Improve C10c lower bound to 5/3

[bangoz]

Summary

This PR updates constants/10c.md with a finite-dimensional certificate giving

$$ C_{10c} \ge 5/3 \approx 1.666667. $$

The certificate uses a $9$ by $9$ sign matrix $A$ with exact discrepancy

$$ \mathrm{disc}(A)=5. $$

Thus

$$ \frac{\mathrm{disc}(A)}{\sqrt{9}}=\frac{5}{3}. $$

This improves the lower bound $4/\sqrt{6}\approx 1.632993$.

Certificate

Use the following $9$ by $9$ sign matrix:

 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

Let $T_i\subseteq{1,\ldots,9}$ be the set of columns where row $i$ has entry $-1$:

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 = {}

The ninth row is a duplicate of the first row; duplicate rows are allowed, and here it simply makes the displayed certificate square.

For $x\in{\pm 1}^9$, let $S\subseteq{1,\ldots,9}$ be the set of columns where $x_j=-1$. Then

$$ (Ax)_i = 9-2\lvert S\triangle T_i\rvert. $$

Hence $\lvert (Ax)_i\rvert\ge 5$ if either

$$ \lvert S\triangle T_i\rvert\le 2, $$

or

$$ \lvert S\triangle T_i^{c}\rvert\le 2, $$

where complements are taken in ${1,\ldots,9}$. It is therefore enough to prove that every vertex $S$ of the full $9$-cube is within Hamming distance $2$ of some $T_i$ or some complement $T_i^{c}$.

Full 9-cube covering proof

Although the useful structure is six-dimensional, the covering statement below is for all $2^9$ vertices of the $9$-cube.

Write

$$ S=U\cup W, $$

where

$$ U\subseteq{1,\ldots,6},\qquad W\subseteq{7,8,9}. $$

Let $w=\lvert W\rvert$. Since all the sets $T_i$ above are contained in ${1,\ldots,6}$, for $i=1,\ldots,8$ we may write $T_i=C_i$, where

C = {}, 123, 145, 2345, 246, 1346, 1256, 356

Let $C_i^{\ast}$ denote the complement of $C_i$ inside ${1,\ldots,6}$. Then the complement of $T_i$ inside ${1,\ldots,9}$ is

$$ T_i^{c}=C_i^{\ast}\cup{7,8,9}. $$

Therefore the two relevant distances in the full $9$-cube are exactly

$$ d_9(S,T_i)=d_6(U,C_i)+w $$

and

$$ d_9(S,T_i^{c})=d_6(U,C_i^{\ast})+(3-w). $$

So the last three coordinates are not being ignored; they appear through the terms $w$ and $3-w$.

We now record the finite $6$-cube fact used in the four cases below. The eight radius-1 balls around

C = {}, 123, 145, 2345, 246, 1346, 1256, 356

miss exactly the eight complements

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 $C^{\ast}$ 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 $8$ missed vertices.

The missed vertices are nevertheless within radius $2$ of $C$; respectively, one may use the following centers:

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 $d_9(S,T_i^{c})=d_6(U,C_i^{\ast})+2$. 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 $U=C_i^{\ast}$ 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 $C_i^{\ast}$, giving $d_9(S,T_i^{c})\le 2$.

If $w=3$, then $d_9(S,T_i^{c})=d_6(U,C_i^{\ast})$. By applying the radius-2 covering of the $6$-cube by $C$ to the complement of $U$, the complements $C_i^{\ast}$ 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 some $T_i^{c}$. Consequently every sign vector $x\in{\pm 1}^9$ satisfies

$$ |Ax|_\infty\ge 5. $$

Equality is attained. For example, with

x = (-1,-1,1,-1,1,1,1,1,1)^T

one obtains

Ax = (3,5,5,3,5,3,3,-3,3)^T.

Therefore $\mathrm{disc}(A)=5$, and consequently

$$ C_{10c}\ge\frac{5}{\sqrt{9}}=\frac{5}{3}\approx 1.666667. $$

Verification

The certificate was also checked by exhaustive enumeration of all $2^9=512$ sign vectors. A minimal verification script is:

const n = 9;
const centers = [
  [],
  [1,2,3],
  [1,4,5],
  [2,3,4,5],
  [2,4,6],
  [1,3,4,6],
  [1,2,5,6],
  [3,5,6],
  [],
];

function mask(set) {
  return set.reduce((m, j) => m | (1 << (j - 1)), 0);
}

function popcount(x) {
  let c = 0;
  while (x) {
    x &= x - 1;
    c++;
  }
  return c;
}

function setString(m) {
  const out = [];
  for (let j = 1; j <= n; j++) {
    if (m & (1 << (j - 1))) out.push(j);
  }
  return `{${out.join(",")}}`;
}

const T = centers.map(mask);
const fullMask = (1 << n) - 1;
let best = Infinity;
let bestS = null;
let bestVals = null;
let uncovered = [];

for (let S = 0; S < (1 << n); S++) {
  const vals = T.map(t => n - 2 * popcount(S ^ t));
  const discrepancy = Math.max(...vals.map(v => Math.abs(v)));
  if (discrepancy < best) {
    best = discrepancy;
    bestS = S;
    bestVals = vals;
  }
  const covered = T.some(t =>
    Math.min(popcount(S ^ t), popcount(S ^ (t ^ fullMask))) <= 2
  );
  if (!covered) uncovered.push(setString(S));
}

console.log({
  best,
  bestS: setString(bestS),
  bestVals,
  ratio: best / Math.sqrt(n),
  uncoveredByRadius2AntipodalBalls: uncovered,
});

It returns

best: 5
bestS: {1,2,4}
bestVals: [3, 5, 5, 3, 5, 3, 3, -3, 3]
ratio: 1.6666666666666667
uncoveredByRadius2AntipodalBalls: []

Changed file

• constants/10c.md

AI assistance disclosure

Codex was used to search for and verify the certificate and to draft this PR text.