>> Nikhil Devanur: So hello, and welcome everyone. It's my pleasure to
welcome Christos Papadimitriou. Christos, I'm sure, needs no introduction to us.
I think this is his fifth or sixth talk here just this year. But I guess in case you
don't know already, Christos taught Bill Gates, and they have a paper together.
So without much ado, over to Christos.
>> Christos Papadimitriou: Okay. Thanks. Thanks. So most of this talk will be
a set of some very interesting stuff that are sort of from long time ago, like 20 -the past 20 years. But so, you know, it's very interesting both from the
computation and mathematics point of view.
So let me write a few theorems on the board. So what Nash's theorem. Every
game has an equilibrium. Another one is Brouwer's fixed point theorem, which
says that if you have a compact, convex set and a mapping, continuous mapping
from the center itself, then there's going to be a fixed point, okay. So we know -so I mean if you have a disk and you -- no matter how you map it to itself, unless
you tear it apart, okay, so you know it's going to have -- one point is going to be
fixed. All right?
So every continuous and so on has a fixed point. All right.
>>: [inaudible].
>> Christos Papadimitriou: Sorry?
>>: These are not the same problem?
>> Christos Papadimitriou: No, no. Well, actually as of three years ago, they are
the same problem, yes. So let -- but let me tell you a few -- a few -- a few more
well-known problems. Factoring. Every integer has a factorization in the primes.
Okay?
So here is another one. Equal sums. Suppose that I give you N integers, okay?
I claim that I can always find two subsets such that if I add the -- sorry. This -this is weird, okay? I give you N integers, all right? I claim that I will be able to
find two subsets that have the same sum. The modular two to the N minus one,
sorry. Okay. Yeah. I always forget the important thing.
>>: [inaudible].
>> Christos Papadimitriou: Sorry?
>>: [inaudible].
>> Christos Papadimitriou: No, no, no. No. But you could say that they are
disjoint, yes. They, in fact, are going to be two disjoint substance. Because you
can always subtract the intersection and they will still -- yeah.
>>: One and two, how do you [inaudible].
>> Christos Papadimitriou: Two to the N -- sorry?
>>: [inaudible]. Allow the empty set?
>> Christos Papadimitriou: Yes, yes, that was empty set, yes. Of course, yeah.
Sorry. You know, you found a counter example to all my theorems. So this one
is safe, okay. [laughter].
So here is another interesting one. You are given a graph that has all the green
nodes. All vertices have node degree, okay. And then you are given the
Hamiltonian cycle. Sorry. And you fix an edge of the Hamiltonian cycle. The
theorem says that there's going to be another Hamiltonian cycle to the research;
in fact there's going to be an even number of Hamiltonian cycles to every edge,
possibly zero. Okay? So that's called Smith's theorem.
How many of you know Chevalley's theorem? Oh, okay. You should know
Chevalley's theorem. So imagine that I give you this. I give you two equations.
Okay? I give you two equations modular two. Okay? It so happens that the
number of variables is larger than the sum of the degrees. The number of
solutions to the equations is even. Okay? It's the same if you want modular P is
divisible by P. Okay? Okay. [Inaudible] is too much. I'll show you proof of this.
Okay. Actually -- yeah?
>>: [inaudible].
>> Christos Papadimitriou: Yeah. So the only hypothesis is that the number of
variables is larger than the sum of the degrees. The sum of the degree of these
two polynomial, of these two equations is one and two. Okay? So three and
there are four variables. That's all you need. So this equations always have the
homogenous solution, everything zero. Well, you know you have another one,
because they have an even medium of solutions. Okay. So you want to see the
proof of that. It's you know, probably one or two minutes. Okay.
So here is -- here is -- you create a bipartite graph. Here you put everything in
01 to the N. N is the number of variables. So all possible solutions. Okay? All
-- the whole solution space. Here you do something really weird. You put all
terms of the following horrible function. So the product of one plus the
polynomial.
So you take the polynomials, you add one to them, and you take the whole
product. Okay? Now, this product, every polynomial is going to be like this,
right. So if you multiply this, you are going to have a train which is X square Y
and so on. Okay? Everybody with me? So basically everything is going to be
there. Okay? The point is that something is the root of this -- of these two
polynomials, even if it is not a root of this. Okay? So it's a funny thing about
modular two. Okay?
Now, what you do here is you say you draw an edge, even though F is a root of
-- this is a root of this, okay, of the particular term. Okay? Two observations.
One, the degrees here are even on this side. Why are all degrees even? Here is
why. Now I have to use my hypothesis, right? I have all new hypothesis, I better
use it. The only hypothesis I made is that the degree is smaller than the number
of variables. Okay? So consider a term here, all right? Its degree is smaller
than number of variables, okay, because therefore a variable is missing.
Therefore this variable, no matter what other roots this one has, can be changed
again so random, right. So the degree is even above that.
Another claim. The possible solutions, the true solutions are the ones that have
all degree, okay? All degree means what? Means that it is not the root of this,
and therefore it is the root of this. Okay? So okay. So this is -- okay. True
story. You know, the legend is the following, that there is -- there is a
seven-page thesis somewhere whose title is a short proof of Chevalley's
theorem, and it's actually what I showed you. Okay. So I have not checked it.
It's apparently the mathematics department at Princeton, okay, so it should be
easy to go there and see. Okay? All right. So that's -- that's Chevalley's
theorem.
Let me tell you another one. It's called Hopfield's theorem. Which is the
following. Image that you have a graph, these four people, a social network, let's
say, and what you have -- the graph is weighted so you have three, one, two
minus one minus seven. Okay? And basically these numbers say how much
these two nodes like each other, okay? And here the questionable assumption is
that it's symmetric, okay? Everybody with me?
So here's what Hopfield says. That there is a way to subdivide this -- these
social networks into two classes in such a way -- so my happiness depends on
how many of my friends weighted are in the same cluster as me and how many
of my enemies, again weighted, are in the opposite cluster, okay?
So there is a way that -- to subdivide in the two classes so that everybody is
happy. Okay? And the reason is the following: That if you subdivide it into -- in
a way that is not good, let's say, suppose you subdivide it this way, all right, then
this node is happy, okay, because his friend is in the other cluster but his ex-wife
is also in the other cluster, so it's okay, right? But this node, for example, is -this one is happy. Wait a minute. Okay. We found the second one. Okay.
Right here. You know. But the point is that if you make something that is not
good, so let's find something that is not good, maybe I made too easy an
example, okay, let's say this, okay. Then you can make -- you can improve on
the total happiness by having somebody who is unhappy switch.
So basically the basic fact about this system is the following. That if somebody is
unhappy there is a number to this unhappiness, and if this person switches
clusters the total unhappiness -- the total happiness increases by this amount,
okay? To put it otherwise, there is a potential function here. Okay? So we know
that -- and the -- and the rational moves of the people sort of basically improves
this potential function. Okay?
All right. So what do these seven theorems have in common? And, incidentally,
here's I use only equal sums but there is -- there are two other theorems that I
could mention which is Minkowski's theorem and [inaudible] theorem which are
sort of -- which are less elementary versions of the same phenomenon. Okay.
So what's my point? These are seven theorems. Some are extremely well
known, famous, some are not. Some are synthetic. What do they have in
common? Okay. Here's what they have in common. All of them are existence
theorems. Okay? For all that exists. That's what they say. All of them what
they have in common -- so, you know, so there are even more famous existence
theorems. For example, the fundamental theorem of algebra, every polynomial
has a solution in the complex numbers. Okay.
Well, the point is that these guys are different. These theorems are essentially
non-constructive existence theorems because their proof is such that we don't
know how to -- how to find -- so, you know, it's not evident how to find the proof,
you know, like, for example, okay, I'll give you the proof of Chevalley's theorem.
Okay. Doesn't tell you sort, you know, how to -- well, it gives you actually an
algorithm to find the second solution. But this algorithm is exponential. Is
everybody following what I'm talking about?
So what I'm saying is that all these have in common is that they are exponential,
exponentially non-constructive theorems. Okay? And it turns out that these and
many other theorems like them ->>: [inaudible] might just keep increasing [inaudible] by one?
>> Christos Papadimitriou: Yeah, yeah, yeah. Exactly. You could -- in other
words, these numbers could be millions and sort of you know it's such a terrible
thing that you keep increase by one. Actually, it's an intractable problem to find -okay, to find the Hopfield partition.
Okay. So the point is that there are very few arguments, okay, that all of these,
all of these theorems -- and actually all exponential non-constructive theorems,
mathematics, and this version, mathematic sophisticated [inaudible] so I want
your correction, okay, so, you know, the things that I'm leaving out. But all of
them rely on the following four principles, okay? [Inaudible] is that okay?
Yeah. So, number one, every graph has an even number over degree nodes. I
mean, that's trivial, right? Well, it turns out that if you look under microscope,
Smith's theorem goes here and also Chevalley's theorem goes here. Okay? So
what I'm saying -- I'm not saying that that's a difficult -- I'm not saying that this is
a difficult principle. I'm not saying that these theorems are easy, okay? I'm
saying the following, that all the theorems, non-constructive theorems that I
know, what do they do, their proofs, okay? They do hard math, hard math, hard
math, hard math, then they reduce the problem to a trivial graph theoretic
situation and then they invoke one of these, and then they do hard math, hard
math, hard math, and they continue. Okay? You understand what I'm saying?
That's -- so that's one.
Here's another one. Every -- if a directed graph has a -- an unbalanced node,
then it has another. Okay? You see -- so this is again trivial. I mean you have a
directed graph, finite graph of course, okay. And this directed graph has a node
where the N degree and L degree are different, okay. Somewhere in the graph
there must be another unbalanced node, okay? That's all this thing is saying.
The third one is the Pigeonhole principle. And as many of you suspected, this
guy is the Pigeonhole principle. As none of you suspected, factoring is the
Pigeonhole principle. And if you want I can tell you why. So, you know, factoring
is essentially randomized reductions reduced to the Pigeonhole principle. That's
a result by Yatserius [phonetic], an observation by Yatserius so, you know, if you
know anything about it.
And the fourth is every dog has a sink. That's Hopfield. Okay? If you have a
potential argument, it has to end at some point.
>>: [inaudible].
>> Christos Papadimitriou: Okay. Mode P version is more sophisticated, okay.
So mode -- okay. You see, I mean I say that these are the only four principles,
okay. The mode P [inaudible] version, there is a mode P of the first one, there is
a mode P variant, okay? The mode P variant is the following.
If you have a bipartite graph, okay, and the -- and the -- one, it has a node, either
side, that -- whose degree is not a multiple of P, then there must be another node
whose degree is not a multiple of P. But you have to have a bipartite graph.
Because, for example, for a non-bipartite graph it's not, okay. You know, it's -you know, for two it's true for non-bipartite graph. But for P -- for N greater than
two, you need to go to bipartite graph. Okay?
Okay. So let's say also mode P, also, yeah, mode P. Mode M. There is another
-- there is another little variant of this. The graph has an unbalanced node, then
it has another of the opposite kind, okay? So both theorems are true. This one
is slightly stronger, okay? It tells you that it's going to be -- you know, you
actually know something, something a little more, okay?
Okay. Now, here comes -- you know, I don't know how many of you know this,
but there is a complexity theory based on this, okay? And the complexity theory
is this, is the following. It says basically that you have NP, then you have
something called TFNP, which, roughly speaking, it means the following. It's
search problems in [inaudible] where you are looking for something, you are
looking to find something that constraint satisfaction problems you wish that are
always guaranteed to always have a solution. Total functions in NP, okay?
They're always guaranteed to have a solution. Everybody with me so far?
So now the point is that there are subclasses. You know, we love to study this
class, okay, because this class contains all these beautiful, interesting, important
problems, all right? But we have to now -- we have to partition them, okay, so we
know. So there are some -- there are other classes. This is called PLS, for
example. It corresponds to the potential argument. It is something called PPP,
which is the Pigeonhole principle. Subclass of it is something called PPAD.
Actually PPADS and PPAD. PPAD is the class that contains Brouwer, Nash, and
in fact, these are complete for PPAD. In other words, you can't do -- PPADS is
the variant that corresponds to of the opposite kind, okay? So it's slightly more
sophisticated, so it's slightly both.
PPA is the parity argument. Okay? And the -- basically, this is the world as you
know it. And so, you know, downwards, downwards inclined I just mean
inclusion, okay?
Any questions? And of course it's humbling to remember always that these two
could be the same, okay? In which case you know we are talking hot air, okay.
Any questions?
>>: Can you give an example of [inaudible].
>> Christos Papadimitriou: Okay. Excellent question. The question is is there
an example of TFNP complete problem. The answer is no. And for a sort of like
robust reasons. The TFNP is a kind of class that is called semantic sometimes.
And this means that it cannot have N complete problems. You know, of course if
T constant P, then, yeah, but I mean if the world is as we think, it cannot have
complete problems.
>>: [inaudible].
>> Christos Papadimitriou: It's -- you know, it's not in TFNP because two graphs
may or may not have a -- may or may not be isomorphic, right? So in this we
have restricted ourselves to problems that always have a solution, okay?
>>: I have a basic question. If you already know the answer, how come there's
a problem here?
>> Christos Papadimitriou: Okay. That's an amazing -- that's a -- okay. I'm
delighted, I'm delighted to ask -- you asked this question, okay.
So the question is, you know, since we know the answer -- since we know there
is an answer, okay, what's the problem? Okay? The problem is for example,
okay, for factoring, okay, of course we know the answer, okay. But we don't
know how to find it. Okay? So I mean I know it's -- it's tempting to assume that if
we know there is an answer, we are motivated to find it, right, I mean, it's like
when you know the theorem, that's the difference between proving a theorem
and solving a problem in math class, okay?
>>: So I understand that. But you are sort of comparing NP to a class of
decision problems to something else where ->> Christos Papadimitriou: No, no, no. So okay. Good point. Yes. So I'm more
interested in search problems, okay. So NP for me is given a boolean formula,
find me [inaudible] term it doesn't exist. Okay? So I'm more interested in -- yes,
so I have sort of switched my point of view, okay. So these are like -- yeah?
>>: So TFNP doesn't have a complete ->> Christos Papadimitriou: Doesn't have a ->>: [inaudible].
>> Christos Papadimitriou: No, no, it's a semantic class. Semantic class means
it's -- so when -- you know, when in complexity do we always have a complete
problem, okay? We have a problem when -- a typical problem in this class.
When we see it, we know that -- you know, for example, okay, why do NP? Why
does NP have complete problems, okay? Because you can give me a
nondeterministic polynomial bounded Turing machine, okay. And I look at it and
say, yeah, it's non-deterministic, it's polynomially bounded because it has a cloak
which is standard [inaudible]. So, yeah, you know, that's a problem with NP.
Okay?
And starting from this, from this technology, you can find immediately complete
problem. For TFNP you give me a little bit of polynomial bounded Turing
machine, but how do I know that it's a total function? How do I know that it
always has, no matter what the input is, that it always has ->>: It doesn't have an obvious reason.
>> Christos Papadimitriou: There is no obvious reason. Or if you wish, whether
a given machine is in this class is undecidable, okay. Yeah?
>>: [inaudible].
>> Christos Papadimitriou: Precisely, yes. So this is the motivation, okay, you
know, whoever asked me if TFNP complete program. This is exact motivation
why we started this, okay, because we want complete problems, okay?
Complete problems basically are the completeness is the fabric that takes -abstracts concepts and problems and puts them together, right? For example, all
of these have complete problems, okay? So which ones have natural problems,
complete problems, okay? This has natural complete problems. Hopfield, for
example, is complete. But a dozen more. I'll mention a couple.
This has natural complete problems like Brouwer, Nash, and many others. This
has natural complete problems, which means basically find me a Brouwer fixed
point of the right polarity or a Nash equilibrium that is the right sign. I mean, so
it's -- so the rest they have natural complete problems, but -- they have complete
problems, but they're not natural, okay. So I'll tell you, you know, their complete
problem is how these are defined, okay?
So I'll tell you, for example, how PPP is defined, okay? The way you define a
class like this is by presenting a complete problem and then say, okay, now
anything that is reduced to it, that's the class. Okay? So what is the problem for
PPP? Here is the problem for PPP. Okay. So it's good to go through this
exercise once. And for PPP is particularly straightforward.
What I give you -- so this is a complete problem for PPP, for the Pigeonhole
principle, right? I give you a [inaudible]. I give it to you explicitly. I'll give you all
the gates and everything, sort of a huge mess, okay, separate. It has N input, N
input bits, N input wires and N output wires. Okay? So this is X and this is C of
X. Okay?
And I ask, given C, find me either an X, such that C of X is 0, or otherwise X and
Y -- X different from Y, such that C of X equals C of Y. If you think about it, that's
the Pigeonhole principle, right? So by the first clause I disallow you zero. So
basically now you have 2 to the N minus one pigeons and 2 to the N pigeons and
2 to the N minus 1 holes, and of course there has to be a collision. Okay? Now
what is PPP? It's everything that is -- that is reduced to that. Okay?
Incidentally, you can modify this. You can say I want the first B to be zero or I
want -- and so on, the first two bits to be zero. Or I want it to be either one of
these five outputs, okay? This creates sort of easier subclasses of the
Pigeonhole principle, okay, where basically you have way fewer pigeons -pigeonholes than you have pigeons, okay, and presumably it's that easy, okay,
people have been studying those.
Okay. So is it clear? So for each one of these 1, 2, 3, 4, 5, we have -- we have a
complete problem like that, and basically the definition is that [inaudible] but
these three are particularly gifted because they have natural complete problems
like Brouwer, Nash, Hopfield, and a couple of dozen more. Yeah?
>>: You going to tell us what natural means?
>> Christos Papadimitriou: No. Okay. I'll tell you what natural means. Natural
means the problem that I knew before I thought about this class. Okay. For
somebody new, okay, and so on. I mean, so it's something that -- that is lying
around legitimately. It's not something you create just for the purpose of proving
a completeness. Sorry?
>>: Reductions are polynomial time.
>> Christos Papadimitriou: The reductions are polynomial time. And of course
they have to be have a kind because these are such problems where I give you
an instance, you give me another instance and also a way for a solution of this
instance to recover a solution for that instance, okay? So it has to be of this sort.
Yeah. Polynomial time inductions, yeah. Any more questions?
So I've been covering a lot of ground. Let's see. All right. So this is something
I've been doing for the past 20 years, all right? So but today I want to tell you
something that I'm doing now with my former student Costos Lascalagis
[phonetic] who is teaching at MIT now. And we looked at some other problems.
And we couldn't classify them there. In fact, we could, but we could classify in a
quite unsatisfactory way which you realize immediately, okay? So let me give
you a few.
So the contractual map fixed point. What's contractual map fixed point. Brouwer
fixed point says any continuous function, okay? Now, suppose that I -- further I
require that the function is such that F of X minus X, in other words -- sorry. The
way to write it better is F of F of X minus F of X is smaller than F of X minus X. In
other words, if you follow -- so you know, the things get closer and closer and
closer together, okay? So first of all this has an easier proof of you don't need to
go Brouwer to do -- go through Brouwer's proof, okay, it has an easier proof.
Second, it's considered easier to find, okay?
>>: [inaudible].
>> Christos Papadimitriou: Sorry. Yes, right. Even though people -- yes, but
you are right. Any compact space with a metric, that's all you need, yes, exactly.
Yeah. I mean in Brouwer you don't even need convexity, I mean, you need
something else. Something weaker. In any event [inaudible].
>>: [inaudible].
>> Christos Papadimitriou: Yeah, yeah, yeah. Not to have holes. Yeah. So -right. So contractual map fixed point. So this -- so in some sense it is a subset -well, where does this lie, okay? It is a subset of -- let me write it, in fact.
[Inaudible] C where C is less than one. Okay?
>>: [inaudible].
>> Christos Papadimitriou: That's even -- yeah ->>: But the other one isn't compact. This is two incomplete ->> Christos Papadimitriou: Incompletes. Yes. So let's take this, all right? So in
some sense it is a subset of -- you know, it is somewhere below PPAD. But it's
also somewhere below PLS because there is a -- there is a potential argument
here, okay. And the potential is the distance, okay? It gets closer and closer and
closer, all right. Okay? So that's one problem.
Another problem is the simple stochastic games. How many of you know about
the simple stochastic games? So they are -- they are -- there is a subculture that
is completely obsessed with the simple stochastic games, okay. And the reason
is that they are very simple, okay.
So if they were defined by Shopley [phonetic] back in the '60s and then they were
studying from computational standpoint earnestly by Ann Condor [phonetic] and
essentially sort of after serious work done by Carpen Hoffman [phonetic] you can
reduce it to the following -- to the following situation, all right? Imagine that you
have a graph, okay, directed graph. Our degree is two. Okay. Except you have
two node -- special node zero in one of those things, okay? But L degrees to
everywhere. N degrees is once.
And the point is, for example, this guy is between this and this, this guy is
between this and this, this guy is between this and this, and there are many other
nodes, okay. I'm not sure where. The point is that some of them are max, some
of them are min, and some of them are average. Okay? So I give you this
graph, and I give you the labeling, and then I tell you find the value. Find a set of
values that is -- that is consistent. Is this clear?
So I won't -- I want you to find values for example, I don't know. For example,
this perhaps is what, is maybe this is .3 and this is .2 and this is .5. Okay. It's an
average of these two. And then this is between this and this, which is .9. It's .9,
okay, because it's the maximum of .2 and .9. Is this clear? So it's incredibly
simple.
Okay. It's known to have a rational solution actually. It's known that if you start
with all zeros except of course for this poor node one who cannot do otherwise,
okay, and you start -- you start sort of, you know, increasing, it will converge. It's
known that it can take exponential time. There are ->>: [inaudible].
>> Christos Papadimitriou: Sorry. There is a unique solution, yes. What else do
we know? We know that -- we know that it's -- that -- we also know another six
or seven other algorithms that -- and, in fact, there are three or four algorithms
that were eventually proven incorrect, okay? But all of them are exponential.
Okay? So it's -- you know, for such a simple problem, it's very intriguing, okay?
Yeah?
>>: On the previous problem it completes [inaudible] for NP.
>> Christos Papadimitriou: You need to put -- yes, yes, exactly. So how do you
put it to NP. Okay. Good point. So what I give you is I give you a -- okay. Why
is it in NP, okay? Okay. So I didn't say everything. So I omitted stuff. I give you
actually a -- how do you call it? An arithmetic circuit. Circuit. Like a straight line
problem with operations plus times minus constants, min-max, so continuity. So
continuity is guaranteed.
And then I ask you if that confuse the function, okay? Okay? And what I'm
asking you -- but I mean how do you know it is contracting, okay? Then I ask
you, either find me a fixed point or embarrass me. Show me a point where it
doesn't contract. Okay? I'll accept both solutions. Is this clear?
So simple stochastic games. There is something else. Okay? There are a few
other problems. There's sort of like total of, you know, more than half a dozen
problems. So let's -- let's -- something else is stationary points. I don't want to -that's a wild goose chase, I mean, you know, or if you wish, KKT. That's for
those who do multidimensional calculus, the KKT is for those who do
optimization, okay?
So basically I give you a function. Let's say polynomial, okay. I give you a
polynomial, multi-variable polynomial, and ask you to find a stationary point,
okay, a point where the gradient vanishes. Well, let's say this is a polynomial X.
The gradient one says nowhere, okay. All right. So I give you also a -- I give it to
you in the box. Okay? Like let's say in the hypercube. And I'm asking you to
either find me a KKT point or something in the boundary which is like a
maximum. So you can define it, so it's the total function. Okay?
So here is another one. I think I -- I think you'll like this one. Okay? Image
network called coordination game Nash. Find a Nash equilibrium and network
coordination game, okay? So what's a network coordination game. A network
coordination game is the following. That it's a huge network, okay? And every
node is playing -- has the same set of choices, let's say two choices. [inaudible]
okay? All right. These are his friends. All right?
And basically the idea is that for every -- for every edge there is a coordination
game that they play which is basically says if they both choose Mac they get
three, if they both choose PC they get four, if for the first one chooses Mac and
the second PC, one and zero. Okay? But the point is that it's a coordination
game because they both get the same amount. Okay?
And basically I choose Mac or PC here and my benefit is the sum of the benefits
I get from all my friends. Okay? That's the game. It turns out that this always
has a pure Nash equilibrium. And you can do it by potential argument, okay,
that's sort of that if you don't like it switch, okay. The same, you know, Hopfield
like. So it's a generalization of Hopfield likes if you wish. Okay?
>>: [inaudible] everybody to be the same.
>> Christos Papadimitriou: Right. Right. Yeah, yeah. Yes. Well, suppose that
it's -- you are penalized so if you get the same thing -- you don't impress
anybody, you have everything, okay, so you -- no telling you ->>: [Inaudible].
>> Christos Papadimitriou: No, no, no. I'm assuming that they are diagonal at
all. General game. General metrics, okay? Except it's coordination. Except it's
the same metrics for both. Okay?
So then it's a generalization of Hopfield, okay? But the point in -- in fact it's
complete. It's PLS complete, okay? It's up here. But it's intractable problem.
But then it's a game, so it has a mixed strategy, okay. So it has a mixed strategy
solution. And this mixed strategy is here.
So it's also that PLS intersect NP, okay? Because PLS intersect PPAD. Okay?
So it's both in the PPAD because it's a Nash equilibrium problem and you could
take -- but it's also MPLS because you can always find through PLS a good
solution, an unknown mixed solution.
Another one is congestion games, which is sort of I'm not going to define it
exactly, but so, you know, you have a network, everybody chooses a path
through the network, according to how many chooses -- choose -- you know, go
through every edge and it has a congestion, everybody -- everybody suffers as
much as the sum of the congestion over the path that they chose, okay?
So this is a game, a game that's in the -- you know, it always has through a
potential argument a solution, a pure solution. It's PLS complete to find,
intractable, but it's also PPAD. So it's somewhere in between, okay?
>>: [inaudible] Nash equilibrium when you say PLS complete do you mean just a
pure Nash equilibrium?
>> Christos Papadimitriou: Pure Nash equilibrium, yeah. But I'm interested in
any Nash equilibrium. Okay. So it's both.
And then there is something called PLCP. And as you can guess from the name,
it's an esoteric problem, except that it's -- again there is a very fanatical
subculture, sort of, you know, who -- the whole life, okay? This is for [inaudible]
linear programming optimization. It's called the linear complimentary problem,
okay, generalization of linear programming. Okay? And the -- and the -- I could
define it, but sort of, you know, I'm running out of time.
And PLCP is the linear complimentary problem which is a classical problem was
defined by Dancer [phonetic] in the '60s. And it has a -- it generalizes in a very
interesting direction of programming, except that it's semi-complete in general.
But when a matrix is a P matrix, and P matrix means that all -- all principal majors
are positive, have positive determinants, all [inaudible] positive determinant, then
-- then it's easier. Okay? For example, it's PPAD. And, you know, that I knew
for 20 years now. Okay? Well, if you work a little harder, and a few people have,
you can prove that it's also in PLS. So everything is in PPAD, intersect PLS.
And there are a couple more that I did not mention.
So it seems that we have an interesting class down here. PPAD intersect PLS.
[inaudible] okay. And this has a lot of interesting problems. Okay? Except that
that's a -- how should I say? That's a [inaudible] okay, it's in the intersection of
this, so it's in the -- it's not -- it's not a -- you know, so it means that it can be
here. It's both here and here, okay. So it has an argument like this and it has an
argument like that, okay?
And the probability of finding a complete problem is diminimus, okay? All right.
Long story short. The new result is that there is a neat class here which you call
GELS, for geometric local search. Which contains all of this. And has a very
interesting, very interesting definition. And it's quite robust, okay?
And I'm going to give you this definition in close, okay? And the definition is sort
of, you know, very, very natural. Okay. What is PPAD? PPAD is a given F
continuous, find X such that F of X minus X is very small, actually something that
you are given. Or embarrass me, convince me that it's not continuous, okay, it
has a Lipschitz discontinue it.
What is PLS? Okay. Since you are talking about PPAD intersect PLS, I'm going
to -- I'm going to use sort of like real analysis again, real value functions again,
okay? PLS is the following. Given two functions in the cube again, F and P, F
and phi, F is the function and P is the potential, okay?
Not continuous. Agnostic about this, so I know, maybe -- you know, probability
continues. That's not the point here. Because you see in PLS it's a discrete
problem, right? I mean, so it's a graph. So it goes from anywhere it wants, to
anywhere it wants. There's no continuity. There is no metric. I mean, there's no
way to talk about continuity.
Find X such that phi of X is fee -- sorry, of F of X is greater than or equal to phi of
X minus epsilon. That's the natural way to define PLS. Okay? Here is why.
Because that's the potential argument. It means that I start from X, okay? If this
does not hold, I go down by at least one epsilon every time I go through. Okay?
So I mean I start somewhere, sooner or later I'm going to find the fixed point.
That's what I mean in a block -- a fixed point of F with respect to phi. Okay?
So basically what I'm saying is that there are two kinds of fixed points, Brouwer
fixed points and fixed points with respect to potential function. Okay?
>>: The continuity, how do you ->> Christos Papadimitriou: There's no continuity here.
>>: No, no, I mean the first one, the PPAD, do you assume that two points ->> Christos Papadimitriou: And so you're also given a Lipschitz constant. And
you are saying sort of, you know, either find me a fixed point or show me two
points that violate the Lipschitz continuity. Okay? So to make them
computational, you have to simplify a lot of things. You have to make that
assumption. Yeah.
All right. So but do you see what I'm saying here? I mean, so what I'm saying is
there are two -- okay. So here is what GELS is about. Given F and phi
continuous, find me a fixed point of either kind. So I give you -- I give you two
functions. They are both continuous. And I'm telling you, the function F has a
fixed point for two reasons. One, it's continuous; and, two, it has a potential.
Okay?
And it turns out that all of these problems belong to this class. And in all of these
problems, for example, okay, let me show you for KKT, okay, why is this the sort
of a very natural idea? For KK -- for stationary point what I'm saying is that the
function F -- I'm sorry, the function phi is the function you want to find a stationary
point of, and the function F is a gradient descent or some kind of gradient
descent which you have made continuous. And you can't do that.
>>: Why isn't the first one stronger than the second? Because if FX and SX is -[inaudible].
>> Christos Papadimitriou: Sorry. What do you mean? This? They're not
stronger. They're dependent.
>>: No. But if FX is close to F, then [inaudible].
>>: [Inaudible] continuous.
>> Christos Papadimitriou: No, it's not continuous.
>>: No, but given FP continuous, which you are assuming now in GELS, why
doesn't the first kind of fixed point imply the second kind?
>> Christos Papadimitriou: It doesn't. You're right. You're right. Yes. That's the
whole point. Yes. Everything converges. There is only one reason. What I'm
saying, you have a Brouwer function but you also have a potential. On top of it
you have a potential to help you.
>>: What is it just [inaudible] the second kind.
>> Christos Papadimitriou: Sorry? Yes, yes, yes. So it turns out that finding a
fixed point of either kind is the same. So thing about -- think about stationary
point, okay? You know, you find a fixed point, what's a fixed point. It's
something where gradient descent tells you, you're in good place, don't move.
And it's also a point where the potential does not approve. Yeah?
>>: Do you have any evidence that GELS is below the intersection?
>> Christos Papadimitriou: So that's -- okay. I'm writing the paper. That's open
problem number one. There should be a way. There should be a simple way to
use oracles to prove that GELS is properly in the intersection. Okay? Because
intuitively the intersection tells you you have two different functions, okay. Find
either this or this. This tells you I give you one function that is -- that has the
same reason, okay? So they're mathematically very different things, and
therefore there should be an oracle that separates them. But thanks. That was
very perceptive.
Okay. And the second open problem is can any of these be complete, and my -and my -- my talk can't do the subject, but these have enough power and they
seem to be hard enough they could possibly be complete for GELS. Okay. So
that's the whole story. Thank you very much.
[applause].
>> Nikhil Devanur: Questions?
>>: [inaudible].
>> Christos Papadimitriou: Yes, for example, yes. Good point. Many of these
have polynomial time approximation schemes. In other words, you can -- for
example, contractual mappings, mapping has a recent result due to -- by Young
and Brook [phonetic], yes. Has a [inaudible] so we know that solves it in order of
N dimension of [inaudible] square. All right? Right? So because there's a
square root of N. Right?
>>: [inaudible].
>> Christos Papadimitriou: Okay. You're right. Yeah, but I mean presumably
you have to do some kind of [inaudible]. Yeah.
So it will be wonderful if all of these have -- for stationary points there are for
some special cases we know how you have polynomial time approximation
schemes. But only for special occasions. You know the stationary point is
another sort of you know, very -- you know, what's the point? What's the
difficulty? You just show the system of equations, right? I mean, you just take
the [inaudible] show them. Well, we don't know how to show [inaudible]
polynomial equations, right? That's -- yeah?
>>: [inaudible].
>> Christos Papadimitriou: Yes, of course. So this is complete.
>>: There are cases when [inaudible] so I think about [inaudible] for average
case.
>> Christos Papadimitriou: Right, right, yeah.
>>: And then you reconstructed a very complicated example, but eventually you
found ->> Christos Papadimitriou: Exactly, yes, precisely, yes. So all of these started
semantic, but after a lot of work you can make them syntactic and find complete
problems, yes. Great. Thank you.
[applause]