A lower bound on dimension reduction for trees in \ell_1
aa r X i v : . [ m a t h . M G ] F e b A LOWER BOUND ON DIMENSION REDUCTION FORTREES IN ℓ JAMES R. LEE AND MOHAMMAD MOHARRAMI
Abstract.
There is a constant c > ε ∈ (0 , n > /ε , the following holds. Any mapping from the n -point starmetric into ℓ d with bi-Lipschitz distortion 1 + ε requires dimension d > c log nε log(1 /ε ) . Introduction
Consider an integer n >
1. The n -node star is the simple, undirectedgraph G n = ( V n , E n ) with | V n | = n , where one node has degree n − ρ n for the shortest-path metricon G n where each edge is equipped with a unit weight. We use ℓ d to denotethe space R d equipped with the ℓ norm. Our main theorem follows. Theorem 1.
There is a constant c > such that the following holds. Con-sider any ε ∈ (0 , ) and n > /ε . Suppose there exists a -Lipschitzmapping f : V n → ℓ d such that k f ( x ) − f ( y ) k > (1 − ε ) ρ n ( x, y ) for all x, y ∈ V n . Then, d > c log nε log(1 /ε ) . One can achieve such a mapping with d O (cid:16) log nε (cid:17) , thus the theoremis tight up to the factor of c/ log(1 /ε ). In general, de Mesmay and theauthors [11] proved that every n -point tree metric admits a distortion 1 + ε embedding into ℓ C ( ε ) log n where C ( ε ) O (( ε ) log ε ). For the special caseof complete trees where all internal nodes have the same degree (such as the n -star), they achieve C ( ε ) O ( ε ).We recall that given two metric spaces ( X, d X ) and ( Y, d Y ) and a map f : X → Y , one defines the Lipschitz constant of f by by k f k Lip = sup x = y ∈ X d Y ( f ( x ) , f ( y )) d X ( x, y ) . The bi-Lipschitz distortion of f is the quantity dist ( f ) = k f k Lip · k f − k Lip ,which is taken as infinite when f is not one-to-one. If there exists such amap f with distortion D , we say that X D -embeds into Y . A finite tree metric is a finite, graph-theoretic tree T = ( V, E ), whereevery edge is equipped with a positive length. The metric d T on V is givenby taking shortest paths. Since every finite tree metric embeds isometricallyinto ℓ , one can view the preceding statements as quantitative bounds onthe dimension required to achieve such an embedding with small distortion(instead of isometrically).Such questions have a rich history. Perhaps most famously, if X is an n -point subset of ℓ , then a result of Johnson and Lindenstrauss [9] statesthat X admits a (1 + ε )-embedding into ℓ d where d = O (cid:16) log nε (cid:17) . Alon [2]proved that this is tight up to a log(1 /ε ) factor: If X ⊆ ℓ n is an orthonormalbasis, then any D -embedding of X into ℓ d requires d > Ω(log n ) ε log(1 /ε ) .The situation for finite subsets of ℓ is quite a bit more delicate. Ta-lagrand [18], following earlier results of Bourgain-Lindenstrauss-Milman [5]and Schechtman [17], showed that every n -dimensional subspace X ⊆ ℓ (and, in particular, every n -point subset) admits a (1+ ε )-embedding into ℓ d ,with d O ( n log nε ). For n -point subsets, this was improved to d O ( n/ε )by Newman and Rabinovich [15], using the spectral sparsification techniquesof Batson, Spielman, and Srivastava [4].In contrast, Brinkman and Charikar [6] proved that there exist n -pointsubsets X ⊆ ℓ such that any D -embedding of X into ℓ d requires d > n Ω(1 /D ) (see also [12] for a simpler argument). Thus the exponential di-mension reduction achievable in the ℓ case cannot be matched for the ℓ norm. More recently, it has been show by Andoni, Charikar, Neiman, andNguyen [3] that there exist n -point subsets such that any (1 + ε )-embeddingrequires dimension at least n − O (1 / log( ε − )) . Regev [16] has given an elegantproof of both these lower bounds based on information theoretic arguments.Our proof takes some inspiration from Regev’s approach.We note that Theorem 1 has an analog in coding theory. Let U n = { e , e , . . . , e n } ⊆ ℓ . Then any (1 + ε )-embedding of U n into the Hammingcube { , } d requires d > Ω(log n ) ε log(1 /ε ) . This was proved in 1977 by McEliece,Rodemich, Rumsey, and Welch [14] using the Delsarte linear programmingbound [8]. The corresponding coding question concerns the maximum num-ber of points x , x , . . . ∈ { , } d which satisfy (1 − ε ) d/ k x i − x j k (1 + ε ) d/ i = j . Alon’s result for ℓ [2] yields this bound as a specialcase since k x − y k = p k x − y k when x, y ∈ { , } d .On the one hand, the lower bound of Theorem 1 is stronger since itapplies to the target space ℓ d and not simply { , } d . On the other hand, itis somewhat weaker since embedding U n corresponds to embedding only theleaves of the star graph G n , while our lower bound requires an embedding ofthe internal vertex as well. In fact, this is used in a fundamental and crucialway in our proof. Still, in Section 3, we prove the following somewhat weakerlower bound using simply the set U n . LOWER BOUND ON DIMENSION REDUCTION FOR TREES IN ℓ Theorem 2.
There is a constant c > such that for every ε ∈ (0 , , forall n sufficiently large, any (1 + ε ) -embedding of U n ⊆ ℓ n into ℓ d requires d > c log nε log ε . For the case of isometric embeddings (i.e., ε = 0), Alon and Pudl´ak [1]showed that if U n embeds isometrically in ℓ d , then d > Ω( n/ (log n )). Ourproof of Theorem 2 bears some similarity to their approach.Finally, we mention that if B h denotes the height- h complete binary tree(which has 2 h +1 − h > ε > B h admits a (1 + ε )-embedding into ℓ d with d O ( h /ε ). It was asked in [13] whether one could achieve d O ( h/ε )and this was resolved positively in [11]. From Theorem 1, one can deducethat this upper bound is asymptotically tight up to the familiar factor oflog(1 /ε ). This corollary is proved in Section 4. Corollary 3.
For any ε > and k > , the following holds. For h suffi-ciently large, any (1 + ε ) -embedding of the complete k -ary, height- h tree into ℓ d requires d > Ω( h log k ) ε log(1 /ε ) . proof of Theorem 1 We will first bound the number of “almost disjoint” probability measuresthat can be put on a finite set. Then we will translate this to a lower boundfor the dimension required for embedding the n -star into ℓ d with distortion1 + ε .Let X be a finite ground set, and let S be a set of measures X . We saythat S is ε -unrelated if, for all distinct elements µ, ν ∈ S , k µ − ν k T V >
12 ( µ ( X ) + ν ( X )) − ε, where k·k T V denotes the total variation distance. The following lemma is aneasy corollary of a fact from [16]. We include the proof here for completeness.
Lemma 4.
For every ε ∈ (0 , and k ∈ N , if there exists a map f :( V n , ρ n ) → ℓ k with distortion ε , then there exists an ε -unrelated set ofprobability measures on { , . . . , k + 1 } of size n − .Proof. Let r ∈ V n denote the the vertex of degree n −
1. By translationand scaling, we may assume that f ( r ) = 0 and f is 1-Lipschiz. Thus for allvertices v ∈ V n , we have k f ( v ) k
1. For each vertex v ∈ V n \ { r } definethe measure µ v as follows µ v ( { i } ) = max(0 , f ( v ) i ) 1 i k max(0 , − f ( v ) i ) k + 1 i k − k f ( v ) k i = 2 k + 1 , JAMES R. LEE AND MOHAMMAD MOHARRAMI where we use f ( v ) i to denote the i th coordinate of f ( v ).Note that for all u, v ∈ V n \ { r } we have k µ u − µ v k T V = 12 (cid:16) k f ( u ) − f ( v ) k + (cid:12)(cid:12)(cid:12) (1 − k f ( u ) k ) − (1 − k f ( u ) k ) (cid:12)(cid:12)(cid:12)(cid:17) > k f ( u ) − f ( v ) k . Since f has distortion 1 + ε , for any two distinct vertices u, v ∈ V n , we have k f ( u ) − f ( v ) k > (cid:18)
21 + ε (cid:19) > − ε ) . Therefore the collection { µ v : v ∈ V n \ { r }} satisfies the conditions of thelemma. (cid:3) The next lemma is the final ingredient that we need to prove Theorem 1.Let M k be the set of all measures { , , . . . , k } , and let P k be the set of allprobability measures on { , , . . . , k } . Lemma 5.
There exists a universal constant C > such that for ε / ,the following holds. If there is an ε -unrelated set S ⊆ P k , then there exists a -unrelated set T ⊆ P k of size at least |S| such that for all µ ∈ T , we have | supp( µ ) | ⌈ Cε ( ε + n ) d ) ⌉ . Before proving the lemma, we use it to finish the proof of Theorem 1.
Proof of Theorem 1.
Suppose that there is a map from the n -star to ℓ d withdistortion 1 + ε . Then by Lemma 4, there exists an ε -unrelated set ofprobability measures on { d + 1 } of size n −
1. Thus by Lemma 5, theremust exist a -unrelated set S of probability measures on { , . . . , d + 1 } ofsize Ω( n ) such that every measure in S has support size at most (cid:24) C · ε · (cid:18) ε + 1 n − (cid:19) · (2 d + 1) (cid:25) , for some universal constant C > Cε ( ε + |S| )(2 d + 1) <
1, every measure in S is supported on exactly one element,therefore |S| d + 1. Hence, d > Ω( |S| ) > Ω( n ) > Ω(log n ) ε log(1 /ε ) , where we have used the assumption that n > /ε .In the second case, we have Cε ( ε + |S| )(2 d + 1) >
1. Since |S| = O ( ε ),each element µ ∈ S has | supp( µ ) | O ( ε d ). Thus for some constant c > (cid:0) d +1 cε d (cid:1) exp (cid:0) O ( ε d log(1 /ε ) d ) (cid:1) different supports of size O ( ε d ) for the measures in S . LOWER BOUND ON DIMENSION REDUCTION FOR TREES IN ℓ Since S is a -unrelated set of probability measures, for any µ, ν ∈ S , wehave k µ − ν k T V > . In particular, if we fix a set Q ⊆ X , then by a simple | Q | -dimensional volumeargument, | µ ∈ S : supp( µ ) ⊆ Q | | Q | . All together, we have |S| O ( ε d ) · e O ( ε d log(1 /ε )) e O ( ε d log(1 /ε )) . Hence, d > Ω (cid:16) log |S| ε log(1 /ε ) (cid:17) , completing the proof. (cid:3) Remark 6.
We note that there is a straightforward volume lower boundfor large distortions D >
1: Any D -embedding of the n -star into ℓ d requires d > Ω( log n log D ). This is simply because the maximal number of disjoint ℓ balls of radius 1 /D that can be packed in an ℓ ball of radius 2 is (2 D ) d in d dimensions.We are left to prove Lemma 5. We start by recalling some simple prop-erties of the total variation distance. For a finite set S and measures µ, ν : 2 S → [0 , ∞ ), we definemin( µ, ν )( T ) = X x ∈ T min n µ ( { x } ) , ν ( { x } ) o . For k ∈ N , and measures µ, ν ∈ M k , we have k µ − ν k T V = 12 ( µ ([ k ]) + ν ([ k ])) − min( µ, ν )([ k ]) , (1)where we use the notation [ k ] = { , , . . . , k } . We also use the followingpartial order on measures on the set S : µ (cid:22) ν , if and only if for all T ⊆ S , µ ( T ) ν ( T ). The following observation is immediate from (1). Observation 7.
Fix k ∈ N , ε >
0, and measures µ, ν, µ ′ , ν ′ ∈ M k , suchthat µ ′ (cid:22) µ and ν ′ (cid:22) ν . If k µ − ν k T V >
12 ( µ ([ k ]) + ν ([ k ])) − ε, then k µ ′ − ν ′ k T V >
12 ( µ ′ ([ k ]) + ν ′ ([ k ])) − ε. We will require the following fact in the proof of Lemma 5.
Lemma 8.
Consider δ ∈ (0 , and a finite subset S ⊆ [0 , ∞ ) such that δ · ( | S | − · X x ∈ S x > X x,y ∈ S,x = y min( x, y ) . (2) Then there exists a set T ⊆ S , such that P x ∈ T x > P x ∈ S x and | T | ⌈ δ ( | S | − ⌉ . JAMES R. LEE AND MOHAMMAD MOHARRAMI
Proof.
Let n = | S | , and let a > · · · > a n > S indecreasing order. Then, n X i =1 n X j =1 i = j min( a i , a j ) = n X i =1 n X j =1 i = j a max( i,j ) = n X i =1 i − a i . Letting k = ⌈ δ ( | S | − ⌉ , we have n X i =1 n X j =1 i = j min( a i , a j ) > n X i = k +1 i − a i > k n X i = k +1 a i > δ ( | S | − n X i = k +1 a i . Combining this inequality and (2) implies that P ni = k +1 a i P x ∈ S x ,therefore P ki =1 a i > P x ∈ S x. Hence the set T = { a , . . . , a k } satisfies bothconditions of the lemma. (cid:3) Proof of Lemma 5.
We will show that each of the following statements im-plies the next one.I) There exists an ε -unrelated set S ⊆ P k of size n .II) There exists an ε -unrelated set S ⊆ M k of size n such that(a) for all µ ∈ S , µ ([ k ]) P µ ∈S µ ([ k ]) > n/ P µ ∈S | supp( µ ) | < (2 εn + 1) k ;III) There exists an ε -unrelated set S ⊆ M k of size at least n/
14 suchthat(a) for all µ ∈ S , | supp( µ ) | < (cid:0) ε + n (cid:1) k ;(b) for all µ ∈ S , we have µ ([ k ]) > / S ⊆ M k , we define,∆ S = X µ,ν ∈S ,µ = ν min( µ, ν ) . Note that, if for some ε ∈ [0 , S ⊆ P k is ε -unrelated, then (1) impliesthat ∆ S ([ k ]) X µ,ν ∈S ,µ = ν
12 ( µ ([ k ]) + ν ([ k ])) − k µ − ν k T V X µ,ν ∈S ,µ = ν (1 − (1 − ε ))= ε |S| · ( |S| − . (3) LOWER BOUND ON DIMENSION REDUCTION FOR TREES IN ℓ I ⇒ II:
Suppose that S I ⊆ P k is ε -unrelated, and let X be a random variablewith state space { , . . . , k } such that P ( X = i ) = P µ ∈S I µ ( { i } ) |S I | . We have E " ∆ S I ( X ) P µ ∈S I µ ( X ) = 1 |S I | k X i =1 ∆ S I ( { i } ) = 1 |S I | ∆ S I ([ k ]) (3) ε ( |S I | − , Markov’s inequality implies that P ∆ S I ( X ) P µ ∈S I µ ( X ) ε ( |S I | − ! > . So if we let A = ( i : ∆ S I ( { i } ) P µ ∈S I µ ( { i } ) ε ( |S I | − ) , then we have 1 |S I | X µ ∈S I µ ( A ) = X µ ∈S I X i ∈ A P ( X = i ) > . (4)By Lemma 8, for all i ∈ A there exists a set W i ⊆ S I such that | W i | ⌈ ε ( |S I | − ⌉ , and X µ ∈ W i µ ( { i } ) > X µ ∈S I µ ( { i } ) . (5)For µ ∈ S I , let Y µ = { i : µ ∈ W i } . For any Y ⊆ S , define R Y : 2 S → S by R Y ( T ) = T ∩ Y . on elements of Y and zero elsewhere.Let S II = { µ ◦ R Y µ } µ ∈S I . Since µ ◦ R Y µ (cid:22) µ , S II satisfies II(a). Further-more, Observation 7 implies that the collection { µ ◦ R Y µ } µ ∈S I is ε -unrelated.Furthermore, S II satisfies II(b) because X µ ∈S I µ ( Y µ ) (5) > X i ∈ A X µ ∈S I µ ( { i } ) = 12 X µ ∈S I µ ( A ) (4) > |S I | . Finally, condition II(c) holds because X µ ∈S I | supp( µ ◦ R Y µ ) | X i ∈ A | W i | < ε ( |S I | − | A | + | A | (2 ε |S I | + 1) k . II ⇒ III:
Suppose that S II ⊆ M k is an ε -unrelated collection of cardinality n satisfying all the conditions of II. We have max { µ ([ k ]) } µ ∈S II P µ ∈S II µ ([ k ]) > |S II | /
4. Therefore, there exists a subcollection S ′ ⊆ S II such that for all µ ∈ S ′ , we have µ ([ k ]) > /
8, and |S ′ | > (cid:18) / − / − / (cid:19) |S II | > n . JAMES R. LEE AND MOHAMMAD MOHARRAMI
By Markov’s inequality, there exists a collection of measures S III suchthat |S III | > |S ′ | > |S II | , where for all µ ∈ S III ,supp( µ ) P µ ∈S ′ | supp( µ ) ||S ′ | P µ ∈S II | supp( µ ) ||S ′ | P µ ∈ S II | supp( µ ) || S II | / II(c) k (cid:18) ε + 1 n (cid:19) . The set S III has size at least n and by construction satisfies conditions(a) and (b) of III. III ⇒ IV:
Suppose S III ⊆ M k is a an ε -unrelated collection of cardinalityat least n/
14. For each measure µ ∈ S III , let Z µ ⊆ { , . . . , k } be the set of (cid:6) · ε (cid:0) k (2 ε + n ) (cid:1)(cid:7) elements of { , . . . , k } that has the largest measureswith respect to µ (breaking ties arbitrarily). Since ε , for all µ ∈ S III wehave µ ( Z µ ) > · ε (14 k (2 ε + n ))14 k (2 ε + n ) ! = 2 ε. (6)Let S IV = n µ ◦ R Zµ µ ( Z µ ) : µ ∈ S III o . Clearly S IV ⊆ P k , and |S IV | > n . Moreover,by our construction for all ¯ µ ∈ S IV , | supp(¯ µ ) | ⌈ ε (2 ε + n ) k ) ⌉ . Tocomplete the proof we need to show S IV is -unrelated. Note that if µ, ν ∈S III , then Observation 7 implies thatmin( ν ◦ R Z ν , µ ◦ R Z µ )([ k ]) (1) = µ ( Z µ ) + ν ( Z µ )2 − k µ ◦ R Z ν − µ ◦ R Z ν k T V ε. Therefore, (cid:13)(cid:13)(cid:13)(cid:13) µ ◦ R Z µ µ ( Z µ ) − ν ◦ R Z ν ν ( Z ν ) (cid:13)(cid:13)(cid:13)(cid:13) T V (1) = 1 − min (cid:18) µ ◦ R Z µ µ ( Z µ ) , ν ◦ R Z ν ν ( Z ν ) (cid:19) ([ k ]) (6) > − min (cid:18) µ ◦ R Z µ ε , ν ◦ R Z ν ε (cid:19) ([ k ]) (1) > , completing the proof. (cid:3) Nearly equilateral sets in ℓ d We will need the following result of Kahane [10].
Theorem 9.
For every ε ∈ (0 , , there exists a mapping K ε : R → ℓ d suchthat d O (1 /ε ) and the following holds: For every x, y ∈ R , (1 − ε ) p | x − y | k K ε ( x ) − K ε ( y ) k p | x − y | . Proof of Theorem 2.
Suppose that f : U n → ℓ d is a (1+ ε )-embedding scaledso that f is 1-Lipschitz. Consider the mapping g : U n → ℓ O ( d/ε )2 given by g ( x ) = (cid:16) K ε ( f ( x ) i ) , K ε ( f ( x ) i ) , . . . , K ε ( f d ( x ) i ) (cid:17) , LOWER BOUND ON DIMENSION REDUCTION FOR TREES IN ℓ where f ( x ) i denotes the i th coordinate of f ( x ). By Theorem 9, for any x, y ∈ U n , we have k g ( x ) − g ( y ) k k f ( x ) − f ( y ) k . On the other hand, k g ( x ) − g ( y ) k > (1 − ε ) k f ( x ) − f ( y ) k > (1 − ε ) ε , implying that k g ( x ) − g ( y ) k > (1 − ε ) / √ ε > − ε . Thus g is a (1 + 2 ε )-embedding of U n into ℓ O ( d/ε )2 . But now by [2], for n sufficiently large, wehave d > Ω(log n ) ε log ε . (cid:3) Extension to k -ary trees We now prove Corollary 3. Combining the next lemma with Theorem 1yields the desired result.
Lemma 10.
For h, k > , let B h,k be a complete k -ary tree of height h .If B h,k admits a (1 + ε ) -embedding into ℓ d for some ε , then the (1 + k ⌈ h/ ⌉ ) -star admits a (1 + 4 ε ) -embedding into ℓ d .Proof. Suppose that f : B h,k → ℓ d is a (1 + ε )-embedding. We may assume,without loss, that f is 1-Lipschitz. Letting n = (1 + k ⌈ h/ ⌉ ), we constructan embedding g : V n → ℓ d of the n -star as follows.Let r ∈ V n denote the vertex of degree n −
1. We put g ( r ) = 0. Let S be the set of vertices in B h,k at height ⌈ h/ ⌉ (we use the convention thatroot has height zero). For any vertex v ∈ S , pick an arbitrary leaf x v inthe subtree rooted at v . Associate to every vertex w ∈ V n \ { r } a distinctelement ˜ w ∈ S and put g ( w ) = f ( x ˜ w ) − f ( ˜ w ) h − ⌈ h/ ⌉ . Since f is 1-Lipschitz, the same holds for g . Moreover for any two distinctelements u, v ∈ S we have2( h − ⌈ h/ ⌉ ) + d B h,k ( u, v ) = d B h,k ( u, v ) + d B h,k ( x v , v ) + d B h,k ( x u , u )= d B h,k ( x u , x v ) (1 + ε ) k f ( x u ) − f ( x v ) k (1 + ε ) k ( f ( x u ) − f ( u )) − ( f ( x v ) − f ( v )) k + (1 + ε ) k f ( u ) − f ( v ) k (1 + ε ) k ( f ( x u ) − f ( u )) − ( f ( x v ) − f ( v )) k + (1 + ε ) d B h,k ( u, v ) . Therefore,(1 + ε ) k ( f ( x u ) − f ( u )) − ( f ( x v ) − f ( v )) k > h − ⌈ h/ ⌉ ) − εd B h,k ( u, v ) > h − ⌈ h/ ⌉ ) − ε ⌈ h/ ⌉ > h − ⌈ h/ ⌉ ) − ε ( h − ⌈ h/ ⌉ ) > (2 − ε )( h − ⌈ h/ ⌉ ) . Since ε /
8, the preceding inequality bounds the distortion of g by ε − ε ε , completing the proof. (cid:3) Acknowledgments.
We are grateful to Jiri Matouˇsek for suggesting theapproach of Section 3. This research was partially supported by NSF grantCCF-1217256.
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