Round elimination in exact communication complexity Teresa Piovesan - - PowerPoint PPT Presentation

Round elimination in exact communication complexity

Teresa Piovesan

Joint work with Jop Bri¨ et, Harry Buhrman, Debbie Leung+ & Florian Speelman (+University of Waterloo) 20th Combinatorial Optimization Workshop Aussois, 6 January 2016

Communication complexity

x y

Communication complexity

x y f (x, y) ?

Communication complexity

x y f (x, y) ?

Communication complexity

x y f (x, y) ? . . . . . . . . .

Communication complexity

x y f (x, y) ? . . . . . . . . . f (x, y)

Communication complexity

x y f (x, y) ? . . . . . . . . . f (x, y)

• Communication complexity = min # bits exchanged

Communication complexity

x y f (x, y) ? . . . . . . . . . f (x, y)

• Communication complexity = min # bits exchanged
• Exact communication complexity: Bob learns f (x, y) without error

Communication complexity

x y f (x, y) ? . . . . . . . . . f (x, y)

• Communication complexity = min # bits exchanged
• Exact communication complexity: Bob learns f (x, y) without error
• Quantum communication complexity = min # qubits exchanged

Bit vs Qubit

Classical bit:

• basis state: 0, 1
• mutually exclusive

Bit vs Qubit

Classical bit:

• basis state: 0, 1
• mutually exclusive

Quantum bit (qubit):

• basis state: e1, e2 ∈ C2 where e1 =

1

• , e2 =

1

• superposition

αe1 + βe2 = α β

• ∈ C2 s.t. |α|2 + |β|2 = 1

Bit vs Qubit

Classical bit:

• basis state: 0, 1
• mutually exclusive

Quantum bit (qubit):

• basis state: e1, e2 ∈ C2 where e1 =

1

• , e2 =

1

• superposition

αe1 + βe2 = α β

• ∈ C2 s.t. |α|2 + |β|2 = 1

Catch: to extract information from

• α

β

• we need to perform a measurement,

i.e., we get e1 with probability |α|2, we get e2 with probability |β|2

Bit vs Qubit

Classical bit:

• basis state: 0, 1
• mutually exclusive

Quantum bit (qubit):

• basis state: e1, e2 ∈ C2 where e1 =

1

• , e2 =

1

• superposition

αe1 + βe2 = α β

• ∈ C2 s.t. |α|2 + |β|2 = 1

Catch: to extract information from

• α

β

• we need to perform a measurement,

i.e., we get e1 with probability |α|2, we get e2 with probability |β|2 ℓ-qubit state: 2ℓ

i=1 αiei ∈ C2ℓ s.t. 2ℓ i=1 |αi|2 = 1.

Promise equality

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . .

Promise equality

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = k

Promise equality

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = n/2 Exponential separation quantum vs classical [Buhrman et al.’98]

Promise equality

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = n/2 Exponential separation quantum vs classical [Buhrman et al.’98]

• Classical: at least Ω(n) (combinatorial result [Frankl and R¨
• dl ’87])

Promise equality

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = n/2 Exponential separation quantum vs classical [Buhrman et al.’98]

• Classical: at least Ω(n) (combinatorial result [Frankl and R¨
• dl ’87])
• One-round quantum protocol that uses O(log n) qubits

Promise equality

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = n/2 Exponential separation quantum vs classical [Buhrman et al.’98]

• Classical: at least Ω(n) (combinatorial result [Frankl and R¨
• dl ’87])
• One-round quantum protocol that uses O(log n) qubits
• Max separation possible [Kremer ’95]

Promise equality: classical

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = k

Promise equality: classical

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = k Graph: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = k

Promise equality: classical

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = k Graph: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = k Classical one-round promise equality = log (chromatic number)

Promise equality: classical

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = k Graph: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = k Classical one-round promise equality = log (chromatic number)

Lemma

Classical communication complexity is attained with a single round.

Promise equality: quantum one-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = k

Promise equality: quantum one-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = k φ : {0, 1}n → Cd

Promise equality: quantum one-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = k φ : {0, 1}n → Cd Quantum model: Collection of ℓ-qubits messages is perfectly distinguishable iff pairwise orthogonal

Promise equality: quantum one-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = k

• r

φ : {0, 1}n → Cd s.t. φ(x) ⊥ φ(y) if ∆(x, y) = k Quantum model: Collection of ℓ-qubits messages is perfectly distinguishable iff pairwise orthogonal

Promise equality: quantum one-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = k

• r

min d φ : {0, 1}n → Cd s.t. φ(x) ⊥ φ(y) if ∆(x, y) = k

• rthogonal rank

Graph: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = k

Promise equality: quantum one-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = k

• r

min d φ : {0, 1}n → Cd s.t. φ(x) ⊥ φ(y) if ∆(x, y) = k

• rthogonal rank

Graph: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = k Quantum one-round promise equality = log (orthogonal rank) [de Wolf ’01]

Promise equality

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/2 Exponential separation quantum vs classical [Buhrman et al.’98]

Promise equality

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/2 Exponential separation quantum vs classical [Buhrman et al.’98] Graph: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = n/2

Promise equality

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/2 Exponential separation quantum vs classical [Buhrman et al.’98] Graph: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = n/2

• Classical: log(chromatic number) is at least Ω(n)

[Frankl and R¨

• dl ’87]

Promise equality

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/2 Exponential separation quantum vs classical [Buhrman et al.’98] Graph: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = n/2

• Classical: log(chromatic number) is at least Ω(n)

[Frankl and R¨

• dl ’87]
• Quantum one-round:

φ : {0, 1}n → Cn s.t. φ(x) =

1 √n

n

i=1(−1)xiei

φ(x) ⊥ φ(y) if ∆(x, y) = n/2

Promise equality

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/2 Exponential separation quantum vs classical [Buhrman et al.’98] Graph: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = n/2

• Classical: log(chromatic number) is at least Ω(n)

[Frankl and R¨

• dl ’87]
• Quantum one-round:

φ : {0, 1}n → Cn s.t. φ(x) =

1 √n

n

i=1(−1)xiei

φ(x) ⊥ φ(y) if ∆(x, y) = n/2 = ⇒ need at most log n qubits

Promise equality: classical

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = αn For any α constant, log(chromatic number) is at least Ω(n) [Frankl and R¨

• dl ’87]

Promise equality: quantum

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = αn Quantum:

• α = 1/2: one-round at most log n

[Buhrman et al.’98]

Promise equality: quantum

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = αn Quantum:

• α = 1/2: one-round at most log n

[Buhrman et al.’98]

• α > 1/2: one-round at most log(n + 1)

[Gruska et al.’14]

Promise equality: quantum

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = αn Quantum:

• α = 1/2: one-round at most log n

[Buhrman et al.’98]

• α > 1/2: one-round at most log(n + 1)

[Gruska et al.’14]

• α < 1/2? Is one-round optimal?

Promise equality: quantum

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = αn with α < 1/2

Theorem

• One-round protocol at least Ω(n) qubits
• Quantum communication complexity O(log n)

Promise equality: quantum

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = αn with α < 1/2

Theorem

• One-round protocol at least Ω(n) qubits

(Lov´ asz theta number)

• Quantum communication complexity O(log n)

Promise equality: quantum

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? . . . . . . . . . Promise: x = y or ∆(x, y) = αn with α < 1/2

Theorem

• One-round protocol at least Ω(n) qubits

(Lov´ asz theta number)

• Quantum communication complexity O(log n)

(Grover’s algorithm)

Promise equality: quantum one-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4

Promise equality: quantum one-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4 min d φ : {0, 1}n → Cd s.t. φ(x) ⊥ φ(y) if ∆(x, y) = n/4

Lower bound to orthogonal rank

Graph G: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = n/4

Lower bound to orthogonal rank

Graph G: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = n/4

• Orthogonal rank is lower bounded by ϑ(G)

ϑ(G) = max

• i,j∈V (G)

Xij : X 0, Tr(X) = 1, Xij = 0 if ij ∈ E(G)

Lower bound to orthogonal rank

Graph G: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = n/4

• Orthogonal rank is lower bounded by ϑ(G)

ϑ(G) = max

• i,j∈V (G)

Xij : X 0, Tr(X) = 1, Xij = 0 if ij ∈ E(G)

• Using properties of graph G,

ϑ(G) = 1 − λmax/λmin where λmax, λmin are eigenvalues of the adjacency matrix

Lower bound to orthogonal rank

Graph G: V = {0, 1}n, (x, y) ∈ E if ∆(x, y) = n/4

• Orthogonal rank is lower bounded by ϑ(G)

ϑ(G) = max

• i,j∈V (G)

Xij : X 0, Tr(X) = 1, Xij = 0 if ij ∈ E(G)

• Using properties of graph G,

ϑ(G) = 1 − λmax/λmin where λmax, λmin are eigenvalues of the adjacency matrix

• Derive a bound on λmin using properties of Krawtchouk polynomials

Promise equality: quantum two-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4 Grover’s algorithm:

• z ∈ {0, 1}n with has exactly n/4 ones.
• quantum query: ei → (−1)ziei,

Promise equality: quantum two-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4 Grover’s algorithm:

• z ∈ {0, 1}n with has exactly n/4 ones.
• quantum query: ei → (−1)ziei, i.e.

1 √n

n

i=1 ei → 1 √n

n

i=1(−1)ziei

Promise equality: quantum two-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4 Grover’s algorithm:

• z ∈ {0, 1}n with has exactly n/4 ones.
• quantum query: ei → (−1)ziei, i.e.

1 √n

n

i=1 ei → 1 √n

n

i=1(−1)ziei

• finds index of a 1

Promise equality: quantum two-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4 Grover’s algorithm:

• z ∈ {0, 1}n with has exactly n/4 ones.
• quantum query: ei → (−1)ziei, i.e.

1 √n

n

i=1 ei → 1 √n

n

i=1(−1)ziei

• finds index of a 1

→ z = x ⊕ y

Promise equality: quantum two-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4

1 √n

n

i=1(−1)yiei

Grover’s algorithm:

• z ∈ {0, 1}n with has exactly n/4 ones.
• quantum query: ei → (−1)ziei, i.e.

1 √n

n

i=1 ei → 1 √n

n

i=1(−1)ziei

• finds index of a 1

→ z = x ⊕ y

Promise equality: quantum two-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4

1 √n

n

i=1(−1)yiei 1 √n

n

i=1(−1)yi⊕xiei

Grover’s algorithm:

• z ∈ {0, 1}n with has exactly n/4 ones.
• quantum query: ei → (−1)ziei, i.e.

1 √n

n

i=1 ei → 1 √n

n

i=1(−1)ziei

• finds index of a 1

→ z = x ⊕ y

Promise equality: quantum two-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4

1 √n

n

i=1(−1)yiei 1 √n

n

i=1(−1)ziei

Grover’s algorithm:

• z ∈ {0, 1}n with has exactly n/4 ones.
• quantum query: ei → (−1)ziei, i.e.

1 √n

n

i=1 ei → 1 √n

n

i=1(−1)ziei

• finds index of a 1

→ z = x ⊕ y

Promise equality: quantum two-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4

1 √n

n

i=1(−1)yiei 1 √n

n

i=1(−1)ziei

finish Grover gets i∗ s.t. xi∗ = yi∗ if x = y Grover’s algorithm:

• z ∈ {0, 1}n with has exactly n/4 ones.
• quantum query: ei → (−1)ziei, i.e.

1 √n

n

i=1 ei → 1 √n

n

i=1(−1)ziei

• finds index of a 1

→ z = x ⊕ y

Promise equality: quantum two-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4

1 √n

n

i=1(−1)yiei 1 √n

n

i=1(−1)ziei

finish Grover gets i∗ s.t. xi∗ = yi∗ if x = y i∗, xi∗ Grover’s algorithm:

• z ∈ {0, 1}n with has exactly n/4 ones.
• quantum query: ei → (−1)ziei, i.e.

1 √n

n

i=1 ei → 1 √n

n

i=1(−1)ziei

• finds index of a 1

→ z = x ⊕ y

Promise equality: quantum two-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4

1 √n

n

i=1(−1)yiei 1 √n

n

i=1(−1)ziei

finish Grover gets i∗ s.t. xi∗ = yi∗ if x = y i∗, xi∗ checks if xi∗ = yi∗ Grover’s algorithm:

• z ∈ {0, 1}n with has exactly n/4 ones.
• quantum query: ei → (−1)ziei, i.e.

1 √n

n

i=1 ei → 1 √n

n

i=1(−1)ziei

• finds index of a 1

→ z = x ⊕ y

Promise equality: quantum two-round

x ∈ {0, 1}n y ∈ {0, 1}n x = y ? Promise: x = y or ∆(x, y) = n/4

1 √n

n

i=1(−1)yiei 1 √n

n

i=1(−1)ziei

finish Grover gets i∗ s.t. xi∗ = yi∗ if x = y i∗, xi∗ checks if xi∗ = yi∗

Lemma

Two-round protocol that uses 2 log n+1 qubits

Promise equality: open problem

Question: What is the orthogonal rank of the graph G=(V,E)

V = {0, 1}n, (x, y) ∈ E if ∆(x, y) ≥ n/2?

Promise equality: open problem

Question: What is the orthogonal rank of the graph G=(V,E)

V = {0, 1}n, (x, y) ∈ E if ∆(x, y) ≥ n/2?

Conjecture: 2Ω(n)

Promise equality: open problem

Question: What is the orthogonal rank of the graph G=(V,E)

V = {0, 1}n, (x, y) ∈ E if ∆(x, y) ≥ n/2?

Conjecture: 2Ω(n)

Implication: no quantum round elimination

Promise equality: open problem

Question: What is the orthogonal rank of the graph G=(V,E)

V = {0, 1}n, (x, y) ∈ E if ∆(x, y) ≥ n/2?

Conjecture: 2Ω(n)

Implication: no quantum round elimination

Promise equality: open problem

Question: What is the orthogonal rank of the graph G=(V,E)

V = {0, 1}n, (x, y) ∈ E if ∆(x, y) ≥ n/2?

Conjecture: 2Ω(n)

Implication: no quantum round elimination

1

Promise equality: open problem

Question: What is the orthogonal rank of the graph G=(V,E)

V = {0, 1}n, (x, y) ∈ E if ∆(x, y) ≥ n/2?

Conjecture: 2Ω(n)

Implication: no quantum round elimination

1 ℓ

Promise equality: open problem

Question: What is the orthogonal rank of the graph G=(V,E)

V = {0, 1}n, (x, y) ∈ E if ∆(x, y) ≥ n/2?

Conjecture: 2Ω(n)

Implication: no quantum round elimination

1 ℓ 2ℓ

Promise equality: open problem

Question: What is the orthogonal rank of the graph G=(V,E)

V = {0, 1}n, (x, y) ∈ E if ∆(x, y) ≥ n/2?

Conjecture: 2Ω(n)

Implication: no quantum round elimination

1 ℓ 2ℓ

Thanks!