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>> Kamal Jain: Today we have Christos Papadimitriou, who will be giving an informal talk on optimal
options. And Christos, these are your true fans that are [indiscernible] [laughter].
>> Christos Papadimitriou: Some of them, yes. [laughter].
Okay. That was funny. All right. Okay. So Yuval asked me to give a talk, and I thought I should tell you
about something I'm working with a student. The student's name is George Perakis (phonetic).
And it's about optimal options. It's an interesting problem. It's a problem that was sort of in a famous
problem when I was in grad school in the late '70s, and then in 1981 Roger Myerson solved it, and he got
the Nobel Prize 30 years later.
And so is there something else to say? So we'll find out. Okay. So the problem is the following. It's very
simple. You have an auctioneer has an item. Let's say a painting. And there are several other people who
are interested in the painting. And he wants to conduct an auction.
So the assumption is that the auctioneer has a prior about -- so the assumption is the following, first of all.
That every individual, every player, bidder, has in their mind a particular number, particular dollar value for
the object, number one.
Number two is that these numbers are drawn from a distribution. So basically there's a solution. And the
Myerson assumes that they come from independent distributions. So in other words that there are every
bidder has one particular distribution. And there are some assumptions about the distributions.
I'm going to assume that -- so let me tell you -- so Myerson assumes that all distributions are always
positive, and as you are going to see I'm going to assume also that. And I'm going to also, for other
reasons, going to assume that the ellipses continues, but let's not jump ahead.
So this is the problem, then. You have these distributions. And how do you design an auction? That's
how this paper was called. Optimal auction design. How do you design an auction in expectation that
maximizes autioneers.
>>: So every bidder has their own contribution?
>> Christos Papadimitriou: Yes.
>>: And as this painting, say, in front of them. And they value that painting according to this distribution.
>> Christos Papadimitriou: No, no. They value every bidder values this painting with one single monetary
value. But the auctioneer does not know this value. All he knows is the distribution.
>>: He knows distribution, the paper so N.
>> Christos Papadimitriou: For N, I know the different distributions. Okay.
>>: But they're not influenced by each other?
>> Christos Papadimitriou: You can actually, in Myerson's original paper, there is -- he has a way to
incorporate his model several things, including influences. And also suppose that I value also the painting
a little. So how should this -- I don't have to give it away. If they offer me $1 for this, I'll keep it.
>>: So Kamal has distribution, right?
>> Christos Papadimitriou: Yes.
>>: So if you poke him he comes out with a number X, say.
>> Christos Papadimitriou: But the point is that Kamal would lie through his teeth to get a painting cheap.
>>: Sure, I know that. But the painting, does the painting determine the value that Kamal will value it at or
does it just determine the probability of value to a certain -- you've got a painting, which is fixed, right. It's
not random.
>> Christos Papadimitriou: The painting is fixed.
>>: Kamal has a distribution. How do they interact.
>> Christos Papadimitriou: They don't interact.
>>: He has a value and a distribution on what Kamal's value is.
>>: For that painting?
>> Christos Papadimitriou: Yes, for the painting. So there is only one painting in the world. And
everybody has --
>>: So basically the action is not sure about how Kamal values the painting. He just has some distribution
in mind.
>> Christos Papadimitriou: He does not have a distribution. He has -- I have a distribution in mind about
Kamal.
>>: Yeah.
>> Christos Papadimitriou: Kamal has a number.
>>: Yeah. So just trying to sort of do it to maximize what your expectation.
>> Christos Papadimitriou: My expectation. What I'm going to get. So basically what I'm going to do, I'm
going to design an auction, which means I'm going to design a protocol, which without generality means
everybody will give me a number and I will have published an algorithm that says what's going to happen
after that.
The algorithm will be deterministic. So let me tell you what ->>: [indiscernible].
>> Christos Papadimitriou: I want to be something that is called ex-post individual rational, which means
the following: That this is ruling out auctions, protocols of the following kind. So I'm filling them out. You
pay me. I'm offering you the following offer. Give-or-take offer. You give me a thousand bucks. And I flip
a coin.
And if it comes heads, you get a painting. Otherwise I keep the painting, or something like that. So this is
not ex-post individual rational, because a lot of people will come out of this auction saying, ah, I should
have stayed home. So I don't want this to happen.
I want everybody to go, to come back home and say all right, that's good. Either I paid nothing but I got
nothing or I paid something but I got something that was even more worth.
>>: The probabilistic thing that you can toss a coin and decide whether to take a thousand dollars from me
or not.
>> Christos Papadimitriou: Right. So but even that, that I'm not allowing. So Myerson allows it. So I'm not
allowing it. So Myerson's theorem is that the solution is deterministic. Good. And I want it to be truthful.
And how many of you are not familiar with what this means? Okay. So -- okay. So let's talk about that.
Truthful means the following: That when at some point I ask you how much do you value the painting,
okay? Each one of you are going to, being very rational. Very mathematically inclined, and very selfish,
are going to think through it. I could tell the truth. I could tell -- I could tell a number which is too high. I
could tell a number which is too low.
Let's see what is the optimum. Then you realize the best thing to do is to tell the truth. Okay. There is
no -- there is absolutely no advantage for you to lie.
Because the algorithm is such. Okay. So I want the algorithm to be such that now this may sound a little -why do economies always assume truthfulness. There are two reasons. One so the real reasons -- the
real, the cynical reasons are two. One is that it's very difficult to do any mathematics in substance,
because if you have to say -- I mean, you've got to speculate about how people are going to think, how
people are going to lie, then there's no end to it.
I'm not saying there could not be an interesting mathematical theory there, but we sort of don't know how to
do. The number two reason is essentially it comes for free. You might as well without lots of generality
assume that it is truthful. Let me -- so a lot of people say, okay, you can't solve that, but what if you remove
this truthfulness assumption could you do better, could there be a better -- the answer is no. There can be
no -- let me give you sort of the -- there is sort of a folk theorem that says that's impossible. Then you just
prove it to you.
So what's the proof? It's very simple. Imagine that you had the protocol, an auction, but it's not truthful.
So people sort of could skim it. Could look at the auction, look at everybody else, and then come up with a
lie, which is the best.
So these are the bidders here. And they could possibly lie to the auctioneer. So now I'm modifying this.
I'm saying, listen, instead of lying, why don't you let a professional lie for you. So basically I'm going to
create a professional liar L-1 for liar, lawyer, whatever. And basically what you do you tell him the truth,
and he's going to lie for you. He's going to represent your interests perfectly, impeccably.
Is this clear? So you tell the truth. And similarly, for everybody, that there is a liar who is going to
represent them. And now the point is that this thing here is a new auction which is truthful.
Make sense?
>>: Provided such a liar exists.
>> Christos Papadimitriou: Provided, yeah. Right, right. So if you assume sort of that lying is, that optimal
lying is not something which is fuzzy and entangled in emotion and so on.
>>: B 1 trusts L 1.
>> Christos Papadimitriou: Exactly.
>>: Might not trust auction prime.
>> Christos Papadimitriou: Might not trust auction prime. So it should trust auction prime, because auction
prime is basically the same as auction unprime with the trusted liar.
>>: The mechanism is verifiable, let's say. They can look and see what option prime is and he can say,
oh ->> Christos Papadimitriou: The mechanism is completely transparent. So in other words if you -alternatively, there's a very detailed contract that says what's going to happen. So you know that. So
either you'll be treated fairly or you are a rich man, you'll go to the courts.
>>: So that's assuming that these lawyers exist, right?
>> Christos Papadimitriou: These lawyers -- so I'm assuming it. But these lawyers essentially are nothing
else but its individual bidders sort of best shot at manipulating this auction. So no matter what is the -- so
this is an interesting result it's called a relational principle. It basically says let's assume truthfulness.
And there is going to be a way to -- so without lots of generality, we can assume that the mechanism is
truthful.
>>: Generality, suppose we have some possibility of repeated auction on the same kind of ->> Christos Papadimitriou: All bets are off. All bets are off. Important, very important and overlooked
often problem. Yes. So basically what Yuval is saying the following. These relation principles say you can
tell, empty or so, to a lawyer. The point is if this is a repeated auction, and not only that, but it's a repeated
auction so that every day it's a new auction.
And every day, of course, the last day, yesterday, you want the auction, and you looked at the painting.
Sort of yesterday's painting, and maybe you have decided you have updated. And also you open in
general you saw how the market is going.
If there are dynamic aspects, okay, then this picture, it's still true. But it's come computationally suspect
because this means exponential communication. So it could be that sort of that polynomial auction. So
there can be that optimal polynomial auctions are not truthful. That's a possibility that has not been looked
at. Okay. But you're back in the world where there is sort of one item, one time and so on. So let me tell
you what this says. This says the following: That basically so already there's severe restrict -- let's look at
the following.
So let me summarize first the result of Myerson. Myerson's result is a beautiful design for auctions.
Basically says the following: Every bidder looks at his distribution. And what he does, he does a very
complicated sort of calculation, sort of mathematical calculation with integrals and so on and quite
complicated. Quite sophisticated calculation.
And out of that comes something which is called the ironed virtual valuation. It's a number. And now this
bidder, this number basically is -- this bidder -- basically what's going to happen he's going to submit this
number. Not a true number. Not the thing that he has in his mind.
And it's sort of a very surprising result. And then the auctioneer will give to the one who has highest
number there, and will charge him the second higher.
So you are going to have the auction with this ironed virtual valuation.
>>: Everybody knows all the distributions?
>> Christos Papadimitriou: Everybody knows all the distributions, yeah. So that's the result of Myerson.
What I'm going to tell you is what I'm working on is the case where you don't have independent
distributions.
So let's -- and in fact without lots of generality, let's suppose that the valuations are between 0 and 1.
Okay. So you are playing -- so you have distribution phi of XY in the unit square.
So I'm going to assume that this is possibility. And for the longer -- Myerson assumes in his paper that it's
positive. And for the longest time I said this is such a silly assumption. Why do you need this?
So I note it. And sort of several weeks later, trying to write my proof, I adopted this assumption. So it's
very -- and also lambda, ellipses continues. Because you think about it, if you have a continuous
distribution, and you are going to at the end of it you're going to analyze by algorithm, you don't want to be
sort of to have arbitrary spikes because sort of you don't want to, no matter how much you probe to miss
the action.
So that would be terrible. That would make a distribution.
>>: With that exponent.
>> Christos Papadimitriou: What?
>>: Is that XY?
>> Christos Papadimitriou: XY are the true valuations. This is X. This is Y, and these are the valuations of
let's say between 0 and one million of the two bidders. So let's -- because you gave me two dimensional
board, let's not talk about the two bidders.
>>: Just in the previous discussion it was the ironed values, how is that still truthful?
>> Christos Papadimitriou: It's still truthful because you don't -- first of all, because this is a faction of -- it's
still truthful.
>>: It's a known function. So the auctioneer ->> Christos Papadimitriou: No, no. It's a function also of what these people have. So basically they look
up now in the table what they should bid. But it's to their advantage to bid truthfully. And also ->>: Bid truthfully. But what they bid is not their own value.
>> Christos Papadimitriou: Not their own value.
>>: What do you mean by truthful?
>> Christos Papadimitriou: Truthfully means the truthful entry of the table they get.
>>: But the auctioneer can look at, can compute the same table.
>> Christos Papadimitriou: No. No. Because you need to have -- you need to have your value.
>>: But he can go back and compute the value from the ->> Christos Papadimitriou: Yes, yes. Sorry. But the easiest way to present it is this is G on the ironed
valuations.
>>: Now I understand.
>> Christos Papadimitriou: Okay. But, what I'm going to do here it's going to be completely different. And
actually I don't know how to get Myerson's result from this without doing again Myerson's work. So from
what I'm going to tell you.
Okay. So now these assumptions already tell me something very -- restricted immensely the valuations,
the solution. Because basically what I want is imagine that I decide to give this to bidder one. In other
words, you understand what I mean? It's going to be deterministic. So basically I'm going to say give me -so I decide that my is going to be give your values.
This is the determinism point here. Every point is painted either blue or red and I'm going to either give it to
the blue guy -- here's the blue guy, or the red guy.
Okay? Now, here's the point: That because I want to be truthful, basically this means the following: That
this guy would have no incentive to bid higher. What does it mean when it says to bid higher? That means
if he gets it here then he gets it all the way here.
Right? It's sort of -- it's not trivial. It's very -- it's very classical. Proves first in Myerson's paper. Then
something else. And then what am I going to charge him? What I'm going to charge him is, of course, if Y
does not change. If Y changes then all bets are off.
But what I'm going to charge him, the point I'm going to charge him the smallest amount by which he would
have gotten it. Because this gives incentive of not to lie below, because he's protected from below.
There's no incentive to lie. So he will tell me the truth, if this is the case.
So everybody understands that? That's very crucial. That once I assume this, this means the deterministic
means that every point is either blue or red. And the ex-post individual rational basically means that
nobody's going to pay more than his value.
>>: So I know my number X.
>> Christos Papadimitriou: And so I know something -- I know something about your distribution.
>>: I assume that ->> Christos Papadimitriou: Yes, of course. Yeah, yeah, everybody knows. I know how you are, yeah.
>>: That means but for the auctioneer we're doing simultaneous [indiscernible].
>> Christos Papadimitriou: Right.
>>: That means the auctioneer doesn't know this factor.
>> Christos Papadimitriou: Yeah. Okay. So basically what this is saying is that so there is such a
function. Let's call it F of Y. Basically it says that all this I give is going to be blue. Okay? And then there
is another function, maybe beta of X.
Basically it says all these I give to Y, to red. That's what it means. These functions maybe doesn't
continue. They could jump around, do very crazy things. What I'm saying for every X there is a beta of X
so that from then up you give it to this guy.
Is this clear? So that's what I'm going -- so basically what I want to do is I want to create these two
functions. That's what the optimal design problem is from now on.
>>: Why is there a gap between the red and the blue.
>> Christos Papadimitriou: There could be a gap. Very excellent question. So why would anybody in their
right mind, especially since, right? Why would in their right mind somebody want to leave this out? Say if
this is then, I keep it and burn it, okay. That sounds so defeating, right? Nobody gets it.
>>: Nobody gets it.
>> Christos Papadimitriou: Nobody gets it. And it's not -- it's no value to me. So I burn it. I throw it away.
>>: You're allowed to not sell it?
>> Christos Papadimitriou: Yeah.
>>: Myerson is the same thing.
>> Christos Papadimitriou: Myerson had it but in his case it doesn't arise too much, only because it was a
value here. He had the reserve price. And that's only because he assumed that the auctioneer may have
value.
>>: Even if auctioneer has 0 value. So he lets it -- it feels uniform Myerson would put value at least half.
>> Christos Papadimitriou: That's right. Yes, right. Right. You're right. Yes. So basically in Myerson's
case, this could look sort of there would be a white square here. There would be a white square. But here
you can have everything. You can have anything. Here let me explain to you why.
Imagine that I'm thinking of sort of whom should I give it? Should I give this to red or to white. And then I
think the following: Imagine that my temptation is to let's say tentatively give it to red -- to blue.
And imagine that this is the lowest point at which I give it to blue. Okay? How much am I getting out of this
decision of deciding that F of Y is this? Think about it. What I'm getting is X times R of Y to 1 P of XY/DX.
P of XY prime DX prime DY. No. DX prime. Does that make sense?
Basically sort of in divided, sort of normalizing by DY, the amount of -- the decision because I'm going to
get X the amount here for everything that is bigger than X for all the mass that's bigger than X.
>>: If you move R of Y to be equal of X.
>> Christos Papadimitriou: If R of Y equals X. Yeah. So I should say X, F from X to 1.
But the point is, okay, maybe there's so little population here. This is so sparsely populated, that by
deciding to giving it to somebody, I'm lowering the amount I'm going to get so much that it's not worth it to
me. Mathematically, what you want to do is you want to have -- exactly. So what you want, by deciding to
tentatively give this to blue, you are getting maximum over all X prime greater than X of X prime times
integral from X prime up to 1 of P of XW prime YDXW prime. Okay. Because it might be -- so this is -- if
you think about this, this is a function that goes sort of like this, okay? It sees max for all this. It's
increasing and then it becomes constant for some time. Constant.
Okay. It's a function like that where the other function sort of if you don't take the max is -- this function as
a function of X prime is something like this, okay? It dips below.
So this is what I'm getting by tentatively deciding when Y is Y to give -- to make alpha Y equals 2 X, okay.
>>: Is that what you meant when you truth X.
>> Christos Papadimitriou: When I.
>>: When you truth the maximizing XY?
>> Christos Papadimitriou: Right. But I don't know what X prime is. The point is at this point that's what
I'm getting. If I decide to give this point to blue, potentially ->>: Aren't you just leaving X times -- X and ->> Christos Papadimitriou: No, because then I may do a sort of -- I may think about it better and do an
optimization. Why do I need to do this. So I'm going to choose the best. That's my point. Yeah.
>>: Is this the same as the auction you think in selling the thing off too cheaply?
>> Christos Papadimitriou: He's selling the thing off too cheap, and sort of and without the benefit of a big
mass here, yes. That's the intuitive what it means. Maybe you want to sell it too cheap if all the action
happens there. So you don't mind if there are some freaks who like it a lot. You want to get the big mass
which is around the half million mark, okay. Even though there are some people sort of who are going to
underpay. That's fine. But there are very few, sparsely later. Maybe you want to sort of -- but if there is
nothing there, you don't want to do it.
>>: The auctioneer is to maximize the bid.
>> Christos Papadimitriou: Is to maximize what you get in there, the expectation.
>>: That's forcing them to bid up.
>> Christos Papadimitriou: Exactly. So what's my point? My ultimate point is that if you take the -actually, the value of this, this is essentially you get something like this, okay?
Okay. These are 0 here where these are horizontal. This is sort of for DX/DY how much it's worth it to you
to give tentatively this point to blue. Okay. And let's call this F of XY. This is basically what you get. It's a
two-dimensional function that tells you sort of what you get per unit here.
Okay. So in other words what I'm saying is that your expected income from blue is going to be Y equals
from 0 to 1. X from alpha of Y up to one of F of XYDXDY. This is what the auctioneer is going to get from
blue.
>>: A few questions. Why isn't the derivative negative?
>> Christos Papadimitriou: So it's -- sorry?
>>: Minus sign.
>> Christos Papadimitriou: Right. So it's always positive. You're right. Yes. So it's always positive
because -- sorry. There's a minus sign, right. Yeah.
>>: And so the ->> Christos Papadimitriou: Should be negative, yes.
>>: On the op side.
>> Christos Papadimitriou: Op side, right.
>>: Doesn't the auctioneer, did you say it gets the second highest price?
>> Christos Papadimitriou: No. So in all this literature, this comes out as a corollary. So the auctioneer,
the auctioneer now is going to say give me your values, valuations, and this is -- so in BCG the protocol is
give your valuations and the winner is going to pay the second highest here. Here is a completely different
protocol, which also has the property of compel to tell the truth but it's a competitive protocol.
>>: This is what you defined to be what the auctioneer gets?
>> Christos Papadimitriou: I'm going to define that, yes. So remember my goal is to fathom the optimum
auction. So I have to find this objective. So this is one part of the objective. And there is a similar for the
second. There is a GXY defined exactly for the red player. DY/DX. And you add those. And, of course,
what you want to do is you want to find the soup of this, subject to alpha and beta this joint. In other words,
you don't want the red and the blue things to intersect. That's all there is to it. Is this clear?
Okay. So that's it. So this is the problem I want to solve. And of course there's a similar completely
analogous for three bidders, four bidders, five bidders, any number of bidders. Except the integrals get
much ground.
All right. Now, I have done nothing except define the problem so far. So what makes me think that I can
solve this problem? So here because I have -- there is -- here comes now sort of the result. And so what's
the idea? The idea is the following: That I have this unit square. I'm looking at a different problem.
With two functions. The function F and the function G. So I'm trying sort of to make, visualize a
two-dimensional landscape. Two two dimensional landscapes. A blue and a red one that exist in the
same. These are the two functions, F and G. The point is here is the idea. Imagine that this is your
garden, okay? You have like a Japanese garden.
And the blue function, F, is what you have. The Japanese garden has this today. And the G function is a
Japanese garden as you want it to be.
Okay. So you have the vision in the morning I want to get done of this garden. I want this new garden,
which is G. So everybody with me? And basically you want to transform this into that. This sounds a little
like Montscolovich [phonetic], but it's not. It's easier. There's a twist.
So you want to transform F into G. And for how do you transform F into G? You have three operations
that you are allowed. One is truck, how do you call it? How do you call it? Whole mash away. So you
take some of the blue mass and you haul it out. The second part of the thing is whole-in mass. So you go
to a quarry and you haul in red mass and you put it where the red thing should be. Plus you have a third
option which sort of saves a lot. You could increase all these two operations but there is a third option
which is going to save you a lot of hauling. And this is haul mass in the garden, only in the southeast
direction.
So you take a grain of blue sand, okay, and you decide where you're going to put it. You can put it here,
for example, but this should be in the southeast direction.
Does that make sense? And basically what you want is to minimize, you want to find [indiscernible] overall
let's call them hauling schedules, of these things, of the things that are hauled in, plus hauled out, plus
hauled southeast.
And I guess you guessed it. These two are the same. So that's duality here. And this makes it tractable
and computable and all that.
Does that make sense?
>>: In the infinite haul in, doesn't that assume there's an F that you're actually trying to find? No ->> Christos Papadimitriou: No, F and G are given. So you're given these two hill configurations. You want
to transform one into the other. Distance doesn't matter. The only thing that matters is how much mass
you haul. So how many trucks in total do you need to do this.
Okay. And so I can give you quickly sort of the core of what's a conclusion from this. There is.
I can then sketch -- once I sketch the proof you'll say, okay, it will make sense to you okay?
So what is -- that's the main theorem.
>>: Both F equals G?
>> Christos Papadimitriou: Yeah.
>>: Then what -- just to understand it ->> Christos Papadimitriou: It should be 0, right?
>>: What?
>> Christos Papadimitriou: It should be 0.
>>: It's 0 but why is the top 0?
>> Christos Papadimitriou: Why is the top 0. What am I saying here? I'm confused. I don't know how to
answer this question. So maybe what I -- obviously what I wrote is not right.
>>: What if [indiscernible] kicks in.
>> Christos Papadimitriou: Yeah, but still you want to -- if F equals G, then you can do anything you want.
You can give all of it to -- give nothing to ->>: [indiscernible].
>> Christos Papadimitriou: You don't have it all in. So it could be that the constant is missing here. Okay?
Which I don't see. So which right now I don't see. But I'll basically give you the proof. And you'll see
what's going on. So now I'm confused.
So it could be -- instead of speculating I'll tell you the corollaries and then tell you the proof. Sketch the
proof. So the corollary is the following, that you can compute the optimal auction for the two bidders two
ways.
Either exactly if you have -- how do you call it -- if you have discrete, if the bid is discrete. So there is an
incongruity between my assumption that P is also known 0 and P is discrete. But never mind that. Or F
bidders if P is continuous, the lambda continuous.
So there is an F bidders. You can approximate it arbitrarily closely if it continues. And another proposition
to complement this it's NP hard for three bidders. And in fact for N bidders, it is next hard.
And there is an old -- and there is a half approximation that's due to Amironin [phonetic], 2000. You can
approximate it by half for any number of bidders, even though for small numbers of bidders, this approach
gives you a better approximation.
So let me now give you 10 minutes worth of the idea. So then we'll see how to fix this. So the idea is the
following: So imagine again forgetting for a moment that P is positive and all that, okay, that your only
masses are in a few points in these six points. Okay. Let's say one-sixth on each point. All the masses -so you can still compute F and G. So that's fine.
And I want this to be the same, maybe not. Okay. Cool. Now, basically what you want, what this tells you
is the following: Now, what you want to do is you want to connect each one of these points either with this
or with that.
That's your dilemma. Okay. And you want to connect this either up or down. The same thing with this.
Right or up. Okay. And the point is that you have computed your F. So every one of these sticks, if you
wish, has a weight, which is the weight it's going to bring to my final calculation, okay. So all the sticks has
a weight. So this is a weight, the G weight of this stick. This is the weight of that stick.
So every stick has a weight. What's your only constraint? The two sticks should not intersect. Okay.
Because if two sticks intersect, you are in trouble. This means that your A and B region are not disjoint.
Make sense? All right. So the easy thing about it is maximum weight independent set for a bipartite graph.
You have a bipartite graph, the bipartite graph are two sticks. The red sticks are one color, the boys. The
blue sticks are the girls. Each boy and girl has a weight. And you want to find an independent set.
Maximum weight independent set in this bipartite graph.
Okay? And the point is that this has a dual max flow. Min cost max flow. Min cost flow. Okay. And late
last night I decided that the min cost flow, you can -- so that -- but there is a min cost flow problem you can
write sort of explicitly and that's what this is what -- okay. There's actually a min cost flow you can write
with integrals. And obviously my interpretation has it back. So my garden interpretation has a back, okay?
Do I make sense now?
Okay.
>>: The nodes in the bipartite graph?
>> Christos Papadimitriou: Excuse me.
>>: What are the nodes in the bipartite graph.
>> Christos Papadimitriou: The nodes in the graph are the sticks.
>>: They're what?
>> Christos Papadimitriou: The sticks, the red and blue sticks. Okay?
>>: You can ->> Christos Papadimitriou: I'm sorry.
>>: A conflict graph.
>> Christos Papadimitriou: And the conflict graph says I should not -- the red nodes are the girls. The blue
nodes are the boys. And this girl -- if they intersect, this means there's an edge between these two. And
this means that it's a conflict graph. You cannot take both. That's why you want an independent set.
Make sense?
And now what this part NP hard for three bidders means, it means the following intuitively: That -- so this
says that there is one -- so there is one little detail which sort of proves that in this rare instance, Gordon
mathematics was on our side, that there is sort of like an interesting detail that this is the incremental cost,
the incremental weight. So the point is that if you take this, you should also take this for that count to be
correct. Because this happened to be the same horizontal line. So if you take this, it's only incremental
cost. And you should also take this.
But optimality gives it to you immediately, and the reason is so that's a little lemma, that since this is a
subset of this, a sub bidder of all this, it intersects fewer red, okay?
So there is no reason why -- if you throw this in, there's no reason to not also throw this in with optimality.
All the weights are the same. So there's no reason -- so modular this, it's exactly the independence
problem for bipartites graph and then you have the dual and then you have algorithm. And the algorithm,
both of these algorithms are sort of immediate.
I mean, to prove this theorem, basically what you do is you go to discrete -- this has an exact dual. Then
you go here. Okay? And basically you have to argue that these are equal plus or minus lambda epsilon
where epsilon is the grid and also plus or minus lambda.
So these are equal plus or minus lambda epsilon. And after that sort of you take the limit as epsilon goes
to 0. And so when I was here last time, I asked James how should -- I want to prove this theorem, how
should I prove it? And he told me you should read about the Hahn-Banach theorem, which is exactly on
the menu. Duality for dimensional spaces. So I told him every time I ask you a question, James, I pray
that the word Banach is not contained in the answer.
Okay. So is there -- I believe that's it, yeah.
>> Kamal Jain: Do we have questions?
>>: So your dual, is that sort of a min cost maximum matching or something?
>> Christos Papadimitriou: Min cost, so -- so it's sort of -- no. It's sort of a maximum match. Here's what it
is. It's a weighted maximum matching. So basically ->>: Minimum cost.
>> Christos Papadimitriou: Minimum cost here.
>>: [indiscernible].
>> Christos Papadimitriou: It means the following; that you should put weights on the edges so that for
every node, the sum of the weights on this edge should be at least as big as the node.
So it's a kind of strange max flow if you subtract the constant. That's where I'm losing this, I guess. So
that's ->>: I'm just thinking if the cost is 1 this would be sort of co-anything's theorem?
>> Christos Papadimitriou: Yes.
>>: It looks like a very special bipartite graph. Maybe you have some sort of greedy algorithm?
>> Christos Papadimitriou: Yes, I actually thought about that. Good point. There is actually literature
about something called convex graph, yeah. So I tried to remember. So if ->>: [indiscernible].
>> Christos Papadimitriou: I tried to remember if this is relevant. But I haven't got it yet.
>>: Saw it last night.
>> Christos Papadimitriou: Really?
>>: Did this last night.
>> Christos Papadimitriou: Yeah.
>>: And if you go for three you get a three-dimensional matching.
>> Christos Papadimitriou: Basically what this NP hardness result says is that if you try, if you want to find
a maximum weighted independent set in where the nodes are three dimensional also segments, and the
intersection is basically, the intersectional graph of this. So a maximum weighted matching in the
intersection graph of three dimensional also segment, meaning parallel to the X axis or the Y axis or the Z
axis, that's NP hard problem. And that's one of these difficult proofs, at least, the way it is now.
So for more than three, you can do it. So there is -- when I interpret this as -- I think that's an accurate
interpretation, modular constant. So I should rethink it, what it is. Thanks for pointing it out.
>>: So do you feel this is a truthful auction?
>> Christos Papadimitriou: It's a truthful auction only because I'm seeking regions. So the blue region is
close to the right and the red region is closed upwards. If that's the only thing -- that's a theorem that
Myerson 30 years ago, that this is all you need in order to have a truthful auction.
So this means if you think about it, that this bidder is not tempted to bid low and is not tempted to bid high.
So he might as well tell the truth. He's not tempted to bid low because he's protected from below. So he's
only going to -- if he bids low, he's going to lose it, which is terrible. Or he's going to follow it in this region,
which he will pay the same thing. So he's not going to gain anything.
And if he bids too high, he's not going to -- he's still -- if he's ->>: [indiscernible].
>> Christos Papadimitriou: Bidding too high is actually risky, because what if he gets it and has to pay
such a price? So if the regions are like that, then -- and precisely then, that's Myerson's theorem, you are
guaranteed to be truthful.
So we started from that, and this gave us this geometric way of looking at things and the schedule covers
the rest.
>>: So it's assumed that everybody knows P of XY.
>> Christos Papadimitriou: It's assumed that P of XY is public knowledge.
>>: Public knowledge. So ->> Christos Papadimitriou: It's not. Yeah. Yeah.
>>: What -- what's the sort of idea to overcome this? Because it seems like you need to know -- even
[indiscernible] what it's worth to X, right?
>> Christos Papadimitriou: Right. Yeah. Yeah. How do you deal with that?
>>: XY becomes common.
>> Christos Papadimitriou: I'm sorry?
>>: If it becomes truthful, it need not be common. The auctioneer has a model of PXY he's charging
according to this curve. It could be suboptimal but truthfulness it does not do ->> Christos Papadimitriou: I see. I see. Just one sec. No, truthful is very common knowledge. So only -sorry. So, yes, Kamal is saying he did not bid common knowledge. It need not be known to the
auctioneer. And I think that's right. So it makes perfect sense, yeah.
>>: So the auctioneer is like a third bidder here, in the sense that if the bids come out it makes one too
small, you won't sell.
>> Christos Papadimitriou: No, he might still want to sell.
>>: Even if he gets into the other region?
>> Christos Papadimitriou: If he has a value to the item, then, of course, yes. Then there is, then there
should be some kind of ->>: So in that region they do ->> Christos Papadimitriou: I'm sorry.
>>: Now you're repeating the talk, but if you're in this middle region he may or may not sell.
>> Christos Papadimitriou: No he won't sell.
>>: He won't sell?
>> Christos Papadimitriou: He won't sell because he's bound by contract to follow this.
>>: Even if the value to him is 0, let's say.
>> Christos Papadimitriou: He will not sell, even though he might get something, and the value -- he won't
sell in deep discount even if the value is 0, because this binding contract that makes him able to extract this
much value. So he doesn't want to play with that.
It's the threat of not selling that makes bidders sort of behave well. It's the threat of burning the item.
>>: I see.
>>: Trying to sell it on Priceline.
>> Christos Papadimitriou: Right. All right. Thank you. [applause]