Richard Cleve DC 3524 cleve@cs.uwaterloo Course web site at:

1. Introduction to Quantum Information ProcessingCS 467 / CS 667Phys 467 / Phys 767C&O 481 / C&O 681 Richard Cleve DC 3524 cleve@cs.uwaterloo.ca Course web site at: http://www.cs.uwaterloo.ca/~cleve/courses/cs467 Lecture 17 (2005)

2. Contents • The Bell inequality and its violation • Physicist’s perspective • Computer Scientist’s perspective • The magic square game • Communication complexity • Equality checking • Appointment scheduling (quadratic savings) • Are exponential savings possible? • Lower bound for the inner product problem • Simultaneous message passing and fingerprinting

3. The Bell inequality and its violation • Physicist’s perspective • Computer Scientist’s perspective • The magic square game • Communication complexity • Equality checking • Appointment scheduling (quadratic savings) • Are exponential savings possible? • Lower bound for the inner product problem • Simultaneous message passing and fingerprinting

4. s t input: a b output: Bell’s Inequality and its violation Part II: computer scientist’s view: Rules: • No communication after inputs received • They win if ab= st With classical resources, Pr[ab=st] ≤0.75 But, with prior entanglement state 00 –11, Pr[ab=st] =cos2(/8) =½+¼√2 = 0.853…

5. st = 11 st= 01 or 10 3/8 /8 -/8 st=00 The quantum strategy • Alice and Bob start with entanglement  = 00 – 11 • Alice: if s = 0 then rotate by A=/16 else rotate by A=+3/16 and measure • Bob: if t = 0 then rotate by B=/16 else rotate by B=+3/16 and measure cos(A–B) (00 – 11) + sin(A –B) (01 + 10) Success probability: Pr[ab=st] =cos2(/8) =½+¼√2 = 0.853…

6. ! classically, communication is needed Nonlocality in operational terms information processing task quantum entanglement

7. The Bell inequality and its violation • Physicist’s perspective • Computer Scientist’s perspective • The magic square game • Communication complexity • Equality checking • Appointment scheduling (quadratic savings) • Are exponential savings possible? • Lower bound for the inner product problem • Simultaneous message passing and fingerprinting

8. even even even odd odd odd Success probabilities: classical and 1 quantum 8/9 Magic square game Problem: fill in the matrix with bits such that each row has even parity and each column has odd parity IMPOSSIBLE Game: ask Alice to fill in one row and Bob to fill in one column They win iff parities are correct and bits agree at intersection (details omitted here) [Aravind, 2002]

9. The Bell inequality and its violation • Physicist’s perspective • Computer Scientist’s perspective • The magic square game • Communication complexity • Equality checking • Appointment scheduling (quadratic savings) • Are exponential savings possible? • Lower bound for the inner product problem • Simultaneous message passing and fingerprinting

10. Classical communication complexity [Yao, 1979] x1x2  xn y1y2  yn f (x,y) E.g. equalityfunction:f (x,y)=1ifx = y,and 0 ifxy Any deterministic protocol requires n bits communication Probabilistic protocols can solve with only O(log(n/)) bits communication (error probability ), via random hashing

11. x1x2  xn x1x2  xn y1y2  yn y1y2  yn qubits f (x,y) f (x,y)   entangled qubits bits Quantum communication complexity Qubit communication Prior entanglement

12. The Bell inequality and its violation • Physicist’s perspective • Computer Scientist’s perspective • The magic square game • Communication complexity • Equality checking • Appointment scheduling (quadratic savings) • Are exponential savings possible? • Lower bound for the inner product problem • Simultaneous message passing and fingerprinting

13. 1 2 3 4 5 . . . n 1 2 3 4 5 . . . n 0 1 1 0 1 … 0 1 0 0 1 1 … 1 Appointment scheduling x= y= i (xi=yi=1) Classically, (n)bits necessary to succeed with prob.  3/4 For all >0, O(n1/2logn)qubits sufficient for error prob. < [KS ‘87] [BCW ‘98]

14. 1 2 3 4 5 6 . . . n 0 0 0 0 1 0 … 1 x= i x i i i log n b  xi b  xi b b 1 Search problem Given: accessible via queries 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]

15. xy y x x y i i 0 0 0 0 b b Bob Alice Bob 1 2 3 4 5 6 . . . n x= 0 1 1 0 1 0 … 0 Alice y= 1 0 0 1 1 0 … 1 Bob xy= 0 0 0 0 1 0 … 0  Communication per xy-query:2(logn+ 3) = O(log n)

16. Bit communication: Cost:θ(n) Bit communication & prior entanglement: Qubit communication & prior entanglement: Cost:θ(n1/2) Cost:θ(n1/2) Appointment scheduling: epilogue Qubit communication: Cost:θ(n1/2)(with refinements) [R ‘02] [AA ‘03]

17. The Bell inequality and its violation • Physicist’s perspective • Computer Scientist’s perspective • The magic square game • Communication complexity • Equality checking • Appointment scheduling (quadratic savings) • Are exponential savings possible? • Lower bound for the inner product problem • Simultaneous message passing and fingerprinting

18. 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]

19. Classical lower bound Theorem: If S {0,1}n has the property that, for all x, x′S, their intersection size is notn/4 then S < 1.99n Let some protocol solve restricted equality with k bits comm. ● 2k conversations of length k ● approximately2n/n input pairs (x, x), where Δ(x)=n/2 Therefore, 2n/2kn 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 notn/4 Theorem implies 2n/2kn<1.99n , so k> 0.007n [Frankl and Rödl, 1987]

20. For each x {0,1}n, define Quantum protocol • Protocol: • Alice sends x to Bob (log(n) qubits) • 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 xy=0, so result is definitely noty Question: How much communication if error ¼ is permitted? Answer: just 2 bits are sufficient!

21. Exponential quantum vs. classical separation in bounded-error models O(log n) quantum vs. (n1/4 / log n) classical : a log(n)-qubit state (described classically) M: two-outcome measurement U: unitary operation on log(n) qubits Output: result of applying M to U [Raz, 1999]

22. The Bell inequality and its violation • Physicist’s perspective • Computer Scientist’s perspective • The magic square game • Communication complexity • Equality checking • Appointment scheduling (quadratic savings) • Are exponential savings possible? • Lower bound for the inner product problem • Simultaneous message passing and fingerprinting

23. Inner product IP(x,y) = x1y1+x2y2+ +xnyn 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]

24. a1 0 x1 x1 H H f a2 0 x2 x2 H H     H H xn 0 an xn H H 1 1 b b  f(x1, x2, …, xn) H H Recall the BV problem Let f(x1, x2, …, xn) =a1x1+a2x2+ +anxn mod 2 Given: Goal: determine a1, a2,…, an Classically, n queries are necessary Quantum mechanically, 1 query is sufficient

25. Lower bound for inner product 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 zIP(x,y) IP(x,y) = x1y1+x2y2+ +xnyn mod 2 Proof:

26. Lower bound for inner product H H H H H H H H IP(x,y) = x1y1+x2y2+ +xnyn mod 2 0 0 0 1 x1 x2 xn Proof: Alice and Bob’s IP protocol Alice and Bob’s IP protocol inverted x1 x2 xn x1 x2 xn 1 Since n bits are conveyed from Alice to Bob, n qubits communication necessary (by Holevo’s Theorem)

27. The Bell inequality and its violation • Physicist’s perspective • Computer Scientist’s perspective • The magic square game • Communication complexity • Equality checking • Appointment scheduling (quadratic savings) • Are exponential savings possible? • Lower bound for the inner product problem • Simultaneous message passing and fingerprinting

28. Equality function: f (x,y)=1ifx = y 0 ifxy Equality revisitedin simultaneous message model x1x2  xn y1y2  yn f (x,y) Exactprotocols: require 2n bits communication

29. Pr[00] = Pr[11] = ½ random k Equality revisitedin simultaneous message model x1x2  xn y1y2  yn classical classical f (x,y) 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

30. Equality revisitedin simultaneous message model x1x2  xn y1y2  yn Bounded-error protocols without a shared key: f (x,y) Classical: θ(n1/2) Quantum:θ(logn) [A ‘96] [NS ‘96] [BCWW ‘01]

31. 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 |xy|≤ ) Answer: 22am, for some constant a > 0 The states can be constructed via a suitable (classical) error-correcting code, which is a function e:{0,1}n {0,1}cn where, for all x ≠ y,dcn≤Δ(e(x),e(y)) ≤ (1−d)cn (c, dare constants)

32. Set x for each x{0,1}n (log(cn) qubits) Then xy Result:m = rlog(cn) qubits storing 2n= 2(1/c)2m/r different states Construction of almost orthogonal states Since dcn≤Δ(e(x),e(y))≤(1−d)cn, we have |xy| ≤1−2d By duplicating each state, xx … x, the pairwise inner products can be made arbitrarily small: (1−2d)r≤

33. 0 H H x S W A P y Intuition: 0xy +1yx Quantum fingerprints Let 000,001, …,111 be 2n states on O(log n) qubits such that |xy|≤ for all x≠y Given xy, one can check if x=y or x≠y as follows: if x=y, Pr[output= 0] = 1 if x≠y, Pr[output= 0] = (1+2)/2 Note: error probability can be reduced to((1+2)/2)r

34. Equality revisitedin simultaneous message model x1x2  xn y1y2  yn Bounded-error protocols without a shared key: f (x,y) Classical: θ(n1/2) Quantum:θ(logn) [A ‘96] [NS ‘96] [BCWW ‘01]

35. Orthogonality test Quantum protocol for equality in simultaneous message model x1x2  xn y1y2  yn x y x y Recall that, with a shared key, the problem is easy classically ...

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