Coauthors Adam Bene Watts Robin Kothari Avishay Tal January 30, - - PowerPoint PPT Presentation

A Separation Between QNC0 and AC0

Adam Bene Watts, Robin Kothari, Luke Schaeffer, Avishay Tal January 30, 2019

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Coauthors

Adam Bene Watts Robin Kothari Avishay Tal

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Introduction

Section 1 Introduction

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Introduction

Quantum Advantage

Broad Goal Prove quantum computers are more powerful than classical computers.

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Introduction

Quantum Advantage

Broad Goal Prove unconditionally that quantum computers are more powerful than classical computers.

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Introduction

Quantum Advantage

Broad Goal Prove unconditionally that quantum computers are more powerful than classical computers.

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Introduction

Quantum Advantage

Broad Goal Prove unconditionally that quantum computers are more powerful than classical computers. Previous work: Shor’s algorithm.

What if factoring is easy classically too?

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Introduction

Quantum Advantage

Broad Goal Prove unconditionally that quantum computers are more powerful than classical computers. Previous work: Shor’s algorithm.

What if factoring is easy classically too?

Boson Sampling. Several hardness assumptions.

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Introduction

Quantum Advantage

Broad Goal Prove unconditionally that quantum computers are more powerful than classical computers. Previous work: Shor’s algorithm.

What if factoring is easy classically too?

Boson Sampling. Several hardness assumptions. Grover’s algorithm.

O( √ N) oracle calls. How do you implement the oracle?

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Introduction

“Quantum Advantage with Shallow Circuits”

Theorem (Bravyi, Gosset, K¨

• nig)

The hidden linear function (HLF) problem can be solved by constant depth quantum circuits but not constant depth classical circuits.

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Introduction

“Quantum Advantage with Shallow Circuits”

Theorem (Bravyi, Gosset, K¨

• nig)

The hidden linear function (HLF) problem can be solved by constant depth quantum circuits but not constant depth classical circuits. Features: ✓ Completely unconditional!

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Introduction

“Quantum Advantage with Shallow Circuits”

Theorem (Bravyi, Gosset, K¨

• nig)

The hidden linear function (HLF) problem can be solved by constant depth quantum circuits but not constant depth classical circuits. Features: ✓ Completely unconditional! ✓ Fair comparison

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Introduction

“Quantum Advantage with Shallow Circuits”

Theorem (Bravyi, Gosset, K¨

• nig)

The hidden linear function (HLF) problem can be solved by constant depth quantum circuits but not constant depth classical circuits. Features: ✓ Completely unconditional! ✓ Fair comparison ✓ Simple quantum circuit

A variant (2DHLF) uses only local gates on a 2D grid

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Introduction

“Quantum Advantage with Shallow Circuits”

Theorem (Bravyi, Gosset, K¨

• nig)

The hidden linear function (HLF) problem can be solved by constant depth quantum circuits but not constant depth classical circuits. Features: ✓ Completely unconditional! ✓ Fair comparison ✓ Simple quantum circuit

A variant (2DHLF) uses only local gates on a 2D grid

✗ Doesn’t use any complexity theory

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Introduction

“Quantum Advantage with Shallow Circuits”

Theorem (Bravyi, Gosset, K¨

• nig)

The hidden linear function (HLF) problem can be solved by constant depth quantum circuits but not constant depth classical circuits. Features: ✓ Completely unconditional! ✓ Fair comparison ✓ Simple quantum circuit

A variant (2DHLF) uses only local gates on a 2D grid

✗ Doesn’t use any complexity theory Open Problem Can we improve this result using ideas from circuit complexity?

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Introduction

Relevant Circuit Classes

“The class of problems solved by constant depth classical/quantum circuits (of poly size) with constant/unbounded fan-in gates.” Constant Fan-In Unbounded Fan-In Classical NC0 AC0 Quantum QNC0 QAC0?

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Introduction

Relevant Circuit Classes

“The class of problems solved by constant depth classical/quantum circuits (of poly size) with constant/unbounded fan-in gates.” Constant Fan-In Unbounded Fan-In Classical NC0 AC0 Quantum QNC0 QAC0? Technicality Actually, these classes (NC0, QNC0, AC0) are for decision problems with 1 bit of output. This talk is about relation problems with multiple bits of output and multiple answers.

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Introduction

Relevant Circuit Classes

“The class of problems solved by constant depth classical/quantum circuits (of poly size) with constant/unbounded fan-in gates.” Constant Fan-In Unbounded Fan-In Classical NC0 AC0 Quantum QNC0 QAC0? Technicality Actually, these classes (NC0, QNC0, AC0) are for decision problems with 1 bit of output. This talk is about relation problems with multiple bits of output and multiple answers. And this is necessary!

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Introduction

Relevant Circuit Classes

“The class of problems solved by constant depth classical/quantum circuits (of poly size) with constant/unbounded fan-in gates.” Constant Fan-In Unbounded Fan-In Classical NC0 AC0 Quantum QNC0 QAC0? Theorem (BGK Result) The Hidden Linear Function Problem (HLF) is in QNC0 but not NC0.

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Introduction

Relevant Circuit Classes

“The class of problems solved by constant depth classical/quantum circuits (of poly size) with constant/unbounded fan-in gates.” Constant Fan-In Unbounded Fan-In Classical NC0 AC0 Quantum QNC0 QAC0? Theorem (BGK Result) The Hidden Linear Function Problem (HLF) is in QNC0 but not NC0. Theorem (Our Result) The Relaxed Parity Halving Problem (RPHP) is in QNC0 but not AC0.

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Introduction

Outline

Main Result Extensions

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Introduction

Outline

Main Result Parity Halving Problem (separate QNC0/qpoly and AC0) Relaxed Parity Halving Problem (separate QNC0 and AC0) Extensions

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Introduction

Outline

Main Result Parity Halving Problem (separate QNC0/qpoly and AC0)

Quantum circuit with advice (PHP ∈ QNC0/qpoly) Classical hardness (PHP / ∈ AC0)

Relaxed Parity Halving Problem (separate QNC0 and AC0) Extensions

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Introduction

Outline

Main Result Parity Halving Problem (separate QNC0/qpoly and AC0)

Quantum circuit with advice (PHP ∈ QNC0/qpoly) Classical hardness (PHP / ∈ AC0)

Hard as a game, Hard against NC0 (via locality), Hard against AC0 (via Switching Lemma).

Relaxed Parity Halving Problem (separate QNC0 and AC0) Extensions

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Introduction

Outline

Main Result Parity Halving Problem (separate QNC0/qpoly and AC0)

Quantum circuit with advice (PHP ∈ QNC0/qpoly) Classical hardness (PHP / ∈ AC0)

Hard as a game, Hard against NC0 (via locality), Hard against AC0 (via Switching Lemma).

Relaxed Parity Halving Problem (separate QNC0 and AC0)

Quantum algorithm (RPHP ∈ QNC0) Classical hardness (RPHP / ∈ AC0)

Extensions

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Introduction

Outline

Main Result Parity Halving Problem (separate QNC0/qpoly and AC0)

Quantum circuit with advice (PHP ∈ QNC0/qpoly) Classical hardness (PHP / ∈ AC0)

Hard as a game, Hard against NC0 (via locality), Hard against AC0 (via Switching Lemma).

Relaxed Parity Halving Problem (separate QNC0 and AC0)

Quantum algorithm (RPHP ∈ QNC0) Classical hardness (RPHP / ∈ AC0)

Extensions Better parameters, Geometric locality, Relation to HLF

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Parity Halving Problem

Section 2 Parity Halving Problem

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Parity Halving Problem

Parity Halving

Notation: For x ∈ {0, 1}n define the Hamming weight |x| :=

i xi.

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Parity Halving Problem

Parity Halving

Notation: For x ∈ {0, 1}n define the Hamming weight |x| :=

i xi.

Parity Halving Problem Given x ∈ {0, 1}n with even Hamming weight (|x| ≡ 0 (mod 2)), output y ∈ {0, 1}n such that |y| ≡ 1 2|x| (mod 2)

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Parity Halving Problem

Parity Halving

Notation: For x ∈ {0, 1}n define the Hamming weight |x| :=

i xi.

Parity Halving Problem Given x ∈ {0, 1}n with even Hamming weight (|x| ≡ 0 (mod 2)), output y ∈ {0, 1}n such that |y| ≡ 1 2|x| (mod 2) Example (n = 3) 000 → 000, 011, 101, 110 (even) 011 → 001, 010, 100, 111 (odd) 101 → 001, 010, 100, 111 (odd) 110 → 001, 010, 100, 111 (odd)

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Parity Halving Problem

Parity Halving Problem Given x ∈ {0, 1}n such that |x| ≡ 0 (mod 2), output y ∈ {0, 1}n such that |x| ≡ 0 (mod 4) = ⇒ |y| ≡ 0 (mod 2), |x| ≡ 2 (mod 4) = ⇒ |y| ≡ 1 (mod 2).

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Parity Halving Problem

Parity Halving Game Nonlocal n player game: each player gets one input bit xj, responsible for one output bit yj. The players win if |y| ≡ 1 2|x| (mod 2)

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Parity Halving Problem

Parity Halving Game Nonlocal n player game: each player gets one input bit xj, responsible for one output bit yj. The players win if |y| ≡ 1 2|x| (mod 2) Special case n = 3 is the GHZ game.

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Parity Halving Problem

Parity Halving Game Nonlocal n player game: each player gets one input bit xj, responsible for one output bit yj. The players win if |y| ≡ 1 2|x| (mod 2) Special case n = 3 is the GHZ game. General case independently discovered by Mermin (1990) and Brassard, Broadbent, Tapp (2005).

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Parity Halving Problem

Quantum Strategy

PHP: Given even parity x, find y such that |y| ≡ 1

2|x| (mod 2).

Theorem Given the state | =

1 √ 2 (|0 · · · 0 + |1 · · · 1), quantum players can always win.

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Parity Halving Problem

Quantum Strategy

PHP: Given even parity x, find y such that |y| ≡ 1

2|x| (mod 2).

Theorem Given the state | =

1 √ 2 (|0 · · · 0 + |1 · · · 1), quantum players can always win.

Proof. Each player applies S = ( 1 0

0 i ) to their qubit if xj = 1. State is |0 · · · 0 + i|x| |1 · · · 1.

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Parity Halving Problem

Quantum Strategy

PHP: Given even parity x, find y such that |y| ≡ 1

2|x| (mod 2).

Theorem Given the state | =

1 √ 2 (|0 · · · 0 + |1 · · · 1), quantum players can always win.

Proof. Each player applies S = ( 1 0

0 i ) to their qubit if xj = 1. State is |0 · · · 0 + i|x| |1 · · · 1.

|0 · · · 0 + |1 · · · 1 |0 · · · 0 − |1 · · · 1

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Parity Halving Problem

Quantum Strategy

PHP: Given even parity x, find y such that |y| ≡ 1

2|x| (mod 2).

Theorem Given the state | =

1 √ 2 (|0 · · · 0 + |1 · · · 1), quantum players can always win.

Proof. Each player applies S = ( 1 0

0 i ) to their qubit if xj = 1. State is |0 · · · 0 + i|x| |1 · · · 1.

|0 · · · 0 + |1 · · · 1 H⊗n − − →

• |x| even

|x |0 · · · 0 − |1 · · · 1 H⊗n − − →

• |x| odd

|x

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Parity Halving Problem

Quantum Strategy

PHP: Given even parity x, find y such that |y| ≡ 1

2|x| (mod 2).

Theorem Given the state | =

1 √ 2 (|0 · · · 0 + |1 · · · 1), quantum players can always win.

Proof. Each player applies S = ( 1 0

0 i ) to their qubit if xj = 1. State is |0 · · · 0 + i|x| |1 · · · 1.

|0 · · · 0 + |1 · · · 1 H⊗n − − →

• |x| even

|x |0 · · · 0 − |1 · · · 1 H⊗n − − →

• |x| odd

|x Therefore all players apply a Hadamard, measure, and output the result.

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Parity Halving Problem

QNC0/qpoly circuit

x1 x2 x3 |

• S

S S H H H x1 x2 x3 y1 y2 y3

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Parity Halving Problem

Classical Strategy

PHP: Given even parity x, find y such that |y| = 1

2|x| (mod 2).

Theorem (Game Hardness – Brassard, Broadbent, Tapp) Any deterministic strategy wins on a random input with probability at most 1

2 + 2−⌈n/2⌉.

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Parity Halving Problem

Classical Strategy

PHP: Given even parity x, find y such that |y| = 1

2|x| (mod 2).

Theorem (Game Hardness – Brassard, Broadbent, Tapp) Any deterministic strategy wins on a random input with probability at most 1

2 + 2−⌈n/2⌉.

Vague Intuition Output parity depends on input HW modulo 4.

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Parity Halving Problem

Classical Strategy

PHP: Given even parity x, find y such that |y| = 1

2|x| (mod 2).

Theorem (Game Hardness – Brassard, Broadbent, Tapp) Any deterministic strategy wins on a random input with probability at most 1

2 + 2−⌈n/2⌉.

Vague Intuition Output parity depends on input HW modulo 4. Any one bit is almost completely independent of HW mod 4.

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Parity Halving Problem

Classical Strategy

PHP: Given even parity x, find y such that |y| = 1

2|x| (mod 2).

Theorem (Game Hardness – Brassard, Broadbent, Tapp) Any deterministic strategy wins on a random input with probability at most 1

2 + 2−⌈n/2⌉.

Vague Intuition Output parity depends on input HW modulo 4. Any one bit is almost completely independent of HW mod 4. Fraction of strings with HW i (mod 4) is 1

4 + O(2−n/2).

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Parity Halving Problem

Locality in circuits

Definition A circuit is ℓ-local if each output bit depends on at most ℓ input bits.

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Parity Halving Problem

Locality in circuits

Definition A circuit is ℓ-local if each output bit depends on at most ℓ input bits. Fact A strategy for the game implies a 1-local circuit for PHP.

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Parity Halving Problem

Locality in circuits

Definition A circuit is ℓ-local if each output bit depends on at most ℓ input bits. Fact A strategy for the game implies a 1-local circuit for PHP. Can improve game hardness to 1-local hardness. Theorem (1-local hardness) A 1-local classical circuit solves PHPn on a random input w.p. ≤ 1

2 + 2−⌈n/2⌉.

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Parity Halving Problem

Locality ℓ > 1

Idea Reduce to 1-local circuit

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Parity Halving Problem

Locality ℓ > 1

Idea Reduce to 1-local circuit by restricting some input bits.

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Parity Halving Problem

Locality ℓ > 1

Idea Reduce to 1-local circuit by restricting some input bits. How do we reduce locality to 1? What problem does a circuit for PHP solve after restriction?

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Parity Halving Problem

Locality ℓ > 1

Lemma Consider a circuit with n inputs, n outputs, and locality ℓ. We can find a subset of Ω( n

ℓ2 ) input bits such that restricting all other inputs gives a 1-local

circuit.

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Parity Halving Problem

Locality ℓ > 1

Lemma Consider a circuit with n inputs, n outputs, and locality ℓ. We can find a subset of Ω( n

ℓ2 ) input bits such that restricting all other inputs gives a 1-local

circuit. Proof. Consider a graph with a vertex for each input bit, an edge if both inputs affect some common output bit.

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Parity Halving Problem

Locality ℓ > 1

Lemma Consider a circuit with n inputs, n outputs, and locality ℓ. We can find a subset of Ω( n

ℓ2 ) input bits such that restricting all other inputs gives a 1-local

circuit. Proof. Consider a graph with a vertex for each input bit, an edge if both inputs affect some common output bit. Choose an independent set of vertices to get locality 1.

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Parity Halving Problem

Locality ℓ > 1

Lemma Consider a circuit with n inputs, n outputs, and locality ℓ. We can find a subset of Ω( n

ℓ2 ) input bits such that restricting all other inputs gives a 1-local

circuit. Proof. Consider a graph with a vertex for each input bit, an edge if both inputs affect some common output bit. Choose an independent set of vertices to get locality 1. Tur´ an’s theorem: Largest independent set has size Ω(n2/|E|).

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Parity Halving Problem

Locality ℓ > 1

Lemma Consider a circuit with n inputs, n outputs, and locality ℓ. We can find a subset of Ω( n

ℓ2 ) input bits such that restricting all other inputs gives a 1-local

circuit. Proof. Consider a graph with a vertex for each input bit, an edge if both inputs affect some common output bit. Choose an independent set of vertices to get locality 1. Tur´ an’s theorem: Largest independent set has size Ω(n2/|E|). Each output is responsible for at most O(ℓ2) edges = ⇒ |E| = O(nℓ2).

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Parity Halving Problem

Inputs Outputs

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Parity Halving Problem

Restrictions of PHP

What happens when we take a circuit for PHP and fix some input bits?

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Parity Halving Problem

Restrictions of PHP

What happens when we take a circuit for PHP and fix some input bits? E.g., xn = 1 = ⇒ remaining inputs have odd parity.

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Parity Halving Problem

Restrictions of PHP

What happens when we take a circuit for PHP and fix some input bits? E.g., xn = 1 = ⇒ remaining inputs have odd parity. Parity Halving Problem (Original) Given x ∈ {0, 1}n such that |x| ≡ 0 (mod 2), output y ∈ {0, 1}n such that |x| ≡ 0 (mod 4) = ⇒ |y| ≡ 0 (mod 2), |x| ≡ 2 (mod 4) = ⇒ |y| ≡ 1 (mod 2).

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Parity Halving Problem

Restrictions of PHP

What happens when we take a circuit for PHP and fix some input bits? E.g., xn = 1 = ⇒ remaining inputs have odd parity. Parity Halving Problem (Variant 1) Given x ∈ {0, 1}n such that |x| ≡ 1 (mod 2), output y ∈ {0, 1}n such that |x| ≡ 3 (mod 4) = ⇒ |y| ≡ 0 (mod 2), |x| ≡ 1 (mod 4) = ⇒ |y| ≡ 1 (mod 2).

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Parity Halving Problem

Restrictions of PHP

What happens when we take a circuit for PHP and fix some input bits? E.g., xn = 1 = ⇒ remaining inputs have odd parity. Parity Halving Problem (Variant 2) Given x ∈ {0, 1}n such that |x| ≡ 0 (mod 2), output y ∈ {0, 1}n such that |x| ≡ 2 (mod 4) = ⇒ |y| ≡ 0 (mod 2), |x| ≡ 0 (mod 4) = ⇒ |y| ≡ 1 (mod 2).

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Parity Halving Problem

Restrictions of PHP

What happens when we take a circuit for PHP and fix some input bits? E.g., xn = 1 = ⇒ remaining inputs have odd parity. Parity Halving Problem (Variant 3) Given x ∈ {0, 1}n such that |x| ≡ 1 (mod 2), output y ∈ {0, 1}n such that |x| ≡ 1 (mod 4) = ⇒ |y| ≡ 0 (mod 2), |x| ≡ 3 (mod 4) = ⇒ |y| ≡ 1 (mod 2).

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Parity Halving Problem

Restrictions of PHP

What happens when we take a circuit for PHP and fix some input bits? E.g., xn = 1 = ⇒ remaining inputs have odd parity. Parity Halving Problem (All Variants) Given x ∈ {0, 1}n such that |x| ≡ b (mod 2), output y ∈ {0, 1}n such that |x| ≡ b (mod 4) = ⇒ |y| ≡ 0 (mod 2), |x| ≡ b + 2 (mod 4) = ⇒ |y| ≡ 1 (mod 2). where b ∈ {0, 1, 2, 3}.

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Parity Halving Problem

Restrictions of PHP

What happens when we take a circuit for PHP and fix some input bits? E.g., xn = 1 = ⇒ remaining inputs have odd parity. Parity Halving Problem (All Variants) Given x ∈ {0, 1}n such that |x| ≡ b (mod 2), output y ∈ {0, 1}n such that |x| ≡ b (mod 4) = ⇒ |y| ≡ 0 (mod 2), |x| ≡ b + 2 (mod 4) = ⇒ |y| ≡ 1 (mod 2). where b ∈ {0, 1, 2, 3}. Claim All problems are equivalent.

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Parity Halving Problem

Locality-ℓ Hardness Result

Theorem An ℓ-local classical circuit solves PHPn on a random input w.p. ≤ 1

2 + 2−Ω(n/ℓ2).

Proof. Find Ω(n/ℓ2) inputs with non-overlapping light cones. Fix the rest. The remaining circuit solves a variant of PHP on Ω(n/ℓ2) bits.

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Parity Halving Problem

Locality-ℓ Hardness Result

Theorem An ℓ-local classical circuit solves PHPn on a random input w.p. ≤ 1

2 + 2−Ω(n/ℓ2).

Proof. Find Ω(n/ℓ2) inputs with non-overlapping light cones. Fix the rest. The remaining circuit solves a variant of PHP on Ω(n/ℓ2) bits. Corollary Since NC0 circuits have locality ℓ = O(1), they solve PHP w.p. 1

2 + 2−Ω(n).

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Parity Halving Problem

AC0 hardness

Problem Unbounded fan-in gates make it easy to have locality n.

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Parity Halving Problem

AC0 hardness

Problem Unbounded fan-in gates make it easy to have locality n. Solution Finally some circuit complexity theory: the Switching Lemma!!

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Parity Halving Problem

AC0 hardness

Problem Unbounded fan-in gates make it easy to have locality n. Solution Finally some circuit complexity theory: the Switching Lemma!! Switching Lemma (Intuition) Consider an AC0 circuit. With high probability, restricting a (large) random subset of bits produces a circuit with no(1) locality.

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Parity Halving Problem

Switching Lemma in Action

Theorem AC0 circuits solve PHP w.p. at most 1

2 + o(1).

Proof. Apply the switching lemma. Locality is reduced, and the resulting circuit solves a variant of PHP, so hardness for local circuits implies 1

2 + o(1) probability of success.

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Relaxed Parity Halving Problem

Section 3 Relaxed Parity Halving Problem

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Relaxed Parity Halving Problem

We don’t want to use an advice state, but we can’t construct it ourselves. Theorem The state

1 √ 2 (|0n + |1n) cannot be constructed in QNC0.

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Relaxed Parity Halving Problem

We don’t want to use an advice state, but we can’t construct it ourselves. Theorem The state

1 √ 2 (|0n + |1n) cannot be constructed in QNC0.

But we can construct a poor man’s cat state! 1 √ 2 (|z + |z)

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Relaxed Parity Halving Problem

We don’t want to use an advice state, but we can’t construct it ourselves. Theorem The state

1 √ 2 (|0n + |1n) cannot be constructed in QNC0.

But we can construct a poor man’s cat state! 1 √ 2 (|z + |z) = X z 1 √ 2 (|0n + |1n)

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Relaxed Parity Halving Problem

We don’t want to use an advice state, but we can’t construct it ourselves. Theorem The state

1 √ 2 (|0n + |1n) cannot be constructed in QNC0.

But we can construct a poor man’s cat state! 1 √ 2 (|z + |z) = X z 1 √ 2 (|0n + |1n) A cat state with some bits flipped.

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Relaxed Parity Halving Problem

Poor Man’s Cat State

Q: If we can construct |z + |z = X z | in QNC0, then why can’t we apply X z to get | ?

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Relaxed Parity Halving Problem

Poor Man’s Cat State

Q: If we can construct |z + |z = X z | in QNC0, then why can’t we apply X z to get | ? A: We don’t know what z is!!

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Relaxed Parity Halving Problem

Poor Man’s Cat State

Q: If we can construct |z + |z = X z | in QNC0, then why can’t we apply X z to get | ? A: We don’t know what z is!! Theorem In QNC0 we can construct

1 √ 2 (|z + |z) for some uniformly random z ∈ {0, 1}n,

with information d ∈ {0, 1}n−1 from which z can be recovered (up to complement) in AC0.

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Relaxed Parity Halving Problem

Poor Man’s Cat State Example

Consider a tree G = (V , E). Let there be a |+ qubit for each vertex. z1 z2 z3 z4 z5

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Relaxed Parity Halving Problem

Poor Man’s Cat State Example

For each edge, measure the parity of the two endpoints. z1 z2 z3 z4 z5 z1 ⊕ z5 = 1 z2 ⊕ z5 = 0 z3 ⊕ z5 = 1 z4 ⊕ z5 = 1

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Relaxed Parity Halving Problem

Poor Man’s Cat State Example

Two vectors, z and z, are consistent with these measurements. 1 1 0 ⊕ 1 = 1 1 ⊕ 1 = 0 0 ⊕ 1 = 1 0 ⊕ 1 = 1

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Relaxed Parity Halving Problem

Poor Man’s Cat State Example

To construct z, let zi be the parity of the path from z1 to zi. z1 = 0 zi = 1 1 1 1

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Relaxed Parity Halving Problem

Poor Man’s Cat State Example

To construct z, let zi be the parity of the path from z1 to zi. z1 = 0 zi = 1 1 1 1 z2 = z1 ⊕ z2 = (z1 ⊕ z5) ⊕ (z2 ⊕ z5) = 1 ⊕ 0 = 1

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Relaxed Parity Halving Problem

Poor Man’s Cat State Example

Final output is state |01001 + |10110 (vertex qubits) and d = 1011 (edge measurements). 1 1 1 1 1

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Relaxed Parity Halving Problem

What kind of tree to use?

Line graph!

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Relaxed Parity Halving Problem

What kind of tree to use?

Line graph! We want low diameter, so it is easier to compute z from d.

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Relaxed Parity Halving Problem

What kind of tree to use?

Line graph! We want low diameter, so it is easier to compute z from d. Star graph!

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Relaxed Parity Halving Problem

What kind of tree to use?

Line graph! We want low diameter, so it is easier to compute z from d. Star graph! We want low degree, since edges incident at the same vertex cannot be simultaneously measured.

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Relaxed Parity Halving Problem

What kind of tree to use?

Line graph! We want low diameter, so it is easier to compute z from d. Star graph! We want low degree, since edges incident at the same vertex cannot be simultaneously measured. Balanced binary tree! Max degree ∆ = 3, diameter d = Θ(log n). (AC0 can compute O(log n) size parities)

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Relaxed Parity Halving Problem

Relaxed Parity Halving Problem

Q: What do we do with the poor man’s cat state?

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Relaxed Parity Halving Problem

Relaxed Parity Halving Problem

Q: What do we do with the poor man’s cat state? A: Pretend it’s a cat state and run the same algorithm!!

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Relaxed Parity Halving Problem

Relaxed Parity Halving Problem

Q: What do we do with the poor man’s cat state? A: Pretend it’s a cat state and run the same algorithm!!

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Relaxed Parity Halving Problem

x1 x2 x3 |

• X

X S S S H H H x1 x2 x3 y1 y2 y3

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Relaxed Parity Halving Problem

x1 x2 x3 |

• S

S S Z Z X X H H H x1 x2 x3 y1 y2 y3

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Relaxed Parity Halving Problem

x1 x2 x3 |

• S

S S Z Z H H H Z Z x1 x2 x3 y1 y2 y3

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Relaxed Parity Halving Problem

x1 x2 x3 |

• S

S S Z Z H H H x1 x2 x3 y1 y2 y3

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Relaxed Parity Halving Problem

x1 x2 x3 |

• S

S S H H H X X x1 x2 x3 y1 y2 y3

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Relaxed Parity Halving Problem

Relaxed Parity Halving Problem Given an even parity input x ∈ {0, 1}n, output y ∈ {0, 1}n such that |y| ≡ 1 2|x| + x, z (mod 2) .

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Relaxed Parity Halving Problem

Relaxed Parity Halving Problem Given an even parity input x ∈ {0, 1}n, output y ∈ {0, 1}n and d ∈ {0, 1}n−1 such that |y| ≡ 1 2|x| + x, z (mod 2) where z ∈ {0, 1}n is either vector consistent d.

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Relaxed Parity Halving Problem

Relaxed Parity Halving Problem Given an even parity input x ∈ {0, 1}n, output y ∈ {0, 1}n and d ∈ {0, 1}n−1 such that |y| ≡ 1 2|x| + x, z (mod 2) where z ∈ {0, 1}n is either vector consistent d. RPHP ∈ QNC0 is clear.

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Relaxed Parity Halving Problem

Classical Hardness

Theorem Any AC0 circuit for RPHP succeeds with probability < 1

2 + o(1).

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Relaxed Parity Halving Problem

Classical Hardness

Theorem Any AC0 circuit for RPHP succeeds with probability < 1

2 + o(1).

Proof. Suppose we have a circuit for RPHP.

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Relaxed Parity Halving Problem

Classical Hardness

Theorem Any AC0 circuit for RPHP succeeds with probability < 1

2 + o(1).

Proof. Suppose we have a circuit for RPHP. In AC0, we can compute z from d because each zi is an O(log n)-bit parity.

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Relaxed Parity Halving Problem

Classical Hardness

Theorem Any AC0 circuit for RPHP succeeds with probability < 1

2 + o(1).

Proof. Suppose we have a circuit for RPHP. In AC0, we can compute z from d because each zi is an O(log n)-bit parity. “Correct” for z. Remove x, z.

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Relaxed Parity Halving Problem

Classical Hardness

Theorem Any AC0 circuit for RPHP succeeds with probability < 1

2 + o(1).

Proof. Suppose we have a circuit for RPHP. In AC0, we can compute z from d because each zi is an O(log n)-bit parity. “Correct” for z. Remove x, z.

Compute wi = xizi for all i. XOR in corrections: y ′

i := yi ⊕ wi.

Note |y ′| = |y| + x, z = 1

2|x| (mod 2).

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Relaxed Parity Halving Problem

Classical Hardness

Theorem Any AC0 circuit for RPHP succeeds with probability < 1

2 + o(1).

Proof. Suppose we have a circuit for RPHP. In AC0, we can compute z from d because each zi is an O(log n)-bit parity. “Correct” for z. Remove x, z.

Compute wi = xizi for all i. XOR in corrections: y ′

i := yi ⊕ wi.

Note |y ′| = |y| + x, z = 1

2|x| (mod 2).

New circuit solves PHP (with the same probability)

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Extensions

Section 4 Extensions

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Extensions

Extensions – Better Parameters

State of the art switching lemma results (Hastad and Rossman) give Theorem AC0 circuits solve RPHP on a random input w.p. at most 1

2 + 2−n0.999.

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Extensions

Extensions – Better Parameters

State of the art switching lemma results (Hastad and Rossman) give Theorem AC0 circuits solve RPHP on a random input w.p. at most 1

2 + 2−n0.999.

Theorem AC0 circuits of depth d and size exp(n1/2d) solve RPHP w.p. at most 1

2 + 2−n0.999.

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Extensions

Extensions – Better Parameters

State of the art switching lemma results (Hastad and Rossman) give Theorem AC0 circuits solve RPHP on a random input w.p. at most 1

2 + 2−n0.999.

Theorem AC0 circuits of depth d and size exp(n1/2d) solve RPHP w.p. at most 1

2 + 2−n0.999.

Note: RPHP or PHP are solved exactly by exp(n1/d) size AC0 circuits.

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Extensions

Parallel Copies

Parallel Parity Halving Problem Given inputs x1, . . . , xk ∈ {0, 1}n, output y1, . . . , yk ∈ {0, 1}n such that for all i, |xi| ≡ 1 2|yi| (mod 2). In other words, make the circuit solve k copies of the problem at once.

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Extensions

Parallel Copies

Parallel Parity Halving Problem Given inputs x1, . . . , xk ∈ {0, 1}n, output y1, . . . , yk ∈ {0, 1}n such that for all i, |xi| ≡ 1 2|yi| (mod 2). In other words, make the circuit solve k copies of the problem at once. Theorem AC0 circuits of depth d and size exp(n1/2d) solve Parallel-RPHP w.p. at most 2−n0.999.

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Extensions

2D Locality

Any tree in the grid has diameter Ω(√n). A different reduction works for trees with diameter d = o(n). Theorem There exists a constant-depth quantum circuit for Grid-RPHP which is local on a 2D grid.

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Extensions

2D Locality

Any tree in the grid has diameter Ω(√n). A different reduction works for trees with diameter d = o(n). Theorem There exists a constant-depth quantum circuit for Grid-RPHP which is local on a 2D grid. Parallel-Grid-RPHP . . .

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Extensions

Extensions – HLF and Geometric Locality

RPHP reduces to HLF Theorem RPHP ≤ HLF. There is no AC0 circuit for HLF.

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Extensions

Extensions – HLF and Geometric Locality

RPHP reduces to HLF Theorem RPHP ≤ HLF. There is no AC0 circuit for HLF. Theorem Parallel-Grid-RPHP ≤ 2DHLF

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Extensions

HLF Problem

Hidden Linear Function Problem Given a symmetric matrix A ∈ {0, 1}n×n and vector b ∈ {0, 1, 2, 3}n, output any string y that may be output by the following circuit. |b |A |0n H⊗n CZ(A) S(b) H⊗n |b |A y

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Extensions

HLF Problem

Hidden Linear Function Problem Given a symmetric matrix A ∈ {0, 1}n×n and vector b ∈ {0, 1, 2, 3}n, output any string y that may be output by the following circuit. |b |A |0n H⊗n CZ(A) S(b) H⊗n |b |A y Suppose we fix all of A, part of b.

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Extensions

x1 x2 x3 |0 |0 |0 |0 |0 H H H H H S S S H H H H H x1 x2 x3 y1 d1 y2 d2 y3

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Extensions

x1 x2 x3 |0 |0 |0 |0 |0 H H H H H S S S H H H H H x1 x2 x3 y1 d1 y2 d2 y3

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Extensions

x1 x2 x3 |0 |0 |0 |0 |0 H H H S S S H H H x1 x2 x3 y1 d1 y2 d2 y3

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Extensions

x1 x2 x3 |+ |0 |+ |0 |+ S S S H H H x1 x2 x3 y1 d1 y2 d2 y3

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Extensions

Open Problems

Can improve the classical hardness to more powerful circuit classes? AC0[2], TC0, NC1

Problem: Best circuit lower bounds stop around TC0. Would need to be conditional. Partial result: QNC0/qpoly vs. AC0[2].

Can we get the same separation with 1D local circuits?

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Extensions

Thank You!

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