to intersect, because $$\begin{aligned}
    \abs{c(X)} + \abs{c(v,Y\setminus \{v\})} &= \abs{X} + \abs{Y} - 1\\
    &\geq 3n-3t - 3a -17 \\
    &> t-1\\
    &= \abs{c(C)}\text.
  \end{aligned}$$ Therefore, there exists an edge $e\in X$ with $c(e) = c(v,u)$ for some $u\in V(C)^c$. By definition, $e$ is incident to some vertex $x\in V(C)$ with $c(x,v){\not\in}c(C)$. Then the rainbow path $(C-e)\cup (x,v,u)$ is a rainbow path of length $t$, but this contradicts the definition of $t$.

This suggests that one could choose $P$ in such a way that $\abs{A(P)\cap B(P)}$ is small. Indeed, we can do this, as we will show next.

\[cons:a-b\] Suppose that there are no nice maximum rainbow paths. Then there is a maximum rainbow path $P$ such that $\abs{A(P)\cap B(P)}\leq \epsilon n$.

Let $k = \ceil{1+2/\epsilon}$ and let $P$ be an arbitrary maximum rainbow path. Now we define a sequence of maximum rainbow paths $(P_0,\dotsc,P_k)$ as follows:

-   $P_0 = P$.

-   $P_l$ is defined in terms of $P_{l-1}=(p_1,\dotsc,p_t)$. Because $P_{l-1}$ is a maximum rainbow path, we have $\abs{A(P_{l-1})} \geq n-t > k$. Then choose a vertex $p_i \in A(P_{l-1})$ that is *not* the starting vertex of any of the paths $P_0,\dotsc,P_{l-1}$, and let $P_l = \rho_i \cdot P_{l-1}$. Then, 