The Catalan multiset algebra
Wild Egg Maths
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We investigate a geometric formulation of the algebra underlying the Catalan sequences, following Exercise 22 of Chapter 7 of Concrete Mathematics: A Foundation for Computer Science by Graham, Knuth and Patashnik. We prefer to frame this in terms of an algebra of multisets formed from triangular (roofed) decompositions of polygons, with a natural triangle multiplication operator so that the corresponding generating function satisfies a quadratic functional equation.
We also look at this from the point of view of planar rooted complete binary trees.
Naturally we get a nice geometric interpretation of the usual recursive formula for the Catalan numbers.
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hello everyone i'm norman wahlberger and
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thank you for joining me
00:03
on this adventure as we try to connect
00:07
polygon subdivision with solution
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of polynomial equations so i want to
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start by delving into one of these
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exercises
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from concrete mathematics that i talked
00:21
about in my last video
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and we're looking at some formal object
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which we don't quite understand what it
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really is but it's like a
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sum a formal sum of all the polygons
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with
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all of their subdivisions into triangles
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okay so we're calling this ongoing sum p
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and it consists of a two gone with just
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like a single kind of edge
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almost a degenerate case and then we
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have our triangle and then we have our
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quadrilateral into two
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uh subdivisions and then the five ways
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of subdividing a
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pentagon into triangles and then the 14
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ways of subdividing a hexagon into
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triangles and so on okay
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and this entire thing we're calling p
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now
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knuth and company ask us to somehow come
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up with a new
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operation on these subdivided
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polygons so that this um
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the sum p is going to have the equation
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that p
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equals the the identity which is that
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that single stroke that starts
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um the formula for p plus
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p triangle p okay so today we want to
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we want to tackle this because this is
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really really relevant
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to understanding the connection between
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subdivisions of a polygon
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and solutions to
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general polynomial equation okay that's
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what's coming into focus here
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okay so we want to really
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delve into this simplest case and really
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try to understand
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what's going on and this you will see um
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is is a very lovely starting point so
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we're creating a bridge here okay
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but we're gonna see that we'll have to
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to make things a little bit rigorous
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we'll have to adopt a kind of a
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multi-set point of view which i like to
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do anyway multi-sets are
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very important kinds of data collections
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that fit into
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my lom's structure lists ordered sets
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multisets sets those are the four basic
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types of data structures in mathematics
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to start with
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okay and uh and the multiset point of
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view i think is the best one to try to
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get at
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describing this algebra or setting it up
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properly
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okay so uh it's not too hard to figure
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out what this triangle operation is and
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so let's uh
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define it i'm sure if you thought about
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it for half an hour you'd come up with
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the same thing okay
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maybe you might want to do that okay so
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let's define
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the triangle operator
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and i'm going to show you how to do this
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via an example
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all right because it's nice to start
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with something specific so we can sort
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of see
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what's going on and we can capture the
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idea without having to use a lot of
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words just a picture being worth
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a thousand words okay so let's take a
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subdivided polygon let's say a pentagon
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and let's choose the second one here all
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right so there's the second pentagon in
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our list
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and now here's our operator our triangle
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operator
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and why don't we multiply say with a
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hexagon
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okay there's 14 ways of subdividing a
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hexagon let's choose one of them
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how about this way with an n something
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like this
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okay there's a particular subdivision
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now i should remind you that
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the way we're visually describing these
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subdivisions
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there's always for us a distinguished
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top edge which is horizontal okay
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so we could make that more explicit by
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putting a bar
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over that top edge
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and in fact to be sort of consistent it
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makes sense actually
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also to put it on the triangle too so
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it's a kind of a
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an oriented you know triangle operator
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and you'll see why in a second
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okay so i have to define what this thing
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is okay what is what is this thing
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okay so i'm going to put the triangle
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here okay there's this bar it has two
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free edges there's a left edge on this
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side
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there's a right edge on this side what
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we're going to do is we're going to take
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the
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pentagon which is the left entry here
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and we're going to stick it
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along here in such a way that
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the top edge is rotated if necessary
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to fit or glue onto that
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free left edge okay so we're kind of
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gluing it along here
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okay so how does that position the
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subdivision uh
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the subdivision will then i guess be
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emanating from this point here so so
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it'll have these two lines
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do you agree so i've rotated that
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hexagon by i guess a fifth of a turn
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and uh and then and glued it onto that
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central
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triangle on the left side the left one
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because this is the left one in the
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product this is the right one in the
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product
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okay and then we're gonna do the same
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thing with this hexagon first we have to
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rotate it
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so that it ends up being glued
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uh again with with the distinguished
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top face now abutting the free right
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edge of our triangle
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okay so we have to kind of rotate now by
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a sixth of a turn and the other
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direction
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uh so let's see how that goes there's
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the edge
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um i guess there should be one like this
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this one goes there i guess it's like
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this
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um does that look right to you
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hope i've done that right uh oh yeah i
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think so we just rotated that hexagon
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sorry
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so we we have glued
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we have glued
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the top edges
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of the left
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and right
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polygons
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to the free
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left and right edges
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of the triangle
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and what is the result uh so let's
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rewrite the results so it looks a little
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bit perhaps cleaner
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so we have how many um
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how many uh edges all together um one
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two
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three four from the the pentagon five
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from the
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um the hexagon four plus five is uh
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is nine and then the top one from the uh
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the triangle so i guess that's ten so
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actually we get a decagon
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or a ten gun a circle
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um there will be our top edge one
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two there and then one
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two three four five how's that
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one two three four five six seven eight
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nine ten looks like ten okay
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so um
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what is a subdivision okay so there's
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our
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top and that uh triangle that's fitting
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in
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has one two three
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one two three so i think we're going
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over like this that's that's the sort of
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the triangle that we started with
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uh originally and then on one side there
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is a
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the pentagon on the left side and let's
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see the pentagon
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looks like this there's
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an edge coming here and there's an edge
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coming here
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okay that's the pentagon side of things
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and then the hexagon on the other side
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um let's see uh from this side
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we should go over like this
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and then we should go over like this and
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then maybe over like this how does that
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sound
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we've got now a subdivision
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into triangles there was that sort of
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starting
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central triangle and we've just glued
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the pentagon and the hexagon to the free
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sides of it
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in such a way that the
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the distinguished edges of the pentagon
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and hexagon
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are glued to the actual free edges okay
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so let's make a note of that note that
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the distinguished
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top edges
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of the two polygons
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get glued get glued
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to the free edges
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of the triangle
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this is now an operation on
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polygons given any two well actually
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unsubdivided polygons which have a
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distinguished
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top edge okay so
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this is a multiplication
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on what on
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polygons
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with a triangular
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subdivision
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and distinguished
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say top edges
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so it's a binary operation
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great now i want to show you this again
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um but in a
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not a polygon way but
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rather in terms of trees
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okay and uh i think that's really
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instructive because that's
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all right so let's have a look at this
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same operation now from a tree point of
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view
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okay so here are the two polygons that
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we want to multiply the pentagon and the
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hexagon
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and i have to remind you how we create
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trees from these
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rather special kinds of trees okay so
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what we do is we
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put a node in each of the
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triangles in the subdivision so three in
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that case and
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four in this case and then we connect
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them according to adjacency of the
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triangles in our polygon
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okay so we get those things and then we
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also
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uh put in the extra
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the extra edges of the polygon that have
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not yet been crossed that sort of
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correspond to going to the outside
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okay and in particular this sort of top
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distinguished one we'll put a little bar
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on it
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um you know to indicate what it is and
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so over here
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we have uh okay those two there that
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one's got another one coming out here
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this one's got another one coming out
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here this one's got one here and one
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here
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okay what about the the triangle in the
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middle well we could do the same kind of
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thing for that it's much simpler there's
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just
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one node in the middle two coming out
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like this and one coming up and we'll
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put a bar on that
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okay so this um
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maybe you know in terms of trees
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in terms of actually planar
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rooted planar rooted
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trees this is
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what does that actually look like so i'm
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going to
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draw these trees uh okay
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i suppose i can do it in redstone okay
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i'm going to draw these trees in in
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the more standard way that computer
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scientists like where we
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we have our top distinguished or rooted
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node up there and then we we come down
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to
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our first node of the pentagon and then
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it's a binary tree so
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to the left we get the output outside
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and to the right we go
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down to that second node then from that
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node if we go left we go outside and if
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we go down
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we get to the third node and then
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there's um
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two uh emanating from that so that is a
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kind of a standard view
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like a computer scientist kind of view
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all that
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planar tree that that up there was just
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wrapping itself around okay now we've
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kind of organized it in a binary way so
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that all the branches are going down
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there's a definitely a top and we can go
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down in this tree
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okay what about the triangle uh well
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that's already pretty well in standard
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form
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we just need to copy that let's copy
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that okay
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and what about the hexagon okay there's
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that top thing there going down to that
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first
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note to the right we go out to the left
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we go to another
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node to the right of that we go out and
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then to the left of that we go
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to another node and then from there if
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we go left we go
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out uh hang on one two three
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uh sorry one two or three
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yes and uh and to the right we we get to
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that that last node
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and that last node then has these two um
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things animating from it
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okay so this is the the planar rooted
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tree
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version of the polygonal subdivision
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and you know there's advantages in both
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ways but i think it's particularly
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pleasant
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to make contact with this because a few
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scientists like these binary
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trees they're very nice okay
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so what does the multiplication amount
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to
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in this case well it's almost uh
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even simpler really okay i mean what are
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we really doing we're just taking this
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pentagon and we're you know attaching it
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to the triangle on the left hand side
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um via the the top note so it's this top
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node here
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this top note here is is simply you know
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going in that position there
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well actually actually no that's maybe
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not the right way of saying it um that
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top node
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sorry i shouldn't say it that way what's
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what's happening is that top
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node is that top note is actually
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sort of going to be glued to that
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central
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node which corresponds to the triangle
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in such a way that
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that um you know this this free edge
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coming out
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is going to be uh attached to uh
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this first note of the pentagon
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okay and similarly uh over here uh this
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thing here is going to sort of be
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attached there and and this first note
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is going to be attached there so
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basically
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we're just gluing the two
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binary trees to to the triangle in in
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what's really probably the simplest
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possible way so let me see if i can
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actually um
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do this okay um what is the result
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so um there's our triangle okay and
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there
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there's the the first note coming out so
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it will go left to
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the pentagon so what i just have to do
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is basically just copy uh
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the
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the pentagon uh graph so there's the
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first one that's like a okay
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that's a there okay and then there's one
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to the left and then there's
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a second one one to the left there's a
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third one and there's two below it
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okay so i've just copied the pentagon
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um by attaching it to our central
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triangle on the left
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okay now we have to attach the hexagon
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hopefully i have enough room here
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so um it's going to go down here to the
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to the right and then we're going to get
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this node here maybe
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call that one b just so we know there's
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b okay
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what happens with b um i may have run
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out of room okay but i'm going i'm going
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down here
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uh first there's one there and then
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going down one and then
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um how am i going to do this
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i'm going to go over there and now
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over from here i have to okay allow me
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this is
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very bad but i'm going to go down here
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that's an extension
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there so that's c let's see and that's d
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okay
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sorry um and c has one coming out and
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then c
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goes to d going sort of down to the left
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and then from d we can go out or we can
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go to the next one and
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there's two more like this
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okay so um
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actually this is
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at the level of at the level
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of planar
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rooted trees
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this is very natural
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okay because because our
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planetary rooted i guess these are
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binary trees for us
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yes they're binary i should put that in
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there too
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at the level of planar rooted binary
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trees this is very natural
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okay we're just sort of attaching the
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the two trees in the obvious way to the
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to the central little triangle tree
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okay so that's great now um
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how do we get this equation let's look
18:45
at the equation that we want satisfied
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so we're trying to convince ourselves
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that p is equal to
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the the identity plus p
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triangle bar p
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why is this true
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well this is true because of the
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following reason that
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if you have if you have
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a i think let's
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let's go back to the to the level of
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polygons if you have a triangular
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polygon subdivision
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with a distinguished
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top edge
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there is a distinguished
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interior
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triangle
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with that top edge
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okay so so we have some some big polygon
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all the way around right and there's our
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top okay
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that that that edge there that
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distinguished top edge
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will some will be part of a triangle
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somewhere
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okay so there's a distinguished triangle
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this triangle could be anywhere it could
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be
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you know leaning to the left it could be
20:29
leaning to the right it could actually
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also be
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um you know a
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sharing two edges with the outside it
20:36
could also be
20:38
uh something you could also have
20:39
something like this
20:41
where that distinguished edge uh looks
20:43
like that
20:44
okay but in any case what you're going
20:47
to get is that there's going to be a
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left edge and a right
20:50
edge of this distinguished triangle and
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the left edge if you look
20:54
at the subdivision of what you have on
20:57
the left
20:58
you have a subdivision of some smaller
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smaller polygon
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into triangles and on the right you also
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have a subdivision of a smaller polygon
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on the right okay so over here there is
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a
21:10
smaller polygon smaller
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polygon subdivided
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and that polygon out actually has a
21:21
distinguished edge namely this one here
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so with distinguished
21:30
edge
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and similarly over here it's exactly the
21:36
same thing you have another smaller
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polygon subdivided with a distinguished
21:39
edge namely this one here
21:42
now what about in this extreme case when
21:44
your triangle
21:46
is is like up here okay so
21:49
in this case you would have a you know a
21:51
big polygon on one side but what you
21:52
have on the other side
21:53
is you really just have the single edge
21:56
which is also part of our story because
22:00
we've agreed
22:00
that this sort of two-sided polygon this
22:03
one here
22:04
is part of our collection of polygons so
22:07
we're allowed to consider also gluing
22:09
that kind of thing
22:10
to the left side or to the right side
22:12
and it's not going to have any effect
22:13
it's not going to actually add anything
22:15
onto that side
22:16
that edge will still be a boundary edge
22:20
what does this show us so this shows us
22:22
that that
22:23
every so every
22:29
um one of these
22:32
every polygon
22:35
subdivided
22:38
into triangles
22:42
with a distinguished top edge
22:51
is the product in exactly the way that
22:53
we've said
22:54
is the product
22:57
of two unique
23:02
such polygons
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one on the left
23:11
and one on the right
23:19
except except there's one polygon which
23:22
is not of this form
23:22
there's one polygon for which this is
23:25
not true
23:26
except the polygon which is just like
23:28
this one here
23:30
right this this this two-sided polygon
23:33
cannot be created
23:34
by starting with a triangle and gluing
23:36
stuff to the triangle because the
23:37
triangle is already too big
23:40
so what what we end up what what we've
23:42
just said is that
23:44
every polygon which is a sort of
23:47
ingredient
23:47
in the in this sum
23:51
p is of the form
23:55
something in p some unique thing in p
23:58
multiplied with our triangular
24:00
operation times some other thing which
24:01
is uniquely in p
24:03
except for the
24:07
two-sided thing which we have to add
24:08
separately so when we take this product
24:10
p
24:12
triangle p we're getting all possible
24:15
subdivided polygons with unique
24:17
distinguished edges at the top except
24:19
for
24:20
the um the two-sided one
24:25
okay so that's
24:29
sort of an immediate beautiful equation
24:33
that arises from this
24:37
multiplicative structure
24:42
now um
24:46
so how do we make sense of this okay so
24:49
how do we make sense of this we have to
24:50
talk a little bit about
24:51
multisets
24:57
now multisets are very flexible data
24:59
structures they don't
25:00
get near the attention and glamour that
25:04
they deserve
25:05
but they are actually very important and
25:07
and for this
25:10
line of argument i think multisets are a
25:12
very natural way of thinking
25:14
so what we have right now is we have
25:16
we've just defined this binary operation
25:18
binary operation on
25:23
triangle triangle bar on our
25:27
are political subdivisions with
25:29
distinguished top edges
25:30
okay so let's say on
25:34
okay on
25:40
planar plane i mean there's
25:44
an orientation planar polygon
25:48
subdivisions
25:52
or if we prefer to work with trees
25:55
planar
25:58
rooted
26:01
i guess complete binary trees probably
26:03
the official
26:08
terminology
26:10
now whenever you have a binary operation
26:13
you know there's all kinds of properties
26:14
that binary operation can have it could
26:16
be commutative
26:17
it could be associative um
26:20
this particular operation is not going
26:22
to be commutative okay it's not
26:24
commutative
26:30
but it's sort of obvious and it's also
26:33
not
26:33
associative either so it certainly
26:36
doesn't
26:37
you know fit into our usual kind of uh
26:41
algebra that we like to do but
26:43
nevertheless commutative
26:45
it's not community if it's not
26:46
associative but nevertheless it's still
26:48
interesting
26:49
so one way of
26:53
extending this is to
26:56
extend it to multisets in fact whenever
26:58
you have a binary operation on some
27:00
kind of object you can extend that
27:03
binary operation to multisets
27:06
okay so
27:09
it extends however
27:18
automatically
27:20
to multisets and i like to just say m
27:23
sets okay m says same as multisets
27:25
and i want to just explain why that is
27:27
so let me uh
27:28
do this in the context of
27:32
some some some sets so let's um
27:36
let me choose okay there is uh
27:42
there's a uh
27:46
a binary tree okay and it's
27:50
it's planar it's got a distinguished top
27:52
and it's a complete binary tree
27:56
and um okay and i'm going to make a
27:58
multi-set
27:59
in fact i'm going to make two i'm going
28:00
to put two of these guys in here
28:02
okay because with multisets we're
28:06
allowed to
28:07
duplicate things and maybe over here
28:11
i'll put in the third one maybe
28:17
going down this way okay
28:21
because there's a multiset with three
28:23
elements two of them happen to be
28:24
the same the other one's different now
28:26
suppose i want to multiply so here's the
28:28
operation
28:29
okay all right let's multiply that by
28:32
the multiset consisting of
28:39
how about
28:43
how about this one
28:45
[Music]
28:53
okay so
28:56
if we have a product of these tree-like
28:59
objects then we automatically have a
29:01
product of multisets of them and what we
29:03
do is we just form all possible products
29:07
so just form all possible products
29:16
all possible products between the uh
29:19
elements of the first multi-set and the
29:20
elements in the second multi-set
29:22
all right so let's see what we would
29:23
actually get here let's practice our
29:26
multiplication of trees so we have to
29:28
take the first
29:29
one here on the left and multiply it by
29:32
this this one on the right over here
29:33
so how do we do that well we create a
29:40
beginning of a triangle and then coming
29:42
over here we're going to have to stick a
29:44
copy of
29:45
of this this one on the left so this is
29:48
a we're not going to stick a here
29:50
so there's a node here and then we're
29:52
going down here and then there's another
29:53
node here and then there's
29:55
two of those
29:59
okay and on the on the right over here
30:01
we're going to stick a copy of this
30:02
thing so what's that look like
30:04
well we're going down to this node over
30:07
here and then there's
30:09
these possibilities and then there's
30:13
something like that okay
30:16
so that's the result of this combination
30:20
now when i do this one here okay well
30:23
because
30:24
these two are the same i'm going to get
30:26
exactly the same thing but i'll
30:28
just put another copy of it there open
30:32
closed open open um
30:36
closed closed open open open open
30:40
okay now there's still the third one
30:42
where we multiply these two
30:44
okay so we start with our triangle and
30:47
then on the left we have to put
30:48
this one here which gives us going over
30:52
there
30:52
going over here going over here and then
30:56
down down and then putting this one on
31:00
this side
31:03
you have these two going down to the
31:05
solid nodes and then these
31:06
ones here
31:13
okay so what what we have
31:16
here then is is a it's a kind of an
31:19
algebra that automatically has
31:22
addition and natural number scale or
31:26
multiplication
31:28
inherent in it because we're dealing
31:29
with m sets so addition
31:33
is just
31:37
addition of m sets
31:41
m sets have the beautiful property that
31:43
you know if you have two m sets you add
31:45
them just by combining
31:46
the the two collections
31:49
and scalar multiplication or
31:52
multiplication by
31:53
uh by a natural number
31:56
multiplication by a scalar
32:00
is just duplication
32:06
of the m set
32:12
so um to illustrate that let me let me
32:15
write this thing sort of in in
32:17
in vector form language okay so what we
32:20
really have here is we have two times
32:22
the three which looks like this
32:28
i'm just copying that thing out
32:34
okay two times that one that covers this
32:37
one and this one
32:37
and then this third one then i can i can
32:39
just express that with an addition
32:42
um like this
32:49
those are solid and then we go solid
32:52
solid solid opens
32:57
okay so you see the m set framework
33:01
allows us to talk about additions and
33:03
and scalar multiples as long as the
33:05
scalars are just natural numbers
33:07
and so we get a kind of an algebra okay
33:09
we get an algebra
33:12
an algebra
33:15
of m sets
33:19
of well if we like uh these
33:22
these uh planar
33:26
rotate complete
33:31
binary
33:37
of trees there's still some question
33:39
about how to deal with the uh the
33:41
unbounded aspect okay
33:42
um but that's something that we can we
33:46
can think about there's a kind of a nice
33:49
grading of this thing so we will be able
33:52
to
33:52
to uh eventually you know understand
33:54
that that's not going to be at such a uh
33:57
an overwhelming point um the point is
34:00
that we now have an algebraic
34:02
kind of um thing that that gives us some
34:06
kind of meaning to
34:07
this equation found in concrete
34:09
mathematics
34:10
and um and now we can you know connect
34:14
with the catalan story quite directly
34:16
and we can
34:17
crucially we can extend this to the by
34:20
try situation
34:21
especially i think it's especially
34:23
convenient to adopt this tree-like point
34:25
of view that's why i've moved in this
34:26
direction because
34:27
as we want to generalize i think this
34:29
this tree-like aspect
34:31
has some advantages over making bigger
34:33
and bigger polygons
34:35
with finer and finer subdivisions that
34:36
are hard to keep track of