We have now come to the final (two-part) chapter of this two year journey, exploring the maximum \u201cdensity\u201d 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.<\/p>\n
With that exhaustive list of mathematics, one would think that our last chess piece \u2014- the Queen \u2014- 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 \u201cfriendly\u201d 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.<\/p>\n
But as we shall see, the Queen takes us on one final trip, through some of the deepest mathematics that we have yet encountered.<\/p>\n
Unguarded Queens on the Countable Board<\/h3>\n
There is one catch with the finite $N \\times N$ board $B_N$, which is that while a maximal unguarded Rook \u00a0position 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).<\/p>\n
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.<\/p>\n
For example, let\u2019s start with a solution to the unguarded Queen problem on the 5-board $B_5$:<\/p>\n
![\"\"](\"https:\/\/www.nilesritter.com\/wp\/wp-content\/uploads\/2021\/06\/9AAAF4DA-3F2C-4F20-8D4B-761990E2CD7B-150x150.jpeg\")
<\/a>Unguarded Queens on 5-Board<\/p><\/div>\n
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:<\/p>\n
![\"\"](\"https:\/\/www.nilesritter.com\/wp\/wp-content\/uploads\/2021\/06\/FD1CDBB6-CE81-49B6-9C63-03A324E3C32D-300x297.jpeg\")
<\/a>Unguarded Queens on 25-Board<\/p><\/div>\n
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:<\/p>\n
![\"\"](\"https:\/\/www.nilesritter.com\/wp\/wp-content\/uploads\/2021\/07\/queen_diagonals-1-1024x246.jpg\")
<\/a>Diagonal Queen Attacks<\/p><\/div>\n
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.<\/p>\n
One can then extend this generally to the countably infinite board \u00a0$B_{\\omega}$ 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 $Q_{\\omega}$\u00a0 is given by:<\/p>\n
$$Q_{\\omega} = \\bigcup^{\\infty}_{i=0} Q_i$$<\/p>\n
To make this infinite solution explicit, we need to give for each horizontal integer coordinate $x = \\{0,1,2 \u2026\\}$ 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 $x=4$:
\n$$\\begin{cases}
\n\\phi(0) = 0 \\\\
\n\\phi(1) = 2 \\\\
\n\\phi(2) = 4 \\\\
\n\\phi(3) = 1 \\\\
\n\\phi(4) = 3 \\\\
\n\\end{cases}\\tag{PHI}
\n$$
\nNote 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,
\n$$x_5 = d_1 d_2 \u2026 d_n,$$
\nwhere $d_i$ are base 5 digits, and then define
\n$$\\phi(x)_5 = \\phi(d_1) \\phi(d_2) \u2026 \\phi(d_n)$$
\nThat 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 $x$.<\/p>\n
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 $B_{625}$:<\/p>\n
![\"\"](\"https:\/\/www.nilesritter.com\/wp\/wp-content\/uploads\/2021\/06\/08CC96B8-9077-43E0-8E67-5EBD7F3F74F8-300x300.jpeg\")
<\/a>625-Board with Queens<\/p><\/div>\n
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:<\/p>\n
![\"\"](\"https:\/\/www.nilesritter.com\/wp\/wp-content\/uploads\/2021\/07\/Board7Queens.png\")
<\/a>Queen Solution on 7-Board<\/p><\/div>\n
However, this approach does not always work. Not all 7-board solutions work for example, and no queen solutions on the 4-board work:<\/p>\n
![\"\"](\"https:\/\/www.nilesritter.com\/wp\/wp-content\/uploads\/2021\/07\/queen_diagonals_4.jpg\")
<\/a>![\"\"](\"https:\/\/www.nilesritter.com\/wp\/wp-content\/uploads\/2021\/06\/7355D886-F968-4653-BB57-C5799393B7F0-1024x318.jpeg\")
![\"\"](\"https:\/\/www.nilesritter.com\/wp\/wp-content\/uploads\/2021\/06\/409F6F42-CDFC-449A-9A30-F442742A72F8-1014x1024.jpeg\")
<\/a>