particular, $\rho(X) \ge r+1$. Also if $\rho(X) \le 9$ then $\#\Gamma \le 9$.

We have only to prove the first assertion. By taking a subgraph, we may assume that $\#\Gamma \le 10$.

If $\Gamma$ is a linear chain, then it has negative definite intersection matrix, and we are done. Thus we may assume that there exists an irreducible component which meets more than two other irreducible components. Let $C_0$ be the irreducible component that meets $k$ other components with the largest $k$. Then $\Gamma - C_0$ has exactly $k$ connected components $\{\Delta_i\}$. We may assume that $k \ge 3$. Let $C_i$ be the irreducible component of $\Delta_i$ that meets $C_0$.

By Lemma \[negdef\], if $\#\Delta_i \le 5$ for all $i$ then each $\Delta_i$ is negative definite. By taking $\Gamma'=\sum
\Delta_i$, we are done.

The remaining cases of $(\#\Delta_1, ..., \#\Delta_k)$ are $\{
(1,1,6), (1,1,7), (1,2,6), (1,1,1,6)\}$. For the case $(1,1,1,6)$, we take $\Gamma'= \Gamma-C_4$, then now $\Gamma'$ has at least two connected components: $C_0+C_1+C_2+C_3$ and others. It is clear that each connected component has at most $5$ irreducible components. Hence $\Gamma'$ is negative definite. For the cases $(1,1,6)$ and $(1,2,6)$, similar argument works.

It remains to work with the case 