The number of touch points (or returns) of a Dyck path.
This is the number of points, excluding the origin, where the Dyck path has height 0.

Return the Dyck path corresponding to the parallelogram polyomino obtained by applying Delest-Viennot's bijection.
Let $D$ be a Dyck path of semilength $n$. The parallelogram polyomino $\gamma(D)$ is defined as follows: let $\tilde D = d_0 d_1 \dots d_{2n+1}$ be the Dyck path obtained by prepending an up step and appending a down step to $D$. Then, the upper path of $\gamma(D)$ corresponds to the sequence of steps of $\tilde D$ with even indices, and the lower path of $\gamma(D)$ corresponds to the sequence of steps of $\tilde D$ with odd indices.
The Delest-Viennot bijection $\beta$ returns the parallelogram polyomino, whose column heights are the heights of the peaks of the Dyck path, and the intersection heights between columns are the heights of the valleys of the Dyck path.
This map returns the Dyck path $(\gamma^{(-1)}\circ\beta)(D)$.

The bounce path determined by an integer composition.

The zeta map on Dyck paths.
The zeta map $\zeta$ is a bijection on Dyck paths of semilength $n$.
It was defined in [1, Theorem 1], see also [2, Theorem 3.15] and sends the bistatistic (area, dinv) to the bistatistic (bounce, area). It is defined by sending a Dyck path $D$ with corresponding area sequence $a=(a_1,\ldots,a_n)$ to a Dyck path as follows:

• First, build an intermediate Dyck path consisting of $d_1$ north steps, followed by $d_1$ east steps, followed by $d_2$ north steps and $d_2$ east steps, and so on, where $d_i$ is the number of $i-1$'s within the sequence $a$.
  For example, given $a=(0,1,2,2,2,3,1,2)$, we build the path
  $$NE\ NNEE\ NNNNEEEE\ NE.$$
• Next, the rectangles between two consecutive peaks are filled. Observe that such the rectangle between the $k$th and the $(k+1)$st peak must be filled by $d_k$ east steps and $d_{k+1}$ north steps. In the above example, the rectangle between the second and the third peak must be filled by $2$ east and $4$ north steps, the $2$ being the number of $1$'s in $a$, and $4$ being the number of $2$'s. To fill such a rectangle, scan through the sequence a from left to right, and add east or north steps whenever you see a $k-1$ or $k$, respectively. So to fill the $2\times 4$ rectangle, we look for $1$'s and $2$'s in the sequence and see $122212$, so this rectangle gets filled with $ENNNEN$.
  The complete path we obtain in thus
  $$NENNENNNENEEENEE.$$