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Peter Høyer www.cpsc.ucalgary.ca/~hoyer Quantum Searching August 1, 2005, Université de Montréal The Fifth Canadian Summer School on Quantum Information.

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Presentation on theme: "Peter Høyer www.cpsc.ucalgary.ca/~hoyer Quantum Searching August 1, 2005, Université de Montréal The Fifth Canadian Summer School on Quantum Information."— Presentation transcript:

1 Peter Høyer www.cpsc.ucalgary.ca/~hoyer Quantum Searching August 1, 2005, Université de Montréal The Fifth Canadian Summer School on Quantum Information

2 Quadratic speed-up Cost was O(M) * Actual running time on a quantum computer might be incomparable to running time on a classical computer. Availability of quantum computers is limited. Additional error correction not included. Not applicable in conjunction with measurements. Verifier required. * √ Now only O( M) Cost: O(M)

3 Grover’s Problem 010010000100 1N Input: Problem: f:{1,2,,N}{0,1} Find integer i such that f(i)=1 You can ask questions of the form: “What is f(j)?”

4 A query j f(j) “What is f(j)?” Query

5 Remember the input j f(j) “What is f(j)?” Query Remember the input: j f(j) Query j

6 Reversible j f(j) “What is f(j)?” Query Remember the input: j f(j) Query j Reversible: j bf(j) Query j b 0000 = 0 0101 = 1 1010 1111 = 0 b  {0,1}

7 Unitary Query j f(j) “What is f(j)?” Query Remember the input: j f(j) Query j Reversible: j bf(j) Query j b Unitary: U f |j |b |j |bf(j) b  {0,1}

8 Unitary Query Unitary: U f |j |b |j |bf(j) |j|b |j|bf(j) |j|0 ∑ j=1 N ∑ N 1 √N 1 |j|f(j) “Apply function f in superposition”

9 Computational Cost U f |j |b |j |bf(j) 1 query  1 unit cost

10 Alternative query model S f |j (-1) f(j) |j U f |j |b |j |bf(j)  U f |j |1 (-1) f(j) |j |1 H H Proof of  : |b H |0+(-1) b |1 “Compute the value of f in the phases”

11 Grover’s Algorithm 010010000100 1N Suppose t solutions (here t=3) Grover:1/p √ = N/t queries √ Repetition:1/p = N/t queries The “classical” algorithm: N 1 √N Success probability = p = t/N A|0| 0 ∑ i=1 |i

12 Standard Analysis 010010000100 1N 1/N 2/N 3/N Success probability Classical repetitionGrover 4/N 5/N 6/N 7/N 8/N 9/N N 1 √N A|0| 0 ∑ i=1 |i (1)(2)(3)(1)(2)(3)

13 Grover’s Algorithm  Rotation

14 Amplitude Amplification A = some algorithm p = success probability of A m repetitions  success probability ≈ mp (provided mp ≤ 2/3, say)

15 Amplitude Amplification A = some algorithm p = success probability of A m repetitions  success probability ≈ mp (provided mp ≤ 2/3, say) (Solutions must be verifiable) √

16 General Setting Algorithm A, Verifier f, f:{1,2,,N}{0,1} A|0| 0 ∑ i=1  i |i N 010010000100 1N f:{1,2,,N}{0,1} A|0| 0 ∑ i=1 |i N 1 √N Example: Grover

17 General Setting Algorithm A, Verifier f, f:{1,2,,N}{0,1} A|0| 0 ∑ i=1  i |i N  1 application of A 1 query to U f 1 unit cost U f |j |b |j |bf(j) A |0 |0|0

18 General Setting Algorithm A, Let |Good~∑ i:f(i)=1  i |i |Bad~∑ i:f(i)=0  i |i Success prob. of A = p = sin 2 θ Verifier f, f:{1,2,,N}{0,1} A|0| 0 ∑ i=1  i |i N A|0| 0 sinθ|Goodcosθ|Bad 

19 General Setting Algorithm A, Let |Good~∑ i:f(i)=1  i |i |Bad~∑ i:f(i)=0  i |i Success prob. of A = p = sin 2 θ Verifier f, f:{1,2,,N}{0,1} A|0| 0 sinθ|Goodcosθ|Bad A|0| 0 ∑ i=1  i |i N

20 2-dimensional subspace A|0| 0 sinθ|Goodcosθ|Bad |Good |Bad θ |0|0 2θ2θ || Q| Operator Q rotates by angle 2θ 3θ3θ 5θ5θ QA|0 QQA|0

21 Amplitude Amplification Algorithm A, Let |Good~∑ i:f(i)=1  i |i |Bad~∑ i:f(i)=0  i |i Success prob. of A = p = sin 2 θ Verifier f, f:{1,2,,N}{0,1} A|0| 0 sinθ|Goodcosθ|Bad A|0| 0 ∑ i=1  i |i N 2θ2θ || Q| |Good |Bad

22 Amplitude Amplification |Good |Bad θ A|0| 0  A|0sinθ|Goodcosθ|Bad QQA|0sin5θ|Goodcos5θ|Bad 5θ5θ QA|0sin3θ|Goodcos3θ|Bad 3θ3θ QQQA|0sin7θ|Goodcos7θ|Bad 7θ7θ Q m A |0 = sin((2m+1)θ) |Good + cos((2m+1)θ) |Bad

23 Maximizing the success prob. |Good |Bad Q m A |0 = sin((2m+1)θ) |Good + cos((2m+1)θ) |Bad θ A|0| 0  4 π √p Theorem: Set m=, then Prob[Bad] ≤ sin 2 θ = p π2π2 After m rotations, angle is (2m+1)θ. We want (2m+1)θ ≈. Since sin 2 θ = p, then θ ≈ √p. □ 2θ2θ Proof: Note: Classically, m=. A quadratic speed-up! 1p1p

24 “De-randomization” (p known) |Good |Bad Q m A |0 = sin((2m+1)θ) |Good + cos((2m+1)θ) |Bad θ A|0| 0  Theorem: If prob. p is known, we can “de-randomize”. 2θ2θ π2π2 If (2m+1)θ is slightly more than, then choose slightly smaller angle θ’<θ such that (2m+1)θ’ IS equal to. □ π2π2 Proof: θ’θ’

25 If m is not known? |Good |Bad θ 2θ2θ A|0| 0 

26 Guessing when p unknown θ A|0| 0  |Good |Bad A random vector yields success prob. = 1/2 Solution: obtain a near-random vector by classically guessing m.

27 Bizarre example Viewer discretion is advised...

28 Kiil, the cat  2 3 4 5 6 7 8 9 10 11   13 1 cat 2 cats  11 trials 7 trials What is the highest floor from which the cat will survive, if it jumps out the window from that floor?

29 Guessing when p unknown θ A|0| 0  |Good |Bad Set M=1, λ=1.99 Repeat forever: Set M=λM Let m  R [1,M] Do Q m A|0 Measure, say |i If f(i)=1, then break Output i 1) 2) 3) Algorithm:

30 Q  Rotation |Good |Bad 2θ2θ || Q| S 0 |0 = |0 S 0 |i = - |i (i≠0) S f |i = |i (f(i)=1) S f |i = - |i (f(i)=0) Q = - SfSf AS 0 A -1

31 Q  Rotation |Good |Bad S 0 |0 = |0 S 0 |i = - |i (i≠0) S f |i = |i (f(i)=1) S f |i = - |i (f(i)=0) S f  reflection about |Good AS 0 A -1  reflection about | 0   Q rotation by 2θ θ || 2θ2θ 2θ2θ |0|0 Q = SfSf AS 0 A -1 -

32 Cost of Q Q = -AS 0 A -1 S f  1 application of A 1 query to U f 1 unit cost U f |j |b |j |bf(j) A |0 |0|0 1 application of Q2 applications of U f + 2 applications of A + 1 application of S 0 5 units cost 

33 The Recipe Algorithm A, Verifier f, f:{1,2,,N}{0,1} A|0| 0 ∑ i=1  i |i N Q = -AS 0 A -1 S f S 0 |0 = |0 S 0 |i = |i (i≠0) S f |i = |i (f(i)=1) S f |i = |i (f(i)=0) Define reflections: Amplitude Amplification operator: - -

34 Complete Grover 000010000000 1N f:{1,2,,N}{0,1} Q = -AS 0 A -1 S f S 0 |0 = |0 S 0 |i = - |i (i≠0) S f |i = |i (f(i)=1) S f |i = - |i (f(i)=0) N A|0=∑ i=1 |i 1 √N Apply A on |0, gives | Repeat O(√N) times: Apply Q on | Measure | Return outcome |i 1) 2) 3) 4) Grover’s algorithm: Theorem: Prob[f(i)=1] ≥ 2/3

35 General rotation Q = -AS 0 A -1 S f S 0 |0 = |0 S 0 |i = |i (i≠0) S f |i = |i (f(i)=1) S f |i = |i (f(i)=0) - - ΦsΦs ΦfΦf Φ s,Φ f  C* - - Suitable choices of phases: Φ s = Φ f = -1 Φ s = Φ f = i = √-1 Φ s = Φ f = e π i/3 Φ s = Φ f such that Re(Φ s )>0

36 General Analysis A|0sinθ|Goodcosθ|Bad Q(Φ s,Φ f )A|0 α sinθ|Good β cosθ|Bad -Φ s Φ f -1+Φ f -Φ s Φ f 1-Φ s -Φ f -Φs-Φs sin 2 θ cos 2 θ α β =

37 General Analysis A|0sinθ|Goodcosθ|Bad Q(Φ s,Φ f )A|0 α sinθ|Good β cosθ|Bad -Φ s Φ f -1+Φ f -Φ s Φ f 1-Φ s -Φ f -Φs-Φs sin 2 θ cos 2 θ α β = Φ s = Φ f = -1 -3 3 1 sin 2 θ cos 2 θ α β =

38 General Analysis A|0sinθ|Goodcosθ|Bad Q(Φ s,Φ f )A|0 α sinθ|Good β cosθ|Bad -Φ s Φ f -1+Φ f -Φ s Φ f 1-Φ s -Φ f -Φs-Φs sin 2 θ cos 2 θ α β = Φ s = Φ f = i 1 i 1-2i -i sin 2 θ cos 2 θ α β =

39 General Analysis A|0sinθ|Goodcosθ|Bad Q(Φ s,Φ f )A|0 α sinθ|Good β cosθ|Bad -Φ s Φ f -1+Φ f -Φ s Φ f 1-Φ s -Φ f -Φs-Φs sin 2 θ cos 2 θ α β = Φ s = Φ f = e π i/3 -e 2π i/3 0 1-2e π i/3 -e π i/3 sin 2 θ cos 2 θ α β =

40 General Analysis A|0sinθ|Goodcosθ|Bad Q(Φ s,Φ f )A|0 α sinθ|Good β cosθ|Bad -Φ s Φ f -1+Φ f -Φ s Φ f 1-Φ s -Φ f -Φs-Φs sin 2 θ cos 2 θ α β = Φ s = Φ f such that Re(Φ s )>0 <1 sin 2 θ cos 2 θ α β = ? ? ?

41 The Recipe Algorithm A, Verifier f, f:{1,2,,N}{0,1} A|0| 0 ∑ i=1  i |i N Q = -AS 0 A -1 S f S 0 |0 = |0 S 0 |i = |i (i≠0) S f |i = |i (f(i)=1) S f |i = |i (f(i)=0) Define reflections: Amplitude Amplification operator: - -


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