Gillat Kol (IAS)
joint work withRan Raz (Weizmann + IAS)
Interactive Channel Capacity
Binary Symmetric Channel with crossover probability (-BSC): Flips every bit w.p. , independently
Channel Capacity
1βπ
1βπ
0
1 1
0
ππ
Alice wants to send an bit message to Bob. How many bits does Alice need to send over the-BSC, so Bob can retrieve w.h.p. ()? Answer [Shannonβ48]:
bitsnoiseless channel
bits-BSC
Channel Capacity
Channel Capacity
π» (π )=βππππ πβ (1βπ ) πππ(1βπ)
Alice and Bob want to have an bits long conversation. How many bits do they need to send over the -BSC, so both can retrieve the transcript w.h.p.?
Interactive Channel Capacity
bitsnoiseless channel
bits-BSC
Alice has input , Bob has input .They want to compute ( public). How many bits do they need to exchange?
Communication Complexity [Yaoβ79]
π₯
π (π₯ , π¦ ) !
π¦π1 (π₯ )π2 ( π¦ ,π1 )
π3 (π₯ ,π1 ,π2 ) . .adaptive!Protocol:
Players may use private and public randomnessNoiseless CC: = the least number of bits players need to exchange over the noiseless channel to compute w.h.p. for every Noisy CC: = the same when players communicate over the -BSC
Randomized CC
π₯ ,π π¦ ,π
π (π₯ , π¦ ) !
. .
Example 1: of Equality
π₯β π¦ !
. .
R R
are -bit stringsEquality function: iff
NaΓ―ve protocol: exchanged bitsAlice sends to Bob
are -bit stringsEquality function: iff
NaΓ―ve protocol: exchanged bitsAlice sends to Bob
Efficient Protocol: exchanged bits; error By hashing (binning). E.g., take inner products with random strings
Example 1: of Equality
Underlying tree: complete binary tree
Example 2: of Pointer Jumping
Underlying tree: complete binary tree: an edge going out of every node in odd layers
Example 2: of Pointer Jumping
π₯
Underlying tree: complete binary tree: an edge going out of every node in odd layers: an edge going out of every node in even layersPointer Jumping function: is the leaf reached by βchasing the pointersβ
Example 2: of Pointer Jumping
π₯ π¦
Underlying tree: complete binary tree: an edge going out of every node in odd layers: an edge going out of every node in even layersPointer Jumping function: is the leaf reached by βchasing the pointersβ
Example 2: of Pointer Jumping
π₯ π¦
Example 2: of Pointer Jumping
(left edge)
(left edge)( right edge)
( left edge)
π₯ π¦
(left edge)
(left edge)( right edge)
( left edge)
input size
protocol is fully adaptive, very susceptible to noise
π₯ π¦
Def: Interactive Channel Capacity
: Noiseless CC: Noisy CC
Interactive Channel Capacity
The interactive setting is more general:1. Allows interaction and adaptivity.
Shannonβs setting is one-way2. Allows any computation (function ). Shannonβs setting corresponds to interactive channel capacity channel capacity
In the interactive setting, error in the first round may cause the conversation to be meaningless. Need to correct βonlineββ!
Why is the Interactive Case More Challenging?
[Schulman β92]: β Theorem: If then
Corollary: β interactive channel capacity channel capacity
Many other works [Sch,BR,B,GMS,BK,BN,FGOSβ¦]: β Simulation of any communication protocol with adversarial noiseβ Large constants, never made explicit
Previous Worksπππ© (π )=π₯π’π¦π’π§ππ§ββ
π¦π’π§{π :ππ ( π )=π§ }( ππ (π )
πππ (π ))
?
Our Results
Theorem 1 (Upper Bound):
For small : interactive channel capacity is strictly smaller than channel capacity
Theorem 2 (Lower Bound):
(for the case of alternating turns)
πππ© (π )=π₯π’π¦π’π§ππ§ββ
π¦π’π§{π :ππ ( π )=π§ }( ππ (π )
πππ (π ))
In our model, exactly one player sends a bit at each time step
β The order of turns in a protocol is pre-determined (independent of the inputs, randomness, noise).
β Alternating turns is a special case
Channel Types & Communication Order
Lower Bound: A simulation protocol
πππ© (π )=π₯π’π¦π’π§ππ§ββ
π¦π’π§{π :ππ ( π )=π§ }( ππ (π )
πππ (π ))
SimulationGiven a communication protocol P, we simulate P over the -BSC
Simulation Idea: Run P for a few rounds and check for errors. If an error occurred, backtrack
Parameter: s.t.
The simulation protocol is recursive:β The basic step simulates steps of Pβ The th inductive step simulates steps of P
πππ© (π )=π₯π’π¦π’π§ππ§ββ
π¦π’π§{π :ππ ( π )=π§ }( ππ (π )
πππ (π ))
Simulating Protocol: Basic StepPlayers run steps of P. Alice observes transcript , and Bob transcript
Players run an bit consistency check:Compute by hashing (binning)
A player that finds an inconsistency starts overand removes this stepβs bits from his transcript
bits of protocol P bitconsistency check
inconsistency
list of all massages exchanged
Players run the Basic Step consecutive times. Alice observes transcript , and Bob transcript
Simulating Protocol: 1st Interactive Step
times
Simulating Protocol: 1st Interactive StepPlayers run the Basic Step consecutive times. Alice observes transcript , and Bob transcript
Players run an bit consistency check: Compute by hashing (binning)
A player that finds an inconsistency starts over and removes this stepβs bits from his transcript
timesinconsistency
bits
Analysis: CorrectnessThe final protocol simulates P w.p. : If an error occurred or the players went out of sync, they will eventually fix it, as the consistency check checks the whole transcript so far and is done with larger and larger parameters
timesinconsistency
bits
Analysis: Waste in Basic StepLength of consistency check: bitsProbability to start over: Total waste (in expectation): + bits was chosen to balance the two losses!Fraction of bits wasted:
π=ππππ /π2
bits of protocol P bitconsistency checkinconsistency
Analysis: Waste in Basic StepLength of consistency check: bitsProbability to start over: Total waste (in expectation): + bits was chosen to balance the two losses!Fraction of bits wasted:
The waste in the inductive steps is much smaller!
π=ππππ /π2
Upper Bound: Example of with
πππ© (π )=π₯π’π¦π’π§ππ§ββ
π¦π’π§{π :ππ ( π )=π§ }( ππ (π )
πππ (π ))
Underlying tree: -ary tree, depth Parameter (as before): s.t. : an edge going out of every node in odd layers: an edge going out of every node in even layers-Pointer Jumping: is the leaf reached by βchasing the pointersβ
Example: -Pointer Jumping
dept
h =
deg=
Observe, We prove - involved! Such lower bounds are typically up-to a constant.
The capacity comes from the second order terms
Upper Bound Outline π=ππππ /π2
πππ© (π )=π₯π’π¦π’π§ππ§ββ
π¦π’π§{π :ππ ( π )=π§ }( ππ (π )
πππ (π ))
deg=
dept
h =
deg=
Bounding : IntuitionβAny good protocol does the following:βAlice starts by sending the first edge ( bits)
w.p. a bit was flipped
Case 1: Alice sends additional bits to correct first edge Even if a single error occurred and Alice knows its index, she needs to send the index bit waste
Case 2: Bob sends the next edge ( bits) w.p. Bob had wrong first edge and these bits are wasted In expectation, bit waste
In both cases, sending the first edge costs ! was chosen to balance the two losses
deg= π=ππππ /π2
Thank You!