; (v1)–(v2); at (2,-2) [$H(1,1,2)$]{};

In the blow-up $H(2,1,1)$ vertex 1 splits into vertices $v_{1,1}$ and $v_{1,2}$; vertex 2 becomes $v_2$ and vertex 3 becomes $v_3$. In the blow-up $H(1,1,2)$ vertex 3 splits into vertices $v_{3,1}$ and $v_{3,2}$; vertex 1 becomes $v_1$ and vertex 2 becomes $v_2$.

[**(Blowing up)**]{} \[blowup\] Let $H$ be any hypergraph and let $s\geq 2$. Then $\pi(H(s))=\pi(H)$.

[**Proof:**]{} Let $R:=R(H)$. By the supersaturation lemma, for any $a>0$ there is a $b>0$ and an $n_0$ so that any $R$-graph $G$ on $n\geq n_0$ vertices with $h_n(G)>\pi(H)+a$ contains at least $b{n \choose v(H)}$ copies of $H$. Consider an auxiliary $v(H)$-uniform hypergraph $U$ on the same vertex set as $G$ where edges of $U$ correspond to copies of $H$ in $G$. For any $S>0$, there is a copy of $K=K^{v(H)}_{v(H)}(S)$ in $U$. This follows from the fact that $\pi(K^{v(H)}_{v(H)}(S))=0$ since it is $v(H)$-partite, and $h_{n}(U)=b>0$. Now color each edge of $K$ by one of $v(H)!$ colors, each color corresponding to one of $v(H)!$ possible orders the vertices of $H$ are mapped to the parts of $K$. The pigeon-hole principle gives us that one of the color classes contains at least $S^{v(H)}/v(H)!$ edges. For large enough $S$ there is a 