Well if you.

Actually think thanks for the invitations
It seems like a really great meeting.

OK so.

So this is going to be
the summary of three fake papers.

But that doesn't mean it's
going to take a long time.

So the first paper was with.

That and
my commit some else it is now in Iceland.

Second paper was with Tony Hanson and
Tony is.

Booked solid in Sweden and Samantha Petit
is here and she's been advised by.

Santa.

I'm Phyllis is fill in because I
gave the last time I gave this talk

it only took twenty minutes.

OK let's go OK So
what's the problem so you.

There's a underline the whole thing
there's a bar to a bipartite graph G..

With two sides Al and left and right L.

and R..

And R.

is bigger than L..

And you don't see you don't see the you
don't see the graph it's given to you

online OK so
what in fact what happens is you

are presented with the vertices of
the left hand side L M When you.

Actually look for it and when you're given
a vertex on the left you're given all

its neighbors so you just see vertex
comes along given all its neighbors.

And.

We assume that everybody has at least
two well actually receive everybody

on the left has L.

neighbors or
where we're at as at least the L.

is always the number of vertices.

Is the number of neighbors
of a sex on the left of L.

for left.

OK And so what's the aim for the i me
as to maintain a some sort of matching.

And the it's a matching on the left
hand side in the sense that each

that he first takes on the left
hand side sees one edge by.

The verges on the right hand
side can see up to our edges.

But our far right hand side.

OK So remember this Allen left and
right OK.

OK so his.

Is a picture of the.

So here's the current state
where we have a matching.

Let's say I think here L.

is three and along comes.

The fourth vertex and unfortunately

it chooses three neighbors and
they're all matched So

Woody So what the vertex does is
it picks one of these at random.

And so you pick this one and

this guy is for now forced out and.

So this one was force that chooses from
its set it choose another one at random.

OK So basically what it's doing is

finding an alternating path on
the actually find in a mentoring path and

is doing it by a random walk OK so

it's just looking for a moment in path and
it's just in a random walk to find it.

So wisely what is going to do it could
use on and of cookies do exactly this.

So this.

So this guy here is supposed
to be a cook OK and.

Picks a random nest this one and kicks out

know which one ethic that this guy
around but this somehow survives and

goes on and pick something else so
that's the one cooking passion.

Story.

Yet all the ages around so
these guys here when it comes along and

picks three random edges
uniformly at random.

Yes I was the what is the problem
in a moment is no problem so

the problem is to sort of control

arm the lengths of the omentum path
What's the expected length of.

Or of the omentum parties died.

And we will consider two cases.

So in the first case we have a large L.

is large.

And but you're only allowed one in a nest.

OK.

And the other case is
Ellie's true exactly to.

But you are allowed day in the nest and
they are specially very different

algorithms by the same out rhythm
the analysis is different and here.

So if you would use if you wanted to
do this in practice you would have D.

small you really would like say D.

not much more than you'd like D.

to be three.

OK.

If there's any sort of interest in
the talk it will be that there's a nice

question is to do some sort of
analysis for the case the three.

Hour.

Day has two articles articles two articles
one is sort of fairly easy it's just.

You just wait for so
we'll talk about that later.

OK so they will be enlarged so

there's some criticism about deviating
you know me better than nothing right OK.

So natural questions are how large should.

So fix D.

How big should M.B.A. or if I fix them
how large should be with this stuff is

completely solved it's not this messy but
you just basically

applying holes there to see when
there's holes theorem kick in.

And.

And how long is it take
to construct this thing.

Is want talking about.

OK so this was it so they know the name
cook who has Xing came from power and

earth and they consider this case and
are two seconds about it later.

Not this yeah.

There are lots of papers on the subject
which are always I haven't read

any of them but this one.

This one the main contribution here is
I think that they show that a sort of

breadth first search breath first search
can be done in sub linear time and

sub linear spies OK so
you don't need to need to do you know

they do the full breadth
first you sort of stop short.

And they also mentioned that the random
walk thing would be of some sort

of interest to figure
out what happens OK So

this is the I've already said this so
I will skip that round already said yes.

So in cookoo hashing
the L A G.'s are the results of.

Pash functions.

And we're going to assume that
they're completely uniform this is

a close you know and should be called
a hash a big Because then you're saying

we assume we've got completely uniform
random hash functions which are not so

easy to produce so another open question
would be how do you do this with.

Hash functions which are.

Not completely random may say
How would you do this with

if you using Linux a linear
congruence would generate.

Something like that OK So

the first attempt I made
with this was when Mark.

Was power males than a Michael MIT's
America and we didn't we aren't we show

that the expected time to insert
was probably logged so before this.

I don't so I suppose only sort of entity
epsilon was none so we surface what we

showed prolly long so how was the argument
he the Army's quotes him Paul.

So M.S.
will be the matching after you inserted S.

elements.

And so now we're going to M.K.
minus one so

the basic idea was the following You
could use you insert the new V..

And if you saw if you
did a bread first raw

what you would find Is there be
a sort of a relatively large set S.

one.

Of an over poly log.

Which can be reached in poly log steps so
this large set of side N.

of poly log.

Such that are that.

So that the probability
you can reach it from V.

in poly log steps is say into a log
steps A is one of a poly log So

this set of size N.

of a poly log and

the expected time to reach it is
poly log that's what I'm saying.

Yes relate to maintain power yeah so

here's this says you can argue by
just say no matter what the moment or

what the perfect matching
is there's a set of size N.

over poly log such it expected time
to reach it is probably logged.

I waited.

For M.

and N.
are Amy's we were seen in the M.

is one perception on times an OK And
we're assuming that M.

is sufficiently large so that with high
probability there is a perfect match or

there's a matching from the left or
the right.

And then you have to argue that.

There not that many as you sort of there's
this say what you're trying to do is

you're trying to reach a set which
has neighbors which are unoccupied

OK And then you argue that So
there's always a so because M.

is one percent silent times an.

There's always a linear sized
set which is unoccupied.

And if you back off from that there's a
sort of a bigger set which is the neighbor

of somebody as I occupied and
you go back further and

further eventually you only have
to go back log log in steps

until you get to a set of
SA is such that all but

something over poly log is within log log.

So So to summarize what it says in
expected time poly log you really

assess a vertex which is
if you only log log or Y.

and if you only log log away and the.

Degree is constant.

OK then you expected time to reach
the end results are only poly log.

And then that's it.

So you this there's this set.

Of a large set which is bigger then

such that everything in this set
is within log log of the end.

When you reach than
expected time Polly log and

you've got a one over poly long
shot now reaching the end and so

the expected time is probably log and
it sort of probably logs high probability

around OK silence means it's Oran OK.

OK So is that OK So
then I was a paper after so

the problem here was I think we had
five day at least eight in that to be

very large they had to be at least
eight and I think these guys.

Show that you could do roughly
the same amount of time.

But you could do for
savings three as well so

that's OK So that was the state
of the art a few years ago and

spent a lot of time trying to get
from Poly log down to constant and

lots of tried lots of ideas
talked to lots of people.

Who turned out to be simpler than
I thought than I thought OK so.

So this is work I did with my grad
student turn a year Hanson which

basically says which made a large enough
you can make the expected time as close

as you like to one so this is.

So how do we do that so we tried using
sort of Markov chains mixing time canals

work but nothing seemed to
work until we had this idea.

To get it where we are yes so

we said this is this is really what
happens for the last for the last one.

So here is the Yeah so some not
only these have been changed to L.

So here L.S.D. and
there's only room for one cook.

Up yet only room for
one bird in the nest inside.

How big is what.

L.S.D. and day is a large constant
Yeah large consonants the point so

it is always a large constant question is
how long does it take to insert a credit.

Or.

Where you go where this analysis doesn't
matter just do it again just keep going

I mean has a side called when
the bird then then what happens is

they will start again.

Don't go back I just assume you just keep.

I mean you can still choose you know I
would have not got kicked out on the R.

though I'm dizzy and I was just added
the beginning I just got kicked out

well I just choose another
random place to go

eventually our farm I work eventually
will find a way to insert run.

One now and then this is Ellie cause D.

left on the left hand
side the articles one.

Redoing the whole thing where.

You're.

Yeah if I was if I was as if it was
a sock or if I came back and I kicked

out the first one I put in just keep going
I just do differ in our Choose another

random number where that remember that so
I might do exactly the same thing for

a long time but
very lightly OK I can skip this so long.

B.K. are those vertices
it on the left right

which have a neighbor that's
not that hasn't been filled.

OK.

R.K. is the rest.

To R.K. month one year or so because
of those can if I gets a big a far if I

if more random walk reaches a vertex
him be undone Rocker's I can see

the outside and I can just put the I
can put the item into that location.

On so that.

Well no actually the algorithm
says R R look to see where the I

mean I could just say will pick something
at random and keep going but it's more

natural to say well I can see this I can
see a place asked the kid in there and.

Only randomize of it if
everything is blocked OK.

Yeah OK so
now our list are come back to this.

Earth and I can live to do this OK so
I'm not an aging Mensing path.

Will look like this something on the left
followed by something that's not in there

because if Y.

one was in B.K.
there and I would be finished.

OK so this one is or I keep going so X.

two is so why one was matched to X.

two and now X.

two chose Y.

those X.

two must not be in here otherwise
you'd be done so no alternates and

no mention part goes left something
not in here than the matching edge and

if it's if if if I'm not in the match
if I don't see if I'm not in here or

keep going so I keep going until
I reach something not in here.

So the sort of powers that
are I'm seeing are of the from

left not in B.K. left not in B.

when actually the other
way round is not in B.K.

up to the top not in B.K. not in B.

can't tell I find something in B.K.
then I'm out.

OK so let's call yes or

there's not a very good choice of words
but let's cause such a path interesting

another set in the sense that any
other part is an interesting OK So

these are the interesting parts the parts
of interest now not all interesting parts

are augmenting that's the point there
are many more interesting parts and

augmenting OK So that's the basic
I was the kid sort of done

thing it's not hard to estimate the number
of interesting powers of a given length

but it is hard to estimate the number
of interesting alternating powers

because somehow this
depends on the matching.

OK And that was a problem which is always
trying to figure out what was the how what

was the distribution of the matching live
you know which is is is constructed by

some process and it's not an adversarial
process but it's quite difficult to

actually understand but if you sort
of forget about the matching and

just think about interesting part somehow
the matchings of this doesn't really

matter what matching you've got so
interesting men.

So it be B.K. with the bad guys right B.K.

can see the outside or
is it the other way around with a look.

Because a because the bad
guys you can't see the side.

So interesting bad guy bad guy bad guy.

Good Guy OK.

And the point is that B.K. is small.

And naive calculation which he would
say that big way is OK says Caleb dom.

And what's the probability
don't see the outside

when it's about one
minus Epsilon to the D..

I mean if it was everything was on
condition that we were minus Epsilon to

the D.

and that's any of the minus EPS less
than any to my stepson on day and

if he is big enough current fastener want
to reps along this is pretty small and

it's not correct this obviously but
you can make something look like that.

The case so B.K. is small so
right means you should get B.K.

bar quite quickly OK so

then you have to argue then you want a
number of interesting parts or with two S.

minus one vertices and then you have
to argue that the expected number are.

Or how do you choose them where they
go the choices are you could achieve

those big hey of the side it's big case
of the S.S. then you go to choose the.

So this is on the left
these are on the right and

then you have to do something about what's
the probability the ages are there.

OK And again you sort of can argue that
this is sort of it's sort of close to been

random and
this is a good approximation OK So

this is the number of paths now
there are possible powers and

this is the roughly the probability
that all the ages of the path exist

OK so a little bit of conditioning is not
easy when you think about it isn't that

there isn't too much conditioning.

And then you can use some
concentration inequalities

three simple applications of Azuma.

And you can show this is highly
concentrated and there and then S.

where you just are the only other
trick mallees So you're right so

you've got the probability that
this is large is pretty small and

then you use the expectation is the sum
of the probabilities have been large and

then you certainly have to
do this up to log log again.

So that we're only looking at really sure
interesting part log log in and then we

can kill it with the previous result
plus the basics so the key there is.

Replacing old Mensing powers by
interesting problems that's the key.

Where we could It is easier with
Well one thing is you know you and

I took out a log log in.

Yeah there is a certain you do you
do need to think about sure parts.

But not log log and
log will again is but you know but

Logan is like a constant so you say.

OK so that's that page OK so
now the one with Samantha Sam Petit.

Are OK So here now we've got Alice two.

And the right hand side are is big.

So it's a completely different
completely different ballgame.

So let's.

Are it's OK So the point is so you see.

Well along comes a vertex the case vertex
and it has two choices P.K. and Q K.

And so that's like an age joining
P Cain Q.K. and one of them has to be

one of them has to be chosen so
that's like orienting it from P.K. to Q K.

or Q kind of Pekin.

So or so let's let's top pick.

OK so so at the beginning when you
have an eight you add it in a sort of

a random order without thinking you
just choose randomly that is D.

and.

There's another thing which we get
by after the ins and what's the list

look at the Insert the insertion algorithm
is quite simple suppose the arm so

what are you going to do if you have to
add the ages then orient them to alter

the you have to order in the edge and
it points to the location

points to the location where
the item is to be placed so

you have to are in the age and
we have to orient the edges so

that no vertex has in
the green more than our

own so here's a case everybody here is
going degree at most or are is two.

So we want to add this guy but
this makes this in degree three so

what happens is we have to reverse one of
these ages so then we reverse this age and

this ages go in degree three so
we reverse that and we've done.

To the algorithm is basically you
have this random die a graph.

OK initially some of the edges some of the
edges which is completely random some of

the edges of been altered and
you add an age you put a give you an R.

and then you what you do is OK if
there's an easy orientation you

do it was an easier in Taishan Suppose R.

is ten and the A's joins two things in one
lead in degree five none other in degree

seven so doesn't really matter which one
you do you just orange it randomly but

if both of if both are in degree
ten then you're in trouble so

what you have to do is you have to sort of
find a path reversing orientations until

you find a vertex which has indeed agree
less than ten and then you can do it.

OK.

OK so.

OK so his in our study so

I is the set of vertices within
degree at least are minus one in D..

Today's other ones are likely
to get saturated very quickly

OK so we're trying lots of complicated
arguments trying to figure this out and

then somewhere Samantha
noticed the following

orders happening again the point I first
of all is these going to be very small.

It's going to be exponentially small.

In our run

OK so this.

Samantha notices that this set contains
all the saturated vertices there so

this is the segment you get as
follows You start with the guys as

the ones with very large in degree.

And then you add you start to follow it

you take a vertex if you've got
two neighbors in a as you add it.

And then you keep doing that and you keep
did it until every vertex is going to

most one neighbor in this and
then the point is that this set S.

contains all the saturated vertices and

that's pretty simple because you have
to have you need to flip the gate

because you weren't in an eye you have
to have two neighbors to get in there.

And the point is that because A is small.

S.
is small.

There's some sort of standard random
graphs way of doing things and

it turns out that if S.

is small a is small basically small sets.

Have a very low density.

And if you keep adding the disease
a degree to and get big

that means you have some or you you've got
to dense at some stage and very nice trip.

And basically that's the end of
the game once you notice this

once you notice that this set S.

is small contains all the saturated
vertices What does that mean

what it means that all that the graph
induced by the small vertices

is basically little trees and
little uni cyclic compounds.

It's just subcritical And that's just
a sort of first moment calculation

and then you can basically it's done

right because you only got to do is
do random walk for little trees.

And that doesn't take very long and
these are most of the trees are constant

size arrives.

Yeah definitely there knowing this and
calculation.

OK So that said What's the weather I think
I've been through the question is that

the real questions are.

Reduced the independents from large
day to three only say some so

it's obviously for small obscene lawn
you're not going to get away with three.

So one thing you might want to do is
say replace the grows one over epsilon

two The grows like log one of reps
alone or give me something for three.

Not when are when thirty thing were.

So when Ellie's to R.

and R.
is one you need.

To N.

You need to air the M.

has to be two in.

Mind because this is just
sort of subcritical A-T.

in the random graph process.

And the other thing is if you could
make it in you could do deletions So

it's toward tempting to think that if
you do random deletions It's like they

never were there.

And so sometimes that mark that
might be so are fairly easy and then

handle the case where the hash functions
are only or not already K Y Z depended on.

What might be more interesting and

never saw the other the one of them and
OK thank you very much.

Thank you.

I know if you do prefer search
it takes up too much space.

Than That was the original paper.