Transcript The Postselection Principle

Le Principe de la
Postselection
Scott Aaronson
Institut pour l'Étude Avançée
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 P
doesn’t have circuits of size nk
NP
NP
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
NP
NP
P
NP
NP
and we win again!
Bshouty et al.’s Improvement (1994)
If a function f:{0,1}n{0,1} has a polynomialsize circuit, then we can find the circuit in
ZPPNP, provided we can somehow compute f
(ZPP: Zero-Error
Probabilistic
Polynomial-Time)
Corollary:
ZPPNP
not have
Idea: Iterative does
learning.
Repeatedly find an
k
circuits
of
size
n
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!
But what about
quantum anthropic
computing?
I hereby define a new
complexity class…
PostBQP
(Postselected BQP)
Class of languages decidable by a boundederror 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
Another Important Animal: PP
Class of languages decidable by a
nondeterministic poly-time Turing machine
that accepts iff the majority of its paths do
PSPACE
P#P=PPP
PP
NP
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
Detour
The “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,
 1 , ,  2 ,
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-andsuch a circuit with high probability?”
Unlike NP, QMA doesn’t seem to be “selfreducible”—we don’t know how to construct
| given an oracle for QMA problems
But we can construct | in PostBQP. (Why?)
Quantum Advice
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?”
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
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)
Closely related: for all (partial or total) Boolean
functions f : {0,1}n  {0,1}m  {0,1},
D
1
 f   O mQ   f  log
1
Q   f .
1
Alice’s Classical Advice
x1
x2
Bob, suppose you used the
maximally mixed state in place of
your quantum advice. Then x1 is the
lexicographically first input for which
you’dGiven
output an
theinput
right answer
with
x,
probability
lessBob
than ½.
clearly lets
But suppose you succeeded on x1,
decide in PostBQP
and used the resulting reduced state
whether xL
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...
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
U: Picks a size-nk quantum
circuit uniformly at random
and runs it
Does U accept x0 w.p.  ½?
If yes, set x0L
Conditioned on deciding x0
If no, set x0L
correctly,
does
accept xx1
Conditioned
onUdeciding
0
w.p.
and x1 ½?
correctly, does U
If
yes, set
x1L ½?
accept
x2 w.p.
If
If no,
yes,set
setxx1L
2L
If no, set x2L
 x0
 x1
x2
x3
x4
x5
For any k, defines a language L that does not
have quantum circuits of size nk
Even works for
Why? Intuitively,
each
iteration cuts the
quantum
circuits
number of potential
circuits in half, but there
with quantum
k
n
were at most ~ advice!
2 circuits to begin with
On the other hand, clearly L  PPP
Quantum Karp-Lipton Theorem
If PP  BQP/qpoly, then the counting
hierarchy—consisting of PP , PP PP , PP
etc.—collapses to PP
PP
PP
,
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
2
From
 
2
n /2

x 0 ,1
n
x
2  s  s
n
x
we can easily prepare
n
 s 0  s 1
2
f
,
H 

1 / 2 2
n
0

1 / 2 2  2s  1
n
2  s  s
2
n
2
2
Goal: Decide if these amplitudes have the
 s 0signs
  1 / 2 2  2s 1
same
or
opposite
:
Yields 
in first qubit
n
 /
 s    / 2 for
 2 s  ,.
  2 some
Prepare |0|+|1H|
Then postselect on second qubit being |1
2
2
2
n
2
To Show PP  PostBQP…
1
n-2s
Suppose
s
and
2
On the other hand, if
are
both
positive then
2n-2s
is negative,
we won’t.
QED/ = 2i
Then
by trying

for all i{-n,…,n}, we’ll
eventually get close to
0
 
0  1
2
 s 0   1 / 2 2  2s 1
n
Yields
  /  :
 s   / 22  2s
2
2
2
n
2
in first qubit
Beigel, Reingold, Spielman 1990: PP is
“closed under intersection”
Solved a problem that was open for 18 years…
Observation: PostBQP is trivially closed
under intersection  PP is too
Given L1,L2PostBQP, to decide if xL1 and
xL2, postselect on both computations
succeeding, and accept iff they both accept
Other classical results proved with
quantum techniques:
Kerenidis & de Wolf, A., Aharonov
& Regev, …