Metric Embeddings and Expanders

Metric Embeddings and Expanders

Grigory Yaroslavtsev
(based on Chapter 13 of a survey
“Expander graphs and their applications” by Hoory, Linial and
Wigderson)

Pennsylvania State University

December 8, 2011

Grigory Yaroslavtsev (PSU) December 8, 2011 1 / 11

Metric embeddings
A ﬁnite metric space is a pair (X , d), where X is a set of n points and
d : X × X → R+ is a distance function (three axioms).
Let f : X → Rn be an embedding of (X , d) into (Rn , 2 )
n 2
(d 2 (x, y ) = ||x − y || = i=1 (xi − yi ) ).
expansion(f ) = max ||f (x1 ) − f (x2 )||/d(x1 , x2 )
x1 ,x2 ∈X

contraction(f ) = max d(x1 , x2 )/||f (x1 ) − f (x2 )||
x1 ,x2 ∈X

distortion(f ) = expansion(f ) · contraction(f )
Example, that requires distortion (shortest-path metric for unit-length
edges):


Background on 2 -embeddings

If (X , d) is 2 -embeddable ⇒ it is p -embeddable for 1 ≤ p ≤ ∞.
Let c2 (X , d) denote the least possible distortion of an embedding of
(X , d) into (Rn , 2 ) (dimension n is suﬃcient).
For any n-point metric space c2 (X , d) = O(log n) [Bourgain’85].
We will see how to compute such an embedding later (via SDP),
together with a Ω(log n) lower bound for expanders (via dual SDP).
Theorem (Johnson-Lindenstrauss ’84)
log n
Any n-point 2 -metric can be embedded into an O 2 -dimensional
Euclidean space with distortion 1 + .

The bound on dimension was shown to be optimal by Jayram and
Woodruﬀ (SODA’11), previous Ω( 2 log n ) was by Alon ’03.
log 1/
Such dimension reduction is impossible for 1 (Brinkman, Charikar
’03, Lee, Naor ’04, . . . ?).

Computing minimal distortion
Theorem (Linial, London, Rabinovich ’95)
Given a metric space (X , d), the minimal 2 -distortion c2 (X , d) can be
computed in polynomial time.

Proof.
Scale f : X → Rn , so that contraction(f ) = 1, so distortion(f ) ≤ γ iﬀ:

d(xi , xj )2 ≤ ||f (xi ) − f (xj )||2 ≤ γ 2 d(xi , xj )2 ∀i, j

A symmetric matrix Z ∈ Rn×n is positive semideﬁnite (PSD), if (all
four are equivalent):
1 v T Zv ≥ 0 for all v ∈ Rn .
2 All eigenvalues λi ≥ 0.
3 Z = WW T for some matrix W .
n
4 Z= λi wi wiT for λi ≥ 0 and orthonormal vectors wi ∈ Rn .
i=1


Computing minimal distortion (continued)

Proof.
Any embedding f : X → Rn can be represented as a matrix
U ∈ Rn×n , where row ui = f (xi ).
Let Z = UU T , so we need to ﬁnd a PSD Z , such that:

d(xi , xj )2 ≤ zii + zjj − 2zij ≤ γ 2 d(xi , xj )2 , ∀i, j,

since ||ui − uj ||2 = zii + zjj − 2zij .
Linear optimization problem with an additional constraint that a
matrix of variables is PSD ⇒ solvable by ellipsoid in polynomial time.


Characterization of PSD matrices

Lemma
A matrix Z is PSD if and only if ij qij zij ≥ 0 for all PSD matrices Q.

Proof.
⇐: For v ∈ Rn let Qij = vi · vj . Then Q is PSD and
v T Zv = ij (vi · vj )zij = ij qij zij ≥ 0.
⇒: Let Q = k λk wk wk for λi ≥ 0, or equivalently Q = k Ak ,
T

where Ak = λk wk wk .
T
kz = λ T
Because ij Aij ij k ij wki wkj zij = λk wk Zwk ≥ 0, we have
k k
ij qij zij = ij k Aij zij = k ij Aij zij ≥ 0.


Lower bound on distortion
Theorem (Linial-London-Rabinovich ’95)
The least distortion of any ﬁnite metric space (X , d) in the Euclidean
space is given by:

2
pij >0 pij d(xi , xj )
c2 (X , d) ≥ max 2
.
P∈PSD,P·1=0 − pij <0 pij d(xi , xj )

Proof.
Primal SDP:

qij zij ≥ 0 ∀Q ∈ PSD
ij

zii + zjj − 2zij ≥ d(xi , xj )2 ∀i, j
2 2
γ d(xi , xj ) ≥ zii + zjj − 2zij ∀i, j

Lower bound on distortion via a dual SDP solution

qij zij ≥ 0 ∀Q ∈ PSD (1)
ij

zii + zjj − 2zij ≥ d(xi , xj )2 ∀i, j (2)
γ 2 d(xi , xj )2 ≥ zii + zjj − 2zij ∀i, j (3)

Take Q ∈ PSD, such that j qij = 0.
If qij > 0, add corresponding inequality (2) multiplied by qij /2.
If qij < 0, add corresponding inequality (3) multiplied by −qij /2.
2 2
qij <0 (zii + zjj − 2zij )qij /2 − γ qij <0 d(xi , xj ) qij /2 ≥
2
qij >0 d(xi , xj ) qij /2 − qij <0 (zii + zjj − 2zij )qij /2
Because ij qij zij ≥ 0 and i qij = 0, we get a contradiction, if
γ 2 qij <0 d(xi , xj )2 qij + qij >0 d(xi , xj )2 qij > 0.

Example: Hypercube with Hamming metric
Let’s denote r -dimensional hypercube with Hamming metric as Qr .
√
Identity embedding gives distortion r .

2
pij >0 pij d(xi , xj )
c2 (Qr ) ≥ max 2
.
P∈PSD,P·1=0 − pij <0 pij d(xi , xj )
r ×2r
Deﬁne P ∈ R2 , such that P1 = 0 as:

−1
 if d(i, j) = 1


r − 1 if i = j
P(x, y ) =
1
 if d(i, j) = r


0 otherwise
P ∈ PSD: eigenvectors χI (J) = (−1)|I ∩J| for I , J ⊆ {1, . . . , n}.
Because pij >0 pij d(xi , xj )2 = 2r · r 2 and
√
− pij <0 pij d(xi , xj )2 = 2r · r , we have c2 (Qr ) ≥ r .

Embedding expanders into 2

√
For the hypercube we’ve got a Ω( log n) lower bound.
Now we will get a Ω(log n) lower bound for expanders.
Take k-regular expander G with n vertices and λ2 ≤ k − for > 0.
√
Embedding vertex i to ei / 2: expansion = 1, contraction = O(log n).

Theorem (Linial-London-Rabinovich ’95)
For G as above c2 (G ) = Ω(log n), constant depends only on k and .

If H = (V , E ) is the graph on the same vertex set, where two vertices
are adjacent if their distance in G is at least logk (n) , then H has a
perfect matching (by Dirac’s theorem has Hamiltonian cycle).


Proof of the lower bound for embdedding expanders
Let B be the adjacency matrix of a perfect matching in H and
P = kI − AG + 2 (B − I ), so P1 = 0.

x T (kI − AG )x ≥ (k − λ2 )||x||2 ≥ ||x||2
x T (B − I )x = (2xi xj − xi2 − xj2 ) ≥ −2 (xi2 + xj2 ) = −2||x||2 .
(i,j)∈B (i,j)∈B

P is PSD, because x T Px = x T (kI − AG )x + x T 2 (B − I )x ≥ 0.

− d(i, j)2 pij = kn
pij <0

d(i, j)2 pij ≥ · n logk n 2 ,
2
pij >0

because distances of edges in B are at least logk n .
Thus, c2 (G ) = Ω(log n).

Metric Embeddings and Expanders

More Related Content

What's hot

Viewers also liked

Similar to Metric Embeddings and Expanders

Recently uploaded

Metric Embeddings and Expanders