New quantum lower bound method, with applications to direct product theorems

Andris Ambainis, U. WaterlooRobert Spalek, CWI,

AmsterdamRonald de Wolf, CWI,

Amsterdam

Query model

Input x1, …, xN accessed by queries.

Complexity = the number of queries.

0 1 0 0...x1 x2 xNx3

i0

ixi

Grover's search

Is there i such that xi=1? Queries: ask i, get xi. Classically, N queries required. Quantum: O(N) queries [Grover, 1996]. Speeds up any search problem.

0 1 0 0...

x1 x2 xNx3

Quantum counting [Boyer et al., 1998]

Is the fraction of i:xi=1 more than ½+ or less than ½- ?

Classical: queries.

Quantum: queries.

0 1 0 0...

x1 x2 xNx3

2

1

1

O

Element distinctness

Are there i, j such that ij but xi=xj? Classically: N queries. Quantum: O(N2/3).

3 1 17 5...

x1 x2 xNx3

Lower bounds Search requires N) queries

[Bennett et al., 1997]. Counting: 1/) [Nayak, Wu,

1999]. Element distinctness: (N2/3) [Shi,

2002].

Lower bound methods Adversary: analyze algorithm,

prove it is incorrect on some input. Polynomials: describe algorithm by

low degree polynomial.

Which method?

Problem Adversary Polynomials

Search Yes Yes

Counting Yes Yes

Element dist. ? Yes

Limits of adversary method Certificate for f on input (x1, x2, …,

xN):

set of variables xi which determine

f(x1, x2, …, xN). Search: is there i:xi=1?

0 1 0 0...

x1 x2 xNx3

Limits of adversary method Certificate for f on input (x1, x2, …,

xN):

set of variables xi which determine

f(x1, x2, …, xN). Search: is there i:xi=1?

0 0 0 0...

x1 x2 xNx3

Certificate complexity Cx(f): the size of the smallest

certificate for f on the input x.

)(max)( 0)(:0 fCfC xxfx )(max)( 1)(:1 fCfC xxfx

Search: C0=N, C1=1.

Limits of adversary method

Theorem [Spalek, Szegedy, 2004] Any quantum adversary lower

bound is at most

NfCfCO ))(),(min( 10

Example:element distinctness

Are there i, j:xi= xj? 1-certificate: {i, j}, xi= xj.

Adversary bound: Actual complexity: O(N2/3).

NONO 2

3 1 17 5...

x1 x2 xNx3

Example: triangle finding Graph G, specified by N2 variables xij:

xij=1, if there is edge between i and j. Does G contain a triangle? 1-certificate:{ij, jk, ik}, xij= xik= xjk=1. Adversary lower bound: at most

The best algorithm: O(N1.3) [MSS 03].

NONO 23

Previous adversary method

Quantum query model

Fixed starting state. U0, U1, …, UT – independent of x1, x2,

…, xN. Q – queries. Measuring final state gives the result.

U0 Q Qstart U1 UT…

Queries Basis states for algorithm’s

workspace: |i, z, i{1, 2, …, N}. Query transformation:

Example: |i, z|i, z, if xi=0; |i, z-|i, z, if xi=1;

zQiziQix

,

|

Adversary framework

Quantum algorithm A x1 x2 …

xN

NxxN xxxQxxxN

...... 21...21 1

Two registers: HA, HI.Query Q:

Example:Grover search Start state: |start|0,

End state

1...00...0...010...101

0 N

1...00...0...0120...1011

NN

Density matrices Measure HA, look at density matrix

of HI

N

N

N

end

100

01

0

001

NNN

NNN

NNN

start

111

111

111

Density matrices Sum of off-diagonal entries. N(N-1) entries. Sum for starting state: Sum for end state: 0. Query changes the sym by at most

2N. (N) queries needed.

11

)1( NN

NN

Limits of this approach (end)x, y measures the possibility of

distinguishing x from y. If every (end)x, y small, we can,

given x, y: f(x)f(y), distinguish x from y.

Limits of this approach It might be that:

Every x can be distinguished from every y;

There is no measurement that distinguishes all x from all y.

f(x)=0 f(y)=1

Adversary method fails

quantum algorithm

New method

K-fold search

K items i:xi=1, find all of them. O(NK) queries: O(N/K) for each

item. This is optimal.

0 1 0 0...

x1 x2 xNx3

Direct product theorem Theorem [KSW 04] Solving K-fold

search with success probability c-K, c>1 requires NK queries.

Easy to prove for success probability c.

Difficult for probability c-K.Why is this useful????

Application:sorting Theorem [KSW04] A quantum

algorithm for sorting x1, x2, …, xN with S qubits of workspace must use

queries.

S

N 5.1

Proof

Divide algorithm into stages: first K items sorted, next K items sorted, …

Suffices to show each stage requires (NK) queries.

Each stage reduces to K-fold search.

Proof At the beginning of ith stage, we get

S qubits from the previous stage. Theorem K-fold search requires (NK) queries, even if we allow K/C qubits of advice.

Proof Theorem K-fold search requires (NK) queries, even if we allow K/C qubits of advice.

Proof Replace advice by completely mixed state.

Success probability p with advice => Success probability p2-K/C, no advice.

Direct product theorem Theorem Solving K-fold search with

success probability c-K, c>1 requires NK queries.

[KSW 04]: proof by polynomials method.

This talk: (new) adversary method.

Proof sketch “Know-0”, “Know-1”, …, “Know-k”

states. Describe quantum state as

K

jj jKnowp

0

""

|

Proof Adversary framework

Start state for input:

Kxi

N

i

xxx|}1:{|

21

Quantum algorithm A x1 x2 …

xN

Proof State of HI if we know

Subspace Tj spanned by all

1...

|}1:{|21...

1

1

jii

i

j

xx

KxiNii xxx

1...1

jii xx

jii ...1

Proof T0T1 … TK. T0 – starting

state. TK – entire HI.

T0 T1 …. TK

Tj – “know at-most j” subspace

Proof Sj=Tj(Tj-1).

T0 T1 … TK

Proof Sj=Tj(Tj-1).

T0 S1 … SK

Sj is “know-j” subspace.

Proof | - state of algorithm including

the input register |x1 … xN.

|j belongs to HA Sj. Probability of “know-j”:

,0

K

jj

2

jjp

Proof Start state: p0=1, p1=…=pK=0. Change in one query:

After NK queries, pK/2+1, …, pK are exponentially small.

Success probability exponentially small.

111' jjjj ppN

Kcpp

Threshold functions

F(x1, x2, …, xN)=1 if xi=1 for at least t values i{1, 2, …, N}.

F(x1, x2, …, xN)=0 if xi=1 for at most t-1 values i{1, 2, …, N}.

Query complexity: (Nt).

0 1 0 0...

x1 x2 xNx3

Threshold functions

F(x1, x2, …, xN)=1 if xi=1 for at least t values i{1, 2, …, N}.

F(x1, x2, …, xN)=0 if xi=1 for at most t-1 values i{1, 2, …, N}.

Query complexity: (Nt).

0 1 0 0...

x1 x2 xNx3

Threshold functions K instances of threshold function. (KNt) queries. Theorem Solving all K instances

with probability at most c-K requires KNt queries.

Proof

K input registers. Each input register initially

,|0, |1 - uniform over |x1 … xN with t-1 and t values i:xi=1.

10

Algorithm11

211 Nxxx K

NKK xxx 21

…

Proof For each instance, states “solved”,

“know-0”, “know-1”, … “know-(t-1)”. For K instances, vector of K states. Progress of a state:

“solved” – progress t/2. “know-t/2”, … “know-(t-1)” – progress

t/2. “know-j”, j<t/2 – progress j.

Proof If progress of final state less than tK/4,

the probability of getting all K correct answers is c-K.

Decompose current state

Potential function

j

j jprogressp ""

,j

jjqpP

tNq

11

Proof

Start state: P()=1. For pj, jtK/4 to be more than c-K,

One query increases P() by at most a factor of

j

jjqpP

KtKK CqcP 4/

tN

O1

1

Proof F(x1, x2, …, xN)=0, “know-j”:

F(x1, x2, …, xN)=1, “know-j”:

1...

1|}1:{|21

0

1

...1

jii

ijii

xx

txiNxxx

1...

|}1:{|21

1

1

...1

jii

ijii

xx

txiNxxx

Proof Starting state:

“Solved”:

“Know-j”

10

10

...1...1 jiijii

10

...1...1 jiijii

Application: testing linear inequalities

aij known, xi, bj accessed by queries.

Which inequalities are true?

NNNNNN

NN

bxaxaxa

bxaxaxa

2211

11212111

Our result

Memory limited to S (qu)bits. Classically: (N2/S) queries. Quantum: (N3/2t1/2/S1/2) queries. Lower bound follows from threshold

function lower bound.

NNNNNN

NN

bxaxaxa

bxaxaxa

2211

11212111

Conclusion New quantum lower bound

method, by eigenspace analysis. Direct product theorems for K-fold

search and threshold functions. Consequences for time-space

tradeoffs.

More details A. Ambainis. A new quantum lower

bound method, with application to direct product theorem for search, quant-ph/0508200.

A. Ambainis, R. Spalek, R. de Wolf, Quantum direct product theorems for symmetric functions and time-space tradeoffs , quant-ph/0511200.

Open problems AND-OR tree: best

lower bound O(N), N – number of variables.

Algorithm: O(N.753).

x1 x2 x3 x4

AND

OR OR

Adversary lower bound for element distinctness?