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Introduction to Quantum Information Processing Lecture 17 Richard Cleve 1 Overview of Lecture 17 • Introduction to communication complexity • Intersection problem (a.k.a. appointment scheduling) • Restricted equality problem • Exponential separation in bounded-error setting • Inner product problem • Simultaneous message passing model and fingerprinting 2 communication complexity 3 Classical communication complexity [Yao, 1979] x1x2 xn y1y2 yn f (x,y) E.g. equality function: f (x,y) = 1 if x = y, and 0 if x y Any deterministic protocol requires n bits communication Probabilistic protocols can solve with only O(log(n/)) bits communication (error probability ), via random hashing 4 Quantum communication complexity x1 x2 xn y1 y2 yn Qubit communication qubits f (x,y) entangled qubits x1 x2 xn y1 y2 yn Prior entanglement bits f (x,y) 5 Appointment scheduling 1 2 3 4 5 ... n 1 2 3 4 5 ... n x= 01101…0 y= 10011…1 i (xi = yi = 1) Classically, (n) bits necessary to succeed with prob. 3/4 For all > 0, O(n1/2 log n) qubits sufficient for error prob. < [KS „87] [BCW „98] 6 Search problem 1 2 3 4 5 6 ... n Given: x= 000010…1 accessible via queries log n i x i 1 b b b xi b Goal: find i{1, 2, …, n} such that xi = 1 Classically: (n) queries are necessary Quantum mechanically: O(n1/2) queries are sufficient [Grover, 1996] 7 1 2 3 4 5 6 ... n Alice x= 011010…0 Bob y= 100110…1 xy = 0 0 0 0 1 0 … 0 i xy i y x x y 0 0 0 0 b b Bob Alice Bob Communication per xy-query: 2(log n + 3) = O(log n) 8 Appointment scheduling: epilogue Bit communication: Qubit communication: Cost: θ( n) Cost: θ( n1/2 ) (with refinements) Bit communication Qubit communication & prior entanglement: & prior entanglement: Cost: θ( n1/2) Cost: θ( n1/2) [R „02] [AA „03] 9 Restricted version of equality Precondition (i.e. promise): either x = y or (x,y) = n/2 Hamming distance (Distributed variant of “constant” vs. “balanced”) Classically, (n) bits communication are necessary for an exact solution Quantum mechanically, O(log n) qubits communication are sufficient for an exact solution [BCW „98] 10 Classical lower bound Theorem: If S {0,1}n has the property that, for all x, x′ S, their intersection size is not n/4 then S < 1.99n Let some protocol solve restricted equality with k bits comm. ● 2k conversations of length k ● approximately 2n/n input pairs (x, x), where Δ(x) = n/2 Therefore, 2n/2kn input pairs (x, x) that yield same conv. C Define S = {x : Δ(x) = n/2 and (x, x) yields conv. C } For any x, x′ S, input pair (x, x′ ) also yields conversation C Therefore, Δ(x, x′) n/2, implying intersection size is not n/4 Theorem implies 2n/2kn < 1.99n , so k > 0.007n 11 [Frankl and Rödl, 1987] Quantum protocol n define ψ x (1) xj For each x {0,1}n, j j 1 Protocol: 1. Alice sends x to Bob (log(n) qubits) 2. Bob measures state in a basis that includes y Correctness of protocol: If x = y then Bob‟s result is definitely y If (x,y) = n/2 then xy = 0, so result is definitely not y Question: How much communication if error ¼ is permitted? Answer: just 2 bits are sufficient! 12 Exponential quantum vs. classical separation in bounded-error models O(log n) quantum vs. (n1/4 / log n) classical : a log(n)-qubit state U: unitary operation (described classically) on log(n) qubits M: two-outcome measurement Output: result of applying M to U [Raz, 1999] 13 Inner product IP(x, y) = x1 y1 + x2 y2 + + xn yn mod 2 Classically, (n) bits of communication are required, even for bounded-error protocols Quantum protocols also require (n) communication [KY „95] [CNDT „98] [NS „02] 14 Recall the BV problem Let f(x1, x2, …, xn) = a1 x1 + a2 x2 + + an xn mod 2 Given: 0 x1H x H 1 a1 0 x2H x H 2 a2 H f H 0 xnH x H n an 1 bH b f H 1(x1, x2, …, xn) Goal: determine a1, a2 , …, an Classically, n queries are necessary Quantum mechanically, 1 query is sufficient 15 Lower bound for inner product IP(x, y) = x1 y1 + x2 y2 + + xn yn mod 2 Proof: x1 x2 xn y1 y2 yn z Alice and Bob‟s IP protocol Alice and Bob‟s IP protocol inverted x1 x2 xn y1 y2 yn zIP(x, y) 16 Lower bound for inner product IP(x, y) = x1 y1 + x2 y2 + + xn yn mod 2 0 0 0 1 Proof: x1 x2 xn H H H H Alice and Bob‟s IP protocol Alice and Bob‟s IP protocol inverted H H H H x1 x2 xn x1 x2 xn 1 Since n bits are conveyed from Alice to Bob, n qubits communication necessary (by Holevo‟s Theorem) 17 quantum fingerprints 18 Equality revisited in simultaneous message model x1x2 xn y1y2 yn Equality function: f (x,y) f (x,y) = 1 if x = y 0 if x y Exact protocols: require 2n bits communication 19 Equality revisited in simultaneous message model x1x2 xn y1y2 yn classical classical f (x,y) Pr[00] = Pr[11] = ½ Bounded-error protocols with a shared random key: require only O(1) bits communication Error-correcting code: e(x) = 1 0 1 1 1 1 0 1 0 1 1 0 0 1 1 0 0 1 e(y) = 0 1 1 0 1 0 0 1 0 0 1 1 0 0 1 0 1 0 20 random k Equality revisited in simultaneous message model x1x2 xn y1y2 yn Bounded-error protocols without a shared key: f (x,y) Classical: θ(n1/2) Quantum: θ(log n) [A „96] [NS „96] [BCWW „01] 21 Quantum fingerprints Question 1: how many orthogonal states in m qubits? Answer: 2m Let be an arbitrarily small positive constant Question 2: how many almost orthogonal* states in m qubits? (* where |xy| ≤ ) am Answer: 22 , for some constant a > 0 To be continued during next lecture … 22 23