Introduction to
Quantum Information Processing
QIC 710 / CS 667 / PH 767 / CO 681 / AM 871
Lectures 17-19 (2011)
Richard Cleve
DC 2117
cleve@cs.uwaterloo.ca
1
Complexity class NP
2
Complexity classes
Recall:
• P (polynomial time): problems solved by O(nc)-size
classical circuits (decision problems and uniform circuit
families)
• BPP (bounded error probabilistic polynomial time):
problems solved by O(nc)-size probabilistic circuits that
err with probability  ¼
• BQP (bounded error quantum polynomial time):
problems solved by O(nc)-size quantum circuits that err
with probability  ¼
• PSPACE (polynomial space): problems solved by
algorithms that use O(nc) memory.
3
Summary of previous containments
P  BPP  BQP  PSPACE  EXP
We now consider further
structure between P and
PSPACE
EXP
PSPACE
Technically, we will restrict
our attention to languages
(i.e. {0,1}-valued problems)
BQP
Many problems of interest can
be cast in terms of languages
P
BPP
For example, we could define
FACTORING = {(x,y) :  2  z  y, such that z divides x}
4
NP
Define NP (non-deterministic polynomial time) as
the class of languages whose positive instances have
“witnesses” that can be verified in polynomial time
Example: Let 3-CNF-SAT be the language consisting of all
3-CNF formulas that are satisfiable
3-CNF formula:
f x1 ,..., xn   x1  x3  x4   x2  x3  x5     x1  x5  xn 
f x1 ,..., xn  is satisfiable iff there exists b1 ,...,bn  0,1
such that f b1 ,...,bn   1
No sub-exponential-time algorithm is known for 3-CNF-SAT
But poly-time verifiable witnesses exist (namely, b1, ..., bn)
5
Other “logic” problems in NP
• k-DNF-SAT:
f x1 ,..., xn   x1  x3  x4   x2  x3  x5     x1  x5  xn 
 But, unlike with k-CNF-SAT, this one is known to be in P
• CIRCUIT-SAT:
1
Λ

0

Λ
1
Λ
Λ

1


Λ
0
Λ
Λ



Λ
Λ
Λ

Λ
Λ
Λ

Λ
output
bit
 All known
algorithms
exponentialtime
6
“Graph theory” problems in NP
• k-COLOR: does G have a k-coloring ?
• k-CLIQUE: does G have a clique of size k ?
• HAM-PATH: does G have a Hamiltonian path?
• EUL-PATH: does G have an Eulerian path?
7
“Arithmetic” problems in NP
• FACTORING = {(x, y) :  2  z  y, such that z divides x}
• SUBSET-SUM: given integers x1, x2 , ..., xn, y, do there exist
i1, i2 , ..., ik {1, 2,... , n} such that xi1+ xi2 + ... + xik = y?
• INTEGER-LINEAR-PROGRAMMING: linear programming
where one seeks an integer-valued solution (its existence)
8
P vs. NP
All of the aforementioned problems have the property that
they reduce to 3-CNF-SAT, in the sense that a polynomialtime algorithm for 3-CNF-SAT can be converted into a polytime algorithm for the problem
Example:
algorithm for 3-COLOR
algorithm for
3-CNF-SAT
If a polynomial-time algorithm is discovered for 3-CNF-SAT
then a polynomial-time algorithm for 3-COLOR follows
And this holds for any problem X  NP
9
P vs. NP
For any problem X  NP
algorithm for X
algorithm for
3-CNF-SAT
A problem that has this property is said to be NP-hard
Polynomial-time algorithm any NP-Hard problem implies P=NP
NP-hard problems: 3-CNF-SAT, CIRCUIT-SAT, 3-COLOR ,
k-CLIQUE, HAM-PATH, SUBSET-SUM, INT-LIN-PROG, …
Some problems in P:
10
k-DNF-SAT, 2-COLOR, 3-CLIQUE, EUL-PATH, ...
FACTORING vs. NP
Is FACTORING NP-hard too?
If so, then every problem in
NP is solvable by a poly-time
quantum algorithm!
3-CNF-SAT
But FACTORING has
not been shown to be
NP-hard
PSPACE
NP
co-NP
FACTORING
Moreover, there is “evidence”
that it is not NP-hard:
FACTORING  NPco-NP
If FACTORING is NP-hard then NP = co-NP
P
11
FACTORING vs. co-NP
FACTORING = {(x, y) :  2  z  y, s.t. z divides x}
co-NP: languages whose negative
instances have “witnesses” that
can be verified in poly-time
Question: what is a
good witness for the
negative instances?
PSPACE
NP
co-NP
FACTORING
P
Answer: the prime factorization
p1, p2 , ..., pm of x will work
Can verify primality and compare
p1, p2 , ..., pm with y, all in poly-time
12
Grover’s quantum
search algorithm
13
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)
14
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*
Subtlety in that the search space might
have to be redefined to achieve this
PSPACE
3-CNF-SAT
NP
co-NP
FACTORING
P
15
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
16
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
17
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)
18
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
xA
xB
Consider the space spanned by A and B
A  goal is to get close to this state
H0...0
B

1
N
x

a
N
A 
b
N
B
x{ 0 ,1} n
Interesting case: a << N
19
Grover’s algorithm: analysis
II
A
Algorithm:
(HU0 HUf )k H 0...0
H0...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
H0...0
Partial proof:
H U0 HH0...0 = H U0 0...0 = H( 0...0) = H 0...0
20
Grover’s algorithm: analysis
III
A
Algorithm:
2
2
2
2

(HU0 HUf )k H 0...0
H0...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
22
Optimality of
Grover’s algorithm
23
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
24
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
25
Optimality of Grover’s algorithm III
Consider
|0
U0
I
U1
ki
I
U2
fr
U3
fr
Uk |ψr,i
i
Note that
|ψr,k  |ψr,0 = (|ψr,k  |ψr,k1) + (|ψr,k1  |ψr,k2) + ... + (|ψr,1  |ψr,0)
which implies
|| |ψr,k  |ψr,0 ||  || |ψr,k  |ψr,k1 || + ... + || |ψr,1  |ψr,0 ||
26
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 || = |2i,r|, since query only negates |r
Therefore, || |ψr,k
 |ψr,0 || 
k 1
2α
i 0
i ,r
27
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

k 1
( )
1 k -1
£ n å2 2n
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
28