For alternating (p, q, r) pretzel knots, pq + pr + qr gives the number of Trees

[onye]

PROOF

According to a paper by Raquel Lopez, a (p, q, r) pretzel is
alternating if and only if p, q, and r all have the same sign. Lets
assume they are all positive. (If they are all negative, the argument
is similar.) So p, q, and r represent the number of crossings in the
three parts of the pretzel. We are interested in creating a planar
graph from which we can derive its trees. If we follow
the procedure given above for finding trees, we can obtain the planar
graph shown in Figure ii). This graph has p edges at left, q edges
down the center, and r at right. Deleting any two edges, one each from
the left and center of the graph will leave a tree as there will
remain no closed cycles. There are pq ways to choose the two edges and
pq trees. By following the same procedure, we can obtain pr trees by
removing edges at left and at right and, finally, qr trees by removing
edges at center and at right. Hence, pq + pr + qr = total number of
maximal trees for alternating pretzel knots.


Please see this link for more details and graphic illustrations by
Raquel Lopez:

http://www.math.jmu.edu/~taal/OJUPKT/lopez.pdf

[Alex]

Hey Onye,

Do you think you could clarify what a tree is? Lopez says a tree is
simply a connected graph, so in what sense does a tree have multiple
trees inside of it?

-Alex

[marg...@math.utah.edu]

Hi Alex,

A tree is not "simply a connected graph", rather it is "a simply
connected graph". A space is said to be "simply connected" if its
fundamental group is trivial. What kinds of graphs have this
property? Why do we call them trees? What other simply connected
spaces do we know?

Dan