Yes. When a group $G$ acts geometrically on a metric space $X$, by choosing a basepoint $x_0 \in X$ you can construct its Dirichlet domain
$$ D_{x_0} = \{x \; | \; d(x, x_0) \leq d(x, g \cdot x_0) \; \forall g \in G\} $$
When the action of $G$ is sufficiently nice, this domain has finitely many sides and geodesics which are perpendicularly bisected by each face form a finite generating set for $G$.

Since the mapping class group acts geometrically on the (labelled) flip graph (with the graph metric) we can do a similar process starting at a triangulation $\mathcal{T}_0$.

- Let $X_1$ be the set of mapping classes which move $\mathcal{T}_0$ by the smallest non-zero amount.
- Let $X_2$ be the set of mapping classes which move $\langle X_1 \rangle \cdot \mathcal{T}_0$ by the smallest non-zero amount.
- Let $X_3$ be the set of mapping classes which move $\langle X_1, X_2 \rangle \cdot \mathcal{T}_0$ by the smallest non-zero amount.
- Let $X_4$ be the set of mapping classes which move $\langle X_1, X_2, X_3 \rangle \cdot \mathcal{T}_0$ by the smallest non-zero amount.

$\vdots$

Then each $X_i$ is finite and for some $N$ the elements of $X_1 \cup X_2 \cup \cdots \cup X_N$ generate $G$. Since each of generator $g$ can be represented by a path in the flip graph from $\mathcal{T}_0$ to $g(\mathcal{T}_0)$ the relations between these generators can then be understood from the 2--cells of the flip graphs which give:

- the square relation - that disjoint flips commute, and
- the pentagon relation - that two flips which share a common triangle form a 5--cycle

Since there are explicit descriptions of the action of the mapping class group on the flip graph this entire process can be done on a computer (although as far as I am aware no one has actually done this).