Algorithmic Game Theory and TFNP

Algorithmic Game Theory and TFNP
Samuel Schlesinger
April 14, 2017

What is Algorithmic Game Theory?
Algorithmic game theory (AGT) is essentially the study of
game theory augmented with problems, assumptions, and
techniques from theoretical computer science, particularly
algorithm design and complexity theory.
I assume we all know a bit about theoretical computer science,
but not about game theory, so we’ll do a quick review.

What is Game Theory?
Game theory in the way we understand it today, as a
mathematical object, was originally put forth as a candidate
for a rigorous theory of microeconomics by Von Neumann and
Morgenstern [2] in the early 20th century.
In this work, a description of the ﬁeld and an argument for the
validity of the underlying assumptions is made.
The sort of games we will consider in this talk are called
simultaneous move games.

Simultaneous Move Games
A simultanous move game (from now on referred to just as a
game) needs the following ingredients:
A set P = {1, 2, ..., n} of players
For any player i ∈ P a set Si of strategies
For each i ∈ P, a way to assess your cost or utility given a
choice of strategy from every other player,
ui , ci : i∈P Si → R, ui = −ci .
For my sanity, we’ll say S = i∈P Si . To play such a game, each
player chooses si ∈ Si and thus we have some s ∈ S which
comprises each of these choices. Each i ∈ P then receive their
outcomes ui (s) and the game is over.

Representing Games
To represent a game we’ll usually think of the utility function
written out in tabular form or given succinctly. In most of the
games one wants to consider, the tabular form will be quite
prohibitive, and so doing analysis on the computational properties
of these games when they’re represented succinctly is quite
important. In the example in the next slide, we’ll see a game which
is much more naturally represented succinctly.

Example: Environmental Policy Game
We define a game where P is interpreted as a set of countries and
∀i ∈ P, Si = {L, N}. We will say that if a country chooses L they
will draft legislation to fight pollution, which will cost them $3,
and if they do not, they add $1 in cost to each other country,
including themselves. Then the cost for player i is
ci (s) = 3[si = L] + k∈P[sk = N], where [p] = 1 if p is true and
[p] = 0 if p is false. The first question we will ask is how we
believe these players will play this game, but first we need to
introduce our first solution concept.

Dominant Strategies and their Solutions
If you’re playing a game as I’ve described them, and you are
thinking between strategies si and si , and you know that for any
strategy vectors s, s , containing si , si respectively but otherwise
completely identical, ui (s) ≥ ui (s ), then you should certainly
choose the strategy si over si . We will say that si dominates si . A
dominant strategy solution is then one in which ∀i ∈ P, si
dominates every single other strategy si which i has. Let’s now
consider our example again in this light.

Example: Environmental Policy Game
I claim that for any player i, N dominates L. How can we see this?
Well, lets assume that the rest of the players’ strategies are ﬁxed
as s−i . We have the cost of i, ci (s) = 3[si = L] + k∈P[sk = N]
and thus:
ci = 3[si = L] + [si = L] +
k∈P−{i}
[sk = N]
If si = L, we pay $2 more than otherwise, so clearly N dominates L
for every given player. This means that we have a dominant
strategy solution s where ∀i ∈ P, si = N. Its also easy to see that
it is unique. That being said, this also maximizes the collective
cost for |P| > 2, so maybe this game wasn’t the best one to use
this solution concept for.

Nash Equilibria
Though we’ve ﬁrst discussed dominant strategy solutions, it should
be clear that not all games will have these. The next solution
concept we will discuss is called the Nash equilibrium. We will say
that a strategy s ∈ S is a Nash equilibrium if:
∀p ∈ P, sp ∈ Sp, up(sp, s−p) ≥ up(sp, s−p)
This means that, given that everybody else behaves as they have
chosen, there is nothing better that anyone can do.

Example: Coordination Game
The simplest game which lacks dominant strategies but has Nash
equilibria is called the coordination game. We imagine two friends,
Lucille and Markus, are trying to decide on their plans for the
night. Lucille wants to spend the night at the house, and we’ll say
her utility will be 1 in this case, as she is being fully satisﬁed,
whereas Markus wants to spend the night at a party and will
receive 1 utility in this case. If either of them end up at the place
they don’t like, then they will be quite sad and receive 1
2 utility,
but they will receive 0 if they don’t end up together, as they really
want to spend some time together. If we give this a little thought,
it’s clear that there are two Nash equilibria, one where they go to
the party and another where they spend the night at the house.

Mixed Nash Equilibria
Sadly, there will as well be games wherein there are no Nash
equilibria, and for this we need to allow the players to randomize
and instead of each choosing a strategy, they will sample a
strategy from some distribution of their choice, which will be what
we call their mixed strategy. We will say a vector of such solutions
m ∈ MS = p∈P Dist(Sp) is a mixed Nash equilibrium if the
expected cost of each player given their distribution is as good as it
could be given that the other players stay the same. Formally, we
have: ∀p ∈ P, mp ∈ Dist(Sp), Esp∼mp,s−p∼m−p up(sp, s−p) ≥
Esp∼mp,s−p∼m−p up(sp, s−p).

Example: Keeper and Shooter
Imagine that Shooter is running up to the net and can either shoot
in the left or right side of the goal, while Keeper is standing there
about to dive left or right. If Shooter shoots left and Keeper dives
left, then Shooter gets 1 utility and Keeper −1. If Shooter shoots
left and Keeper dives right, then Keeper will block the goal and
they will each receive 0 utility. We can represent this game as the
following bimatrix:
left right
left -1/1 0/0
right 0/0 -1/1

Example: Keeper and Shooter
Giving this a little bit of thought, there are no pure Nash equilibria.
On the other hand, in what scenarios will both of these players not
want to change? Well, I bet it is clear by the symmetry of this
situation that they should each choose their strategies totally
randomly, i.e. by choosing left with probability 1
2 as well as right.
Let’s do some thinking together to show how to prove that this is
the unique mixed Nash equilibrium. We will ﬁnd that if Shooter is
working with this strategy, Keeper will be okay with doing
anything, and otherwise they will bias towards one side, and the
same will be true for Shooter. Thus, we know that the unique
mixed Nash equilibrium is when they each choose uniformly at
random.

Mechanism Design
Mechanism design is the art/science of manipulating the
mechanics of situations which actors are in so that we achieve
some desired outcome. Our ﬁrst experiment in this domain will be
in designing auctions to meet various criteria.

Single Item Auction
The single item auction (in a single parameter environment) is a
slight variation on a simultaneous move game wherein the
auctioneer gets to see the bidders moves moves before they make
their own. This game can be described in the following way, given
a set of bidders B and an auctioneer a, and valuations for each
player i, vi ∈ R+.
P = {a} ∪ B
Si∈P = R+
Sa = (B, R+)
ui∈B(s) = (vi − snd(sa))[fst(sa) = i]
ua(s) = snd(sa)

Vickrey Auction
Given a vector of bids b such that ∀i ∈ B, bi = si , the strategy
that the auctioneer takes is to choose the highest bidder and
charge him the second highest bidder’s bid. I claim that this
auction maximizes what is called the social surplus for the bidders
if they play their dominant strategies, that in this situation it is a
dominant strategy to bid honestly, and that no honest player will
ever receive negative utility. First let’s show the latter.

Truth Telling Gives Nonnegative Utility
Assume that bi = vi and bj=i is set arbitrarily. Then if i loses, this
is clearly true, as the utility is 0. If i wins, then they will pay
maxj=i bj , which by deﬁnition must be less than or equal to vi , and
thus i’s utility will be positive or 0 if there was a tie.

Truth Telling in the Vickrey Auction is a Dominant
Strategy
We ﬁx some unknown vector of bids b and assume that bi = vi for
some player i ∈ P. If bi = maxj∈Bbj , then the utility received by
player i is this valuation vi minus second highest bidder’s bid, let’s
say bk. Otherwise the utility received is 0. If the utility is 0, then
by decreasing their bid, their utility cannot increase because they
already are not the max. By increasing their bid to the point where
they win, their utility becomes negative, as the second price will
still be higher than their valuation as it will be the former ﬁrst
price. Assuming that the utility is vi − bk, increasing bi will not do
anything for you, as this utility is independent of bi . Decreasing it,
on the other hand, will risk loss, as if you accidentally go below the
second highest bid suddenly they win and your utility goes from
some positive number to 0.

The Vickrey Auction Maximizes Social Surplus
The social surplus in such an auction is deﬁned as i∈B vi xi ,
where xi is an indicator being 1 if i won and 0 if i lost. Clearly this
auction maximizes this quantity given that players play their
dominant strategies.

What Does the Vickrey Auction Give Us?
1. Dominant Strategy Incentive Compatibility (DSIC)
2. Maximizes the Social Surplus
3. Implementable in Polynomial Time
The question we will now touch upon is: when can we (as the
auctioneer) get these nice properties for other sorts of auctions?

Proposed Methodology
1. Create a polynomial time algorithm for maximizing the social
surplus, assuming truthful bidders
2. Given what we’ve done in step 1, how do we assign payments
to make our assumption true
We will ﬂesh this methodology out for a set of situations called
single parameter environments.

Single Parameter Environments
We still have our set of bidders B and we’ll say that n = |B|. Each
player i ∈ B has a valuation vi ∈ R+ which is the only thing we
don’t know about the bidder as the auctioneer. We also have a set
of feasible allocations X which the auctioneer can choose, where
X ⊆ Rn.

Sealed Bid Auctions
1. Collect the bids b ∈ Rn
2. Choose x(b) ∈ X where xi (b) is allocated to bidder i
3. Choose p(b) ∈ Rn where pi (b) is what bidder i pays
ui (b) = vi xi (b) − pi (b)

Implementable Allocation Rules
Given a single parameter environment for a sealed bid auction, an
allocation rule x : Rn → X is called implementable if
∃p : Rn → Rn, a payment rule, such that x and p together make
our sealed bid auction DSIC. Clearly to use our methodology, we
can only use implementable allocation functions.

Monotone Allocation Rules
An allocation rule x is monotone if ∀i ∈ B, b−i , xi (bi , b−i ) is
monotone increasing in bi .

Myerson’s Lemma [7]
1. x monotone ⇔ x implementable
2. Assuming bi = 0 ⇒ pi (b) = 0, x monotone ⇒ ∃!p which is
explicit and makes x and p DSIC

Proof of Myerson’s Lemma [3]
Given some x, , we’ll try to understand what p must look like in
order for x, p to be a DSIC mechanism. We ﬁx i, b−i . Assuming
that we have such a p, we will take two values 0 ≤ z ≤ y, and
imagine two scenarios:
1. z is i’s true value but they are scheming to bid y.
2. y is i’s true value but they are scheming to bid z.

x implementable ⇒ x monotone
I will invoke the DSIC constraint on each of these scenarios, writing
x(z) = xi (z, b−i ), p(z) = pi (z, b−i ) and see what I can retrieve.
1. zx(z) − p(z) ≥ zx(y) − p(y)
2. yx(y) − p(y) ≥ yx(z) − p(z)
From these two relationships, we can write
z(x(y) − x(z)) ≤ p(y) − p(z) ≤ y(x(y) − x(z))
If x not monotone, ∃0 ≤ z < y, x(z) ≥ x(y). We thus have that
−q = x(y) − x(z) < 0 and we have that −qz ≤ −qy ⇒ z ≥ y,
which is a contradiction. Thus we have shown that x
implementable ⇒ x monotone.

x monotone, piecewise constant ⇒ x implementable
Assuming x monotone, piecewise constant, we take the limit
y → z from above of our inequality and realize that, if there is a
jump of magnitude h at z, the left and the right hand sides both
tend to zh. This means that for any z, the increment of p at z is
equal to z times the jump in x at z. Assuming p(0) = 0, we can
see that we have an explicit payment function
pi (bi , b−i ) =
l−1
j=0
zj · jump(zj )
where zi∈[l] are the breakpoints of our piecewise constant function
xi (·, b−i ) in the range [0, bi ] and jump(zi ) are the associated
jumps.

x monotone, differentiable ⇒ x implementable
Assuming x is monotone and differentiable, we take that same
inequality and divide it by y − z:
z(x(y) − x(z))
y − z
≤
p(y) − p(z)
y − z
≤
y(x(y) − x(z))
y − z
. Taking the limit as y → z, it is clear by the definition of the
derivative that p (z) = zx (z). Using our assumption that
p(0) = 0, we have that:
pi (bi , b−i ) =
bi
0
z
d
dz
xi (z, b−i )dz

Vickrey Auction from Myerson
Given the former discussion, we can now derive the Vickrey auction
easily. We decide we want to allocate our item to the highest
bidder and the jump takes place at the second to highest bidder
and is of magnitude 1 so the payment is that of the second highest
bidder.

Nash’s Existence Theorem and a Detour
John Nash showed in his thesis that for any ﬁnite game, there exist
mixed Nash equilibria. The obvious question for us to tackle here
is: given such a game, can we compute a mixed Nash equilibrium
eﬃciently? If not, more generally, how can we capture the
complexity of this problem?

Capturing the Complexity of Problems like Nash
This question turns out to be very interesting and ends up inspiring
an approach to capturing the complexity of search problems like
Nash which are guaranteed to have a solution. Generally, when one
has an existence proof for some class of objects and one would like
to compute them, they might go to the proof and try to extract an
algorithm which will allow one to compute the existent object
eﬃciently. In the case of computing mixed Nash equilibria however,
it seems that the known proofs do not yield eﬃcient algorithms.

Why Might These Problems Be Hard?
Normally in TCS, when one runs up against a seemingly hard
problem, they would like to understand why it is so. To do this, we
rely upon the paradigms of reductions and completeness. Let’s
say we were to try and reduce SAT to Nash. Well, we’d try to
produce a game such that there is a Nash equilibrium if and only if
the formula is satisﬁable. On the other hand, Nash’s theorem tells
us that if such a reduction were possible, then every formula would
be satisﬁable, which of course is not true. Clearly SAT does not
reduce to Nash at least in the way described.

FNP
We say that a relation R ⊂ (Σ∗)2 is polynomially balanced if
∃c ∈ N, ∀(x, y) ∈ R, |x|c ≥ |y|. We deﬁne FNP to be the class of
relations which are polynomially balanced and that are
recognizable in polynomial time.

TFNP
We deﬁne TFNP = {R ∈ FNP | ∀x∃y, xRy}. This class is clearly
extremely related to F(NP ∩ coNP), which is the class of search
problems which involve two relations R1, R2 which are polynomially
balanced such that ∀x, ∃y, (x, 1y) ∈ R1 ∨ (x, 2y) ∈ R2. If we have
R ∈ TFNP, clearly R1 = R, R2 = ∅ is in F(NP ∩ coNP) and if
R1, R2 are in F(NP ∩ coNP) then we have R = R1 ∪ R2 clearly in
TFNP [5].

TFNP Reductions
Say we have relations R, S ∈ TFNP and we want to reduce R ≤ S.
Given any x, our goal in the search problem defined by R is to find
y such that (x, y) ∈ R. To do this using S, we can map input x to
rx (x), find y such that (rx (x), y ) ∈ S, and then map this solution
back such that (x, ry (x, y )) ∈ R. Thus each reduction is witnessed
by two maps, one mapping the input space of R to the input space
of S, and the other mapping the output space of S to the output
space of R.

TFNP Reductions
Given one of these reductions (rx , ry ) : R ≤ S, if we did not know
prior to having this reduction that R ∈ TFNP, but we did know
that S ∈ TFNP, then this reduction completes a proof that
R ∈ TFNP. Thus, whatever proof methods we used to show that
S ∈ TFNP, as well as whatever was used to show the correctness
of the reduction, will be that which was used to show that
R ∈ TFNP.

Complete Problems in TFNP
As of yet, no complete problems have been found for TFNP,
essentially for the reason that there is no known recursive
enumeration of the machines which solve these problems. For this
reason, we are forced to construct subclasses of TFNP which do
have complete problems. About a week ago, I would have gone on
to tell you about PPAD, the class of relations which ﬁnding Nash
equilibria is complete for, but because of a new paper on April 6th,
I think there are more interesting avenues to go down. I’ve
included the PPA, PPAD slides but I will skip them.

PPA
All problems in PPA are defined in the following way, given some
polytime deterministic machine M: given an input x, for any string
in the configuration space c ∈ C(x) = Σ[p(|x|)], where p is some
polynomial, M outputs in time O(p(n)) a set of at most two
configurations. We say that c, c are neighbors ({c, c } ∈ G(x)) if
c ∈ M(x, c ) ∧ c ∈ M(x, c). This generates a symmetric graph of
degree at most 2. We define our machines M in such a way that
M(x, 0...0) = {1...1} and 0...0 ∈ M(x, 1...1), so 0...0 is always a
leaf which we will call the standard leaf. Our question is to find a
leaf which is nonstandard [6].

PPAD
We deﬁne PPAD in a very similar way, except that
(c, c ) ∈ G(x) ≡ M(x, c) = ( , c ) ∧ M(x, c ) = (c, ).

Towards a Uniﬁed Theory of Total Functions [4]
In a recent paper by Goldberg and Papadimitriou, there is a
uniﬁcation of the known syntactic subclasses of TFNP into the
class PTFNP. The classes in particular which are contained inside
of it are listed below with the lemmas which embody them:
PPP: f : [n] → [n − 1] has at least one collision
PPAD: G with one unbalanced node has another
PPADS: same as PPAD, looking for oppositely unbalanced
node
PPA: G with one vertex of odd degree has another
PLS: every DAG G has a sink
The new class PTFNP embodies the fact that in a consistent
proof system, if you have a proof where the last two lines are A,
¬A, then there is some error somewhere in the proof.

A Connection with Model Theory
There is an interesting way that each of the lemmas which embody
these classes are connected: they are true in finite models but not
infinite. Given that we believe there are hard problems in each of
these classes, this makes sense, as by Herbrand’s theorem, any
existential sentence which is true in all models is a finite
disjunction of quantifier free formulas and thus the related search
problem will be in P.

PTFNP
We define the problem Wrong Proof to be one wherein we receive
an succinctly represented proof in a propositional proof system
similar to extended Frege but with the ability to define new
function symbols which, in the last two lines, proves A and ¬A.
Given that our theory is consistent, we then want to find where the
misstep in the proof takes place, which is clearly verifiable in
polynomial time but not clearly discoverable. The class PTFNP is
the set of problems which are reducible to Wrong Proof . In the
new paper, we find that:
PPAD, PPA, PPADS, PLS, PPP ⊆ PTFNP

Sources I
N. Nisan. T. Roughgarden. E. Tardos. V. Vazirani.
Algorithmic Game Theory
J. Von Neumann. O. Morgenstern.
Theory of Games and Economic Behavior
T. Roughgarden.
CS364A: Algorithmic Game Theory (Fall 2013)
P. Goldberg. C. Papadimitriou.
Towards a Uniﬁed Theory of Total Functions
C. Papadimitriou. N. Megiddo.
A Note on Total Functions, Existence Theorems, and
Computational Complexity
C. Papadimitriou.
On the Complexity of the Parity Argument and Other
Ineﬃcient Proofs of Existence

Sources II
R. Myerson
Optimal Auction Design

Algorithmic Game Theory and TFNP

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