Capacity Region of the Gaussian Arbitrarily-Varying Broadcast Channel Fatemeh Hosseinigoki and Oliver Kosut A RIZONA S TATE U NIVERSITY ISIT 2020 1 / 12
Gaussian Arbitrarily-Varying Broadcast Channel Transmitter power constraint � X � 2 ≤ nP 2 / 12
Gaussian Arbitrarily-Varying Broadcast Channel Transmitter power constraint � X � 2 ≤ nP Jammer power constraints � S j � 2 ≤ n Λ j , j = 1 , 2 2 / 12
Gaussian Arbitrarily-Varying Broadcast Channel Transmitter power constraint � X � 2 ≤ nP Jammer power constraints � S j � 2 ≤ n Λ j , j = 1 , 2 We want to decode messages no matter what the jammers do 2 / 12
Gaussian Arbitrarily-Varying Broadcast Channel Transmitter power constraint � X � 2 ≤ nP Jammer power constraints � S j � 2 ≤ n Λ j , j = 1 , 2 We want to decode messages no matter what the jammers do Assumptions: Oblivious adversary, but knows the code 2 / 12
Gaussian Arbitrarily-Varying Channel Transmitter power constraint � X � 2 ≤ nP Jammer power constraint � S � 2 ≤ n Λ 3 / 12
Gaussian Arbitrarily-Varying Channel Transmitter power constraint � X � 2 ≤ nP Jammer power constraint � S � 2 ≤ n Λ [Csisz´ ar-Narayan 1991]: Capacity is � � P C Λ < P , where C ( x ) = 1 C = Λ + N 2 log(1 + x ) 0 , Λ ≥ P 3 / 12
Gaussian Arbitrarily-Varying Channel Transmitter power constraint � X � 2 ≤ nP Jammer power constraint � S � 2 ≤ n Λ [Csisz´ ar-Narayan 1991]: Capacity is � � P C Λ < P , where C ( x ) = 1 C = Λ + N 2 log(1 + x ) 0 , Λ ≥ P As long as jammer’s power is less than legitimate transmitter’s, then the worst the jammer can do is send noise 3 / 12
Symmetrizability If Λ ≥ P , then the jammer can choose a false message m ′ and transmit S = X ( m ′ ) 4 / 12
Symmetrizability If Λ ≥ P , then the jammer can choose a false message m ′ and transmit S = X ( m ′ ) Receiver gets Y = X ( m ) + X ( m ′ ) + V Cannot tell which message is correct 4 / 12
Gaussian AVC with Common Randomness 5 / 12
Gaussian AVC with Common Randomness [Hughes-Narayan 1982]: If transmitter/receiver have common randomness (unknown to adversary) of at least O (log n ) bits, then � P � C = C N + Λ 5 / 12
Gaussian AVC with Common Randomness [Hughes-Narayan 1982]: If transmitter/receiver have common randomness (unknown to adversary) of at least O (log n ) bits, then � P � C = C N + Λ Symmetrizability is not a problem because two possibilities can be distinguished via the common randomness 5 / 12
Gaussian AVBC Capacity Region Let C no-adv ( S 1 , S 2 ) be the capacity region for the no-adversary BC with SNRs S 1 , S 2 6 / 12
Gaussian AVBC Capacity Region Let C no-adv ( S 1 , S 2 ) be the capacity region for the no-adversary BC with SNRs S 1 , S 2 i.e., if S 1 ≥ S 2 , then R 1 ≤ C ( αS 1 ) , � for some α ∈ [0 , 1] C no-adv ( S 1 , S 1 ) = ( R 1 , R 2 ) : � αS 2 ¯ R 2 ≤ C αS 2 + 1 6 / 12
Gaussian AVBC Capacity Region Let C no-adv ( S 1 , S 2 ) be the capacity region for the no-adversary BC with SNRs S 1 , S 2 i.e., if S 1 ≥ S 2 , then R 1 ≤ C ( αS 1 ) , � for some α ∈ [0 , 1] C no-adv ( S 1 , S 1 ) = ( R 1 , R 2 ) : � αS 2 ¯ R 2 ≤ C αS 2 + 1 Theorem ( R 1 , R 2 ) ∈ C GAV BC if and only if � � P P ( R 1 , R 2 ) ∈ C no-adv , N 1 + Λ 1 N 2 + Λ 2 If Λ 1 ≥ P , then R 1 = 0 If Λ 2 ≥ P , then R 2 = 0 6 / 12
Gaussian AVBC Capacity Region Let C no-adv ( S 1 , S 2 ) be the capacity region for the no-adversary BC with SNRs S 1 , S 2 i.e., if S 1 ≥ S 2 , then R 1 ≤ C ( αS 1 ) , � for some α ∈ [0 , 1] C no-adv ( S 1 , S 1 ) = ( R 1 , R 2 ) : � αS 2 ¯ R 2 ≤ C αS 2 + 1 Theorem ( R 1 , R 2 ) ∈ C GAV BC if and only if � � P P ( R 1 , R 2 ) ∈ C no-adv , N 1 + Λ 1 N 2 + Λ 2 If Λ 1 ≥ P , then R 1 = 0 If Λ 2 ≥ P , then R 2 = 0 Converse is straightforward: (a) adversaries send Gaussian noise, (b) symmetrizability 6 / 12
Gaussian AVBC with Common Randomness Easy to show that, with common randomness, the capacity region is � P P � C no-adv , N 1 + Λ 1 N 2 + Λ 2 7 / 12
Gaussian AVBC with Common Randomness Problem: How to send common randomness. . . 7 / 12
Gaussian AVBC with Common Randomness Problem: How to send common randomness. . . 1 without the adversary preventing it, 7 / 12
Gaussian AVBC with Common Randomness Problem: How to send common randomness. . . 1 without the adversary preventing it, 2 without conflicting with superposition coding 7 / 12
Idea that doesn’t work #1 [Ahlswede 1978] 8 / 12
Idea that doesn’t work #1 [Ahlswede 1978] Works for unconstrained discrete channels, e.g. [Jahn 1981] 8 / 12
Idea that doesn’t work #1 [Ahlswede 1978] Works for unconstrained discrete channels, e.g. [Jahn 1981] Doesn’t work in the power-constrained setting, because the adversary can inject more power during the short common randomness transmission 8 / 12
Idea that doesn’t work #2 Use the common message as common randomness 9 / 12
Idea that doesn’t work #2 Use the common message as common randomness Codebooks X 1 ( m 1 , m 2 ) ∼ N (0 , αP ) , X 2 ( m 2 ) ∼ N (0 , ¯ αP ) 9 / 12
Idea that doesn’t work #2 Use the common message as common randomness Codebooks X 1 ( m 1 , m 2 ) ∼ N (0 , αP ) , X 2 ( m 2 ) ∼ N (0 , ¯ αP ) X = X 1 ( m 1 , m 2 ) + X 2 ( m 2 ) 9 / 12
Idea that doesn’t work #2 Use the common message as common randomness Codebooks X 1 ( m 1 , m 2 ) ∼ N (0 , αP ) , X 2 ( m 2 ) ∼ N (0 , ¯ αP ) X = X 1 ( m 1 , m 2 ) + X 2 ( m 2 ) Decoder 2 decodes m 2 , treats X 1 as noise 9 / 12
Idea that doesn’t work #2 Use the common message as common randomness Codebooks X 1 ( m 1 , m 2 ) ∼ N (0 , αP ) , X 2 ( m 2 ) ∼ N (0 , ¯ αP ) X = X 1 ( m 1 , m 2 ) + X 2 ( m 2 ) Decoder 2 decodes m 2 , treats X 1 as noise Decoder 1 decodes m 2 , then decodes m 1 using m 2 as common randomness 9 / 12
Idea that doesn’t work #2 Use the common message as common randomness Codebooks X 1 ( m 1 , m 2 ) ∼ N (0 , αP ) , X 2 ( m 2 ) ∼ N (0 , ¯ αP ) X = X 1 ( m 1 , m 2 ) + X 2 ( m 2 ) Decoder 2 decodes m 2 , treats X 1 as noise Decoder 1 decodes m 2 , then decodes m 1 using m 2 as common randomness Problem: To avoid symmetrizability, we need Λ 1 < ¯ αP, Λ 2 < ¯ αP 9 / 12
Idea that doesn’t work #2 Use the common message as common randomness Codebooks X 1 ( m 1 , m 2 ) ∼ N (0 , αP ) , X 2 ( m 2 ) ∼ N (0 , ¯ αP ) X = X 1 ( m 1 , m 2 ) + X 2 ( m 2 ) Decoder 2 decodes m 2 , treats X 1 as noise Decoder 1 decodes m 2 , then decodes m 1 using m 2 as common randomness Problem: To avoid symmetrizability, we need Λ 1 < ¯ αP, Λ 2 < ¯ αP This idea does work for the interference channel 1 1 F. Hosseinigoki and O. Kosut, “The Gaussian Interference Channel in the Presence of a Malicious Jammer,” Allerton, 2016. 9 / 12
Idea that works 10 / 12
Idea that works Divide length- n block into √ n segments of length √ n 10 / 12
Idea that works Divide length- n block into √ n segments of length √ n In each segment, transmit at either full power or 0 10 / 12
Idea that works Divide length- n block into √ n segments of length √ n In each segment, transmit at either full power or 0 The main BC code is contained in the full power segments 10 / 12
Idea that works Divide length- n block into √ n segments of length √ n In each segment, transmit at either full power or 0 The main BC code is contained in the full power segments The on/off sequence contains the common randomness 10 / 12
Idea that works Divide length- n block into √ n segments of length √ n In each segment, transmit at either full power or 0 The main BC code is contained in the full power segments The on/off sequence contains the common randomness Wait. . . can’t the adversary mess up the on/off sequence? 10 / 12
Common randomness coding At decoders, determine if the received power in each segment is high or low 11 / 12
Common randomness coding At decoders, determine if the received power in each segment is high or low Transmitter power Adversary power Received power P anything high 0 high >P 0 <P low 11 / 12
Common randomness coding At decoders, determine if the received power in each segment is high or low Transmitter power Adversary power Received power P anything high 0 high >P 0 <P low This is essentially a binary “or” channel 11 / 12
Common randomness coding At decoders, determine if the received power in each segment is high or low Transmitter power Adversary power Received power P anything high 0 high >P 0 <P low This is essentially a binary “or” channel If Λ < P , the adversary can only send power >P so often 11 / 12
Common randomness coding At decoders, determine if the received power in each segment is high or low Transmitter power Adversary power Received power P anything high 0 high >P 0 <P low This is essentially a binary “or” channel If Λ < P , the adversary can only send power >P so often Looks like a discrete AVC with cost-constrained adversary 11 / 12
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