For the last talk today.

David good morning.

You tell us about maximum current
problems on sports thank you and

thank you the organizers for
the invitation and

the opportunity here just for
a long time talk.

Talk title towards the end of the day
maybe it's not a good idea so

I think I actually like sound of his
version better than my own version.

And I also I guess I want to apologize
to some of you who have heard

this story before the other hand
looking at the program I realize that

some of you who've heard to stop before
approaching your talk as well so maybe.

Maybe maybe not I don't
feel that there's enough.

Anyways.

So before I get to the I before I
get to the one concrete model and

concrete question that I want to explore.

Today I want to set it
in the broader context.

And the sort of big picture question
here is the this is infamous.

Gap that appears between.

Optimisation problem up to musician
value optimal values that we can often

compute by nonconstructive
means in the random

graphs in the random graph settings
versus what can be achievable by.

Putting normal time algorithms and

there's a growing list of such problems
including the largest clinker and the K.

said Mark Scott and so on it's a really
long least of problems not only in

the context of random graphs but
also in the context of machine learning.

Models and high dimensional statistics and

one perhaps the most infamous problem
in this setting is the problem of

finding the largest peak in
the density of the strain the graph.

No they lived in the dense of the Sunni
Graf the largest clique is of this size

twice the rate them of the number of notes
but the best algorithm that we know of

that can find a lot of this
clique that around simple enough

time is only half of that and since one
thousand nine hundred eighty six when

the carp falls this is an open challenge
since then this has not been improved.

So when you when you sort of experience
this repeated failure to to to

to bridge the bridge the algorithmic
gap the new search for

answers you want to understand
why why is it the case that

these problems are hard
apparently hard algorithmically.

And a lot of insight was.

Offered proposed by.

Statistical physicists and

people in this is working on spin glass
model that essentially for the first time

suggested that maybe we should be looking
at some kind of a face that incision.

That might explain it's not proof
that this problems are hard

in some sense so they suggested that

that may be the geometry of near optimal
solutions in this problem change is

scum placated enough to prohibit at
least certain classes of algorithms and

maybe even all pull in nominal time
algorithms were quite far from that but

but that's that's a potentially
useful insight but

I'm not going to discuss any
heavy duty statistical physics.

Methods here but one particular
insight that incredibly useful is this

certain overlap gap property of near
optimal solution that I will discuss

today in the context of one problem and
one problem only and

the problem is of finding him mocks him
caught in the random sparse hyper graph

that about introduced So here's the.

Here's the problem.

So we have a random hyper random
hyper graph on endnotes we

have D N hyper edges and they'll just
refer to them as edges from this point on

all uniform see each edge
is the collection of K.

on the order to.

Note So we have and this edges
are chosen uniformly at random and

we think of Dia's constants each edge
is unordered sequence of keynotes.

OK whether you generate
them with repetition or

not the repetition that's usually besides
the point it doesn't matter doesn't change

the big questions at the scale
that we were interested in.

So let me introduce
the maximum cut problem in.

It's defined as follows We want

to partition the set of notes into two
groups in coded by minus one and one.

And the value of the cut
associated with this partition.

Is written out in the way that you might
object to and I don't mind if you're just

to that I might I object to that as well
but it's convenient way to introduce marks

marks got problems so here's here's
the value of the cut with big negative.

Some over all edges and on each edge would
take the product of the of the values so

it's really will look at
the parity of this product.

So the good and

our goal is to our goal is to maximize
find a card that maximizes this value.

Maximizes this very soon other
words we we want to hope it was me.

In other words we want to we want to find
the petition that makes the number of

the number of edges where this is
product as negative as large as possible

in the number of edges when
this is supported piece

positive is little spots of all but
in fact.

OK So this is not the canonical Mark Scott
problem the you're familiar with it's in

a way it's a center of Mark Scott problem
and it turns out that the marks caught in

the case one k's equal to two you can read
off from this version by just adding and

subtracting been a number of edges
the inside two partitions so

what in fact what isn't being
measured here is not a cut for

the case k's equal to two
is not exactly the cot but

the cot minus the number of
internal edges in two parts.

So we want to maximize the use
in that essentially if we learn

how to weave know how to maximize this we
know how to maximize the original cut.

Point.

For Capel to.

Finding the marks so we have a good
approximation for the values in L.

also all talk about those
even in the case for K.

you for the hyper graphs but
as far as they all get it and

so are concerned we don't know
whether we can get as close to the.

Optimality as as we want for
graphs for hyper graphs We will

show some negative results that one of
the goals of this is the growth factor

No well the get factor of
between the optimality.

No it's it's.

It's it's not constant time.

For growing a constant degree but
growing you can make the gap smaller and

smaller I'll get to that's all I'll
give some in facts I think now

let me give some background maybe this
will answer some of the questions.

Sort of.

Yes.

This is in other.

OK.

OK good two to two and
one answer to two questions first.

So this is this is the goal and

we want to say something
about the objective function.

Sort of say something about how
does it scale with the number of

notes with the degree.

And also say something about very quickly
sort of can we get near optimality or

not just Sanders was asking for
the case of for the graphs.

Sort of jumping ahead the reason we're
looking at the hyper graphs is because

we can say something about hyper graphs
that we cannot say about graphs and

that's sort of the limitation
of the methods and

OK so when one key is in because it's
as you observed it's a because it's X.

or set when case are this problem is

trivial because making it minus
one everywhere is makes it.

In the car to every in on the.

Hyper if you have very negative once or
you quote unquote cut every action

objective function is the sense not
interesting still the smallest normal

graph setting would be one case
even that means at least four and

that's the setting we consider.

It's it's not hard to show.

That this scaling of the largest
got as N goes to infinity and as D.

goes to infinity but in a different or
the first you take end to infinity and

then make deal larger and larger is of the
order of magnitude it's linear in then and

a square root N N A degree OK and
that's it's not hard to do and.

OK then the question is What
can can can one say something

about the constant in front of this of
four now it's an existential question.

So.

We will look at this question in
context of graphs in two thousand and

four got some.

On.

That bound was that bound is.

Algorithmic and there is a fairly
straightforward algorithm greedy type I'll

get it and that will get you to the right
skinny I'm not sure it's exactly this one

but they will get you to the right scaling
or squirt of the this worth of the appears

because if you do a random partition then
the number of edges cutting across is

roughly Gaussian with parameter Gaussian.

With its poor song with mean
with mean value of these so

with playing by playing with fluctuations
with standard deviation squared of the you

can take advantage of that and create a
cut of the size roughly scored of the and

it turns out that just by first
moment method you cannot do much

better than that then so
that's the right skin.

And that scaling prevails even if
you have your in the hyper graphs

even when case known to but
what's to constant.

You can get some bond here and.

We've improved this bound
somewhat in the more recent paper

using existential mean and
explore as far as existential

result our concern and
case equal to two for the graph setting.

Pretty much a complete answer was given in
the paper by Demba mountain are incensed

who's who obtained the factor up in
the constant in front of square root

of the times then and that constant turns
out to be numerically about one point zero

seven and
the origin of this constant I'm going to.

Comment on.

Assume.

OK this is this is a existence or
algorithmically for

the graph so we can get.

Some numbers definitely
smaller smaller than this so

the thing is that going
back to the square.

If if these increases then the gap
between the algorithmic and

existential big gap drops at the rate
of one of a square root of D.

so it's not exactly.

In some type setting it's the game
is slightly different here.

OK so.

So.

First and since then I'm pretty sure that
the that is improved algorithmic result

although I was not able to quite the a
nail that in the literature they all get

at them is actually quite simple
that you go note by note and

it states you have some partition
that already is there you just

simply put a note that into group that
maximizes the implied caught by this

by this assignment for this for
a for the current node

at such a greedy type algorithm and
turns out that every time essentially what

you're doing you picking a maximum of
two Gaussians and maximum of two Gulshan

is roughly like standard deviation that's
a good deal and that's how you get.

You know any other questions.

OK so where is this constant coming
from it comes from this theory.

And since the work of them will mount an
audience send this was extended to quiet

several other papers
throughout the models.

In the end the idea is that
you relate the model and

sparse of the shiny graph to a model which
is called the Sherrington Kirkpatrick or

in general case been glossed more than
the more those various The analysis

of the models involved but the model
itself is very simple so let me let me.

Tell you what it is to remind you what it
is if you've seen it before so we are now

looking at sensually at the complete cave
uniform hyper graph what on every key.

On every edge which is.

A set of notes you have put
their weight which is just

a standard normal Gaussian So
now it's like a complete K.

uniform hyper graph and

every edge is associated with
a standard normal Gaussian.

And what you look at is that you
look essential at them largest.

Mark Scott in this model where
you partition the notes into two

into two groups.

And but for
every product of the once and minus one

you multiplied by the associated weight so
that it's even less of a familiar

kind of a Mark Scott problem because the
Gazans could be positive and negative and

natural This products are minus one and
one in any event you want to arrange plus

and minus once in such a way that this
total sum is as large as possible.

And that's a that's a canonical
object in this spring was theory.

That has been of interest.

And thanks to the long serious so

for their paper starting from
the really groundbreaking work of.

And then telegram and punching and
others they have managed to nail

the the value of this
optimization problem.

As well as the scaling to result in this
case you need to scale it by and by the.

By this amount that turns out
to be the right value for

the scaling and
the limiting constant exists and

can be in principle solved numerically and
that value.

That live it is precisely the limit
that the tears in the in the marks cut

in the key for the Capel to two and
as it turns out for the general case well.

So in summary what what what we know
is we more or less know the value of

the marks cut problem on hyper graphs for.

Degree selfish smarts could ask for
degrees sufficiently large.

But the focus is on the algorithms and
so on and no doctor.

Program.

Or how the does it grow with K..

I'm not I'm not sure they have
analyzed because even computing that

involves solving a certain partial
differential equation numerically and

that's a good question so.

But not not sure anybody looked at that.

I don't want to get into that turns out
that I'll turn to two way of doing that is

that it is a like a countable sequence
of measures over measures and

you need to do some more of its.

Not but we're stuck it's complex.

It and I frankly I don't have
a very good in intuition and

understanding of that I
had a few slides about.

About the method of connecting.

How do you actually get from

model this sharing can can Partick
speak with us model to the sparser and

then graphs this short one of
these the in the pollution method.

But I'm not going to spend a lot of
time on that the idea is to build a mix

model which on the same set
of notes has a partially

the original sparse random
graph more though and

at the same time populates is with this
mean field model with a galaxy in.

Hyper weight hyper edges and

introducing some intercalation parameter
that the that continuously sort of.

Interplay between the random graph
model and this being with us model and

then somehow controlling what's happening
with the ground state and so on and so

forth let me in the interest
of time not do that

offline like happy to talk about that yes.

Yet this.

Yes so so this is the Interpol lation this
is the sum of two her Milton Yes this

is simply Mark Scott on the graph this is.

This is the Hamiltonian associated
with this last model so it's.

This Yeah.

Yeah yeah I me other questions about
the slice that I was not planning to.

And that's part of the into pollution
now I want to I want to turn in to

stay focused on the algorithms and
the first idea that I.

Want to introduce is the idea of
a Kells that goes under the name

coast which is if quite a fascinating
phenomena again discovered and

started primarily in the context of
spin glass models but it's equally easy

to describe it in the context of
random graphs So what's the cause.

And how's the described in
context of random graphs consider

Cup two versions of our
of the shiny graph G.

one G two on the same set of nodes.

And and only with only.

A certain number of shared fractions.

In Specifically suppose this two random
graph share a one line is dealt a fraction

have one one is dealt a fraction of common
at it's its is it to generate this graph

but the but so that each of them
marginally just looks like a graph so

says it to generate the graph you start
with just one version of this can often do

you simply dissemble each edge with
probability Delta independently OK and

then some of the edges
are from your old graph and

some of the edges are brand new and
the one minute fraction of them is common.

So these are like two
graphs kind of alike.

Sisters when when one Delta
is equal to zero then

it's natural is just two core piece of an
identical graph and one Delta is equal to

one then these are just too
independent core piece of them graphs

both of the same collection of notes but
of course it's edges that make graphic.

Interesting no no no not notes So
here's the serum

that was in the came as again again
a culmination of serious the worst but

in the sort of last form was proven by
Chen and punch Enco in two thousand and

seventeen called The Kills this
is what it says now because

I it might be hard to parse what it is
let me just say informally what it is and

then connected to the formal statement

basically it says is full that this
is the following fix any delta.

Even Dell love Dell to be any small as

small as you want point zero
zero zero zero zero zero one.

And considering though this couple.

Versions So in other words the two graphs
share a ninety nine point nine nine

percent of the edges but

there is a little bit of idiosyncratic
edges that each of the graphs has

consider Mark Simone caught on the first
graph and consider the maximum

caught in the second graph each of them
is a partition of notes into two groups.

And look at their inner product
the inner product basically

how how much they overlap
how much the grid.

And naive really want would expect
that because you perturbed the graph

just a little bit only one zero point
one percent of just have been perturbed

you there should be a card
which is new almost optimal and

second graph which agrees almost
everywhere with the first go but

that turns out to be
extremely far from being true

it turns out that as soon as your birth or
fraction of edge is

the optimal every nearly optimal solution
in the second graph is nearly our fog and

all to the every nearly optimal
solution in the first graph.

Just.

Rut.

So for.

Every it for every Epsilon and
dealt a civil suit I mean I.

Know you can but
you know because you by making this.

May create a little
careless in my computer so

I don't I don't I don't believe so
what what what Depends or

both of the parameters is how close to
optimality you want there is another

there is another complication
it's a sort of come complexity in

the statement is that you need to make
the degree large enough so that exist for

every absolute Delta did
exist large enough but

I'm going to gamma gamma in cause
the proximity to optimality.

And lodging of these such that is as
long as you look at the graphs with

degree larger than the zero.

And two solutions in two
graphs one nearly optimal in

the first graph one nearly optimal in the
second graph there in a product will be

there will be epsilon orthogonal.

So.

OK So that's the.

Result.

And I already said it in sort of.

Human terms what it means any questions.

OK so.

Just a quick word about the proof of
this is the proof to show the something

like this happens in the spin class
model itself and then Interpol it from

us model two to the hyper
graphs a random type of graphs.

The second property that
I want to share with you.

Which is also geared towards
the stablish a negative

result is the overlap get property
that I have alluded to earlier.

So for now let's just fix
one copy of the graphs and

when I'm going to look at it this SR
is just one of the friendly graph.

And.

It turns out.

The following is the case and for

that we need the really the graph
to be hyper graphs and.

Therefore the number of notes
per should be at least four.

That exist to two numbers tower one and
tell two between zero and one and

a certain parameter which is typically
smaller than the up optimal parameter.

Such that the following holds for
all sufficiently large degree D..

So what holds is that if you take.

Every two nearly optimal.

Cut where loose and nearly optimal
means that sigma on the supported that

normalized appropriately gives you
at least a value that much which is

still below the optimality but
may be close enough and

the any other solution and if you look
at the their overlap in a product.

Then there overlap is either
at most one or at least two.

In human words what it means
is is the following so

take two nearly optimal cuts in this.

Graph and see what percentage
of notes the egregious.

And what this says is that the percentage
of knows their greed is either

extremely large is large or small.

So if let's say toe to is like point eight
or Tower One is point one It means that

either the agree on the least eighty
percent of notes or they disagree.

On or they agree on at most ten percent
of notes but nothing in between this in.

There's nothing in between you
cannot find two nearly optimal cuts

that agree on thirty percent of the notes.

This value style one and
toe to depend on how close you

get to the up the military but
importantly for what I asked for

the algorithmic implications they
do not depend on the on this.

Degree so they.

Overlap the overlap gap that is
created by dis to barely one told to

survive somehow no matter how large
large and large a degree you take.

Yes.

Me.

A Yes By making dumb a closer and
closer to gum a star.

Then.

It's not.

A Yes A Yeah it otherwise none of that
none of this is that curious right.

Right none of this is vacuous first
always I can start with the cart and

stupidity of just a small
percentage of notes so

that I still stay nearly optimum So
that gives me this part and that

is sufficiently many of them which are
nearly orthogonal to give me this even.

In non Backus.

The high probability is for that with

respect to the randomness of the graph
there is no other the randomness here.

Yes with high probability with respect to
just the randomness of the graph every

Gama optimal solution and
every other gum up the most solution.

Satisfy the property that there overlap
is has this discontinue support.

Why.

In the various I try I don't have
an intuitive Explain of that but but

I have some references to the literature
and that that phenomenon is not unique to

Mark Scott It has been showed
up in some other models as well.

But in other what models
is more straightforward.

To show that as far as why is that the
happens I don't have an explanation maybe

others do but big but
this phenomena helps.

Establish the negative results for
algorithms and

that what we want to that's
what I want to discuss next OK.

It turns out that you can put results
together so that we have a sort of

combined version basically it says
that even if you take now to couple.

To random graphs models which
are coupled any fraction of edges and

that could be zero or zero percent or
one percent that could be completely

independent of completely identical
This overlap get properties survive so

even the couple of versions take nearly
optimal solution in the first graph and

in nearly optimal solution
in the second graph

their overlap is either more star one or
at least out to.

It and that turns out to be the case
the the proof is by combining

the overlap get property in one copy of
the graph with a careless property so

the care combined with this gives us
this and that turns out maybe for

the reasons that are not there not
clear now that turns out to be

what we needed for
the algorithmic implications yes.

So the ODP theorem here is not chaos chaos

usually refers to perturbing the graph
itself this is about one graph but

it turns out by combining with chaos and
this essentially get their G.P.

in when the coupled model and
the end it's I highlighted the word

all here and for us it was important that
this is this literally means all for all

Delta starting from zero all the way to
including zero and including one could be

identical to graphs or could be completely
independent two graphs that property holds

any other questions getting later
in the days of you you got that.

Ninety percent happy but
my the stuff OK So so let me continue.

So while I don't have a good
explanation proof technique.

Let let me let me skip what I wanted to
say is that while I don't have a good

explanation why something like this should
happen this phenomena was suggested and

then proven by a statistical
physicist earlier.

In the different context and different
models in the different versions of

the statement usually it's was
described in the form of clustering or

shattering for the so
called random constraint K.

set model but the idea is.

Very similar Why is it the case maybe
physicists have good explanation I I

don't what it perhaps not surprising that
when you have something like that that

that complexity of the geometry of optimal
solution should indicate some kind

of a hardness when you have the behavior
like this when your optimal solution for

extremely sensitive to the even tiny
percentage of the edges of your

graph perhaps it's not surprising that
it's hard to find nearly optimal cuts.

So that's a that's a guess.

And now I want to introduce a class
of algorithms for which the.

Emergence of the overlap their property
implies some negative results.

So this this class of algorithms sort
of loosely speaking we would refer

to it as local algorithms formally
These are called the factors of ID.

Practice or by id type algorithms.

A question like.

That with like you story
if something goes.

Up.

Easily.

None.

In a sense what the shattering
the overlap get property sort of suggest

kind of the opposite in the sense that you
might be pretty close to the solution but

to get to even close it let alone
the optimal solution you almost have to go

down quite a bit before it gets
to be better quality solutions so

it's like extremely extremely.

Complicated landscape of solutions
there's lots of local optimums

with the various qualities some close and
closer but every one of them is still

quite significantly separated from higher
quality solution and the optimal solution

so that and that with this picture it's
going to kind of becomes almost immediate

that something like Mark of Cain want
to call a method would not be a good.

Method to solve this problem or
it so let me introduce the class of

algorithms it is it is not Markov chain
Monte Carlo method it's something else.

So the the.

Idea of the algorithms or
called Factor society is to code

algorithms which make decisions entirely
based on the local neighborhoods

of a graph so
it's it's an attempt to construct for

example cuts where you decide
whether the no there lose one or

minus one just by starting a small
neighborhood around your note and

just observing this neighborhood
then maybe with some additional

in the musician decide that this no
this one and that no this minus one.

And two go that let's call that more
formally define it more formally

we fix a certain radius R..

And first we consider the set of all
trees rooted to release with depth R..

And suppose on this set of trees we.

Have a function.

That maps the following things so
we take the interest from.

Internal from zero to one to
the cardinality of the number of nodes and

the function maps each element
here to the minus one and one.

So what US dysfunction stand for and
what is this meaning the meaning

of it is as follows If the notes
of my tree are associated with

some randomly generated
uniform uniformly randomly

generated weights then this
function will tell me based on

those weights whether the answer should
be decision should be minus one or one.

So if I generate this nodes uniformly at
random you can think of it as the random

seeds in your randomized Elgar written
that OP one observing as a radius R.

neighborhood of a fixed note makes
a determination whether the notice one or

a minus one you could have a function
which it completely ignores

the completely ignores the random
season just makes a determination

makes a decision just by looking at the.

Graph structure on your note what
will make it the more powerful class

a fellow get at them by letting
it to have that in the MS Asian.

And no.

We going to use we use this
function to make a cut

decision so the cut decision is done
as follows We make associate uniform

random weights with every
note of the entire graph so

generate an end of variables
uniformly at random I.E. D..

And then for

every note we'll look at the our
neighborhood depths neighborhood around

this node we typically observe a tree
because there's a sparse random graph.

We observe the three on which nodes
are associated with some random weights

generated here we use this function and
that function tells us who is one and

who is minus one.

So that class a fellow
get it them's is called.

Factor of a DA or
show local algorithms for short.

Any questions or confusions about the L.

that yes the function can depend
on that it could be that if I have

three like that I use this function if I
had three like that I use that function.

A Not in this setting so
we're sure that it's somewhat iffy isn't

equivalent that as long as the only
thing that matters is the policy of

the three itself and the random
the creation of the nodes Yes yes.

Just so you.

Mean you know.

Say F.

is fixed yet.

So suppose we have this this function
this function just maps the decorated

set of waited three waited
with values into minus one and

one this is just some fixed function
function so let me describe how we use

this function to make a cut in
the random at the shiny graph so

the way we do that is we for even every
note we need to decide is minus one or

one with for every node we look at
the art of Depp's neighborhood and

on this note we observe a tree
that trees decorated with

a random weights we apply this
function and it gives minus one or

one and we do that for all nodes of
the graph some altering its Lee.

So that creates kind of a sort of
like a correlated set of decisions

on the entire graph.

So I.

Think my what when did
that start actually.

Five.

Minutes to five and you have about
five minutes five minutes OK so that.

I have some.

Kind of a historical bullet points
where this algorithm comes from

interestingly perhaps it's originated
not in the sirup algorithm but

in the regarded theory but then
through such a line of work on Gaspard

graph on the graph limits percolated
the field of and in the graphs.

Given time limitations let me jump to the.

Jump to the conclusions.

The conclusion and
put put put to sort of informally

says that the local algorithms
defined to light like that they cannot

construct a nearly largest caught
in the random hyper graphs.

With the uniformity at least for.

Me.

To be more precise that exists a value

which called gamma algorithmic which is
strictly smaller than optimal Gumma star.

And there exists a large in of
the zero such that as longest

Well naturally this number
of nodes go to infinity and

there's long as the degree of
the graph is at least does you know.

Every cut producible by
the local algorithm.

Normalized appropriately will get a value

at most Gumma algorithmic
that's not me I think.

OK.

At most gum augury it make and

that constant historically
smaller than gamma star.

And that now putting that formally
involves several soups Unfortunately

the first is just a set of those going
to infinity the degree going to infinity

that's natural perhaps the more most least
natural object here is taking the supremum

of the radio of the action
of the local out.

Rate them as well as the any function
that is used within the class of.

Trees with depth are so
that says that even if you decide

that you're going to use the best
rule out there for trees with depth

thousand and among those you will
find the best somehow find the best

decision to you half this
will still come short

of the melody even approach
upon the proper normalisation.

So.

I do have a proof but it's not a good
idea to do to go through the proof

sketch no given that I have what's
known three minutes two minutes left.

Or it's two point five minutes so instead.

Let me just share with you
some some some kind of key key

key things that are used one of
the things that we use is that it

turns out that optimal cuts in
nearly optimal cuts turns out and

this model are nearly balanced and
there are nearly those in the sense that

the number of ones versus minus one says
is about the same one can quantify that

by giving a more precise bound on
the imbalance turns out to be a poor

normalization by the number of notes
the ratio of one versus minus one D.

case at the rate like this as degree
grows so that turns out to be important.

And the idea of the proof is.

To use Interpol ation on
the starting from completely.

So we fix a particular incarnation
of the local algorithm for

particular edges
are a particular function F..

And what we do is we started with two
independent sources of random the X.

X.
and Y.

as if we were running this
algorithm twice on completely

independent copies of two graphs so
we have one graph at yes and the other

thing is that where is that we're going
to run this algorithm in analyzing.

On couple versions of two grabs at the
beginning suppose we have two independent

versions of the graph and two independent
rounds of this algorithms and.

Because everything is completely
independent the two carts that

are produced are completely independent
themself both of them are nearly balanced

so you almost create like two
completely orthogonal cuts that way

each of them fairly balanced and

they are in the independent In other
words nearly orthogonal to each other.

And then what we do is remember this
couple version of graphs when they start

over they share certain fraction of edges
we start moving these two independent

copies and merge them to each other by
making them forcing them to share more and

more fraction of edges by was sampling
let's say one edge at a time but

while we do that we also started
resembling the sources of

the random seeds that we used to
generate the cuts so at the beginning X.

and Y.
are independent but

now we're one of the copies we keep X.

but for for the second one.

Where is it we used a certain fraction
from the first source in the certain

fraction from the seconds or so
that in the end we have a one that on of

the algorithm on one copy of the graph
is the beginning they are completely

orthogonal to each other in the end
they are identical by construction

it turns out that that Interpol
ation is scantiness enough in the in

the intercalation parameter Delta so
that when you look at the how this

overlap between two solutions evolves as a
function of the of the commonness of these

two graphs it's a continuous function
starts completely orthogonal independent

ends up in completely identical to
solutions as a result it should.

But is that.

The have a good as as the result
it go through all the stages of

the overlap starting from
zero all the way to one and

therefore at some point it should hit
the region which is prohibited to region

the overlap get Bridgette where
we know the nearly optimal cuts.

Not exist and that that gives the that's
the give the contradiction argument.

And.

This is this is it I have some
open questions in the work in

progress that perhaps better to
take offline in the thinking.

Yes turns out the care
theorem is also holds for

cable to the overlap get
property does not hold for

Cakewalk to to at least not
conjecture to hold that if you

look at the overlaps it turns out to
have a continuing continue continues to

support on the overlaps and that implies
that means that that line of argument

cannot be used for the Scots and
graphs as opposed to high progress.

Yes And in fact

this is not the first argument to to show
the overlap gap implies nonexistence

of local algorithms first we use that in
the context of largest independent sets.

I see right so you're bringing X.

or because it's solvable.

Yet so X.

or write X.

or is solvable and then and

it still would mean it it did thus
exhibit overlap get property in

in a certain sense it just means that you
cannot use local algorithms to do that but

Gaussian elimination that you would use
to solve the problem is non-local it's.

That that fine you're only optimal
solution or what you haven't mind.

I think.

You know.

Thank you thanks.