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>> David Wilson: Today we're pleased to have Johan Wastlund from Gothenburg. He's going to
tell us about the replica symmetry method and combinatorial optimization.
>> Johan Wastlund: Thank you. It's really a pleasure to be here. So what I'm going to talk about
is what I call the mean feed model of distance, which is a random model where we have a
complete graph end points for each power point there's a number which we can think of distance
or weight or length of the edge. Cost of buying the edge, maybe.
So you should imagine that there's a nonnegative real number written on each of these edges.
And I would assume that these are identical distributed and independent, but they have some
kind of distribution which can be, well, some different distributions. So say uniform 01 to begin
with and then so this is what's called quenched disorder in physics. So we have some fixed
random disorder and then you would like to solve some optimization problem, find some ground
state energy or whatever.
The problems that I'm going to talk about are all of this form that I want to minimize the total
length of an edge set that satisfies some criteria. For instance, the minimum spanning tree. This
is a problem of this form. Or maybe perfect matching, which would require a number of vertices
to be even. Just want to pair them up and so on.
The TSP, of course. And for each of these problems I just look at the theta solutions and I pick
the one where the sum of the edge weights are smallest. Also have the two factor which is
relaxation of the TSP where we apply each vertex to have two edges but it could be disjoint
cycles.
Also, one of my favorites, which I mentions in yesterday's teaser, edge cover. So I want each
vertex to have at least one edge. But unlike matching I can, I allow for several edges from the
same vertex.
So this differs from the first four problems in the sense that the number of edges in a solution in
the optimum solution is dependent on the sample. So in this case it could be anything from three
to five, if I have six vertices.
So there's something called a replica method which originated in physics and it's a bit difficult to
understand what the replica symmetric inset is. It doesn't seem to be based on some reasonable
assumptions that can be verified. It's just a formal thing.
So what I claim here is that this inset would be justified in some cases that I will show you here.
But it's a little bit difficult to say really what that means. So anyway but I would look at the optimal
solution for large number of points. And some background here.
So I'm not going to say very much about this. So the number of physicists have been looking at
matching in TSMP and also some other problems. And they have been -- they were able to
obtain some quite spectacular but nonrigorous results that were sort of tested and, well, people
agreed that they must be true. But the method was not rigorous.
Also I mentioned in this context but this is much later but also it has to do with algorithms like
belief propagation and so on, but I'll not say so much about that.
So this is a picture of Joe Parisi, an Italian physicist, and this is one of my favorite pictures. He
looks so happy here. [laughter]
He computed something. So if you look at the blackboard, there are some nice computations
there. And so what he's talking about is the minimum matching problem, actually. And if you
look here, there's some number that comes out of this calculation, and it's pi squared over 12.
And he even has three digits there.
And what this is is the limit costs of minimum matching on the complete graph, with uniform 01
edge costs. So what happens is that there's no renormalization needed. You can just look at the
cost, because as the number of vertices grows, the typical edges in the minimum solution will be
cheaper.
So the cost actually concentrates at this limit value. So this is basically what I like about this
model compared to spatial models that we don't just prove concentration or tail bounds or
convergence or scaling windows or anything like that. We actually get answers like pi squared
over 12 and so on.
And this -- so this was in 1986, I think. And this was proved rigorously by David, something like
15 years later.
And I'll mention some of these limits that have been established for various problems. The
matching pi squared over 12. Spanning tree was proved by Allen Freeze already in the '80s and
the limit is theta three the refon theta function.
The traveling salesman problem. The limit is some number, 2.04 and it can be expressed as an
integral of the function which Y is a function of X, Y is a function of X and it's given by this
equation. So it can be characterized in terms of calculus, but it's a number that, well, it doesn't
have a nice name.
Edge cover. There the limit cost can be mostly described as half of the minimum of X squared
plus E to the minus X where X goes over all real numbers. And I was able to prove this quite
recently. The bipartite case was done with joint work with Martin Hessler, who is one of my Ph.D.
students.
Okay. So I'm going to look at some other distributions of edge costs now. So if you want to
model D dimensional space in this mean field model, what you would like to do is to have nearest
neighbor distances behave roughly like in D dimensional space.
So the probability that an edge length is shorter than a given small number R should be
proportional to R to the D when R is going to 0 and one way of doing this is to take -- so if I want
dimension D, I take the length to be N is number of points here times an exponential 1 variable
and then take the Dth root of this. It turns out that this will satisfy this requirement. So this is a
simple model of D dimensional space. But, of course, in this graph all distances are independent
so there's no way of assigning a geometry to it. It doesn't satisfy the triangle import and so on.
And this factor N tells us that it's like taking the unit of distance to be the distance at which your
expected number of neighbors is 1. If you look at a particular vertex and what neighbors you
have within distance 1, they expect the number is 1.
And with this rescaling we expect the typical edges in the solutions to these optimization
problems to be of order one. So here's a theorem. For D, at least equal to one, the cost of
perfect matching and now we rescale by dividing by N converges in probability to some number
which depends on D.
And so for D equals 1, this is pi squared over 12 and for larger D, well, it's some number that
occurs in some solution to some integral equation I'll show you later.
>>: Its expotential assumption is key. It depends on more than 0.
>> Johan Wastlund: No, it's a behavior near 0 that's relevant. Because if N is large -- but thanks
for the question. If N is large it will only be using extremely cheap edges. So it's just a simple
way of getting this D dimensional behavior but really at the large scale it doesn't matter the D
distance distribution.
And then the corresponding statement for the PSP. Okay. So there's some kind of deeper
meaning to this that it's not just that I get convergence to some number, but it's also that this
replica symmetric predictions are somehow proved to be correct. But I'm not going to tell you
exactly what these predictions are. It's a bit complicated.
Anyway, here's a two-person game, which has some connection to minimum matching.
So there are two players which you call Alice and Bob. And they take turns choosing the edges
of the self-avoiding walk. So here's a starting point and there's some parameter feet which is
greater than 0 but has something to do with ->>: So these results were for the complete bipartite graph or for ->> Johan Wastlund: What I'm talking about here is the complete graph. From the teaser
yesterday. No?
>>: The previous slide?
>> Johan Wastlund: The previous slide.
>>: The geometry.
>> Johan Wastlund: The complete graph. But the actual whole doesn't for the bipartite case. If
you go back to the previous slide. If you take the bipartite graph there would be an effect of two
because you have twice as many edges compared to how many neighbors you have.
So it doesn't really matter if you take the complete bipartite or complete graph with this approach.
Okay. So Alice and Bob are playing this game. So what you do when it's your turn is you choose
an edge and you pay the length of that edge to the opponent.
Okay. Or you finish the game by paying this prescribed penalty for quitting. So it's theta over
two. And the reason I say theta over 2 is it means that edges that are longer than theta are
irrelevant, because if you play according to an edge which is longer than theta, then it better
would have definitely been better to have quit the game because well if the opponent quits right
after, then you would have been in better shape if you quit yourself.
So if they play this game, then we will start here and Alice chooses this red edge and then Bob
goes from there to a new vertex and so on. So they take turns choosing this path. And then at
some point the game stops.
So in this case it would be Alice who makes the last move and then Bob chooses to quit because
there was no short enough edge from this vertex. So he quit the game.
And payoff then is the sum of all these payments, the payments of the edges and then this
penalty for quitting the game.
>>: Doesn't have to be because there was an edge, immediate, just because he looked ahead ->> Johan Wastlund: Right.
>>: Focus on ->> Johan Wastlund: Right. So, yes. So if you look ahead, you could perhaps see that in a
million moves or so I would be in trouble so I better quit right now.
Okay. And here's another optimization problem. Now I'm going to give away the solution to this
game. So I call it the diluted matching problem. So instead of looking at perfect matching, I want
a partial matching. I choose any set of edges that have no vertex in common. It could be an
intercept or perfect matching.
And the cost of this partial matching is the total lengths of the edges plus the punishment of theta
over 2 for each vertex that I leave out. So in this setting for each value of theta there's an optimal
solution and it doesn't matter whether N is wrong because they're always these solutions.
Okay. So now the solution to the game is the following: So suppose fix all the parameters of the
game and then we let -- here's some notation. Let M of G be the cost of this diluted matching
problem. And let F of G and V be Bob's pay off. So Bob goes second. Bob's pay off under
optimal play if you start from B.
So here's the lemma. This pay off under optimal play is the difference of the cost of the dilutive
matching problem in G and G with the starting point removed.
And this is quite easy to prove. So the proof goes like this. So for F, which is a pay off in the
game, we can write down a recursion. So here we take the minimum. So we look at Bob's pay
off, so Bob's pay off is the minimum over all move options by Alice. Alice makes a move she
either quits right away in the case the payoff is theta over 2, or she chooses an edge length LI
that goes to vertex VI. And then her payoff in the rest of the game would be F minus G in the
starting point because you can't go back there and the new game starts at TI. And this minus
sign because Bob's payoff is minus that.
And, on the other hand, this cost of the diluted matching problem also satisfies some recursion
where you look at the vertex V and what you do with this. Either you pay the punishment plus
what it costs you to match the rest of the graph, or you match it to one of its neighbors, and then
you have to pay a cost of the diluted matching problem in the remaining graph here.
Okay. So if we look at this and we subtract this term from both sides, because we want to obtain
this, then you see that we can subtract this in here and if we put the parentheses here, we can
write with a minus sign here because well minus and minus is plus, and then -- [chuckling] -- and
then we see that these two things here, which I claim to be equal, they satisfy the same
recursion, actually. So it follows by induction that they are equal.
Okay. So what this means is that if we look at the game under optimum play, then this path will
describe the symmetric difference between the optimum diluted matchings on G and G minus the
starting point.
And also the moves of the players are -- so if you play this game, if you go first, you should just
solve the diluted matching problem on the whole graph and then play according to that, which
means that as long as you're at a vertex which is matched in this matching, you play according to
that matching, and when you reach a vertex, which is not covered by matching, you quit.
>>: Are we supposed to be playing the matching or the game?
>> Johan Wastlund: You mean when I -- when I came up with this?
>>: Yeah. [laughter].
>> Johan Wastlund: Well, it's a good question. So I was actually looking at the paper which had
to do with analysis of computer algorithms for games like chess and so on. And I found an
equation which looked very similar to an equation that I had seen in David Alice's paper on this pi
squared over 6 limit which I recognized from the physics papers.
So somehow I thought that there should be a formulation of this replica symmetric answer in
terms of games. And then I came up with this game which was described by this equation. And
then I realized that the game could be solved in terms of matching theories. So I got back to
optimum matchings and so actually I came up with the game first.
Okay. So the Poisson way to the infinite tree is introduced by David. So this is another random
structure, random process, which goes like this: That's a root and each vertex in this tree has
infinitely many children. And there's a cost for each edge. And from a given vertex, the costs of
the children are given by a Poisson process on the positive real line.
So the first cheapest edge to a child is exponential 1 and then the increments are independent
exponentials. And it goes on recursively. So each vertex on the next level has its own children
and so on.
And so I look at what I call a theta cluster, which means take away all edges that cost more than
theta. So what remains is the Gulta [phonetic] process with edges and Poisson distributed
offspring.
And I like to play this game on the theta cluster, graph exploration game.
So why is this? Well, this Poisson infinite tree, the PWIT is in some sense a local weak limit of
the mean field model. And there's a simple way of making this precise, namely the following.
If we look at some -- I want to compare this Poisson infinite tree with the mean field model which
is KN and I pick a root at random in KM. And then I look at what I call K theta neighborhoods,
which means that I have the right to go K steps and each edge should have cost at most theta.
And then I look at the neighborhood which I can reach in that way.
And then I claim that there exists a coupling of the PWIT and this mean field model, which is such
that if I look at the roots the probability that there are K theta neighborhoods are amorphic, which
means they all look the same and all edges have the same cost. This probability is at least this
number. So if I fix theta and then let M be large this would be close to 1.
So what it means if I ask some question which depends only on the K theta neighborhood, then I
can approximate the mean field model by the PWIT.
>>: So this is the two dimensional work?
>> Johan Wastlund: Yes, thanks for the question. And of course if I want to do this in two
dimension D, then if you remember how I chose these distances, you can just take the PWIT and
then take the Dth root of all the edge costs and you get a similar model of the D dimensional
mean field model.
But actually what I'm going to talk about in the rest of this talk doesn't really depend on D. So you
can think of D as being equal to 1 if you like, or, yeah, right. So anyway so what this means is
that I want to study this game, graph exploration on the PWIT.
And then the problem is, of course, that theta cluster can be infinite and I haven't said anything
about how the game is played. The game is well defined. You can play it. But what happens if
the game just goes on forever. There's no well defined outcome. So I can't even define optimal
play unless I know what the outcome is.
And so on. So I get into some situation where it's not really clear what the players want to
achieve.
But what you can do is to look at what I call optimistic K look ahead values. So what you do is
you look at K moves ahead. And then you just assume that if the game ever reaches this far
without anyone quitting, then I will be in good shape. So the opponent will pay me theta over 2
right away because I just assume that everything will be as good as it can be when I'm behind the
horizon, so to speak.
And then so this would give a well-defined strategy, of course. I can play according to this
strategy.
But what I want to do here is then let K go to infinity so that I get two different views of the game.
One is Bob's view and one is Alice's view. So both of them are optimistic and seem to think that
infinity, so to speak, they will be rewarded in some sense.
>>: Any relation in some way to having a discount factor, purchase 0? You could also play with
an infinite time horizon with a discount factor.
>> Johan Wastlund: Discount factor.
>>: So money K generation time costs vertically.
>> Johan Wastlund: I don't know. I haven't thought of that. Maybe that's another way of doing
this.
>>: Just a ->> Johan Wastlund: Oh well. So anyway, what it boils down to is I want to -- so we get this
infinite look at values, which are by FA and FB so F is in some sense the value having moved to
a certain [inaudible] which means it's the same value of the game from the second player's point
of view if the game would start at that vertex and just go down. And I can define this optimistic or
pessimistic, if you like, if you see it from the other point of view, values of this game from Alice's
and Bob's point of view. And the interesting question is whether these are equal or not.
So the theorem is that almost all of these are equal. So this means in some sense if you put
arbitrary values at level K moves ahead and then you propagate this upwards and get some sort
of approximation of the value at the point where you are, then if you want to make K large, it will
not matter what you put at the Kth -- at the horizon, so to speak. What you get here will be
roughly the same anyway.
So the sketch of proof here is that suppose these are not equal, then we can let Alice and Bob
play this game optimistically. So we put two optimistic players against each other. And of course
if they don't agree on the value of the root, then nobody can terminate the game, because both
will think that they will be better off than what the opponent thinks. So the game will go on
forever.
Now, so this is something I'd like to take a couple of minutes to explain. So now let's look at it
from Bob's point of view. So Bob has this view of the values of the theta cluster. And when Alice
and Bob play this game Bob will think he plays optimally. And from his point of view he will
perceive it as if Alice possibly at some points makes mistakes.
So Alice will play according to her view of the value and from Bob's point of view he will see as if
he gets some sort of advantages that accumulate. And each time Alice makes a move, which is
not according to Bob's, what he thinks is optimal, there's a simple way of quantifying how far it is
from being optimal. You just sum the costs that have actually been paid with the difference of the
values of this vertices. So there's a certain -- each time Alice makes what Bob will think is a
mistake there's a certain amount of -- there's a certain how far it is from being optimal from Bob's
point of view.
And the sum of all these mistakes cannot be arbitrarily large, because that would be much larger
than theta, then Bob could sort of speak cash his gain by terminating the game and get enough
payoff, which is better than Alice would have thought from the start that it would be.
So is this somehow -- it looks suspicious here, I didn't explain this very well.
>>: How do you --
>> Johan Wastlund: By terminating the game. So if Bob thinks that Alice has made all these
mistakes that have given him a billion dollars -- well it depends on what theta is, but has given
him more than theta, more than what he was supposed to get, then he could terminate even if
this would not be optimal from his point of view, he could terminate and reach a payoff which is
better for him than FA.
>>: By terminating even when he thinks he shouldn't, it's going to cost him theta.
>> Johan Wastlund: Yes.
>>: And he's already been given 2 theta.
>> Johan Wastlund: Yes. But the point is that these mistakes will add up to some finite amount.
So that means that eventually in this game, which goes on forever, after a while Alice will make
no large mistakes. All her mistakes will be very, very small. So let's say a move is delta
reasonable, if it's within delta of being optimal, okay, then we can look at this as a branching
process.
We look at all the optimal moves and all of Alice's delta reasonable moves.
>>: Why is delta ->> Johan Wastlund: It means that it's not necessarily optimal but if you look at this value from
Bob's point of view -- if you look at the length of the edge minus this F of the next vertex it's within
delta of the optimal thing. So if we look at the branching process ->>: Averaging over optimal strategies from now on so he makes a move and he's busy, from
now on they're both going to play optimally and you measure the delta.
>> Johan Wastlund: I just looked at this from Bob's optimistic point of view. So there's a function
FV that gives a value to each vertex.
>>: Which is the function.
>> Johan Wastlund: I mean, delta is some number which is positive and this is ->>: I said the same thing.
>> Johan Wastlund: I'm not sure. Okay. So the point here is that if we just choose delta small
enough, then the branching due to this delta reasonable moves will not be able to counter the
effect of -- the fact that the game terminates at certain points, whenever you reach a point which
has value theta over 2.
So this will involve some computation that I will not show you. But perhaps you can believe this,
since we can pick delta as small as we please. So given theta, which is some delta, which is
extremely small, so that this branching effect is cancelled so that this delta reasonable lines do
not percolate. So the lines of play that we must consider is just some finite thing.
So that actually gives a contradiction, because if the probability that FA is equal to FB, if that's
not -- I mean it is one or it's somewhere between 0 and 1 and it's the same everywhere in the
tree.
So if you cannot have this infinite delta reasonable paths anywhere in the tree, then you cannot
have FA equal to -- different from FB.
Okay. So what does this mean in terms of optimum matching and so on? Well, it means that
what the physicist would call replica symmetry holds. In something like this, if we look at the
complete graph of N vertices now N is some enormous number. It's like theta is large and N is
exponential in theta and we have some vertex V in this graph.
What this means is that if we look at the dilutive matching problem, then with high probability we
can decide what to do with this vertex V by just looking at some neighborhood which is big over 1,
but it depends on theta but not on N. So by just looking at this game, because if the game
terminates within this neighborhood, then we know what to do with V in the optimum diluted
matching. And this is precisely what we need to prove in order to prove that this cost converges
as N goes to infinity, because it means essentially the average cost of an edge in a solution will
not depend on N.
>>: Somehow what other people might call measurability.
>> Johan Wastlund: Yeah. I guess. So what we arrive at is that for each theta the cost of this
dilutive matching problem converges in the sense probability to something which depends on
theta. But this is increasing with theta.
And actually it's not hard to show that it's upper bounded. So it has a limit. And then we have to
show in some way which is nontrivial but anyway it's not that hard.
So the proof of this convergence on perfect matching involves interchanging the limits. We let N
tend to infinity and at the same time theta tends to infinity. And at the same time we have to
show sometimes it doesn't matter in which order we do this. But this can be done.
So I already mentioned this result. This was David's result. And also for D equals 2 I made some
numeric integration and got some decimals. So this is -- well, this is just a number but at least I
can prove convergence to probability to this number. And for the TSP, this is the limit cost.
Now, I didn't tell you how to get these results for the TSP. But there's a game associated to the
[inaudible] assessment problem as well. Now, refusal chess. So refusal chest is a chest variant
which I think has been played, was invented somehow in the '50s, I think. It has been played at
chest -- well not well known, but perhaps -- people have actually played this game.
>>: [indiscernible].
>> Johan Wastlund: Yes [laughter].
So the rules are as follows. When you make a move or when you want to make a move, you
make the move and then your opponent has the right to forbid one move. So if your opponent
doesn't like this move he or she will say you have to make another move. You make another
move and the opponent does not have the right to refuse more than once per turn. But one
refusal per move.
So, for instance, in this position, so this is position which arrives in a game of Jonathan Yidyia
when he was in the conference he was not in the audience at my talk but he was at the
conference where I gave this talk last time.
So suppose that black would play this move. This is actually quite strong move in ordinary chess
but anyway.
>>: [indiscernible].
>> Johan Wastlund: Yes. So here it goes. Okay. Now if white doesn't like this move, he will
say: No, make another move. Moves back this bishop. And black has to play another move. So
let's say captures this opponent, the queen.
Now, white doesn't have the right to refuse another time. So now it's white's turn. So let's
suppose that white makes this move, captures this bishop with this queen. Now in ordinary chess
this would perhaps not be a very good move because black can just take back here.
And of course in refusal chess it's not obvious. So now black can either accept this move or
refuse it. And suppose black accepts it, then black can, well, let's say black gives a check here
and looks very dangerous, now white's problem is of course that if he refuses this move by black
then black can capture white's queen here and so on. So I'm not going to analyze this position in
detail. This is just to give you some idea of this game.
So this is similar to a game that we can call my tour is better than yours. So suppose both Alice
and Bob play a game that we assume for some reason Bob is claiming that this edge which goes
from -- so you can think of this as being some large graph and this is some sub graph where Alice
and Bob are fighting over who has the best traveling salesman tour. Now, just a warning, I'm
going to sweep some details under the rug here. So don't ask too many detailed questions about
this. But for some reason Bob says that this is a good edge to use in a traveling salesman tour
and Alice says no, it's not. Alice said no, I use this edge instead.
Okay. Well, the difference is that in the traveling salesman problem each vertex has two edges.
So if Bob says I'm going to use this edge and Alice says, well, I'm going to use this one, then Bob
actually has the option to say that, okay, good but I also use this.
Okay. So then we have some sort of cancellation here that they agree on using this edge in their
tours, but they still fight over whether this edge was good or not. So Bob says I'm going to use
these two edges and Alice says she uses this one. Bob asks what else are you going to do with
this vertex because this vertex has to have two edges from it. So Alice says okay I use this edge
instead.
Now Bob, of course, cannot use this edge because he has already said that he is using these two
edges. So Bob has to use some other edge from this vertex. So Bob says, okay, I use this edge.
And Alice just says okay good for you but so will I.
Now the roads are reversed. Now Alice has two edges to this vertex. And Bob only has one.
And Bob does not have the right to use this one. So Alice can ask Bob, okay, what are you going
to do with this vertex, because you only have one edge here.
And Bob goes, okay, I use this edge and so on. So what happens here is that it's like graph
exploration but with this refusal option. So each time a player makes a move the other one can
refuse it at once or accept it.
So if Alice does this move -- well, sorry. Yeah. Bob can either accept or refuse it. So I'm not
going to say anything more about the TSP now because everything is basically similar to the
matching problem. You analyze this game and get this replica symmetry and convergence and
probability to some limits.
What I haven't told you about is how to actually get from this FA equals FB to this limit constants.
So now I thought maybe I have a few minutes left. I can show you a little toy example where I
can actually work out quite easily the resulting computation.
So this is quite simple problem edge cover on the reline. Suppose we have some reline and
some point set a Poisson process, right one. And what we want to do is to find edges between
this that cover all vertices. So you can think of this as a street where people live and we have to
construct some telephone connections and something like that and currently everybody should
have someone to speak to.
So each one should be connected to either the right or the left neighbor. So solution perhaps
looks like this. So we want to use short edges, of course. So this looks okay. But then if we
have, for instance, three vertexes quite close to each other, perhaps we want to use both edges
adjacent to this vertex and so on.
So we'd like to -- it's going to be roughly a matching, but at some points we want to use both
edges adjacent to a certain point. Okay. So if we use our friends Alice and Bob now, let's say
that Bob is going to use this particular edge and Alice isn't. Okay?
So what happens then is that we can look at how much Bob gains from using this edge. So
without looking at the length of this edge, just ask what's Bob's advantage on the right-hand side
and on the left-hand side respectively. So let F then be Bob's gain from using this particular edge
in the right-hand side of this line, then this is, of course, well, Bob can always say that he will use
the same cover as Alice on this side. So it's at least 0. Bob can always cancel Alice's move.
Otherwise, because Alice has to use this edge. And if Bob finds that maybe he doesn't want to
use this edge, then he can let -- he can try to obtain an advantage here by -- so Bob's advantage
will be the cost of this edge which I call X. It's an exponential then, minus F 1 where this F 1 is
just distributed as F. So this is Alice's gain from using this edge compared to Bob in covering the
rest of the line here.
>>: So I don't understand what is the gain?
>> Johan Wastlund: Yes, so now we have this equation we can interpret this as a gain where -so the gain is played on, say, the right-hand side of this edge. And --
>>: But maybe before that what's the definition of -- what do you mean by Bob's gain?
>> Johan Wastlund: Oh. Okay. So we're talking about -- okay. So the line is infinite so we can't
really talk about total length. But Bob is using this edge and Alice is not, then the optimal thing
now I run into some vicious circle of how to define things.
But in the optimal edge cover, eventually they will become equal at some point here. I'm sure
you believe this deep in your hearts.
So.
>>: The optimal one that uses this one and the optimal that doesn't use it?
>> Johan Wastlund: Yes. And the game is -- Bob says I'm going to use this edge and Alice says
okay then I'm going to use this one. And then Bob can either, after that, either say that, okay, I
too use this edge and then they have reached some sort of equilibrium where the rest of their
covers must be equal or he says I'm not going to use this edge, I'm going to use this one instead.
And then Alice has the option of saying, okay, I'm going to use this too against ops or going on
and saying no, I'm going to use this one instead.
>>: What's adjacent to the game analyzed by -- analyzing the corresponding problem? You start
to cover and then you cover the first end point and the end point is a cover from the left or not?
And continue to have just two states and there's some distribution of what the cost of the next two
stages so we can probably ->> Johan Wastlund: Yes. So in an attempt to sort of cancel Andrew's question here. Let's say in
this case it's not clear what the point is in looking at it as a game but we reach this kind of
equation which would arise also from the game point of view.
Anyway, so from this equation we can just sort of untangle the distribution of this F. Well, so from
here we get the probability that F is larger than some fixed number is equal to -- if this number is
greater than 0 then we only look at this thing. So then essentially we have F distributed like X
minus F where X is exponential.
And this is -- well, quite straightforward calculation. I think you believe me if I say that from this
you can work out what the unique solution is.
So actually -- well, let me skip this calculation details of it. But turns out that the probability that F
is greater than some number T if T is greater than 0 is two-thirds of E to the minus T. So Bob's
advantage, if you like, you can think of this as we take the -- we look at the finite line, not a very
long one, and then we look at what's, how much does Bob gain from using this edge if we
compare how Alice and Bob's, how they cover the rest of the vertexes here. And then this would
be 0 with probability one-third and probability two-thirds it would just be an exponential.
So then the question is when does Bob -- when does this edge actually belong to the optimum
cover? So the point is that it does -- if it's smaller than what Bob will gain -- so if his cost of using
this edge is smaller than what he will gain from the left hand and from the right-hand side, then
his cover is better than Alice's and vice versa.
And if we know the distribution of F1 and F2 the independent caucus of this distribution, then we
can just compute what this thing is. So it turns out that the probability of using a particular edge a
priori without conditioning on its cost is 5 over 9 if you look at the expectation of this contribution,
so it's cost times the indicator variable of it appearing in the optimum cover is one-third.
So the conclusion would be then that the optimal solution covers one-third of the reline, and then
one vertex out of nine is connected to both, its left and right neighbors.
So in this mean field model this is also what kind of condition that occurs that an edge is used if
its cost is cheaper than the sum of two independent samples from the game theoretical value of
corresponding game.
Okay. I think this is about it. Some comments on future work. Well, of course there are some
technical reasons why I haven't been able to prove these things for D between 0 and 1. But
there's no reason to think it shouldn't work. So this is something that I would like to do.
Gains for other optimization problems, this would be very interesting for problems like graph
coloring K set where replica symmetry is known not to hold. So I don't know what the results
would be. But maybe there's some way of -- so belief propagation. And also I'd like to talk at
some point to people here about something called social networks. I don't think I'll go into that
now.
Computer analysis and gains. So this is a replica symmetry seems to be somehow linked to this
idea that you can play a game efficiently by looking K moves ahead if K is some large number.
So you think then that games like chess where computers now beat any human being they sort of
represent this Kth symmetry where there are games that are more difficult where you have some
sort of long-term factors that are invisible to a fixed search.
Okay. So there's this chess problem. So this chess problem is -- there are two versions of it and
two different solutions. So if someone is interested in this, go ahead. It's made into an ordinary
chess and then there's a different solution in refusal chess. So thank you very much.
[applause].
>>: So are you constructing a different interpretation of the word replica or can you build a bridge
stronger back to the physics and make the connection in some form of words to what physicists
are doing? If you took replica symmetry entirely intellectually, it would be a perfectly good
lecture. I don't understand at this point what replica symmetry has to do with it. Maybe I don't
know what replica symmetry is.
>> Johan Wastlund: I used this in order to more or less invite people in the audience to answer
questions like that. [laughter].
>>: So you are playing a game, as I understand it.
>> Johan Wastlund: So if I go back to this picture. Let me see where it was. Okay. Right back
here. Here. Okay. So this is the sort of essence of it, that what you do to a certain vertex when
you solve this optimization problem, you can work it out just by looking at a close neighborhood.
Well, this is some kind of symmetry [chuckling].
>>: But somehow they want to know what happens to then some small F.
>> Johan Wastlund: Yes.
>>: So it's an infinite thing, so that's just saying that the number you're interested in is
measurable with respect to this data; it's a function of it.
>> Johan Wastlund: Yes.
>>: Rather than given the data, the edge length or whatever, it's some sort of more than one
possible value.
>> Johan Wastlund: Yes. So if you look at the physics literature, there's some introduction that
talks about magnetization and temperature and a lot of other things that I don't understand. And
then suddenly there are some equations.
And if you then try to understand how to justify these equations, it's something like this that you
want this to behave like some physical system that if you add something down here it shouldn't
affect -- well, it should be extremely unlikely that it has a global effect.
>>: This is a character on the screen?
>> Johan Wastlund: Is it?
>>: I don't know. [laughter] two characters. Three.
>> Johan Wastlund: Okay.
>>: So is the replica, asking anyone [laughter], so there is the replica somehow you imagine
building two copies of this full system?
>> Johan Wastlund: Yes.
>>: Where the edge comes or whatever they are are the same.
>> Johan Wastlund: Yes.
>>: Otherwise there's independent condition on that and you're asking whether the ->> Johan Wastlund: Yes, it's like if you have one of them over here and the other replica at the
other side of the room and the temperature and so on is slightly different, will you get the same
solution or not. Something like that.
>> David Wilson: All right. Let's thank Johan again.
[applause]