Solution
= 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 impose
$$
ab^{-1}cd^{-1}=1.
$$
Using 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>.