Scott Aaronson Institut pour l' É tude Avançée

Scott AaronsonInstitut pour l'Étude Avançée

Le Principe de la Postselection

Could you ever learn enough about a person to predict his or

her future behavior reliably?

Examples

Good novels don’t just put their characters in random situations—they repeatedly subject the characters to “crucial tests” that reveal aspects of their personalities we didn’t already know

(I guess)

The Karp-Lipton Theorem (1982)

Suppose NP-complete problems were solvable in polynomial time, but only nonuniformly—that is, with polynomial-size circuits

Then we could use those circuits to collapse the Polynomial-Time Hierarchy down to the seocnd level, NPNP

This would be almost as shocking as if P=NP!

“If pigs could whistle, then donkeys could fly”

We want to exploit a small circuit for solving NP-complete problems… but all we know is that it exists!

Does there exist a circuit C of size nk, such that

for all Boolean formulas of size n,

C correctly decides whether is satisfiable, and

C outputs “yes” on whatever problem we wanted to solve originally?

But why should we care?

CIRCUIT LOWER BOUNDS

Theorem (Kannan 1982): For every k, there exists a language in NPNP that does not have circuits of size nk

Proof: It’s not hard to show that doesn’t have circuits of size nk

NPNPP

So either

• NP doesn’t have circuits of size nk, in which case NPNP doesn’t either, or

• NP does have circuits of size nk, in which case

and we win again!NPNPNP PNP

Bshouty et al.’s Improvement (1994)

If a function f:{0,1}n{0,1} has a polynomial-size circuit, then we can find the circuit in ZPPNP, provided we can somehow compute f

(ZPP: Zero-Error Probabilistic Polynomial-Time)

Idea: Iterative learning. Repeatedly find an input xt such that, among the circuits that correctly compute f on x1,…,xt-1, at least a 1/3 fraction get xt wrong

This process can’t continue for long!

Corollary: ZPPNP does not have

circuits of size nk

But what about quantum anthropic

computing?

PostBQP

Class of languages decidable by a bounded-error polynomial-time quantum computer, if at any time you can measure a qubit that has a nonzero probability of being |1, and assume the outcome will be |1

I hereby define a newcomplexity class…

(Postselected BQP)

Another Important Animal: PP

Class of languages decidable by a nondeterministic poly-time Turing machine that accepts iff the majority of its paths do

NP

PP

P#P=PPP

PSPACE

P

Theorem (A., 2004):PostBQP = PP

Unexpectedly, this theorem turned out to have an implication for classical complexity: the simplest known proof of the “Beigel-Reingold-Spielman Theorem,” that PP is closed under intersection

DetourThe “maximally mixed state” In is just the uniform distribution over n-bit strings

But a key fact about quantum mechanics is that, given any orthogonal basis of n-qubit quantum states,

,,,21 n

we could just as well write In as the uniform distribution over that basis

Quantum Proofs

QMA (defined by Kitaev and Watrous) is the quantum version of NP: “Does there exist a quantum state | accepted by such-and-such a circuit with high probability?”

Unlike NP, QMA doesn’t seem to be “self-reducible”—we don’t know how to construct | given an oracle for QMA problems

But we can construct | in PostBQP. (Why?)

Quantum Advice

BQP/qpoly: Class of languages decidable by polynomial-size, bounded-error quantum circuits, given a polynomial-size quantum advice state |n that depends only on the input length n

Mike & Ike: “We know that many systems in Nature ‘prefer’ to sit in highly entangled states of many systems; might it be possible to exploit this preference to obtain extra computational power?”

How powerful is quantum advice?

Could it let us solve problems that are not even computable given classical advice of similar size?!

Limitations of Quantum Advice

NP BQP/qpoly relative to an oracle(Uses direct product theorem for quantum search)

BQP/qpoly PostBQP/poly( = PP/poly)

.log 111 fQfmQOfD

Closely related: for all (partial or total) Boolean functions f : {0,1}n {0,1}m {0,1},

Alice’s Classical Advice

Bob, suppose you used the maximally mixed state in place of your

quantum advice. Then x1 is the lexicographically first input for which you’d output the right answer with

probability less than ½.But suppose you succeeded on x1,

and used the resulting reduced state as your advice. Then x2 is the

lexicographically first input after x1 for which you’d output the right answer

with probability less than ½...

x1

x2

Given an input x, clearly lets Bob

decide in PostBQP whether xL

But how many inputs must Alice specify?

We can boost a quantum advice state so that the error probability on any input is at most (say) 2-100n; then Bob can reuse the advice on as many inputs as he likes

We can decompose the maximally mixed state on p(n) qubits as the boosted advice plus 2p(n)-1 orthogonal states

Alice needs to specify at most p(n) inputs x1,x2,…, since each one cuts Bob’s total success probability by at least half, but the probability must be at least ~2-p(n) by the end

PPP Does Not Have Quantum Circuits of Size nk

Does U accept x0 w.p. ½?If yes, set x0LIf no, set x0L

U: Picks a size-nk quantum circuit uniformly at random

and runs it

x0

x1

x2

x3

x4

x5

Conditioned on deciding x0 correctly, does U accept x1 w.p. ½?If yes, set x1LIf no, set x1L

Conditioned on deciding x0 and x1 correctly, does U accept x2 w.p. ½?If yes, set x2LIf no, set x2L

For any k, defines a language L that does not have quantum circuits of size nk

Why? Intuitively, each iteration cuts the number of potential circuits in half, but there were at most circuits to begin with

kn2~

On the other hand, clearly L PPP

Even works for quantum circuits

with quantum advice!

Quantum Karp-Lipton Theorem

If PP BQP/qpoly, then the counting hierarchy—consisting ofetc.—collapses to PP

,,,PPPPPP PPPPPP

Also:

PP does not have quantum circuits of size nk

PEXP requires quantum circuits of size f(n), where f(f(n))2n

Concluding Thought: What Makes Science Possible?

That which we can observe, we can understand

That which we can observe, and then observe in a new situation where we can’t predict what it will do even given the earlier observation, and so on for a polynomial number of steps, we can understand (provided we can postselect a description consistent with our observations)

To Show PP PostBQP…Given a Boolean function f:{0,1}n{0,1},let s=|{x : f(x)=1}|. Need to decide if s>2n-1

From

/ 2

0,1

2n

n

x

x f x

2 22 2

2 0 1 1/ 2 2 0 1/ 2 2 2 1,

2 2

n n n

n n

s s sH

s s s s

we can easily prepare

Goal: Decide if these amplitudes have the same or opposite signs

Prepare |0|+|1H| for some ,.Then postselect on second qubit being |1

/ 22 2 2

0 1/ 2 2 2 1:

/ 2 2 2

n

n

s s

s s

Yields in first qubit

To Show PP PostBQP…

/ 22 2 2

0 1/ 2 2 2 1:

/ 2 2 2

n

n

s s

s s

Yields in first qubit

1

0

Suppose s and 2n-2s are both positive

Then by trying / = 2i for all i{-n,…,n}, we’ll eventually get close to

0 1

2

On the other hand, if 2n-2s is negative, then we won’t. QED

Beigel, Reingold, Spielman 1990: PP is “closed under intersection”Solved a problem that was open for 18 years…

Other classical results proved with quantum techniques: Kerenidis & de Wolf, A., Aharonov & Regev, …

Observation: PostBQP is trivially closed under intersection PP is too

Given L1,L2PostBQP, to decide if xL1 and xL2, postselect on both computations succeeding, and accept iff they both accept