The Complexity of AgreementA 100% Quantum-Free Talk

Scott Aaronson

MIT

People disagree. Why?

• Because some people are idiots?• Because people care about things other than

truth (winning debates, not admitting error, etc.)?• Because even given the same information,

people would reach different honest conclusions?

Can rational agents ever “agree to disagree” solely because of differing information?

Aumann 1976: No.Suppose Alice and Bob are Bayesians with a common prior, but different knowledge.

Let EA be Alice’s estimate of (say) the chance of rain tomorrow, conditioned on all her knowledge.

Let EB be Bob’s estimate, conditioned all his knowledge.

Suppose EA and EB are common knowledge (Alice and Bob both know their values, both know that they both know them, etc.)

Theorem: EA = EB

“Standard Protocol”Alice’s initial expectation EA,0

Bob’s new expectation EB,1

Alice’s new expectation EA,2

Bob’s new expectation EB,3

Geanakoplos & Polemarchakis 1982: Provided the state space is finite, Alice and Bob will agree after a finite number of rounds

What’s Going On?

00 01 10

00 01 10 11 Alice knows the first of 2 bits

11 Bob knows their sum

We can represent an agent’s “knowledge” by a partition of the state space…

If Alice and Bob don’t agree with certainty, then one of them can learn something from the other’s message—meaning its partition gets refined

But the partitions can get refined only finitely many times

MESSAGE

Problems• Huge number of messages might be needed

before Alice and Bob agree

• Messages are real numbers

Conjecture• For some function f of Alice’s input x{0,1}n

and Bob’s input y{0,1}n, and some distribution over (x,y), Alice and Bob will need to exchange (n) bits even to probably approximately agree about EX[f(x,y)]

Intuition comes from communication complexity

Main Result: Conjecture Is False

2

1

O

For all f:{0,1}n{0,1}n[0,1], and all prior distributions over (x,y) pairs, Alice and Bob can (,)-agree about the expectation of f, by exchanging only bits.

Moral: “Agreeing to disagree” is problematic, even for agents subject to realistic

communication constraints

Independent of n

Say Alice and Bob “(,)-agree” if they agree within with probability at least 1- over their prior

Intuition

In the standard protocol, as long as Alice and Bob disagree by , their expectations EA and EB follow an unbiased random walk with step size . But since EA,EB[0,1], such a walk hits an absorbing barrier after ~1/2 steps.To prove, use (EA)2,(EB)2 as progress measures

A bit more formally…

Let EA,t(), EB,t() be Alice and Bob’s expectations at time t, assuming the true state of the world is

For any function F:[0,1], let||F|| = EXD[F()2]

Let ={0,1}n{0,1}n be the state space, and let D be the shared prior distribution

Lemma: Suppose Alice sends the tth message. Then ||EB,t|| - ||EA,t-1|| = ||EB,t - EA,t-1||

Proof: Crucial observation: if Alice has just sent Bob a message, then her expectation of Bob’s expectation equals her expectation.

1,,

1,1,,

1,,1,,1,,

2

2

tAtB

tAtAtB

tAtBD

tAtBtAtB

EE

EEE

EEEXEEEE

Hence

Theorem: The standard protocol causes Alice and Bob to (,)-agree after 1/(2) messages

Proof: Suppose Alice sends the tth message, and suppose .Pr 1,,

tAtB

DEE

Then .21,1,,1,, tAtAtBtAtB EEEEE

Likewise, after Bob sends Alice the (t+1)st message, ||EA,t+1|| > ||EB,t|| + 2.

But max{||EA,t||,||EB,t||} is initially 0 and can never exceed 1.

Weren’t we cheating, since messages in the standard protocol are unbounded-precision real numbers?

Discretized standard protocol: Let Charlie (C) be a “monkey in the middle” who sees Alice and Bob’s messages but doesn’t know their inputs. At the tth step, Alice tells Bob whether EA,t>EC,t+/4, EA,t<EC,t-/4, or neither. Bob does likewise.

Theorem: The discretized standard protocol causes Alice and Bob to (,)-agree after at most 3072/(2) messages.

Two interesting questions:

1. Does the standard protocol ever need ~1/2 messages?2. Is there ever a protocol that does better?

Example answering both questions: Let Alice’s input x=x1…xn and Bob’s input y=y1…yn be uniform over {-1,1}n. Then let

1, if 1

0, if 0

1,0, if ,

,

22

1,

11

yxF

yxF

yxFyxF

yxf

yxxyyxFn

iiiii

Theorem: The standard protocol requires

/1log

/1 2

messages before Alice and Bob’s expectations of f agree within with constant probability.

On the other hand, there’s an “attenuated protocol” that causes them to agree within with constant probability after only O(1) messages

Three or More Agents

Could a weak-willed Charlie fail to bring Alice and Bob into agreement with each other?

Theorem: On a strongly-connected graph with N agents and diameter d, messages

suffice for every pair of agents to “(,)-agree”

2

2

Nd

O

Computational ComplexityFine, few messages suffice for agreement—but what if it took Alice and Bob billions of years to calculate their messages?

Result: Provided the agents can sample from their initial partitions, they can simulate Bayesian rationality using a number of samples independent of n (but alas, exponential in 1/6)

Meaning: By examining their messages, you couldn’t distinguish these “Bayesian wannabes” from true Bayesians with non-negligible bias

Computational ComplexityProblem: An agent’s expectation could lie on a “knife-edge” between two messages

10

Pr[

mes

sage

]

tiE

Solution: Have agents “smooth” their messages by adding random noise to them

Open Problems• Do Alice and Bob need to exchange (1/2) bits to agree within with high probability?

Best lower bound I can show is the trivial (log1/)

• Is there a scenario where Alice and Bob must exchange (n) bits to (,0)-agree?

• Can “Bayesian wannabes” agree within with high probability using poly(1/) samples, rather than 2poly(1/)?

Open Problems (con’t)• Suppose Alice and Bob (,)-agree about the expectation of f:[0,1]. Then do they also have “approximate common knowledge”?

(Alice is pretty sure of Bob’s expectation, Alice is pretty sure Bob’s pretty sure of her expectation, Alice is pretty sure Bob’s pretty sure she’s pretty sure of his expectation, etc.)