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A sublinear Time Approximation Scheme for Clustering in Metric Spaces Author: Piotr Indyk IEEE FOCS 1999

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組員名單 R90922058 李秉憲 R90725031 陳柏安 R90725045 張緒遠 R90725052 鄭安巽

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outline Introduction Preliminaries The Algorithms The analysis of BC The analysis of UC Sublinear time algorithm

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Introduction k clustering problem: Input:Given a weighted graph G = (X,d) on N vertices. Output:Partition X into k sets S 1 …S k such that the value of is minimized.

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Introduction This problem is NP-complete (for k >=2) and can’t be approximated up to any constant. A standard way to reduce the complexity of clustering problems is to assume that the weight function d is a metric.

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Introduction Facts: –Guttman-Beck showed a 2-approx algorithm –Vega and Kenyon gave a PTAS for metric max cut. –Unfortunately, it does not imply a PTAS for the 2-clustering problem.

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Introduction The result of this paper focus on the case when the de la Vega-Kenyon PTAS doesn't work. If done correctly, this procedure yields a (1+ε) approximate solution.

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Preliminaries Let (X,d) be a metric space.For any two sets A,B X, we define We use d(A) to denote d(A,A)/2 We also use d(u,B) or d(B,u)

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Preliminaries We define Observe, that is a metric.

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Preliminaries For sets A,B of equal cardinality, we define and Notice that both d M and satisfy metric properties.

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Preliminaries For any α [0,1], u and A X, we define d α (u,A) to be the sum of the α|A| smallest values of {d(u,a)|a A}. We also define The algorithms which we give in this paper are randomized and producing an (1+ε)-approximate solutions with high probability.

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The Algorithms The result is obtained by running three algorithms in parallel: MAXCUT,BC,UC We will assume that |S 1 |=m and |S 2 |=n are given to us, as otherwise we use all N possible combinations.

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The Algorithms MAXCUT: –This is the algorithms of another paper for (1-ε)- approximate MAX-CUT –The algorithm is useful if the cut/clustering ratio is smaller than a constant value c. –Thus we need another algorithm for the case when the cut/clustering ratio exceeds c.

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The Algorithms BC: –Balance ratio: –If the cut/clustering ratio is greater than c and the balance ratio is smaller than ρ,we will run BC

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The Algorithms BC: –Uniformly chooses set T of t =O(ρlogn) points –Guesses and, s.t. |T1| = |T2| = λ = O(logn) –It checks for each point u X-T 1 -T 2 if Back

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The Algorithms BC(contd.): –If the above inequality holds, then u is added to R 1 ; otherwise we add it to R 2. –The pair is returned as a solution.

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The Algorithms T T1T1 T2T2 X S1S1 S2S2

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UC: –Use this algorithm when b(S 1,S 2 ) > ρ,assume |S 1 | > ρ|S 2 | –Obtain a set T of λrandom points from S 1. –Sort all points u X-T by (in ascending order) –The first |S 1 |-λ points from the list are added to R 1, the remaining points are added to R 2. –Output (R 1,R 2 ) Back

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The Algorithms X T Sort u by d 1-α (u,T) S1S1 S2S2

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The Algorithms MAXCUT B(S 1,S 2 )<ρ BU UC Solution Yes No Cut/clustering <= c Cut/clustering > c

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The Algorithms Q1:polynomial? –Yes. (BC, UC)BCUC Q2:feasible solution? –Yes.

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The analysis of BC Relating the outcome of the (randomized) comparison of d(u,T 1 ) and d(u,T 2 ) to a certain (deterministic) property of d Lemma1 Consider any u S 1. If for every set S 2 ’ which is a subset of S 2 such that |S 2 ’|>=(1-2α)|S 2 | we have d(u,S 2 ’)>=d(u,S 1 ), then with high probability we have n·d 1-α (u,T 1 )>m·d 1-α (u,T 2 )

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The analysis of BC(cont’d) Proof: –Without loss of generality we can consider S 2 ’ which contains smallest (1-2α)n elements from S 2 –Moreover, we can assume d(u,S 2 ’)=d(u,S 1 )=1 –Finally, we will assume that the largest 2αn elements of S 2 are all equal –For λ large enough we have a significant gap in the expected values of d 1-α (u,T 1 ) and d 1-α (u,T 2 ) E[d 1-α (u,T 1 )]/(1-α)|T 1 | <= E[d 1-0 (u,S 1 )]/(1-0)|S 1 | E[d 1-α (u,T 2 )]/(1-α)|T 2 | >= E[d 1-0 (u,S 2 ’)]/(1-0)|S 2 ’|

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The analysis of BC(cont’d) (m/λ)E[d 1-α (u,T 1 )] <= 1-α <= 1 (n/λ)E[d 1-α (u,T 2 )] >= (1-α)/(1-2α) >= 1+α/(1-2α) >= 1+α/2 –(m/λ)E[d 1-α (u,T 1 )] <= 1 (n/λ)E[d 1-α (u,T 2 )] >= 1+α/2 –We want to convert the expectation bounds into bounds holding with high probability –Applying standard tail inequalities, we obtain that n·d 1-α (u,T 2 ) >= m·d 1-α (u,T 1 ) with high probability if λ = Ω(log n/α ) 4

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The analysis of BC(cont’d) We will upper bound the additional cost incurred by assigning u S 1 to R 2 ; the opposite case can be handled in the same way W(C) <= (1+ε)OPT W(C) – OPT <= ε·OPT From the above Lemma, we can assume that for every u S 1 which has been included in R 2 (i.e. such that n·d 1-α (u,T 2 ) <= m·d 1-α (u,T 1 ) there exists a set S 2 of cardinality (1-2α)|S 2 | such that d(u,S 2 ) < d(u,S 1 ) (1) Thus we need only to bound d(u,S 2 -S 2 ) u u u

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The analysis of BC(cont’d) d(S 1 ’)+d(S 2 ’) d(S 1 )+d(S 2 ) d(S 1 -U)+d(S 1 -U,V)+d(V)+ d(S 1 -U)+d(S 1 -U,U)+d(U)+ d(S 2 -V)+d(S 2 -V,U)+d(U) d(S 2 -V)+d(S 2 -V,V)+d(V) d(S 1 -U,V)+d(S 2 -V,U) d(S 1 -U,U)+d(S 2 -V,V) d(S 1,V)-d(U,V)+d(S 2,U)-d(V,U) d(S 1,U)-d(U)+d(S 2,V)-d(V) d(u,S 2 )+d(u,S 2 -S 2 )-d(u,V) d(u,S 1 )-d(u,U) ∵ d(u,S 2 ) < d(u,S 1 ) ∴ d(u,S 2 -S 2 )-d(u,V)-(d(u,S 1 )-d(u,S 2 ))+d(u,U) <= d(u,S 2 -S 2 ) uu u uu VS1-US1-US1-US1-UUUVS 2 -V S1’S1’S1S1 S2’S2’S2S2

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The analysis of BC(cont’d) We will be only interested in u’s such that d(u,S 2 ) >= d(u,S 1 )(1+ε) (2) as otherwise the difference in the cost can be easily bounded (i.e. d(u,S 2 )-d(u,S 1 ) < d(u,S 2 )-d(u,S 2 ) < ε·d(u,S 1 )) From (1) and (2) we obtain that d(u,S 2 )-d(u,S 2 ) >= ε·d(u,S 1 ) >= ε·d(u,S 2 ) which can be rewritten as đ(u,S 2 -S 2 ) >= ε((1-2 α) /2α)·đ(u,S 2 ) u uu uu

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The analysis of BC(cont’d) By triangle inequality we have đ(,S 2 - ) >= đ(u,S 2 - )-đ(u,S 2 ) >= (1-2 α/ ε(1-2 α)) đ(u,S 2 - ) (3) The above give a bound for đ(,S 2 - ). In the following we show that the number of such us is also not very large if (as we assume) d(S 1,S 2 ) >= c(d(S 1 )+d(S 2 )) Firstly, observe that đ(S 1,S 2 ) <= đ(u,S 1 )+đ(u,S 2 ) u S 2 -S 2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u u S1S1 S2S2 S2S2 u

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The analysis of BC(cont’d) ∵ đ(u,S 2 )= d(u,S 2 )/n=d(u, )/n+d(u,S 2 - )/n =(1-2 α) đ(u, )+2 α đ(u,S 2 - ) đ(S 1,S 2 ) <= đ(u,S 1 )+(1-2 α) đ(u, )+2 α đ(u,S 2 - ) <= đ(u,S 1 )+(1-2 α) đ(u, )+ 2α(đ(u, )+đ(,S 2 - )) = đ(u,S 1 )+đ(u, )+2αđ(,S 2 - ) We can rewrite it as: S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u u S 2 -S 2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u

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The analysis of BC(cont’d) d(S 1,S 2 )/|S 1 ||S 2 | <= d(u,S 1 )/|S 1 |+d(u, )/(1-2 α)| S 2 | + 2αd(,S 2 - )/(1-2 α) 2α|S 2 ||S 2 | d(S 1,S 2 ) <= |S 2 |d(u,S 1 )+|S 1 |d(u, )/(1-2 α)+ |S 1 |d(,S 2 - )/(1-2 α) |S 2 | <= |S 2 |d(u,S 1 )+|S 1 |d(u, )/(1-2 α)+ ρd(S 2 )/(1-2α) Therefore c(d(S 1 )+d(S 2 )) <= d(S 1,S 2 ) <= |S 2 |d(u,S 1 )+|S 1 |d(u, )/(1-2 α)+ ρd(S 2 )/(1-2α) S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u

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The analysis of BC(cont’d) Alternatively (c-ρ/(1-2 α ))(d(S 1 )+d(S 2 )) <= c(d(S 1 )+d(S 2 ))- ρd(S 2 )/(1-2 α) <= |S 2 |d(u,S 1 )+|S 1 |d(u, )/(1-2 α) <= ( |S 2 |+|S 1 |/(1-2 α)) d(u,S 1 ) ( ∵ d(u, ) <= d(u,S 1 )) The upper bound for the number of u satisfying the above inequality can be obtained as follows. Assume that this number is equal to γ|S 1 | S2S2 u S2S2 u

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The analysis of BC(cont’d) Σ u U d(u,S 1 ) <= d(S 1 ). By plugging in the lower bound for d(u,S 1 ) we get Σ u U (c-ρ/(1-2 α ))(d(S 1 )+d(S 2 ))/(|S 2 |+|S 1 |/(1-2 α)) <= d(S 1 ) γ|S 1 |(c-ρ/(1-2 α ))(d(S 1 )+d(S 2 ))/(|S 2 |+|S 1 |/(1-2 α)) <= d(S 1 ) γ(c-ρ/(1-2 α ))(d(S 1 )+d(S 2 ))/(1/ρ+1/(1-2 α)) <= d(S 1 ) γ(c-ρ/(1-2 α ))/(1/ρ+1/(1-2 α)) <= 1 Therefore γ <= (1/ρ+1/(1-2 α)) /(c-ρ/(1-2 α ))

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The analysis of BC(cont’d) Denote the set of u’s as above by U. We can bound the total cost difference by Σ u U d(u,S 2 - ) <= Σ u U 2 αn đ(u,S 2 - ) <= Σ u U 2 αn đ(,S 2 - )/(1-2 α/ ε(1-2 α)) ( ∵ (3)) = γ|S 1 |·2 αn đ(,S 2 - )/(1-2 α/ ε(1-2 α)) = γ|S 1 |·2 αnd (,S 2 - )/[(1-2 α/ ε(1-2 α)) 2 α |S 2 |(1-2 α) |S 2 | ] <= γρd(S 2 )/[(1-2 α/ ε(1-2 α)) (1-2 α)] = A d(S 2 ) S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u S2S2 u

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The analysis of BC(cont’d) The factor A becomes smaller than ε when we set c = Ω(ρ /ε) and α = O(ε), for sufficient constants 2

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The analysis of UC

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Lemma 2

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notes For every included by mistake to R2 there is included to R1 Let U denote the set of mistaken u’s and let V denotes the set of mistaken v’s. We will bound the differences d(V,S 1 )-d(U,S 1 ) and d(U,S 2 )-d(V,S 2 ) It is sufficient to bound

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fact1 To bound right hand side therefore

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We also use the fact therefore thus

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In this way we bounded the first component by setting we make the value 1/F-1 smaller than ε

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To Bound the second part

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Observe that by setting we make B smaller than ε

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Sublinear time algorithm We improve the running time of the above algorithm to

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Sublinear time algorithm The running time of UC is bounded by the time needed sampling t points from large cluster.We improve UC by using random sampling, we can perform sampling in time roughly We improve MAXCUT running time of [2] by using the techniques of[1].

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Sublinear time algorithm The main time bottleneck is the time needed for exhaustive partitioning of the set T in BC. We divide T into C 1 and C 1, choosing T 1 and T 2 from C 1 C 1, from lemma below,we show they are good enough for our algorithm.

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Sublinear time algorithm Lemma3: Let (S 1, S 2 ) and (S 1 ’, S 2 ’ ) be two partitions of the metric space over S.There exists a constant B such that for any A and any β<1/B if d(S 1 )+d(S 2 ) <= βA/Bd(S 1, S 2 ) and d(S 1 ’)+d(S 2 ’ ) <= A/B(d(S 1 )+d(S 2 )), then (S 1, S 2 ) and (S 1 ’, S 2 ’ ) differ on at most βn points.

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Sublinear time algorithm Select a sample R of r points.It can be split into R 1 and R 2 such that d(R 1, R 2 )/d(R 1 )+d(R 2 ) and d(S 1, S 2 ) /d(S 1 )+d(S 2 ) are comparable. Find T 1 ’ R 1 and T 2 ’ R 2,such that | T 1 ’ |= | T 2 ’ |=t and | T 1 ’ - S 1 |= | T 2 ’ - S 2 | <= β t It turns out T 1 ’ and T 2 ’ are almost as good as T 1 and T 2 obtained by exhaustive search.

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Sublinear time algorithm For any two equal size sets A’ and A,if max(| A’ |-| A |,| A |-| A’ |)<= β | A | then for any α and u, d 1- α- β (u, A) <= d 1- α (u, A’ ) <= d 1- α+β (u, A) So we can replace Lemma 1 nd 1- α (u, T 2 ) <= m d 1- α (u, T 1 ) by nd 1- 3α/2 (u, T 2 ) <= m d 1- α/2 (u, T 1 )

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