diff --git a/constants/10c.md b/constants/10c.md index 1ba449e..21a2f39 100644 --- a/constants/10c.md +++ b/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