Thirty-six Officers and their Code

TL;DR

This paper provides a short proof of Euler's 36 officer conjecture, demonstrating the non-existence of an affine plane of order 6 through both indirect and direct methods.

Contribution

It offers a concise proof of Euler's 36 officer conjecture and confirms the non-existence of an affine plane of order 6 with two different approaches.

Findings

Confirmed no affine plane of order 6 exists

Provided a short proof of Euler's 36 officer conjecture

Presented a direct proof method

Abstract

This note presents a short proof of Euler's 36 officer conjecture. This implies that there is no affine plane of order $6$, but we also give a direct proof.

Equations

$$a\alpha$$

$$\left[\begin{array}[]{cccc}0&1&1&1\\ 1&0&1&1\\ 1&1&0&1\\ 1&1&1&0\end{array}\right]$$

$$\sum _{\lambda \in {\Pi }_i}{v}^{\lambda }-\sum _{\lambda \in {\Pi }_j}{v}^{\lambda }=0,i\ne j,$$

$$6l$$

$$m$$

$$p_1$$

$$p_2$$

$$\mathrm{wt}(c(\Lambda ))=p_1+p_3=6l-2m+4p_3+8p_4.$$

$$2p_2$$

$$4p_4$$

$$2222,2226,3330,3332.$$

$$xy00\text{-}x0z00\text{-}y\text{-}z0$$

$$x\text{-}y00\text{-}x0\text{-}z00yz0$$

$$\text{-}xy00x0\text{-}z00\text{-}yz0$$

$$\text{-}x\text{-}y00x0z00y\text{-}z0.$$

$$0011$$

$$a:1100,0\text{-}10\text{-}1,\text{-}1001$$

$$e\epsilon$$

$$\times$$

$$R$$

$$D$$

$$y=6$$

$$\left\{xy|1\le x\le 3,1\le y\le 3\right\}$$

$$11$$

$$R$$

$$D$$

$$y=6$$

Full text

Introduction

Leonhard Euler published his famous 36 officers problem in 1782 (a translation of the paper is listed under [7]): “This question concerns a group of thirty-six officers of six different ranks, taken from six different regiments, and arranged in a square in a way such that in each row and column there are six officers, each of a different rank and regiment.” Euler thought there was no such arrangement and conjectured that analogous ones for rank-regiment counts that leave a remainder of 2 when divided by 4 were all impossible. His conjecture for the original six was proved correct by G. Tarry in 1900 [12]. But in 1959, E. T. Parker [10] showed that the problem is solvable for an infinite subset of those counts, including 10. Then in 1960, Parker, R. C. Bose, and S. S. Shrikhande [4] proved the problem solvable for *all *rank-regiment counts other than two and six.

There is a combinatorial proof of the 36 officer impossibility by D. R. Stinson [11] and a coding-theory one by S. T. Dougherty [6] (incorporating part of Stinson’s proof). The present note follows the general outline of these two papers, but it contains some different ways of setting up details. They will be itemized in steps.

It is well-documented [9, Section III.3] that the officer problem is equivalent to one involving nets (among other structures). An $(n,k)$ net is a combinatorial design $(\mathcal{P},\mathcal{L})$ consisting of a set $\mathcal{P}$ of $n^{2}$ points and a collection* *$\mathcal{L}$ of lines that are $n$-subsets of $\mathcal{P}$. The lines have these properties:

$\mathcal{L}$ is the disjoint union of $k$ parallel classes. Each class is a partition of $\mathcal{P}$ into $n$ lines. 2. 2.

Two lines from different parallel classes meet at exactly one point.

A solution to the 36 officer problem is equivalent to the existence of a $(6,4)$ net, the positions in the square being the points. The rows and columns provide two parallel classes, the locations of officers by regiment form the lines of the third class, and the locations by rank form the fourth. Here is an illustration of a square given by Euler in [7] that almost works. Latin letters denote the regiments and Greek the ranks, Euler’s traditional symbols leading to the name “Graeco-Latin square.” Unfortunately, the pairs $b\zeta$ and $d\varepsilon$ are duplicated (and $b\varepsilon$ and $d\zeta$ left out).

[TABLE]

Step One

The code of a net $(\mathcal{P},\mathcal{L})$ over a field $\mathbb{F}$ is the subspace of the $\mathbb{F}$-space $\mathbb{F}^{\mathcal{P}}$ of $\mathbb{F}$-valued functions on $\mathcal{P}$ spanned by the characteristic functions of the lines [6]. Following Assmus and Key [1, Definition 1.2.5], we denote the characteristic function of a subset $X$ of $\mathcal{P}$ by $v^{X}$. If $\Pi$ is a parallel class, $v^{\mathcal{P}}=\sum_{\lambda\in\Pi}v^{\lambda}$, and $v^{\lambda}v^{\lambda^{\prime}}=0$ for $\lambda$ and $\lambda^{\prime}$ different members of $\Pi$. The weight $\mathrm{wt}(f)$ of $f\in\mathbb{F}^{\mathcal{P}}$ is the number of points $P$ with $f(P)$ nonzero; and the standard dot product on $\mathbb{F}^{\mathcal{P}}$ is given by $f\cdot f^{\prime}=\sum_{P\in\mathcal{P}}f(P)f^{\prime}(P)$. Let $\mathcal{N}$ be a $(6,4)$ net $(\mathcal{P},\mathcal{L})$ and let $\mathcal{C}$ be its binary code ($\mathbb{F=F}_{2}$). As in [6] and [11], we eventually show that $\mathcal{N}$ cannot exist by establishing contradictory information about the dimension of $\mathcal{C}$.

The hull $\mathcal{H}$ of $\mathcal{N}$ at $\mathbb{F}_{2}$ is $\mathcal{C}\cap\mathcal{C^{\perp}}$ (the orthogonal space $\mathcal{C^{\perp}}$ taken with respect to the dot product) [1, Definition 2.4.3]. It contains the differences (sums in this binary case) of parallel lines. If $\Pi_{1},\ldots,\Pi_{4}$ are the parallel classes and $\lambda_{i}\in\Pi_{i}$, then the span $\left\langle\lambda_{1},\ldots,\lambda_{4}\right\rangle$ has dimension 4, because the Gram matrix $\left[\lambda_{i}\cdot\lambda_{j}\right]$ is

[TABLE]

which is nonsingular. Thus $\mathcal{C}=\left\langle\lambda_{1},\ldots,\lambda_{4}\right\rangle\perp\mathcal{H}$, since for $\lambda\in\Pi_{i}$, $\lambda-\lambda_{i}\in\mathcal{H}$.

Lemma 1

We have $\dim\mathcal{C}\leq 20$.

Proof. Since $\mathcal{C}\subseteq\mathcal{H}^{\perp}$, $\dim\mathcal{C}\leq 36-\dim\mathcal{H}=36-(\dim\mathcal{C}-4)$, making $\dim\mathcal{C}\leq 20$.

Because $v^{\mathcal{P}}=\sum_{\lambda\in\Pi_{i}}v^{\lambda}$, there are parallel class dependencies

[TABLE]

spanned by the three with $j=4$. The next step in the proof of Euler’s conjecture is to show that there are no further independent dependencies. That will imply that $\dim\mathcal{C}\geq 24-3=21$ and contradict the Lemma.

Step Two

To that end, consider sets of lines, and ascribe to such a set $\Lambda$ its parallax $\pi(\Lambda)=l_{1}l_{2}l_{3}l_{4}$, with $l_{i}=\left|\Lambda\cap\Pi_{i}\right|$. Call the points on the lines of $\Lambda$ the points of $\Lambda$, and let $p_{j}$ be the number of $j$-points, those that appear on $j$ lines of $\Lambda$. Put $l=l_{1}+\ldots+l_{4}$ and $m=\sum_{i<j}l_{i}l_{j}$. Then double-counting incident point-line pairs gives

[TABLE]

Consequently

[TABLE]

Let $c(\Lambda)$ be the binary sum $\sum_{\lambda\in\Lambda}v^{\lambda}$, the member of $\mathcal{C}$ spanned by the lines of $\Lambda$. It is a consequence of the class dependencies (1) that if we change $\Lambda$ to $\Lambda^{\prime}$ by switching the lines of $\Lambda$ in $\Pi_{i}$ with those in $\Pi_{i}$ not in $\Lambda$, for an even number of parallel classes $\Pi_{i}$, then $c(\Lambda^{\prime})=c(\Lambda)$. The corresponding line counts $l_{i}$ and $l_{i}^{\prime}$ of $\Lambda$ and $\Lambda^{\prime}$ satisfy $l_{i}+l_{i}^{\prime}=6$. A line dependency corresponds to a line set $\Lambda$ with $c(\Lambda)=0$. For such a set, what are the possibilities for $\pi(\Lambda)$? For instance, the parallel class dependencies have $\pi(\Lambda)=6600,6060,\ldots,6666$, with an even number of 0s and 6s.

In searching for the parallaxes corresponding to line sets $\Lambda$ with $c(\Lambda)=0$, we may assume that in $\pi(\Lambda)$, $3\geq l_{1}\geq l_{2}\geq l_{3}$, by switchings involving $\Pi_{4}$ and then renumbering. Moreover, if $l_{1}=3$, we can also take $3=l_{1}\geq l_{2}\geq l_{3}\geq l_{4}$. The 1’s of any $c(\Lambda)$ appear at the 1-points and the 3-points of $\Lambda$, and

[TABLE]

If $c(\Lambda)$ is to be 0, we need $p_{1}=p_{3}=0$. That gives

[TABLE]

Demanding that $p_{2}$ and $p_{4}$ be nonnegative integers for parallaxes satisfying the inequalities listed gives four possibilities, as a short Maple computation shows:

[TABLE]

All but $2222$ can be ruled out. Switching will change $2226$ to $2240$. But if $\pi(M)=2240$, then $\mathrm{wt}(c(M))=4p_{3}+8p_{4}+8$, by (2), which cannot be 0. If $\pi(\Lambda)=3330$ or $3332$ and $c(\Lambda)=0$, then adding or removing a line in $\Pi_{4}$ gives a line set $M$ with $\pi(M)=3331$ and $\mathrm{wt}(c(M))=6$. But $\mathrm{wt}(c(M))=4p_{3}+8p_{4}-12$, a multiple of 4; so $3330$ and $3332$ are out, too.

When $c(\Lambda)=0$ and $\pi(\Lambda)=2222$, $p_{1}=p_{3}=0$, $p_{2}=24$, and $p_{4}=0$. This line set comes up in [6], where it is ruled out. We shall present another argument to exclude it in the next step. Before that, begin by labeling the two lines in $\Lambda\cap\Pi_{i}$ with 1 and $-1$. Then the following facts are all consequences of the intersection properties of the $(6,4)$ net. Each of the 24 points of $\Lambda$ can be tagged by the two lines of $\Lambda$ through it, in a quadruple $x_{1}x_{2}x_{3}x_{4}$. Two of the $x_{i}$ are [math] and the other two $\pm 1$. For example, $10\text{-}10$ means the point on line $1$ of $\Lambda\cap\Pi_{1}$ and line $-1$ of $\Lambda\cap\Pi_{3}$. (If these quadruples are interpreted as points in $\mathbb{R}^{4}$, they are the vertices of a regular $24$-cell [5, Section 3.7].) Each line not in $\Lambda$ goes through three of the 24 points (the six intersections with the lines of $\Lambda$ not parallel to it, doubled up), and it can be labeled by them. If, say, the line is in $\Pi_{4}$, its labeling points will be $xy00,\text{-}x0z0,0\text{-}y\text{-}z0$, where $x,y,z\in\left\{1,-1\right\}$. This and the other three lines of $\Pi_{4}$ not in $\Lambda$ will then be

[TABLE]

It follows that there are just two possible line lists, one containing $1100$, $\text{-}1010$,

$0\text{-}1\text{-}10$, and the other, $1100$, $\text{-}10\text{-}10$, $0\text{-}110$. Moreover, the second is obtained from the first by negating all entries. On the other hand, upon exchanging the signs for $\Lambda\cap\Pi_{4}$, the displayed list will not change, but the new line lists for each $\Pi_{i}-(\Lambda\cap\Pi_{i})$, $i<4$, will be obtained by negating all entries. What this implies is that all the possible collections of the four parallel class line lists for the lines not in $\Lambda$ are equivalent under sign changes, that is, label changes of the members of the $\Lambda\cap\Pi_{i}$.

Step Three

Set up the standard layout for a six by six Graeco-Latin square. As before, the 36 small squares represent the points of the net, and rows and columns correspond to the first two parallel classes. The lines of the third are $a,b,c,d,e,f$, and those of the fourth, $\alpha,\beta,\gamma,\delta,\varepsilon,\zeta$. The lines in $\Lambda$ are the right two columns, $x_{1}=\pm 1$; the bottom two rows, $x_{2}=\pm 1$; $e$ and $f$, $x_{3}=\pm 1$; and $\varepsilon$ and $\zeta$, $x_{4}=\pm 1$. Because of the labeling flexibility for the other lines, outlined above, we can include the 24 points of $\Lambda$ in the diagram:

[TABLE]

Continuing with the flexibility, we assign the points of $\Lambda$ on the remaining lines:

[TABLE]

The result, on filling in the line names, is two-thirds of a Graeco-Latin square:

[TABLE]

Now the challenge is to fill in the twelve blank squares with pairs from $\left\{a,b,c,d\right\}\times\left\{\alpha,\beta,\gamma,\delta\right\}$ with the desired non-repetition properties. So $a\alpha,b\beta,c\gamma,d\delta$ are excluded, being present in the lower right, and each row and column has further exclusions coming from the bottom two rows and the right two columns. We abbreviate the layout this way, showing the row and column exclusions at the right and bottom sides:

[TABLE]

For instance, none of the three pairs in the top row can involve $b$ or $\alpha$; and none in the left column $a$ or $\gamma$. (The four squares with $\times$’s are to be left blank.)

The second row and the third column both have the same exclusions, $c$ and $\beta$. So the pairs available for the “cross” of the five squares of that row and that column are $a\gamma,a\delta,b\alpha,b\gamma,b\delta,d\alpha,$ and $d\gamma$. The pair $b\gamma$ can go only in the center square of the cross, because of the exclusions governing its other squares. But that means the four pairs $a\gamma,b\alpha,b\delta,$ and $d\gamma$ cannot appear in the cross, and that leaves only two pairs for the other four positions. So $b\gamma$ cannot appear in the cross. Now whatever pair is in the center rules out two other pairs, leaving only three pairs for the four remaining squares of the cross. Thus the challenge cannot be met.

In conclusion, the assumption of a further dependency on the lines of a $(6,4)$ net beyond the parallel class dependencies, which necessarily involves a line set with parallax $2222$, has been shown to be untenable. Thus there is no $(6,4)$ net and Euler’s 36 officer conjecture is indeed correct!

No affine plane of order $6$

Since four parallel classes of an affine plane of order $6$ would constitute a $(6,4)$ net, there can be no such plane. A. Bichara [2] gave a direct proof from the incidence properties of such a plane,, relating them to arcs. Here we give a different direct proof.

Let $\mathbf{A}$ be a hypothetical affine plane of order $6$. We first show that the diagonals of any parallelogram in $\mathbf{A}$ are parallel. Suppose not, and set up a coordinate system for $\mathbf{A}$ using $1,\ldots,6$ for the coordinates and arranging things so that the sides of the offending parallelogram are the lines $x=1,x=3,y=1,y=3$, the line $R:y=x$ is one of the diagonals, and it and the other diagonal $D$ meet in $11$ (we shall abbreviate $(x,y)$ to $xy$). By permuting $4,5,6$, we can arrange $R$ and $D$ to contain these points:

[TABLE]

Here’s a schematic diagram of the coordinate grid, with the points of $R$ and $D$ indicated:

[TABLE]

Divide $\mathbf{A}$ into four quadrants:

[TABLE]

So $R$ and $D$ lie entirely in LL and UR

Thinking about how a line $L$ not parallel to a grid line $x=c$ or $y=d$ meets these grid lines, one sees that $L$ must have as many points in LL as in UR. Moreover, $L$ cannot lie entirely in UL and LR , because it must meet at least one of $R$ and $D$, since they cannot both be parallel to $L$. So the five lines parallel to $R$ other than $R$ each have a point in LL. As there are six points in LL not on $R$, four of these parallels have one point in LL and one has two. The same thing holds for the parallels of $D$.

Now suppose that the points $45,56,64$ in UR not on $R$ or $D$ are collinear, on $L$, say. Then the part of $L$ in LL would be one of these four possibilities:

[TABLE]

The fourth entry tells which of $R$ or $D$ the line $L$ meets. But each choice presents a parallel to the other of $R$ or $D$, and that gives $L$ too many points in LL.

Thus one of the three sides of the triangle with vertices $45,56,64$ is not parallel to either of $R$ or $D$. Such a side $S$ cannot meet either $R$ or $D$ in UR without being a grid line. So $S$ meets both $R$ and $D$ in LL. The only way to do that without being a grid line is to go through $22$; but now there is no second point available for $S$ in LL that is not already on a line through $22$.

Therefore the diagonals of every parallelogram in $\mathbf{A}$ are parallel. Keep the grid layout above and the line $R$. Work with grid parallelograms with vertices $xy,xy^{\prime},x^{\prime}y,x^{\prime}y^{\prime}$ to determine points on lines, as follows: if we know one diagonal and know that a parallel to it goes through a third vertex, then that parallel must go through the fourth vertex. Line $R$ is a diagonal of the parallelogram $11,12,21,22$, and we now take $D$ to be the other diagonal, parallel to $R$. If $D$ goes through $3z$, then by parallelogram $33,3z,z3,zz$ we find that $z3\in D$. Renumber to make $z=4$. Then $R$ and $D$ have these points, $56$ and $65$ now being forced to be on $D$:

[TABLE]

Take $A$ and $B$ to be parallels to $R$ and $D$ with $13\in A$ and $14\in B$ initially. The further points of $A$ and $B$ shown in the diagram here are obtained from the parallelogram rule.

[TABLE]

For instance, parallelogram $11,13,31,33$ with diagonal $R$ and $13\in A$ implies that $31\in A$. Then $21,24,31,34$ with diagonal $D$ and $31\in A$ makes $24\in A$. Continuing this way, we fill in all the positions $xy$ with $1\leq x,y\leq 4$. But now we’re stuck – there’s no place for two more $A$’s and $B$’s. We conclude that $\mathbf{A}$ does not exist!

Comments

The section of the Handbook of Combinatorial Designs cited [9, Section III.3] presents general results on mutually orthogonal Latin squares. There is a recent disproof of Euler’s conjecture using certain combinatorial matrices in the paper by K. Wang and K. Chen [13].

Once it is known that the diagonals of parallelograms in affine planes of order $6$ are parallel, it follows that the diagonal points of any quadrangle in a projective plane of order 6 are collinear. Such a plane is a Fano plane in the terminology of A. Gleason [8]. But then it is Desarguesian, by Theorem 3.5 of [8]. So its order would have to be a power of 2.

The famous Bruck-Ryser theorem [3] rules out infinitely many orders for affine planes, $6$ being one of them.