Interactive Channel Capacity
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Transcript of Interactive Channel Capacity
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Gillat Kol (IAS)
joint work withRan Raz (Weizmann + IAS)
Interactive Channel Capacity
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Binary Symmetric Channel with crossover probability (-BSC): Flips every bit w.p. , independently
Channel Capacity
1−𝜀
1−𝜀
0
1 1
0
𝜀𝜀
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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 [Shannon‘48]:
bitsnoiseless channel
bits-BSC
Channel Capacity
Channel Capacity
𝐻 (𝜀 )=−𝜀𝑙𝑜𝑔 𝜀− (1−𝜀 ) 𝑙𝑜𝑔(1−𝜀)
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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
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Alice has input , Bob has input .They want to compute ( public). How many bits do they need to exchange?
Communication Complexity [Yao‘79]
𝑥
𝑓 (𝑥 , 𝑦 ) !
𝑦𝑚1 (𝑥 )𝑚2 ( 𝑦 ,𝑚1 )
𝑚3 (𝑥 ,𝑚1 ,𝑚2 ) . .adaptive!Protocol:
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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
𝑥 ,𝑅 𝑦 ,𝑅
𝑓 (𝑥 , 𝑦 ) !
. .
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Example 1: of Equality
𝑥≠ 𝑦 !
. .
R R
are -bit stringsEquality function: iff
Naïve protocol: exchanged bitsAlice sends to Bob
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are -bit stringsEquality function: iff
Naïve 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
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Underlying tree: complete binary tree
Example 2: of Pointer Jumping
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Underlying tree: complete binary tree: an edge going out of every node in odd layers
Example 2: of Pointer Jumping
𝑥
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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 “chasing the pointers”
Example 2: of Pointer Jumping
𝑥 𝑦
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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 “chasing the pointers”
Example 2: of Pointer Jumping
𝑥 𝑦
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Example 2: of Pointer Jumping
(left edge)
(left edge)( right edge)
( left edge)
𝑥 𝑦
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(left edge)
(left edge)( right edge)
( left edge)
input size
protocol is fully adaptive, very susceptible to noise
𝑥 𝑦
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Def: Interactive Channel Capacity
: Noiseless CC: Noisy CC
Interactive Channel Capacity
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The interactive setting is more general:1. Allows interaction and adaptivity.
Shannon’s setting is one-way2. Allows any computation (function ). Shannon’s 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 “online’’!
Why is the Interactive Case More Challenging?
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[Schulman ‘92]: – Theorem: If then
Corollary: – interactive channel capacity channel capacity
Many other works [Sch,BR,B,GMS,BK,BN,FGOS…]: – Simulation of any communication protocol with adversarial noise– Large constants, never made explicit
Previous Works𝐂𝐚𝐩 (𝛆 )=𝐥𝐢𝐦𝐢𝐧𝐟𝐧→∞
𝐦𝐢𝐧{𝐟 :𝐂𝐂 ( 𝐟 )=𝐧 }( 𝐂𝐂 (𝐟 )
𝐂𝐂𝛆 (𝐟 ))
?
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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)
𝐂𝐚𝐩 (𝛆 )=𝐥𝐢𝐦𝐢𝐧𝐟𝐧→∞
𝐦𝐢𝐧{𝐟 :𝐂𝐂 ( 𝐟 )=𝐧 }( 𝐂𝐂 (𝐟 )
𝐂𝐂𝛆 (𝐟 ))
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In our model, exactly one player sends a bit at each time step
– The order of turns in a protocol is pre-determined (independent of the inputs, randomness, noise).
– Alternating turns is a special case
Channel Types & Communication Order
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Lower Bound: A simulation protocol
𝐂𝐚𝐩 (𝛆 )=𝐥𝐢𝐦𝐢𝐧𝐟𝐧→∞
𝐦𝐢𝐧{𝐟 :𝐂𝐂 ( 𝐟 )=𝐧 }( 𝐂𝐂 (𝐟 )
𝐂𝐂𝛆 (𝐟 ))
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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:– The basic step simulates steps of P– The th inductive step simulates steps of P
𝐂𝐚𝐩 (𝛆 )=𝐥𝐢𝐦𝐢𝐧𝐟𝐧→∞
𝐦𝐢𝐧{𝐟 :𝐂𝐂 ( 𝐟 )=𝐧 }( 𝐂𝐂 (𝐟 )
𝐂𝐂𝛆 (𝐟 ))
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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 step’s bits from his transcript
bits of protocol P bitconsistency check
inconsistency
list of all massages exchanged
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Players run the Basic Step consecutive times. Alice observes transcript , and Bob transcript
Simulating Protocol: 1st Interactive Step
times
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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 step’s bits from his transcript
timesinconsistency
bits
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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
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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:
𝜀=𝑙𝑜𝑔𝑘 /𝑘2
bits of protocol P bitconsistency checkinconsistency
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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!
𝜀=𝑙𝑜𝑔𝑘 /𝑘2
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Upper Bound: Example of with
𝐂𝐚𝐩 (𝛆 )=𝐥𝐢𝐦𝐢𝐧𝐟𝐧→∞
𝐦𝐢𝐧{𝐟 :𝐂𝐂 ( 𝐟 )=𝐧 }( 𝐂𝐂 (𝐟 )
𝐂𝐂𝛆 (𝐟 ))
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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 “chasing the pointers”
Example: -Pointer Jumping
dept
h =
deg=
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Observe, We prove - involved! Such lower bounds are typically up-to a constant.
The capacity comes from the second order terms
Upper Bound Outline 𝜀=𝑙𝑜𝑔𝑘 /𝑘2
𝐂𝐚𝐩 (𝛆 )=𝐥𝐢𝐦𝐢𝐧𝐟𝐧→∞
𝐦𝐢𝐧{𝐟 :𝐂𝐂 ( 𝐟 )=𝐧 }( 𝐂𝐂 (𝐟 )
𝐂𝐂𝛆 (𝐟 ))
deg=
dept
h =
deg=
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Bounding : Intuition“Any good protocol does the following:”Alice 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= 𝜀=𝑙𝑜𝑔𝑘 /𝑘2
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Thank You!