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\section{Lower bounds for the Moser problem}\label{moser-lower-sec}

In this section we discuss lower bounds for $c'_{n,3}$. Clearly we have $c'_{0,3}=1$ and $c'_{1,3}=2$, so we focus on the case $n \ge 2$. The first lower bounds may be due to Koml\'{o}s \cite{komlos}, who observed that the sphere $S_{i,n}$ of elements with exactly $n-i$ 2 entries (see Section \ref{notation-sec} for definition), is a Moser set, so that $c'_{n,3}\geq \vert S_{i,n}\vert$

holds for all $i$. Choosing $i=\lfloor \frac{2n}{3}\rfloor$  and 

applying Stirling's formula, we see that this lower bound takes the form \begin{equation}\label{cpn3} c'_{n,3} \geq C 3^n / \sqrt{n} \end{equation} for some absolute constant $C>0$. In particular $c'_{3,3} \geq 12, c'_{4,3}\geq 24, c'_{5,3}\geq 80, c'_{6,3}\geq 240$. Asymptotically, the best lower bounds we know of are still of this type, but the values can be improved by studying combinations of several spheres or

semispheres or applying elementary results from coding theory.

Observe that if $\{w(1),w(2),w(3)\}$ is a geometric line in $[3]^n$, then $w(1), w(3)$ both lie in the same sphere $S_{i,n}$, and that $w(2)$ lies in a lower sphere $S_{i-r,n}$ for some $1 \leq r \leq i \leq n$. Furthermore, $w(1)$ and $w(3)$ are separated by Hamming distance $r$.

As a consequence, we see that $S_{i-1,n} \cup S_{i,n}^e$ (or $S_{i-1,n} \cup S_{i,n}^o$) is a Moser set for any $1 \leq i \leq n$, since any two distinct elements $S_{i,n}^e$ are separated by a Hamming distance of at least two. (Recall Section \ref{notation-sec} for definitions), this leads to the lower bound $$ c'_{n,3} \geq \binom{n}{i-1} 2^{i-1} + \binom{n}{i} 2^{i-1} = \binom{n+1}{i} 2^{i-1}.$$ It is not hard to see that $\binom{n+1}{i+1} 2^{i} > \binom{n+1}{i} 2^{i-1}$ if and only if $3i < 2n+1$, and so this lower bound is maximised when $i = \lfloor \frac{2n+1}{3} \rfloor$ for $n \geq 2$, giving the formula \eqref{binom}. This leads to the lower bounds $$ c'_{2,3} \geq 6; c'_{3,3} \geq 16; c'_{4,3} \geq 40; c'_{5,3} \geq 120; c'_{6,3} \geq 336$$ which gives the right lower bounds for $n=2,3$, but is slightly off for $n=4,5$.

The work of Chv\'{a}tal \cite{chvatal1} already contained a refinement of this idea which we here translate into the usual notation of coding theory: Let $A(n,d)$ denote the size of the largest binary code of length $n$

and minimal distance $d$.

Then \begin{equation}\label{cnchvatal} c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right). \end{equation}

With the following values for $A(n,d)$: {\tiny{ \[ \begin{array}{llllllll} A(1,1)=2&&&&&&&\\ A(2,1)=4& A(2,2)=2&&&&&&\\ A(3,1)=8&A(3,2)=4&A(3,3)=2&&&&&\\ A(4,1)=16&A(4,2)=8& A(4,3)=2& A(4,4)=2&&&&\\ A(5,1)=32&A(5,2)=16& A(5,3)=4& A(5,4)=2&A(5,5)=2&&&\\ A(6,1)=64&A(6,2)=32& A(6,3)=8& A(6,4)=4&A(6,5)=2&A(6,6)=2&&\\ A(7,1)=128&A(7,2)=64& A(7,3)=16& A(7,4)=8&A(7,5)=2&A(7,6)=2&A(7,7)=2&\\ A(8,1)=256&A(8,2)=128& A(8,3)=20& A(8,4)=16&A(8,5)=4&A(8,6)=2 &A(8,7)=2&A(8,8)=2\\ A(9,1)=512&A(9,2)=256& A(9,3)=40& A(9,4)=20&A(9,5)=6&A(9,6)=4 &A(9,7)=2&A(9,8)=2\\ A(10,1)=1024&A(10,2)=512& A(10,3)=72& A(10,4)=40&A(10,5)=12&A(10,6)=6 &A(10,7)=2&A(10,8)=2\\ A(11,1)=2048&A(11,2)=1024& A(11,3)=144& A(11,4)=72&A(11,5)=24&A(11,6)=12 &A(11,7)=2&A(11,8)=2\\ A(12,1)=4096&A(12,2)=2048& A(12,3)=256& A(12,4)=144&A(12,5)=32&A(12,6)=24 &A(12,7)=4&A(12,8)=2\\ A(13,1)=8192&A(13,2)=4096& A(13,3)=512& A(13,4)=256&A(13,5)=64&A(12,6)=32 &A(13,7)=8&A(13,8)=4\\

\end{array} \] }}

Generally, $A(n,1)=2^n, A(n,2)=2^{n-1}, A(n-1,2e-1)=A(n,2e), A(n,d)=2$, if $d>\frac{2n}{3}$. The values were taken or derived from Andries Brower's table at\\$\sim$aeb/codes/binary-1.html \textbf{include to references? or other book with explicit values of $A(n,d)$ }

For $c'_{n,3}$ we obtain the following lower bounds: with $k=2$ \[ \begin{array}{llll} c'_{4,3}&\geq &\binom{4}{0}A(4,3)+\binom{4}{1}A(3,2)+\binom{4}{2}A(2,1) =1\cdot 2+4 \cdot 4+6\cdot 4&=42.\\ c'_{5,3}&\geq &\binom{5}{0}A(5,3)+\binom{5}{1}A(4,2)+\binom{5}{2}A(3,1) =1\cdot 4+5 \cdot 8+10\cdot 8&=124.\\ c'_{6,3}&\geq &\binom{6}{0}A(6,3)+\binom{6}{1}A(5,2)+\binom{6}{2}A(4,1) =1\cdot 8+6 \cdot 16+15\cdot 16&=344. \end{array} \] With k=3 \[ \begin{array}{llll} c'_{7,3}&\geq& \binom{7}{0}A(7,4)+\binom{7}{1}A(6,3)+\binom{7}{2}A(5,2) + \binom{7}{3}A(4,1)&=960.\\ c'_{8,3}&\geq &\binom{8}{0}A(8,4)+\binom{8}{1}A(7,3)+\binom{8}{2}A(6,2) + \binom{8}{3}A(5,1)&=2832.\\ c'_{9,3}&\geq & \binom{9}{0}A(9,4)+\binom{9}{1}A(8,3)+\binom{9}{2}A(7,2) + \binom{9}{3}A(6,1)&=7880. \end{array}\] With k=4 \[ \begin{array}{llll} c'_{10,3}&\geq &\binom{10}{0}A(10,5)+\binom{10}{1}A(9,4)+\binom{10}{2}A(8,3) + \binom{10}{3}A(7,2)+\binom{10}{4}A(6,1)&=22232.\\ c'_{11,3}&\geq &\binom{11}{0}A(11,5)+\binom{11}{1}A(10,4)+\binom{11}{2}A(9,3) + \binom{11}{3}A(8,2)+\binom{11}{4}A(7,1)&=66024.\\ c'_{12,3}&\geq &\binom{12}{0}A(12,5)+\binom{12}{1}A(11,4)+\binom{12}{2}A(10,3) + \binom{12}{3}A(9,2)+\binom{12}{4}A(8,1)&=188688.\\ \end{array}\] With $k=5$ \[ c'_{13,3}\geq 539168.\]

It should be pointed out that these bounds are even numbers, so that $c'_{4,3}=43$ shows that one cannot generally expect this lower bound gives the optimum.

The maximum value appears to occur for $k=\lfloor\frac{n+2}{3}\rfloor$, so that using Stirling's formula and explicit bounds on $A(n,d)$ the best possible value known to date of the constant $C$ in equation \eqref{cpn3} can be worked out, but we refrain from doing this here. Using the Singleton bound $A(n,d)\leq 2^{n-d+1}$ Chv\'{a}tal \cite{chvatal1} proved that the expression on the right hand side of \eqref{cnchvatal} is also $O\left( \frac{3^n}{\sqrt{n}}\right)$, so that the refinement described above gains a constant factor over the initial construction only.

For $n=4$ the above does not yet give the exact value. The value $c'_{4,3}=43$ was first proven by Chandra \cite{chandra}. A uniform way of describing examples for the optimum values of $c'_{4,3}=43$ and $c'_{5,3}=124$ is the following:

Let us consider the sets $$ A := S_{i-1,n} \cup S_{i,n}^e \cup A'$$ where $A' \subset S_{i+1,n}$ has the property that any two elements in $A'$ are separated by a Hamming distance of at least three, or have a Hamming distance of exactly one but their midpoint lies in $S_{i,n}^o$. By the previous discussion we see that this is a Moser set, and we have the lower bound \begin{equation}\label{cnn} c'_{n,3} \geq \binom{n+1}{i} 2^{i-1} + |A'|. \end{equation} This gives some improved lower bounds for $c'_{n,3}$:

\begin{itemize} \item By taking $n=4$, $i=3$, and $A' = \{ 1111, 3331, 3333\}$, we obtain $c'_{4,3} \geq 43$; \item By taking $n=5$, $i=4$, and $A' = \{ 11111, 11333, 33311, 33331 \}$, we obtain $c'_{5,3} \geq 124$. \item By taking $n=6$, $i=5$, and $A' = \{ 111111, 111113, 111331, 111333, 331111, 331113\}$, we obtain $c'_{6,3} \geq 342$. \end{itemize}

This gives the lower bounds in Theorem \ref{moser} up to $n=5$, but the bound for $n=6$ is inferior to the lower bound $c'_{6,3}\geq 344$

given above. The lower bound $c'_{6,3} \geq 

353$ was located by a genetic algorithm, see Appendix \ref{genetic-alg}. This suggests that greedily filling in spheres, semispheres or codes is no longer the optimal strategy in dimensions six and higher. The current record In any event, bounds such as \eqref{cnchvatal} or \eqref{cnn} only seem to offer a minor improvement over \eqref{binom} at best, and in particular we have been unable to locate a bound which is asymptotically better than \eqref{cpn3}.