Peter Høyer hoyer Quantum Searching August 1, 2005, Université de Montréal The Fifth Canadian...
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Transcript of Peter Høyer hoyer Quantum Searching August 1, 2005, Université de Montréal The Fifth Canadian...
Peter Høyer
www.cpsc.ucalgary.ca/~hoyer
Quantum Searching
August 1, 2005, Université de MontréalThe Fifth Canadian Summer School on Quantum Information
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)
Grover’s Problem
0 1 0 0 1 0 0 0 0 1 0 0
1 N
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)?”
Reversible
j f(j)
“What is f(j)?”
Query
Remember the input:
j
f(j)Queryj
Reversible:
j
bf(j)Query
j
b
00 = 001 = 110 = 1
11 = 0
b {0,1}
Unitary Query
j f(j)
“What is f(j)?”
Query
Remember the input:
j
f(j)Queryj
Reversible:
j
bf(j)Query
j
b
Unitary:
Uf
|j
|b
|j
|bf(j)
b {0,1}
Unitary Query
Unitary:
Uf
|j
|b
|j
|bf(j)
|j|b |j|bf(j)
|j|0∑j=1
N∑j=1
N1
√N1
√N |j|f(j)
“Apply function f in superposition”
Alternative query model
Sf|j (-1)f(j)|j Uf
|j
|b
|j
|bf(j)
Uf
|j
|1
(-1)f(j)|j
|1H H
Proof of :
|b H |0+(-1)b|1
“Compute the value of f in the phases”
Grover’s Algorithm
0 1 0 0 1 0 0 0 0 1 0 0
1 N
Suppose t solutions (here t=3)
Grover: 1/p√ = N/t queries√
Repetition: 1/p = N/t queries
The “classical” algorithm: N1
√N
Success probability = p = t/N
A|0|
0∑i=1|i
Standard Analysis
0 1 0 0 1 0 0 0 0 1 0 0
1 N
1/N2/N3/N
Success probability
Classical repetition Grover
4/N5/N6/N7/N8/N9/N
N1
√N
A|0|
0∑i=1|i
(1)(2)(3) (1)(2)(3)
Amplitude Amplification
A = some algorithm p = success probability of A
m repetitions success probability ≈ mp (provided mp ≤ 2/3, say)
Amplitude Amplification
A = some algorithm p = success probability of A
m repetitions success probability ≈ mp (provided mp ≤ 2/3, say)
(Solutions must be verifiable)
√
General Setting
Algorithm A,
Verifier f, f:{1,2,,N}{0,1}
A|0|0∑i=1 i|i N
0 1 0 0 1 0 0 0 0 1 0 0
1 N
f:{1,2,,N}{0,1}
A|0|
0∑i=1|iN
1√N
Example: Grover
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 Uf
1 unit cost Uf
|j
|b
|j
|bf(j)
A|0 |0
General Setting
Algorithm A,
Let |Good~∑i:f(i)=1 i|i
|Bad~∑i:f(i)=0 i|i
Success prob. of A = p = sin2θ
Verifier f, f:{1,2,,N}{0,1}
A|0|0∑i=1 i|i N
A|0|0sinθ|Goodcosθ|Bad
General Setting
Algorithm A,
Let |Good~∑i:f(i)=1 i|i
|Bad~∑i:f(i)=0 i|i
Success prob. of A = p = sin2θ
Verifier f, f:{1,2,,N}{0,1}
A|0|0sinθ|Goodcosθ|Bad
A|0|0∑i=1 i|i N
2-dimensional subspaceA|0|0sinθ|Goodcosθ|Bad
|Good
|Badθ
|02θ |
Q|
Operator Q rotates by angle 2θ
3θ5θ
QA|0
QQA|0
Amplitude Amplification
Algorithm A,
Let |Good~∑i:f(i)=1 i|i
|Bad~∑i:f(i)=0 i|i
Success prob. of A = p = sin2θ
Verifier f, f:{1,2,,N}{0,1}
A|0|0sinθ|Goodcosθ|Bad
A|0|0∑i=1 i|i N
2θ |
Q||Good
|Bad
Amplitude Amplification|Good
|Badθ
A|0|0
A|0 sinθ|Goodcosθ|Bad
QQA|0 sin5θ|Goodcos5θ|Bad
5θ
QA|0
sin3θ|Goodcos3θ|Bad
3θ
QQQA|0 sin7θ|Goodcos7θ|Bad
7θ
QmA |0 = sin((2m+1)θ) |Good + cos((2m+1)θ) |Bad
Maximizing the success prob.|Good
|Bad
QmA |0 = sin((2m+1)θ) |Good + cos((2m+1)θ) |Bad
θA|0|0
4 π √p
Theorem:
Set m= ,
then Prob[Bad] ≤ sin2θ = p
π2
After m rotations, angle is (2m+1)θ.
We want (2m+1)θ ≈ .
Since sin2θ = p, then θ ≈ √p. □
2θ
Proof:
Note: Classically, m= . A quadratic speed-up! 1p
“De-randomization” (p known) |Good
|Bad
QmA |0 = sin((2m+1)θ) |Good + cos((2m+1)θ) |Bad
θA|0|0
Theorem:
If prob. p is known,
we can “de-randomize”.
2θ
π2
If (2m+1)θ is slightly more than ,
then choose slightly smaller angle θ’<θ
such that (2m+1)θ’ IS equal to . □
π2Proof:
θ’
Guessing when p unknown
θA|0|0
|Good
|Bad
A random vector yields success prob. = 1/2
Solution: obtain a near-random vector by classically guessing m.
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?
Guessing when p unknown
θA|0|0
|Good
|Bad
Set M=1, λ=1.99Repeat forever: Set M=λM
Let mR[1,M]
Do QmA|0 Measure, say |i If f(i)=1, then breakOutput i
1)
2)
3)
Algorithm:
Q Rotation
|Good
|Bad
2θ |
Q| S0|0 = |0 S0|i = - |i (i≠0)
Sf|i = |i (f(i)=1) Sf|i = - |i (f(i)=0)
Q = - SfAS0A-1
Q Rotation
|Good
|Bad
S0|0 = |0 S0|i = - |i (i≠0)
Sf|i = |i (f(i)=1) Sf|i = - |i (f(i)=0)
Sf reflection about |Good
AS0A-1 reflection about |0
Q rotation by 2θ
θ
|2θ
2θ
|0
Q = SfAS0A-1-
Cost of Q
Q = -AS0A-1Sf
1 application of A
1 query to Uf
1 unit cost Uf
|j
|b
|j
|bf(j)
A|0 |0
1 application of Q 2 applications of Uf
+ 2 applications of A
+ 1 application of S0
5 units cost
The Recipe
Algorithm A,
Verifier f, f:{1,2,,N}{0,1}
A|0|0∑i=1 i|i N
Q = -AS0A-1Sf
S0|0 = |0 S0|i = |i (i≠0)
Sf|i = |i (f(i)=1) Sf|i = |i (f(i)=0)
Define reflections:
Amplitude Amplification operator:
- -
Complete Grover
0 0 0 0 1 0 0 0 0 0 0 0
1 Nf:{1,2,,N}{0,1}
Q = -AS0A-1Sf
S0|0 = |0 S0|i = - |i (i≠0)
Sf|i = |i (f(i)=1) Sf|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
General rotation
Q = -AS0A-1SfS0|0 = |0 S0|i = |i (i≠0)
Sf|i = |i (f(i)=1) Sf|i = |i (f(i)=0)
-
-
Φs
Φf Φs,Φf C*
-
-
Suitable choices of phases:
Φs = Φf = -1
Φs = Φf = i = √-1
Φs = Φf = eπi/3
Φs = Φf such that Re(Φs)>0
General AnalysisA|0sinθ|Goodcosθ|Bad
Q(Φs,Φf)A|0αsinθ|Goodβcosθ|Bad
-ΦsΦf
-1+Φf-ΦsΦf
1-Φs-Φf
-Φs
sin2θ
cos2θ
α
β=
General AnalysisA|0sinθ|Goodcosθ|Bad
Q(Φs,Φf)A|0αsinθ|Goodβcosθ|Bad
-ΦsΦf
-1+Φf-ΦsΦf
1-Φs-Φf
-Φs
sin2θ
cos2θ
α
β=
Φs = Φf = -1
-1
-3
3
1
sin2θ
cos2θ
α
β=
General AnalysisA|0sinθ|Goodcosθ|Bad
Q(Φs,Φf)A|0αsinθ|Goodβcosθ|Bad
-ΦsΦf
-1+Φf-ΦsΦf
1-Φs-Φf
-Φs
sin2θ
cos2θ
α
β=
Φs = Φf = i
1
i
1-2i
-i
sin2θ
cos2θ
α
β=
General AnalysisA|0sinθ|Goodcosθ|Bad
Q(Φs,Φf)A|0αsinθ|Goodβcosθ|Bad
-ΦsΦf
-1+Φf-ΦsΦf
1-Φs-Φf
-Φs
sin2θ
cos2θ
α
β=
Φs = Φf = eπi/3
-e2πi/3
0
1-2eπi/3
-eπi/3
sin2θ
cos2θ
α
β=
General AnalysisA|0sinθ|Goodcosθ|Bad
Q(Φs,Φf)A|0αsinθ|Goodβcosθ|Bad
-ΦsΦf
-1+Φf-ΦsΦf
1-Φs-Φf
-Φs
sin2θ
cos2θ
α
β=
Φs = Φf such that Re(Φs)>0
<1
sin2θ
cos2θ
α
β=
? ?
?