GPT 6 Astra Ultra attempt on Math-6212
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Contents
Erdős Problem 1158 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
For fixed integers $t,r\geq2$, let $K_t(r)$ have $t$ vertex classes of size $r$, with every transversal $t$-set an edge. Is \[ \operatorname{ex}_t(n,K_t(r))\geq n^{\,t-r^{1-t}-o(1)} \qquad(n\to\infty) \] always true?CONTEXT AND SCOPE
This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.The parameter range t,r>=2 is the nondegenerate Turan-problem range; the official wording says 'for all t,r' without spelling it out.
ATTACK PLAN AND WORK
The elementary random construction quantifies a specific exponent gap. Write $v=tr$ and $e=r^t$ for the vertex and edge counts of the forbidden hypergraph. In a random $t$-uniform hypergraph on $n$ vertices with edge probability $p$, the expected numbers of edges and labeled forbidden copies are respectively $\binom nt p$ and at most $n^{tr}p^{r^t}$. Remove one edge per surviving copy. The resulting graph is forbidden-copy-free and retains at least the original edge count minus the original copy count.Put $a=(tr-t)/(r^t-1)$ and $p=c n^{-a}$. The two expectations both have exponent $t-a$, since $tr-ar^t=t-a$. For a sufficiently small constant $c=c(t,r)>0$, the positive term, of order $c n^{t-a}$, dominates the negative term of order $c^{r^t}n^{t-a}$. Some outcome therefore gives \[\operatorname{ex}_t(n,K_t(r))\gg n^{t-(tr-t)/(r^t-1)}.\tag{1}\]
The loss in $1$ is strictly larger than the requested $r^{1-t}$, because $t(r-1)r^{t-1}>r^t-1$ for $t,r\ge2$. For $t=r=2$, for instance, this method gives exponent $4/3$ rather than $3/2$. This comparison is about this construction, not a claim that $4/3$ is the best known bound in that special case.