Lecture #20. n ( x p i )1/p = max

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1 COMPSCI 632: Approximatio Algorithms November 8, 2017 Lecturer: Debmalya Paigrahi Lecture #20 Scribe: Yua Deg 1 Overview Today, we cotiue to discuss about metric embeddigs techique. Specifically, we apply metric embeddigs techique to solve the sparsest cut problem. 2 Embeddig to l p orm l p orm of a vector x is defied by x p = ( i=1 x p i )1/p. I this sectio, we discuss embeddig from ay metric space to R with l p orm. 2.1 l orm Let s first loo at l orm, where x = lim ( p i=1 x p i )1/p = max i x i. We say a embeddig is isometric if the distortio of the embeddig is 1. Theorem 1. l orm is uiversal, i.e., give ay metric space, there exists a isometric embeddig to R ( could be arbitrary) with l orm. Proof. Assume there are poits i the origial metric space M, labelled from 1 to. Let d(i, j) be the distace betwee the i-th poit ad j-th poit i the metric space M. Defie the embeddig fuctio f : [] R to be [ f (i)] = d(i,) The, with l orm, we have f (i) f ( j) = max f (i) f ( j) = max d(i,) d( j,) d(i, j) (Triagular iequality is applied i the last iequality.) Moreover, whe we choose the idex to be either i or j, (say i) we have f i (i) f i ( j) = d(i,i) d(i, j) = d(i, j) sice d(i,i) = 0. Therefore, we ca coclude that f (i) f ( j) = d(i, j) Exercise 1. Show that Euclidea (l 2 ) orm is ot uiversal with the followig star graph: V = {1,2,3} {m} with distace d(i,m) = 1 for all i {1,2,3} ad d(i, j) = 2 for all i j ad i, j {1,2,3}. #20-1

2 2.2 l 1 orm For l 1 orm, we have the followig theorem, which we will use without provig for the sparsest cut problem. Theorem 2 ([Bou85]). Ay metric space o poits ca be determiisticly embedded ito a l 1 orm space with O(log 2 ) dimesio ad distortio 4log. 3 Sparsest cut Let S be the collectio of edges i the cut (S,V \ S): S = {(i, j) E i S, j V \ S} ad deote the capacity of a edge (i, j) E be cap i j ad the capacity of a cut by cap( S) = (i, j) S cap i j Cosider a graph G = (V,E). The sparsity of a cut (S,V \ S) equals ψ(s) = cap( S) mi( S, V \ S ) I the sparsest cut problem, the objective is to fid a cut with miimum sparsity: 3.1 Relate to Flux The flux of a cut G is defied by Notice that for each choice of S V, φ(g) = mi ψ(s) flux(g) = mi cap( S) S V \ S φ flux = S V \ S mi( S, V \ S ) = 1 max( S, V \ S ) [1/2,1] Therefore, if we ca get a α approximatio for flux, we ca obtai a 2α approximatio of φ. 3.2 Demad Let s rewrite the flux fuctio, flux(g) = mi (i, j) S cap i j (i, j) S (V \S) 1 We ca view the costat 1 i the above formula as demad, which idicates that for each pair (i, j) S V \ S, we eed to push a oe-uit flow from i to j. The algorithm we are goig to preset ca wor if we replace the costat 1 by a demad fuctio: dem : V V R. The objective fuctio becomes, f (G) = mi (i, j) S cap i j (i, j) S (V \S) dem i j #20-2

3 3.3 Cut Metric Elemetary Cut Metric. A elemetary cut metric is a metric defied by a cut (S,V \ S) such that the distace d i j betwee i ad j is 1 if ad oly if i ad j are separated by the cut; otherwise, d i j = 0: d i j = 1 iff {i, j} S = 1 With elemetary cut metric, we ca chage the search space from fidig a cut to fidig a elemetary cut metric ad rewrite the objective fuctio as follows: f (G) = mi d (i, j) S cap i j d i j (i, j) S (V \S) dem i j d i j (geeral) Cut Metric. A geeral cut metric is a liear combiatio of some elemetary cut metrics 3.4 LP formulatio d i j = y S S: {i, j} S =1 First otice that the value of the objective is the same up to scalig. Therefore, without loss of geerality, we ca costrai that dem i j d i j = 1 (i, j) S (V \S) ad tur the objective to be mi (i, j) S cap i j d i j. The remaiig costraits mae sure that d comes from a metric space. The etire program is as follows: mi {i, j} S =1 cap i j d i j s.t {i, j} S =1 dem i j d i j = 1 d ii = 0 d i j = d ji d i j + d j di d is a elemetary cut metric i i, j i, j, We chage S to S (V \ S) i the above LP by lettig cap i j = 0 if (i, j) S (V \ S) but (i, j) E ad also rewrite (i, j) S (V \ S) by {i, j} S = 1 for simlicity. Fially, i order to obtai a LP, we drop the last costrait so that we compute the best metric istead of the best elemetary cut metric. 3.5 Aalysis Give the solutio metric d from LP, we first apply Theorem 2 to embed it to metric d l 1 i R log2 with l 1 orm. I the ext step, we tur the d l 1 to a (geeral) cut metric d gc, which we will show that this embeddig is isometric. Fially, we extract a elemetary cut metric d ec from d gc to obtai our solutio. Claim 1. Embeddig d l 1 to d gc is isometric. #20-3

4 Proof. Let s first cosider the case whe d l 1 is i oe dimesio. Therefore, all the poits are located o a lie. Without loss of geerality, assume these poits to be x 1 < x 2 < x. For each 1 i <, we defie a cut betwee x i ad x i+1 such that S i = {1,,i} ad let y Si = x i+1 x i. The, for ay pair (i, j), we have d l 1 i j = y Si = x +1 x = x j x i. i < j i < j Therefore, embeddig d l 1 to d gc is isometric whe d l 1 is i oe dimesio. Notice that l 1 distace betwee x i ad x j is x i x j 1 = (x i ) (x j ) Thus, we ca apply the argumet for oe dimesio case to each dimesio separately to show that the embeddig is isometric eve whe d l 1 is i a higher dimesio space. Recall that after applyig Theorem 2, d l 1 is i O(log 2 ) dimesio. Accordig to the proof of Claim 1, the cut metric d gc is a liear combiatio of O(log 2 ) elemetary cut metrics (deote the set of these cut by EC). Therefore, we ca fid the elemetary cut metric d ec with miimum objective value i poly time. It remais to show that the elemetary cut metric we obtai is a good solutio. {i, j} S =1 cap i j mi S EC {i, j} S =1 dem i j y S {i, j} S =1 cap i j = mi S EC y S {i, j} S =1 cap i j S EC y S {i, j} S =1 dem i j S EC y S {i, j} S =1 dem i j = (i, j) cap i j S EC, {i, j} S =1 y S = l (i, j) cap i j d 1 i j (i, j) dem i j S EC, {i, j} S =1 y S (i, j) dem i j d l 1 (i, j) cap i j 4log d i j (i, j) dem i j d i j 4logLP 4logOPT i j The first step is just to multiply y S o both the deomiator ad the omiator, which still eeps the fuctio value the same. The secod step applies the followig claim Claim 2. If a i > 0 ad b i > 0 for all i, we have a i mi i a i. i b i i b i The third step is a chage of order of summatio while the fourth step uses the fact that S EC, {i, j} S =1 y S = d gc i j ad Claim 1. I the fourth step, we apply Theorem 2 with di j dl 1 i j 4log di j. 4 Summary I this lecture, we discuss embeddig to l p orm ad use the embeddig to l 1 orm to desig a approximatio algorithm for the sparsest cut problem [LLR95]. #20-4

5 Refereces [Bou85] Jea Bourgai. O lipschitz embeddig of fiite metric spaces i hilbert space. Israel Joural of Mathematics, 52(1):46 52, [LLR95] Natha Liial, Era Lodo, ad Yuri Rabiovich. The geometry of graphs ad some of its algorithmic applicatios. Combiatorica, 15(2): , #20-5