Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Optimal classical RAC Optimal decoding ◮ For each bit there are only four possible decoding functions: D ( x ) = 0 , D ( x ) = 1 , D ( x ) = x , D ( x ) = NOT x . ◮ We cannot make things worse if we do not use constant decoding functions 0 and 1 for any bits. ◮ We can always avoid using decoding function NOT x (by negating the input before encoding). Hence there is an optimal joint strategy such that Bob always replays the received bit no matter which bit is actually asked. Optimal encoding Once Alice knows that Bob’s decoding function is D ( x ) = x , she simply encodes the majority of all bits.
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Exact probability of success Counting Let us choose a string from { 0 , 1 } n uniformly at random and mark one bit at a random position.
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Exact probability of success Counting Let us choose a string from { 0 , 1 } n uniformly at random and mark one bit at a random position. What is the probability that the value of the marked bit equals the majority of all bits?
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Exact probability of success Counting Let us choose a string from { 0 , 1 } n uniformly at random and mark one bit at a random position. What is the probability that the value of the marked bit equals the majority of all bits? Answer Exactly: p (2 m ) = p (2 m + 1) = 1 1 � 2 m � 2 + 2 2 m +1 m
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Exact probability of success Counting Let us choose a string from { 0 , 1 } n uniformly at random and mark one bit at a random position. What is the probability that the value of the marked bit equals the majority of all bits? Answer Exactly: p (2 m ) = p (2 m + 1) = 1 1 � 2 m � 2 + 2 2 m +1 m Using Stirling’s approximation: p ( n ) ≈ 1 1 √ 2 + 2 πn
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Probability of success � 2 m Exact probability: p (2 m ) = p (2 m + 1) = 1 / 2 2 m +1 � 2 + m p � n � 15 n 2 3 4 5 6 7 8 9 10 11 12 13 14 0.9 0.8 0.7 0.6
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Probability of success √ Using Stirling’s approximation: p ( n ) ≈ 1 2 + 1 / 2 πn p � n � 15 n 2 3 4 5 6 7 8 9 10 11 12 13 14 0.9 0.8 0.7 0.6
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Probability of success √ √ � n � n e � n � n e 1 1 12 n +1 < n ! < Using inequalities 2 πn 2 πn 12 n e e p � n � 15 n 2 3 4 5 6 7 8 9 10 11 12 13 14 0.9 0.8 0.7 0.6
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Quantum random access codes with shared randomness
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Bloch sphere Alice encodes a classical bit string into a qubit state and sends it to Bob. We will use the Bloch sphere to visualize these states.
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Bloch sphere Alice encodes a classical bit string into a qubit state and sends it to Bob. We will use the Bloch sphere to visualize these states. Bloch vector � cos θ � | ψ � = 2 e iϕ sin θ 2 0 ≤ θ ≤ π , 0 ≤ ϕ < 2 π
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Bloch sphere Alice encodes a classical bit string into a qubit state and sends it to Bob. We will use the Bloch sphere to visualize these states. Bloch vector z � cos θ � | ψ � = 2 � Ψ � e iϕ sin θ 2 0 ≤ θ ≤ π , 0 ≤ ϕ < 2 π Θ � y x
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Bloch sphere Alice encodes a classical bit string into a qubit state and sends it to Bob. We will use the Bloch sphere to visualize these states. Bloch vector z � cos θ � | ψ � = 2 � Ψ � e iϕ sin θ 2 0 ≤ θ ≤ π , 0 ≤ ϕ < 2 π Θ � � y � r = ( x, y, z ) x x = sin θ cos ϕ y = sin θ sin ϕ z = cos θ
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Bloch sphere Alice encodes a classical bit string into a qubit state and sends it to Bob. We will use the Bloch sphere to visualize these states. Measurement � Ψ 0 � |� ψ 1 | ψ 2 �| 2 = 1 2(1 + � r 1 · � r 2 ) d 0 � Ψ � Α d 1 � Ψ 1 �
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Bloch sphere Alice encodes a classical bit string into a qubit state and sends it to Bob. We will use the Bloch sphere to visualize these states. Measurement � Ψ 0 � |� ψ 1 | ψ 2 �| 2 = 1 2(1 + � r 1 · � r 2 ) d 0 � Ψ � Pr[ | ψ 0 � given | ψ � ] Α d 1 � Ψ 1 �
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Bloch sphere Alice encodes a classical bit string into a qubit state and sends it to Bob. We will use the Bloch sphere to visualize these states. Measurement � Ψ 0 � |� ψ 1 | ψ 2 �| 2 = 1 2(1 + � r 1 · � r 2 ) d 0 � Ψ � Pr[ | ψ 0 � given | ψ � ] = |� ψ 0 | ψ �| 2 Α d 1 � Ψ 1 �
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Bloch sphere Alice encodes a classical bit string into a qubit state and sends it to Bob. We will use the Bloch sphere to visualize these states. Measurement � Ψ 0 � |� ψ 1 | ψ 2 �| 2 = 1 2(1 + � r 1 · � r 2 ) d 0 � Ψ � Pr[ | ψ 0 � given | ψ � ] = |� ψ 0 | ψ �| 2 Α = 1 + cos α d 1 2 � Ψ 1 �
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Bloch sphere Alice encodes a classical bit string into a qubit state and sends it to Bob. We will use the Bloch sphere to visualize these states. Measurement � Ψ 0 � |� ψ 1 | ψ 2 �| 2 = 1 2(1 + � r 1 · � r 2 ) d 0 � Ψ � Pr[ | ψ 0 � given | ψ � ] = |� ψ 0 | ψ �| 2 Α = 1 + cos α = d 1 d 1 2 2 � Ψ 1 �
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Bloch sphere Alice encodes a classical bit string into a qubit state and sends it to Bob. We will use the Bloch sphere to visualize these states. Measurement � Ψ 0 � |� ψ 1 | ψ 2 �| 2 = 1 2(1 + � r 1 · � r 2 ) d 0 � Ψ � Pr[ | ψ 0 � given | ψ � ] = |� ψ 0 | ψ �| 2 Α = 1 + cos α = d 1 d 1 2 2 Pr[ | ψ 1 � given | ψ � ] = |� ψ 1 | ψ �| 2 = 1 − cos α = d 0 � Ψ 1 � 2 2
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Known results Pure strategies Only two specific QRACs are known when pure quantum strategies are allowed. That means: 1. Alice prepares a pure state, 2. Bob uses a projective measurement (not a POVM), 3. the worst case success probability must be at least 1 2 . Note: shared randomness is not allowed.
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Known QRACs p �→ 1 code 2 p �→ 1 code with p = 1 1 There is a 2 2 + 2 ≈ 0 . 85 . √ 2 This code is optimal. [quant-ph/9804043] z 11 10 y 01 00 x
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Known QRACs p �→ 1 code 3 p �→ 1 code with p = 1 1 3 ≈ 0 . 79 . There is a 3 2 + √ 2 This code is optimal. [I.L. Chuang] z 110 100 010 000 y x 111 101 011 001
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Known QRACs p �→ 1 code 4 p �→ 1 code for p > 1 There is no 4 2 . Proof idea – it is not possible to cut the surface of the Bloch sphere into 16 parts with 4 planes passing through its center. [quant-ph/0604061]
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Known QRACs p �→ 1 code 4 p �→ 1 code for p > 1 There is no 4 2 . Proof idea – it is not possible to cut the surface of the Bloch sphere into 16 parts with 4 planes passing through its center. [quant-ph/0604061]
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary What can we do about this?
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary What can we do about this? Use shared randomness!
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Different kinds of quantum RACs Definition Pure quantum n �→ 1 RAC is an ordered tuple ( E, M 1 , . . . , M n ) that consists of encoding function E : { 0 , 1 } n �→ C 2 and n �� � ψ i � � ψ i � �� orthogonal measurements: M i = , . 0 1
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Different kinds of quantum RACs Definition Pure quantum n �→ 1 RAC is an ordered tuple ( E, M 1 , . . . , M n ) that consists of encoding function E : { 0 , 1 } n �→ C 2 and n �� � ψ i � � ψ i � �� orthogonal measurements: M i = , . 0 1 Definition Mixed quantum n �→ 1 RAC is an ordered tuple ( P E , P M 1 , . . . , P M n ) of probability distributions. P E is a distribution over encoding functions E and P M i are probability distributions over orthogonal measurements of qubit.
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Different kinds of quantum RACs Definition Pure quantum n �→ 1 RAC is an ordered tuple ( E, M 1 , . . . , M n ) that consists of encoding function E : { 0 , 1 } n �→ C 2 and n �� � ψ i � � ψ i � �� orthogonal measurements: M i = , . 0 1 Definition Quantum n �→ 1 RAC with shared randomness is a probability distribution over pure quantum RACs.
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Finding QRACs with SR Recall Let � r 1 and � r 2 be the Bloch vectors corresponding to qubit states | ψ 1 � and | ψ 2 � . Then |� ψ 1 | ψ 2 �| 2 = 1 r 1 · � 2 (1 + � r 2 ) .
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Finding QRACs with SR Recall Let � r 1 and � r 2 be the Bloch vectors corresponding to qubit states | ψ 1 � and | ψ 2 � . Then |� ψ 1 | ψ 2 �| 2 = 1 r 1 · � 2 (1 + � r 2 ) . Qubit ( C 2 ) ⇒ Bloch sphere ( R 3 )
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Finding QRACs with SR Recall Let � r 1 and � r 2 be the Bloch vectors corresponding to qubit states | ψ 1 � and | ψ 2 � . Then |� ψ 1 | ψ 2 �| 2 = 1 r 1 · � 2 (1 + � r 2 ) . Qubit ( C 2 ) ⇒ Bloch sphere ( R 3 ) ◮ encoding of string x ∈ { 0 , 1 } n : | E ( x ) � ⇒ � r x ,
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Finding QRACs with SR Recall Let � r 1 and � r 2 be the Bloch vectors corresponding to qubit states | ψ 1 � and | ψ 2 � . Then |� ψ 1 | ψ 2 �| 2 = 1 r 1 · � 2 (1 + � r 2 ) . Qubit ( C 2 ) ⇒ Bloch sphere ( R 3 ) ◮ encoding of string x ∈ { 0 , 1 } n : | E ( x ) � ⇒ � r x , �� � ψ i � � ψ i ◮ measurement of the i th bit: � �� , ⇒ { � v i , − � v i } . 0 1
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Finding QRACs with SR Recall Let � r 1 and � r 2 be the Bloch vectors corresponding to qubit states | ψ 1 � and | ψ 2 � . Then |� ψ 1 | ψ 2 �| 2 = 1 r 1 · � 2 (1 + � r 2 ) . Qubit ( C 2 ) ⇒ Bloch sphere ( R 3 ) ◮ encoding of string x ∈ { 0 , 1 } n : | E ( x ) � ⇒ � r x , �� � ψ i � ψ i � ◮ measurement of the i th bit: � �� , ⇒ { � v i , − � v i } . 0 1 Optimize The average success probability is: n 1 + ( − 1) x i � 1 v i · � r x � � p ( { � v i } , { � r x } ) = 2 n · n 2 x ∈{ 0 , 1 } n i =1
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Optimal quantum encoding Probability n 1 + ( − 1) x i � 1 v i · � r x � � p ( { � v i } , { � r x } ) = 2 n · n 2 x ∈{ 0 , 1 } n i =1
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Optimal quantum encoding Probability n 1 + ( − 1) x i � 1 v i · � r x � � p ( { � v i } , { � r x } ) = 2 n · n 2 x ∈{ 0 , 1 } n i =1 Observe n � n � � � � � � � � � ( − 1) x i � ( − 1) x i � max � r x · v i = max v i � � � � { � v i } , { � r x } { � v i } x ∈{ 0 , 1 } n x ∈{ 0 , 1 } n i =1 � i =1 �
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Optimal quantum encoding Probability n 1 + ( − 1) x i � 1 v i · � r x � � p ( { � v i } , { � r x } ) = 2 n · n 2 x ∈{ 0 , 1 } n i =1 Observe n � n � � � � � � � � � ( − 1) x i � ( − 1) x i � max � r x · v i = max v i � � � � { � v i } , { � r x } { � v i } x ∈{ 0 , 1 } n x ∈{ 0 , 1 } n i =1 � i =1 � Optimal encoding Given { � v i } , optimal encoding of x is a unit vector � r x in direction of n � ( − 1) x i � v i i =1
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Optimal quantum encoding Probability n 1 + ( − 1) x i � 1 v i · � r x � � p ( { � v i } , { � r x } ) = 2 n · n 2 x ∈{ 0 , 1 } n i =1 Observe n � n � � � � � � � � � ( − 1) x i � ( − 1) x i � max � r x · v i = max v i � � � � { � v i } , { � r x } { � v i } x ∈{ 0 , 1 } n x ∈{ 0 , 1 } n i =1 � i =1 � Optimal encoding Given { � v i } , optimal encoding of x is a unit vector � r x in direction of n � ( − 1) x i � v i i =1 Note: if all � v i are equal, this corresponds to the optimal classical (majority) encoding.
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Upper bound Success probability using optimal encoding � � n � � v i } ) = 1 1 � � � � p ( { � 1 + a i � v i � � 2 n · n 2 � � a ∈{ 1 , − 1 } n � � i =1
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Upper bound Success probability using optimal encoding � � n � � v i } ) = 1 1 � � � � p ( { � 1 + a i � v i � � 2 n · n 2 � � a ∈{ 1 , − 1 } n � � i =1 Lemma For any unit vectors � v 1 , . . . ,� v n we have: v n � 2 = n · 2 n � � a 1 � v 1 + · · · + a n � a 1 ,...,a n ∈{ 1 , − 1 }
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Upper bound Success probability using optimal encoding � � n � � v i } ) = 1 1 � � � � p ( { � 1 + a i � v i � � 2 n · n 2 � � a ∈{ 1 , − 1 } n � � i =1 Lemma For any unit vectors � v 1 , . . . ,� v n we have: v n � 2 = n · 2 n � � a 1 � v 1 + · · · + a n � a 1 ,...,a n ∈{ 1 , − 1 } Think of this as a generalization of the parallelogram identity v 2 � 2 + � � v 2 � 2 = 2 v 1 � 2 + � � � v 2 � 2 � � � v 1 + � v 1 − � � �
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Upper bound Success probability using optimal encoding � � n � � v i } ) = 1 1 � � � � p ( { � 1 + a i � v i � � 2 n · n 2 � � a ∈{ 1 , − 1 } n � � i =1 Lemma For any unit vectors � v 1 , . . . ,� v n we have: v n � 2 = n · 2 n � � a 1 � v 1 + · · · + a n � a 1 ,...,a n ∈{ 1 , − 1 } To remove the square, use inequality that follows form ( x − y ) 2 ≥ 0 : xy ≤ 1 2( x 2 + y 2 )
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Upper bound Success probability using optimal encoding � � n � � v i } ) = 1 1 � � � � p ( { � 1 + a i � v i � � 2 n · n 2 � � a ∈{ 1 , − 1 } n � � i =1 Lemma For any unit vectors � v 1 , . . . ,� v n we have: v n � ≤ √ n · 2 n � � a 1 � v 1 + · · · + a n � a 1 ,...,a n ∈{ 1 , − 1 }
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Upper bound Success probability using optimal encoding � � n � � v i } ) = 1 1 � � � � p ( { � 1 + a i � v i � � 2 n · n 2 � � a ∈{ 1 , − 1 } n � � i =1 Lemma For any unit vectors � v 1 , . . . ,� v n we have: v n � ≤ √ n · 2 n � � a 1 � v 1 + · · · + a n � a 1 ,...,a n ∈{ 1 , − 1 } Theorem �→ 1 QRAC with shared randomness: p ≤ 1 1 p For any n 2 + 2 √ n .
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Upper bound Success probability using optimal encoding � � n � � v i } ) = 1 1 � � � � p ( { � 1 + a i � v i � � 2 n · n 2 � � a ∈{ 1 , − 1 } n � � i =1 Lemma For any unit vectors � v 1 , . . . ,� v n we have: v n � ≤ √ n · 2 n � � a 1 � v 1 + · · · + a n � a 1 ,...,a n ∈{ 1 , − 1 } Theorem �→ 1 QRAC with shared randomness: p ≤ 1 1 p For any n 2 + 2 √ n . Note: this holds even if Bob can use a POVM measurement.
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Lower bound Success probability using optimal encoding � n � � � v i } ) = 1 1 � � � � p ( { � 1 + a i � v i � � 2 n · n 2 � � a ∈{ 1 , − 1 } n � � i =1
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Lower bound Success probability using optimal encoding � n � � � v i } ) = 1 1 � � � � p ( { � 1 + a i � v i � � 2 n · n 2 � � a ∈{ 1 , − 1 } n � � i =1 Random measurements Alice and Bob can sample each � v i at random. This can be done near uniformly given enough shared randomness. Observe � n � � n � � � = 2 n · E � � � � � a i � v i v i � E � � � � � � � � { � v i } { � v i } a ∈{ 1 , − 1 } n � i =1 � � i =1 �
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Lower bound Success probability using random measurements � n � � � v i } ) = 1 1 + 1 � � � v i } p ( { � n · E E � v i � � 2 � � { � { � v i } � � i =1 Random measurements Alice and Bob can sample each � v i at random. This can be done near uniformly given enough shared randomness. Observe � n � � n � � � = 2 n · E � � � � � a i � v i v i � E � � � � � � � � { � v i } { � v i } a ∈{ 1 , − 1 } n � i =1 � � i =1 �
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Lower bound Success probability using random measurements � n � � � v i } ) = 1 1 + 1 � � � v i } p ( { � n · E E � v i � � 2 � � { � { � v i } � � i =1 Random measurements Alice and Bob can sample each � v i at random. This can be done near uniformly given enough shared randomness. Observe � n � � n � � � = 2 n · E � � � � � a i � v i v i � E � � � � � � � � { � v i } { � v i } a ∈{ 1 , − 1 } n � i =1 � � i =1 � What is the average distance traveled in 3D after n steps of unit length if the direction of each step is chosen uniformly at random?
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Lower bound Success probability using random measurements � n � � � v i } ) = 1 1 + 1 � � � v i } p ( { � n · E E � v i � � 2 � � { � { � v i } � � i =1 Random walk Probability density to arrive at point � R after performing n ≫ 1 steps of random walk [Chandrasekhar 1943]: � 3 � 3 / 2 2 / 2 n e − 3 � � R � W ( � R ) = 2 πn
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Lower bound Success probability using random measurements � n � � � v i } ) = 1 1 + 1 � � � v i } p ( { � n · E E � v i � � 2 � � { � { � v i } � � i =1 Random walk Probability density to arrive at point � R after performing n ≫ 1 steps of random walk [Chandrasekhar 1943]: � 3 � 3 / 2 2 / 2 n e − 3 � � R � W ( � R ) = 2 πn Thus the average distance traveled is � ∞ � 2 n 4 πR 2 · R · W ( R ) · dR = 2 3 π 0
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Lower bound Success probability using random measurements � v i } ) = 1 2 v i } p ( { � 2 + E 3 πn { � Random walk Probability density to arrive at point � R after performing n ≫ 1 steps of random walk [Chandrasekhar 1943]: � 3 � 3 / 2 2 / 2 n e − 3 � � R � W ( � R ) = 2 πn Thus the average distance traveled is � ∞ � 2 n 4 πR 2 · R · W ( R ) · dR = 2 3 π 0
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Lower bound Success probability using random measurements � v i } ) = 1 2 v i } p ( { � 2 + E 3 πn { � Theorem � �→ 1 QRAC with SR such that p = 1 2 p There exists n 2 + 3 πn .
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Quantum upper and lower bounds p � n � 15 n 2 3 4 5 6 7 8 9 10 11 12 13 14 0.95 0.90 0.85 0.80 0.75 0.70 0.65
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Quantum upper and lower bounds p � n � 15 n 2 3 4 5 6 7 8 9 10 11 12 13 14 0.95 0.90 0.85 0.80 0.75 0.70 0.65 Black dots correspond to a lower bound obtained using measurements on orthogonal Bloch vectors.
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Results of numerical optimization
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Results of numerical optimization See our homepage http://home.lanet.lv/ ∼ sd20008/RAC/RACs.htm
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Numerical 2 �→ 1 QRAC p = 1 1 2 + √ 2 ≈ 0 . 8535533906 2
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Numerical 3 �→ 1 QRAC p = 1 1 √ 2 + 3 ≈ 0 . 7886751346 2
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Numerical 4 �→ 1 QRAC √ p = 1 2 + 1 + 3 √ ≈ 0 . 7414814566 8 2
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Numerical 5 �→ 1 QRAC √ p = 1 2 + 1 � 17) ≈ 0 . 7135779205 2(5 + 20
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Numerical 6 �→ 1 QRAC √ √ p = 1 2 + 2 + 3 + 15 √ ≈ 0 . 6940463870 16 6
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Numerical 9 �→ 1 QRAC √ √ √ p = 1 2 + 192 + 10 3 + 9 11 + 3 19 ≈ 0 . 6568927813 384
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Numerical 15 �→ 1 QRAC √ √ √ √ √ √ √ p = 1 2 + 152 3+100 11+50 19+20 35+5 43+2 51+ 59 ≈ 0 . 6203554614 8192
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Symmetric (but not optimal) constructions
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Symmetric 4 �→ 1 QRAC √ p = 1 2 + 2 + 3 ≈ 0 . 7332531755 16 ≤ 0 . 7414814566
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Symmetric 6 �→ 1 QRAC √ √ p = 1 32 + 1 5 � 2 + 75 + 30 5 ≈ 0 . 6940418856 96 ≤ 0 . 6940463870
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Symmetric 9 �→ 1 QRAC p ≈ 0 . 6563927998 ≤ 0 . 6568927813
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Symmetric 15 �→ 1 QRAC p ≈ 0 . 6201829084 ≤ 0 . 6203554614
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Summary
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Summary Classical RACs with SR
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Summary Classical RACs with SR ◮ exact success probability of optimal RAC: p (2 m ) = p (2 m + 1) = 1 1 � 2 m � 2 + , 2 2 m +1 m
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Summary Classical RACs with SR ◮ exact success probability of optimal RAC: p (2 m ) = p (2 m + 1) = 1 1 � 2 m � 2 + , 2 2 m +1 m ◮ asymptotic success probability: p ( n ) ≈ 1 1 √ 2 + 2 πn .
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Summary Classical RACs with SR ◮ exact success probability of optimal RAC: p (2 m ) = p (2 m + 1) = 1 1 � 2 m � 2 + , 2 2 m +1 m ◮ asymptotic success probability: p ( n ) ≈ 1 1 √ 2 + 2 πn . Quantum RACs with SR
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Summary Classical RACs with SR ◮ exact success probability of optimal RAC: p (2 m ) = p (2 m + 1) = 1 1 � 2 m � 2 + , 2 2 m +1 m ◮ asymptotic success probability: p ( n ) ≈ 1 1 √ 2 + 2 πn . Quantum RACs with SR ◮ upper bound: p ( n ) ≤ 1 1 2 √ n , 2 +
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Summary Classical RACs with SR ◮ exact success probability of optimal RAC: p (2 m ) = p (2 m + 1) = 1 1 � 2 m � 2 + , 2 2 m +1 m ◮ asymptotic success probability: p ( n ) ≈ 1 1 √ 2 + 2 πn . Quantum RACs with SR ◮ upper bound: p ( n ) ≤ 1 1 2 √ n , 2 + � ◮ lower bound: p ( n ) ≥ 1 2 2 + 3 πn .
Introduction Classical RACs Quantum RACs Numerical results Symmetric constructions Summary Comparison of classical and quantum RACs with SR p � n � 15 n 2 3 4 5 6 7 8 9 10 11 12 13 14 0.9 0.8 0.7 0.6 White dots correspond to QRACs obtained using numerical optimization. Black dots correspond to optimal classical RAC.
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