PowerPoint Presentationpeople.csail.mit.edu/rrw/6.045-2019/lec8-color.pdf9 Theorem: If 𝑳has a streaming algorithm using Q space, then cc(𝒇𝑳Q4 + 4. Proof Idea: Alice runs streaming

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6.045Lecture 8: More on

Communication Complexity,Start up Turing Machines

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6.045Announcements:

- Dylan’s office hours? Vote!- Midterm: March 19

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Communication Complexity

A theoretical model of distributed computing

• Function f : {0,1}* £ {0,1}* ! {0,1}

– Two inputs, 𝑥 ∈ {0,1}* and 𝑦 ∈ {0,1}*

– We assume |𝒙|=|𝒚|=𝒏. Think of 𝒏 as HUGE

• Two computers: Alice and Bob

– Alice only knows 𝑥, Bob only knows 𝑦

• Goal: Compute f(𝒙, 𝒚) by communicating as few bits as possible between Alice and Bob

We do not count computation cost. We only care about the number of bits communicated.

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A(x,ε) = 0

B(y,0) = 1

A(x,01) = 1

B(y,011) = 0

x y

f(x,y)=0f(x,y)=0

Def. A protocol for a function f is a pair of functionsA, B : {0,1}* × {0,1}* → {0, 1, STOP} with the semantics:

On input (𝒙, 𝒚), let 𝒓 := 0, 𝒃𝟎 = ε. While (𝒃𝒓 ≠ STOP),

𝒓 + +If 𝒓 is odd, Alice sends 𝒃𝒓 = 𝑨 𝒙, 𝒃𝟏⋯𝒃𝒓−𝟏

else Bob sends 𝒃𝒓 = 𝑩 𝒚, 𝒃𝟏⋯𝒃𝒓−𝟏Output 𝒃𝒓−𝟏. Number of rounds = 𝒓 − 𝟏

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A(x,ε) = 0

B(y,0) = 1

A(x,01) = 1

B(y,011) = 0

x y

f(x,y)=0f(x,y)=0

Def. The cost of a protocol P for f on 𝒏-bit strings is

𝐦𝐚𝐱𝒙,𝒚 ∈ 𝟎,𝟏 𝒏

[number of rounds in P to compute f(𝒙, 𝒚)]

The communication complexity of f on 𝒏-bit strings, cc(f),is minimum cost over all protocols for f on 𝒏-bit strings= the minimum number of rounds used by any protocol

computing f(𝒙, 𝒚), over all 𝒏-bit 𝒙, 𝒚

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A(x,ε) = 0

B(y,0) = 1

A(x,01) = 1

B(y,011) = 0

x y

f(x,y)=0f(x,y)=0

Example. Let f : {0,1}* × {0,1}* → {0,1} be arbitrary

There is always a “trivial” protocol:Alice sends the bits of her 𝒙 in odd roundsBob sends whatever bit he wants in even roundsAfter 𝟐𝒏 − 𝟏 rounds, Bob knows 𝒙 and can send 𝒇(𝒙, 𝒚)

Proposition: For every 𝒇, cc(𝒇) ≤ 𝟐𝒏

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x y

f(x,y)=0f(x,y)=0

Example. EQUALS(𝒙, 𝒚) = 1 ⇔ 𝒙 = 𝒚

What’s a good protocol for computing EQUALS?

????

Communication complexity of EQUALS is at most 2𝒏

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x y

Examples:𝑳 = { x | x has an odd number of 1s}

⇒ 𝒇𝑳 𝒙, 𝒚 = PARITY(x,y) = σ𝒊𝒙𝒊 + σ𝒊𝒚𝒊 mod 2𝑳 = { x | x has more 1s than 0s}

⇒ 𝒇𝑳 𝒙, 𝒚 = MAJORITY(x,y)𝑳 = { xx | x ∈ {0,1}*}

⇒ 𝒇𝑳 𝒙, 𝒚 = EQUALS(x,y)

Connection to Streaming and DFAs

Let 𝑳 ⊆ {0,1}*Def. 𝒇𝑳: {0,1}*×{0,1}*→ {0,1}

for 𝒙, 𝒚 with |𝒙|=|𝒚| as:𝒇𝑳 𝒙, 𝒚 = 𝟏 ⇔ 𝒙𝒚 ∈ 𝑳

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Theorem: If 𝑳 has a streaming algorithm using ≤ 𝒔 space, then cc(𝒇𝑳) ≤ 4𝒔 + 4.

Proof Idea: Alice runs streaming algorithm A on 𝒙, reaches a memory state 𝒎. She sends 𝒎 to Bob in 4s+3rounds. Then Bob starts up A from state 𝒎, runs A on 𝒚. Gets an output bit, sends bit to Alice.

(…why 4𝒔+3 rounds?)

Connection to Streaming and DFAs

x y

Let 𝑳 ⊆ {0,1}*Def. 𝒇𝑳: {0,1}*×{0,1}*→ {0,1}

for 𝒙, 𝒚 with |𝒙|=|𝒚| as:𝒇𝑳 𝒙, 𝒚 = 𝟏 ⇔ 𝒙𝒚 ∈ 𝑳

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Theorem: If 𝑳 has a streaming algorithm using ≤ 𝒔 space, then cc(𝒇𝑳) ≤ 4𝒔 + 4.

Corollary: For every regular 𝑳, cc(𝒇𝑳) ≤ O(1).

Example: cc(PARITY) = 2

Corollary: cc(MAJORITY) ≤ O(log n),because there’s a streaming algorithm for {x : x has more 1’s than 0’s} with O(log n) space

What about the Comm. Complexity of EQUALS?

Connection to Streaming and DFAs

Let 𝑳 ⊆ {0,1}* Def. 𝒇𝑳 𝒙, 𝒚 = 𝟏 ⇔ 𝒙𝒚 ∈ 𝑳

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Theorem: cc(EQUALS) = 𝚯(𝒏).In particular, every communication protocol for EQUALS needs to send ≥ 𝒏 bits.

No communication protocol can do much better than “send your whole input”!

Corollary: 𝑳 = {xx | x in {0,1}*} is not regular.

Moreover, every streaming algorithm for 𝑳needs 𝒄 𝒏 bits of memory, for some constant 𝒄 > 0!

Communication Complexity of EQUALS

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Theorem: cc(EQUALS) = 𝚯(𝒏). In particular, everyprotocol for EQUALS needs ≥ 𝒏 bits of communication!

Idea: Consider all possible ways A & B can communicate.

Definition: The communication pattern of a protocol on inputs (𝒙, 𝒚) is the sequence of bits Alice & Bob send.

0

1

1

0

Pattern: 0110𝒙 𝒚

Communication Complexity of EQUALS

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A(x’,ε) = 0

B(y’,0) = 1

A(x’,01) = 1

B(y’,011) = 0

𝒙’ 𝒚’

A(x,ε) = 0

B(y,0) = 1

A(x,01) = 1

B(y,011) = 0

𝒙 𝒚

Key Lemma: If (𝒙, 𝒚) and (𝒙′, 𝒚′) have the same pattern 𝑷in a protocol, then (𝒙, 𝒚′) and (𝒙′, 𝒚) also have pattern 𝑷

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A(x’,ε) = 0

B(y’,0) = 1

A(x’,01) = 1

B(y’,011) = 0

𝒙’ 𝒚’

A(x,ε) = 0

B(y’,0) = 1

A(x,01) = 1

B(y’,011) = 0

𝒙 𝒚

Key Lemma: If (𝒙, 𝒚) and (𝒙′, 𝒚′) have the same pattern 𝑷in a protocol, then (𝒙, 𝒚′) and (𝒙′, 𝒚) also have pattern 𝑷

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Key Lemma: If (𝒙, 𝒚) and (𝒙′, 𝒚′) have the same pattern 𝑷in a protocol, then (𝒙, 𝒚′) and (𝒙′, 𝒚) also have pattern 𝑷

A(x,ε) = 0

B(y’,0) = 1

A(x,01) = 1

B(y’,011) = 0

𝒙’ 𝒚’

A(x,ε) = 0

B(y,0) = 1

A(x,01) = 1

B(y,011) = 0

𝒙 𝒚

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Theorem: The comm. complexity of EQUALS is 𝚯(𝒏).In particular, every protocol for EQUALS needs ≥ 𝒏 bits of communication.

Proof: By contradiction. Suppose cc(EQUALS) ≤ 𝒏 − 𝟏.Then there are ≤ 𝟐𝒏 − 𝟏 possible communication patterns of that protocol, over all pairs of inputs (𝒙, 𝒚) with n bits each.

Claim: There are 𝒙 ≠ 𝒚 such that on (𝒙, 𝒙) and on (𝒚, 𝒚), the protocol uses the same pattern 𝑷.

By the Key Lemma, (𝒙, 𝒚) and 𝒚, 𝒙 also use pattern 𝑷

So Alice & Bob output the same bit on (𝒙, 𝒚) and (𝒙, 𝒙).But EQUALS(𝒙, 𝒚) = 0 and EQUALS(𝒙, 𝒙) = 1. Contradiction!

Communication Complexity of EQUALS

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Randomized Protocols Help!

EQUALS needs ≥ 𝒏 bits of communication, but…

Theorem: For all 𝒙, 𝒚 of 𝒏 bits each, there is a randomized protocol for EQUALS 𝒙, 𝒚 using

only O(log 𝒏) bits of communication, which is correct with probability 99.9%!

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Turing Machines

Turing Machine (1936)

FINITE

STATE

CONTROL

INFINITE REWRITABLE TAPE

I N P U T

q0q1

A …

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In each step:- Reads a symbol- Writes a symbol- Changes state- Moves Left or Right

Turing Machine (1936)

INFINITE REWRITABLE TAPE

I N P U TA …

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https://www.cs.utah.edu/~draperg/cartoons/2005/turing.html

Turing Machines versus DFAs

The TM can both write to and read from the tape,and can write symbols that aren’t part of input

The “tape head” can move right and left

The input is written on an infinite tape

Accept and Reject take immediate effect

with “blank” symbols after the input

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Deciding the language L = { w#w | w {0,1}* }

STATE

0 1 1 # 0 1 10 #1 0 1X XX

1. If there’s no # on the tape (or more than one #), reject. 2. While there is a bit to the left of #,

Replace the first bit with X, and check if the first bit bto the right of the # is identical. (If not, reject.) Replace that bit b with an X too.

3. If there’s a bit to the right of #, then reject else accept

Definition: A Turing Machine is a 7-tuple T = (Q, Σ, Γ, , q0, qaccept, qreject), where:

Q is a finite set of states

Γ is the tape alphabet, where Γ and Σ Γ

q0 Q is the start state

Σ is the input alphabet, where Σ

: Q Γ → Q Γ {L, R}

qaccept Q is the accept state

qreject Q is the reject state, and qreject qaccept

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= “blank”

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0 → 0, R

read write move

→ , R

qaccept

qreject

0 → 0, R

→ , R

This Turing machine decides the language {0}

Σ = {0}

= “blank”

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0 → 0, R

read write move

→ , R

qaccept

0 → 0, R

→ , R

0 → 0, R

→ , L

This Turing machine recognizes the language {0}

Σ = {0}

= “blank”

Turing Machine Configurations

q01101000110 2 (Q [ Γ)*

corresponds to the configuration:

q0

1 0 0 0 0 01 1 1 1

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Turing Machine Configurations

0q1101000110 2 (Q [ Γ)*

corresponds to the configuration:

q1

1 0 0 0 0 00 1 1 1

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Turing Machine Configurations

0000011110q7 2 (Q [ Γ)*

corresponds to the configuration:

q7

0 0 0 1 1 00 0 1 1

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Defining Acceptance and Rejection for TMs

Let C1 and C2 be configurations of a TM MDefinition. C1 yields C2 if M is in configuration C2

after running M in configuration C1 for one step

Example. Suppose (q1, b) = (q2, c, L)Then aq1bb yields q2acbSuppose (q1, a) = (q2, c, R)Then abq1a yields abcq2

Let w Σ* and M be a Turing machine.M accepts w if there are configs C0, C1, ..., Ck, s.t.

• C0 = q0w [the initial configuration]• Ci yields Ci+1 for i = 0, ..., k-1, and • Ck contains the accept state qaccept

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accepting computation

history of M on x

A TM M recognizes a language L if M accepts exactly those strings in L

A TM M decides a language L if M accepts all strings in L and rejects all strings not in L

A language L is recognizable (a.k.a. recursively enumerable)

if some TM recognizes L

A language L is decidable (a.k.a. recursive)if some TM decides L

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A Turing machine for deciding { 0 | n ≥ 0 }2n

1. Sweep from left to right, x-out every other 02. If in step 1, the tape had only one 0, accept3. If in step 1, the tape had an odd number of 0’s,

reject4. Move the head left to the first input symbol.5. Go to step 1.

Turing Machine PSEUDOCODE:

Why does this work?

Observation: Every time we return to step 1, the number of 0’s on the tape has been halved.

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even0’s

Step 1

odd 0’s

0 → , R

→ , R

qacceptqreject

0 → x, R

x → x, R→ , R

x → x, R

0 → 0, Lx → x, L

x → x, R

→ , L→ , R

0 → x, R0 → 0, R

→ , Rx → x, R

q0 q1

q2

q3

q4

{ 0 | n ≥ 0 }2n

Step 2

Step 3

Step 4

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0 → , R

→ , R

qacceptqreject

0 → x, R

x → x, R→ , R

x → x, R

0 → 0, Lx → x, L

x → x, R

→ , L→ , R

0 → x, R0 → 0, R

→ , Rx → x, R

{ 0 | n ≥ 0 }2n

q0 q1

q2

q3

q4

q00000

q1000

xq300

x0q40

x0xq3

x0q2x

xq20x

q2x0x

q2x0x

… 34

q1x0x

xq10x

Multitape Turing Machines

: Q Γk → Q Γk {L,R}k

FINITE

STATE

CONTROLk

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FINITE

STATE

CONTROL

0 01

FINITE

STATE

CONTROL 0 01 # # #. . .

Theorem: Every Multitape Turing Machine can be transformed into a single tape Turing Machine

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FINITE

STATE

CONTROL

0 01

FINITE

STATE

CONTROL 0 01 # # #. . .

Theorem: Every Multitape Turing Machine can be transformed into a single tape Turing Machine

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FINITE

STATE

CONTROL

0 01

FINITE

STATE

CONTROL 0 01 # # #. . .

Theorem: Every Multitape Turing Machine can be transformed into a single tape Turing Machine

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FINITE

STATE

CONTROL

0 01

FINITE

STATE

CONTROL 0 01 # # #. .

Theorem: Every Multitape Turing Machine can be transformed into a single tape Turing Machine

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.

Theorem: Every Multitape Turing Machine can be transformed into a single tape Turing Machine

FINITE

STATE

CONTROL

0 01

FINITE

STATE

CONTROL 0 01 # # #. . .

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Theorem: Every nondeterministic Turing machine Ncan be transformed into a Turing Machine M that accepts precisely the same strings as N.

Nondeterministic Turing Machines

Proof Idea (more details in Sipser p.178-179)Pick a natural ordering on the strings in (Q [ Γ [ #)*

M(w): For all strings D 2 (Q [ Γ [ #)* in the ordering,Check if D = C0# #Ck where C0, …,Ck is an

accepting computation history for N on w. If so, accept.

Have multiple transitions for a state, symbol pair

Fact: We can encode Turing Machines as bit strings

0n10m10k10s10t10r10u1 …

n states

m tape symbols

(first k are input

symbols)

start

state

accept

state

reject

state

blank

symbol

( (p, i), (q, j, L) ) = 0p10i10q10j101

( (p, i), (q, j, R) ) = 0p10i10q10j100142

Similarly, we can encode DFAs and NFAs as bit strings, and w 2 Σ* as bit strings

ADFA = { (B, w) | B encodes a DFA over some Σ,and B accepts w 2 Σ* }

ANFA = { (B, w) | B encodes an NFA, B accepts w }

ATM = { (M, w) | M encodes a TM, M accepts w }

For x 2 Σ* define bΣ(x) to be its binary encodingFor x, y 2 Σ*, define the pair of x and y as

(x, y) := 0|bΣ(x)|1 bΣ(x) bΣ(y)Then we define the following languages over {0,1}:

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ATM = { (M, w) | M encodes a TM over some Σ,w encodes a string over Σ and M accepts w}

ATM = { z | z decodes to (M, w) and M does not accept w }

Technical Note: We’ll use an decoding of pairs, TMs, and strings so that every binary string decodes to some pair (M, w)

If z ∈ {0,1}* doesn’t decode to (M, w) in the usual way, then we define that z decodes to the pair (D, ε)

where D is a “dummy” TM that accepts nothing.

Universal Turing Machines

Theorem: There is a Turing machine Uwhich takes as input:- the code of an arbitrary TM M- and an input string wsuch that U accepts (M, w)M accepts w.

This is a fundamental property of TMs:

There is a Turing Machine that can run arbitrary Turing Machine code!

Note that DFAs/NFAs do not have this property.That is, ADFA and ANFA are not regular.

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ADFA = { (D, w) | D is a DFA that accepts string w }

Theorem: ADFA is decidable

Proof: A DFA is a special case of a TM. Run the universal U on (D, w) and output its answer.

ATM = { (M, w) | M is a TM that accepts string w }

Theorem: ATM is recognizable

ANFA = { (N, w) | N is an NFA that accepts string w }

Theorem: ANFA is decidable. (Why?)

The Church-Turing Thesis

Everyone’s Intuitive Notion

of Algorithms

= Turing Machines

This is not a theorem –it is a falsifiable scientific hypothesis.

And it has been thoroughly tested!

w L ?

accept reject

TM

yes no

w Σ*

L is decidable

(recursive)

w L ?

accept reject or loop

TM

yes no

w Σ*

L is recognizable

(recursively enumerable)

Theorem: L is decidableiff both L and L are recognizable

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Recall: Given L µ Σ*, define L := Σ* \ L

How? Any ideas?M1 always accepts x, when x is in L

M2 always accepts x, when x isn’t in L

Theorem: L is decidableiff both L and L are recognizable

Given:a TM M1 that recognizes L anda TM M2 that recognizes L,

want to build a new machine M that decides L

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Theorem: L is decidableiff both L and L are recognizable

Decider for L:Simulate M1(x) on one tape, while also simulating M2(x) on another tape. Exactly one will accept. If M1 accepts then accept. If M2 accepts then reject

Recall: Given L µ Σ*, define L := Σ* \ L

Given:a TM M1 that recognizes L anda TM M2 that recognizes L,

want to build a new machine M that decides L

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