10c: lower bound 7/√17 → 5/√8 from the perfect [7,4] Hamming code

README.md

@@ -33,7 +33,7 @@ Bounds for which the level of available verification is currently at minimal lev
 | [9](https://teorth.github.io/optimizationproblems/constants/9a.html) | Shannon capacity of the 7-cycle | 3.2578 | 3.3177 |
 | [10a](https://teorth.github.io/optimizationproblems/constants/10a.html) | The real Grothendieck constant | $\frac{6\pi}{11}\approx 1.71360$ | $\frac{\pi}{2\log(1+\sqrt{2})} - 10^{-4} \approx 1.78211$ |
 | [10b](https://teorth.github.io/optimizationproblems/constants/10b.html) | The complex Grothendieck constant | 1.338 | 1.40491 |
-| [10c](https://teorth.github.io/optimizationproblems/constants/10c.html) | Spencer discrepancy constant (“six standard deviations suffice”) | 1.697749 | 3.674235 (3.65*) |
+| [10c](https://teorth.github.io/optimizationproblems/constants/10c.html) | Spencer discrepancy constant (“six standard deviations suffice”) | 1.767767 | 3.674235 (3.65*) |
 | [11a](https://teorth.github.io/optimizationproblems/constants/11a.html) | $L^1$ Poincaré constant on the Hamming cube | $\sqrt{\pi/2} \approx 1.2533$ | $\pi/2 - 0.00013 \approx 1.5707$ |
 | [11b](https://teorth.github.io/optimizationproblems/constants/11b.html) | Critical exponent for isoperimetric inequality on the Hamming cube | 0.5 | 0.5 |
 | [12](https://teorth.github.io/optimizationproblems/constants/12a.html) | The Beardwood–Halton–Hammersley constant | 0.6277 | 0.90304 |
@@ -133,6 +133,7 @@ Bounds for which the level of available verification is currently at minimal lev
 
 ## Recent progress
 
+- [10c](https://teorth.github.io/optimizationproblems/constants/10c.html) **improved lower bound:** $C_{10c} \geq 5/\sqrt{8} \approx 1.767767$ by Y. H., submitted to this repository, 5 Sep 2026.
 - [51](https://teorth.github.io/optimizationproblems/constants/51a.html) **improved lower bound:** $C_{51} \geq 0.5850724$ by [Y. He and Q. Tang](https://arxiv.org/abs/2602.12217), 12 Feb 2026.
 - [11b](https://teorth.github.io/optimizationproblems/constants/11b.html) **solved:** $C_{11b} = 0.5$ by  [P. Durcik, P. Ivanisvili, J. Roos, X. Xie](https://arxiv.org/abs/2602.20462), 24 Feb 2026.
 - [3c](https://teorth.github.io/optimizationproblems/constants/3c.html) **improved lower bound:** $C_{3c} \geq 1.67471$ by T. Astor (paper coming soon).

constants/10c.md

@@ -47,6 +47,7 @@ $$
 | $4/\sqrt{6}\approx 1.632993$ | [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 $5$. |
 | $7/\sqrt{17}\approx 1.697749$ | [L2026] | A 17 by 17 sign matrix with exact discrepancy $7$ ($16$ distinct rows, one repeated to square it). |
+| $5/\sqrt{8}=\tfrac{5\sqrt2}{4}\approx 1.767767$ | [H2026] | An 8 by 8 matrix over $\{-1,0,1\}$ with exact discrepancy $5$; column 8 is identically zero. Its eight rows are the perfect $[7,4]$ Hamming code: the sets $T_i$ below, together with their complements in $\{1,\dots,7\}$, are the 16 codewords, whose radius-1 balls **partition** $\{0,1\}^7$. Optimal at $n=8$; see below. |
 
 ## Certificate for the $4/\sqrt{6}$ lower bound
 
@@ -338,6 +339,83 @@ $$
 
 The centres $T_i$ were found by a greedy set-cover search for a small radius-$5$ covering code of the $17$-cube; only $16\le 17$ balls are needed, which is what allows the matrix to be square. (Note $n=17$ continues the pattern $n=2^d+1$ of the $n=9$ construction)
 
+## Certificate for the $5/\sqrt{8}$ lower bound
+
+Use the following $8$ by $8$ matrix. Entries lie in $\{-1,0,1\}\subset[-1,1]$, which
+is the range the definition of $C_{10c}$ allows; the eighth column is identically
+zero. Every previously recorded lower bound here uses a sign matrix, and that
+restriction is what this entry drops. Note that the upper-bound side quantifies
+over the same set: [PV2022]'s Theorem 4.5 reads "for every $A\in[-1,1]^{n\times n}$".
+
+```text
+ 1  1  1  1  1  1  1  0
+ 1  1  1 -1 -1 -1 -1  0
+ 1  1 -1  1 -1 -1  1  0
+ 1  1 -1 -1  1  1 -1  0
+ 1 -1  1  1 -1  1 -1  0
+ 1 -1  1 -1  1 -1  1  0
+ 1 -1 -1  1  1 -1 -1  0
+ 1 -1 -1 -1 -1  1  1  0
+```
+
+As in the certificates above, for a sign vector $x$ let $S$ be the set of columns
+where $x$ is $-1$, and for row $i$ let $T_i$ be the set of columns $j\le 7$ where
+that row is $-1$:
+
+```text
+T1 = {}          T5 = {2,5,7}
+T2 = {4,5,6,7}   T6 = {2,4,6}
+T3 = {3,5,6}     T7 = {2,3,6,7}
+T4 = {3,4,7}     T8 = {2,3,4,5}
+```
+
+Then $(Ax)_i = 7 - 2\lvert S\triangle T_i\rvert$, so $\lvert (Ax)_i\rvert \ge 5$
+exactly when $S$ is within Hamming distance $1$ of $T_i$ or of its complement in
+$\{1,\dots,7\}$. The sixteen sets $T_i$ and their complements are precisely the
+codewords of the perfect $[7,4]$ Hamming code, whose radius-1 balls partition
+$\{0,1\}^7$ — $16\cdot 8 = 128 = 2^7$ — so **every** $S$ is covered and
+
+$$\min_{x\in\{\pm1\}^{8}}\lVert Ax\rVert_\infty \;=\; 5 .$$
+
+The eighth coordinate of $x$ is irrelevant, so this is an exhaustive check over
+$2^7=128$ cases; over all $2^8=256$ sign vectors the distribution of
+$\lVert Ax\rVert_\infty$ is $5$ (224 times) and $7$ (32 times). Equality is
+attained, e.g. at $x=(-1,-1,-1,-1,-1,-1,1,-1)^T$. Hence
+$\mathrm{disc}(A)=5$ and
+
+$$C_{10c}\ \ge\ \frac{5}{\sqrt{8}}\ =\ \frac{5\sqrt2}{4}\ \approx\ 1.767766953 .$$
+
+### $5$ is optimal at $n=8$
+
+For $a\in[-1,1]^8$ and $t\ge 6$, if $\langle a,x\rangle\ge t$ and
+$\langle a,y\rangle\ge t$ then $2t\le\langle a,x+y\rangle\le\sum_i|x_i+y_i|
+=2\,(8-d(x,y))$, so $d(x,y)\le 8-t\le 2$. By Kleitman's diameter theorem a
+subset of the $8$-cube of diameter $2$ has at most $\sum_{i\le 1}\binom{8}{i}=9$
+elements, so each row certifies at most $2\cdot 9=18$ of the $256$ sign vectors,
+and $8\cdot 18=144<256$. Hence $\mathrm{disc}(A)\le 5$ for every
+$A\in[-1,1]^{8\times 8}$, and this certificate is best possible at $n=8$. The
+same argument gives the exact optima $\sqrt2$ at $n=2$ and $3/2$ at $n=4$; the
+three cases $n=2,4,8$ are the zero-slack rows of this counting bound and are now
+all attained.
+
+### How this was found
+
+The reduction above turns the problem into one about covering codes: a matrix
+with $\mathrm{disc}\ge d$ at size $n$ is exactly a binary covering code of length
+$n$ and radius $r=(n-d)/2$, of size at most $2n$ and closed under
+complementation, giving $C_{10c}\ge (n-2r)/\sqrt{\max(n,P)}$ for $P$ centres.
+Evaluating that over every entry of Kéri's table of bounds on $K(n,R)$
+(`old.sztaki.hu/~keri/codes/2_tables.pdf`), the best value obtainable is
+$5/\sqrt8$, from the row Kéri marks as a perfect code; the second is $7/4$ from
+$K(15,4)\le 32$ and the third is $7/\sqrt{17}$, the bound recorded above, from
+$K(17,5)\le 32$. A search using this framing rediscovers the $n=17$ certificate
+of [L2026] from scratch in under a second.
+
+Zeros are what make $n=8$ reachable. In a sign matrix at even $n$ every
+$(Ax)_i=n-2\lvert S\triangle T_i\rvert$ is even, so the discrepancy is even;
+beating $4$ at $n=8$ would need $\mathrm{disc}\ge 6$, i.e. $K(8,1)\le 16$,
+whereas $K(8,1)=32$. A single zero column breaks that parity constraint.
+
 ## Further remarks
 
 - For large $n$, the best asymptotic lower bound remains $1$ [Band2024].
@@ -356,6 +434,7 @@ The centres $T_i$ were found by a greedy set-cover search for a small radius-$5$
 - [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).
 - [L2026] Li, Youhua. 17 by 17 sign-matrix certificate for C10c, submitted to this repository (2026).
+- [H2026] Y. H. 8 by 8 $\{-1,0,1\}$-matrix certificate for C10c from the perfect $[7,4]$ Hamming code, submitted to this repository (2026). Verification notes: [howiehwong.github.io/poolish/constants.html](https://howiehwong.github.io/poolish/constants.html).
 
 ## Contribution notes