Download koutofn

A Hypercontractive Inequality for
Matrix-Valued Functions
with Applications to Quantum Computing
Avraham Ben-Aroya
(Tel Aviv University)
Oded Regev
(Tel Aviv University)
Ronald de Wolf
(CWI, Amsterdam)
Outline
• The new hypercontractive inequality
• Main application: k-out-of-n random access
•
codes
Another application: direct product theorem
for one-way communication complexity
The New Inequality
The Parallelogram Law
b
•
a
For any two vectors a,bRd,
• Or equivalently,
The Parallelogram Law
b
•
•
a
This was for the 2 norm
What happens in the p norm, for 1p<2?
• The equality no longer holds, take, e.g.,
a=(1,0),b=(0,1) and p=1
• But, we have the following powerful inequality
for all a,bRd and 1p2:
The Extended Parallelogram Law
•
•
This inequality was proven by [Tomczak-Jaegermann74,
BallCarlenLieb94]
• Originally used to prove the ‘sharp uniform convexity’ of
p spaces
• Implies the Bonami-Beckner hypercontractive inequality
• An extremely useful inequality in computer science
(analysis of Boolean functions, hardness of
approximation, learning theory, communication
complexity, percolation, etc.)
• Recently used by [LeeNaor04] to prove a lower bound on
the distortion of embeddings into 1 spaces
Amazingly, the same inequality also holds with a,b being
matrices and norms being matrix p-norms (i.e., Schatten pnorms)
Prelims: Fourier Transform
•
•
Let f be a function from {0,1}n to Rd (or ℂd×d)
Then we define its Fourier transform as
•
So, e.g.,
The New Hypercontractive Ineq.
•
Thm: For any vector- or matrix-valued f on {0,1}n
and 1p2,
•
Remark: This is the extension of the BonamiBeckner inequality to vector/matrix-valued
functions
The New Hypercontractive Ineq.
•
Thm: For any vector- or matrix-valued f on {0,1}n
and 1p2,
•
Proof: By induction on n.
• The case n=1 is exactly the [BCL94] inequality
with a=f(0), b=f(1)
•
For simplicity, let’s see how to get the n=2 case.
• This involves four matrices, a=f(00), b=f(01),
c=f(10), d=f(11)
The New Inequality (cont.)
•
Using the case n=1 we get
•
By averaging the two inequalities, we get
The New Inequality (cont.)
•
Using the case n=1 again, the left side is at least
Application 1:
Random Access Codes
Compressing Information?
•
Assume we are trying to store n (random) bits into n/8 bits
or qubits
•
Recovering all of the n original bits is ‘clearly’ impossible
• The best success probability is obtained by storing, say,
the first n/8 bits and is only 2-(n)
n/8
1 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ?
•
n
Proving this is easy, both in the classical and quantum
cases
•
•
Random Access Codes
But assume we wish to recover only 1 bit of the original n
bits. Such a primitive is called a random access code (RAC).
•
‘Clearly’ impossible classically… what happens quantumly?
More formally:
n/8
n
2
• A RAC is a function f:{0,1} R mapping each x{0,1}n
to a probability distribution on n/8 bits,
with the property that for all i{1,…,n}
•
Using entropy-based arguments one can show that RACs
don’t exist [AmbainisNayakTa-ShmaVazirani99, Nayak99]
•
Quantum entropy behaves a lot like classical entropy, so
same proof applies also for quantum RAC
•
•
k-out-of-n Random Access Codes
Now assume we wish to recover some arbitrary k
bits of x (think of k=logn)
One would expect the success probability to
behave like 2-(k)
n/8
1 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ?
•
•
n
Entropy-based arguments no longer work!
• For instance, consider the encoding that given
x{0,1}n outputs x with probability 10% and
000…0 with probability 90%. Then it has low
entropy (roughly 0.1n) yet we can recover all of
x prefectly with probability 10%
We therefore have to use the fact that the
dimension of the encoding is low (2n/8)
k-out-of-n Random Access Codes
Thm: For any k-out-of-n quantum random access code
on n/8 qubits, the success probability is 2-(k).
Remark: The classical case can be proven by ‘brute-force’
Proof:
• For simplicity, let’s prove the classical k=1 case
• The quantum case is identical (using matrices
instead of vectors)
• k>1 case is similar
• Recall that the RAC is described by a function
•
n/8
n
2
f:{0,1} R
Let us apply the inequality to f
k-out-of-n Random Access Codes
•
Since f(x) is a probability distribution, we have
•
therefore the RHS is at most 1
The LHS is at least
k-out-of-n Random Access Codes
•
By rearranging, we get
•
Choosing p=1+4/n yields
in contradiction.
Application 2:
Communication Complexity
Direct product theorem for one-way
quantum communication complexity
•
•
•
Alice
Bob
Consider the Disjointness function:
• Alice and Bob are each given a subset of {1,…,n}
and need to decide whether their subsets are
disjoint
A naïve one-way protocol requires n bits of oneway communication (Alice just sends her subset)
• This is essentially optimal (even quantumly)
We show that if Alice and Bob try to solve k
independent instances of the problem with less
than kn/2 (qu)bits of one-way communication, then
their success probability is 2-(k)
Open Questions
•
Find other applications of the inequality