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

[HowieHwong]

An 8×8 matrix over ${-1,0,1}$ with exact discrepancy 5, giving $C_{10c} \ge 5/\sqrt{8} = 5\sqrt2/4 \approx 1.767766953$, against the recorded $7/\sqrt{17} \approx 1.697749$.

The one thing a reviewer should look at first. Every lower bound recorded on this page so far uses a sign matrix. This one does not — column 8 is identically zero. That is inside the definition of $C_{10c}$ as stated at the top of the entry, and the upper-bound side quantifies over the same set: [PV2022]'s Theorem 4.5 reads "for every $A\in[-1,1]^{n\times n}$". If the intent of the entry is that lower bounds be restricted to sign matrices, then this submission does not belong and I would rather that be said than have the row stand — within sign matrices, $7/\sqrt{17}$ is untouched by this.

The zero is not a technicality; it is the mechanism. In a sign matrix at even $n$, $(Ax)_i = n - 2|S \triangle T_i|$ is always even, so the discrepancy is even, and beating 4 at $n=8$ would require $K(8,1)\le 16$ whereas $K(8,1)=32$. One zero column breaks that parity constraint.

Checkable by hand, per the contributing guidelines. The eight rows are the perfect $[7,4]$ Hamming code. The sets $T_i$ and their complements are its 16 codewords, and their radius-1 balls partition ${0,1}^7$ ($16\cdot 8=128=2^7$) — so the covering condition holds with zero slack. The entry prints the matrix, the $T_i$, and the full distribution of $\lVert Ax\rVert_\infty$ over all $2^8$ sign vectors (5 for 224 of them, 7 for 32). Nothing needs downloading.

Also included: 5 is optimal at $n=8$. By Kleitman's diameter theorem each row certifies at most 18 of the 256 sign vectors, and $8\cdot 18 = 144 < 256$, so $\mathrm{disc}(A)\le 5$ for every $A\in[-1,1]^{8\times 8}$. The same counting gives the exact optima at $n=2$ and $n=4$; those three are the zero-slack cases and are now all attained.

How it was found, since it may be reusable: the reduction in this entry turns the problem into a covering-code question, and evaluating $(n-2r)/\sqrt{\max(n,P)}$ over Kéri's published table of $K(n,R)$ bounds returns $5/\sqrt8$ as the best value available by lookup — from the row Kéri marks as a perfect code. The second-best is $7/4$ and the third is $7/\sqrt{17}$, the bound recorded here. A search in the same framing rediscovers [L2026]'s $n=17$ certificate from scratch in under a second, which is the sensitivity check for the negative results at larger $n$: no published $K(n,R)$ upper bound for $18\le n\le 33$ is small enough to beat $5/\sqrt8$, so going further would mean improving a covering-code bound by 20–50%.

README table cell and Recent progress line updated in the same commit, per CONTRIBUTING.

Separately, I have filed #168 about the upper-bound row on this same page recording a value its source has since corrected by erratum; I have deliberately not touched it here.

[curious7-web]

Ran this $8 \times 8$ construction through the Bansal-style Gram SDP relaxation tracked in #67 to check how the continuous relaxation handles the covering code structure without the discrete parity constraint.

Numerical findings ($n=8$):

• Discrete discrepancy: $5.0000$ ($\approx 1.7678\sqrt{n}$)
• SDP optimal $t$: $3.2087$
• SDP bound ($\sqrt{t}$): $1.7913$ ($\approx 0.6333\sqrt{n}$)

For comparison, an $8 \times 8$ Sylvester Hadamard baseline yields $t = 8.0000 \implies 1.0000\sqrt{n}$ in the same relaxation.

The large integrality gap ($5 / 1.7913 \approx 2.79$) cleanly illustrates that the strength of this certificate relies heavily on the discrete covering geometry and the parity obstruction broken by the zero-column; once relaxed to the unit sphere ($X \succeq 0, \text{diag}(X)=1$), the continuous solver exploits the slack to decouple the row correlations down to $0.6333\sqrt{n}$. Consistent with what @whyihaveyou observed on the $17 \times 17$ certificate in #67.