Download presentation

Presentation is loading. Please wait.

Published byAlisha Chace Modified over 3 years ago

1
Trees and Markov convexity James R. Lee Institute for Advanced Study [ with Assaf Naor and Yuval Peres ] RdRd x y

2
Distortion: Smallest number C ¸ 1 such that: the Euclidean distortion problem Given a metric space (X,d), determine how well X embeds into a Euclidean space. Why study this kind of geometry (in CS)? - Applicability of low-distortion Euclidean embeddings - Understanding semi-definite programs - Optimization, harmonic analysis, hardness of approximation, cuts and flows, Markov chains, expansion, randomness… Euclidean embedding: An injective map f : X ! R k (or L 2 )

3
Distortion: Smallest number C ¸ 1 such that: the Euclidean distortion problem Given a metric space (X,d), determine how well X embeds into a Euclidean space. Euclidean embedding: An injective map f : X ! R k (or L 2 ) Actually, then the distortion is A ¢ B.

4
the problem for trees One of the simplest families of metric spaces are the tree metrics. graph-theoretic tree T = (V,E) + edge lengths len : E ! R + len(e) x y d(x,y) = length of shortest geodesic

5
the problem for trees [Bourgain 86]: The complete binary tree B k of height k has Euclidean distortion [Matousek 99]: Every n-point tree metric embeds with distortion at most [Gupta-Krauthgamer-L 03]: A tree metric T embeds with constant distortion into a finite-dimensional Euclidean space if and only if T is doubling. When does does a tree embed into some Euclidean space (arbitrary dimension) with bounded distortion?

6
the problem for trees [Bourgain 86]: The complete binary tree B k of height k has Euclidean distortion [Matousek 99]: Every n-point tree metric embeds with distortion at most [Gupta-Krauthgamer-L 03]: A tree metric T embeds with constant distortion into a finite-dimensional Euclidean space if and only if T is doubling. e1e1 e2e2 e3e3

7
why don’t trees embed in Hilbert space? O NE A NSWER : (EQUILATERAL) FORKS If both of these paths of length 2 are embedded isometrically in a Euclidean space, then A and B must conincide! A B Quantitative version holds: If both 2-paths are embedded with distortion 1 + , then

8
uniform convexity A B W Z paralellogram identity: for any pair of vectors a,b 2 R 2, f(W)=0 a=f(A) b=f(B) 4 ± O( ) O( )

9
forks in complete binary trees RdRd Ramsey style proof: If B k is embedded into L 2 with distortion then there exists some almost-isometric fork. [Matousek] C ONTRADICTION!

10
on forks Natural question: Are forks the only obstruction? The problem isn’t forking; it’s forking, and forking, and forking… T HEOREM: For a tree metric T, the following conditions are equivalent. -- T embeds in a Euclidean space with bounded distortion -- The family of complete binary trees {B k } do not embed into T with bounded distortion. In other words, a tree embeds into a Euclidean space if and only if it does not “contain” arbitrarily large binary trees!

11
quantitative version T HEOREM: Let c 2 (T) be a tree’s Euclidean distortion, then (up to constants), D EFINITION: For a metric space (X,d), i.e. the height of the largest complete binary tree that embeds into T with distortion at most 2. Let’s prove this…

12
monotone edge colorings If T=(V,E) is a (rooted) tree, then an edge-coloring of T is a map The coloring is monotone if every color class is a monontone path in T (monotone path = continguous subset of root-leaf path) A coloring is -good if, for every u,v 2 T, at least an -fraction of the u-v path is monochromatic. u v

13
monotone edge colorings If T=(V,E) is a (rooted) tree, then an edge-coloring of T is a map The coloring is monotone if every color class is a monontone path in T (monotone path = continguous subset of root-leaf path) A coloring is -good if, for every u,v 2 T, at least an -fraction of the u-v path is monochromatic. u v

14
good colorings ) good embeddings We associate to every color class j 2 {1, 2, …, C}, a unit vector j 2 R C. Given a vertex x whose path from the root uses edges e 1, e 2, …, e k, we define our embedding f : T ! R C by e1e1 e2e2 e3e3 e4e4 x f(x) = [len(e 1 )+len(e 2 )] 1 + len(e 3 ) 2 + len(e 4 ) 3

15
good colorings ) good embeddings We associate to every color class j 2 {1, 2, …, C}, a unit vector j 2 R C. Given a vertex x whose path from the root uses edges e 1, e 2, …, e k, we define our embedding f : T ! R C by e1e1 e2e2 e3e3 e4e4 x Claim: f is non-expansive, i.e. (triangle inequality)

16
good colorings ) good embeddings We associate to every color class j 2 {1, 2, …, C}, a unit vector j 2 R C. Given a vertex x whose path from the root uses edges e 1, e 2, …, e k, we define our embedding f : T ! R C by e1e1 e2e2 e3e3 e4e4 x Claim: For every x,y 2 T, lca(x,y) y x Monotonicity ) disjoint colors lca(x,y) x

17
good colorings ) good embeddings L EMMA: If T admits an -good coloring, then the Euclidean distortion of T is at most 2/ . The hard part comes next… T HEOREM: If * is the biggest for which T admits an -good coloring, then

18
good colorings ) good embeddings L EMMA: If T admits an -good coloring, then the Euclidean distortion of T is at most 2/ . The hard part comes next… T HEOREM: If * is the biggest for which T admits an -good coloring, then

19
good colorings ) good embeddings L EMMA: If T admits an -good coloring, then the Euclidean distortion of T is at most 2/ . The hard part comes next… T HEOREM: If * is the biggest for which T admits an -good coloring, then C OROLLARY: [ stronger embedding technique gives ]

20
good colorings ) good embeddings L EMMA: If T admits an -good coloring, then the Euclidean distortion of T is at most 2/ . The hard part comes next… T HEOREM: If * is the biggest for which T admits an -good coloring, then Proof outline:1. Give some procedure for coloring the edges of T. 2. If the procedure fails to construct an -good coloring, find a complete binary tree of height O( 1 / ) embedded inside T.

21
constructing a good coloring First, we define a family of trees {M k }: These are just {B k } with an extra “incoming” edge… M k = BkBk M0M0 M1M1 M2M2 Given a tree T, we say that T admits a copy of M k at scale j if… 1. M k embeds into T with distortion at most 4. 2. The root of M k maps to the root of T. 3. The edges of M k have length ¼ 4 j.

22
constructing a good coloring Now, suppose we have a “scale selector” function g : T ! Z which assigns a “scale” to every vertex in T. We produce a coloring as follows… T1T1 T2T2 T3T3 T4T4 v How to continue a coloring: Continue toward the T i which admits the largest copy of M k at scale g(v)… (break ties arbitrarily) j = g(v) 4j4j

23
constructing a good coloring Suppose we failed to produce an -good coloring… u v D · D [ assume ¼ D ¼ 4 j ] Assume that g(w) = j for every breakpoint w on the u-v path. w In this manner, we construct a complete binary tree of height ¼ 1 / inside T. But what about our assumptions on g(w)?

24
constructing a good coloring Suppose we failed to produce an -good coloring… u v D Can define g so that every sufficiently dense set of breakpoints contains a large subset with the “right” g-values using hierarchical nets. j j+2 j+1 j+3 Points with g(w) ¸ k form a 4 k -net. At most a ¼ fraction of the 4 k -net points have label higher than k (geometric sum). Now reconstruct a complete binary tree of height 1 / ) just using the green nodes.

25
cantor trees So we have these bounds: this upper bound is tight There exists a family of trees {C k } for which [ so the “branching” lower bound only gives ]

26
cantor trees Spherically symmetric trees (SST): Every path with marked vertices yields a binary SST.

27
cantor trees The Cantor trees are binary SSTs based on inductively defined paths… P 0 = P2P2 P2P2 P k+1 = PkPk PkPk length 2 k+1 P 1 = P 2 = P 3 = len(P k ) = 2 len(P k-1 ) + 2 k = k ¢ 2 k log log |C k | ~ k br(C k ) ~ k Claim: c 2 (C k ) ~ √k

28
strong edge colorings A monotone edge coloring is -strong if, for every u,v 2 T, at least half of the u-v path is colored by classes of length at least ¢ d(u,v). T HEOREM: If * is the biggest for which T admits a -strong coloring, then Proof sketch: 1. Show that -strong colorings yield good embeddings. 2. Give some procedure to construct a monotone coloring. 3. If the coloring fails to be -strong, show that T must contain a Cantor-like subtree. 4. Show that every Cantor-like subtree requires large distortion to embed in a Euclidean space.

29
cantor trees The Cantor trees do not have (good) strong colorings… P 0 = P2P2 P2P2 P k+1 = PkPk PkPk length 2 k+1 P 1 = P 2 = P 3 = 1/k 1/2k 1/4k ) best coloring is 2 -k/2 strong!

30
Markov convexity Idea: Look at Markov chains wandering in a Euclidean space; must satisfy special properties, e.g. symmetric random walk on Z, Z 2, … t=0t=0 t=kt=k

31
Markov convexity Idea: Look at Markov chains wandering in a Euclidean space; must satisfy special properties, e.g. symmetric random walk on Z, Z 2, … A metric space (M,d) is Markov 2-convex if, for every Markov chain {X t } taking values in M, and every number m 2 N, we have for some constant C ¸ 0.

32
Markov convexity T HEOREM: Every Euclidean space is Markov 2-convex. (with some universal constant C) A metric space (M,d) is Markov 2-convex if, for every Markov chain {X t } taking values in M, and every number m 2 N, we have for some constant C ¸ 0.

33
discrepancy with Euclidean space ) distortion ~ √m ~ √ log k lower bounds from Markov convexity If {X t } is the downward random walk on B k, then… 2m2m 1)1) m¢2m)m¢2m) (with the leaves as absorbing states)

34
lower bounds from Markov convexity P2P2 P2P2 P k+1 = PkPk PkPk length 2 k+1 P 3 = Let {X t } be the downward random walk on C k.

35
lower bounds from Markov convexity P 3 = Let {X t } be the downward random walk on C k. Key fact: At least a j/k fraction of P k is covered by segments whose length is at most 2 j.

36
conclusion M AIN T HEOREM: For every tree T, we have and -- Markov convexity is a notion for general metric spaces (X,d). Can we relate non-trivial Markov convexity to the non-containment of arbitrarily large complete binary trees? -- What about other Markov-style lower bounds for Hilbert space? -- Can we use reversible Markov chains to construct NEG metrics? -- Are these techniques useful for studying the bandwidth of trees? Q UESTIONS?

Similar presentations

OK

Topics in Algorithms 2007 Ramesh Hariharan. Tree Embeddings.

Topics in Algorithms 2007 Ramesh Hariharan. Tree Embeddings.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on advertisement Ppt on data collection methods for organizational diagnosis Ppt on andhra pradesh tourism Ppt on preservation of public property search Ppt on series and parallel circuits worksheets Ppt on ipad 2 Ppt on regional trade agreements definition Ppt on acid-base indicators lab Ppt on teachers day free download Ppt on recent earthquake in india