Bikash DeySidharth Jaggi

Causal

Michael Langberg

OPEN UNIVERSITY OF ISRAEL

(adversarial)channel codes

pn R=m/n

ûmum xn yn

en

=um

Alphabet q

Noisy Channelsq=2 (binary)

R

p

1

0.5

“large” q

R

p

1

1

[Sha48]“Folklore”

ChannelsNoisyAdversarial

Adversarially chosen

q = 2 (binary)

R

p

1

0.5

“large” q

R

p

1

10.25 0.5

[McERRW77] [Reed-Solomon]

[Gilbert

Varshamov]

ChannelsAdversarialCausal (online)

Causally adversarially chosen

1. 0-delay

R

p

1

10.5

Large-q results

1. 0-delay

R

p

1

1

2. d-delay additive

d

Large-q results

1. 0-delay2. d-delay additive3. d-delay overwrite R

p

1

1

d

d

Tight rate-regions,Poly-time algorithms.

Large-q results

This talk…q=2 (binary)0-delay

Intriguingly intricate…

Turan’s theoremBayes’TheoremPlotkinboundRandomCoding

Adversary’s attack

Wait-and-push

Initial simplifying assumption: Alice’s codebook uniform and non-random

Suppose R=(1-4p)++2ε

Set of possible xn from Calvin’s perspective as he waits

n[(1-4p)++ε]

Plotkin bound: largest set of remaining codewords with pairwise dmin > 2pn is O(1/ε)

Remaining codeword length = n[4p-ε] Wait-

Turan’s theorem: W.p. Ω(ε), two randomly chosen remaining codewords have dH < 2pn

and-push

fake xn

(flip a differing bit w.p. ½)

Adversary’s attack

Wait-and-push

Initial simplifying assumption: Alice’s codebook uniform and non-random

Suppose R=(1-4p)++2ε

Plotkin bound: largest set of remaining codewords with pairwise dmin > 2pn is O(1/ε)

Remaining codeword length = n[4p-ε]

Turan’s theorem: W.p. Ω(ε), two randomly chosen remaining codewords have dH < 2pnBayes’ theorem: W.p. 1/2, Bob can’t determine which of the two was transmitted ??

Overall Pr(error) Ω(ε)

Can still show overall Pr(error) Ω(1/(poly(n)))

Summary/thoughts

Large alphabets (now) understood, small not…

Causal adversary still pretty strong…

… but delays can weaken him. Types of error matter

(additive/overwrite/…?)

Questions?