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