I did my work on talking functions and

basically before I can begin speaking
about any of this I was hoping to perhaps

go over graphs graphs just so
everybody sort of has some prerequisite

information about what I'm talking about
because otherwise nobody really cares.

So we can think of it
as just a collection.

And a collection of edges and

these edges little dots on the edges or
lines of between these dots.

We can also think of it like a computer
network so perhaps some of these.

Some edges connect
between some vertices and

vice versa we can just draw something
like this and maybe we don't

necessarily have to have that all these
voters are connected by edges and things

like that we can think of it also like a
computer network where sees our computers.

And the edges are cables between them and

so we could actually have
a computer out here by itself.

And that's OK.

Exactly.
But really doesn't make any sense to just

throw a cable here and
say that's part of the network.

So in essence when you have these with
these vertices can stand alone but

we don't really have edges standalone So

it has to be between two very serious
in the sense two computers so

that these things probably productions
are actually with respect to a graph.

So you take some general graph whatever
and you have a problem function on those

so I like to motivate what I
want to be speaking about.

So first things first.

We have an open problem in comment or
X. and now

the end cube is a graph it starts
out actually we can actually

start out with the actually
the cube is actually just a line.

So it's.

He would be in one dimension and
then we can actually extend that by by

making another copy of this and

actually connecting it by
adjacent edges like this and

this actually so this is a square and
this is like the two dimensional cube and

we can go up to three and four dimensions
and everything like that from here.

So we just do the same thing we just did.

And that ends up being and

we can continue that until we
have an arbitrary N.Q. so.

So that's just what that means.

Now one question always is because
a lot of times these problems do apply

to computer networks spanning
tree spanning the network so

in essence we want to connect all
the computers to one another but

we don't want to use up any excess cables.

So for instance if this was our
graph here let me get rid of this we

have to have a connected graph which means
that from any vertex we can run around and

get to another this actually this is
a bad example because this already.

Is a spanning tree so.

Actually to have it not be a Spain tree
let's say we can connect this here.

So then we have an extra edge somewhere
in here because we can cycle all the way

around and actually we could have
deleted we could delete this.

Or something like that.

And so
really the question often out comes up.

How many spending terms of graph are there
certainly depends on the graph and

there's actually a nice
result that tells us this.

It's called The Matrix tree theorem.

And so but this sort of gives you sort
of a haphazard way of completely.

It's a very good way of
computing these things.

However you don't really get any
information about what are these

spanning three so you just get a number
basically a second in direct way of

getting the number of
Frenchies of the graph and so.

Anyway we have in terms of the end
we have this number actually for

arbitrary ends and just being

essentially the number of times that we do
this process that I showed you guys there.

And as you can probably imagine
it gets arbitrarily ugly

in terms of the structure of the graph.

This is number Spain trees of N. and
obviously it's and it's arbitrarily large

family will become very very
not very nice to work with and

so the first problem being is we want to
find a common torpor for this family.

We just work on a total proof means is
that we can actually find encounter each

one of these as opposed to just plugging
something into this forum and getting how

many Spanish trees now we actually know
the Spain trees are another problem

really where my work comes in is actually
connecting these two problems together.

So this is actually not.

This is what you usually see in the
literature but really what this just means

is that when you have your graph or
something like this a directed graph

let's say this we can put a direction
on each one of these edges.

So it's one of two ways it's
toward one vertex or the other and

so just an Taishan just means that
there is a cycle in the graph and

in just a cycle the graph means that
you have something like this so.

You have a subset of the graph to
look something like this it's a bunch

of bunch of.

This way connected with edges and you can
just cycle around it just like a clock or

something like this.

So you always go around in one way and
this in general is a very difficult

problem and it ends up pops
up a lot in computer science.

So what do we do saying we.

We sort of found a link
between these two problems and

we were able to do a lot between
the literature with these things and

we did this actually using
the parking functions so the move on.

These were actually defined
by Alex person got them for

Shapiro relatively new back
in two thousand and three but

they were motivated by things like and
what about advantage by the billion

sandpile model which is what Doris paper
was on and a few others sense that.

And this is actually the definition of
a parking function parking function.

Now I want to say that this definition
doesn't really it's not really very

straightforward to be honest and a lot of
times we didn't really use this there's

actually and I still equivalent condition
that you can check which is much easier

to see which I actually
I just want to go ahead.

Since I'm limited on time and
speak on that for

a second in a moment.

Actually one of the reasons why we
actually do this is because there's

an explicit by Jackson between
the parking functions and

the spending freeze that
I was just talking about.

And so perhaps if we can understand
Spanish These are actually

kind of hard to understand in
terms of their structure but

probably a function actually has
a little bit more structure to it.

So really what a party function is going
back we can think of it very simple sense.

We take each one of these vertices and
we actually put a Johnny Grant here.

Just something very random.

And now we can actually give
each one of these viruses and

integer value number where all but

one of them is non-native and
they have to actually fit.

This is actually the main criterion here.

For each vertex The Q.
is a vertex and for each vertex it isn't.

Q It actually has a criterion.

And then we just cannot actually
arbitrarily pick you to be any one of

these vertices and the number of functions
is actually the same no matter what.

So now the idea Why are we looking it.

So actually a valid function is actually
something like this and in the second.

I'll show you why.

So the idea basically
the overall idea is that we want

to be able to we can't count
necessarily the spankers physically.

So perhaps we can count the parking
functions which we have by Jackson I

want to own correspondence between these
two objects that we automatically get that

we're counting the spanning trees so
that actually helps us out.

It just gives us a simple
algorithm that we can follow and

makes it like a game in order
to check if this is a first

thing that we do is we just start
marking off and if we can mark all

over the series then it's a function if we
can't then it's not applying a function.

So the first thing we can do is we
look at the number of Mark neighbors.

So initially negative one doesn't
have any more neighbors but negative

one is strictly last and zero which is
number of my neighbor so we can mark.

That's what we're looking for
and we're looking to have its

integer value to be strictly last than the
number of my neighbors it has two guys or

both zero and Mark neighbors neighbors as
being they have an edge between them and

this and this vertex here so

each one of these has one mark neighbor so
we can mark each one of these and

then one neighbors we could have
made this too but it's wrong.

It doesn't really matter.

So we can.

So this is in fact you probably infection
now if we were to go through this process

and we were able to mark everything off
there would be a G. problem for you.

So this is actually in
a critical condition.

And the proof was originally
it is actually pretty

well explosive from DARS work and then how
the coffin Shapira to find these things.

So anyway this is this is
another on one Q three.

This is an example of a parking function
for the graph Q three we can first Mark.

Q Which is negative one.

Then we can mark this zero then that
zero because each of those has one.

My neighbor has one.

My neighbor.

Now this one has this this guy's marked
this guy's mark so we can mark this.

Now this one also has the mark
neighbor here and my neighbor here so

he can be marked.

OK so now we have everything but
these two marked this.

So this one and this one is Mark so
then we can mark him and

then this guy has this neighbor's
mark here here and here.

And so we can actually mark him.

So that's just so
that as a being she probably function so

moving from what time did I start for
a moment.

OK I just to try to give you guys an
overview of where we really did in terms

of a maximal there's

actually a natural the natural
ordering in a parking function is that

we can compare we have to probably
function on the same graph.

And we think compare the vertices sort
of like term by term so if I have

let's say we had this tracking function
and if I draw the equivalent graph.

Then.

This is probably a function.

Actually greater than this one because
if we look these are equal here but

then this is larger than this.

So we just look for Rotex if if the one

parking function it isn't the first my
impression is the same parting function.

We look at it.

Vertex by vertex and if this condition
is then then we can say that

this ordering is true and for
a particular party a function

where there are no parting functions where
this suffices we say that's maximal.

And then we can actually say that the dome

of it is all the part in functions that
that satisfy there are less than that.

So actually if I have a maximum function
which actually the sense of being I can

actually arbitrarily scale down all the
values that aren't zero so in this case

it's just a two I can scale down to one or
zero and there is still a parking function

because if we follow the dark right here
throughout it then we can still this guy.

So one of the things we didn't work.

I guess I could show you guys is that
we were able to explain this factor.

We actually were able to come up with
a parking function a very very sort of

canonical parking function
that was Max there was

a maximum maximal actually we show in
our paper that maximum and maximal

are actually equivalent which maximum just
means that if you sum up all the values.

It is a taking the highest
value in Richard Stanley's

problem the original problem
that I first showed you find

a very canonical maximum value
function where we take the dome and

they don't actually end up being this
explains one a part of the factor now if

you go back to I think I have
another slide of that that's this.

So if you go and you go back and
you try to understand the original.

Factor that actually becomes quite hard
again even if you know what this is this

is still so still a problem we
were actually able to solve it but

nothing we were able to do to show
that every again being the same thing.

Partly function is actually and being

unique source at Q So

a search just being that this is actually
a source of this graph because all of

the edges are oriented away
from that particular vertex so

we were actually able to show that there
is a wonder one correspondence between

the unique source and

maximum parking functions but
again this problem is very very hard.

So now we can't even count the number
of maximum parking functions because

this because we actually get no

it's almost easier to look at everything
from the standpoint of the deploying

function spanning trees and trying to
figure stuff out about the patient.

So that was another
corollary of we did but

again we weren't necessarily able to take
out either one of these problems but

more or less link the two and that ends up
being anyway about as much as I can talk

to you guys about right now fast but

hopefully anyone has any
insightful questions.

There

is there is that there are a couple
students that will graduate

students at MIT that gave the sensors then
I think there's two explicit by Jackson's

exactly they're not particularly
pleasant and what happened with.

Stanley actually actually Shapiro actually
when they originally family these things

but they felt like it be easier
to do objection between this and

between these and the spending trees
as opposed to doing when you look at

them I mean

yeah that's a big

very well in paper but there are still
some there's a couple guys from Minnesota.

Tried to tried to do this
they try to extract exactly

figure out what they mean
in terms of the comet and.

They get stuck on a very like they're not
able to factor the way they were doing

a very algebraically and at some point
there's quite a few difficulties

to the paper but and actually a comment

actually when I was doing
this Cambridge University.

Maybe I think it was a small
Smith of The Matrix and

see if perhaps if we could get some
sort of because perhaps it doesn't

mean much for the Spain or perhaps I mean
something for the parking functions and so

we haven't got a chance to really
see if there is anything but

again it's you know it's sort of
mysterious the way that Laplacian works

because you sort of kill one of
the rows and the columns and the.

And then you just take the determinant and
all of a sudden it's you know so it's very

interact in terms of how you're getting
you know exactly the Spanish phrase from

this but yeah we were very surprised to
find out the not much is really known

between that but then when you
try it you know you can't do it.