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

Round elimination in exact communication complexity Teresa Piovesan Joint work with et, Harry Buhrman, Debbie Leung + Jop Bri¨ & Florian Speelman ( + University of Waterloo) 20 th Combinatorial Optimization Workshop Aussois, 6 January 2016

Communication complexity y x

Communication complexity f ( x , y ) ? y x

Communication complexity f ( x , y ) ? y x

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

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

Communication complexity f ( x , y ) ? y x . . . . . . . . . f ( x , y ) • Communication complexity = min # bits exchanged

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

Communication complexity f ( x , y ) ? y x . . . . . . . . . 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): � 1 � � 0 � - basis state: e 1 , e 2 ∈ C 2 where e 1 = , e 2 = 0 1 - superposition � α � ∈ C 2 s.t. | α | 2 + | β | 2 = 1 α e 1 + β e 2 = β

Bit vs Qubit Classical bit: - basis state: 0, 1 - mutually exclusive Quantum bit (qubit): � 1 � � 0 � - basis state: e 1 , e 2 ∈ C 2 where e 1 = , e 2 = 0 1 - superposition � α � ∈ C 2 s.t. | α | 2 + | β | 2 = 1 α e 1 + β e 2 = β Catch: � � α to extract information from we need to perform a measurement, β i.e., we get e 1 with probability | α | 2 , we get e 2 with probability | β | 2

Bit vs Qubit Classical bit: - basis state: 0, 1 - mutually exclusive Quantum bit (qubit): � 1 � � 0 � - basis state: e 1 , e 2 ∈ C 2 where e 1 = , e 2 = 0 1 - superposition � α � ∈ C 2 s.t. | α | 2 + | β | 2 = 1 α e 1 + β e 2 = β Catch: � � α to extract information from we need to perform a measurement, β i.e., we get e 1 with probability | α | 2 , we get e 2 with probability | β | 2 i =1 α i e i ∈ C 2 ℓ s.t. � 2 ℓ ℓ -qubit state: � 2 ℓ i =1 | α i | 2 = 1.

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

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

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

Promise equality Promise : x = y or x = y ? ∆( x , y ) = n / 2 x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n . . . . . . . . . Exponential separation quantum vs classical [Buhrman et al.’98] • Classical: at least Ω( n ) (combinatorial result [Frankl and R¨ odl ’87])

Promise equality Promise : x = y or x = y ? ∆( x , y ) = n / 2 x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n . . . . . . . . . Exponential separation quantum vs classical [Buhrman et al.’98] • Classical: at least Ω( n ) (combinatorial result [Frankl and R¨ odl ’87]) • One-round quantum protocol that uses O (log n ) qubits

Promise equality Promise : x = y or x = y ? ∆( x , y ) = n / 2 x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n . . . . . . . . . Exponential separation quantum vs classical [Buhrman et al.’98] • Classical: at least Ω( n ) (combinatorial result [Frankl and R¨ odl ’87]) • One-round quantum protocol that uses O (log n ) qubits • Max separation possible [Kremer ’95]

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

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

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

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

Promise equality: quantum one-round Promise : x = y or x = y ? ∆( x , y ) = k x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n φ : { 0 , 1 } n → C d

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

Promise equality: quantum one-round Promise : x = y or x = y ? ∆( x , y ) = k x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n or φ : { 0 , 1 } n → C d 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 Promise : x = y or x = y ? ∆( x , y ) = k x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n or φ : { 0 , 1 } n → C d s.t. φ ( x ) ⊥ φ ( y ) if ∆( x , y ) = k min d orthogonal rank Graph: V = { 0 , 1 } n , ( x , y ) ∈ E if ∆( x , y ) = k

Promise equality: quantum one-round Promise : x = y or x = y ? ∆( x , y ) = k x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n or φ : { 0 , 1 } n → C d s.t. φ ( x ) ⊥ φ ( y ) if ∆( x , y ) = k min d orthogonal 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 Promise : x = y or x = y ? ∆( x , y ) = n / 2 x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n Exponential separation quantum vs classical [Buhrman et al.’98]

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

Promise equality Promise : x = y or x = y ? ∆( x , y ) = n / 2 x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n 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¨ odl ’87]

Promise equality Promise : x = y or x = y ? ∆( x , y ) = n / 2 x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n 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¨ odl ’87] • Quantum one-round: φ : { 0 , 1 } n → C n s.t. φ ( x ) = � n 1 i =1 ( − 1) x i e i √ n φ ( x ) ⊥ φ ( y ) if ∆( x , y ) = n / 2

Promise equality Promise : x = y or x = y ? ∆( x , y ) = n / 2 x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n 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¨ odl ’87] • Quantum one-round: φ : { 0 , 1 } n → C n s.t. φ ( x ) = � n 1 i =1 ( − 1) x i e i √ n φ ( x ) ⊥ φ ( y ) if ∆( x , y ) = n / 2 = ⇒ need at most log n qubits

Promise equality: classical Promise : x = y or x = y ? ∆( x , y ) = α n x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n . . . . . . . . . For any α constant, log(chromatic number) is at least Ω( n ) [Frankl and R¨ odl ’87]

Promise equality: quantum Promise : x = y or x = y ? ∆( x , y ) = α n x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n . . . . . . . . . Quantum: • α = 1 / 2: one-round at most log n [Buhrman et al.’98]

Promise equality: quantum Promise : x = y or x = y ? ∆( x , y ) = α n x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n . . . . . . . . . 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 Promise : x = y or x = y ? ∆( x , y ) = α n x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n . . . . . . . . . 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 Promise : x = y or x = y ? ∆( x , y ) = α n with α < 1 / 2 x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n . . . . . . . . . Theorem - One-round protocol at least Ω( n ) qubits - Quantum communication complexity O (log n )

Promise equality: quantum Promise : x = y or x = y ? ∆( x , y ) = α n with α < 1 / 2 x ∈ { 0 , 1 } n y ∈ { 0 , 1 } n . . . . . . . . . Theorem - One-round protocol at least Ω( n ) qubits (Lov´ asz theta number) - Quantum communication complexity O (log n )