princeton u. sp’02 cos 598B: algorithms and complexityLecture 2: Geometric Embeddings (continued)Lecturer: Sanjeev Arora Scribe:Michael Dinitz1Focusonthe1normIn the last lecture we defined metric spaces, normed spaces, and considered the distortion resultingfrom certain embeddings. In particular, we proved that l1norms cannot always be embeddedisometrically into l2by considering a specific four-point l1norm and showing that it requires atleast√2 distortion.Today’s lecture further explores the 1norm. We see a couple of interesting examples of 1spaces. We try to understand the distortion required to embed 1into 2.Wealsoseethatthis apparently simple norm (“Manhattan distance”) is computationally very interesting. We willexplore this further in future lectures.2 Lowerbound for embedding into 2.The simplest example of an n-point 1metric is the k-dimensional hypercube {−1, 1}k, assumingn =2k.Herethe1distance between two points is 2 × number of coordinates they differ on.-1, 1, -1-1, -1, 1-1, 1, 11, 1, 11, -1, -11, 1, -11, -1, 1-1, -1, -1ABCDEFGHFigure 1: The hypercube in three dimensionsToday we show that this simple example of 1requires large distortion for embedding into 2.Theorem 1 (Enflo 1969)Ev ery embedding of the hypercube {−1, 1}kin to l2has distortion at least√k. (Thus denoting thenumber of points by n =2k, the distortion is√log n.)Proof: Recall the proof technique that we used previously: we come up with two weight functions,w1: X × X → R+∪{0} and w2: X ×X → R+∪{0}, such that for all embeddings f,w1(x, y)d(x, y)2w2(x, y)d(x, y)2(1)12is different by a factor of k fromw1(x, y)f(x) − f(y)2w2(x, y)f(x) − f(y)2(2)Here w1will put a weight 1 on all diametrically opposite pairs, and w2will put a weight 1 onall edges (i.e., adjacent pairs). Then (1) has value2k/2 · k22k−1· k · 1= k.We show by induction on k that for every embedding f, the value of (2) is at most 1, which willprove the theorem.The base case k = 2 is exactly the problem that we considered last time. Assuming truthfor k − 1weproveitfork. A dimension k hypercube can be thought of as two k − 1dimensionhypercubes that have the corresponding points attached. By the quadrilateral inequality, the sumof the squares of the diagonals (where a diagonal refers to the distance between a point i and −i)of the k dimensional hypercube is less than or equal to the sum of the squares of the diagonals ofthe k − 1 dimensional hypercube plus the sum of the squares of the edges between correspondingpoints of the two k − 1 dimensional hypercube. In Figure 1, k = 3 and the inequality we are usingis AG2+ CE2≤ AC2+ EG2+ AE2+ GC2.NotethatAC, EG are diagonals of k −1-dimensionalhypercubes. So equation (2) becomes (in shorthand)diagonals from k−1+new adjacenciesadjacencies from k−1+new adjacencies(3)By the inductive hypothesis,diagonals from k−1≤adjacencies from k−1, so equation (2) is atmost 1. 23 Understanding the l1normWe saw in the last lecture that we can test a metric to see if it come from an l2space in polynomialtime. This is more difficult in l1, though, and is in fact NP-hard. The l1norm, while simple (anyonecan understand the concept of Manhattan distance), seems to actually be more complicated thanthe l2norm. While we will not give a formal proof of this, we will show some of the intuition behindit.3.1 Tree metricsOne class of metrics are the tree metrics, which are metrics that come from the shortest path metricon a weighted tree.Theorem 2Every tree metric embeds isometrically into l1Proof: We do this by using induction on the number of vertices of the tree. The base case, whenthere is only one vertex, is obvious. For the inductive step, we assume that all trees with fewer3that k vertices can be embedded isometrically into l1.LetT be a tree with k vertices, and let iand j be any two adjacent vertices with dijthe weight of the edge between them. If we remove theedge {i, j} then we have two smaller trees T1and T2with i ∈ T1and j ∈ T2, so by the inductivehypothesis each of them can be embedded into l1. They may require different dimensions, though,so say that T1is embedded in Rmand T2is embedded in Rkwith the l1norm. We assign eachvertex to an element of Rm+k+1.Letv be a vertex of T .Ifv is in T1, then its first m coordinatesare the coordinates assigned to v by the embedding of T1into Rm, the next coordinate is 0, andthe last k coordinates are the coordinates assigned to j by the embedding of T2into Rk. Similarly,if v is in T2, then the first m coordinates are the m coordinates assigned to i by the embedding ofT1into Rm, the next coordinate is dij,andthelastk coordinates are the coordinates assigned tov by the embedding of T2into Rk. For any two vertices u and v,iftheyarebothinT1or both inT2then the distance between them is clearly equal to the distance between them in the embeddingof their subtree, which we know is the same as in the tree metric. If u ∈ T1and v ∈ T2then thedistance between them in the l1norm is clearly equal to the distance from u to i plus the distancefrom i to j (which we know is dijby the way we assigned the vectors) plus the distance from j tov. Since each of the subtrees embedded isometrically, this is also the distance between them in thetree metric. 23.2 A characterization of 1:ConeofcutmetricsIn this section we think of an n-point metric space as a subset of R(n2), since the space can becompletely described by describing all n2pairwise distances.A convex cone in Rkis a subset S ⊆ Rkwhere (i) if x1,x2∈ S then λ1· x1+ λ · x2∈ S for allλ1,λ2≥ 0.Note that the set of n-point l1metrics form a convex cone. (If d1,d2are two finite 1metricson an n-point set, and if they are realizable in l, m dimensionions respectively then λ1d1+ λ2d2isrealizable in l + m dimensions, where the first l dimensions correspond to a copy of d1scaled by λ1and the last m dimensions correspond to a copy of d1scaled by λ2.)Definition 1 A cut metric is a subset (i.e. a cut) S ⊆ [n] with dS(i, j)=1if i and j are not onthesamesideofthecutanddS(i, j)=0if they are.Note that a cut metric is not an actual metric but a pseudo-metric, since dS(i, j)=0doesnotimply that i = j.Thecone of cut pseudo-metrics consists of any d ∈ R(n2)such that d is expressibleas d =S⊆[n]αSdS,whereαS≥ 0. Now we see that the two cones defined above are exactly thesame. (As a sanity check