interactive channel capacity ran raz weizmann institute joint work with gillat kol technion

Interactive Channel Capacity

Ran RazWeizmann Institute

Joint work withGillat KolTechnion

[Shannon 48]:A Mathematical Theory of

Communication

An exact formula for the channel

capacity of any noisy channel

-noisy channel:Each bit is flipped with prob

Alice wants to send bits to Bob. They

only have access to an -noisy channel.

How many bits Alice needs to send, so

that Bob can retrieve the original bits,

with prob ?

1-

1-

0

1 1

0

𝛆𝛆

Channel Capacity [Shannon 48]: 1) are sufficient (using error correcting codes)2) are needed

channel capacity:

Communication Complexity [Yao 79]:

Player gets . Player gets They need to compute ( is publicly known)How many bits they need to

communicate?

deterministic CC of (for worst case ) probabilistic CC of (with negligible error for

every ) (with shared random string)

CC over the -noisy channel: Assume: How many communication bits are

neededto compute over the -noisy channel?

deterministic CC of CC of over -noisy channel (with negligible error for

every ) (with shared random string)

Interactive Channel Capacity: deterministic CC of CC of over -noisy channel

(note: is not the input size)

Interactive Channel Capacity: deterministic CC of CC of over -noisy channel

Can use instead of All the results hold for both

Interactive Channel Capacity: deterministic CC of CC of over -noisy channel

Assumption: Order of communication in

all protocols is pre-determined(for simplicity)Justification: Otherwise both playersmay try to send bits at the same

time

Types of Channels:

1) Synchronous: At each time stepexactly one player sends a bit2) Alternating: The players alternatein sending bits3) Asynchronous: If both send bits atthe same time these bits are lost4) Two channels: Each player sends abit whenever she wants

Previous Work: [Schulman 92]: Hence, [Sch,BR,B,GMS,BK,BN]:Simulation of any CC protocol in

the presence of adversarial noise[Shannon 48]: [Schulman 92]: Is ?

𝑪 (𝜺 )=𝐥𝐢𝐦𝒏→∞

𝐦𝐢𝐧{ 𝒇 :𝑪𝑪 ( 𝒇 )=𝒏 }( 𝒏

𝑪𝑪𝜺( 𝒇 ))

Our Results:

Upper Bound: In particular, for small enough , (with strict inequality)

Lower Bound: in the case of alternating channel

𝑪 (𝜺 )=𝐥𝐢𝐦𝒏→∞

𝐦𝐢𝐧{ 𝒇 :𝑪𝑪 ( 𝒇 )=𝒏 }( 𝒏

𝑪𝑪𝜺( 𝒇 ))

Upper Bound:

We give a function that proves this

We prove a lower bound on

𝑪 (𝜺 )=𝐥𝐢𝐦𝒏→∞

𝐦𝐢𝐧{ 𝒇 :𝑪𝑪 ( 𝒇 )=𝒏 }( 𝒏

𝑪𝑪𝜺( 𝒇 ))

Pointer Jumping Game:

ary tree, depth , owns odd layers owns even layers

Each player gets an edge going out ofevery node that she ownsGoal: Find the leaf reached

deg=

depth=

Pointer Jumping Game:

Our main result:

Hence,

deg=

depth=

High Level Idea: starts by sending the first edge ( bits)With one of these bits was flipped

Case I: sends the next edge ( bits)With these bits are wasted (since had the wrong first edge)

In expectation: wasted bits

deg=

depth=

High Level Idea: starts by sending the first edge ( bits)With one of these bits was flipped

Case II: sends additional bits, tocorrect the first edge.

Needs to send bits to correct one error

deg=

depth=

High Level Idea: starts by sending the first edge ( bits)With one of these bits was flipped

In both cases bits were wasted (in expectation). was chosen to be to balance the losses in the two cases

deg=

depth=

Lower Bound:

Given a communication protocol , we

simulate over the -noisy channel

𝑪 (𝜺 )=𝐥𝐢𝐦𝒏→∞

𝐦𝐢𝐧{ 𝒇 :𝑪𝑪 ( 𝒇 )=𝒏 }( 𝒏

𝑪𝑪𝜺( 𝒇 ))

The Basic Step:

Fix . run steps of and observe the transcripts , , resp. run a Consistency Check. If an inconsistency was found they start over

bits bitsconsistency checkinconsistenc

y

, run steps of and observe the transcripts , , resp.Consistency Check: choose random functions: . sends 100 times each takes majority vote of each and compares to . sends 100 times each takes majority vote of each and compares to .A player that finds inconsistency starts over

, run steps of and observe the transcripts ,Consistency Check: choose random functions: . sends 100 times each. takes majority vote of each and compares to . sends 100 times each. takes majority vote of each and compares to .A player that finds inconsistency starts over

bits bitsconsistency checkinconsistenc

y

Good: No player starts over Bad: Both players start over Very Bad: One player starts over

bits bitsconsistency checkinconsistenc

y

Inductive Protocol:

Consistency check:Done with random functions, sent times each ()In the protocol: random functions, sent times each

times bitsconsistency checkinconsistenc

y

Analysis: If an error occurred or the players wentout of split, eventually they will fix it, since the consistency check is done with larger and larger parameters. Thus, the final protocol simulates withprobability close to .How many communication bits are wasted?

times bitsconsistency checkinconsistenc

y

Analysis of Wastes in the Basic Step:

Length of consistency check: bitsProbability to start over: Total waste (in expectation): bitsFraction of bits wasted:

bits bitsconsistency checkinconsistenc

y

Wastes in First Inductive Step:

Length of consistency check: Probability to start over: Total waste (in expectation): Fraction of bits wasted: (negligible compared to the basic step)

times bitsconsistency checkinconsistenc

y

Bound on the Channel Capacity:

Thank You!