Consequences and Limits of Nonlocal Strategies
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Transcript Consequences and Limits of Nonlocal Strategies
Introduction to
Quantum Information Processing
CS 667 / PH 767 / CO 681 / AM 871
Lecture 18 (2009)
Richard Cleve
DC 2117
[email protected]
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Grover’s quantum
search algorithm
2
Quantum search problem
f : {0,1}n {0,1}
f is satisfiable (if x {0,1}n s.t. f(x) = 1)
Given: a black box computing
Goal: determine if
In positive instances, it makes sense to also find such a satisfying
assignment x
Classically, using probabilistic procedures, order 2n queries are
necessary to succeed—even with probability ¾ (say)
Grover’s quantum algorithm that makes only O(2n) queries
Query:
[Grover ’96]
x1
xn
y
x1
Uf
xn
y f(x1,...,xn)
3
Applications of quantum search
The function f could be realized as a 3-CNF formula:
f x1 ,..., xn x1 x3 x4 x2 x3 x5 x1 x5 xn
Alternatively, the search could
be for a certificate for any
problem in NP
The resulting quantum
algorithms appear to be
quadratically more efficient
than the best classical
algorithms known
PSPACE
3-CNF-SAT
NP
co-NP
FACTORING
P
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Prelude to Grover’s algorithm:
two reflections = a rotation
Consider two lines with intersection angle :
reflection 2
2
2
1
1
reflection 1
Net effect: rotation by angle 2, regardless of starting vector
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Grover’s algorithm: description I
Basic operations used:
x1
xn
y
x1
Uf
x1
xn
y
xn
y f(x1,...,xn)
x1
U0
H
Uf x = (1) f(x) x
xn
y [x = 0...0]
Hadamard
Implementation?
X
X
X
X
X
X
U0 x = (1) [x = 0...0]x
H
H
H
6
Grover’s algorithm: description II
iteration 1
iteration 2
...
0
0
H
Uf
H
U0
H
Uf
H
U0
H
1. construct state H 0...0
2. repeat k times:
apply H U0 HUf to state
3. measure state, to get x{0,1}n, and check if
f(x) =1
(The setting of k will be determined later)
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Grover’s algorithm: analysis I
A = {x {0,1}n : f (x) = 1} and B = {x {0,1}n : f (x) = 0}
and N = 2n and a = |A| and b = |B|
Let
Let
A
a
1
x
x
b
B
and
1
xA
xB
Consider the space spanned by A and B
A goal is to get close to this state
H0...0
B
1
N
x
a
N
A
b
N
B
x{ 0 ,1} n
Interesting case: a << N
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Grover’s algorithm: analysis II
A
Algorithm:
(HU0 HUf )k H 0...0
H0...0
B
Observation:
Uf
is a reflection about B:
Uf A = A
Question: what is HU0 H ?
U0
and
Uf B = B
is a reflection about
H0...0
Partial proof:
H U0 HH0...0 = H U0 0...0 = H( 0...0) = H 0...0
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Grover’s algorithm: analysis III
A
Algorithm:
2
2
2
2
(HU0 HUf )k H 0...0
H0...0
B
Since HU0 HUf is a composition of two reflections, it is a rotation
by 2, where sin()=a/N
When a = 1, we want
a/N
(2k+1)(1/N) /2 , so k (/4)N
More generally, it suffices to set k
(/4)N/a
Question: what if a is not known in advance?
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Unknown number of solutions
1 solution
2 solutions
3 solutions
6 solutions
100 solutions
success
probability
1
0 number of iterations
√N/2
4 solutions
success probability
very close to zero!
Choose a random k in the range to get success probability > 0.43
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Optimality of
Grover’s algorithm
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Optimality of Grover’s algorithm I
Theorem: any quantum search algorithm for
(2n) queries to
f
f : {0,1}n {0,1} must make
(if f is used as a black-box)
Proof (of a slightly simplified version):
|x
Assume queries are of the form
(1) f(x)|x
f
and that a k-query algorithm is of the form
|0...0
U0
f
U1
f
U2
f
U3
where U0, U1, U2, ..., Uk, are arbitrary unitary operations
f
Uk
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Optimality of Grover’s algorithm II
Define fr : {0,1}n {0,1} as fr (x) = 1 iff x = r
Consider
|0
U0
fr
U1
fr
U2
fr
U3
fr
Uk |ψr,k
I
U1
I
U2
I
U3
I
Uk |ψr,0
versus
|0
U0
We’ll show that, averaging over all r {0,1}n, || |ψr,k
|ψr,0 || 2k / 2n
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Optimality of Grover’s algorithm III
Consider
|0
U0
I
U1
ki
I
U2
fr
U3
fr
Uk |ψr,i
i
Note that
|ψr,k |ψr,0 = (|ψr,k |ψr,k1) + (|ψr,k1 |ψr,k2) + ... + (|ψr,1 |ψr,0)
which implies
|| |ψr,k |ψr,0 || || |ψr,k |ψr,k1 || + ... + || |ψr,1 |ψr,0 ||
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Optimality of Grover’s algorithm IV
query
|0
U0
I
U1
I
query
|0
U0
I
U1
i
I
query i+1
fr
U2
i
U3
|ψr,i
fr
Uk
fr
Uk |ψr,i-1
query i+1
I
U2
α
i ,x
U3
x
x
|| |ψr,i |ψr,i-1 || = |2i,r|, since query only negates |r
Therefore, || |ψr,k
|ψr,0 ||
k 1
2α
i 0
i ,r
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Optimality of Grover’s algorithm V
Now, averaging over all r {0,1}n,
1
2n
r
ψ r ,k ψ r ,0
1
n
2
k 1
2 α i ,r
r
i 0
1
n
2
2 α i ,r
i 0 r
1
n
2
k 1
k 1
n
2
2
i 0
(By Cauchy-Schwarz)
2k
2n
Therefore, for some r {0,1}n, the number of queries k must be (2n),
in order to distinguish fr from the all-zero function
This completes the proof
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