Improve C10c lower bound to 5/3

constants/10c.md

@@ -45,6 +45,7 @@ $$
 | $1$ | Trivial | $A=[1]$. Also achieved by Hadamard matrices [Band2024]. |
 | $\sqrt{2}\approx 1.414214$ | [Band2024] | The 2 by 2 sign matrix with rows $(1,1)$ and $(1,-1)$. |
 | $4/\sqrt{6}\approx 1.632993$ | [G2026] | A 6 by 6 sign matrix with exact discrepancy $4$. |
+| $5/3\approx 1.666667$ | [X2026] | A 9 by 9 sign matrix with exact discrepancy $5$. |
 
 ## Certificate for the $4/\sqrt{6}$ lower bound
 
@@ -148,6 +149,134 @@ $$
 C_{10c}\ge \frac{4}{\sqrt{6}}\approx 1.632993.
 $$
 
+## Certificate for the $5/3$ lower bound
+
+Use the following sign matrix:
+
+```text
+ 1   1   1   1   1   1   1   1   1
+-1  -1  -1   1   1   1   1   1   1
+-1   1   1  -1  -1   1   1   1   1
+ 1  -1  -1  -1  -1   1   1   1   1
+ 1  -1   1  -1   1  -1   1   1   1
+-1   1  -1  -1   1  -1   1   1   1
+-1  -1   1   1  -1  -1   1   1   1
+ 1   1  -1   1  -1  -1   1   1   1
+ 1   1   1   1   1   1   1   1   1
+```
+
+All entries are $\pm 1$. The ninth row is a duplicate of the first row; duplicate rows are allowed, and here it simply makes the displayed certificate square. This is a padding step in the row direction only: the construction still has nine columns, so it is not an $8$ by $8$ certificate.
+
+For a sign vector $x$, let $S$ be the set of columns where the corresponding coordinate of $x$ is $-1$. For row number $i$, let $T_i$ be the set of columns where that row has entry $-1$:
+
+```text
+T1 = {}
+T2 = {1,2,3}
+T3 = {1,4,5}
+T4 = {2,3,4,5}
+T5 = {2,4,6}
+T6 = {1,3,4,6}
+T7 = {1,2,5,6}
+T8 = {3,5,6}
+T9 = {}
+```
+
+Then
+
+$$
+(Ax)_i = 9 - 2\lvert S\triangle T_i\rvert.
+$$
+
+Therefore the absolute value of the $i$th coordinate of $Ax$ is at least $5$ whenever $S$ is within Hamming distance $2$ of either $T_i$ or the complement of $T_i$.
+
+We now prove this covering condition for the full $9$-cube. Write
+
+$$
+S=U\cup W,
+$$
+
+where
+
+$$
+U\subseteq\{1,\dots,6\},\qquad W\subseteq\{7,8,9\}.
+$$
+
+Let $w=\lvert W\rvert$. Since all the sets $T_i$ above are contained in $\{1,\dots,6\}$, for $i=1,\dots,8$ write $T_i=C_i$, where
+
+```text
+C = {}, 123, 145, 2345, 246, 1346, 1256, 356
+```
+
+Let $C_i^*$ denote the complement of $C_i$ inside $\{1,\dots,6\}$. Then the complement of $T_i$ inside $\{1,\dots,9\}$ is
+
+$$
+T_i^c=C_i^*\cup\{7,8,9\}.
+$$
+
+Thus the relevant distances in the full $9$-cube are
+
+$$
+d_9(S,T_i)=d_6(U,C_i)+w
+$$
+
+and
+
+$$
+d_9(S,T_i^c)=d_6(U,C_i^*)+(3-w).
+$$
+
+The last three coordinates are therefore accounted for by the terms $w$ and $3-w$.
+
+The following finite fact about the $6$-cube will be used. The eight radius-$1$ balls around
+
+```text
+C = {}, 123, 145, 2345, 246, 1346, 1256, 356
+```
+
+miss exactly the eight complements
+
+```text
+C* = 123456, 456, 236, 16, 135, 25, 34, 124
+```
+
+Indeed, the codewords in $C$ have mutual Hamming distance at least $3$, so their radius-$1$ balls are disjoint and contain $8(1+6)=56$ vertices. Each listed vertex in $C^*$ has distance at least $2$ from every codeword in $C$, so none is in these balls. Since the $6$-cube has $64$ vertices, these are exactly the eight missed vertices.
+
+The missed vertices are nevertheless within radius $2$ of $C$; respectively, one may use the following centers:
+
+```text
+missed vertex:    123456  456  236  16    135  25    34    124
+radius-2 center:  2345    246  123  1346  356  1256  2345  123
+```
+
+Now consider the four possible values of $w=\lvert W\rvert$.
+
+If $w=0$, then $d_9(S,T_i)=d_6(U,C_i)$. The radius-$2$ covering of the $6$-cube by $C$ gives some $i$ with $d_9(S,T_i)\le 2$.
+
+If $w=1$, then $d_9(S,T_i)=d_6(U,C_i)+1$ and $d_9(S,T_i^c)=d_6(U,C_i^*)+2$. By the radius-$1$ statement above, either $U$ is within distance $1$ of some $C_i$, which gives $d_9(S,T_i)\le 2$, or $U=C_i^*$ for some $i$, which gives $d_9(S,T_i^c)=2$.
+
+If $w=2$, apply the previous $w=1$ case to the complement of $S$. Equivalently, either $U=C_i$ for some $i$, giving $d_9(S,T_i)=2$, or $U$ is within distance $1$ of some $C_i^*$, giving $d_9(S,T_i^c)\le 2$.
+
+If $w=3$, then $d_9(S,T_i^c)=d_6(U,C_i^*)$. By applying the radius-$2$ covering of the $6$-cube by $C$ to the complement of $U$, the complements $C_i^*$ also form a radius-$2$ covering of the $6$-cube. Hence some $i$ satisfies $d_9(S,T_i^c)\le 2$.
+
+Thus every vertex $S$ of the full $9$-cube is within Hamming distance $2$ of some $T_i$ or of some complement $T_i^c$. Consequently every sign vector satisfies
+
+$$
+\|Ax\|_\infty\ge 5.
+$$
+
+Equality is attained. For example,
+
+```text
+x  = (-1,-1,1,-1,1,1,1,1,1)^T
+Ax = (3,5,5,3,5,3,3,-3,3)^T
+```
+
+Hence $\mathrm{disc}(A)=5$, and consequently
+
+$$
+C_{10c}\ge \frac{5}{\sqrt{9}}=\frac{5}{3}\approx 1.666667.
+$$
+
 ## Further remarks
 
 - For large $n$, the best asymptotic lower bound remains $1$ [Band2024].
@@ -164,6 +293,7 @@ $$
 - [MO175826] MathOverflow. [*Spencer’s “six standard deviations” theorem – better constants?*](https://mathoverflow.net/questions/175826/) Question 175826 (2014).
 - [PV2022] Pesenti, L.; Vladu, A. *Discrepancy Minimization via Regularization.* [arXiv:2211.05509](https://arxiv.org/abs/2211.05509)
 - [Spe1985] Spencer, J. *Six standard deviations suffice.* Trans. Amer. Math. Soc. **289** (2) (1985), 679–706.
+- [X2026] Xie, Chuhan. 9 by 9 sign-matrix certificate for C10c, submitted to this repository (2026).
 
 ## Contribution notes