Asymmetric Communication Complexity And its implications on Cell Probe Complexity Slides by Elad...

Asymmetric Communication Asymmetric Communication ComplexityComplexity

And its implications on Cell Probe And its implications on Cell Probe ComplexityComplexity

Slides by Elad Verbin

Based on a paper of Peter Bro Miltersen, Noam Nisan, Muli Safra and Avi Wigderson

PurposePurpose

We want to get Lower Bounds.We want to get Lower Bounds.

Best known lower bounds:Best known lower bounds:Sorting is Sorting is ΩΩ(nlogn) in the comparison (nlogn) in the comparison

modelmodelTrivial lower bounds. i.e. MAX is Trivial lower bounds. i.e. MAX is ΩΩ(n)(n)What can we really do, i.e. for RAM?What can we really do, i.e. for RAM?

OutlineOutline

Yao decided to strengthen the model – Yao decided to strengthen the model – Considered the Cell Probe model.Considered the Cell Probe model.

Lower bounding Cell Probe is hard too. Lower bounding Cell Probe is hard too. We strengthen even more – We strengthen even more – Communication ComplexityCommunication Complexity

OutlineOutline

We show the relationship between Cell We show the relationship between Cell Probe and Communication Complexity.Probe and Communication Complexity.

We show how to get lower bounds for We show how to get lower bounds for Communication Complexity using two Communication Complexity using two techniques:techniques:

1.1. The Richness TechniqueThe Richness Technique

2.2. The Round Elimination TechniqueThe Round Elimination Technique

Communication ComplexityCommunication Complexity

The problem : f:XThe problem : f:XYY{0,1}{0,1}Alice gets xAlice gets xX, Bob gets yX, Bob gets yY, their goal is Y, their goal is

to exchange messages to to exchange messages to decidedecide f(x,y). f(x,y).

A solution is a communication protocol A solution is a communication protocol that can compute f(x,y) for all x,y.that can compute f(x,y) for all x,y.

f(x,y)

Asymmetric Communication Asymmetric Communication ComplexityComplexity

A Communication Protocol that computes A Communication Protocol that computes function f in which Alice sends at most function f in which Alice sends at most a a bits and Bob sends at most bits and Bob sends at most bb bits is called bits is called a [a [aa,,bb]-protocol for f.]-protocol for f.

f(x,y)Pink<=a Blue<=b

Randomized ProtocolsRandomized Protocols

If the protocol is allowed to flip (public) If the protocol is allowed to flip (public) coins, and gives the correct answer with coins, and gives the correct answer with probability > 2/3 it is called a randomized probability > 2/3 it is called a randomized protocol.protocol.

If it always correctly identifies a 0-instance If it always correctly identifies a 0-instance it is called a one-sided error protocol.it is called a one-sided error protocol.

ExampleExample

For any problem, there are trivial For any problem, there are trivial deterministic protocols:deterministic protocols:

[log|X|,1]-protocol[log|X|,1]-protocol [1,log|Y|]-protocol.[1,log|Y|]-protocol.

The Problem DISJ(N,k,l)The Problem DISJ(N,k,l)

We work on the universe U={0,1,…,N-1}We work on the universe U={0,1,…,N-1}Alice gets x, a set of k elementsAlice gets x, a set of k elementsBob gets y, a set of l elementsBob gets y, a set of l elementsThey must decide if x∩y=ØThey must decide if x∩y=Ø

x∩∩y=Ø=Ø?

x y

A one-sided error randomized A one-sided error randomized [O(k),O(l)]-protocol for DISJ(N,k,l)[O(k),O(l)]-protocol for DISJ(N,k,l)

If we say that x and y are disjoint then we If we say that x and y are disjoint then we want to have complete confidence. If we want to have complete confidence. If we say they intersect, we want to be say they intersect, we want to be reasonably certain.reasonably certain.

Flip public coins to get a sequence of Flip public coins to get a sequence of random subsets of the universe: Rrandom subsets of the universe: R11, , RR22, …, …

R3

x y

R3

x y

y=y∩R3

y

R3

x y

y=y∩R3

yR8

x=x∩R8

AND SO ON….


We don’t really send the index of RWe don’t really send the index of R88, we , we just send the distance from the last setjust send the distance from the last set (8-3=5). This means that the expected (8-3=5). This means that the expected numbers of bits sent by a player is equal numbers of bits sent by a player is equal to the size of his setto the size of his set

If at some point one of the sets becomes If at some point one of the sets becomes empty, then the originals were disjoint – empty, then the originals were disjoint – say so. Otherwise, after a long time, say say so. Otherwise, after a long time, say that there is an intersection.that there is an intersection.


If x and y were indeed disjoint, the If x and y were indeed disjoint, the sizes of x and y decrease by a factor sizes of x and y decrease by a factor of 2 each round. Therefore the total of 2 each round. Therefore the total communication is [O(k),O(l)].communication is [O(k),O(l)].

If the sets were disjoint, what is the If the sets were disjoint, what is the chance that we say that there is an chance that we say that there is an intersection? Very low.intersection? Very low.

Fixed-round protocolsFixed-round protocols

If t alternating messages are sent and If t alternating messages are sent and each message is of size a or b it is called each message is of size a or b it is called a [t,a [t,aa,,bb]-protocol. ]-protocol.

f(x,y)

a

ab

bt

Static Data Structure ProblemsStatic Data Structure Problems

A static data structure problem is a function A static data structure problem is a function f:Df:DQQRR

D – the dataD – the data Q – the queriesQ – the queries R – Possible answers. Typically, R={0,1}R – Possible answers. Typically, R={0,1}

queryData

DS

The Problem MEMBERSHIP(N)The Problem MEMBERSHIP(N)

INPUT: a set SINPUT: a set S[N][N]QUERIES: of the form “xQUERIES: of the form “xS?”S?”

D={SD={S[N]}, |D|=2[N]}, |D|=2NN

Q=[N]Q=[N]R={0,1}R={0,1}

The trivial solution is optimal.The trivial solution is optimal.

The Problem MEMBERSHIP(N,n)The Problem MEMBERSHIP(N,n)

INPUT: a set SINPUT: a set S[N] of size n[N] of size nQUERIES: of the form “xQUERIES: of the form “xS?”S?”

D={SD={S[N] | |S|=n}, |D|=choose(N,n)[N] | |S|=n}, |D|=choose(N,n)Q=[N]Q=[N]R={0,1}R={0,1}

The Cell Probe ModelThe Cell Probe Model

Parameter w – word sizeParameter w – word sizes cells, each containing w bits.s cells, each containing w bits.Each query probes at most t cells to get Each query probes at most t cells to get

answeranswer

A query is a decision tree of depth t and A query is a decision tree of depth t and degree 2degree 2ww

MEMBERSHIP(N,n)MEMBERSHIP(N,n)

Solutions:Solutions:Keep every possible answer. s=N, t=1 Keep every possible answer. s=N, t=1

(better – s=N/w, t=1)(better – s=N/w, t=1)Keep a nonredundant representation. Keep a nonredundant representation.

s=log(choose(N,n)), t=log(choose(N,n))s=log(choose(N,n)), t=log(choose(N,n))

MEMBERSHIP(N,n)MEMBERSHIP(N,n)

Solutions:Solutions:Keep a sorted list of all elements. Keep a sorted list of all elements.

s=nlog(N)/w , t=log(n)*log(N)/ws=nlog(N)/w , t=log(n)*log(N)/wThere is a randomized solution with There is a randomized solution with

s=(n/w)s=(n/w)cc, t=O(1), for some constant c., t=O(1), for some constant c.

What is the connection What is the connection between Cell Probe and between Cell Probe and

Asymmetric Asymmetric Communication Communication

Complexity?Complexity?

ACC <-> Cell ProbeACC <-> Cell Probe

The communication problem related to a The communication problem related to a static data structure problem f:Dstatic data structure problem f:DQQ{0,1} {0,1} if the problem where Alice gets a query, if the problem where Alice gets a query, Bob gets the data, and they should decide Bob gets the data, and they should decide if this is a “yes” instance or a “no” instanceif this is a “yes” instance or a “no” instance

Communication Problem MEM(N,n) MEM(N,n)

Alice gets xAlice gets x[N], Bob gets y[N], Bob gets y[N], |y|=n, [N], |y|=n, they should decide if xthey should decide if xy.y.

Trivial protocols: [1,nlogN] , [logN,1]Trivial protocols: [1,nlogN] , [logN,1]

Lemma CP->AACLemma CP->AAC

If there is a solution to the data If there is a solution to the data structure problem with word size w structure problem with word size w taking s cells and with query time t, taking s cells and with query time t, then there is a [2t,log(s),w]-protocol then there is a [2t,log(s),w]-protocol for the communication problemfor the communication problem

Therefore a lower bound on ACC gives us Therefore a lower bound on ACC gives us a lower bound on Cell Probea lower bound on Cell Probe

Finer points of CP->AACFiner points of CP->AAC

How is the communication complexity How is the communication complexity model stronger than the Cell Probe model stronger than the Cell Probe Model?Model?

Answer: In its adaptivityAnswer: In its adaptivity

Which form of Cell Probe lower bounds Which form of Cell Probe lower bounds can we get from the CP->AAC Lemma?can we get from the CP->AAC Lemma?

Answer: the bound on space is up to a Answer: the bound on space is up to a polynomialpolynomial

Restricted AAC->CPRestricted AAC->CP

If there is a [O(1),a,b]-protocol for the If there is a [O(1),a,b]-protocol for the communication problem then the data communication problem then the data structure problem has a solution with word structure problem has a solution with word size w=b, t=O(1) and s=2size w=b, t=O(1) and s=2O(a)O(a)

Proof: The Data Structure for input y Proof: The Data Structure for input y contains the message Bob should send contains the message Bob should send next for every possible history of next for every possible history of messages Alice can send, for any query.messages Alice can send, for any query.

Lower Bounding The Lower Bounding The Communication ComplexityCommunication Complexity

Communication Problem MEM(N,l) MEM(N,l)

Alice gets xAlice gets x[N], Bob gets y[N], Bob gets y[N], |y|=l, [N], |y|=l, they should decide if xthey should decide if xy.y.

NONMEM(N,l) is the same problem, when NONMEM(N,l) is the same problem, when Alice and Bob want to decide if xAlice and Bob want to decide if xyy

Trivial protocols: [1,l*logN] , [logN,1]Trivial protocols: [1,l*logN] , [logN,1]

Problem <-> MatrixProblem <-> Matrix

We identify a communication problem We identify a communication problem f:X×Yf:X×Y{0,1} with a |X|×|Y| Matrix where {0,1} with a |X|×|Y| Matrix where M[x][y]=f(x,y).M[x][y]=f(x,y).

The matrix of NONMEM(N,l) has N rows The matrix of NONMEM(N,l) has N rows and and columns. Each column has N-l 1-columns. Each column has N-l 1-entriesentries

l

N

Problem <-> MatrixProblem <-> Matrix

A problem (matrix) is (u,v)-rich if at least v A problem (matrix) is (u,v)-rich if at least v columns contain at least u 1-entries.columns contain at least u 1-entries.

NONMEM(N,l) is (N-l, )-rich. NONMEM(N,l) is (N-l, )-rich.

l

N

(4,3)-rich

The Richness LemmaThe Richness Lemma

Let f be a communication problem that:Let f be a communication problem that:

1.1. is (u,v)-richis (u,v)-rich

2.2. has a randomized one-sided error [a,b]-has a randomized one-sided error [a,b]-protocol.protocol.

Then f contains a u/2Then f contains a u/2a+2a+2 over over

v/2v/2a+b+2a+b+2 submatrix of 1-entries. submatrix of 1-entries.

Randomized Lower Bound for Randomized Lower Bound for MEM(N,l)MEM(N,l)

Say MEM(N,l) has a negative-one-sided Say MEM(N,l) has a negative-one-sided error [a,b]-protocol. Let a<log(l), l<N/2.error [a,b]-protocol. Let a<log(l), l<N/2.

Then NONMEM(N,l) has a one-sided error Then NONMEM(N,l) has a one-sided error [a,b]-protocol[a,b]-protocol

NONMEM(N,l) is (N-l, )-richNONMEM(N,l) is (N-l, )-rich

Therefore it has a 1-submatrix of Therefore it has a 1-submatrix of dimensions at least (N-l)/2dimensions at least (N-l)/2a+2a+2 over over

/2/2a+b+2a+b+2

Randomized Lower Bound for Randomized Lower Bound for NONMEM(N,l)NONMEM(N,l)

l

N

l

N

However, if there is a 1-submatrix of However, if there is a 1-submatrix of dimensions r on s then s≤dimensions r on s then s≤

By substituting for s and r, simplifying and By substituting for s and r, simplifying and bounding we get 2bounding we get 2aa(a+b)=(a+b)=ΩΩ(l)(l)

Randomized Lower Bound for Randomized Lower Bound for NONMEM(N,l)NONMEM(N,l)

l

rN

Randomized Randomized [O(a),O(l/2[O(a),O(l/2aa)] )] Upper Upper Bound for NONMEM(N,l)Bound for NONMEM(N,l)

On the other hand, NONMEM(N,l) has a On the other hand, NONMEM(N,l) has a [O(a),O(l/2[O(a),O(l/2aa)]-protocol, for all a<log(l):)]-protocol, for all a<log(l):

Alice sends Bob the first Alice sends Bob the first aa indices of R’s indices of R’s that contain x. This allows Bob to reduce y that contain x. This allows Bob to reduce y to expected size l/2to expected size l/2aa..

Then Bob sends a couple indices that Then Bob sends a couple indices that contain y.contain y.

If we are not yet sure that they are disjoint, If we are not yet sure that they are disjoint, we say that they intersect.we say that they intersect.

Tightness for NONMEM(N,l)Tightness for NONMEM(N,l)

NONMEM(N,l) has a [O(a),O(l/2NONMEM(N,l) has a [O(a),O(l/2aa)]-protocol, for )]-protocol, for all a<log(l)all a<log(l)

22aa(a+b)= 2(a+b)= 2aa(a+l/2(a+l/2aa)=)=??O(l) Therefore the last result O(l) Therefore the last result is tight?is tight?

There are constants c,c’>0, so that for any a,There are constants c,c’>0, so that for any a, b=l/2b=l/2caca is enough. is enough. b=l/2b=l/2c’ac’a is not enough. is not enough.

The Richness LemmaThe Richness Lemma

Let f be a communication problem that:Let f be a communication problem that:

1.1. is (u,v)-richis (u,v)-rich

2.2. has a randomized one-sided error [a,b]-has a randomized one-sided error [a,b]-protocol.protocol.

Then f contains a u/2Then f contains a u/2a+2a+2 over over

v/2v/2a+b+2a+b+2 submatrix of 1-entries. submatrix of 1-entries.

Proof of the Richness LemmaProof of the Richness Lemma

First let us prove a weaker result: if f has a First let us prove a weaker result: if f has a deterministic [a,b]-protocol then it contains deterministic [a,b]-protocol then it contains a contains a u/2a contains a u/2aa over over

v/2v/2a+ba+b submatrix of 1-entries. submatrix of 1-entries.

We prove this by induction on a+b:We prove this by induction on a+b:


For a+b=0 – |X|For a+b=0 – |X|≥u, |Y|≥v, ≥u, |Y|≥v, and f(x,y)=1 for and f(x,y)=1 for all x,y, so this is trivial.all x,y, so this is trivial.

Now, if Alice send the first bit:Now, if Alice send the first bit:XX00 – inputs for which she sends 0 – inputs for which she sends 0

XX11 – inputs for which she sends 1 – inputs for which she sends 1

Let fLet f00, f, f11 be the restrictions of f to X be the restrictions of f to X00Y, Y,

XX11Y.Y.


At least one of them is (u/2,v/2)-rich, and At least one of them is (u/2,v/2)-rich, and both have a [a-1,b] protocol.both have a [a-1,b] protocol.

By the induction it contains a (u/2)/2By the induction it contains a (u/2)/2a-1a-1 over over (v/2)/2(v/2)/2a+b-1 a+b-1 1-submatrix.1-submatrix.

In the other case, Bob send the first bit. In the other case, Bob send the first bit. Define YDefine Y00,Y,Y11,f,f00,f,f11. At least one of them is . At least one of them is

(u,v/2)-rich, and proceed similarly.(u,v/2)-rich, and proceed similarly.


Now let us prove the general case:Now let us prove the general case:Let S be the set of u*v rich-positions in the Let S be the set of u*v rich-positions in the

matrixmatrixLet us look at some coin-flip sequence.Let us look at some coin-flip sequence.


Let X = #{1s in S}Let X = #{1s in S}E[X]>=2/3 * uvE[X]>=2/3 * uv=> There exists such a sequence for => There exists such a sequence for

which X>=2/3 * uvwhich X>=2/3 * uv

Fix the sequence, to get a deterministic Fix the sequence, to get a deterministic algorithm. This algorithm computes a algorithm. This algorithm computes a function f’ that is close to f.function f’ that is close to f.


By a counting argument, f’ is (u/4,v/4)-rich, By a counting argument, f’ is (u/4,v/4)-rich, and so it has a 1-submatrix of the required and so it has a 1-submatrix of the required size ( u/2size ( u/2a+2a+2 over v/2 over v/2a+b+2a+b+2 ) )

This is a 1-submatrix in f too, because the This is a 1-submatrix in f too, because the error is one-sided.error is one-sided.

Q.E.D.Q.E.D.

A Richness Results for two-sided A Richness Results for two-sided errorerror

Let d,e>0, and let f:X×YLet d,e>0, and let f:X×Y{0,1} be a {0,1} be a communication problem with at least a d-communication problem with at least a d-fraction of 1s. If f has a randomized two-fraction of 1s. If f has a randomized two-sided error [a,b]-protocol then f has a sided error [a,b]-protocol then f has a submatrix M of dimensions at least |X|/2submatrix M of dimensions at least |X|/2O(a)O(a) over |Y|/2over |Y|/2O(a+b)O(a+b) with at least a (1-e)- with at least a (1-e)-fraction of 1s.fraction of 1s.

The SPAN(n) ProblemThe SPAN(n) Problem

In SPAN, Alice gets xIn SPAN, Alice gets x{0,1}{0,1}n n and Bob gets and Bob gets a vector subspace ya vector subspace y{0,1}{0,1}n n

y can be represented using a basis of k≤n y can be represented using a basis of k≤n vectors – O(nvectors – O(n22) bits) bits

Alice and Bob must decide if xAlice and Bob must decide if x∈∈y.y.

Trivial Protocols:[n,1] , [1,nTrivial Protocols:[n,1] , [1,n22]]

Lower bounds for SPANLower bounds for SPAN

Let’s prove that in any [a,b] randomized Let’s prove that in any [a,b] randomized one-sided error protocol for SPAN, either one-sided error protocol for SPAN, either a=a=ΩΩ(n), or b=(n), or b=ΩΩ(n(n22))

We will assume that y is of dimension We will assume that y is of dimension n/2.n/2.

We will prove that:We will prove that:1.1. SPAN is (2SPAN is (2n/2n/2,2,2nn22/4/4)-rich, and)-rich, and2.2. SPAN does not contain a 1-submatrix of SPAN does not contain a 1-submatrix of

dimensions 2dimensions 2n/3n/3 over 2 over 2nn22/12/12

SPAN is (2SPAN is (2n/2n/2,2,2nn22/4/4)-rich)-rich

Each subspace contains exactly 2Each subspace contains exactly 2n/2n/2 vectors => each column contains 2vectors => each column contains 2n/2n/2 1s. 1s.

How many subspaces of dimension n/2 How many subspaces of dimension n/2 are there?are there?

Lets choose a basis: we have 2Lets choose a basis: we have 2nn-1 -1 possibilities for the first vector, 2possibilities for the first vector, 2nn-2 for the -2 for the second, 2second, 2nn-4 for the third, etc.-4 for the third, etc.

SPAN is (2SPAN is (2n/2n/2,2,2nn22/4/4)-rich)-rich

We chose each basis (n/2)! timesWe chose each basis (n/2)! timesHow many basis does a subspace has?How many basis does a subspace has?We have 2We have 2n/2n/2-1 options to choose the first -1 options to choose the first

vector, 2vector, 2n/2n/2-2 for the second, etc.-2 for the second, etc.We again chose each basis (n/2)! times.We again chose each basis (n/2)! times.

Thus, there are at least 2Thus, there are at least 2nn22/4/4 subspaces of subspaces of dimension n/2.dimension n/2.

SPAN does not contain a 1-submatrix SPAN does not contain a 1-submatrix of dimensions 2of dimensions 2n/3n/3 over 2 over 2nn22/12/12

Lets look at a 1-submatrix with at least 2Lets look at a 1-submatrix with at least 2n/3 n/3

rows. The subspace spanned by them is rows. The subspace spanned by them is of dimension at least n/3.of dimension at least n/3.

How many subspaces of dimension n/2 How many subspaces of dimension n/2 can include this entire subspace?can include this entire subspace?

22nn22/12/12

And we’re done.And we’re done.

The Round Elimination The Round Elimination LemmaLemma

f->Pf->Pmm(f)(f)

Let f:X×Y->{0,1} be a communication problemLet f:X×Y->{0,1} be a communication problem

PPmm(f) is:(f) is:

Alice gets m elements from X, xAlice gets m elements from X, x11, …, x, …, xmm

Bob gets 1≤i≤m, yBob gets 1≤i≤m, yY and also xY and also x11, …, x, …, xi-1i-1

They want to compute f(xThey want to compute f(x ii,y),y)

How meaningful can Alice’s first message be?How meaningful can Alice’s first message be?

The Round Elimination LemmaThe Round Elimination Lemma

Let C=99, R=4256.Let C=99, R=4256.Say that PSay that PRaRa(f) has a randomized two-(f) has a randomized two-

sided error [t,a,b]-protocol in which Alice sided error [t,a,b]-protocol in which Alice sends the first message.sends the first message.

Then there is a randomized two-sided Then there is a randomized two-sided error [t-1,Ca,Cb]-protocol with Bob error [t-1,Ca,Cb]-protocol with Bob sending the first message.sending the first message.

General framework for LB proofs using General framework for LB proofs using the Round Elimination Lemmathe Round Elimination Lemma

[t,a,b]-protocol for F(n)[t,a,b]-protocol for F(n)

[t,a,b]-protocol for P[t,a,b]-protocol for Pmm(F(n’)) (typically n’=n/m)(F(n’)) (typically n’=n/m)

[t-1,Ca,Cb]-protocol for F(n’)[t-1,Ca,Cb]-protocol for F(n’)

[1,Ct-1a,Ct-1b]-protocol for F(n[1,Ct-1a,Ct-1b]-protocol for F(n(t-1)(t-1)) )

The problem GT(n)The problem GT(n)

Alice and Bob each gets an n-bit integer.Alice and Bob each gets an n-bit integer.They want to decide if x<y.They want to decide if x<y.

x<y?

x y

The problem GT(n)The problem GT(n)

Deterministic communication complexity is Deterministic communication complexity is linearlinear

Randomized comm. complexity with two-Randomized comm. complexity with two-sided error is O(logn) (using a logarithmic sided error is O(logn) (using a logarithmic number of rounds)number of rounds)

When limited to t rounds: There is a When limited to t rounds: There is a [t,n[t,n1/t1/tlogn,nlogn,n1/t1/tlogn]-protocol logn]-protocol

GT(n) does not have aGT(n) does not have a[t,n[t,n1/t1/tCC-t-t,n,n1/t1/tCC-t-t]-protocol ]-protocol

Theorem: Let C=99.Theorem: Let C=99.

There does not exist a randomized two-There does not exist a randomized two-sided error [t,nsided error [t,n1/t1/tCC-t-t,n,n1/t1/tCC-t-t]-protocol for ]-protocol for GT(n)GT(n)

GT(n) does not have aGT(n) does not have a[t,n[t,n1/t1/tCC-t-t,n,n1/t1/tCC-t-t]-protocol]-protocol

By induction on tBy induction on t

Say there was a [t,nSay there was a [t,n1/t1/tCC-t-t,n,n1/t1/tCC-t-t]-protocol for ]-protocol for GT(n).GT(n).

Then there is a [t,nThen there is a [t,n1/t1/tCC-t-t,n,n1/t1/tCC-t-t]-protocol for ]-protocol for PPmm(GT(n’)) for m=n(GT(n’)) for m=n1/t1/t, n’=n, n’=n(t-1)/t(t-1)/t

From the round Elimination Lemma, there From the round Elimination Lemma, there is a [t-1, nis a [t-1, n1/t1/tCC-(t-1)-(t-1),n,n1/t1/tCC-(t-1)-(t-1)]-protocol for ]-protocol for GT(n’). Contradiction.GT(n’). Contradiction.

GT(n) does not have aGT(n) does not have a[t,n[t,n1/t1/tCC-t-t,n,n1/t1/tCC-t-t]-protocol]-protocol

[t,n[t,n1/t1/tCC-t-t,n,n1/t1/tCC-t-t]-protocol for GT(n)]-protocol for GT(n)

[t,n[t,n1/t1/tCC-t-t,n,n1/t1/tCC-t-t]-protocol for P]-protocol for Pmm(GT(n’)) for m=n(GT(n’)) for m=n1/t1/t, , n’=nn’=n(t-1)/t(t-1)/t

Alice constructs a n-bit integer x’: She Alice constructs a n-bit integer x’: She concatenates xconcatenates x11…x…xmm

Bob constructs a n-bit integer y’: He Bob constructs a n-bit integer y’: He concatenates xconcatenates x11…x…xi-1i-1 then y and the rest is 1s then y and the rest is 1s

We get x’>y’ We get x’>y’ x xii>y>y

THE ENDTHE END

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