A Chess Puzzle, Part VII Episode 1: The Fractal Queen

Part VII Episode 1 of A Mathematical Chess Puzzle:

We have now come to the final (two-part) chapter of this two year journey, exploring the maximum “density” of various chess pieces on an infinite chess board, with the rule that no piece is attacking any other. The journey has taken us through many realms of mathematics, including geometry, set theory, group theory, differential calculus, real analysis, combinatorics, topology, measure theory, and transfinite numbers.

With that exhaustive list of mathematics, one would think that our last chess piece —- the Queen —- would have little more to offer us. After all, in terms of attacks, the queen is simply a Rook that can also attack like a Bishop, and so any position of “friendly” queens is also a position of friendly rooks and/or Bishops. In terms of set theory, the set of unguarded queen positions on a board is the intersection of unguarded Bishop and Rook positions.

But as we shall see, the Queen takes us on one final trip, through some of the deepest mathematics that we have yet encountered.

Unguarded Queens on the Countable Board

There is one catch with the finite $N \times N$ board $B_N$, which is that while a maximal unguarded Rook  position consists of $N$ rooks, a maximal unguarded Bishop position will have $2N-2$ bishops, and so (on the finite boards at least) maximal unguarded queen solutions will not be maximal as bishops, merely “Rook maximal” (to coin a phrase).

There is a simple way in which one can take finite solutions to the unguarded queen problem and turn them into solutions on larger boards, and by extension to the countably infinite board $B_\omega$. We we show one such construction here.

For example, let’s start with a solution to the unguarded Queen problem on the 5-board $B_5$:

Unguarded Queens on 5-Board

If we now go to the 25-board $B_{25}$, we can think of it as a $5 \times 5$ array of $B_5$ boards. So suppose we place one of the above solutions on each of the array locations corresponding to the queen solutions, like this:

Unguarded Queens on 25-Board

it turns out that this is a Rook-maximal unguarded Queen position on the 25 board. It is not obvious that it is an unguarded Bishop position; even though it is clearly an unguarded rook position you have to verify that none of the queens in one block attack diagonally any other queens in the other blocks. Amazingly, none of them do. To see this, we only have to consider the diagonal attacks between the “blocks” of queens. As can be seen, when two such blocks are placed adjacent to each other, the queens in each block control separate diagonals from the other block:

Diagonal Queen Attacks

In the exact same manner one can take this solution on $B_{25}$ and place five copies on a $B_{125}$ board to produce an unguarded queen solution there, and continue in with all larger boards that are powers of 5.

One can then extend this generally to the countably infinite board  $B_{\omega}$ by simply noting that each finite solution $Q_i$ contains all smaller solutions, and so you can simply take the union of all of them to cover $B_{\omega}$, that is. our position $Q_{\omega}$  is given by:

$$Q_{\omega} = \bigcup^{\infty}_{i=0} Q_i$$

To make this infinite solution explicit, we need to give for each horizontal integer coordinate $x = \{0,1,2 …\}$ the vertical location $y = \phi(x)$ where the queen in that column is located. There is a neat trick for doing this. We first explictly define the first five $y$ values of $\phi$ from $x=0$ to $x=4$:
$$\begin{cases}
\phi(0) = 0 \\
\phi(1) = 2 \\
\phi(2) = 4 \\
\phi(3) = 1 \\
\phi(4) = 3 \\
\end{cases}\tag{PHI}
$$
Note that $\phi$ is a permutation of the five digits $\{0,1,2,3,4\}$. We then define $\phi(x)$ for all other positive integers $x$ by writing its value in base 5, ie,
$$x_5 = d_1 d_2 … d_n,$$
where $d_i$ are base 5 digits, and then define
$$\phi(x)_5 = \phi(d_1) \phi(d_2) … \phi(d_n)$$
That is, we permute each of the input digits of $x$ by $\phi$ to get the base 5 representation of the output $y$. Note that this also shows that all rows contain a queen, because to find the column $x$ where a queen resides on row $y$, just write $y$ in base 5 and apply the inverse of $\phi$ on the digits to get $x$.

A close approximation of the infinite board $B_{\omega}$ solution can be seen below, with a scaled down version of the solution on the very large board $B_{625}$:

625-Board with Queens

We should remark here that there is nothing magical about the number 5. We could just as easily have found another solution for the omega board using a $7 \times 7$ board queen solution as a template, such as this:

Queen Solution on 7-Board

However, this approach does not always work. Not all 7-board solutions work for example, and no queen solutions on the 4-board work:

Colliding Queens on the 4-board

The Troublesome Transfinite Queen

As with the Rook and Bishop, we now turn to unguarded queen problem on the board of the continuum $B_c$. Unlike the omega board whose “squares” are located at integer coordinate 0,1,2…, this board is transfinite, and has an uncountably infinite number of “squares” whose coordinates are real numbers. As with the Rook and Bishop, we will look strictly at the open bounded case where the board is the unit square (technically, that would be $B_I$, where $I$ is the open unit interval $I = (0,1) = \{ x: 0 \lt x\lt 1\} $. We have selected the open interval to eliminate the corner points and boundary.

As a naïve first attempt at constructing a solution, we might try the same trick as above, with an iterative process. The difference here, however, is that instead of a union of successively larger sets of discrete queens on the integer board, we create an infinite sequence of shrinking compact nested sets $Q_i, i=0,1,2,…$ where $Q_0 \supset Q_1 \supset Q_2  … $ and whose intersection forms the solution $Q$, that is,

$$Q = \bigcap^{\infty}_{i=0} Q_i$$

The first three iterations are shown below:


And the resulting fractal set $Q$ looks very much like the board $B_{625}$ shown above. It can be shown that this set has a fractal dimension of 1, and a 1-dimension fractal measure of $\sqrt 2$

Now one would think that this “solution” $Q$ works, because at each iteration the cells do not “attack” each other as Queens. However, the actual “squares” on this transfinite board are points, so we need to make sure that there are no collisions in he limiting case.

Failure!

… In fact, the solution $Q$ we have constructed is a failure.

Let’s first try to characterize the point set $Q$ mathematically, as we did in the case of the omega board.

In the omega board, whose squares had integer coordinates going off to infinity, we represented the values in Base 5, that is, multiples of powers of 5. In the case of the unit square, all coordinates are real numbers between 0 and 1, and we are dividing them by powers of 5.

We have the following theorem:

THEOREM: a point $(x,y) \in B_c$ is in $Q$ if and only if $x$ can be expressed as a Base 5 decimal, ie

$$x_5 = 0.d_1d_2d_3 ….$$

in such a way that the value $y$ can be expressed in Base 5 as

$$y_5 =0.\phi(d_1)\phi(d_2)\phi(d_3) …$$

And $\phi$ is the permutation on the digits $\{0,1,2,3,4\}$ defined by equation (PHI).

PROOF: The proof consists of observing that on the $nth$ iterative step $Q_n$ the points $(x,y)$ which are removed are exactly those where in their Base 5 representation the $nth$ digit of $y$ is not equal to $\phi(d_n)$. Conversely, if the $\phi()$ relationship holds for all digits of $x$ and $y$, then no iteration of $Q_n$ will eliminate the point $(x,y)$. QED

So if this is all good, what is the problem?

The problem is, there are many real numbers which can be expressed as infinite Base 5 decimals in two ways. In fact, any number whose Base 5 expansion terminates (ie, ends with 000…) has this property. For example, the (Base 10) value $x=0.2$ can be expressed (in Base 5) as either

$$x_5 = 0.100000000….$$

or

$$x_5 = 0.044444444….$$

(In the same way that in base ten 0.9999… = 1).

But if we apply $\phi()$ to these two decimal expansions we get two distinct values for $y$, ie

$$y_5 = 0.200000…$$

and

$$y_5 = 0.0333333…$$

…with the result that we have two distinct queens in the position $Q$ sharing the same $x$  value:

$$ (x,y)_5 = (0.1, 0.2)_5 = (\frac{1}{5},\frac{2}{5})$$

and

$$ (x,y)_5 = (0.0444…, 0.0333…)_5 = (\frac{1}{5},\frac{3}{20})$$

The two points can be seen here, along with (in red) many other queens in $Q$ which are mutally attacking, both by rook and by bishop moves:

Attacking queens

And so we see that $Q$ is a total failure. Not only are there many members of this position that are hostile, but there are an uncountably many such failures!

As with the Rook and Bishop, the root difference between $B_{\omega}$ and $B_c$ has to do with topology. In the omega board, though infinite, it has a discrete topology with no limit points. In all of the cases with this continuum board, the queens which are attacking are limit points, belonging to members of the $Q_i$ sequence which bound the (rook or bishop) attack line from opposite sides of the line.

An Unsolved Problem

Although I have tried a number of variations on this “fractal” approach, I have not found any explicit (fractal or otherwise) constructions of a queen position on the continuum board $B_c$ which is maximal unguarded one. That is, a set of points $Q \subset B_c$ defined by a closed-form computable function, such that for every horizontal and vertical line $\lambda$ intersecting $B_c$, there is exactly one point in $\lambda \cap Q$, and for every diagonal line, there is no more than one point in the intersection.

A Solved Problem

As with many such questions about exotic properties of infinite point sets, it possible that no such closed form solution may even exist for the unguarded queens on the continuum board.

For example, a very similar problem to the unguarded queens is about the existence of “Two Point Sets”. That problem has been solved and can be stated like this:

TWO POINT SET THEOREM: There exists a set of points $T \subset \mathbb{R}^2$ such that every line in the plane contains exactly two distinct points of $T$.

This is a tantalizing statement, as it is about points and lines in the plane which satisfy a rule about which lines contains which points. And as with our unguarded queen problem, the quirky aspect of this (proven) theorem is that it only shows that a solution exists, but does not give any clue as to what it would look like.

In order to prove a theorem like this, however, requires some rather exotic techniques which depend on some deep hypotheses in set theory such as the Axiom of Choice and others. In particular, what is needed is a technique called Transfinite Induction.

This is a mathematical tool of which I had not been familiar during my younger days, and which most folks would have difficulty understanding at all, given that the word “transfinite” itself starts with Cantor’s discovery that there are sets which are so large that they are “uncountable” by the infinite set of numbers 1,2,3 etc.

The Theorem

Using the techniques mentioned above, it turns out that not only will we be able to prove the existence of an unguarded queen solution as defined above, but an even stronger result, which has no equivalent statement on either the finite boards or even the countably infinite omega board.

We first begin with a definition:

DEFINITION: a chess position $Q \subset B$, satisfies the Super Queen property on $B$ if every horizontal, vertical AND (45-degree) diagonal line $\lambda$ intersecting $B$ contains exactly one member of $Q$.

Note that this is property is called super, because the property is not satisfied by any queen positions on the finite or countably infinite chess boards. On those boards, a queen position can be rook-maximally guarded positions, but not Bishop-maximally guarded. There will always be diagonals without queens controlling them.

The theorem we will prove in the next (and final) post on the topic will be the following:

SUPER QUEEN THEOREM: On the open continuum board $B = B_{I}$, where $I = (0,1) = \{ x: 0 \lt x\lt 1\} $, there exists a queen position $Q$ having the Super Queen property.

In other words, there is a position $Q$ on the board such that for every horizontal, vertical and diagonal line that intersects the board $B_{I}$, there is exactly one Queen on that line that belongs to $Q$ !

Stay tuned.

 

A Chess Puzzle, Part VI: The Bishop

Part VI of A Mathematical Chess Puzzle:

We are now approaching the end of the story, with the final tale of the Fractal Queen approaching. But for now we must deal with with the Bishop, a close cousin to the Rook.

Recall that our puzzle revolves around the maximum “unguarded” positions of a chess piece on a board; in this case Bishops. In some sense, the Bishop can be thought of a Rook turned sideways, because both pieces can only move in two orthogonal directions, but the Bishop’s directions are along the 45 degree diagonals. The difference this makes though is that Bishops have a “color” while Rooks do not. Rooks can move to any square on the board while a Bishop on black squares can only attack other black squares.

A Bishop is a sideways Rook

This limited ability to only attack half of the squares that a Rook can, makes Bishops slightly “weaker”, and so it should not be surprise (as we showed in part I), that on an $n \times n$ board, we can place $2n – 2$ Bishops, giving them a density of roughly $\frac{2}{n}$ for large $n$, and asymptotically going to zero on the infinite board. So in some sense the friendly Bishop’s density on the board is roughly twice that of the slightly more powerful Rook.

But as with the rook, this is the two-dimensional density, while the unguarded Bishop positions clearly have a fractal dimension of one on the board. And as we saw with the Rook, once we go to the unit square board $\mathbb{B}_I$, with a continuum of “squares” at each real coordinate, the rigid shape of the solutions can loosen, so that instead of just the diagonal set with linear measure of $\sqrt{2}$ we can have a set of “friendly” Rooks on the board with 1-dimensional content of nearly 2:

Almost maximal Rooks

and so, turning everything “sideways”, we can see that instead of a set of Bishops on the top and bottom row of the board with measure 2, we can stretch them toward each other, approaching a linear measure of almost $2\sqrt{2}$:

Almost Maximal Bishops

(Note that we have placed black and white circles in the corners to indicate that only the bottom corners are occupied with Bishops, which are in turn attacking the top corner locations). Again, as with the Rook solutions, there are no analogous version of these “continuum” solutions in the finite/discrete boards, as their topology is not dense.

Now one would think that the remaining piece — the Queen — would not be much different from the Rook or Bishop, as it simply has the capturing power of both. But as we will see, the problem suddenly become *very* different, and goes into some very deep and fascinating mathematics on the infinite boards.

A Chess Puzzle, Part V: Rooks and Topology

Part V of A Mathematical Chess Puzzle:

Review

We — that is, I — have been studying the (classic) “unguarded” chess puzzle, which is to say, how to pack as many of one chess piece on a chess board without any of the pieces attacking any others. But with the twist, that the board may be infinite, and possibly uncountably infinite. In the last post we saw that the set of maximal unguarded solutions for the Rook on the board $B_A$ (for any set $A$) had an interesting mathematical property, that they could be represented as the permutation group on the set $A$. In later posts we will see that the Queens also have some interesting properties, some of which can already be seen in our current focus on the Rook.

What is Topology and Why does it Matter?

The glib definition of topology is “rubber sheet geometry”, which is to say, the study of what properties are preserved in an object even if they are made of rubber and can be twisted and stretched but not “torn”. For example, a solid ball and a solid cube can be deformed into each other just by stretching, but to deform a ball into a 1-holed doughnut you would need to puncture the ball to make the hole — which is not allowed.

cube turned to sphere

One way in which a “topology” can be defined on a space is by specifying the way in which two points are “close” to each other, or whether two points can be “connected” by a continuous path.

We have been defining “chess boards” as two-dimensional things (denoted $B_A$) whose “squares” are identified with coordinates (x,y), where the x and y are in some subset $A$ of the real line $\mathbb{R}$. For example, the standard 8×8 chessboard we called $B_8$ uses the coordinates in the subset $\mathbb{Z}_8 = \{ 0, 1, 2, … 7 \}$ (though we could just as easily have taken {1, … 8} ). Among the other coordinate sets $A$ we could consider are the whole numbers, the rationals, the algebraic numbers, and the non-negative real numbers, or for that matter the whole real line.

Topological Density

To see why questions about chess puzzles on various boards $B_A$ may have different answers depending on topology, consider that one of the properties of a subset of $\mathbb{R}$ is called density. A set $A \subseteq \mathbb{R}$ is said to be dense at a point $x \in A$ if for any other point $y \in A$ where $y \neq x$, there exists another point $z \in A$ strictly between them, ie, such that $(z-x)*(z-y) < 0$. The set A is everywhere dense if it is dense at every point of A, and nowhere dense (also called “discrete”) if it is not dense at any point of A.

For example, the whole numbers $\mathbb{W}$ is nowhere dense because there are no other integers between n and n+1. Meanwhile, the rationals $\mathbb{Q}$ are dense, because if $p$ and $q$ are two distinct rational numbers (ie, can be expressed as an integer fraction $m/n$), then it can be shown that $(p+q)/2$ is another rational strictly between them.

And so, to see the reason why density is relevant for chess, simply note that the moves and attack regions for both the King, Pawn and Knight are defined by the squares *adjacent* to their current location, and on a board $B_A$ where $A$ is everywhere dense, there is no such thing as an adjacent square, because there is always another square in A in-between. In other words, the “unguarded” puzzle for Kings, Pawns and Knights does not make sense for dense sets $A$ such as rationals, algebraics, or reals. Meanwhile, because the attacks of Bishops, Rooks, and Queens is defined in terms of directions on the board (horizontal, vertical, diagonal), they have well-defined meanings even on everywhere dense boards.

Connectedness

Strictly speaking, a set is topologically connected if it cannot be represented by a union of two disjoint open sets. By that definition, a set like the integers or the rationals is nowhere connected. For example, the rationals can be represented as $\mathbb{Q} = \mathbb{Q}^+ \cup \mathbb{Q}^-$, where $\mathbb{Q}^+$ and $\mathbb{Q}^-$ are the sets of rationals greater and less than $\sqrt{2}$, respectively. These are both open sets because for any rational number $x \in \mathbb{Q}^+$, all other rationals (sufficiently close) to $x$ are also in the same set.

Nevertheless, we can still define “weaker” forms of connectedness on the rationals, in the sense that the closure of $\mathbb{Q} \subset \mathbb{R}$ is connected. There are also other more general definitions. A set is considered connected, intuitively, if you can get from any place in the set to any other by a “connected” path lying wholly within the set (whose definition needs to be made clear).

For example, suppose we take the set of whole numbers $\mathbb{W}$, and define that two elements $m, n \in \mathbb{W}$ to be “weakly connected” if they differ by no more than 1. So, 2 is connected to 3 and 1, as well as itself, but 2 is not connected to 4. More generally, we can say that two subsets A and B of $\mathbb{W}$ are weakly connected to each other, if there exists a member a in A and b in B such that a and b are connected. Finally, we can define a single subset $A \subseteq \mathbb{W}$ to be weakly connected, if for any two elements $a,b \in A$, there exists a finite sequence of elements $\{ a_i \} \subset A$ such that $a = a_0, a_n = b$, and where $a_i$ and $a_{i+1}$ are connected for all $i = 0…n-1$.

With this definition, we can see that the set $A = \{0,1,2,3,4\}$ is a weakly connected set, but that $A=\{0,1,3,4\}$ is not connected, because there is no way to to get from 1 to 3 by connected elements in $A$ (2 is missing and forms a gap).

Connectedness

The reason all this matters is that we will see that for the solutions to the unguarded rook problem, the topology of those solutions differs when we go from the infinite integer board $B_{\omega}$ to the infinite rational board $B_{\mathbb{Q^{+}}}$ or real board.

The Topology of the Integer Board

Just as with the rationals, we can define “weakly connected” on the integer board $B_{\mathbb{W}}$, by first specifying that two squares a and b are “weakly connected” if they are a Kings move away from each other, that is, one square away horizontally, vertically, or diagonally. And by extension, we can say that any subset $A$ of $B_{\mathbb{W}}$, is connected, if any two members $a$ and $b$ of $A$ can reach each other, by a king moving strictly within the set $A$. From this point on, we will use “connected” to mean “weakly connected”, unless we say otherwise.

Connected Rook Theorem 1: On any finite board $B_n$, the only connected maximal solutions to the unguarded rook problem are the two diagonal solutions, ie the graphs of the function $y=\phi(x)$, where

$$\phi(x) = x$$

or

$$\phi(x) = n – 1 -x$$

The two connected rook solutions on the standard board.

PROOF

We first claim that if two squares in an unguarded rook solution are both connected to a third one, then all three must lie on the same diagonal (northwest or northeast). For, consider the case where square $a$ is connected to square $b$ along the north-east diagonal. We can go through and eliminate with an x all of the squares which are attacked by either $a$ or $b$. And what we see is that the only remaining square which a third rook can be placed and still be attached to $a$ is the lower-left corner.

By repeated application of this claim, what we have just shown is that all of the rooks in a connected component of an unguarded rook solution must lie on the same diagonal. Since this is a maximal unguarded solution, there must be a rook on the first row somewhere. We now claim that it must either be on the leftmost square (forming the northeastern diagonal solution), or else the rightmost square (forming the northwestern diagonal solution). For suppose that the rook is anywhere else along the bottom row, and that (without loss of generality) that it is part of a northeasterly diagonal.

Once again, by a process of eliminating all other squares under attack by the northeasterly diagonal line of rooks, we can see that this leaves a disconnected square in the upper left corner as the only available locations of the rooks that must occupy the remaining rows and columns. But none of these squares are connected to the northeast diagonal, and so a maximal unguarded solution cannot be constructed in this manner that is also connected.

QED

Connected Rook Theorem 2: On the infinite integer board $B_{\omega}$, the only connected maximal solution to the unguarded rook problem is the main diagonal solution, $\phi(x) = x$.

Proof: Clearly the above solution is maximal. Suppose that there is another maximal solution where there is a rook located at a position $P = (m,n)$, where (without loss of generality) we can assume $m > n > 0$. By eliminating all cells attacked by this rook, we now have a board segmented into four non-empty regions, (I, II, III, IV),  each of which would be disconnected from each of the others, save for their connection to the rook at $P$.

By the same argument previously, either $P$ connects regions I to IV by the north-west diagonals, or else it connects regions II and III, by the northeast diagonals. If we choose to join regions II and III, then region III (including P) is an $m x n$ rectangle, in which we would have to fit $m$ rooks, even though we only have $n$ rows. This would force at least two rooks to occupy the same rows, which is not allowed. This leaves us with P joining up regions I and IV, by a northwest diagonal.  Now region I only has $m-1$ columns, while region IV has only $n-1$ rows, and so we have only a finite number of additional rooks we can add to this solution (region II not being connected to P). Thus neither option allows us to define a maximal solution to the unguarded rook problem, meaning that we must conclude that only the case where $m=n$ is valid.

QED.

Unguarded Rooks on a Rational Board

While we could take the entire set $\mathbb{Q}$ of rational numbers, for our purposes it suffices to look at a much smaller but equivalent space, which is the unit interval of rational numbers, ie

$$ \mathbb{Q}^* = \mathbb{Q} \cap [0,1] $$

The set of rationals is countably infinite, and so our board $B_{\mathbb{Q}^*}$ has the same number of “squares” as the infinite integer board. Nevertheless, the unguarded rook problem on this board has a surprising answer in store for us, given by the following theorem:

Connected Rook Theorem 3: There are an infinite number of connected maximal unguarded Rook solutions on the rational unit board $B_{\mathbb{Q}^*}$.

Comment: Just so you can see how on earth this could be possible, consider just one valid solution, defined on $\mathbb{Q}^*$ by the function:

$$\phi(x)=
\begin{cases}
\frac{x}{2}, & \text{if $x \in [0, \frac{2}{3})$} \\
2x – 1, & \text{$x \in [\frac{2}{3}, 1]$}
\end{cases}
$$

The function $\phi(x)$ is a piecewise-linear function, which is a bijection on the set $\mathbb{Q}^*$.

The graph on $B_{\mathbb{Q}^*}$ shows that the placement of the rooks on the board looks like this:

what makes this work is that linear functions are invertible on the rational numbers, and that the rational numbers (unlike the integers) are *dense*. The same trick would not work on the integers, because you can’t apply y = x/2 to the odd integers and get another integer back.

Just to remind the reader of what this picture is about, that thin line is a chess position, consisting of a set of infinitely many rooks along that line, so an (infinite) magnification would reveal them, like this:

This example is just one of many other such piecewise-linear functions which do the trick. The only requirement is that the functions must be strictly monotonic (increasing or descreasing) on the interval $\mathbb{Q}^*$, and (to ensure it is maximal), covers the entire interval, either by $\phi(0) = 0, \phi(1) = 1$ or else $\phi(0) = 1, \phi(1) = 0$.

It turns out that the number of such functions is infinite. This can be seen easily by just noting that there was nothing special about the bend at $x=2/3$, and you could create a similar function with a bend on any other rational number between zero and one, of which there are countably infinite many of them.

Fractal Content

The (two-dimensional) density of the Rook solutions is zero, since lines have zero area. This is the case for all three pieces Rook, Bishop and Queen. But we can still compare the relative density of each of these pieces, by measuring the (one-dimensional) “fractal” content of the positions. For example, the fractal content of the solution shown above is simply the length of the two line segments, which is $2\sqrt{5}/3 \approx 0.75$. But we can actually do much better, by pushing the “corner” of the two lines toward the bottom right like this:

in which case the fractal density is $\sqrt{82}/10 \approx 1.82$. As can be seen, by pushing further the rook solutions can attain fractal densities as close to 2 as we like.

Uncountably Many Solutions

We showed above that there are an infinite number of connected maximal Rook solutions on the rational interval board $B_{\mathbb{Q}^*}$. But you can actually say a lot more; in fact the number of such solutions can be shown to be uncountably infinite, which is to say, there are at least as many of these solutions as there are real numbers (the infinity of the continuum) !

One way you can prove that the number of solutions is uncountable, is to construct a mapping between the (uncountable) set of all infinite decimals between 0 and 1, and a piecewise-linear function with rational coefficients, whose graph is a maximal connected solution of the unguarded Rook problem.

To do this, we first define for each digit from $i = 0 . . . 9$, a different piecewise linear function $f_i()$ on the unit square, simply by mapping the digits to different “bend locations” in the square (having rational coordinates):

Next, we take the unit square and along the main diagonal, lay out an infinite number of smaller squares, each half the size of the previous:

We now show how to map each infinite decimal fraction $z$ (e.g. $z = 0.2739 . . .$ ) to a specific piecewise-linear function $F = F_z(x)$ on the unit square. For each of the smaller squares we have placed inside the unit square, we go through the decimal digits in $z$ and for each digit $d$ define the corresponding function $F_z(x)$ restricted to that range by $F_z(x) = f_d(x)$, corresponding to that digit $d$.

As these infinite number of squares converge to the point $(1,1)$, we define $F_z(1) = 1$, which gives a complete definition on the whole unit interval. Since this mapping is injective, this means that the number of piecewise linear functions with rational coefficients is at least as large as the number of reals on the unit interval, which is uncountably infinite. This completes the proof.

So yes, there are a lot of them. And again, the reason this is so different from the integers has to do with the topological density of the sets.

Just to close the loop, if we look at the full real interval $I  = [0,1] \subset \mathbb{R}$, it is actually much easier to construct topologically connected (in fact, compact) maximal unguarded Rook solutions. To construct such solutions, all that we require is that it be the graph of a function $y = \phi(x)$, where $\phi$ is a bijection on $[0,1]$ such that it is continuous and monotonically increasing or decreasing. That is, if $x > y$ then $\phi(x) > \phi(y)$. A simple set of examples of this are the powers of x, such as $\phi(x) = x^2, \phi(x) = x^3$, and so on.

As with the rational board case, with higher powers of x the curve begins to hug the bottom and right side of the board, with a curve length approaching 2. But can we do any better, or is 2 the largest 1 dimensional fractal content we can attain for connected unguarded rook solutions? The answer is given by the following theorem:

Fractal Content Rook Theorem: The path length of  a connected maximal unguarded Rook solution on the unit interval is not more than 2. That is, the length of any curve $y = \phi(x)$ on the unit interval $[0,1]$ where $\phi(x)$ is a continuous monotonically increasing function from 0 to 1 on the interval is less than or equal to 2.

Proof: We will need to use a little bit of calculus to do this. First, we parameterize the curve by $c(t) = (x(t),y(t))$ where the parameter $t$ ranges from 0 to 1, and

$$
\begin{cases}
x = x(t) = t \\
y = y(t) = \phi(x(t))
\end{cases}
$$

From this we can compute the path length $s$ given by the integral (ie, sum):

$$s = \int_{t=0}^1 \frac{ds}{dt} dt$$

where $ds$ is the incremental increases in the path distance by $(dx,dy)$, given by

$$ds = \sqrt{dx^2  + dy^2}$$

For those unversed in calculus, think of $ds$ as just very small line segments into which the curve has been broken, and use Pythagorean Theorem). Now $dx/dt = 1$ since $x=t$, and $dy/dt \ge 0$, given that $y = \phi(x)$ is monotonically increasing. We can thus take advantage of the inequality

$$ \sqrt{a^2 + b^2} \le \sqrt{a^2 +2ab + b^2} = (a + b) $$

when $a,b \ge 0$ to assert that (formally)

$$\frac{ds}{dt} = \sqrt{(\frac{dx}{dt})^2  + (\frac{dy}{dt})^2} \le \sqrt{(\frac{dx}{dt} + \frac{dy}{dt})^2} = \frac{dx}{dt} + \frac{dy}{dt}$$

and so we can conclude (since x(1) = 1 and y(1) = 1):

$$s \le \int_{t=0}^1  \frac{dx}{dt} + \int_{t=0}^1\frac{dy}{dt}dt = 1 + 1 = 2 $$

which was to be proved. QED

We have now pretty much exhausted the study of unguarded Rooks, so it’s time to touch base briefly on the Bishop, before we move on (at last) to the unguarded Queens.

Sedona Journal – 2020

Preface

With the prospect of lengthy election-related noise in front of us, Gigi and I decided to take a week-long vacation from the world and head down to Sedona, Arizona, far from technology, news, internet, phone, or other channels by which the world would tell us more than we want to know.

Phoenix, Arizona

Our first leg of the trip would be a drive down to Phoenix Arizona to visit my younger brother John and his wife Sheri for the weekend. On Sunday we plan on heading north to Sedona for the rest of the trip. 

Friday, October 30, 2020

Sunny, warm

We drove down the 89A along the North Rim of the Grand Canyon, and from there straight down the 17 into Phoenix and John and Sheri’s house. They gave us a tour of their house; they had moved back to Arizona from Sacramento, where John had been working for a while.

Saturday, October 31, 2020

Desert Botanical Garden

Sunny, shorts weather

Desert Botanical Garden

In the morning we visited the Desert Botanical Garden, whose exhibits were accented by large colorful plastic animals.

 

 

 

 

Mimosas at O.H.S.O Barking Bar (L-R: John, Niles, Sheri, Gigi)

 

 

Barking Bar

Later that day we took bikes over to the “canal” and stopped at the O.H.S.O “Barking Bar” Brewery which served good beer, nachos, breakfast burritos and other fare. Also Mimosas. The bikes in the background have fans in the wheels. Camelback Mountain in the distance.

 

 

Happy Dog

 

 

 

 

This was a dog-friendly venue and many dogs hung out with their masters and had lively conversations.

 

 

 

Arizona Falls

 

Arizona Falls

After brunch we rode further down the canal to Arizona Falls, a dam whose design was inspired by Frank Lloyd Wright, and includes poetry about water and life.

 

 

 

 

 

 

 

 

Trilobite (upper left) found by Nick Edwards

 

Halloween and Fossils

That evening was Halloween, and we sat out front and ate chili while we handed out candy to the neighborhood kids. I noticed that John had framed a set of fossils, including the trilobite that our grandfather Norvis (“Nick”) Edwards had found during a road construction job, and sent to us.

 

 

Sedona, Arizona

Sunday, November 1, 2020

Corn Maze (“Maize Maze”)

The Corn Maze at Tolmachoff Farms

John and Sheri joined us on our way out of Phoenix at Tolmachoff Farms, to explore a Corn Maze. The maze folks gave us a blank map of the maze, which we could fill out with patches of the maze map that we may run across along the way. As a mathematician, my brief glance at the overview map told me that we could “solve” the maze using the simple “right hand rule” (run your right hand along the wall and follow it where ever it goes). We spotted several of the patches (and some riddles) on our path to the exit, but did not find all the patches. So instead of going out the exit I followed the right hand back into the maze, which took us to other segments that (mostly) recovered the other patches. Fun!

Postscript to Maze: After returning home, I suddenly noticed that the shape of the maze was a pickup truck. I pointed this out to Gigi, who replied that she and Sheri noticed that it was a truck all along and were even commenting on being along the “fender” etc but (me being obsessed with solving the maze) had never noticed this obvious fact.

Arrival at Briar Patch Inn

Sunny, cool but pleasant

The “Heron” Cabin at Briar Patch Inn (Click to enlarge)

After leaving Phoenix we drove up to Sedona and checked in at the Briar Patch Inn bed and breakfast. We had reserved their “Heron” cabin, which had a secluded patio and was next to Oak Creek. We were married at the Briar Patch back in 2002, and when visiting we usually stay in the Casa Piedra cabin, but it was booked. The weather was pleasant, sunny but cool. By this time I had put my phone on Airplane mode, and there are no phones or TV’s in the cabins. The world outside was going away.

“The Swimming Hole” At Briar Patch at Sunset (click to enlarge)

The Briar Patch Inn has a dozen or so small rustic cabins scattered about the tree-lined grounds, some large enough for whole families with full kitchens, others just enough for one person with few needs but privacy.

The grounds surround a grass-lined lawn where two sheep graze. When we first came to the Briar Patch, their names were Woolly and Bully. They have since moved to greener pastures, and we are now hosted by two sheep named Rosie and Posie.

The swimming hole is at a place where the creek slows down, bounded on each side by small cascades. It is a bit too cold now to swim in, but is very pretty and peaceful.

 

 

 

 

We settled in for the night and lit a fire in the fireplace, with a large stack of books to read in cozy settings.

 

 

Monday, November 2, 2020

We drove out to “Montezuma’s Castle” and “Montezuma’s Well”, both ancient anasazi settlements notable for having nothing to do with Montezuma (who lived centuries later and never came through here). We reconnected to the world just long enough to get word about our cat Desert Kitty and our dog Blue. Kitty had been feeling bad when we left, not eating, but we got good news that he (whom our cat-sitter renamed “D-Cat”) was eating again.  At least now I would sleep better.

Montezuma’s Castle

Tuesday, November 3, 2020

Cloudy, thunder and lightning. Rain.

Election Day in the US. We are offline.

We were heading to a train ride through red rock and ancient native ruins. We bought “first class” tickets, which gave us a nice train car with a champagne toast at the beginning, along with snacks. I understand that the $800 deal would give one a caboose and no other people. Tempting; I would prefer no exposure to other people, bringing with them news, virii and political notions — or, god forbid, baseball caps.

Our car is the “Sycamore”, though its original name could still be seen on the end exit. We are seated on the “engineer’s left,” which is a double-seater facing another pair of seats (unoccupied for COVID). After looking at the map it was clear that we would have preferred the right-side of the train, which were pairs of singles with a view out to the river gorge and fall colors. Fortunately, we had the option of getting up (with masks) and going to the open-air car where one can look out and over views, and take pictures.

Scene: One of the host/guides came into our car and she was encouraging everyone to get out and take in the view of the open car. Gigi had already done so, but a woman across the aisle from us had not, and started crying inconsolably. The host noticed the woman and came over to find out what was the matter. The woman was older, gray-haired, and sat alone. She explained through the tears that she and her husband had planned to go on this trip for years now, and that just before they began the trip he died unexpectedly. The woman had decided to go on the trip by herself anyway, and now here she was seeing all these beautiful sights, but without her partner to share them.

Along the way, the loudspeakers were playing songs with train themes. For the record, among the songs that we were able to identify included:

  • (Doin’) The Locomotion (Little Eva / Carole King and Gerry Goffin)
  • Peace Train — Cat Stevens
  • City of New Orleans — Arlo Guthrie
  • Petticoat Junction theme — Curt Massey / Paul Henning
  • Midnight Train to Georgia — Gladys Knight & the Pips
  • Last Train to Clarksville – Monkees
  • Blue Train – Johnny Cash
  • Gotta know when to Hold ’em – Kenny Rogers
  • Homeward Bound – SImon & Garfunkle
  • (Join the) Love Train – The O’Jays
  • Nobody Cares About the Railroads Anymore – Harry Nilsson
  • Canadian Railroad Trilogy – Gordon Lightfoot

Wednesday, November 4, 2020

Gigi’s phone is now off, and mine is on Airplane mode, turned on only for the flashlight and clock. Light and Time, in other words.

The agenda for the day: stroll the grounds of the Briar Patch, sit on the wooden swing by the creek, build another stack of logs in the fireplace, read a few chapters of a John LeCarre novel (Russia House), make plans to begin assembling at least the foundation of a Lego block model of Frank Lloyd Wright’s “Robie House”.

3pm: We have finished the foundation of the little house and are declaring victory for the day.

New people have arrived at the Briar Patch; we can hear them talking. We closed our door, lest they carry with them news of the outside world, for good or bad.

Home cooked dinner: Grilled curry chicken, squash and grilled onions with rice. A Negroni digestiv cocktail. Played recorded music by Lyra, who often plays here.

Thursday, November 5, 2020

Drove down to look at Arcosanti, a project by Paolo Soleri, visionary architect who lived from 1919 to 2013. Long time. “Arcology” was his name for a merging of Architecture and Ecology, and began the Arcosanti development based on his ideas.

The Journey Home

Friday, November 6, 2020

A quick trip down to the V-bar-V Ranch, to view some ancient petroglyphs, then hit the road for home. Blue Dog and D-Cat were pleased to see us. Good to be home. Wonder what has been happening in the world. The cat doesn’t seem to care. Neither do I, all I’m thinking of is the afternoon sky in Sedona.

The Sky in Sedona

 

 

A Chess Puzzle, Part IV: A Group of Rooks

Part IV of A Mathematical Chess Puzzle:

Review

We have been studying the (classic) “unguarded” chess puzzle, which is to say, of packing as many of one chess piece on a chess board without any of the pieces attacking any others. What we have been doing is to measure the “density” $\delta$ of the pieces on the $n \times n$ chess board, as $n \rightarrow \infty$. So far we have shown that the densities for the Pawn, Knight and King are 1/2, 1/2, and 1/4 respectively, and also shown what such solutions look like on the infinite board.

In this post we will start to look at the other three pieces, Rooks, Bishops, and Queens, and will see how different things become. The reason things are different is that, unlike the Pawn, Knight and King, these remaining pieces have a “control region” which is unlimited, and extends out infinitely on an infinite board.

Our first clue that things are different may be seen by this analogous result from the previous post:

The Chess Density Theorem – Rooks, Bishops and Queens

DENSITY THEOREM 2: For Rooks, Bishops and Queens, their asymptotic density $\delta$ on the infinite board ($B_{\omega}$) is given by:

$\delta = 0$.

In other words, on an infinite chess board, the fraction of squares you can cover with “friendly” Rooks, Bishops or Queens is vanishingly small. As a reminder, by “friendly” I mean that none of the pieces are threatening each other by standard chess rules. Also, if you were to throw a dart at the board with a densest packing of friendly pieces, the odds of hitting a nonempty square is zero.

PROOF: Unlike the Pawns, Knights and Kings, the proof of this theorem is practically a one liner, as we can easily show that for each piece X in the set {R,B,Q}, there is a constant C>0 so that if $\Phi \in \mathscr{U}(B_n,X)$ is an unguarded chess position of that piece on the n-Board, then its weight (ie, the number of pieces in the position) $\Vert \Phi \Vert$  is bounded by

$$\Vert \Phi \Vert \le C*n$$

And so the density $\delta_n$ on the n-board is bounded by

$$ \delta_n = \frac{\Vert \Phi \Vert}{n^2} \le C/n$$

and so, since $C$ is a constant, then the quotient $C/n \rightarrow 0$ as $n \rightarrow \infty$.

For Rooks and Queens, the constant $C=1$ will do, because the n board has n rows, and neither rooks nor queens can share a row. For Bishops, the constant $C=2$ will do, because there are exactly $2n-1$ northwest diagonals on the n-board, and friendly bishops cannot share a diagonal. $\blacksquare$

Okay, so friendly Rooks, Bishops, and Queens have to be very sparse on the infinite board. But what do their solutions look like?

The Rook

From part one we already know at least what one unguarded Rook solution looks like on the 8-board:

8 Friendly Rooks

Which immediately suggests a general solution on the infinite board:

$$\Phi_R(x,y)=
\begin{cases}
R, & \text{if $x = y$} \\
\varnothing, & \text{otherwise}
\end{cases}\tag{Rook}
$$

On the infinite board the position looks like this (the diagonal line is infinitely thin):

Friendly Rooks on the Infinite Board

We can readily see that this is indeed a solution, and it is also apparent that except for that infinitely thin line of rooks the board is empty. But are there any other solutions? There are, and they have a very interesting characteristic. At least for me…

Let’s start with a definition:

DEFINITION: A chess position $\Phi$ on a board B is said to have the Rook Property if and only if every row and column of B has exactly one non-empty square.

Obviously, any maximally unguarded Rook positions have the Rook property. But this is also a property of maximally unguarded Queens, so the definition is a bit more general.

If you think about it, if a position $\Phi$ has the Rook Property on the n-Board, then this means for any integer $x$ between 0 and $n-1$, there is exactly one integer $y$ in that same range so that $\Phi(x,y)$ is a non-empty piece. In other words, $y$ is a function of $x$, so that we can write $y = \phi(x)$. And not only that, but since no two non-empty squares of $\Phi$ can have the same row (y-value), this means that if $\phi(x_1)= \phi(x_2)$, then $x_1= x_2$. In other words, the function $\phi(x)$ is also an invertible function (bijection) from the set {0,1,….n-1} to itself.

Such functions $\phi(x)$ have a more common name, which is that they are permutations (rearrangements) of the set of n elements. The set of all permutations form what is called the symmetry group Sym(n), and it appears all over the place in group theory.

A group is simply a set $G$, together with a “group product” often denoted $a{\circ}b$, with the property that (1) there is an “identity” element $e$ for which $e{\circ}a = a{\circ}e = a$ for all $a \in G$, (2) for every $a \in G$ there is an “inverse” $a’$, so that $a’{\circ}a = a{\circ}a’ = e$, and (3) the product is associative, ie $(a{\circ}b)\circ c= a{\circ}(b \circ c)$. The integers and addition are an example. Group theory has a lot of uses.

In any case, it can (easily) be proved that this relationship between $\Phi(x,y)$ and $\phi(x)$works both ways, on any board:

THE ROOK GROUP THEOREM: If $A$ is any set (finite or infinite), $\Phi$ is a position on the board $B_A$, then $\Phi$ has the Rook Property if and only if there is a permutation $\phi \in Sym(A)$ such that the position $\Phi$ satisfies the following:

$$ \forall x,y \in A: \Phi(x,y) \neq \varnothing \text{ if and only if } y=\phi(x)$$

In particular, if consider only rook positions on a board $B_A$, then there is an exact one to one correspondence between the maximally unguarded positions and elements of the symmetry group $Sym(A)$. In other words, the set of maximal friendly Rook positions forms a group, and you can actually define a group operator (*) on solutions so that the group product of two such solutions forms another solution!

In fact, if you inspect the one example we showed for eight rooks on the standard 8-board, you can see that it corresponds to the identity element of Sym(8), ie, $\phi(x) = x$.

This may all sound like gobbledegook, but it gives us some simple ways to cook up new examples of friendly Rook positions. For the eight board, all we need to do is write down a couple of permutations of the set {0,1,…8}.

For example consider the following two permutations:

(3,2,1,0,7,6,5,4) and  (1,3,5,7,0,2,4,6).

These can be thought of as two functions, each being the ordered list of values you get by plugging in 0, then 1, then 2, and so on. The chess positions that correspond to them, are simply the graph of the functions on the board, like this:

Friendly Rooks as Permutations

To show how to “multiply” these two solutions to get a new one, you just compose their two permutations as functions. So for example, the first permutation takes 0 to 3, while the second permutations takes 3 to 7, so the composition of the two takes 0 to 7, and so on. We can show their group product graphically like this:

Group Product of two Rook Positions

So how many solutions on the n-board are there? From our theorem above we know the answer, which is the same as the number of permutations of $n$ objects, which is to say $n! = 1*2*3*4*….*n$ (n factorial). For the 8×8 board, this works out to 8! = 40320 solutions.

Now how about $B_\omega$, the omega Board? How many solutions? We know from our theorem the friendly Rook positions correspond to the elements of the permutation group $Sym(\mathbb{W})$, where $\mathbb{W}=\{0,1,2,3,…\}$, the set of whole numbers. Now the set of whole numbers is infinite, and has cardinality $\aleph_0$, so is countably infinite. But how many permutations of this set are there?
We will have to do some transfinite arithmetic. To specify a permutation $\phi$ on $\mathbb{W}$ we have to chose values for $\phi(n)$, for every value of n. For n=0 we can pick any of an countably infinite ($\aleph_0$) number of values for $\phi(0)$. Suppose we pick 7. Now how about the next value for n=1? We can’t choose 7, but we still have an infinite number of choices left. And so on. Altogether, we have to make a (countably) infinite number of choices among a countably infinite number of values. This gives us a total count of

$$\aleph_{0} \times \aleph_{0} \times \aleph_{0} \times … =\aleph_{0}^{\aleph_0}$$

solutions. This is an uncountably infinite number of solutions, equal to c, the infinity of the continuum, the number of points on the real line.

Wow. That’s a lot of solutions.

As we will see later, the queen also has a similarly large collection of unguarded solutions, but the Rook is unique in being the only piece of the six whose solutions have a natural group structure.

In the later posts we’ll take a look at the Bishop and queen, and also start to examine fractals and even more infinite boards.

A Chess Puzzle, Part III: The Pawns, Knights and Kings

Part III of A Mathematical Chess Puzzle:

This is part III of my series of posts on the “Chess Density Problem”. In Part I, I introduced the problem and laid out some definitions to clarify what I meant by some of the terms such as “Board” and “Density”. In Part II I gave everyone a breather.  Hope you’ve rested up. In this one the math is going to be a bit thick, but the purpose of it all will be to prove this first theorem:

The Chess Density Theorem – Pawn, Knight and King

DENSITY THEOREM 1: For Pawns, Knights, and Kings, their asymptotic density $\delta$ on the infinite board ($B_{\omega}$) is given by:

1.1) $\delta(P) = 1/2$.

1.2) $\delta(N) = 1/2$.

1.3) $\delta(K)= 1/4$.

In other words, on an infinite chess board, you can pack at most half of the squares with “friendly” Pawns or friendly Knights, and only a quarter of the board with friendly Kings. As a reminder, by “friendly” I mean that none of the pieces are threatening each other by standard chess rules.

Plan of Attack

As a starting point, we can note that the solutions found on the standard 8-board can be seen as built up from simple “tile” patterns made of identical small boards:

 

These tiles each have the densities that are claimed by the Density Theorem above. If we can prove that no such tiles can have a greater density, we can show that this must hold for larger boards. The way which we will do this is to define a precise way to “measure” the density of any arbitrary block of squares, and prove some useful properties of that measure-function.

Definitions

It always helps to define things precisely. There are a lot of fancy symbols, but the general ideas should be clear.

The Pieces

The six standard Chess pieces are King, Queen, Bishop, Knight, Rook, and Pawn, which we will denote with the initials {K,Q,B,N,R,P}, respectively. To this set we will also add the “Empty” piece, denoted $\varnothing$, which indicates that a square is empty.

The set of all pieces including the empty piece will be denoted

$$\mathscr{P} = \{K,Q,B,N,R,P,\varnothing \}$$

We include that “empty” piece to allow us to explicitly specify that a square is empty.

Control Zones

DEFINITION: The control zone of a piece X at position $x$ on a board B is the set of all other locations $y \in B$ which (by the standard rules of chess) are subject to attack by X (with no other pieces on board).

The Control Zones of the Pawn, Knight and King are shown in blue below:

Pawn

Knight

King

 

 

 

 

 

 

One thing to note about these three pieces is that their control zones can be defined by a specific finite set of integer offsets from the location of the piece itself. The same cannot be said for the other three pieces (Rook, Bishop, Queen), as their Control Zones are not defined by a fixed set of offsets, but a set of directions on the board, by which they can attack (again, assuming that there are no other pieces blocking the way).

The Control Zone directions for the Rook, Bishop and Queen are shown below.

Rook

Bishop

Queen

 

 

 

 

 

 

As we begin to work with the infinite boards we will see that this difference between the two sets of pieces profoundly affects how “dense” they can be.

In particular, on infinite boards the control zones of the Rook, Bishop and Queen are also infinite.

Chess Positions

A lot of the definitions below are just stuff I made up. If they correspond to standard chess terms I will mark them with (standard). If I feel that the definition may not be clear, I will give an example as well.

DEFINITION: For a given board $B$ and any subset $\Omega \subseteq B$, a chess position $\Phi$ on $\Omega$ is simply a function $\Phi$ that for each location $x \in \Omega$ there is a piece $p \in \mathscr{P}$ given by $p = \Phi(x)$ at that location. That is, it is a function

$$\Phi: \Omega \rightarrow \mathscr{P}$$

For all of the chess positions we are considering for this puzzle, we only have one kind of piece on the board at the time. And so we will call them “Pawn-positions”, “Queen-positions” etc. Thus if $\Phi$ is a King-position on $\Omega$, then it is a function $\Phi: \Omega \rightarrow \{K, \varnothing\}$.

DEFINITION (standard): A Position is called Unguarded if no piece in the position is in the control zone of another piece. In other words, unguarded positions are positions where the pieces are all “mutually friendly”.

DEFINITION: If $n > 0$ is an integer and $X$ is a (non-empty) chess piece, define $\mathscr{U}(n,X)$ to be the set of all “Maximally Unguarded” X-positions using only piece $X$, on the n-Board. That is, if there are no other unguarded positions with a larger number of pieces.

Measurement of Positions

DEFINITION: For any position $\Phi$ on $\Omega$, we can define the size ($\vert \Phi \vert$) and weight ($\Vert \Phi \Vert$) of that position by the number of elements in its domain $\Omega$ and the number of non-empty locations in its range, ie

$$\vert \Phi \vert = \vert \Omega \vert$$

$$\Vert \Phi \Vert = \vert \{ x \in \Omega: \Phi(x) \neq \varnothing \} \vert$$

EXAMPLE: In the little 4-board below, there are four queens, and the rest are empty.

Four friendly Queens

So, following the defintion above, the board $\Omega$ is the 4×4 board, and its position function $\Phi$ maps every square to the empty piece $\varnothing$ except for the four squares that are mapped to the Queen. The size of the position $\Phi$ is simply the number of squares in $\Omega$, ie $\vert \Phi \vert = \vert \Omega \vert = 16$, while the weight of the position is the number of squares mapped by $\Phi$ to the Queen, ie $\Vert \Phi \Vert$ = 4.

DEFINITION: The for any position $\Phi$ on some board $\Omega$ its Density (denoted $\delta(\Phi)$ ) is the ratio of its weight divided by the size of the position , ie

$$\delta(\Phi) =  \frac{\Vert \Phi \Vert}{ \vert \Phi \vert }$$

DEFINITION: The for any integer n>0, the n-Density (denoted $\delta_n(X)$ ) of a piece $X$ on the n-Board is the maximum density of all possible unguarded positions $\Phi \in \mathscr{U}(n,X)$ ie

$$\delta_n(X) =  \max_{\Phi \in \mathscr{U}(n,X) }\delta(\Phi)$$

EXAMPLE: So, in the previous example for the unguarded queens on the 4-board, it can be seen that:

$$\delta_4(Q) = 4/4^2 = 4/16 = 1/4$$

With all of these preliminaries out of the way, we can now define what we really mean by the density of a piece on an infinite board:

DEFINITION: The Asymptotic Density (denoted $\delta(X)$ ) of a piece $X$ on the infinite board $B_\omega$ is the limit (if it exists) of the n-Density $\delta_n(X)$ of piece X as $n \to \infty$ ie

$$\delta(X) =  \lim_{n \to \infty }\delta_n(X)$$

The Linear Density Measure

For each of the pieces mentioned in the theorem above, their claimed density is of the form $\delta = 1/m$, where $m$ is some positive integer. It turns out that working with density ratios directly is tricky, so we will look at a different “measure” of density that turns out to be easier to work with, as we break a large board into tiles.

So, suppose $w$ is the weight of an X-position $\Phi$ on a set $\Omega$ of size $s$. Then the density $\delta$ is given by

$$\delta = \frac{w}{s}$$

and so if indeed $\delta=1/m$ then

$$1 = m\delta = m\frac{w}{s}$$

and therefore $mw = s$, which is to say that whenever a position has density 1/m, then we can also write

$$mw – s = 0$$

Using this fact, for any sub-position $\Upsilon \subseteq \Phi$ let’s define the “m-linear density” measure function $\Lambda_m(\Upsilon)$ by

$$\Lambda_m(\Upsilon) = m \Vert\Upsilon\Vert  – |\Upsilon|$$

Intuitively, what this measure value tells you is how much over or under the expected density a position is from $1/m$. if the value is exactly zero then the density is exactly $1/m$.  We now present a lemma (ie, a small theorem we can prove now and use later):

LEMMA 1: If $\Phi$ and $\Upsilon$ are two positions defined on disjoint subsets $S$ and $T$ of a board $B_n$, and $m > 0$ is an integer then

$$\Lambda_m(\Phi \cup \Upsilon) = \Lambda_m(\Phi) + \Lambda_m(\Upsilon)$$

that is, the m-linear density of the union of two disjoint positions is the sum of the m-linear densities of each.

PROOF: Follows immediately from the fact that the size of the union of disjoint sets is the sum of the sizes, and regrouping of terms:

$$\Lambda_m(\Phi \cup \Upsilon) = m\Vert\Phi \cup \Upsilon\Vert – \vert\Phi \cup \Upsilon\vert$$

$$=m(\Vert\Phi\Vert + \Vert\Upsilon\Vert) – (\vert\Phi\vert + \vert\Upsilon\vert)$$

and so, regrouping like terms:

$$=(m\Vert\Phi\Vert – \vert\Phi\vert) + (m\Vert\Upsilon\Vert -\vert\Upsilon\vert)$$

$$= \Lambda_m(\Phi) + \Lambda_m(\Upsilon)$$

QED (Quod Erat Demonstrandum = which was to be proved).

$\blacksquare$

LEMMA 2: There exists a positive constant $C > 0$ such that the 2, 2, and 4 linear densities $\Lambda(\Omega)$ of maximal unguarded positions $\Phi$ of Pawns, Knights and Kings on the n-board are bounded by the following expressions:

$$ \forall \Phi \in \mathscr{U}(n,P): 0 \le \Lambda_2(\Phi) <  C * n \tag{2.1} $$

$$ \forall \Phi \in \mathscr{U}(n,N): 0 \le \Lambda_2(\Phi) <  C * n \tag{2.2} $$

$$ \forall \Phi \in \mathscr{U}(n,K): 0 \le \Lambda_4(\Phi) <  C * n \tag{2.3} $$

PROOF: We will start with the formula for the Pawn; the other proofs will be similar. We first note that for the $2 \times 2$ tile $T$ for the Pawn case, the linear density is exactly $\Lambda_2(T) = 2*2 – 4 = 0$. Is it possible to place any more than two “friendly” pawns in the tile $T$? No, because each cell in the first row attacks the diagonal cell in the second row, so you can choose at most one pawn from each “attack pair”. This means that for any unguarded pawn position $\Phi$ on the $2 \times 2$ board, its linear density is bounded by $\Lambda_2(\Phi) \le 0$.

Attack Pairs

So now what about the 2-linear density for the $n \times n$ board $B_n$? Note that if $n$ is divisible by two we can completely fill up the entire board with these 2-tiles, like this:

Tiled 8-board

Now note that since this n-board (for $n$ = 2k even) is the disjoint union of these $2 \times 2$ tiles $T_1, T_2, …T_{k^2}$, then for any unguarded position $\Phi \in \mathscr{U}(n,P)$, its 2-linear density must be the sum of the densities of these tiles (this follows from Lemma 1) But the positions on those tiles must themselves be unguarded positions, and so for each tile, $\Lambda_2 \le 0$. Therefore, we must have that:

$$\Lambda_2(\Phi) = \Lambda_2(T_1) + … + \Lambda_2(T_{k^2}) \le 0 $$

since all of the terms are known to be $\le 0$. Note that we have also shown that the maximum value of 0 can be attained.

We have just shown that for all even n, and unguarded positions $\Phi$ on the n-board, the value of $\Lambda_2(\Phi)$ is bounded by 0. What about odd n, where $n = 2k+1$? In that case, we can break up the n-board into the disjoint union of $k^2$ tiles which are $2 \times 2$, $k$ tiles which are $1 \times 2$, $k$ tiles which are $2 \times 1$, plus a $1 \times 1$ square, like this:

Odd nxn Board, tiled

So now we can play the same game here, and compute the largest possible values of $\Lambda_2$ for each of these little blocks. For both the 1×2 and 2×1 tiles, we have no attack-pairs, and so we can fill both squares with pawns, giving them a maximal 2-density of $\Lambda_2 = 2*2 -2 = 2$. Similarly, we can fill the single 1×1 square with a pawn, for a 2-density of $\Lambda_2 = 2*1 -1 = 1$.

Now there are $k$ of each of the 1×2’s and 2×1’s, and $k^2$ of the 2×2’s and so altogether we can bound the 2-density of the odd n-Board by the sum of these bounds, ie

$$\Lambda_2(\Phi) = k^2 \Lambda_2(2×2) + k\Lambda_2(1×2) + k\Lambda_2(2×1) + \Lambda_2(1×1)$$

$$ \le k^2*0 + k * 2  + k * 2 + 1 $$

$$ \le 4k + 1 \le 4k + 2 $$

$$ \le 2 * (2k + 1) \le 2n $$

And so, what all this means is that for the Pawns, given any unguarded position $\Phi$ on any n-Board, we can bound the largest possible value for its 2-density with a value of C=2, ie,  $\Lambda_2(\Phi) \le 2n$. This proves part (2.1).

Similarly, for the King, we can play the same game, but now using the 4-density $\Lambda_4$. In this case, we can only place one king on the 2×2 tile (since each corner attacks all the others, forming an “attack quadruple”). This gives the 2×2 tile the maximal 4-density of $\Lambda_4 = 4*1 – 4 = 0$. And in the same way, we can show that for even n, the 4-density is bounded by 0, and for the odd case where $n=2k+1$, we can place one king each on the 1×2 and 2×1’s (for a density of $4*1 – 2 = 2$), and one king on the single cell (for a density of $4*1 -1 = 3$). Putting it all together, the King’s 4-density on the odd n-board is given by

$$\Lambda_4(\Phi) = k^2 \Lambda_4(2×2) + k\Lambda_4(1×2) + k\Lambda_4(2×1) + \Lambda_4(1×1)$$

$$ \le k^2*0 + k * 2  + k * 2 + 3 $$

$$ \le 4k + 3 \le 6k + 3 $$

$$ \le 3 * (2k + 1) \le 3n $$

And so for the King we can use the constant $C=3$, so $\Lambda(\Phi) \le 3n$.  This proves (2.3).

Finally, for the Knight, we have to work a bit harder. The Control Area around the knight extends beyond just a single square, and so we need to use a 4×4 tile to build up a solution. In the 4×4 tile we showed above, we were able to fill exactly half (8) with friendly knights, giving the 4×4 tile a 2-density of $\Lambda_2 = 2*8 – 16 = 0$, just like the Pawn. To prove that this is the best we can do with the 4×4 tile and knights, consider the following “attack pairs” of knights on the tile:

Knight attack-pairs

We see that there are eight pairs of mutually-attacking squares by knight moves, meaning we can choose at most one from each pair on which to put a knight. This means there can be at most eight friendly knights on any 4×4 tile. And so, just as with pawns we have $\Lambda_2 \le 0$ on any 4×4 tile.

By the same arguments we gave before, if $n$ is divisible by 4, then we can show that an unguarded knight position on an n-board has density $\Lambda_2 \le 0$, and for all other $n = 4j + k$, with $k \le 3$, we can cover the n-board by $j^2$ 4×4 tiles, $2j$ tiles on the border of size $k \times 4$, which can be shown to never have 2-linear densities greater than $\Lambda_2 = 4$, plus a $k \times k$ corner tile, with density no greater than 1. Putting this altogether, it can be shown that a value of $C=1$ is good enough for knights to bound $\Lambda_2(\Phi) \le C*n$ for any unguarded $\Phi \in \mathscr{U}(n, N)$. This proves inequality (2.2) and the Lemma.

$\blacksquare$

PROOF OF THEOREM 1

The theorem is in three parts, for the Pawn, Knight, and King. We will only show the proof for the Pawn, the other proofs are identical in structure. From Lemma 2, formula (2.1), there is a positive constant C such that for all positive integers $n$ and maximally unguarded Pawn positions $\Phi$ on the n-Board, we have

$$ 0 \le \Lambda_2(\Phi) <  C * n$$

which by definition of $\Lambda_2$ means that

$$ 0 \le 2*\Vert \Phi \Vert – \vert \Phi \vert < C*n $$

Dividing both sides by $\vert \Phi \vert$ we have

$$  0 \le \frac{2*\Vert \Phi \Vert}{\vert \Phi \vert} – 1 < \frac{C*n}{\vert \Phi \vert}$$

But since on the n-Board $\vert \Phi \vert = n^2$, we can use the definition of the position density $\delta(\Phi)$ to obtain

$$ 0\le 2\delta(\Phi) – 1 < C/n$$

in other words (adding one and dividing by 2):

$$ \frac{1}{2} \le \delta(\Phi) < \frac{1}{2} + C/2n$$

And so, since $C$ is a positive constant, as $n$ goes to infinity, the value of $C/2n$ becomes vanishingly small, and in the limit goes to zero. By the definition of the asymptotic density, then, the value of $\delta(Pawn)$ is exactly 1/2, which was to be proved.

$\blacksquare$

Unguarded Positions on the Infinite Board

So far, we have been talking about the density of unguarded positions on large but finite boards $B_n$, and their limit as $n \to \infty$ (“n goes to infinity”). But what happens when $n$ gets there? Is there a position on the infinite board $B_\omega$ that is still “unguarded”, and what do we mean by “maximal”?

If we use the finite solutions as a guide, we can quickly propose positions $\Phi_p$ on $B_\omega$ for p = Pawn, Knight, and King. Recall that a “position” on any board B is simply a function mapping every element $(x,y) \in B$ to a member of $\mathscr{P}$, the set of all chess pieces including the “empty” piece $\varnothing$. The omega board consists of all integer pairs $(x,y)$ such that x and y are greater than or equal to zero (so, the bottom left corner square is $(0,0)$).

The solutions are:

$$\Phi_P(x,y)=
\begin{cases}
P, & \text{if $y$ is even} \\
\varnothing, & \text{otherwise}
\end{cases}\tag{Pawn}
$$

 

$$\Phi_N(x,y)=
\begin{cases}
N, & \text{if $(x+y)$ is even} \\
\varnothing, & \text{otherwise}
\end{cases}\tag{Knight}
$$

 

$$\Phi_K(x,y)=
\begin{cases}
K, & \text{if $x$ is even AND $y$ is even} \\
\varnothing, & \text{otherwise}
\end{cases}\tag{King}
$$

It is relatively straightforward to see that when restricted to a finite n-Board these positions are the same as the ones using the “tiles” we defined above, and so they are still unguarded positions. But in what sense are they “maximal”? The weight of these positions are infinite, so how can one infinite position have a greater weight than another? The answer is, they are maximal in that expanding the position to include additional pieces in locations where the original position does not, must also have a hostile piece in the set.

The proof that these positions are indeed maximal, is left as an exercise to the reader.

What the infinite solutions look like

Here are the Knight and King Positions for the infinite boards, as viewed from a great distance (the white squares are the empty ones):

King Position on the 300 Board

Knight Position on the 300-Board

 

 

 

 

 

 

 

 

 

 

This is not actually an infinite board, of course, but a 300 x 300 one. As you can see, as the number of squares on a side increases to infinity, the individual pieces blur out to form a shade of grey, about 50% black for the Knight, and 25% black for the King. Another way to define the density, is that if you were to throw a dart at one of the boards (say, the King), the odds would be 1/4 that your dart would hit a King.

In the next post, we will look at the Bishops, Rooks, and Queens, and see just how strange and different the situation becomes, especially as we venture into larger transfinite boards.

A Chess Puzzle, Part II: Ruminations

Part II of A Mathematical Chess Puzzle:

This is part II of my series of blog pieces on a mathematical chess problem. In part one I tried to keep the math to a minimum, but after looking back at the post I see that it may still be a bit too much, and in the next couple of posts the math is only going to get more dense, but there is not much I can do about it.

So for this post I decided to give you a breather, and talk a bit about what the point of all this is.

After digging through my phone I just realized that it has been eight months since I was in the doctor’s waiting room, staring up at the ceiling, and visualizing an infinite chess board with chess pieces sitting on it, none of which were under attack. I took a photo of the ceiling with my phone and included it in my previous post. The date was May 15, 2019.

That fact surprised me; I was sure it was only a few weeks ago that I started scribbling little chess positions on scraps of paper, making notes for a blog post or maybe an article in the American Mathematical Monthly. Gigi’s family will tell you I spent much of the Christmas holiday sitting in a chair, with scraps of paper filled with diagrams and chess boards and incomprehensible hieroglyphics.

What appears to be the case is that this question, of how to fit pawns or queens on an infinite board without attack, spun off numerous other questions and math problems, and some of them were tricky, and some led to questions that seemed difficult. And in the process of thinking about these problems I became lost in a sort of reverie, and lost track of time. Days turned to Weeks turned to months. Eventually I found easy ways to answer many of the problems, and later posts will show those results. But you should realize that it took many tries before I found what now seems to be the obvious approach.

One of the problems was hard enough that I felt that I would need to just leave it in the article is as a conjecture, and hope that somebody smarter than me would be able to solve it.

But just this past Sunday, while out on a training run for the Zion Half Marathon, I came up with a way to solve the problem, and within an hour after getting home I was able to write QED at the end of a proof. Now it too seems obvious, but even so might have made a good Putnam math problem.

In any case, here is the thing. There is a reason why there are all these math symbols and arrows. Each symbol has a precise meaning, and in order to solve a problem it is often the case that the hard part is figuring out how to state the problem in a clear and precise manner. In writing down the math, you are creating a machine, and once done the machine can do a lot of the heavy lifting.

And here is the other thing: why do I do this? Because, believe it or not, it’s fun.

Here’s another preview:

 

A Chess Puzzle, Part I: The Infinite Chess Board

First part of A Mathematical Chess Puzzle:

A while ago I was in the waiting room of the doctor’s office, staring at the ceiling, and looking for something to keep myself amused before the appointment. (The visit was nothing serious, just wanting to update my vaccinations.)

Ceiling at doctor’s office

The ceiling was covered with those white acoustic panels, forming a very large colorless and seemingly infinite chess board. I stared at the tiles, trying to think up some kind of puzzle to keep me amused while waiting. I came up with this:

The Chess Density Problem

Suppose you have an infinite chessboard, and an infinite number of one of the six kinds of chess pieces (King, Queen, Rook, Bishop, Knight, and Pawn). What is the densest packing of “friendly” pieces you can have on that board?

Just asking the question raises more questions:

  • What do I mean by “infinite” ?
  • What do I mean by “densest” ?
  • What do I mean by “friendly” ?
  • And how would one prove such things?

So far I’ve spend a fair amount of time and scratch paper on this — quite a bit while sitting around with Gigi’s family for Christmas — and have some things to report. The first version of this post was very math-heavy. Now it is a bit lighter. You’re welcome.

Disclaimer

Friends who have suffered through games with me will tell you that I am terrible at playing chess, and spend far too long to come up with a bad move that in standard chess notation would probably be denoted “Nf3 Nc6??”. Not just Chess, but Go, Twixt, Bridge or other strategic games. In defense, I can only say that I am spending all of that time trying to solve the problem, based on the game-theoretic fact that a “best” move exists. The decision tree is so deep, however, that no computer yet exists that can explore it completely more than a few moves ahead.

Good chess players acquire skill through experience and study, making decisions with imperfect knowledge. They stand at the edge of the Grand Canyon of moves and can walk down trails blazed by others. I can only stand at the edge, frozen by the awesome infinitude of the depths. Then I fall in.

Anyway, this isn’t the organic chemistry of Chess, but the simpler math of atoms and fundamental particles, Boards and pieces. Moving on…

Density: A Simple Example

Here is a simple (finite) example of four Queens on a $4 \times 4$ board, where no Queen is attacking any other, either on rank, file, or either diagonal:

 

Four friendly Queens

So for the $4 \times 4$ board, with its 16 squares, you can only “pack” at most four queens, meaning the “density” (which I denote $\delta$) of Queens on this board is

$$\delta = 4/16 = 1/4$$

Is this the “best” we can do in this case? The answer is yes, for the simple reason that at most one Queen can be on each row (or they would not be friendly), and there are four rows.

What is Already Known

This “packing” puzzle is what’s known as a Mathematical Chess Problem, and falls in the category of “recreational mathematics”. That is, mathematics done for amusement rather than “professional” work on “serious” problems. It should be noted, however, that often these problems lead to questions which are actually “deep” and lead to more important work.

In any case, from the literature, the task of arranging “friendly” chess pieces is called an Independence Problem (or an “Unguard” problem), and a chess position in which all of the pieces are friendly is called “Unguarded”. If an unguarded position is the “best” we can do (ie, there is no unguarded position with more pieces), we will call it maximally unguarded.

So, for example, on the standard $8 \times 8$ chessboard, here are examples of how to arrange “friendly” Pawns, Knights and Kings:

32 Friendly Pawns

32 Friendly Knights

16 Friendly Kings

 

 

 

 

 

 

 

We will show (later) that these are “maximally unguarded” positions, and so the density of these pieces on the standard 8-board are 1/2, 1/2, and 1/4, respectively.

And here is how to arrange the remaining three “friendly” pieces, the Rooks, Bishops and Queens:

8 Friendly Rooks

14 Friendly Bishops

8 Friendly Queens

 

 

 

 

 

 

 

These last three are maximally unguarded positions, and so the density for these three pieces on the standard board are 1/8, 7/32, and 1/8, respectively.

To see this in the case of the Rooks and the Queens, we can use the same argument as we did above with the Queens on the 4-board, by noting that there are exactly 8 rows and only one Rook or Queen can be on any row.

For the 14 Bishops, you can make a similar argument, by noting two things:

  1. There are exactly 15 diagonals in one direction (shown below in blue), and
  2. For each diagonal numbered 1 and 15, there is only one square available for placing a Bishop, and those two squares are “hostile” along the red diagonal, so you can only choose one.

Fifteen Bishop Diagonals

We have separated out the pieces into these two sets of three, because each set have their own unique characteristics, unlike the other set; and the differences in density become more pronounced as the size of the boards becomes larger and becomes infinite. Not only that, but the kind of “density” has to be defined differently, and even the size of the “infinite” board becomes important; as we’ll see, some chess boards are more infinite than others.

Chess Boards: Finite and Infinite

All of the “chess boards” we will be working with will be two-dimensional, and each “square” on which a chess piece can be placed can be indicated by a pair of numbers $(x,y)$, where $x$ and $y$ are coordinates in some set $A$, which we will call the Address space. A chess board that uses coordinates in Address space $A$ will be denoted $B_A$. Mathematically, the board $B_A$ can be identified with the Cartesian product $A \times A$.

For example, the squares of our standard $8 \times 8$ chess board can be specified using $A = \{0,1,2,3,4,5,6,7\}$. The set of integers from 0 to 7 is sometimes denoted $\mathbb{Z}_8$, and so the standard chess board is $B_{\mathbb{Z}_8}$.  To make life simple(r), we will also call this board $B_8$, or simply the “8-Board”, and similarly an $n \times n$ board (where n is an integer) will be called the “n-Board”, or $B_n$.

Here for example is the 5-board $B_5 = B_{\mathbb{Z}_5}$:

The 5-Board

Okay, so far so good (you still with me, here?). Now the point of all this mathematical gobbledegook (a scientific term), is that I wanted to look at the “Unguarded” chess problem on an infinite chess board, meaning that I want to use for a “coordinate space” $A$ an infinite set. The first one that comes to mind is the set of whole numbers, ie

$$\mathbb{W} = \{0,1,2,. . .\}$$

And so we can define our first infinite chess board as $B_{\mathbb{W}}$, consisting of all $(x,y)$ where $x$ and $y$ are non-negative integers.

Intuitively, the board can be visualized like this:

The ω-Board

This picture looks a lot like the ceiling at my doctor’s office, which is what started this whole thing.

Now the “size” (cardinality) of the whole numbers is “infinite”, but the specific mathematical name for this “infinite” value is $\aleph_0$ (pronounced aleph-null). Unlike the conventional symbol for infinity ($\infty$), $\aleph_0$ has a more precise meaning, and it refers only to “countably” infinite sets, ie, those which you can count with the integers 1,2,3, etc. A related infinite “ordinal” number is called $\omega$ (small omega), and with some “abuse of notation” we will refer to $B_{\mathbb{W}}$ as $B_{\mathbb{\omega}}$, or simply the $\omega\text{-Board}$.

Chess Boards: Infinity and Beyond

There are many other ways to define infinite boards, but for now the only other board we will look at uses as its coordinate space the closed unit interval on the real line, that is,

$$I = \{ x \in \mathbb{R}: 0 \le x \le 1 \}$$

The board $B_I$ can be visualized as the unit square, including every single point on the boundary and in the interior as a distinct “square” on which you could place a chess piece:

The c-Board

 

 

 

 

 

 

 

 

The size of the set of real numbers in $I$ is a transfinite number which is infinitely larger than $\aleph_0$, and is called simply $c$, the infinity of the continuum. If we embrace “The Continuum Hypothesis“, this number c can be identified as $\aleph_1$, the next largest infinity. We shall therefore also call $B_I$ the c-Board, or $B_c$. The number of “squares” on $B_c$, unlike the $\omega\text{-Board}$, is uncountably infinite.

Sneak Preview

The “density” results for each infinite board and piece will be shown (in later posts) to be as follows.
The pieces are in order of their decreasing (2d or fractal) density, and thus their increasing “power”:

$$\omega\text{-Board}$$

Piece Board 2d-Density Fractal Dimension 1d-Measure (**)
Pawn $B_\omega$ 1/2 2 *
Knight $B_\omega$ 1/2 2 *
King $B_\omega$ 1/4 2 *
Rook $B_\omega$ 0 * *
Bishop $B_\omega$ 0 * *
Queen $B_\omega$ 0 * *

$$c\text{-Board}$$

Piece Board 2d-Density Fractal Dimension 1d-Measure (**)
Pawn $B_c$ * * *
Knight $B_c$ * * *
King $B_c$ * * *
Bishop $B_c$ $0$ $1$ $2\sqrt 2$
Rook $B_c$ $0$ $1$ $2$
Queen $B_c$ $0$ ? ?

(*) Not Defined for this piece on this board.

(?) Unknown whether a Hausdorff-measurable solution exists.

(**) 1d-Measure = (Hausdorff Content)

As you can see from the two tables, there is something very different between the two groups of pieces, and how they “pack” on those infinite boards. In following posts, I will try to explain what the differences are, what a Fractal Dimension is, and what other mathematical issues and questions arise in the process.

 

 

 

 

All Possible Worlds

There are more things in Heaven and Earth, Horatio,
than are dreamt of in your Philosophy. –Hamlet

The Bottom Line

After many years of thought, here is what I think is really going on:

All Possible Universes Exist, and You are in Many of Them.

I’ve been thinking about this sort of thing for some time now, and it seems like the simplest explanation for everything we see, and most mysteries of the universe and life.

What Exactly Does This Mean?

Let’s explain the picture in detail. Many ancient philosophers such as Plato and Socrates claimed that some mathematical concepts such as “circle”, “line”, “three”, “truth”, and so on have an existence which do not depend on the particular “kind” of universe we happen to be in, and that those mathematical “objects” are valid in all possible worlds, such as ours. At the same time, over centuries we have seen how our particular universe obeys mathematical laws, and the deeper we look, the more fundamental the mathematics seems to become.

But in no case do we see any of the mathematical laws being logically inconsistent with each other. Nor do we see any particular reason for some of the laws and constants (such as the fine-structure constant) to be the particular laws and constants that they are. Why not some other laws? Why not inverse-cube law for gravity? Why not?

So let’s define a “Possible Universe” to be any universe, governed by laws, which are not self-contradictory. We already know that — logically — contradictions cannot exist within consistent systems. It is quite easy to make up simple universes in which the laws are very simple. In fact, these days, many  people spend much of their time playing in simulations of worlds in computer games, which operate by various strange yet logical rules, and in some ways those games become very “real” to them.

What I am claiming is that all possible universes of all types exist. And that following the “weak anthropic principle“, the only ones of which we humans are aware of are the ones whose laws have allowed humans to come into existence. All of the other universes also exist, it is just that we do not live in them, because we cannot emerge from them.

So What?

There are some interesting, some very strange and even some disturbing consequences of this theory.

Let’s take one of the really strange and extreme ones first, and lay it out as another story:

The reason You exist is that there exists a Universe in which you eventually came to be in it.

This is a statement that goes beyond the weak (and even the strong) Anthropic Principle, and says that, almost certainly, the Universe that You see around you, including not just its laws, and its physics, and its cosmology, but its peculiar chain of events which resulted in the Milky Way Galaxy, the formation of our Solar System, the complex interplay over several billion years of Jupiter and Saturn to permit the formation of a life-sustaining Earth, the Extinction of the Dinosaurs permitting the rise of the mammals, the evolution of Primates, the development of the powers of speech and language, the splitting of the supercontinents into the continents, the recorded history of all the wars, philosophies, saints, martyrs, attrocities, art, science, and even the rise of modern western civilization, is all part of one very specific collection of universes whose flows of time closely parallel each other, and which still remain in a superposition now, as a small yet vastly uncountable number of universes in which You

— and I mean YOU, not me, and not your kids —

came to be conceived by your parents, and still be alive now to read this paragraph, as we speak.

The “So What” is exactly that. This universe in which we live (including You) has your name all over it. At the moment it also has my name all over it, but when I’m gone, if you are still here, this is most certainly only yours.

Everyone is George Bailey

Now here is where it goes from strange to downright disturbing. Think of any awful historical event, whether the Holocaust in World War II, or the Black Plague in the dark ages, or any other person or event in the past that you would have wished never happened. Let’s assume that any one of these things could have turned out different, had only slight changes years previous happened.

It could have happened, for example, that just a slight change in breeze could have caused Adolph Hitler’s parents to not marry and have a child, and the whole middle of the 20th Century played out differently. Life in America now would certainly be different, and quite possibly the disruption to America and Europe and Japan and Africa would not have occurred (at least the same way). Would your parents have still met, and had the same life ? Probably not, but if my theory is correct, that whole collection of universes not only was possible, but they exist now, but elsewhere, and — the main point — you are not in those universes.

In other words, all of those awful things that happened in your past, are there because they are there in every single universe in which You exist. You are not to blame for those things happening, but I am afraid that the theory presented says that those awful things are a necessary part of the universes in which you live.

So this is kind of like “It’s a Wonderful Life”, but it plays out for practically everybody that was in the world’s history that allowed you to become possible. So even with Hitler, if he never existed, then almost certainly, neither would YOU.  Hitler gets no credit for you, it is just that he is one of the awful things that are probably in the superposition of every single Universe in which you have found yourself.

You Have Probably Already Died, Many Times

Think back in your own past. Have you had near-misses? Maybe you were on a narrow hiking trail and almost slipped, but didn’t. Or maybe somebody shot at you, but missed. Or the lab report came back and surprise, you don’t have cancer. Many of those things may have been close calls, or even cases where you made a choice, and it just turned out that going down this alley in a small village in war-torn Afghanistan was not the one with the IED bomb.

It is quite possible that — like Schrödinger’s cat — you could have made the other choice, and that a universe existed (and still does) where you made that choice.

But you didn’t — Miraculously — and you are still alive to tell the tale. But here’s the thing: you are here and still alive, because this is the other possible universe, where you survived. I am sorry to say, you didn’t make it in the other one. And it’s not over, either. If you are still alive there will eventually be something that will get you, and many other choices to make.

The good news is, you will live as long as you (and specifically you) possibly can, making just the right choice at each step. I am sorry to say, it may come at the cost of everybody else you have ever known, and you will only die — for good, when there are no longer any possible universes where you will have survived, and all universes thereafter will be home to all the rest of humanity, who whom those universes are places where they can live.

This at least is what the “All Possible Universes Exist” theory means.

A Simple Example

Let’s consider my life, for example. In 1937, my mother was five years old, and living with her older sister Pat and her parents Nick and Lennye Edwards in Paducah Kentucky. That year, there was a terrible flood of the Ohio, Tennessee and Mississippi rivers which flooded all of Paducah. Nick decided to pack up the family and leave, but had to run the car past a police roadblock and drive right through a raging stream. If things went badly, which was quite possible, the whole family would have been swept away and so Lucille (my mom) would never have grown up, and I would never have been born.

But that didn’t happen. At least in the universe I live in. Perhaps in another one they didn’t make it, but I can’t see it from here.

So anyway, that’s what I’ve got so far.

The Planet Uranus in Opposition

Note: I hope to update this piece when I catch Uranus at the turning point, August 2018. As of right now we are in thunderstorms, so it may take a while.

Okay Let’s Cut to the Chase

Here is an animated GIF of two astrophotographs I took of the planet Uranus from my house in Virgin, Utah. The first one was taken on October 24, 2017 around 2:00 am, the next on November 19,2017 at 10pm (click on the image for full-size animated versions). See if you can spot Uranus. In the course of that month it has moved a bit, near the center of the image, so you should be able to see a blue-green dot jumping back and forth.

One-month TIme-lapse of Uranus. (Nikon D-5000, F9 3 minute exposure, equatorial mount)

If you still can’t catch it, here is an annotated versions, with labels and stuff (again, click on the images for the full screen version):

In addition to the dated labels, I have put in some graphics showing the constellation Pisces, as well as a chart, showing where computer models say Uranus should be in that part of the sky, for various dates between 2016 and 2020. I had to pull all of these other things in, just to convince myself that I really caught the planet, and not just a random earth satellite or other transient object.

It has taken me quite a bit of work to get to this final product, of which I am quite proud, and happy that it came out so cleanly, riding exactly along those predicted lines. In August of 2018 (now) I hope to capture that endpoint of maximum extent. The rest of this blog piece is the retelling of the story of this image, along with the occasional digressions into the geometry of the whole thing.

About the Planet Uranus

Uranus (Wikipedia) Voyager II photo 1986

Here is a picture of the planet Uranus, taken by the Voyager II spacecraft in 1986.

By the time I came to work at the NASA Jet Propulsion labs in ’87 the Voyager II probe had already passed by Uranus and was approaching Neptune, so I never got a chance to see these “live” images coming in. It’s not much to look at, and is best described as a large ice ball (unlike Saturn or Jupiter which are mostly gaseous). Even with a really good earthbound 8″ telescope, you’re not going to see much more than a fuzzy dot.

Though it had been seen before (even in ancient times), the object was not identified as a planet until it was observed and reported by William Herschel in 1781, who thought it might be a comet. However, after reporting it to the Astronomer Royal Nevil Maskelyne (who figures prominently in the quest to measure Longitude), Maskelyne concluded that it was probably a planet.

Other than its name (being the only one in the solar system based on the original Greek gods, and not the later Latin names of the Roman gods), Uranus is notable for having its rotational axis nearly horizontal to the orbital plane, so that for half the Uranian year (about 45 earth years) the “north” pole is in perpetual day, and the other half the year is perpetual night.

Uranus in Opposition

What started all this for me was the announcement last month that the planet Uranus was in what they call “opposition“, meaning that it was on the opposite side of the celestial sphere (as seen from Earth) as the Sun. From the Sun’s perspective, this means that the Earth and Uranus are on the same side of the Sun, and typically on closest approach to each other:

Planetary Opposition (source: Wikipedia)

The news in social media suggested that it would be so close that “it could be seen with the naked eye.” That sounded like hogwash to me, as there have been a lot of viral bogus memes around about being able to see things like the rings of Saturn and such.

Having now tracked down the planet, I can attest that — technically — it would be possible for a young person with excellent sharp eyesight to see the planet Uranus without binoculars … if they knew exactly where to look, and gazed at it out of the corner of their eye, and in a place (like where I live) with extremely dark skies and no cities nearby, but only on a cool clear night. But otherwise, forget about it.

The Plan

Barn Door Equatorial Mount

Anyway, with the announcement of the opposition of Uranus in October I decided that this was a good opportunity to do some amateur astronomy and try to capture Uranus with some very low-tech equipment, which is a Nikon D-5000 camera mounted on a crude “Barn Door” equatorial mount. Using this mount, I can take long-exposures of up to 15 or 20 minutes, without smearing of the stars due to earth’s rotation.

The Barn Door mount is a clever contraption which anybody can build with $20 of parts from Ace Hardware or Walmart. The idea is simple, you just have two boards attached with a hinge, one board fixed to a tripod. You line up the hinge with Polaris (the north start), and mount the camera to the board that moves.

But before setting up my rig, I first had to track down the current position of the planet, which was said to be somewhere inside the constellation Pisces (the fish). Credit here must be given to Martin J Powell’s website NakedEyePlanets.com,  which has this great chart of the path of Uranus:

Navigating the Stars

For anyone interested in astronomy, one way to begin is to learn how to find your way around the sky visible to the naked eye, without aid of telescopes and such. To do this, you need to learn some old-school tricks, in the form of stories. For example, to find Polaris, you first find the Big Dipper (Ursa Major) and follow the line traced by two of the stars in the “pan” of the dipper.

Rant: With the latest GPS-enabled telescopes, it is far too easy to track down stars, planets and other bodies. These days, all you need to do is type in the name of the object (e.g. Uranus), and the telescope’s computer will use its GPS to determine where the telescope itself is located, as well as the current date/time, and then guide the telescope to the place in the sky where the object may be found. Or you can use one of the “Planets” apps on smart phones, which you can hold up in the sky and see what you are looking at.

 

 

The Greatest Women Mathematicians

I have to admit a reluctance to putting the modifier “Women” in this post, because it would seem to imply that on an absolute scale the mathematicians I mention here are not intrinsically great. Perhaps a better title would be The Greatest Mathematicians (who happen to be Women). In any case, it has always bothered me when I see girls and women either discouraged from or outright forbidden from becoming mathematicians. One way or another, it is I think a sign of our times that “mathematicians you’ve never heard of” is kind of redundant.

I present these mathematicians in no particular order, for many reasons. Among those reasons is my personal opinion that the field of mathematics itself is not (again popular notions notwithstanding) like a vertical ladder, where first you learn counting, arithmetic, then algebra, geometry, trig, calculus and so on. In fact Mathematics, like Art, is more of a tree, with many branches, and many ways of thinking and seeing things. Some math is visual, some verbal, even some tactile. The fields these women pursued were likewise in many different areas, and their peculiar genius or accomplishment in each was profound. I won’t talk about all of the women pictured above, just the ones about which I would like to make a point. If you like, google “Greatest Women Mathematicians” for a very long and interesting list.

Maryam Mirzakhani

When I was writing this piece this morning I was shocked and saddened to see that Maryam Mirzakhani had just died last year (2017) of breast cancer. She was only 40, but had already done some profound work in geometry, especially Riemannian geometry — used by physicists in general relativity and elsewhere. She won the Fields Medal for her work in 2014, and became the first woman in history to win this award, described as the “Nobel Prize in Mathematics”. Maryam was born in Iran, and upon news of her death, a number of Iranian newspapers broke the taboo of printing a picture of her (a woman) with her hair uncovered.

Cathleen Synge Morawetz

Just one month after Maryam Mirzakhani died, we also lost Cathleen Morawetz (1923-2017), Professor Emeriti at New York University. Unlike most of the other mathematicians in this list, I had the great fortune to meet and get to know Cathleen in the 1980’s, while doing postdoc work at the Courant Institute in New York, where she at the time was the Director.

I had gone to Courant to continue my studies of nonlinear wave equations, and Cathleen had made much of her own fame in that area, studying compressible fluids and shock waves. She was also the creator of the “Morawetz Inequality(ies)”, which have proven to have many uses, even to the understanding the stability of Black Holes.

Cathleen was a very smart and jovial woman, and I will miss her.

Emmy Noether

Going back a bit, it would be difficult to convey just how profound and far-reaching was the work done by Emmy Noether, who lived from 1882 to 1935, and whose work touched many different branches of the tree of mathematics, including abstract algebra, geometry, and dynamical systems. One of the most profound theorems she proved (actually two with her name) is now known as Noether’s Theorem. What Noether’s (first) Theorem says is that for any conservation Law (such as energy, momentum, charge, etc), there is a fundamental geometric symmetry in the universe that corresponds to it. To express this poetically, Emmy proved that in mathematical physics, Truth (Law) is Beauty (Symmetry). Emmy’s Theorem resolved questions that Einstein had not been able to solve (!), and Einstein lobbied with Göttingen University (where she worked without pay or title) to promote her to a professorship. Eventually she was made professor, but with the rise of Nazi Germany soon had to leave the country for the US, due to her Jewish ancestry.

Sofie Kovalevskaya

Sofia “Sofie” Kovalevskaya lived from 1850 to 1891, and like Emmy Noether made substantial contributions to mathematical physics. She was a true pioneer, the first European female to earn a PhD in modern times. Together with Augustin Cauchy, she proved the Cauchy-Kovalevskaya Theorem, regarding the solutions to many equations in physics, especially those governing waves (light waves, sound waves, matter waves etc). Without her work I likely would not have had a job. Sofie was good in math but unlucky in love, her heart often broken. She had married and had children early on, and occasional star-crossed relationships later, but was also an early radical feminist and maintained a close and possibly romantic relationship with playwright Anne Charlotte Edgren-Leffler, the sister of Gosta Mittag-Leffler.  Besides her main theorem, she was also the discoverer of what is now called the Kovalevskaya Top, an exact solution to a spinning top that completed work begun long ago by Euler and Lagrange. She was also a writer, and wrote “Nihilist Girl”, a semi-autobiographical work.

“It is impossible to be a mathematician without being a poet in soul.”

 

–Sofie Kovalevskaya.

Florence Nightingale

(Yes that Florence Nightingale)

Diagram of Causes of Mortality (click to enlarge)

 

Besides being the founder of modern Nursing, Florence Nightingale had a knack for mathematics and especially statistics, and made great contributions in the visual display of quantitative information, a field which later was made popular by Edward Tufte in his seminal works. Ms. Nightingale was one of the first to make use of the Pie Chart, making clear causes and relationships in mortality among WWI soldiers.

 

Hypatia

There are so many others, such as Maria Gaetana Agnesi (the first woman appointed as full professor, but who died and like Mozart was buried in a pauper’s grave) but on my short list I have saved Hypatia for last. No likeness has ever been found, but she was said to be as beautiful as she was smart.

The first documented female mathematician, Hypatia lived in 400 AD in Alexandria, and is considered by many to be the patron saint of mathematics. And a martyr. She was the daughter of the mathematician Theon, and inherited from him the position of Director of the Library of Alexandria, the ancient repository of world knowledge. Though Theon was considered a great geometer and wrote many treatises on Euclid, Hypatia was said to have surpassed her father in mathematics and astronomy, made astrolabes, and wrote many other works and commentaries on geometry.

None of Hypatia’s works have survived, nor much of Library, whose destruction was considered one of the great tragedies in intellectual history. Hypatia was brutally assassinated by christian extremists, opposed to the “practice of sorcery, witchcraft, and mathematics”. Ironically, she was also a great teacher, and one of her most devoted students was Synesius, who studied under Hypatia as a neoplatonist, but eventually he converted to christianity and became a bishop, and contributed to the understanding of the doctrine of the Trinity.

Hypatia fought hard to save the Library, but the world was changing and she could not stop it. So much was lost when the Library fell. With it was lost much knowledge, and science, and wisdom, that we will never recover. The fall of the Library presaged the Dark Ages. Had the Library stood, some have said, we might have landed on the moon in 1492, not just Florida.

To be a woman. To be a scientist. To be a mathematician. All these things require more of one than any of us could ever know.

Be brave, these women tell us.

Be brave.

 

You Will Be Remembered

On my Sunday training runs for the Zion Half Marathon, I usually go for a run up Utah SR 9, heading towards Zion from the town of Virgin. A few miles up, I pass by this gravestone just off the highway, the only remaining thing of the ghost town called Duncan’s retreat.

I have never seen anybody stop here. Most people are tourists going 70 along this lonely stretch of highway, hell bent for Springdale lodging, and if they were looking anywhere it would be the other way, towards the Virgin River. A rock quarry lies just behind it, and nothing that would draw your attention to it.

The history book has this to say about the long-gone town:

A man named Chapman Duncan settled here in 1861. Shortly after several other families moved here also. In 1862 the Virgin river flooded and destroyed most the town. A lot of the people moved away but new settlers came. By January of 1863 about 70 people lived here. In 1863 a post office was built, a school was built in 1864 along with a meeting house. In 1866 floods took its toll on this town also and over the next few years high water from the Virgin river destroyed the fields and killed the town. By 1891 the town was deserted. All that remains of the town today is a grave of a lady named Nancy Ferguson Ott who died here in town in 1863. Her grave is located on the north side of highway 9. (Submitted by Bob Bezzant.)

My understanding from other locals is that Duncan used to live in our town of Virgin (now population 600), but “retreated” to this place far from Virgin to get away from the bustling metropolis.

When Nancy Ferguson Ott died, Duncan’s Retreat had 70 citizens, a post office, families, friends. It was a town, and they thought it would stick around long enough for there to be a cemetery with more than just Nancy, but Nature and circumstance had other ideas.

Now she is utterly alone, on a lonely stretch of anonymous highway, with no one around, no town, no post office, no friends no family.

I stopped to look at the grave marker. There are fresh flowers, stuffed animals, cards. They are replaced regularly by somebody somewhere. I do not know who.

I take some comfort in this, and I hope you do too. No matter who you are, or how alone you may feel in this world, Nancy Ferguson Ott is here, on State Route 9, miles from anywhere and anyone, to deliver this message to you: No one dies alone. You will be Loved, You will be Remembered.

Relativity, Simplified. Really.

I’ll make this (mercifully) short.

Consider this statement, which I call Ω:

Everything travels through the Cosmos at exactly the speed of light.

Ω is Relativity. That’s it. Really, it is. And Ω is true not only for Special Relativity, but the General one too. All you have to know is that the “Cosmos” means both space and time.

If you were to ask somebody what Relativity says, they would probably say something like “everything is relative.” The problem with that is, it’s not precise. In fact, it’s not even true. Some things in physics (and the world) are thought to be absolute. So I’ve been trying to come up with a good version of Relativity that can be justified mathematically, and that little box is what I came up with. If you understand what each word means exactly, everything follows from this statement, and if you like you can stop reading now and get on with life. Because that really is what relativity says.

The rest of this short post is just me rambling about why I like it. In a later post I’ll defend it. Also, here is a cool picture of a galaxy for no reason. But it’s cool (*).

Why I like This Version of Relativity

One of the things I like about this version (Ω) is that it is simple and precise, sounds a little strange — but just enough strange to be right — and answers immediately a number of other questions people ask about relativity, matter, speeds etc. For example:

Q: Will we ever be able to go faster than light?

A: NO. The reason you can’t go faster than the speed of light is that you can’t go any slower either. Nothing in the universe can go any speed in spacetime but c, the speed of light. The only thing you can ever change is the direction in space-time that you are going.

Q: Why does Relativity say that when you go faster, time slows down?

A: Because you are always going exactly at the speed c, so if you go faster in the space directions, you have to go slower in the time direction so it still adds up to exactly c.

See?  Details later, film (and math) at 11. But really, it is true. Ask a physicist. He’ll scratch his head, then say yeah, that works, then go have a beer.

(*) Photo: Centaurus A, (Image Credit: X-ray: NASA/CXC/CfA/R.Kraft et al.; Submillimeter: MPIfR/ESO/APEX/A.Weiss et al.; Optical: ESO/WFI)

 

An Anti-Christmas Carol

Another useless parasite.

Tomorrow will be June 25, the orbital opposite of December 25, and making us nearly as far as possible astronomically from the spirit of Christmas as we can be. Seems appropriate then, with the current sentiment pervading our leadership these days regarding the poor, the young, the sick and the elderly, and the changes proposed to the way our society treats these people. It therefore seems appropriate to publish here an excerpt from Dicken’s classic story, reflecting what seems to be the mood of the times, which is to tell Tiny Tim that he is a freeloader, and needs to pull himself up by his bootstraps and get a job. Or just go away and die.

Merry Christmas, Everyone.

A Christmas Carol – Marley’s Ghost

“At this festive season of the year, Mr. Scrooge,” said the gentleman, taking up a pen, “it is more than usually desirable that we should make some slight provision for the Poor and destitute, who suffer greatly at the present time. Many thousands are in want of common necessaries; hundreds of thousands are in want of common comforts, sir.”
“Are there no prisons?” asked Scrooge.
“Plenty of prisons,” said the gentleman, laying down the pen again.
“And the Union workhouses?” demanded Scrooge. “Are they still in operation?”
“They are. Still,” returned the gentleman, “I wish I could say they were not.”
“The Treadmill and the Poor Law are in full vigour, then?” said Scrooge.
“Both very busy, sir.”
“Oh! I was afraid, from what you said at first, that something had occurred to stop them in their useful course,” said Scrooge. “I’m very glad to hear it.”
“Under the impression that they scarcely furnish Christian cheer of mind or body to the multitude,” returned the gentleman, “a few of us are endeavouring to raise a fund to buy the Poor some meat and drink, and means of warmth. We choose this time, because it is a time, of all others, when Want is keenly felt, and Abundance rejoices. What shall I put you down for?”
“Nothing!” Scrooge replied.
“You wish to be anonymous?”
“I wish to be left alone,” said Scrooge. “Since you ask me what I wish, gentlemen, that is my answer. I don’t make merry myself at Christmas and I can’t afford to make idle people merry. I help to support the establishments I have mentioned—they cost enough; and those who are badly off must go there.”
“Many can’t go there; and many would rather die.”
“If they would rather die,” said Scrooge, “they had better do it, and decrease the surplus population. Besides—excuse me—I don’t know that.”
“But you might know it,” observed the gentleman.
“It’s not my business,” Scrooge returned. “It’s enough for a man to understand his own business, and not to interfere with other people’s. Mine occupies me constantly. Good afternoon, gentlemen!”


 

The Seven Billion Species of Man

As of the moment of this writing, there were currently 7,509,287,532 people alive on the planet. (source: World Odometer). Each of these people include themselves as members of various subgroups, such as Men, Women, Caucasian, Native American, Bipolar, Russian, Spectrum Autistic Syndrome, LGBTQ, Republican, Democratic, Episcopalian, Atheist, USC Alumni, Rotary Club, ISIS, Down’s Syndrome, and others.

Each of these groups has their own collection of stories, which include origin stories (how they came to be), membership stories (who is qualified to be a member), enemy stories (with whom they have fights and why), and stories about why their group is so much better than all the others.

All of these stories are important to the members of those groups. Some stories are so important to those members that they are willing to fight and die for that story. Even those whose stories are to anybody outside the group clearly arbitrary (such as Professional Football Team fans), people get into mortal combat over a bad referee call, with another human being whose only difference is the color of the shirt favored by that team over the other.

Recently there have been many arguments (especially in academic circles) around the questions of ethnicity, sensitivity to other cultures, and offenses (real or imagined) taken by somebody when someone else “appropriates” the others culture, tribe, race, religion, or other group membership. This includes such things as “whitewashing” a dramatic or historical role (e.g. by a person associated with one “race” portraying an historical character of another “race”). These controversies become especially heated when one of the groups has historically been oppressed, attacked, targeted for genocide etc by another group in power.

The tragedy of all these stories, both of the oppression, the suffering, and of the later push-back by the oppressed groups, is that they are all based on stories, as as stories, they are fiction, and as fiction, they are far from the actual reality on the ground, at least to the best  that science can discern.

Here is the story that evolutionary science suggests may be a more accurate story about the species we call Man:

Every single 'Human' born is a mutation, and may even be a brand new species.

In other words, none of us are the same, and any allegiance to a particular group, religion, ethnicity or other collection of humans is arbitrary, and made up only of stories, nothing more. The earth is not populated with over seven billion humans. It is populated with over seven billion individuals, each a unique experiment of Nature, bearing only incidental similarities in physical, sexual, psychological or other features with others which whom they may share social or familial descendancy.

This is the general perspective I have been gradually coming to embrace. Perhaps it is easier for me, because I have never particularly felt like I belonged to any group, race, or religion. Almost certainly it is also due to what others call “privilege,” in that because of my outward appearance of whiteness, the dominant ethnic group of society in America does not constantly harass or subdue me because of my differences. It is quite possible that I would tell a different story if my skin were a different shade. When under assault, groups of people unite out of self-defense, and have only their shared oppression to bind them together.

And yet, regardless of that fact, the evidence remains that the only story that truly seems to be the case is one, not of races and ethnicities, but of radical individualism, that we are each a member of groups which have only one member in them, and that all other associations are pure fictions, what Kurt Vonnegut Jr. called “Granfalloons“, which are arbitrary and pointless associations that bind people together. To be aware of this absurd  situation may seem depressing, but over the years, I have found it to be liberating, and ironically joins me in brotherhood with all other humans, with the great Granfalloon, based on the fact that we are all utterly alone.

And now at the end of this piece, the population has risen to 7,509,296,070 people. I would like to welcome these new 8538 creatures to the earth, and wish them each well in their unique journey through life, and to remind them not to take the stories told to them too seriously.

On Leadership and the President as CEO

Editor Comment: I wrote this piece as a draft  way back in 2012, during which time the Presidential CEO contender was Mitt Romney, and some had advocated for him (as for Ross Perot) because of his business experience. I never published this piece then, but the words seem even more salient, now that we are in this dubious boat. I therefore present and publish it today, in its original form. You be the judge. -NR


There has been some discussion in the course of the recent [2012] presidential campaign, with the general implication being that the President of the United states is (or should be evaluated as) the CEO of the country, and at any rate must provide leadership in how the country is run.

We are all story tellers. And the archetype of a “Leader” is a profound story, and not always a healthy one in a civilized world. In another essay, I hope sometime to discuss the two metaphorical and psychological ways I observe groups of people organize themselves and others, called Vertical (top-dog, hierarchical, canine, Yang), and Horizontal (egalitarian, all equal, “it takes a village”, feline, Yin). A Leader is a character in the “Vertical” view of the world. I’m not saying either view is entirely good, but each has its hazards.

I tend to agree with the founders of this country who had grown sick of Old World countries that were “run” by somebody, whether their name be King George or Putin or Berlusconi…or Obama if power ever goes to his head. Regardless of how my philosophy has evolved, to this day I remain a profound admirer of Ayn Rand, whose supreme heroes John Galt and Hank Rearden were brilliant, rational and secular men who did not seek leaders and did not seek to be a leader — and all they asked of the government was for it to Get The Hell Out of Their Way and let them pursue their work and happiness, to the best of their ability.

Galt and Rearden are heroes for (as Obama recently said in Rolling Stone) smart 17 year old intravert misfit teenagers, who feel the world does not understand them. And remain so for those who grow up to be smart adult emotionally integrated intravert misfits who name their houses “Anthem” for more reason than one.

Regarding CEO’s, an article from Thomson Reuters’s archives indicates that the three top skills for an entrepreneurial CEO are:

  • Financial Management
  • Communication
  • Motivation of Others

Good god, I must waste half my time at work convincing whoever is my current boss not to promote me to some “leadership” or “management” role involving people skills and such. Communication I can handle, preferably by email and preferably from far away here in Zion Canyon; if they really want eye-contact we can Skype. I’m Scotty in the engine room and dilithium crystals are fragile, let Kirk deal with the damn aliens. I have no interest in learning motivational psychology or finance professionally. But I know myself. I remain a shy but bright mathematician and software architect, who solves difficult problems in the structure of programs and languages that make it possible for those who write the higher-level business logic to express it coherently, and execute it fast and effectively. For me to be promoted to an executive or managerial role would be tragic for all involved, and I state as much explicitly in my annual employee performance review. Just get the hell out of my way and let me do work that makes me happy and makes you a profit. Everybody wins!

The genius of our country is that it was the first on the planet to be designed based on principles, predicated on the idea that its purpose was to establish ground rules preserving the rights of man, and as men are prone to abrogate power, the founders split the government and its powers into three separate branches. Of these three, only congress is charged with the formulation and passage of laws and its leadership role is composed of many men and women to prevent a single ambitious man dominating. The executive branch was originally intended to be purely executive and the president was intentionally weak, and though he has veto power, certainly not a leadership role. Indeed, George Washington refused proposals that he be made King.

The “leaders” of our country were always reluctant ones, and rightfully so.

Alas, as time has passed, the nature of humanity seems to be that some segment who leans toward “The Vertical”  will always seek a strong powerful figurehead to guide them and save them, whether in the form of an all powerful deity, or as a testosterone laden bully/hero such as Mussolini or Stalin (or even FDR, when he tried to pack the Supreme Court). And unfortunately, with a strong powerful leader, you also wind up with a large, powerful government.

I still agree with fiscal conservatives that the proportion of GDP consumed by the current government is far too large, and poses an existential threat to the country. One place of many where I disagree with Ayn Rand and Libertarians now is on the minimal “ground rules” established by a morally defensible government: For example, I would expand the Libertarian “national defense” segment to include defense against natural disasters and disease, from which an argument for a universal baseline healthcare system can be derived, and by my thinking should come directly out of the Defense Budget.

Getting back to leadership, I can only think of a few times where the person in the role of the President has stood up and used the “bully pulpit” to deliver helpful words crafted in thought and grace, to speak to the American people in a way that either united them all as a people, or brought a modicum of closure. FDR’s fireside chats, JFK’s moon speech, and yes Reagan’s address following the Challenger disaster. But there were others such as Martin Luther King Jr, who also served that role, armed only with the power of their own spiritual substance and the confidence in their cause, with no need of elected office.

If we are talking about the president leading the country by putting in place dramatic changes in the economic and regulator power of the state as well as its size, then there are only a few examples of this, one with FDR (who dramatically increased the size and scope) and Reagan (who dramatically decreased it). As president, however, both remained powerless to implement their goal without the fact that congress also had to be in the same party in order to propose and pass those laws, as well as approve members of the Supreme Court to interpret  and not strike down those laws. In general, however, it has been observed  that Americans seem to be more comfortable when the two branches that pass and sign bills are split. Americans don’t seem to care so much who is in which branch, just so long as they are at odds, in line with the “checks and balances” concept.

We are Americans, where each person is the owner of their own destiny, and who seem to mistrust any one man or party to claim power over them, even if we may agree with them. The president’s activities by rights and by the founder’s intent should be as much in the daily news as the day to day activities of the school janitor. The fact that some have come to seek a President to act as a a leader and CEO of the country is a disturbing state of affairs, in my opinion.

From Euclid to Euler to Einstein

It is of course the holiday tradition this time of year, to exchange gifts and ponder over how you would explain modern mathematics to the ancient Greeks.

In line with the latter part of that tradition, I’ve been sketching out a diagram to explain Euler’s number $e$ (2.71828…) to Euclid. It turns out that even though the classic Greek mathematicians knew all about the number π (3.1415…), they never knew about or defined the number $e$. Which is a shame, because they could have. And had they done so, they could have beaten Einstein to the punch 2500 years earlier.

Just a quick note here: for those of you who have not heard of $e$, it pops up all over the place in science, and especially when things are growing or accelerating. For example, suppose you just crossed the state line, and for some reason you thought that the mile-markers were actually speed limits, so that at the one-mile marker you slowed down to go at one mile an hour, and so on. Suppose that there were a lot of mile markers along the way, and so you were continuously speeding up with each marker. Obviously you would be going pretty slow, but at least you are speeding up. It turns out that if you obeyed the signs to the letter, by the end of one hour from mile marker one you will be at the $e$ mile marker, and would be going $e$ miles an hour.

In any case, after much fiddling around and fanfare, here is the diagram I came up with that I think would make Euclid happy. It is a “proof without narrative”, and simply uses the classically understood conic sections (e.g. circles, and hyperbolas) to show how the numbers π and $e$ may be used to relate areas of pie-shaped sectors in two conic sections, to the linear measurements along their respective curves:

One of the things I like about this diagram is that on the one hand it shows how these two numbers are similar, in that they both provide a ratio relating the area of a sector in each type of conic section, with a linear measure, but on the other, we see how these two numbers differ in a fundamental way with successive sectors.

For circles of radius 1, its area compares with its radius squared by a ratio of π (so the pie-slices are each π/8). For the hyperbola, drawing a line from the center to the vertex of the hyperbola, a sector of area one is made by drawing a second line whose x-axis length differs from the area by a ratio of $e$. In both cases we have a ratio relating a linear measure to an area.

But at this point the similarity ends. For as we go to successive circular arcs, the areas remain in fixed linear ratios, so to produce a quarter of a circle, you have an arc-length of π/4, and so on. But for the hyperbola, to produce a sector of area 2, you need to draw a line segments whose x-axis length is not $2 * e$, but $e$ to the power of 2, in other words $e^2$. For an area of three, you need to use $e^3$, and so on.

So what we see is that the number π seems to be most commonly used as a linear factor or ratio, having to do with rotational symmetry in space, while the number $e$ seems to be used as the base of an exponent, and is involved with things that grow exponentially over time.

Which brings us to light, waves, and Einstein’s space-time.

What do cones, planes and conic sections have to do with spacetime? Suppose you turn a flashlight on and off quickly. The light pulse from that event travels out in all directions at the same speed, $c$, the speed of light. Einstein (and Minkowski) suggested that we view the event where time plays the role of a fourth dimension. If we toss out one of our three dimensions, and make the time dimension the z-axis, we can visualize the light propagating out.

So in the picture on the right, the horizontal plane represents space at time $t=0$, and the vertical dimension is time, with the “up” direction representing the future, and “down” representing the past. The flashlight has just gone off at time zero, but now the light wave is expanding out in a circle, getting larger with time. And so as it grows over (upward) time, the expanding circular wave traces out the “future light cone”. Conversely, all of the light from the past that reaches us can only come from the region below the plane, marked by the “past light cone”.

The thing to note is that these “space-like” planes are always horizontal, though they may tilt a little due to relativistic motion of the observer. Space-like planes can be identified by the fact that their “normal” line (the one perpendicular to the plane) are pointing roughly up, in a time-like direction. Space-like planes can only intersect light-cones in circles or ellipses. In no case can an observer’s “plane” ever become vertical, so that its normal vector is pointing in a space-like direction outside of the light cone. Such planes are called “time-like”, and have the property that they always intersect light cones in hyperbolas.

So I am hoping that you are starting to see how I think these two numbers $pi$ and $e$ are related, but also very different. Somehow, the number $pi$ is related more to space, and to circular rotation in space, while $e$ seems to be related to time, hyperbolic curves, and exponential growth over time.

It turns out that we can even be very specific about how $e$ and $pi$ are related to each other, but it requires the introduction of a number that the ancient Greeks would have no concept of, and that is the number $i$, the square root of negative one.

The relationship was itself discovered by Euler himself, and has come to be known as Euler’s Equation, and has also been called (at least by mathematicians), The Most Beautiful Equation in the World. I hope some time in a future post to try to explain what the equation means, but for the moment, we will just display it here and be done with it.

$$ e^{i\pi} + 1 = 0  $$

And yes, this is how I spend my holiday vacations. Having Fun ! Happy new year !

 

The Geometry of Meteor Showers

Whenever a meteor shower is coming up, the news gives details on how to find the constellation in which the “radiant” can be found. Don’t bother trying to find the constellation. Too much work. Here is all you really need to do:

On the night of the shower, go outside around 2am. Look eastward, toward where the sun has been rising, and halfway up the sky, along the path the sun takes. That's the center ('the radiant'). Further away from this point the meteor trails will be longer.

That’s it. The rest of this post is just my rambling about the geometry (or astrometry as it were) that makes this all work. You won’t need it. If you come out sooner, around midnight, the radiant will be close to the horizon, and as it gets closer to sunrise the radiant will be almost overhead.

The Radiant

If you study the pattern of meteors in the picture above, it looks like we are flying through a bunch of stars very quickly, and that the center point where all those stars appear to be streaking from is simply the direction that we are flying.

It turns out that is exactly what you are seeing. The center point (in the upper left quadrant of the picture) is called the Radiant of the meteor shower, and it is the current direction in which the earth is moving, as it travels along its orbit around the Sun.

The Picture

Here is a (simplified) picture describing the general situation. To keep things simple, I’ve put the little guy (who’s supposed to be us) right on the earth’s equator, around 3am his time. We are looking down at the earth from above the North Pole, and the earth is rotating counter-clockwise on its axis. Meanwhile, the earth is travelling around the sun at 18.6 miles a second, going from right to left in the picture.

 

meteor_diagram

The comet dust in the picture was left behind by a comet years before, and now is for the most part not moving much. The earth however is plowing through the dust trail at 18.6 miles/second, and so the relative motion of the dust to the observer is likewise 18.6 miles a second, or about 30km/s.

That speed, by the way, adds a lot of energy to the situation. Many of the comet dust particles are small, some just grains of sand. But if we take a quarter-inch piece of iron, with a mass of one gram say, and compute its kinetic energy when the earth hits it, we get

$$E = \frac{1}{2}mv^2 =\frac{1}{2}(1gm)(30km/s)^2 = 450,000  Joules$$

Now a Joule is the amount of Energy to drive a one Watt light bulb for a second, which is about how long a meteor flare takes. So the light that our quarter inch piece of iron is putting out during that second is close to a half a megawatt of power. Impressive.

Another Picture Closer In

So here is a much closer picture. We’ve now rotated the picture so that the little guy is on “horizontal” ground, and we only see a small slightly curved part of the earth. The atmosphere is a very thin shell not more than 70 miles above the earth (1 percent of the earth’s diameter), and the shaded part is what the little guy can see from where he’s standing. It is a flat lens shaped piece of atmosphere, and the comet dust is coming in at about a 45 degree angle, about to slam into that circular lens. I’ve drawn a cylinder around all of the dust that will hit the part of the sky that the guy can see.

meteor_detail

Now if you look at the cylinder of comet dust coming at you from the little guy’s perspective, the rays of dust look like this:

cylinder2

Which looks just like the photo of the Geminid meteor shower. So, the reason that showers look like rays flying away from the Radiant is simply a matter of perspective, and the Radiant itself is just the direction that we are are flying, along earth’s orbit.

A Bigger Picture, Further Out

Just to tie everything together, here is a diagram showing the geometry of a typical meteor shower, arising from a regularly reappearing comet such as Halley’s comet:

meteor_shower_big_picture

 

In the case of Halley’s comet, the diagram shows how the orbit of the comet may intersect the Earth’s orbit in two places. In the current picture, the Earth is passing through one of the intersections, and is going in the direction of the constellation Orion (bottom left). This is the Orionid meteor shower, which this year (2016) will be visible from October 2 to November 7. The other intersection occurs when the Earth is heading in the direction of Aquarius, which happens around May 5-6, during the Eta Aquarid’s meteor shower. Not all comets have orbits which intersect Earth’s orbit twice, but Halley’s does.

Valid Religion

This monastery was just down the street from us when we lived in Long Beach ten years ago. It has a beautiful view of the ocean and we’d often pass it while out for a walk along the beach. To the left you can see a white shrine in which the Virgin Mary is standing. It is a very pretty installation though we sometimes would call it “Virgin Mary on the Half-shell.” About half the time we’d pass by, there would be someone there, praying to the virgin, or lighting candles or incense.

The monastery used to be a convent owned by the Catholic Church, but was long ago sold to a Vietnamese Buddhist sect, and they have converted the building into a monastery for their monks. Every so often you would see one of the monks leave the building in their saffron robes, but for the most part they keep to themselves and maintain a quiet life of contemplation.

virgin_maryNow here’s the thing: not only did the monks not tear down the Virgin Mary shrine, but they actively keep it intact, tending to the plants around it, the benches were people come to pray, and bring fresh flowers. I have never asked the monks about why they do this, or if it was a condition of the sale, but they never seem to mind. The Buddhist tradition maintains a very ecumenical respect for the beliefs of others, and maintains that there are many paths to enlightenment.

In my personal system of philosophy, I have a name for religions of this type: VALID.

I have to explain what I mean by this. With the possible exception of mathematics, I doubt that any statements of a philosophical nature will or ever can be determined to be TRUE. Thus, it is in my opinion foolish for any religion to declare itself True, as it is the height of arrogance, and proclaims that all of the other fifty thousand religions in the world are False. So, what I mean by a VALID religion is one in which the religion makes an admission of the following things:

  1. Fallibility, that this religion may have made some mistakes
  2. That absolute Truth is unknowable by man,
  3. That other religions may have important and valid points, and be a legitimate source of hope and inspiration in the lives of its believers, and rightfully so,

A VALID religion is therefore one in which it recognizes other religions as potentially equally valid, and there is no Law of Excluded Middle (as with True/False) that gets in the way.

Any religion which declares itself to be the sole possessor of this elusive thing called The Truth, is by my definition INVALID. Alas, this includes almost all religions of the world, including many brands of atheism.

This monastery is, then, to me a triple-shrine. First, to the Virgin Mary herself, secondly to the buddhist monks that respect people’s belief in her, and third to the possibility that a religion could if it so chooses become VALID — a thing which seems to me to be the only hope for peace among humans who choose to believe in the wildest of fairy tales. And to date, this monastery represents the only example to my knowledge of a valid religion.

Rainbow, Part II: Yellow Is An Idea

This is part II of my discussion of color which began with Part I, “The Infinite Piano”. In the first part I explained that the colors of the rainbow are single “notes” on an infinite piano whose keys are pure “tones” of light, and the “sheet music” for a more complex color such as PINK can be written as a 3-note chord composition in RED, GREEN, and BLUE. This composition can be written out over the color piano keyboard with three vertical bars, each indicating the loudness or softness of each of the three keys we need to play, using ranges from 0 to 255, like this:

pink3

We can further shorten this musical notation by saying (Red,Green,Blue) = (255, 192, 203). Now you may think that I just made up those particular numbers, but in fact if you check with Wikipedia, the internet standard for color on computer displays has exactly these three values for the color pink. They chose the range 0 to 255 because it is easy to express using 8 bits — which makes computers happy.

We live in the computer age, and this (R,G,B) system is now used to define all the colors that you can see on a computer monitor. So, it sounds like color is three dimensional, and you can represent any color in nature (or at least in a photo of nature) using just three colors. But is this true ?

Anyone who has tried to match paint colors may doubt this. Each paint manufacturer has their own system of specifying colors, and complex formulas of mixing their “component” pigments into Salmon, Chestnut, or other copyrighted name and color. There are many systems of defining color, such as Munsell and CIE-Lab, which are 3-dimensional, like this:

By SharkD - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=8401562

3D Munsell color space (Wikipedia – credit)

These systems are oriented toward luminance-based applications such as TV’s and computer monitors that emit their own light. There are also CMYK (Cyan-Magenta-Yellow-BlackKey) and Pantone™ systems, which are effectively 4 dimensional dimensional and used mostly in pigment-based applications such as printing and paint. But Pantone also had a six-dimensional version called Hexachrome, which add Orange and Green to form a CMYKOG space (now discontinued), and there is also a CcMmYK system used in six-color inkjet printers. These latter are called “subtractive” systems, because the pigments effectively absorb colors from white light to give you their indicated color.

So clearly something must be going on. Why do we even think color is three dimensional, when there are so many color systems using more than three. What’s up?

The Yellow That Isn’t There

Let’s take a closer look at this Wikipedia computer color thing. If you look up Yellow in wikipedia, you’ll see that standard Yellow is defined in color coordinates by (R,G,B) = (255, 255, 0). But if we plot that “musical chord” out on our piano we get this:

yellow2

Now this is crazy, because there is clearly a “yellow” key halfway between green and red, and we aren’t hitting that key at all. Instead we are leaning with a strong 255 “forte” on both RED and GREEN. Indeed, in the same Wikipedia entry for Yellow, it indicates that the “spectral” coordinates of Yellow is 570–590 nanometers. This is the wavelength of the light which is colored yellow in the rainbow spectrum.

To understand what is going on requires an understanding of human beings more than the color spectrum, and how we evolved. Modern humans perceive color with the use of three kinds of cells in the retina of our eyes, called cones. These cones come in three types, each of which respond only to specific “chords” in the color spectrum. The three chords look something like this (approximately):

cone_spectrum_2What this says is that we have in our eyes three kinds of cells (not counting rods which detect brightness), which respond to “color chords” that are centered (roughly) around the blue, green and red keys. There is no cell that responds just to “yellow” chords, and so the way that we “see” the yellow color is that our brains get strong positive signals from both the Green and the Red cones.

One of the interesting consequences is that it is possible to make a person “see” yellow even if there is no yellow in the light at all. All you have to do is to take a pure green and red light (such as from two distinct lasers), and shine them on the same spot on the wall:

red_green_yellow

Our retinas will report to the brain that where they intersect it is getting a strong green and red signal, and the brain will interpret that as yellow — even though a light spectrometer pointed at the wall will report that there is no yellow there at all. It is a color optical illusion !

Here is the take-away from all this: the color YELLOW is an IDEA, as are all other colors. It is something unique that our brain thinks — a state of mind — in response to what the outside world is doing. In the case above, the YELLOW our brain “sees” is entirely in our own heads. Now most of the time, in nature, there really is a yellow frequency light wave “out there”, and we know from the yellow in the rainbow that this frequency of light actually exists. You can create a pure yellow by simply dropping salt into a flame (sodium ions radiate at that color). But the idea of yellow must be distinguished from the light that usually triggers it.

And so, YELLOW as a specific color of light must be understood as a separate dimension from RED, GREEN, and BLUE. So how many dimensions does color really have? We will explore this further in the next post, “Shadows of The Infinite.”

1 2 3