There are $n$ non-intersecting strings (with ends $x_1,\dots, x_n$ and $y_1,\dots, y_n$). An additional string intersects the first $n$ strings somehow. All the intersections are simple (vertices of degree $4$). The edges incident to an intersection points are divided into two pairs of *opposite* edges (two opposite edges in a pair belong to the same string).

We consider sets of $n$ paths in the graph formed by these $n+1$ strings which have three properties:
1) paths have the same ends as the original $n$ strings (i.e. each path connects $x_i$ with $y_i$ for some $i$);

2) paths have no common edges;

3) paths can have common vertices but their intersections are not transversal. Two path intersect *transversly* if each path contains a pair of opposite edges at the intersection point. In this sense, the intersections of the additional string with the other strings are all transversal.

The original $n$ strings satisfy these three conditions.

Is there a number $N=N(n)$ such that if the number of intersections of the additional string with the other strings is greater than $N$, we can find a set of $n$ paths that satisfies the conditions and differs from the original strings?

The example below presents a configuration where there are two different sets of paths (the original strings (left) and an alternative set of paths (right)).

Warning: the example is braid-like since all the strings are monotonic, but in general case the additional string can go up and down at its discretion.

(*Example image included by J.O'Rourke*.)