# Turing's Invisible Hand

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## Communication Complexity of Reaching Equilibrium

August 18, 2009 by algorithmicgametheory

## Motivation

The dogma of game theory (and economics) is that strategic situations will reach equilibrium — one way or another. How this will happen is often left un-answered, but from a computational point of view this is quite a central question. Indeed, a great amount of work has been done regarding the computational complexity of reaching an equilibrium of a game, reaching a conclusion that an infeasible amount of computation might be needed. In this post I want to discuss another aspect of the complexity of reaching equilibrium: the amount of information that needs to be transferred between the different players. The point of view here is that initially each player “knows” only his own utility function $u_i$ and then the players somehow interact until they find an equilibrium of the game $(u_1 .... u_n)$ defined by their joint utilities. The analysis here abstracts away the different incentives of the players (the usual focus of attention in game theory) and is only concerned with the amount of information transfer required between the players. I.e., we treat this as a purely distributed computation question where the input $(u_1 ... u_n)$ is distributed between the players who all just want to jointly compute an equilibrium point. We assume perfect cooperation and prior coordination between the players, i.e. that all follow a predetermined protocol that was devised to compute the required output with minimum amount of communication. The advantage of this model is that it provides lower bounds for all sorts of processes and dynamics and that eventually should reach equilibria: if even under perfect cooperation much communication is required for reaching equilibrium, then certainly whatever individual strategy each player follows, convergence to equilibrium cannot happen quickly.

The main result that we shall give in this post is that finding a pure Nash equilibrium (or determining that none exists) requires essentially communicating everything about the utility functions — no shortcuts are possible. This is the basic result that appears in two papers that studied the issue: Communication complexity as a lower bound for learning in games by Vincent Conitzer and Tuomas Sandholm that considered two-player games and How Long to Equilibrium? The Communication Complexity of Uncoupled Equilibrium Procedures by Sergiu Hart and Yishay Mansour whose focus is multi-player games. See also the presentation by Sergiu Hart. A future post will discuss the much more complicated case of finding a mixed Nash equilibrium, a question which is still mostly open, despite results given in the H&M paper.

## The Communication Complexity model

The basic communication complexity model considers $n$ players, where each player $i$ holds part of the input which we assume is an $N$-bit string, $x^i \in \{0,1\}^N$. The usual focus is on the two-player case, $n=2$, in which case we simplify the notation to use $x$ and $y$ rather than $x^1$ and $x^2$. The joint goal of these players is to compute the value of $f(x^1 ... x^n)$ where $f : (\{0,1\}^N)^n \rightarrow Z$ is some predetermined function. As $f(x^1 ... x^n)$ depends on all values of $x^i$, the players will need to communicate with each other in order to do so (we will use “broadcasting” rather than player-to-player channels), and they will do so according to a predetermined protocol. Such a protocol specifies when each of them speaks and what he says, where the main point is that each player’s behavior must be a function of his own input as well as what was broadcast so far. We will assume that all communication is in bits (any finite alphabet can be represented in bits). The communication complexity of $f$ is the minimum amount of bits of communication that a protocol for $f$ uses in the worst case. This model has been introduced in a seminal paper by Andy Yao, the standard reference is still the (just slightly dated) book Communication Complexity I wrote with Eyal Kushilevtiz, and online references include the wikipedia article, and the chapter from the Barak-Arora computational complexity book.

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## The Disjointness Function

Our interest will be in lower bounds on the communication complexity. There are several techniques known for proving such lower bounds, the easiest of which is the “fooling set method” which we will use on the “disjointness function” for two players. The disjointness function, introduced in Yao’s original paper, plays a key role in communication complexity, and indeed it will turn out to be useful to us too: once we have a lower bound for it, it will be easy to easily deduce the lower bound for the function of our interest, finding a pure Nash equilibrium, by simple reduction.

The disjointness function: Alice holds $x \in \{0,1\}^N$ and Bob holds $y \in \{0,1\}^N$. They need to determine whether there exists an index $i$ where they both have 1, $x_i=1=y_i$. If we view $x$ as specifying a subset $S \subseteq \{1...N\}$ and $y$ as specifying $T \subseteq \{1 ... N\}$, then we are asking whether $S \cap T = \emptyset$.

Lemma: Any deterministic protocol for solving the disjointness problem requires at least $N$ bits of communication in the worst case.

Before proving this lemma, notice that it is almost tight, as Alice can always send all her input to Bob ($N$ bits) who then has all the information for determining the answer, who then sends it back to Alice (1 more bit). It is known that a $\Omega(N)$ lower bound also applies to randomized protocols, but the proof for the randomized case is harder, while the proof for the deterministic case is easy.

Proof: Let us look at the $2^N$ different input pairs of the form $\{<x,y> | \forall i,\: x_i+y_i=1\}$ (these are the “maximal disjoint” inputs). We will show that for no two of these input pairs can the protocol produce the same communication transcript (i.e. the exact same sequence of bits is sent throughout the protocol). It follows that the protocol must have at least $2^N$ different possible transcripts, and since these are all binary sequences, at least one of them must be of length at least $N$ bits.

Now comes the main point: why can’t two maximally disjoint pairs $<x,y>$ and $<x',y'>$ have the same transcript? Had we been in a situation that $f(x,y) \ne f(x',y')$ then we would know that the transcripts must be different since the protocol must output a different answer in the two cases; however in our case all “maximal disjoint” input pairs give the same answer: “disjoint”. We use the structure of communication protocols — that each player’s actions can only depend on his own input (and on the communication so far) — to show that if $<x,y>$ and $<x',y'>$ have the same transcript then $<x,y'>$ has the same transcript too (and so does $<x',y>$). Consider the players communicating on inputs $<x,y'>$: Alice, that only sees $x$, cannot distinguish this from the case of $<x,y>$ and Bob, that only sees $y'$ cannot distinguish it from the case $<x',y'>$. Thus when Alice communicates she will not deviate from what she does on of $<x,y>$ and Bob will not deviate from what he does on $<x',y'>$. Thus, none of them can be the first to deviate from the joint transcript, thus none of them ever does deviate and the joint transcript is also followed on $<x,y'>$. We now get our contradiction since either $x$ and $y'$ are not disjoint or $x'$ and $y$ are not disjoint, and thus at least one of these pairs cannot not share the same transcript, a contradiction.   (The reason for non disjointness is that since $x \ne x'$ there must be an index $i$ with $x_i \ne x'_i$. If $x_i = 1$ and $x'_i=0$ then $y_i = 0$ and $y'_i=1$ in which case $x$ and $y'$ are not disjoint. Similarly, if $x'_i = 1$ and $x_i=0$ then $y'_i = 0$ and $y_i=1$ in which case $x'$ and $y$ are not disjoint.)

## Finding a Pure Nash Equilibrium

Back to our setting where each player $i$ holds his own utility function $u_i : S_1 \times ... S_n \rightarrow \Re$, where the $S_i$‘s are the commonly known strategy sets, which we will assume have size $m$ each. Thus each player’s input consists of $m^n$ real numbers, which we will always assume are in some small integer range. Perhaps it is best to start with a non-trivial (just) upper bound.  Let us consider the class of games that are (strictly) dominance solvable, i.e. those that iterated elimination of strictly dominated strategies leaves a single strategy profile, which is obviously a pure Nash equilibrium.  A simple protocol to find this equilibrium is the following:

• Repeat $mn$ times:
• For $i = 1 ...n$ do
• If player $i$ has a strategy that is dominated after the elimination of all previously removed strategies, then announce it, else say “pass”.
• Output the (single) profile of non-eliminated strategies.

What is remarkable about this protocol is that number of bits communicated is polynomial in $n$ and $m$, which may be exponentially smaller than  the total size of the input which is $O(m^n)$ numbers for each player.  Can some protocol like this be designed for all games that have a pure Nash equilibrium?  The basic answer in no:

Theorem: Every deterministic (or randomized) protocol that recognizes whether the given game has a pure Nash equilibrium requires $\Omega(m^n)$ bits of communication.

Note that this theorem implies in particular that one cannot design a more efficient protocol that finds an equilibrium just for games that are guaranteed to have one, as adding a verification stage (where each player checks whether the strategy suggested to him is indeed a best reply to those suggested to the others) would convert it to a protocol that recognizes the existence of a pure Nash.

Proof: The proof is by a reduction from the disjointness problem.  We will show that a protocol for determining the existence of a pure Nash equilibrium for games with $n$ players each having $m$ strategies can be used to solve the disjointness problem on strings of length $N = \Omega(m^n)$.

Let us start with the case $n=2$ and solve disjointness for length $N=(m-2)^2$.  The idea is very simple: Alice will build a utility matrix $u_A$ by filling it with the values of its bit string $x$ in some arbitrary but fixed order, and Bob will build a matrix $u_B$ from $y$ using the same order.  Now, any cell of the two matrices where $x_i=y_i=1$ will turn out to be a Nash equilibrium since both players get the highest utility possible in this matrix.  For this idea to formally work, we just need to make sure that no other cell is a Nash equilibrium.  This can be done in one of several ways (and different ones are used in the C&S and H&M papers), the simplest of which is adding two more rows and columns as follows:

With the addition of these rows and columns, each player’s best-reply always obtains a value of 1, and thus only a (1,1) entry may be a Nash equilibrium.

Now for general $n$.  We add to Alice and Bob $n-2$ players that have utilities that are identically 0.  Thus any strategy will be a best reply for these new players, and a pure Nash equilibrium is determined solely by Alice and Bob.  The main change that the new players bring is that the utility functions of Alice and Bob are now $n$-dimensional matrices of total size $m^n$.  Thus Alice and Bob can fold their length $N$ bit string into this larger table (as before keeping two strategies from their own strategy space to ensure no “none-(1,1)” equilibrium points), allowing to answer disjointness on $N=(m-2)^2 \cdot m^{n-2}$-bit long vectors.

## Variants

There are several interesting variants on the basic question:

1. The lower bound uses heavily the fact that the utilities of different strategies may be equal and thus there are many possible best replies.  Would finding a Nash equilibrium be easier for games in “general position” — specifically if there was always a single best reply for each player?  For the case of $n=2$ players, C&S show that the complexity does go down to $\Theta(m \log m)$.  (Exercise: show that the randomized communication complexity in this case is $\Theta(m)$.)  As $n$ grows, the savings are not exponential though and H&M show a $2^{\Omega(n)}$ lower bound for constant $m$.  A small gap remains.
2. We have seen that finding an equilibrium of a dominance solvable game is easy.  Another family that is guaranteed to posses a pure equilibrium is potential games.  However, H&M show that finding a Nash equilibrium in an ordinal potential game may require exponential communication.  I don’t know whether things are better for exact potential games.
3. The efficient protocol for dominance solvable games was certainly artificial.  However, an efficient natural protocol exists as well: if players just best reply in round robin fashion then they will quickly converge to a pure Nash equilibrium.  (This seems to be well known, but the only link that I know of is to one of my papers.)

Posted in Uncategorized | Tagged complexity of equilibria, technical | 6 Comments

### 6 Responses

1. [...] Communication Complexity of Reaching Equilibrium « Algorithmic … [...]

2. I wonder what the communication complexity of finding a Nash equilibrium using replicator dynamics would be. If you start from the interior, the system will converge to a Nash.

In terms of communication each agent will need to transmit their strategy profile each round. To make it rigorous, I suppose we should approximate strategy profiles using fixed point arithmetic and one could look at the average bits of communication to converge to within epsilon of an equilibrium. The epsilon being limited to how finely you wish to represent a strategy profile. A setup like that would allow one to measure communication in bits anyway.

My experience with the replicator dynamics for routing games is that they converge pretty quickly. I am not sure how fast the number of bits transmitted would go up as epsilon shrinks though.

http://en.wikipedia.org/wiki/Replicator_equation

• While I have not talked about convergence to mixed Nash equilibria yet, Replicator dynamics are certainly an example of the natural dynamics that the communication complexity approach aims to rule out as general mechanisms that always converge quickly. The numeric precision is not the problem but rather the high dimensionality.

For the case of two players (even in an evlolutionary single population setting), if the best reply for each strategy i is strategy i+1, then it will certainly take about m steps (m is the size of the strategy space) to “march all the way” to have enough weight on strategy m as required for equilibrium. For n players, it is possible to design utility functions that essentially “fold” an exponential in n sequence of best replies into the product of the strategy spaces. For such utilities we get an exponential convergence time of replicator dynamics as well as other “sequential reply strategies”, indpendantly of epsilon.

3. [...] Complexity of Mixed-Nash Equilibria In a previous post I discussed the communication complexity of reaching a pure Nash equilibrium.  The communication complexity model aims to capture the basic information transfer bottleneck [...]

4. [...] There are also “concrete” models that study explicitly specific parameters such as communication or queries.  One may dislike the fact that this complexity analysis does not restrict attention to [...]

5. In terms of communication each agent will need to transmit their strategy profile each round. To make it rigorous, I suppose we should approximate strategy profiles using fixed point arithmetic and one could look at the average bits of communication to converge to within epsilon of an equilibrium. The epsilon being limited to how finely you wish to represent a strategy profile. A setup like that would allow one to measure communication in bits anyway.

thanks
killing games

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