Adversary Lower Bound for the k -sum Problem Robert Alexander - - PowerPoint PPT Presentation

SLIDE 1

Adversary Lower Bound for the k-sum Problem

Alexander Belov University of Latvia Robert ˇ Spalek Google, Inc.

January 22, 2013 QIP 2013, Beijing, China

This work has been supported by the European Social Fund within the project “Support for Doctoral Studies at University of Latvia”

SLIDE 2

On the Power of Learning Graphs

Alexander Belov University of Latvia Ansis Rosmanis University of Waterloo Robert ˇ Spalek Google, Inc.

Based on arXiv:1206.6528 and arXiv:1210.3279

SLIDE 3

Query Complexity

Query Complexity Certificate Structures Our Results Proof Sketch

SLIDE 4

Problem

Query Complexity Certificate Structures Our Results Proof Sketch 4 / 34

Computational Problem

The amount of resources required to solve it? Ideally: Time necessary for a quantum computer to solve it.

SLIDE 5

Problem

Query Complexity Certificate Structures Our Results Proof Sketch 4 / 34

Computational Problem

The amount of resources required to solve it? Ideally: Time necessary for a quantum computer to solve it. Alas, we don’t know much about it.

SLIDE 6

Problem

Query Complexity Certificate Structures Our Results Proof Sketch 4 / 34

Computational Problem

The amount of resources required to solve it? Ideally: Time necessary for a quantum computer to solve it. Simplification: Number of accesses to the input string

SLIDE 7

Quantum Query Complexity

Query Complexity Certificate Structures Our Results Proof Sketch 5 / 34

Function f : [q]n ⊇ D → {0, 1} Query algorithm: calculate f(x1, x2, . . . , xn), can access individual xj in one query. Quantum query complexity: number of queries the best quantum query algorithm makes on the worst input.

SLIDE 8

Quantum Query Complexity

Query Complexity Certificate Structures Our Results Proof Sketch 5 / 34

Function f : [q]n ⊇ D → {0, 1} Query algorithm: calculate f(x1, x2, . . . , xn), can access individual xj in one query. Quantum query complexity: number of queries the best quantum query algorithm makes on the worst input. Does this make things simpler?..

SLIDE 9

Adversary Bound

Query Complexity Certificate Structures Our Results Proof Sketch 6 / 34

Quantum query complexity admits formulation as an SDP: Adversary Bound maximize Γ subject to Γ ◦ ∆j ≤ 1 for all j ∈ [n]. Here: Γ is an f−1(1) × f−1(0)-matrix with real entries, and ∆j[ [x, y] ] =

• 1,

xj = yj; 0,

• therwise.

SLIDE 10

Certificate Structures

Query Complexity Certificate Structures Our Results Proof Sketch

SLIDE 11

Simplification

Query Complexity Certificate Structures Our Results Proof Sketch 8 / 34

Simplification II: Only consider the positions of certificates inside the input string. Not the values therein.

SLIDE 12

Example/Motivation

Query Complexity Certificate Structures Our Results Proof Sketch 9 / 34

Quantum walk on the Johnson Graph Ambainis developed it to solve k-distinctness: Given (x1, . . . , xn), detect whether there are k equal elements among them. Quantum walk on subsets of [n]. Accept if the values of variables in S ⊆ [n] are enough to deduce f(x) = 1. Runs in O

• nk/(k+1)

quantum queries.

SLIDE 13

Example/Motivation

Query Complexity Certificate Structures Our Results Proof Sketch 9 / 34

Quantum walk on the Johnson Graph Ambainis developed it to solve k-distinctness: Given (x1, . . . , xn), detect whether there are k equal elements among them. Quantum walk on subsets of [n]. Accept if the values of variables in S ⊆ [n] are enough to deduce f(x) = 1. Runs in O

• nk/(k+1)

quantum queries. Childs and Eisenberg: The same algorithm can be used for any function with small certificates: k-distinctness, k-sum, graph collision, matrix product verification... k-sum: Given (x1, . . . , xn) ∈ [q]n, detect whether there are k elements whose sum is divisible by q.

SLIDE 14

Certificate Structure

Query Complexity Certificate Structures Our Results Proof Sketch 10 / 34

Function f : [q]n ⊇ D → {0, 1} For x ∈ f−1(1), write out: Mx = {S ⊆ [n] | S is enough to deduce f(x) = 1 }. The set of all Mx is a certificate structure C. (Interested in inclusion-wise minimal Mx only.)

SLIDE 15

Certificate Structure

Query Complexity Certificate Structures Our Results Proof Sketch 10 / 34

Function f : [q]n ⊇ D → {0, 1} For x ∈ f−1(1), write out: Mx = {S ⊆ [n] | S is enough to deduce f(x) = 1 }. The set of all Mx is a certificate structure C. (Interested in inclusion-wise minimal Mx only.) k-subset certificate structure Mutual certificate structure of k-distinctness and k-sum.

SLIDE 16

2-subset on 4 variables: 1147 1417 1471 ∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮ tttttt

• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏

∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮

• tttttt
• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏

∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮

• tttttt
• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎

• rrrrrrr
• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴
• ♣♣♣♣♣♣♣
• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏

4117 4171 4711 ∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮ tttttt

• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏

∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮

• tttttt
• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏

∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮

• tttttt
• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎

• rrrrrrr
• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴
• ♣♣♣♣♣♣♣
• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏

(Only interested in inclusion-minimal Mx.)

SLIDE 17

Example/Motivation

Query Complexity Certificate Structures Our Results Proof Sketch 12 / 34

Quantum walk on subsets of [n]. Accept if the values of x in S ⊆ [n] are enough to deduce f(x) = 1. Runs in O

• nk/(k+1)

quantum queries. Conjecture (Childs and Eisenberg). Quantum walk on the Johnson graph is

• ptimal for the k-sum problem.

SLIDE 18

Example/Motivation

Query Complexity Certificate Structures Our Results Proof Sketch 12 / 34

Quantum walk on subsets of [n]. Accept if the values of x in S ⊆ [n] are enough to deduce f(x) = 1. Runs in O

• nk/(k+1)

quantum queries. Conjecture (Childs and Eisenberg). Quantum walk on the Johnson graph is

• ptimal for the k-sum problem.

Intuition: Even if we are given k − 1 elements of a k-tuple, we have absolutely no additional information whether the k-tuple forms a certificate. The k-sum problem does not possess any structure.

SLIDE 19

Another Example

Query Complexity Certificate Structures Our Results Proof Sketch 13 / 34

Collision Problem Distinguish between two cases Negative: each symbol in the input string is unique; or Positive: each symbol in the input string has exactly two appearances. E.g., negative input: 2746 and three variants of positive inputs: 1144 1414 1441 ∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮ tttttt

• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏

∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮

• tttttt
• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏

∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮

• tttttt
• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎

• rrrrrrr
• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴
• ♣♣♣♣♣♣♣
• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏

SLIDE 20

Learning graphs

Query Complexity Certificate Structures Our Results Proof Sketch 14 / 34

■

Computational model that relies on the cer- tificate structure by definition.

■

Generalizes quantum walk on the Johnson graph.

SLIDE 21

Learning graphs

Query Complexity Certificate Structures Our Results Proof Sketch 15 / 34

∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮ tttttt

• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏ ✕✕✕✕✕✕✕✕✕✕✕✕✕ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ☞☞☞☞☞☞☞☞☞ ✜✜✜✜✜✜✜✜ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ✮ ✮ ✮ ✮

∅

■

Each edge e of the Hasse diagram is assigned non-negative conductance ce.

■

For each M ∈ C, we connect ∅ to one terminal, and all S ∈ M to the other terminal

• f a current source.

SLIDE 22

Learning graphs

Query Complexity Certificate Structures Our Results Proof Sketch 15 / 34

∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮ tttttt

• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏ ✕✕✕✕✕✕✕✕✕✕✕✕✕ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ☞☞☞☞☞☞☞☞☞ ✜✜✜✜✜✜✜✜ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ✮ ✮ ✮ ✮

∅

■

Each edge e of the Hasse diagram is assigned non-negative conductance ce.

■

For each M ∈ C, we connect ∅ to one terminal, and all S ∈ M to the other terminal

• f a current source.

Learning graph complexity of C is defined as minimize

• e∈E ce

subject to effective resistance from ∅ to M is at most 1 for all M ∈ C

SLIDE 23

Learning graphs

Query Complexity Certificate Structures Our Results Proof Sketch 15 / 34

∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮ tttttt

• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏ ✕✕✕✕✕✕✕✕✕✕✕✕✕ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ☞☞☞☞☞☞☞☞☞ ✜✜✜✜✜✜✜✜ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ✮ ✮ ✮ ✮

∅

■

Each edge e of the Hasse diagram is assigned non-negative conductance ce.

■

For each M ∈ C, we connect ∅ to one terminal, and all S ∈ M to the other terminal

• f a current source.

Learning graph complexity of C is defined as minimize

• e∈E ce

subject to effective resistance from ∅ to M is at most 1 for all M ∈ C Theorem (Belov and Lee). For each f having certificate structure C, there exists a quantum query algorithm with complexity equal to the learning graph complexity of C up to a constant factor.

SLIDE 24

Our Results

Query Complexity Certificate Structures Our Results Proof Sketch

SLIDE 25

Outline

Query Complexity Certificate Structures Our Results Proof Sketch 17 / 34

■

We derive a dual formulation of the learning graph complexity.

■

We use it to give (almost) tight lower bounds for some certificate structures: k-subset, collision, hidden shift, triangle.

SLIDE 26

Outline

Query Complexity Certificate Structures Our Results Proof Sketch 17 / 34

■

We derive a dual formulation of the learning graph complexity.

■

We use it to give (almost) tight lower bounds for some certificate structures: k-subset, collision, hidden shift, triangle.

■

We prove learning graphs are tight for any certificate structure.

■

We prove an analogue of Childs-Eisenberg conjecture for a wide range of certificate structures. (Implies the original conjecture).

SLIDE 27

Learning Graph Revisited

Query Complexity Certificate Structures Our Results Proof Sketch 18 / 34

∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮ tttttt

• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏ ✕✕✕✕✕✕✕✕✕✕✕✕✕ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ☞☞☞☞☞☞☞☞☞ ✜✜✜✜✜✜✜✜ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ❁ ✮ ✮ ✮ ✮

∅ More details (using electric flow): minimize

• e∈E ce

subject to

• e∈E

pe(M)2 ce ≤ 1 for all M ∈ C; for each M ∈ C, pe(M) form a flow from ∅ to M of value 1 The dual formulation (using potentials): maximize

• M∈C α∅(M)2

subject to

• M∈C
• αS(M) − αS∪{j}(M)

2 ≤ 1 for all j / ∈ S ⊆ [n]; αS(M) = 0 whenever S ∈ M;

SLIDE 28

k-subset certificate structure

Query Complexity Certificate Structures Our Results Proof Sketch 19 / 34

Theorem. The learning graph complexity of the k-subset certificate structure is Ω(nk/(k+1)). max.

M∈C

α∅(M)2

• M∈C
• αS(M) − αS∪{j}(M)

2 ≤ 1 αS(M) = 0 whenever S ∈ M ∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮ tttttt

• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t

• tttttt
• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏

SLIDE 29

k-subset certificate structure

Query Complexity Certificate Structures Our Results Proof Sketch 19 / 34

Theorem. The learning graph complexity of the k-subset certificate structure is Ω(nk/(k+1)). max.

M∈C

α∅(M)2

• M∈C
• αS(M) − αS∪{j}(M)

2 ≤ 1 αS(M) = 0 whenever S ∈ M ∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮ tttttt

• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t

• tttttt
• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏

• Proof. Define

αS(M) = n

k

−1/2 max

• nk/(k+1) − |S|, 0
• ,

S / ∈ M 0,

• therwise.

Perform simple calculations.

SLIDE 30

Other Certificate Structures

Query Complexity Certificate Structures Our Results Proof Sketch 20 / 34

We also prove that the learning graph complexity

• f the collision and the hidden shift certificate structures

is Ω( 3 √n) and

• f the triangle certificate structure is ˜

Ω(n9/7). Corollary. The learning graph for the triangle problem from the next presentation is essentially tight.

SLIDE 31

Tightness I

Query Complexity Certificate Structures Our Results Proof Sketch 21 / 34

We prove learning graphs are tight: Theorem. For any certificate structure C, there exists f possessing C such that the quantum query complexity of f is at least the learning graph complexity of C up to a constant factor.

SLIDE 32

Tightness I

Query Complexity Certificate Structures Our Results Proof Sketch 21 / 34

We prove learning graphs are tight: Theorem. For any certificate structure C, there exists f possessing C such that the quantum query complexity of f is at least the learning graph complexity of C up to a constant factor. For the analogue of the Childs-Eisenberg conjecture, we need more notions...

SLIDE 33

Boundedly generated certificate structures

Query Complexity Certificate Structures Our Results Proof Sketch 22 / 34

Definition. A certificate structure C is boundedly generated if, for any M ∈ C, one can find a subset AM ⊆ [n] such that |AM| = O(1), and S ∈ M if and only if S ⊇ AM. The k-subset certificate structure is boundedly generated: ∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁ rrrrrrr

• ✮

✮ ✮ tttttt

• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎ rrrrrrr

• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴

♣♣♣♣♣♣♣

• ✕✕✕
• ✴

✴ ✴ ❏❏❏❏❏❏

The collision certificate structure is not: ∅

• 1
• ❏

❏ ❏ ❏ ❏ ❏

2

• ✴

✴ ✴

3

• ✎✎✎

4

• tttttt
• ✴

✴ ✴

• ❏

❏ ❏ ❏ ❏ ❏

• ✕✕✕
• ▲

▲ ▲ ▲ ▲ ▲ ▲ ♣♣♣♣♣♣♣

• ❇

❇ ❇ ❇ ❇

• ✜✜✜
• ❁

❁ ❁ ❁

• rrrrrrr
• ✮

✮ ✮

• tttttt
• ✎✎✎
• ✎✎✎
• ✮

✮ ✮

• ❇

❇ ❇ ❇ ❇

• t

t t t t t tttttt

• ❁

❁ ❁ ❁

• ▲

▲ ▲ ▲ ▲ ▲ ▲

• ✎

✎ ✎

• rrrrrrr
• ✜✜✜
• ❏

❏ ❏ ❏ ❏ ❏

• ✴✴✴
• ♣♣♣♣♣♣♣
• ✕✕✕
• ✴

✴ ✴

• ❏❏❏❏❏❏

SLIDE 34

Tightness II

Query Complexity Certificate Structures Our Results Proof Sketch 23 / 34

Definition. A certificate structure C is boundedly generated if, for any M ∈ C, one can find a subset AM ⊆ [n] such that |AM| = O(1), and S ∈ M if and only if S ⊇ AM. C-sum problem. Given (x1, . . . , xn) ∈ [q]n, decide whether there exists M ∈ C such that

• j∈AM xj is divisible by q.

Theorem. If C is boundedly generated and f is the C-sum problem with q > 2|C|, then the quantum query complexity of f equals the learning graph complexity of f up to a constant factor.

SLIDE 35

Proof Sketch

Query Complexity Certificate Structures Our Results Proof Sketch

SLIDE 36

Adversary Bound

Query Complexity Certificate Structures Our Results Proof Sketch 25 / 34

We use the adversary bound maximize Γ subject to Γ ◦ ∆j ≤ 1 for all j ∈ [n]. Here: Γ is an f−1(1) × f−1(0)-matrix with real entries, and ∆j[ [x, y] ] =

• 1,

xj = yj; 0,

• therwise.

SLIDE 37

Former Modes of Applications

Query Complexity Certificate Structures Our Results Proof Sketch 26 / 34

Adversary bound has been used as: 1. Non-negative weight adversary Original version by Ambainis. Combinatorial reasoning. Easy to use. Has strong limitations (certificate complexity, property testing barriers). Fails for our applications. 2. Small functions By solving the optimization problem on computer. 3. Tight composition theorems Composing functions from the second point. Formulae evaluation. We use spectral analysis via embedding.

SLIDE 38

Hamming Association Scheme

Query Complexity Certificate Structures Our Results Proof Sketch 27 / 34

Two orthogonal projectors on Cq: E0 =      1/q 1/q · · · 1/q 1/q 1/q · · · 1/q . . . . . . ... . . . 1/q 1/q · · · 1/q      E1 =      1 − 1/q −1/q · · · −1/q −1/q 1 − 1/q · · · −1/q . . . . . . ... . . . −1/q −1/q · · · 1 − 1/q      For S ⊆ [n], define ES =

n

• j=1

ES[

[j] ].

These are orthogonal projectors on Cqn.

SLIDE 39

Action of ∆

Query Complexity Certificate Structures Our Results Proof Sketch 28 / 34

subject to Γ ◦ ∆j ≤ 1 for all j ∈ [n]. For E0 =      1/q 1/q · · · 1/q 1/q 1/q · · · 1/q . . . . . . ... . . . 1/q 1/q · · · 1/q      E1 =      1 − 1/q −1/q · · · −1/q −1/q 1 − 1/q · · · −1/q . . . . . . ... . . . −1/q −1/q · · · 1 − 1/q      we have E0 → E0 E1 → −E0.

SLIDE 40

Embedding Γ into Γ

Query Complexity Certificate Structures Our Results Proof Sketch 29 / 34

• Y
• Γ

XM1 XM2 XMk

• GM1
• GM2
• GMk

GM1 GM2 GMk

• GM1
• GM2
• GMk

. . .

C-sum problem. Given (x1, . . . , xn) ∈ [q]n, decide whether there exists M ∈ C such that

j∈AM xj is

divisible by q.

• GM is [q]n × [q]n-matrix.

XM = {x ∈ [q]n |

j∈AM xj ≡ 0 (mod q)}

|XM| = qn−1 Y is the set of negative inputs q ≥ 2|C| = ⇒ |Y | ≥ qn/2

SLIDE 41

Defining Γ

Query Complexity Certificate Structures Our Results Proof Sketch 30 / 34

• max. Γ

Γ ◦ ∆j ≤ 1

• Y
• Γ

XM1 XM2 XMk

• GM1
• GM2
• GMk

GM1 GM2 GMk

• GM1
• GM2
• GMk

. . .

max.

M∈C

α∅(M)2

• M∈C
• αS(M) − αS∪{j}(M)

2 ≤ 1 αS(M) = 0 whenever S ∈ M

• GM =
• S⊆[n]

αS(M)ES

• GM = √q

GM[ [XM, [q]n] ] GM = GM[ [XM, Y ] ]

SLIDE 42

Transformation

Query Complexity Certificate Structures Our Results Proof Sketch 31 / 34

• Y
• Γ

XM1 XM2 XMk

• GM1
• GM2
• GMk

GM1 GM2 GMk

. . .

• Γ′
• Y
• G′

M1

• G′

M2

• G′

Mk

. . .

∆j

✤

SLIDE 43

Properties of Γ′

Query Complexity Certificate Structures Our Results Proof Sketch 32 / 34

• max. Γ

Γ ◦ ∆j ≤ 1

• Γ′
• Y
• G′

M1

• G′

M2

• G′

Mk

. . .

∆j

✤

• max.

M∈C

α∅(M)2

• M∈C
• αS(M) − αS∪{j}(M)

2 ≤ 1 αS(M) = 0 whenever S ∈ M Due to E0 → E0 and E1 → −E0, we get

• G′

M =

• S∋j

(αS(M) − αS∪{j}(M))ES

• G′

M = √q

G′

M[

[XM, [q]n] ] We prove this does not increase the norm a lot.

SLIDE 44

Summary

Query Complexity Certificate Structures Our Results Proof Sketch 33 / 34

■

We defined the notion of certificate structure.

■

We derived a dual formulation of the learning graph complexity.

■

We used it to give (almost) tight lower bounds for some certificate structures: k-subset, collision, hidden shift, triangle.

■

We proved learning graphs are tight for any certificate structure.

■

We defined boundedly generated certificate structures.

■

We proved an analogue of Childs-Eisenberg conjecture for boundedly generated certificate structures.

SLIDE 45

Thank you!