# Pretzel knot equivalence

How would you go about proving the $(p, q, r)$ -pretzel knot is equivalent to the $(p, r, q)$ -pretzel knot?

By "equivalent" I mean you can change one knot into the other by elementary deformations.

I've found this question/example in several books and papers on knot theory where they state the proof as obvious.

• Have you tried looking at a sequence of Reidemeister moves to get the $(p,r,q)$ pretzel knot from the $(p,q,r)$ pretzel knot? In particular you are trying to find an ambient isotopy. I think one can just use the definition of elementary deformation (e.g. look at the top of the second tangle and choose a point $P$ not collinear and try to deform it to the same section in the third tangle, etc...) So we keep moving down the second tangle. – PrimeNumber Feb 17 '11 at 3:37

You can change the $(p,q,r)$ pretzel link into the $(r,q,p)$ pretzel link by a 180 degree rotation around the midpoint of the central twist box (the one containing $q$ twists).
There is also a way to change the $(p,q,r)$ link into the $(q,r,p)$ link. This is a bit three-dimensional, so takes a bit more work to see. Draw the diagram on a sphere $S$ in $R^3$ instead of on a plane in $R^3$. Now you can drag the twist box containing $r$ twists around the back of $S$ ("past infinity") until it becomes the first twist box. (The arcs connecting the twist boxes rearrange themselves in the most convenient way!)