This relates to the answer to a question "Who wins two player sudoku?" and this awesome blog.

A Sudoku can be $N \times N$ where $\sqrt{N}$ is a natural number because $N \times N / \sqrt{N} \times \sqrt{N} = N$, the number of regions in the Sudoku.

It seems more computationally efficient to describe it as $(m^2)^2 = m^2\;(m^2) = m \times m\: (m \times m)$ because the square root does not have to be derived, and more notationally efficient because the regions are described.

$m \times m\; (m \times m)$ means "an $m$ by $m$ matrix of $m$ by $m$ matrices". $m^2\;(m^2)$ means "$m$ squared $m$ squared", a colloquial description of the geometry.

Coordinates (cells) are $\left[-\frac{m}{2},\ldots, \frac{m}{2}\right]$ for even $m$ and $\left[-\frac{m-1}{2},\ldots, \frac{m-1}{2}\right]$ for odd $m$.

Using these methods, I understand the boundaries between regions in any finite Sudoku.

  • What I don't understand is the common boundaries between regions in an infinite Sudoku.

I find it helpful to think of the surface of the Sudoku as a torus, so the left side connects to the right side, and the top to the bottom, which does not violate the constraints but makes every regional boundary common.


I'm also interested in how to describe an infinite Sudoku using cardinal numbers.

Using the above notation, $\infty = \aleph_0$ is the set of all natural numbers, I'd do it:

$$ (\infty^2)^2 = \infty\times\infty\;(\infty\times\infty) $$

I feel like $\sqrt{\infty}$ is problematic because it seems like $\sqrt{\aleph_0} = \aleph_0$

It also strikes me while the regions in infinite Sudoku can be $\aleph_0$, a Sudoku comprised of an such infinite regions is probably uncountable.

Since I am only confident in my understanding of $\aleph_0$, spending most of my time on finite sets, is the Sudoku comprised of an infinite set of regions with an infinite number of cells an $\Omega$?

There's also the question of $(\infty^n)^n$ where $n = \aleph_0$

  • $\begingroup$ I posted an answer, but I wonder if this question might be better placed at math.stackexchange? If people agree, we can migrate it over. $\endgroup$ – Joel David Hamkins May 16 '18 at 3:50
  • $\begingroup$ @JoelDavidHamkins I'm banned currently (haven't figured out how to ask suitable questions. your work is helpful and should fix the problem:) $\endgroup$ – DukeZhou May 16 '18 at 3:53
  • $\begingroup$ Oh, I see. You mean that you're banned on math.stackexchange? In that case, I'm at a loss. $\endgroup$ – Joel David Hamkins May 16 '18 at 4:02
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    $\begingroup$ This is perhaps not the place for this discussion, but are you sure you're banned? AFAICT Math.SE seems to consider you a member in good standing; your most recent question is on hold, with a handful of downvotes, but I see no indication that there's anything more to it than that... $\endgroup$ – Steven Stadnicki May 16 '18 at 4:43
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    $\begingroup$ While I respect the effort you went to to use purely unicode to encode your mathematics, this is not good practice; there are good reasons to encode mathematics as "LaTeX"/MathJaX. I've edited the question accordingly. $\endgroup$ – David Roberts May 16 '18 at 5:25

Thanks for your kind words about my blog.

In the general-size square Sudoku board, you have an $\kappa\times\kappa$ array of $\kappa\times\kappa$ local blocks for some (possibly infinite) cardinal $\kappa$. Every location on the Sudokou board has an address $(\alpha,\beta)$ in its local block, and the local block itself has coordinates $(\gamma,\delta)$. Thus, every location on the Sudoku board is specified by four coordinates: $$(\alpha,\beta,\gamma,\delta).$$ This is (local address, global block coordinates). The boundaries between the local boards play no meaningful role. Every row, column and local block has size $\kappa^2$.

On the $\mathbb{Z}$-Sudoku board, for example, which I considered on my blog, we have a $\mathbb{Z}\times\mathbb{Z}$ array of $\mathbb{Z}\times\mathbb{Z}$ local boards. For example, each local $\mathbb{Z}\times\mathbb{Z}$ block has its own origin and its own address $(2,17)$ and so on. One needn't imagine any "boundaries" between the sub-boards.

Infinite Z-Sudoku board

Asymmetric Sudoku is somewhat more general, since it works with rectangular sub-boards, which may not be square. Indeed, for any two cardinals $\kappa$ and $\lambda$, one may consider a $\lambda\times\kappa$ array of local boards with shape $\kappa\times\lambda$. It is interesting to consider even the extreme case of a $1\times\kappa$ array of $\kappa\times 1$ blocks. Note that the order of the coordinates must swap, in order that the size of the rows and columns is the same, and the same as the size of the local blocks, since the Sudoku conditions require these to form bijections with the set of labels. Think of a $6\times 6$ board as a $2\times 3$ array of $3\times 2$ matrices. In the general case, every location is specified by the coordinates $(\gamma,\delta)$ of the board, and the address $(\alpha,\beta)$ within that local block, where now $\gamma,\beta<\lambda$ and $\delta,\alpha<\kappa$.

A Sudoku solution is a function $f:(\kappa\times\lambda)\times(\lambda\times\kappa)\to L$, where $L$ is the set of labels (of size $\lambda\times\kappa$) that obeys the three requirements:

  • Every row is a bijection with $L$. So $f(\alpha,\beta,\gamma,\delta)$ is a bijection, if you fix $\delta<\kappa$ and $\beta<\lambda$.
  • Every column is a bijection with $L$. So $f(\alpha,\beta,\gamma,\delta)$ is a bijection, if you fix $\gamma<\lambda$ and $\alpha<\kappa$.
  • Every local board is a bijection with $L$. So $f(\alpha,\beta,\gamma,\delta)$ is a bijection, if you fix $\gamma<\lambda$ and $\delta<\kappa$.

Note that if $G$ is a group of size $\lambda$ and $H$ is a group of size $\kappa$, then we can use these groups as location coordinates, and define $f(h,g,g',h')=(gg',hh')$, taking $L=G\oplus H$ as the set of labels. This is a solution, because if you fix $g$ and $h'$, it is a bijection; and similarly if you fix $h$ and $g'$, or if you fix $g'$ and $h'$. So every assymmetric Sudoku board has a solution arising in this way. This answers a question asked by Gerhard in a comment on an earlier post.

Actually, this way of thinking leads one to a fourth Sudoku-like condition, namely, requiring that $f(\gamma,\delta,\alpha,\beta)$ is a bijection when you fix $\alpha$ and $\beta$, which is the collection of locations, one per local block, each with the same local coordinates in those blocks. For example, the labels appearing in all the locations with local address $(2,17)$, over all the local blocks, would have to be a bijection with $L$.

In the group example, this extra condition corresponds to fixing $g$ and $h$, and this shows that any Sudoku solution arising from groups will indeed obey the extra requirement. However, one can easily find Sudoku solutions not fulfilling the extra requirement, and so these solutions cannot arise from groups in the way I described (e.g. perhaps the operation that they define is not associative). One can also find infinite solutions without the extra requirement, since the argument on my blog shows that any finite position not yet violating the Sudoku rules extends to a full solution on a countably infinite board. So one could start with two labels violating the extra requirement and extend it.

  • $\begingroup$ Thank you for answering! (This is very helpful on many levels:) I noticed the vanishing boundaries in your graphic, and they made me think of the calculus functions for approaching infinity or approaching 0. I tend to think of regional boundaries because the form of m,n,p-games (what we're calling Sudoku games) under the canonical form of the set of games [M] involve element values and numeric "influence" (value/2) that extends from border cells to external regions connected orthogonally or diagonally. I may have actual followups, but will chew on your solution and answers for a while. $\endgroup$ – DukeZhou May 17 '18 at 16:55
  • $\begingroup$ Dr, Hamkins, please let me know what you think of my topological(?) approach to the infinite Sudoku board: mathoverflow.net/a/300521/105789. I'm quite certain it doesn't break any of your solutions for impartial Sudoku, and I need an infinite board with regional connections to demonstrate how your even solution extends into intrinsic, partisan forms of Sudoku. (You are the first thinker I've come across who not only acknowledges the two sets of coordinates (α,β,γ,δ), but, via infinite Sudoku, explains why it must be so.) $\endgroup$ – DukeZhou May 21 '18 at 19:58

In M-games, the elements of Sudoku carry weight $v$, and elements sharing a common boundary or vertex with exterior regions "influence" these regions $v/2$.

Infinite spaces $ℤ$ may be “split” by arbitrarily assigning any cell coordinate 0; the $(c < 0 , 0 < c)$ disjoints will have the same cardinality:

$c < 0 > c$

In order to maintain the common boundaries, make the common boundaries 0.

$> 0$ is $[1,2,…]$ and $[…, -2, -1]$ is $< 0$

$c > 0 < c$ is the boundary

$[c > 0, .., c < 0]$

$[c_{-i} > 0, .., c_{-i} < 0] , [c_{i} > 0, .., c_{i} < 0]$

Now the sets expand infinitely inward—the center of regions can never be reached but connections are maintained.


(i don't know if this breaks anything. naively, i feel like this is "twisting" the paradox and stowing it in the center to solve the problem of the common boundaries. i’m more of an engineer so I can’t help thinking of infinity procedurally—how high i can count in how much time i have. i see coordinates as scalars because the elements have weight, and here there are 2n vectors for any n dimensions. Each pair of opposite vectors points towards the other but can never intersect. My thought is they probably spiral toward infinity so I’m calling this spiral sudoku: "At the still point of the turning world, neither flesh nor fleshless; neither from nor towards; at the still point, there the dance is.";)

Spiral Sudoku can be expressed procedurally with an additional constraint that element expression $p_x$ must start in the greatest $c < 0$ or least $c > 0$, with the least unexpressed element $p > 0$ for any region, and all subsequent $p_x$ must share a cell boundary with a previous $p_x$

The toroidal surface is attractive because it reinforces the idea of "no center"—every region and vertex is the same until the initial $p_x$, and the regions themselves have to expressible center.

Where n is number of dimensions: $p_x \cap n$

Wherever that first $p_x$ is expressed becomes the center vertex of the sudoku.

This becomes a stacking game around a single vertex, or, because spaces can be virtual, players may “jump” across “hyperspace” to the closest uncolonized vertices to express new $p_x$

If new p must only share a vertex with $p_x$ sparser aggregations may be formed with empty spaces between $p_x$. Element weights $v$ may be balanced where $p_x$ can also be the greatest unexpressed $p < 0$.


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