Random Access Codes and a Hypercontractive Inequality for Matrix-Valued Functions Avraham Ben-Aroya (Tel Aviv University) Oded Regev Ronald de Wolf (CWI, Amsterdam)

Outline Main result: k-out-of-n random access codes Proof: A new hypercontractive inequality The proof Other applications of the inequality: Direct product theorem for one-way communication complexity A new approach to lower bounds on locally decodable codes (LDCs)

Random Access Codes

Squeezing Information? Assume we are trying to store n (random) bits into n/8 bits or qubits Recovering all 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) Proving this is easy, both in the classical and quantum cases 1 ? n n/8

Random Access Codes But assume we wish to recover only 1 bit of the original n bits with good probability. Such a primitive is called a random access code (RAC). Seems ‘clearly’ impossible classically Not so clear what happens quantumly Using entropy-based arguments one can show that RACs don’t exist [AmbainisNayakTa-Shma Vazirani99, 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 (say, k=logn) One would expect the success probability to behave like 2-(k) 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) n/8 1 ? n

Main Result Thm: For any k-out-of-n quantum RAC on n/8 qubits, the success probability is 2-(k). Remarks: The classical case can be proven by combinatorial arguments See also this Friday for a related result by Koenig and Renner

The New Inequality

The Parallelogram Law a+b a-b b a For any two vectors a,bRd, Or equivalently,

The Parallelogram Law a+b a-b 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 p- norms) [Tomczak-Jaegermann74, BallCarlenLieb94]

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 Bonami- Beckner 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 induction hypothesis (case n=1) we get By averaging the two inequalities, we get

The New Inequality (cont.) Using the case n=1, the left side is at least

Proof of the Main Theorem

Main Theorem (again) Thm: For any k-out-of-n quantum RAC on n/8 qubits, the success probability is 2-(k). Proof: For simplicity, let’s prove the case k=1 k>1 case is similar So assume by contradiction that there exists a function f:{0,1}nℂ2n/8×2n/8 mapping each x{0,1}n to a density matrix on n/8 qubits, with the property that for all i{1,…,n}

Proof Let us apply the inequality to f Since f(x) is a density matrix, we have therefore the RHS is at most 1, and we obtain Choosing p=1+4/n yields a contradiction.

Further Applications

Direct product theorem for one-way quantum communication complexity Alice Bob Consider the Disjointness problem: Alice and Bob are each given a subset of {1,…,n} and need to decide whether their subsets are disjoint Only one message from Alice to Bob is allowed A naïve protocol requires n bits (Alice just sends her subset) This is essentially optimal (even quantumly) In other words, if Alice sends only, say, n/8 (qu)bits, then their success probability is necessarily <60%.

Direct product theorem for one-way quantum communication complexity Assume now that Alice and Bob try to solve k independent instances of the problem So input consists of k subsets A1,…,Ak for Alice and k subsets B1,…,Bk for Bob, and Bob is supposed to tell for each i whether Ai is disjoint from Bi Clearly kn bits from Alice to Bob are enough We show that if Alice sends less than kn/8 (qu)bits, then their success probability is 2-(k) Such a result is known as a direct product theorem

Lower Bounds on Locally Decodable Codes A q-query locally decodable code (LDC) is a mapping f from n bits into N bits with the property that For any x{0,1}n, i{1,…,n}, and y{0,1}N that differs from f(x) in at most 0.01N locations, we can recover xi by querying only q bits in y For q=2: The Hadamard code is a LDC with N=2n This is essentially optimal due to [Kerenidis-deWolf02] Their proof uses quantum arguments We can give an alternative proof using the hypercontractive inequality For q=3: Best known code uses N=2n1/32582657 [Yekhanin07] Almost no lower bounds are known; a huge open question !

Open Questions Find other applications of the inequality Compare this inequality to entropy-based techniques