I'm very very happy to have been here
today to speak as a joint mouthpiece your

goal of your joining has been one of
the people who is really promoted and

also developed the problems thinking
the method in quite some influential way

he's not only co-founded
including one of the and

giving joining us in the area he's also
reading together his move along with

the thoughts of textbooks on the method
and today I think he's going to tell us

about two gems which connect sort
of the promise the methods with.

So that for they're doing so
OK thank you very much loads and

it's always a real pleasure
to come back are we OK OK.

It's always really a pleasure
to come back to Atlanta and

it's always a pleasure to talk about Paul
because Uncle Paul we all called him

Uncle Paul is now a figure
of legend he was a giant

of twentieth century mathematics and
he worked in so

many fields he had a unique personality
in there a million anecdotes.

About him and they're all true.

I want to talk about a particular
area that he worked in.

And in my prejudiced view it's
it's I won't say it's the most

important one though I I actually think
that because it's my hair here too but

but it's the area that most clearly bears
his stamp because it was something that

that he really invented in the sense
that anyone can invent mathematics and

he worked it and it's the area called
the probabilistic method but more and

more I like to use the term
erudition magic to describe it and.

So let's back up so

ever this magic is a way to prove

the existence of some common
authorial object be it a coloring or

tournaments or a graph or a ranking or
or some kind of thing and

and the way you do it is you create
a random object and how you do that.

A lot of ways to do that so
that's part of it and

then somehow you show that the random
object has a positive chance of

having the of of being the object
with the properties you want and

then the conclusion is not that
the object might exist but

the that the object must exist in the
mathematical sense of existence because

if something has probability greater
than zero it can't be the null set so

there's got to be you know
in the space of objects so

there's got to be an object in this
is what we call air dish magic.

But today I want to talk about modern
Well modern being not that recent but.

More and more.

Mathematicians in general
are looking at algorithms and

here I want to call a modern era dish
magic you create a randomized algorithm

whose object whose Who where you
hope that the algorithm will

create this object with the desired
properties and if you can show that

the randomized algorithm will work
with positive probability once again.

The object must exist so you get
the same conclusion I'm going to give

specific examples in a minute but
not only do you get specific examples but

furthermore you get
an algorithm Now I started out

in the math department and
I'm ending up as a mathematician.

But I'm in the math department and
the computer science department and

to me the great thing about computer
science is that it's led to

such interesting mathematics problems
it's rather parochial view of the world

as Stephen smale said it said it better
he once said that peak was and P.

is computer science is
gift to mathematics and

here you have this beautiful beautiful
mathematical problem but we would have

thought of it as mathematicians except
that we have computer science OK so.

Let me let me I'm going
to look at two examples

where the original results one
is the first one is due to me.

And the second one is due
to the last low low Vos and

in both cases the original proofs
did not involve algorithms and

now we have new proofs that involve
algorithms so a this gives an algorithm.

Which.

A lot of people maybe not you and
me but a lot of people care about

having an album and by algorithm I mean
effective algorithm I won't say much about

just how much time they take but
these are all polynomial time algorithms.

But suddenly in both cases these
give new proofs of the existence

of these objects and it's really quite
interesting to look at the two of them so

let me turn to my own result from back.

Quite a few years ago.

And always get more and
more embarrassing but there it was so

here's the result of first do it
this was a problem of air dish.

Dealing with discrepancies So
you have an sets on and elements and

you want to color the elements plus one
minus one or red and blue if you want and

you want the set to be evenly
balanced as much as possible between

the red in the blue so you red is plus
one Blues minus when you add them up and

look at the absolute value and you want
that to be as close to zero as you can and

the theorem that I proved in one
thousand nine hundred eighty five

was that there is a coloring where.

A coloring of the underlying sets one
thru and so that all of the sets S.

one through S.N.
have discrepancy last the real

to some absolute constant the six
is maybe five point three two but

six standard deviations was the title
of the paper in that sort of stuck.

That all of them are within six
that they're all less than or

equal to six root N.

Now it's not hard Well it's not easy but
it's known that you can't do better than.

Constant root N.

a one example if your income in taxes had
a mark major sees you as a hatemonger

matrix you can show that the discrepancy
if you turn the how to mark major So

you into a bunch of sets on elements
then the discrepancy is greater there to

some constant Ruden I forget the constant
So either way it's known that Ruben is

the Best Buy We don't know
the constant OK but my proof was.

Actually used to pigeonhole principle
I mean this is my best theorem ever

Everybody should have one best theorem
ever and hopefully you don't have it yet

hopefully you have it hopefully it's still
to come OK still in front of you you know

but I think I can say that this is
my best theorem ever was was proving

this to use the pigeonhole principle but
I'm not going to talk about the proof but

immediately pigeonhole principle
is very non algorithmic I mean is

there some classical some logicians
have can make that substantial but

so the proof it just seemed it
was not algorithmic Now of course

you can use brute force but by algorithm
I mean polynomial time algorithm.

So immediately I conjectured that.

There is no polynomial time
algorithm that will give this.

Coloring.

And then in one thousand nine hundred ten

Maciel Bansal using semi
definite programming.

Found a polynomial time algorithm and

then a couple of years later I love it and
make.

Also gave an argument and I want to
give you an idea of their argument so

we want to get this coloring.

And the key thing about it is that all
the sets have discrepancy less than or

equal to squared event constant squared
event Now why is that what's the intuition

suppose you call a randomly OK so
you have a set maybe it has an A and

elements in the bigger
the set the harder but

I suppose there is an element if
you color randomly it's discrepancy

well before you do the absolute value it's
discrepancy is going to be gaussian and

it's going to have standard
deviations rootin so

the chance that it's more than six that
that's why I call it standard deviations

the chance that it's bigger than six
through ten is like one in the million but

the thing is one in the millions
a constant and we've got an sets so if N.

is ten to the one hundred we're going to
have this one in a million we're going to

have these outliers so the question is can
we eliminate the outliers the key word.

Is all we can get most of the descriptive
we get all the one in a million

less than or equal to six wouldn't just by
coloring randomly but we want it we want

to eliminate the outliers OK Let
me switch to a vector formulation.

Where all the work you have N.

vectors in N.

space.

That have L.

infinity norm less than or equal to one.

It's easy to make the connection to
sets these or you look at the sets and

the elements and the matrix and you look
at the row vectors in something like that.

There's an initial value that's
only think of the initial value is

zero because it's only in the proof
you're moving it around and so

we want to do with a general initial
value but whenever you see Z.

just think of zero zero.

And now what we want is the coloring
is a vector of pluses and

minuses and
we dot it with our audience which

is a vector say with all coordinates maybe
plus one or minus one like the Hadamar

matrix then we want the dot product
to be less than or equal to K.

Ruden again for
all of them again forget the Z.

just Z.

is zero.

So this is this is the formulation
I wanna look at and

again like I said the problem is
the outliers that is if you just pick the.

X.
at random it will work for

all but one in a million cases but
that's no good we want is zero

we want all the I's from one to N.

OK So actually I'm only going
to do phase one I'm going to so

I'm going to use a notion that
due to Joseph BECK We want

a coloring so coloring of the elements
that we want a vector of N.

plus ones and minus ones we're going
to start the coloring with all

the colors at zero so think of the
coloring as a point in the end dimensional

cube going from minus one to plus one
it starts at all zero Well that's great

all the discrepancies are zero because you
add up all the colors you get zero except

we haven't done anything so that's so but
now we're going to move it around we're

going to move the coloring around
inside the an dimensional cube so.

So we're going to start
here except this is N.

dimension not two dimension and
we're going to move around until.

We read it this way.

Until we until all the things
get to be plus one and

minus one and
this idea is called Floating colors and

it's due to Joseph Beck
who's now a Rutgers but

we're going to move them around and
this is the love a chick car I idea

in in a Brownian motion but
a controlled Brownian motion if you just.

So for one thing we're never going
to leave the the unit cube so.

The way I think of it is when you reach
minus one or plus one you stay there so

if a VERY of the variables are moving
around when when one of the variables

reaches minus one or plus one it stays
there in terms of the algorithm because

we're taking discrete steps we say if you
get really really close to plus one or

minus one you stay there but if you're
within one over and square then it doesn't

matter you could just you know so so it's
just a technical thing that will will stop

when we get when we freeze it
when we get real close and we're

only going to do phase one we're going to
go until half of them are frozen and L.

say a few words about the other
stuff now we're looking at

again zero zero we're looking at R.J.
got X.

which starts outed zero because X.

starts out at zero.

And we look and
the normalization is divided by RUDIN So

instead of one constant Rudin we want
constant OK so so were we will we want

is that all of the L A J S
are less than or equal to K..

This constant Here's what we do

so we're out a certain point and
but again think of this is N.

dimensions some of the coordinates
are frozen and the others we want to move

in a random direction but
a controlled random direction.

And here is the control first of all
once something is frozen we're not

going to move it OK but we're going to
this is only going to be phase one so

that will only be half
of them will be frozen.

That.

Second technical thing which which
is actually not in the love it

make approach but I like it we move our
thought only to to where we are but

the Imp here's the important one

we look at the current values of L.J.
Remember we want all of the L.

days to be less than or
equal to six in absolute value so

this says how dangerous
each coordinate is we or

sorry how dangerous each are subject is
because the dot product has gotten big and

so the value of remember L.J.

is is in the back up L L J

is the danger it's R.J.
dotted with with the current value of X..

So we look at the.

Top quarter most dangerous our js and
those we freeze

well these we freeze temporarily and
we freeze them by saying that for

those are js we insist on going
perpendicular to those our J's.

OK So when we do that we
have how many things have

how many degrees of freedom that we have
left we've taken we've taken at most and

over to coordinates which were frozen and
we've taken at most and

over for our Jayson insisted
that we be a thousand oil but

we're still in a hyperplane of
dimension and over four and

I could move around the constants and
do two other constants that would be fine.

So the.

Dimension in which were allowed to move

is still of high dimension it's
still a dimension constant N.

and now we make a Gaussian move.

In that and that's our step and
now to make it into an algorithm when

we take the galaxy move their parameters
about just how far we want to go for

because in my mind this is a continuous
algorithm but continuous algorithm is

somewhat of an oxymoron I mean you
know computers don't well anyway.

So so so there's some stuff that
we actually do the Gaussian and

go a certain distance.

And then we were given by this
some little number Delta and

then that are moving OK So

what happens is we're moving along and

Delta was the distance
we move at each step so

since we're going orthogonally we're
adding Delta squared it each time and.

At the end everything's at
most plus one or minus one so

the el to Norm's at most and so
the total time is at most and

Delta squared And
now let's look at a particular L.J.

and and this error L.J.
and what's happening to it

at each step Well it might be frozen OK
well that's great it doesn't move at all

or we're making this Brownian move and
a brownie move with.

This D.

dimensional Brownian move
when you project down to L.J.

It's just a Gaussian move it's just
a gas so you moving it by a Galaxian.

And the Galaxian has variance at
most Delta squared for over and

where the four reflects the price
that instead of making the move in N.

dimensions were making the move in
an over four dimensions but that will

only be a constant and now what happens
is if you look at any particular L.J..

It's a Martin gale at each step it might
be frozen it might not move at all but

if it moves it moves as a Galaxian
with a certain variance so

it's a Martin Gill Now how far it
moves depends on previous history but

that's the whole strength of Martin Gales
is that your expectation is zero but

your move depends on can can depend and
here does depend on previous history so

we have a Martin gale and
the variants of the Martin Gale.

The the ends cancel out.

And the Deltas cancel out as
they should and they do and

the total variance is at most four and
that four is the expense of

going from an dimensions to an over
four dimensions so now you have a mark.

Gale with total variance for
and each step is Gaussian and

so the L.J. are the the the.

The.

Large deviation results
on any particular L.J.

It's just as if it was a galaxy
in with with variance for.

So this is the Martin gales when every
step is Gaussian are particularly pretty

everything works out very nicely So
what happens is that the probability

that L.J. is bigger than K.

is as if it was a Martin gale and
we pick a equal six or ten or

whatever the number is so we make it
less than point one at this stage it's

not at all clear where we've gained but
there's only one line to come after this.

At this stage.

Any particular Elwood today
has probability less than

a tenth of being an outlier let's define

success saying that fewer than
twenty percent of them are outliers.

But if the expected number
of outliers is ten percent.

Then when we run the algorithm.

At least half the time fewer than twenty
percent of them will be outliers so

what have we done with made an algorithm
with at most twenty percent of them

outliers but I started out saying that
just doing a random one would get ten to

the minus six of them outliers so
what's the last line here Whoops.

None of them are outliers

because we were always freezing the top
quarter and offset up the quarter in

the fifth I mean you could adjust
that to make the constants better so

we were always freezing the top quarter of
them and each step is just very tiny and

doesn't have an effect so we're always
freezing the top quarter of them so

if less than a fifth of them are outliers
none of them are ever outliers.

Because as soon as they became
outliers that is bigger than K.

they were frozen.

So it's a funny little piece of
logic here that you go from having

at most twenty percent outliers but since
you were always freezing the top twenty

five percent it means you have no
outliers and this was the argument and

this gives the algorithm and
it turns out it's really

quite a fast algorithm and indeed some
of the technical things about just what

what to choose the Delta I mean
the bigger the Delta you choose.

The faster the algorithm is but

on the other hand you don't
want to jump over the.

So there's thickness of the boundary so

there are a whole bunch of technical
things skip but this was the argument and

I want to add to that this is a completely
different proof from the proof

that I had which was then improved by
many people especially Ravi Bopara.

But still was non algorithmic But now to
me this not only gives an algorithm but

it actually gives a better proof
of the original result data.

No.

Good it's not the same constant you know
to tell the truth I haven't worked it out

yet it's not the same constant.

And that really gets into the technical
things of getting the Delta correctly.

OK.

And then all of the there's other
technical part once you get down to

an over two frozen now you want to
move down to an over four frozen and

you do kind of the same thing but
I'm going to skip that.

And turning to the low Vos
local lemme low of us I

think of him that you know we think of
historical figures like air dish but

you know each generation
has some people and for

my generation it's the last low low VOS
In Budapest who's done tremendous work in

many many fields and in particular
is this result to love us local M.

which has turned out to be a a central
thing in probabilistic methods

it allows us to I like the phrase find
a needle in an exponential haystack.

So I'm rather than give to
the general result if you've.

Taken blitzes courses hopefully
you'll you'll know about it.

But I'm going to give it
with a specific example.

With Boolean case sat OK.

So I want to think of K.

is fixed and I'm going to take boolean N.

variables X.

one through through accent Sorry yes X.

one through accent and
I'm going to look at clauses

case that clauses that
is the disjunction of K.

of the.

X.
js or not X.

js and then an instance of
case at is that you want

all the clauses to hold simultaneously
you want their conjunction to hold so.

For those.

From theory C.S.I.
The three sat is the classical and P.

is the classic N.P.
problem OK to which we contribute nothing

with that with with this OK but
still gives you an idea all

right now I want to add
something I want to say the two.

Was overlap if they have
a common variable X.

of seventeen including if one has AK
seventy nine the other has not except in

they still overlap so
now here is the local lemma.

Assume that each clause overlaps
at most the other clauses.

And the clauses all have size K.

So when you randomly put in truth values
they have probability two to the minus K.

of failing OK And then I want ears
actually two point seven one eight

actually he surprised me with
that if Edi to the minus K.

is less than or equal to one the strange
but then the US local lemma says yes

one can satisfy then all of the clauses
that is the conjunction is satisfiable So

a key thing about this is that
you can put in values of K N D So

for example cake was five in the quiz
ten so you have an instance of five

said not a random instance an arbitrary
instance of five set where

each clause overlaps at most ten
other clauses but you've got.

Ten to the seven variables all right and
yet the last local lemma.

Says that it's possible to satisfy all of
them at the same time the reason why it's

called local lemma Well there's in the
litter a sion I guess that's the reason

sound sounded good but also it fits
because locally you've got a clause and

it's only sort of connected to
a small number of other clauses so

locally things are looking pretty good but
you want something that works globally.

And the original proof.

There was no way to actually get the
algorithm Joseph back I keep mentioning

his name had some key input the question
then came Could you give an algorithm.

Beck gave an algorithm that worked for
some instances of D.M.K.

and then Robin Moser came
along he was a first year.

First year graduate student had a T.

Ha he teaches in Zurich and
he got this idea and

it totally transformed
the entire thing yeah just.

Blew Mazie Yeah yeah he was
student of him a belt so.

Yeah M.-O.
said he was a pretty good student.

All right here's his algorithm.

Interesting.

No go alone and
I have many many people look at and

try to come up with algorithms Here's his
algorithm I call it the fix it algorithm.

First.

You just randomly assign true and

false to everything well if that works
you're done go home OK no problem there.

If it doesn't work all right so if it
doesn't work there's some clause that's

turning out false again I'm going to
leave out there's a lot of data structure

questions here about making the algorithm
efficient that I'm leaving out OK but

this is definitely a polynomial time
album I polynomial time I mean you

fix the five the numbers five in ten and
then you let and go to infinity.

So you take you select one of the bad

clauses and indeed you can
do that in any way you wish.

You really remarkable part
of this algorithm and

you take that bad clause
with its five variables and

you randomly reassigned those
five variables to True or

false but my favorite step.

Is the Fourth Step some of you
have seen this talk before

there is no fourth step that was it that's
the whole louder of them that's all you do

that's all you do end of
algorithm you just start randomly.

If there's a problem there may be many
problems pick one of the problems

any way you want to do it
you can be clever or dumb or

anything anyway you want to pick you
know you can do it lexical graphically

anywhere you want you pick one of the you
don't take all the clauses at once that

are a problem you take one of the clauses
that's a problem and you fix it

that is you reassign course you could
be reassigned the same way you know so

but and that's the loop now of
course what happens is when you fix

this it breaks this and
then you fix this and it breaks this and

then you fix this and maybe at that it
breaks the thing you started with so

it's not at all clear why
this would work indeed.

My memory with no that alone is we sort
of looked at this algorithm and said No.

Can't do this this is
ridiculous you turned out.

This is the algorithm and I want to
give you an idea of why it works.

And there are many approaches
I'll give an approach due to

Don canoes who I know I think of Don.

You know one thing about computer
science is you know a lot of things in

math are like two hundred years old
you know but computer science I mean

Don can do things like the George
Washington of computers I mean theoretical

computer science I mean to me he really
just like you look back at kalian Abo I

mean these figures and math what other
fields get developed in they have to

start at some time and I mean there
are other people carp certainly but

and others but but
to me it's done can do that really

started the subject of
theoretical computer science and

this is his argument for this result so

you look at this algorithm and
you take a log and

if English is not your native language
log has about six definitions and

here what I mean by log is like a ship's
log where you're like a diary that you're

keeping of it and you just keep
a track of of which clause so

so I changed I'm going to use letters for
the clause and

instead of looking A one A two A three
I'm just going to call him a B.

C.
D.

E. F. G.

So maybe I switched clause A and
then I switched Clause D.

and then I switched clause C.

and then I switched clause F and
up priority it could go on forever indeed

if there is no satisfying Sime it will go
on forever because it won't stop right but

this is called the log so
what we want is to show

that the length of the log which operate
already might be infinity but if it's

expected sizes is finite that means it can
stop fact it stops with probability one.

On.

But it can stop and
therefore by modern air dish.

Magic there will be a running
of the algorithm that leads and

the only way you can stop is when you have
the assignment because otherwise you keep

going so so you'll be done and furthermore
again the part that I'm leaving out.

The expected size is
actually quite reasonable so

you get an effective algorithm.

OK.

So we're going to play Tetris.

So we're going to take
this log a D C F D C B I

hear the clauses so when I say the clauses
one two three I mean the clauses on X.

one X.
two X.

three maybe with some knots on it and
they might be X.

one or not X.

two or X.

or and the fact that there consecutive
is only is not part of the theorem

I mean it's just there for
the display purposes OK So I mean

I've made the clauses consecutive numbers
but but but that won't be the case.

So you have to expand your butt
in Tetris usually the sentance.

All right so here's this.

Clause.

A D C B E C F B And
let's start throwing them down and

playing Tetris with them so
we throw down a then we throw down D.

then we throw down C.

then we throw down half then we
throw down he then we throw down C.

then we throw down B.

and then we throw down at the case so

we get this I could have made had you have
to go further down that it doesn't matter

OK I just do so
then we get this Tetris pattern.

Now I want to talk about.

Looking at the last one of the sequence F.

I want to look at what I'll call the
paramedic which is really the things that

are holding that half up all right so
in this case the be in the C..

Are not holding the F.

up but the ears and this a is
because the A is holding up the C.

which is holding up the E.

which is holding up the F.

So we get this notion of the pyramid.

C.
which is just all the stuff that

support the last letter OK.

Now.

What happens is the a little bit of nice
comment torics if you take any sequence

forget about the probability you take any
sequence and you look at the param aids

of the prefixes So you go out one step
two steps three steps forceps and

each step you look at the pyramid you're
not going to get the same pyramid twice

you're not going to get the same pyramid
twice why is that while the last letter

would have to be the same but
if the last letter of this one was a D.

and the last letter of this one was a D.

then this one's permit would include
this once pyramid and have one more so

nice little common to torics
the pyramids are distinct and

now we can say is that the expected value
of the log is just for each pair amid.

We can ask for the probability
that that Param it comes up.

As the pyramid of some prefix and
that's right because if you

have a log of length one hundred they're
going to be one hundred pyramids and

now we turn this around for each
permit we look at the probability that

pyramid comes up that we don't say when
it comes up and then and then it's equal.

Now here's a powerful idea.

In looking at random algorithms and
I think it's very powerful.

I think you're going to take two things
away from this talk of first the phantom

quote and secondly this this idea of
pre-processing randomness because

with randomized algorithms you get into
all sorts of very subtle questions and

mistakes involving conditioning and so

it really helps if you can set
up the randomized algorithm so

that all of the random
This is pre-processed.

And to do that you you put in more
random this than you might use so

how does that work here.

What happens is each variable X.

eleven flips a coin
a countable number of times so

it goes True True False False True
true false.

At the start it uses it zero thing

if it ever needs to reflate
it just goes to the next one.

They see that's exactly the same as
here but somehow at least it helps me

this pre-processing I find
is a helpful idea OK now.

If you look at it as a.

The chance that a is a paramedic Well
if A is a pair made it means that

A has to be false on the zero is calling
flip which is probability one in eight OK.

But now what is the chance and
it's actually less than that

because it might be that that A and B.

were both falls on the zero coin flip and
you actually flipped B.

but it's a most one eighth to most one.

Now what is the probability.

That a C.

is a pyramid.

A and C.

overlap.

So it looks like it doesn't look like
it's an eighth times an eighth but

it is because in order for a C.

to be a paranoid it means that
A is false with the zero three

point flips of coin one point
two in coin three and then C.

is false with the zero The coin
flips of four and five and the.

First point the start I'm a real
mathematician I start counting at zero

the first coin flip of of three so

though they overlap their independent
events so the probability that A C.

is a pyramid is actually
an eighth square actually

Again that's an upper bound because it may
have again it may have been that that B.

was flipped first but it's an upper bound.

You know.

Yeah.

That's right that's right so

because we're doing this now we get
into some nice alledge with this and

beautiful semigroup theory here we have
these strings on a finite set of letters.

And we're looking at them but
let's define a semigroup and

this is been studied by a group theorist
will say that when two letters.

Have no overlap we're going to say they
commute to have a finite set of letters

we look at the free semigroup on that
finite set of letters but then for

certain pairs X.Y.
we say that these pairs X.Y. commute and

that gives an interesting semi group
definitely interesting to study

just by itself I mean what do you
get what are its properties well.

One or two things equal when they have the
same Tetris picture that's the nice thing

so if you don't like Tetris you can talk
about this algebra of semi grooves but

I like Tetris but it turns out this is
this is the equivalent I mean it's a nice

exercise to prove it so just for
the example you see a D.

C.

If instead the sequence was D.

A C.

first put down D.

then put down a then put down C.

we would get the same Tetris picture and
a indeed don't overlap so

they commute so
in the algebra A D.C. and D.

A C..

Are equal and
in the Tetris picture they're the same and

that turns out to be if and
only if it's not not too hard.

To prove and

then we get into a very nice
algebraic common a tour Express and.

I think I'll just say this and
then pretty much leave it there so

let's look at all of the pyramids.

Now you can do this one.

Not everything is size K.

but here everything is size K.

so the probabilities
are all to do the minus K.

So if you have a pyramid of length ass
it has probability to the minus K.

times yes and we know that
the expected length the expected.

Time of the time unquote of the algorithm
is less than or equal to the sum

of the probabilities over each pyramid
that you get the pyramid which is.

This Some here so
if this sum here is finite.

Then by modern Arish magic.

You'll get that fix it takes time
meaning the length of the lot T.

log At most this sum and

now you get into a very nice question
in algebraic common a tour ix.

I give you a finite set
of letters A through Z..

In this case to each letter
I assign to do the minus K.

but that you could give them
different values also and

now I'm in this Tetra semi group where I
say that certain pairs of them commute and

other pairs do not OK and I tell you
which ones commute and which ones don't.

And now you look over each pair amid And
you have so

let me take a simple example because
we're just going to stop with this.

Suppose you only have two variables and
and they they do commute

then everything can be
written as to the ages.

You can put all the A's to the left and

now if each of let's let's assume
they each go to pay though.

So it's a P.

and Q.
which are right.

For the let's give them
different things and so

what we want is the canoeist
property is that the sum of P.

to the a Q.

to the B.

is less than infinity and
that holds if and only if P.

and Q.

are less than one but if they don't
commute if they don't commute

you're looking at all words of length
an and each word is each always given a P.

and each B.

is given a Q And then the condition
becomes that people ask you is less than

one nice not so easy exercise try for
the four cycle suppose you have a B.

C.
and D.

A and B.

commute B.

and C.
commute C.

and D.

commute D.N.A.
commute you know what is the value of P.

for which it's finite P.

is too big it will be infinite How do you
find it out and you get into some nice

some nice generating functions
as a whole nice area here.

Which I won't say but
the partial answer is

that in the case we're each letter only
overlaps with and most other letters.

So that means that everything
commutes with all but D.

other things.

So most things are commuting but
for each thing there said most D.

other things that it
doesn't commute with and

then one can work things out and
one can get the.

Proper even more than that
here the eve for explicit D.

and K.

The E.
is given by an explicit function of D.

and K.
which has limited value D.

so it's even better than that.

I want to add and with a will and

OK on an intriguing thing and I've
talked to some people and in theory C.S.

about this that this algorithm
actually works against depression.

Oppression adversary now.

So imagine that you randomly make
these countable number of coin flips

now your enemy is going to decide
which clause you're going to flip and

he's going to try to make
the algorithm go on forever OK I mean

if there's only one clause that's bad
the adversary can't do anything but

if there are ten clauses the adversary
can the adversary is trying to make

the algorithm go on forever circle around
the round do something like that you know.

It's an adversary.

So imagine

that this adversary actually
sees all of the coin flips.

That is for each X.J. there's a countable
number of coin flips the coin flips were

done at random but the adversary
sees it so when the adversary says.

Flip this clause the adversary
knows what's going to happen

the adversary suppression it
turns out it doesn't help.

The theorem works against oppression
algorithm oppression adversary so

when I say.

That you can do any selection for
which clause to pick it's even stronger

than that it even works against depression
adversary that actually knows what's going

to happen when the adversary says flip
this one the adversary knows what's going

to happen and even against that powerful
pressure and algorithm adversary.

The.

The argument for this argument that I
gave is fine it's the same argument that.

You don't change it at all and but I think
it's really remarkable that you can have

this notion of oppression doubt
adversary and find quite intriguing so

I just leave with that and.

Paul was known for
many things poetry was not one of them.

But in this letter to their show
she is number one certainly

is number one female collaborator
maybe is number one collaborator.

He.

You know his letters always
started you know let S.

One Dot That S.

and be a family of set go on and on and
in fact I did actually see the letter and

that was the whole letter except
that at the very end he added.

This.

Nice line it is six in the morning
the house is asleep nice music is playing

I prune and conjecture Well
he's not known as a poet but

I think I think this is
poetry thank you very much.

Yes yes yes yes yes yes
yes yes it's really

quite striking that I don't
know how one would use it but

the notion seems very striking to me that
you can defeat oppression Dowd with a.

Book if you actually write my.

Own can we do randomize.

In the.

Case of the no outliers we can

do you randomize because when you
look at Brownian motion you can

make Brownian motion the large
deviation results on on Brownian motion

a really depend on a weight
function that you can create and

there it is possible to do randomized
I'm not sure about the full local

lemma whether it will work in that case
I don't know yet that's a good question.

Yes Yes You know that's what was
asked yes the answer's yes in fact

the wall I'm not saying it's
exactly the same on time but

the bounds on the time are the same
between the pressure and algorithm and

the arbitrary algorithm is when you look
at the proof the bounds are the same.

With this that I don't and
it is polynomial time and

in fact not only is a polynomial time but
it's actually quite quick and

if you run this sits very easy to run
this you can say Take instances rant

say you want five sat were
worse say one five sat and.

One way you could do it is just
on a million variables you could

split the million million variables into
two hundred thousand sets of five and so

they'll give you two hundred randomly and

they'll give you two hundred thousand
clauses were each variable as in one.

Thing and then you just.

I guess you do it again I guess you do it
depending on the very way you do it you do

it a certain number of times so it's easy
to create these randomized it's easy to

create instances and then when you run
the album In fact they do run very quickly

there and very quickly Yeah.

OK.

Thank you.