This question - so far as I know - has no broader mathematical significance, but it occurred to me a while ago and I haven't been able to make any headway.


Any knot diagram $D$ splits the plane into a finite number of pieces. For example, in a standard diagram for a trefoil (e.g., <http://en.wikipedia.org/wiki/File:Trefoil_knot_left.svg>), the plane is split into five pieces: one lying "outside the diagram," three "lobes" of the knot, and a central region around which the three "lobes" are arrayed. The outer region has three crossings on its boundary, as does the inner region; each of the lobes only have two crossings each on their boundaries.


A diagram for a more complicated knot may split the plane into many more regions, and one of these regions may have many crossings on its boundary. My question is: Does there exist some $n$ such that for any knot $K$, there is a diagram $D$ of $K$ such that any region created by $D$ has at most $n$ crossings on its boundary?


If the answer is no, a (very soft) question that then arises is: given a knot $K$, how hard is it to find the least $n$ ($=n\_K$) such that for some diagram $D$ of $K$, every region created by $D$ has at most $n$ crossings on its boundary? For example, for any knot $K$, we can easily see that $n\_K>2$.



Polygons arising from knot diagrams
Colin Adams, Reiko Shinjo and Kokoro Tanaka have a paper (<http://arxiv.org/abs/0812.2558>) that shows that for any knot you can find a diagram which has only regions with 2, 4 and 5 sides. 


