Quantum query complexity and the adversary bound Part I: The - - PowerPoint PPT Presentation

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Quantum query complexity and the adversary bound Part I: The adversary bound

Alexander Belov University of Latvia

22nd EWSCS, 5-10 March 2017, Palmse

Introduction

Introduction Settings Why? Sub-question 1 Sub-question 2 Sub-question 3 Short outline Basic Adversary Spectral Adversary Dual Adversary Composition

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Settings

Introduction Settings Why? Sub-question 1 Sub-question 2 Sub-question 3 Short outline Basic Adversary Spectral Adversary Dual Adversary Composition

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A computational problem:

f : [q]n ⊇ D → {0, 1}

A computational device (deterministic, randomised, or quantum) :

Device

• utput
• x1

x2 x3 . . . xn

• Query complexity: number of queries to the input string

(worst-case) required to solve the problem.

• Both upper and lower bounds.

Why?

Introduction Settings Why? Sub-question 1 Sub-question 2 Sub-question 3 Short outline Basic Adversary Spectral Adversary Dual Adversary Composition

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Why quantum query complexity?

Why quantum query complexity? Why quantum query complexity? Why quantum query complexity?

Sub-question 1

Introduction Settings Why? Sub-question 1 Sub-question 2 Sub-question 3 Short outline Basic Adversary Spectral Adversary Dual Adversary Composition

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Why query complexity?

Reason 1: For some problems, this is the right model.

• In hypothesis testing, we are interested in reducing the number
• f experiments. Experiments are expensive, computation is not

so.

• In property testing, complexity is traditionally measured in the

number of samples.

Sub-question 1

Introduction Settings Why? Sub-question 1 Sub-question 2 Sub-question 3 Short outline Basic Adversary Spectral Adversary Dual Adversary Composition

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Why query complexity?

Reason 1: For some problems, this is the right model. Reason 2: Because we can.

• We are interested in time complexity usually, but in most cases

we can only prove lower bounds by proving lower bound on query complexity.

• In cryptography, proofs in the random oracle model are,

essentially, query complexity lower bounds.

Sub-question 1

Introduction Settings Why? Sub-question 1 Sub-question 2 Sub-question 3 Short outline Basic Adversary Spectral Adversary Dual Adversary Composition

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Why query complexity?

Reason 1: For some problems, this is the right model. Reason 2: Because we can. Reason 3: Query algorithm can give insights into the problem.

• Trying to minimise query complexity of an algorithms result in
• ur better understanding of algorithms and the problem.
• One might hope to implement query-efficient algorithm

time-efficiently.

Sub-question 1

Introduction Settings Why? Sub-question 1 Sub-question 2 Sub-question 3 Short outline Basic Adversary Spectral Adversary Dual Adversary Composition

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Why query complexity?

Reason 1: For some problems, this is the right model. Reason 2: Because we can. Reason 3: Query algorithm can give insights into the problem. Reason 4: Query complexity gives impossibility results.

• We can prove that black-box approaches are doomed.
• We can exclude a lower bound via query complexity.

Sub-question 2

Introduction Settings Why? Sub-question 1 Sub-question 2 Sub-question 3 Short outline Basic Adversary Spectral Adversary Dual Adversary Composition

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Why quantum query complexity?

Sub-question 3

Introduction Settings Why? Sub-question 1 Sub-question 2 Sub-question 3 Short outline Basic Adversary Spectral Adversary Dual Adversary Composition

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Why quantum query complexity?

Reason 1: You do not need to understand quantum computation to study quantum query complexity.

• Quantum query complexity is tightly characterised by a

semi-definite optimisation problem: the adversary bound.

Sub-question 3

Introduction Settings Why? Sub-question 1 Sub-question 2 Sub-question 3 Short outline Basic Adversary Spectral Adversary Dual Adversary Composition

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Why quantum query complexity?

Reason 1: You do not need to understand quantum computation to study quantum query complexity. Reason 2: In some sense, we understand quantum query complexity better than randomised query complexity.

• Randomised query complexity is a collection of separated

results.

• Quantum query complexity start resembling a theory.

Sub-question 3

Introduction Settings Why? Sub-question 1 Sub-question 2 Sub-question 3 Short outline Basic Adversary Spectral Adversary Dual Adversary Composition

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Why quantum query complexity?

Reason 1: You do not need to understand quantum computation to study quantum query complexity. Reason 2: In some sense, we understand quantum query complexity better than randomised query complexity. Reason 3: Quantum query complexity provides upper bound on approximate polynomial degree.

• Approximating polynomials are used in various fields (e.g.,

learning theory)

• One can get an approximating polynomial by constructing a

quantum query algorithm.

Short outline

Introduction Settings Why? Sub-question 1 Sub-question 2 Sub-question 3 Short outline Basic Adversary Spectral Adversary Dual Adversary Composition

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We are mostly considering total functions

f : [q]n → {0, 1}.

• For such functions, one can only get a polynomial improvement

(at most 6-th power)

• We will mostly consider sub-quadratic improvements.

Basic Adversary

Introduction Basic Adversary Main idea Quantum Version

k-threshold function

Spectral Adversary Dual Adversary Composition

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Main idea

Introduction Basic Adversary Main idea Quantum Version

k-threshold function

Spectral Adversary Dual Adversary Composition

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Distinguishing inputs

OR function:

000...00 100...00 ❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱ 010...00◗◗◗◗◗◗◗◗◗◗◗◗◗ 001...00

❉❉❉❉❉❉❉❉

······ 000...10

q q q q q q q q q q

000...01

❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ No randomised algorithm can distinguish all these pairs of inputs in

• (n) queries.

Hence, randomised query complexity is Ω(n).

Quantum Version

Introduction Basic Adversary Main idea Quantum Version

k-threshold function

Spectral Adversary Dual Adversary Composition

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000...00 100...00 ❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱ 010...00◗◗◗◗◗◗◗◗◗◗◗◗◗ 001...00

❉❉❉❉❉❉❉❉

······ 000...10

q q q q q q q q q q

000...01

❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ Select:

X ⊆ f −1(1) Y ⊆ f −1(0)

Relation ∼ between X and Y

Quantum Version

Introduction Basic Adversary Main idea Quantum Version

k-threshold function

Spectral Adversary Dual Adversary Composition

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000...00 100...00 ❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱ 010...00◗◗◗◗◗◗◗◗◗◗◗◗◗ 001...00

❉❉❉❉❉❉❉❉

······ 000...10

q q q q q q q q q q

000...01

❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ ❥ Calculate:

m : minimum, over x ∈ X, of the number of y with x ∼ y: 1 m′ : minimum, over y ∈ Y , of the number of x with x ∼ y: n ℓx,j : number of y ∈ Y such that x ∼ y and xj = yj: 1 ℓ′

y,j : number of x ∈ X such that x ∼ y and xj = yj: 1

ℓmax : maximum of ℓx,jℓ′

y,j over x ∼ y and j such that xj = yj: 1

Q(f) = Ω

• mm′

ℓmax

• = Ω(√n)

k-threshold function

Introduction Basic Adversary Main idea Quantum Version

k-threshold function

Spectral Adversary Dual Adversary Composition

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k-threshold function: f(x) = 1 if there are at least k ones in the input. X ⊆ f −1(1): x of Hamming weight k Y ⊆ f −1(0): y of Hamming weight k − 1 ∼: x ∼ y if x and y differ in 1 position x : 111 . . . 11100 . . . 00 y : 111 . . . 11000 . . . 00

k-threshold function

Introduction Basic Adversary Main idea Quantum Version

k-threshold function

Spectral Adversary Dual Adversary Composition

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x : 111 . . . 11100 . . . 00 y : 111 . . . 11000 . . . 00 m : minimum, over x ∈ X, of number y with x ∼ y: k m′ : minimum, over y ∈ Y , of number x with x ∼ y: n − k + 1 ℓx,j : number of y ∈ Y , such that x ∼ y and xj = yj: 1 ℓ′

y,j : number of x ∈ X, such that x ∼ y and xj = yj: 1

ℓmax : maximum of ℓx,jℓ′

y,j over x ∼ y and j such that xj = yj: 1

Q(k-threshold) = Ω

• mm′

ℓmax

• = Ω
• k(n − k + 1)
• This is tight!

Spectral Adversary

Introduction Basic Adversary Spectral Adversary Matrix Representation Example Theorem Proof Spectral Adversary Flavours Dual Adversary Composition

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Matrix Representation

Introduction Basic Adversary Spectral Adversary Matrix Representation Example Theorem Proof Spectral Adversary Flavours Dual Adversary Composition

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We had a relation x ∼ y

000 001 010 100 011

⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦

101

♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦

110

⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧

111

Represent as a 01-matrix:

Γ = 011 101 110 111 000 001 1 1 010 1 1 100 1 1

Matrix Representation

Introduction Basic Adversary Spectral Adversary Matrix Representation Example Theorem Proof Spectral Adversary Flavours Dual Adversary Composition

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We had

ℓx,j : number of y ∈ Y , such that x ∼ y and xj = yj ℓ′

y,j : number of x ∈ X, such that x ∼ y and xj = yj

ℓmax : maximum of ℓx,jℓ′

y,j over x ∼ y and j such that xj = yj

Erase all entries (x, y) such that xj = yj:

Γ ◦ ∆1 = 011 101 110 111 000 001 1 010 1 100

Matrix representation

Introduction Basic Adversary Spectral Adversary Matrix Representation Example Theorem Proof Spectral Adversary Flavours Dual Adversary Composition

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We had

Q(f) = Ω

• mm′

ℓmax

• Spectral version of the adversary bound

maximise

Γ

subject to

Γ ◦ ∆j ≤ 1

for all j ∈ [n]; Where · is the spectral norm:

Γ = max

u=1,v=1 |u∗Γv|

Example: OR function

Introduction Basic Adversary Spectral Adversary Matrix Representation Example Theorem Proof Spectral Adversary Flavours Dual Adversary Composition

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maximise

Γ

subject to

Γ ◦ ∆j ≤ 1

for all j ∈ [n];

Γ = 00 . . . 0 10 . . . 00 1 01 . . . 00 1

. . . . . .

00 . . . 01 1 Γ ◦ ∆2 = 00 . . . 0 10 . . . 00 01 . . . 00 1

. . . . . .

00 . . . 01 Γ = √n Γ ◦ ∆j = 1

Theorem

Introduction Basic Adversary Spectral Adversary Matrix Representation Example Theorem Proof Spectral Adversary Flavours Dual Adversary Composition

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maximise

Γ

subject to

Γ ◦ ∆j ≤ 1

for all j ∈ [n]; implies

m : minimum, over x ∈ X, of number y with x ∼ y m′ : minimum, over y ∈ Y , of number x with x ∼ y ℓx,j : number of y ∈ Y , such that x ∼ y and xj = yj ℓ′

y,j : number of x ∈ X, such that x ∼ y and xj = yj

ℓmax : maximum of ℓx,jℓ′

y,j over x ∼ y and j such that xj = yj

Q(f) = Ω

• mm′

ℓmax

Proof

Introduction Basic Adversary Spectral Adversary Matrix Representation Example Theorem Proof Spectral Adversary Flavours Dual Adversary Composition

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Lemma Assume A, B and C are real matrices such that A = B ◦ C. Then,

A ≤ max

i,j : A[ [i,j] ]=0 ri(B)cj(C),

where

ri(B) is the ℓ2-norm of the i-th row of B, and cj(C) is the ℓ2-norm of the j-th column of C.

Spectral Adversary

Introduction Basic Adversary Spectral Adversary Matrix Representation Example Theorem Proof Spectral Adversary Flavours Dual Adversary Composition

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maximise

Γ

subject to

Γ ◦ ∆j ≤ 1

for all j ∈ [n]; works for any matrix

Γ = 011 101 110 111 000 4.5 3.1 1.2 0.5 001 −2.1 7.8 2.5 6.9 010 9.1 −1.8 1.3 −0.2 100 −7.4 6.2 4.3 5.5 Γ ◦ ∆3 = 011 101 110 111 000 4.5 3.1 0.5 001 2.5 010 9.1 −1.8 −0.2 100 −7.4 6.2 5.5

Flavours

Introduction Basic Adversary Spectral Adversary Matrix Representation Example Theorem Proof Spectral Adversary Flavours Dual Adversary Composition

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Basic adversary

Γ has only 0s and 1s.

Positive-weighted adversary

Γ has only non-negative entries.

• Combinatorial interpretation using weights and the Lemma.
• Subject to some limitations:

certificate complexity and property testing barriers. Negative-weighted adversary

Γ is arbitrary real matrix.

• No nice combinatorial interpretation is known.
• It is tight!

Dual Adversary

Introduction Basic Adversary Spectral Adversary Dual Adversary Duality PSD matrices Composition

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Duality

Introduction Basic Adversary Spectral Adversary Dual Adversary Duality PSD matrices Composition

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The adversary bound maximise

Γ

subject to

Γ ◦ ∆j ≤ 1

for all j ∈ [n]; has dual formulation minimise

max

z

• j∈[n]

Xj[ [z, z] ]

subject to

• j:xj=yj

Xj[ [x, y] ] = 1

whenever f(x) = f(y);

Xj is a p.s.d. D × D matrix

for all j ∈ [n],

Positive Semi-Definite Matrices

Introduction Basic Adversary Spectral Adversary Dual Adversary Duality PSD matrices Composition

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Usual Definition: A complex matrix A is positive semi-definite iff

• it is Hermitian: A∗ = A; and
• all its eigenvalues are non-negative real numbers.

Procedural Definition:

• Every matrix of the form uu∗ is PSD (rank-1).
• Every linear combination of PSD matrices with positive

coefficients is PSD.

Composition

Introduction Basic Adversary Spectral Adversary Dual Adversary Composition Definition Adversary Bound

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Definition

Introduction Basic Adversary Spectral Adversary Dual Adversary Composition Definition Adversary Bound

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Let

f : {0, 1}n → {0, 1}

and

g: {0, 1}m → {0, 1}

be functions. Define the composed function

f◦g(x1,1, . . . , xn,m) = f

• g(x1,1, . . . , x1,m), . . . , g(xn,1, . . . , xn,m)
• f
• ♦♦♦♦♦♦♦♦♦♦

g

• x1,1

③ ③ ③ ③ ③ ③ ③

x1,2

✎✎✎✎· · ·

x1,m

❉ ❉ ❉ ❉ ❉ ❉ ❉ ⑤ ⑤ ⑤ ⑤ ⑤ ⑤ ⑤

· · ·

• ❖

❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖

g

• xn,1

③ ③ ③ ③ ③ ③ ③

xn,2

✎✎✎✎· · ·

xn,m

❉ ❉ ❉ ❉ ❉ ❉ ❉

Adversary Bound

Introduction Basic Adversary Spectral Adversary Dual Adversary Composition Definition Adversary Bound

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We have

Adv±(f ◦ g) ≤ Adv±(f) Adv±(g)

In particular,

Q(f ◦ g) = O

• Q(f) Q(g)
• f
• ♦♦♦♦♦♦♦♦♦♦

g

• x1,1

③ ③ ③ ③ ③ ③ ③

x1,2

✎✎✎✎· · ·

x1,m

❉ ❉ ❉ ❉ ❉ ❉ ❉ ⑤ ⑤ ⑤ ⑤ ⑤ ⑤ ⑤

· · ·

• ❖

❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖

g

• xn,1

③ ③ ③ ③ ③ ③ ③

xn,2

✎✎✎✎· · ·

xn,m

❉ ❉ ❉ ❉ ❉ ❉ ❉

• We save a logarithmic factor.
• On each composition!

Adversary Bound

Introduction Basic Adversary Spectral Adversary Dual Adversary Composition Definition Adversary Bound

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We also have

Adv±(f ◦ g) = Adv±(f) Adv±(g)

If f (k) is the k-th composition, then

Q(f (k)) = Θ(Adv±(f)k)

Or,

Adv±(f) = lim

k→∞

k

• Q(f (k)).

Introduction Basic Adversary Spectral Adversary Dual Adversary Composition Definition Adversary Bound

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Thank you!