 Solution

Give each bounded region $R_i$ a generator $a_i$ and give the unbounded region $R_0$ the identity generator. At every crossing, read the four incident regions cyclically as $a,b,c,d$ and imposeUsing the opposite cyclic convention inverts all such relators and gives the same group. One crossing relation is redundant, leaving $n$ generators and $n-1$ relators; this is the Dehn presentation of a knot group.