile, take a few minutes to think about it yourselves.

At first, it might not seem like the warden’s speech gave any really useful information. Louis now knows that Marie sees either 10 or 12 trees, and Marie now knows that Louis sees either 6 or 8 trees. But it seems that neither can be sure which of the two numbers the other sees.

If we delve deeper into the thought processes of the prisoners, though, and consider some counterfactual situations, a different picture arises. First, suppose that Marie actually saw 19 or 20 trees out her window. Then, on Day 1, Marie could confidently say, “We see 20 trees between us,” because she see more than 18 trees herself. Therefore, back in the real world, after neither prisoner says anything on Day 1, and the prisoners therefore remain in their cells, Louis knows that Marie sees at most 18 trees. You might naturally raise the following objection here: “Of course Louis knows that Marie sees at most 18 trees. He didn’t have to wait until after Day 1 to know this. He already knew this immediately after the warden’s speech, when he knew that Marie sees either 10 or 12 trees!”

But the important point here is not just that Louis knows that Marie sees at most 18 trees, but that this fact is now common knowledge, since it follows immediately from the information given by the warden and by the fact that nobody said anything on Day 1, both of which are themselves common knowledge.

Furthermore, just for the sake of completeness, and at the risk of getting lost in the weeds, let us argue that the fact that Marie sees at most 18 trees was not common knowledge before 5 pm on Day 1. Indeed, Louis knew then that Marie sees 10 or 12 trees. Put Marie does not know that Louis knows this. This is because, as far as Marie knows, Louis could see 6 trees, in which case Louis would know that Marie sees either 12 or 14 trees. Therefore, Marie can only know that Louis knows that Marie sees either 10, 12, or 14 trees. But, in turn, Louis cannot know that Marie knows that, since, as far as Louis knows, Marie could see 10 trees, in which case Marie would only know that Louis knows that Marie sees either 8, 10, or 12 trees. It follows that Louis only knows that Marie knows that Louis knows that Marie sees either 8, 10, 12, or 14 trees. And we can continue in this way. The higher order the knowledge becomes, the more uncertainty is introduced into the range, until we reach the statement: Marie knows that Louis knows that Marie knows that Louis knows that Marie knows that Louis knows that Marie knows that Louis knows that Marie sees 4, 6, 8, 10, 12, 14, 16, 18, or 20 trees. Therefore, before 5 pm on Day 1, the knowledge that Marie sees at most 18 trees is not even ninth order knowledge!

I’m sure that was crystal clear. So let’s continue. We’ve established that, after Day 1, it is common knowledge that Marie sees at most 18 trees. Suppose now that it happened to be the case that Louis saw fewer than 2 trees. Then this fact, combined with the knowledge that Marie sees at most 18 trees, would lead Louis to conclude that the prisoners see fewer than 20 trees between them, so they must see exactly 18. In the real world, when Day 2 goes by and the prisoners are still in their cells, it therefore becomes common knowledge that Louis sees at least 2 trees. Again, this was not common knowledge before 5 pm on Day 2.

It should now be clear how to continue. After Day 3 goes by with no change, it becomes common knowledge that Marie sees at most 16 trees. Then, after Day 4, it becomes common knowledge that Louis sees at least 4 trees. After Day 5, Marie sees at most 14 trees. After Day 6, Louis sees at least 6 trees. After Day 7, Marie sees at most 12 trees.

So far, none of this common knowledge is enough for either prisoner to deduce the total number of trees. In fact, neither of the prisoners has learned anything new about the number of trees seen by the other. However, their higher order knowledge has steadily grown. Indeed, the increasing ranges we saw creeping in to the prisoners’ higher order knowledge on Day 1 have gradually been winnowed by the passage of time. And this pays off after Day 8, when it becomes common knowledge that Louis sees at least 8 trees. Marie then carefully recounts the trees outside her window, confirms that it is 12, and concludes that, together, they see at least (and therefore exactly) 20 trees. She gives the correct answer on Day 9, and the prisoners are released!

There’s another famous puzzle involving common knowledge, initially even more baffling than this one. It is known as the blue-eyed islanders puzzle. It has been written about extensively elsewhere, so let me just point you to one such place, namely, Terence Tao’s blog.

For more reading on common knowledge, see this entry at the Stanford Encyclopedia of Philosophy.

Cover Image: Spock performing a Vulcan mind meld, thereby establishing common knowledge with Dr. Simon Van Gelder.

# Non-Euclidean Geometry and a Goldfish

We’ll be back, probably next week, with a new post about common knowledge. Today, though, a couple of links.

First, coming off of our recent posts about non-Euclidean geometry, a delightful 1970s BBC program on the subject:

Second, a poem, by Sara Baume and published at Granta, about a goldfish.

Enjoy!