Knot theory question: bridge number vs. min generators of fundamental group of complement Given a knot in the 3-sphere in Bridge Position you can find a presentation for the fundamental group of the complement (a Wirthinger presentation) containing $n$ generators and $n-1$ relators, where $n$ is the number of bridges in the presentation. 
The bridge index of the knot is the minimal number of bridges it takes to present the knot.  Is that the same as the minimal number of generators among all presentations of the fundamental group of the knot complement?  
I imagine this is well known but after a while of Googling around and looking through Neuwirth's "Knot Groups" book I haven't found where this question is addressed. 
 A: It is not a problem, but this is a problem.
A question posed by Cappell and Shaneson asks whether the bridge number of a given link equals the minimal number of meridian generators of its group.
http://ccms.claremont.edu/topology-seminar/Bridge-numbers-links-and-meridian-ranks-link-groups
A: The (p,q) torus knot has a presentation with two generators, namely $\langle x,y \mid x^p = y^q\rangle$, but if $p,q>2$ then it's non-alternating and so it must have bridge index greater than 2.
A: The existence of such examples is proved in a preprint of Jesse Johnson. He
proves that for any $n$, there is a tunnel number 1 knot which has bridge number at least $n$.
A tunnel number one knot complement has a Heegaard surface of genus 2, which splits the knot complement up into a genus 2 handlebody and a genus 2 compression body with a torus boundary corresponding to the boundary of a regular neighborhood of the knot. Thus tunnel number 1 knots are 2-generator (and in fact with 1 relation), so this answers your question.
