quasirandomness.

\[thm: chung-graham\] Let $h \in {\mathbb N}$ with $h\geq 4$. Then for any $n$-vertex tournament $T$, the following properties are equivalent:

-   ${\cal P}_1$: $|F_{\sigma ,T}| = 
        \frac {1}{2} \binom {n}{2} \pm  o(n^2)$ for every ordering $\sigma $ of $T$.

-   ${\cal P}_2(h)$: $N^*_T(H) = 2^{-\binom {h}{2}}n^h \pm 
        o(n^h)$ for every $h$-vertex tournament $H$.

In fact, there are nine further equivalences given in this paper (see also [@Griffiths], [@KSh]). Equivalence here is understood in the following sense, focusing on the implication ${\cal P}_1 \implies {\cal P}_2(h)$: given $\varepsilon >0$ there is $\delta >0$ and $n_0 \in {\mathbb N}$ so that if $T$ is an $n$-vertex tournament which satisfies $|F_{\sigma ,T}| = \frac {1}{2} \binom {n}{2} \pm \delta n^2$ for every ordering $\sigma $ of $T$ and $n\geq n_0$ then $N^*_T(H) = 2^{-\binom {h}{2}}n^h \pm \varepsilon n^h$ for every $h$-vertex tournament $H$.

It is easily seen that both properties ${\cal P}_1$ and ${\cal P}_2(h)$ hold for a random $n$-vertex tournament $T$ with high probability. We will say $T$ is *quasirandom* if it satisfies ${\cal P}_1$, that is $|F_{\sigma ,T}| = \frac {1}{2}\binom {n}{2} \pm o(n^2)$. In light of Theorem \[thm: chung-graham\], this notion is equivalent to the analogous notion arising 