Quantum ComputingMAS 725Hartmut KlauckNTU9.4.2012

Today

Lower bounds in the black box model polynomial method adversary method

The black box model

Input x1,…,xn with query access We want to compute some function f(x1,…,xn) Examples:

Deutsch‘s problem: x1,x2 are the values g(0) and g(1) for a function g, we want to know if g is balanced) Compute parity of x1,x2

Grover: Find i mit xi=1, consider f=OR

The black box model

We measure the number of queries an algorithm does in the worst case, cost of an algorithm Not time, space etc.

The query complexity of a function f is the minimal cost of an algorithm computing f

Types of algorithms

Deterministic algorithms: classical, no error, the complexity is denoted D(f)

Randomized algorithms: allow error probability 1/3, over the randomness of the algorithm, complexity R(f)

Quantum algorithms: We count the number of quantum queries. With error 1/3: Q(f) Withour error: QE(f)

Examples

We know some bounds already: QE(XOR2)=1, but R(XOR2)=2 QE(XORn)· d n/2e Q(OR)=O(n1/2) R(OR)=(n) Q(OR)=(n1/2)

How can we show quantum lower bounds in general? How much better can quantum algorithms be compared to

randomized algorithms?

Examples

We showed the lower bound for OR with an adversary argument

Are there other more general techniques?

For certain problems (Simon, Period Finding) we have seen exponential speedups

What is the largest speedup for a total Boolean function?

What is the hardest function?

Consider ID: ID(x)=x How many queries are needed to compute ID?

R(ID)=n We get no information on a position we do not

query Formally: apply the Yao principle: fix random

string, then a deterministic algo with n-1 queries must have error 1/2.

Q(ID)· n/2+O(n1/2)

Ansatz: We use the sign oracle (-1)xi |ii Formula for Hadamard transform:

Another Hadamard maps back to |xi So suffices to generate this superposition Doing this exactly needs n queries Number of queries corresponds to |y| Use only y with|y|· n/2+O(n1/2)

Algorithm

Generate uniform superposition over all y with |y|· n/2+O(n1/2)

On |yi query all bits xi with yi=1 Apply Hadamard

Success probability? 99% of all strings have Hamming weight at most

n/2+O(n1/2) Hence resulting superposition is close to the

desired one and we will have small error For example error 5% with n/2+n1/2 queries

Conclusion

n/2· Q(ID)· n/2+n1/2

...if we can show Q(XOR)¸ n/2

The polynomial method

A quantum black box algorithm is a sequence of unitaries and query unitaries

We construct a low degree polynomial from such an algorithm that represents the computed function

Then we analyze the minimum degree for such polynomials

The polynomial method

We claim that the amplitudes of the final state of a T query algorithm can be written as polynomials of degree T

Proof by induction: T=0: amplitudes do not depend on the input, i.e. are constants T! T+1: i(x) is given by a degree T polynomial Next we apply a unitary that does not depend on x The new ®i(x) is a linear combination of degree T polynomials,

degree unchanged The query transformation: state |ii|ai|ki maps to |ii|a© xii|ki The new iak(x) is xi¢ i, a©1, k + (1-xi)¢ i, a, k Degree is no more than T+1

The polynomial method

The acceptance probability on input x is the sum of squred amplitudes

Hence can be written as a polynomial of degree 2T We may replace xi

k by xi

The result is multilinear

Conclusion

Given a quantum algorithm with T queries that computed f exactly we get a multilinear polynomial of degree 2T that satisfies p(x)=f(x) for all x.

If the quantum algorithm has error 1/, then there is a polynomial with p(x)2[0,1/3] for f(x)=0 and p(x)2[2/3,1] for f(x)=1

Now we have to consider the degree of polynomials representing Boolean functions

Exact quantum algorithms

For every total Boolean function f there is a unique multilinear polynomial that represents f exactly

deg(f) denotes the degree of this polynomial QE(f)¸ deg(f)/2 Example: XOR, the polynomial is

Degree is n, and QE(XOR) =n/2

Multilinear polynomials

Claim: For every Boolean function there is a unique multilinear polynomial, which represents f exactly, i.e. f(x)= p(x) for all Boolean x. Proof: Assume f(x)=p(x)=q(x) for alle Boolean x, yet pq Then p-q is a multilinear polynomial for g(x)=0, and p-q is

not the zero polynomial Take a minimum degree monomial m in p-q with nonzero

coefficient a0. Let z be the string string, that contains all xi in m as ones,

and contains no other ones m(z)=1, and p(z)=a, contradiction

Another example

Polynomial for OR:

Also has degree n

Hence QE(OR)¸ n/2

But actually QE(OR)=n

QE(OR)

We consider the amplitudes of the final state of an optimal algorithm for OR (no error), as polynomials of degree T

Basis states |0yi are rejecting. B is the set of those For i in B we have pi(x)=0 when x is not 0n

There is a j in B with pj(0n) 0 Consider the real part q(x) on

1-pj(x)/pj(0n) Then: deg (q) · T = QE(OR) and q(0n)=0 and q(x)=1 for other x,

hence deg(q)¸ deg(OR) = n

Some facts about polynomials

If a Boolean f depends on n variables, we have deg(f)¸ log n - loglog n

All symmetric, nonconstant f have degree n-o(n) Hence QE(f)¸ (log n)/2 –o(log n) And QE(f)¸ n/2-o(n) for symmetric nonconstant f

Approximating polynomials

Given a quantum algorithm with T queries that we get a multilinear polynomial of degree 2T with p(x)2[0,1/3] for f(x)=0 and p(x)2[2/3,1] for f(x)=1

adeg(f) denotes the minimal degree of such a polynomial

Then: Q(f)¸ adeg(f)/2 Example OR on 2 bits: x1/3 + x2/3 + 1/3,

adeg(OR2)=1

Symmetrization

Let f denote a symmetric function, i.e, f is constant on all x with |x|=k, i.e., we have function values f0,…,fn

Symmetrization turns a multilinear polynomial for f into a univariate polynomial of the same degree

p(k)2[0,1/3] for fk=0 and p(k)2[2/3,1] for fk=1

XOR

We get a polynomial such that p(k)2[0,1/3] for even k and p(k)2[2/3,1] for odd k

k2{0,…,n}

Clearly p(k)-1/2 has n roots! And degree n

Q(XOR)=n/2

Symmetrization

Let be a permutation of [1,…,n], p polynomial in n var. Set psym(x)=

Lemma: If p is a multilinear polynom of degree d, then there is a univariate degree d polynomial with q(|x|) = psym(x) for all x Proof: Let Vj denote the sum of all products of j variables Then psym can be written as i=0...d bi Vi

Value of Vj on x is

The sum is

The approximation degree of OR But what about OR?

We know already Q(OR)=n1/2

But adeg(OR) could be smaller Symmetrization:

p(0) 2[0,1/3] p(1),…,p(n)2[2/3,1]

What is the minimal degree of such a polynomial?

A result from approximation theory Theorem: Let p be a polynomial with 0· p(x)· 1 for

all integers 0· x· nsuch that |p’(x)|¸ c for some real 0· x· n

Then deg(p)¸ (cn/(c+1))1/2

But: p(0)<1/3 and p(1)>2/3 Hence p’(x)¸ 1/3 for some x2 [0,1] adeg(OR)¸ (n/4)1/2

We recover the lower bound for search etc.

Some more facts

For a total Boolean function f we have D(f)=O(adeg(f)6)

Hence also D(f)=O(Q(f)6) This is clearly only true for total functions The best speedup that is known (Grover) is only

quadratic Polynomial method is very useful for functions with

a lot of symmetry, example Element Distinctness

The adversary method

This method actually characterizes Q(f) [in its strongest form]

Leads to a characterization as a semidefinite program

Original idea is to bound the progress achieved by one query in distinguishing pairs of inputs

Certificate complexity

A certificate for x is a set of positions and values that fixes the function value for all x that are consistent with them Example: x1=1 fixes OR XOR has no certificate of length <n

C(f) is the max over all x of the min certificate for x C(XOR)=n C(OR)=n

0n needs a certificate of size n

An observation

There are 1-certificates and 0-certificates xi=1 is 1-cert for OR x1=0,…, xn=0 is 0-cert for OR

For all f: 1-cert and 0-cert need to share at least one variable

Certificate and adversaryC(f ) = min

px2f 0;1gnmax

x

X

jpx(j )

such thatX

j :xj 6=yj

px(j )py(j ) ¸ 1 if f (x) 6= f (y)

Adv(f ) = minpx2R n

maxx

X

jpx(j )2

such thatX

j :xj 6=yj

px(j )py(j ) ¸ 1 if f (x) 6= f (y)

Adversary boundAdv(f ) = min

px2R nmax

xjjpxjj2

such thatX

j :xj 6=yj

px(j )py(j ) ¸ 1 if f (x) 6= f (y)

Example: OR on 0

n 10

n-1 … 0

n-11

px: (1…1) (1 0….) (01 0…) (0…01)

Rescale by 1/n

1/2 n

1/2

Adv(OR)· n

1/2

Adversary bound

Need to show two things: Q(f)=(Adv(f)) How to prove lower bounds on Adv(f)

How to prove lower bounds

Adversary bound as stated is a minimization problem, so we take the dual

Adv(f ) = max¡ 2R D£ D

jj¡ jj

such that

¡ (x;y) ¸ 0

¡ (x;y) = 0 if f (x) = f (y)

8j jj¡ ±X

x;y:xj 6=yj

jxihyj jj · 1

Generalized adversary bound

Adv§ (f ) = max¡ 2R D£ nD

jj¡ jj

such that

¡ (x;y) = 0 if f (x) = f (y)

8j jj¡ ±X

x;y:xj 6=yj

jxihyjjj · 1

This bound is asymptotically equal to Q(f)