tag:blogger.com,1999:blog-5195188167565410449.post7004215305689127158..comments2014-03-09T17:09:45.698ZComments on Haskell for Maths: Combinatorial Hopf Algebras I: The Hopf Algebra YSym of Binary TreesDavidAhttp://www.blogger.com/profile/16359932006803389458noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-5195188167565410449.post-79738046853617920512012-03-22T21:37:39.230Z2012-03-22T21:37:39.230ZI think you're right that all trees are planar...I think you're right that all trees are planar graphs. So I assume that "planar" as applied to trees has the implication that an embedding into the plane is given, and that it therefore makes sense to speak of one child node as being to the left or right of another, as seen from the parent node.DavidAhttp://www.blogger.com/profile/16359932006803389458noreply@blogger.comtag:blogger.com,1999:blog-5195188167565410449.post-35827979344849612932012-03-22T09:45:10.612Z2012-03-22T09:45:10.612Z> The binary trees we're concerned with are...> <i>The binary trees we're concerned with are the familiar binary trees from computer science. In the math literature on this, however, they're called (rooted) planar binary trees. As far as I can tell, that's because in math, a tree means a simple graph with no cycles.</i><br /><br />The planarity requirement seems to be redundant here because it seems to be implied by the absence of cycles. <br /><br />As it is stated by (Pontryagin—)Kuratowsky theorem the graph is not planar if it contains subdivisions of K_5 or K_{3,3} (cf. <a href="http://en.wikipedia.org/wiki/Planar_graph#Kuratowski.27s_and_Wagner.27s_theorems" rel="nofollow">Wikipedia</a>) which implies cycles. So no cycles — no K_5 or K_{3,3} — the graph (tree in our case) is planar. <br /><br />Please, correct me if I'm wrong.Artemhttp://www.blogger.com/profile/16597823518560944008noreply@blogger.com