On strongly walk regular graphs, triple sum sets and their codes

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Problem: The conjecture is formalized as follows. Let $q=p^h$ be a prime power, and let $C\le \mathbb F_q^n$ be a $k$-dimensional linear code. “Projective” means that in a generator matrix no column is zero and no two columns are proportional, equivalently $d(C^\perp)\ge 3$. Suppose the nonzero Hamming weights of $C$ are exactly

$$w_1<w_2<w_3$$

and

$$w_1+w_2+w_3=3(1-1/q)n.$$

The conjecture asserts that

$$w_2=(1-1/q)n$$

and that $w_1=w_2-t,\ w_3=w_2+t$, where $t$ is a power of the characteristic $p$.

This is the standard coding-theoretic reading of the supplied Conjecture 3.4.

Result: The conjecture is false.

Work over $\mathbb F_3$, and let $C$ be the row span of

$$G=\begin{pmatrix} 0&0&0&0&1&1&1\\ 0&1&1&1&0&0&0\\ 1&0&1&2&0&1&2 \end{pmatrix}.$$

The columns are

$$(0,0,1),(0,1,0),(0,1,1),(0,1,2),(1,0,0),(1,0,1),(1,0,2),$$

which are seven distinct projective points in $\mathrm{PG}(2,3)$. Hence no two columns are proportional and none is zero. Also $G$ contains the standard basis columns, so $\operatorname{rank}G=3$. Thus $C$ is a projective $[7,3]_3$ code.

For a message vector $(a,b,c)\in\mathbb F_3^3$,

$$(a,b,c)G=(c,\ b,\ b+c,\ b+2c,\ a,\ a+c,\ a+2c).$$

If $c\ne0$, then among $b,b+c,b+2c$ exactly one entry is zero, and among $a,a+c,a+2c$ exactly one entry is zero, while $c\ne0$. Hence the weight is $5$.

If $c=0$, then

$$(a,b,0)G=(0,b,b,b,a,a,a).$$

If exactly one of $a,b$ is nonzero, the weight is $3$. If both $a,b$ are nonzero, the weight is $6$.

Therefore the nonzero weights are exactly

$$w_1=3,\qquad w_2=5,\qquad w_3=6.$$

They satisfy the conjecture’s hypothesis:

$$w_1+w_2+w_3=3+5+6=14 =3\left(1-\frac13\right)7.$$

But the asserted conclusion would require

$$w_2=\left(1-\frac13\right)7=\frac{14}{3},$$

which is impossible and in particular not equal to $5$. Also the weights are not symmetric about $w_2$, since $5-3=2$ but $6-5=1$.

Thus Conjecture 3.4 is disproved.

The defect is an omitted integrality/divisibility condition: the hypothesis does not force $(1-1/q)n$ to be integral when $q=3$. A natural repair would at least require $q\mid n$, but the conjecture as stated is false.

Verification audit: all hypotheses are satisfied by the displayed projective $[7,3]_3$ code; the weight set is computed exhaustively for all nonzero messages; the required weight-sum identity holds exactly; and the stated conclusion fails exactly.

Citation: Counterexample constructed above. Source of the conjecture: Kiermaier–Kurz–Solé–Stoll–Wassermann, “On strongly walk regular graphs, triple sum sets and their codes,” arXiv:2012.06160 / Designs, Codes and Cryptography 91 (2023).