S. ( G) = min |E(S, S)| |S|. ( G) = min |E(S, S)| |S|. S µ V. S µ V. c. |S| ¸ c ¢ |V|. Sparsest Cut. G = (V, E). S. c- balanced separator. Both NP-hard. Why these problems are important.
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S
(G) = min |E(S, S)|
|S|
(G) = min |E(S, S)|
|S|
S µ V
S µ V
c
|S| ¸ c ¢ |V|
G = (V, E)
S
c- balanced separator
Both NP-hard
3) Embeddings of finite metric spaces into l1
(Linial, London, Rabinovich’94)
2) O(log n) -approximation via multicommodity flows (Leighton-Rao 1988)
log n
log n
1
0
0
1
1
Semidefinite
Min (i, j) 2 E Xij
0 · Xij· 1
Motivation: Every cut (S, Sc) defines a (semi) metric
Xij2 {0,1}
Xij + Xj k¸ Xik
i< j Xij¸ c(1-c)n2
There exist unit vectors v1, v2, …, vn2<n
such that Xij = |vi - vj|2 /4
Min (i, j) 2 E |vi –vj|2/4
|vi|2 = 1
|vi –vj|2 + |vj –vk|2¸ |vi –vk|28 i, j, k
i < j |vi –vj|2¸ 4c(1-c)n2
Unit l22 space
Vi
Vj
Vk
s
s
s
s
Unit vectors v1, v2,… vn2<d
|vi –vj|2 + |vj –vk|2¸ |vi –vk|28 i, j, k
Angles are non obtuse
Taking r steps of length s
only takes you squared distance rs2
(i.e. distance r s)
|u – v|2 = i |ui – vi|2
= 2 i |ui – vi| = 2 |u – v|1
In fact, every l1 space is also l22
Conjecture (Goemans, Linial): Every l22 space is l1 up to distortion O(1)
log n
Two subsets S and T are -separated if
for every vi2 S, vj2 T |vi –vj|2¸
<d
¸
Thm: If i< j |vi –vj|2 = (n2) then there exist two sets S, T
of size (n) that are -separated for = ( 1 )
log n
log n
) |E(R, Rc)| · SDPopt /
· O( SDPopt)
v1, v2,…, vn2<d is optimum SDP soln;
SDPopt = (I, j) 2 E |vi –vj|2
S, T : –separated sets of size (n)
Do BFS from S until you hit T. Take the level of the
BFS tree with the fewest edges and output the cut (R, Rc) defined by this level
(i, j) 2 E |vi –vj|2¸ |E(R, Rc)| £
Next 10-12 min: Proof-sketch of Main Thm
-t2
/2
e
log n
1
1
d
d
Pru[ projection exceeds 2 ] < 1/n2
d
v
<d
u
<u, v> ??
If any vi2 Su and vj2 Tu satisfy
|vi –vj|2·,
delete them
and repeat until no such vi, vj can be found
0.01
“Stretched pair”: vi, vj such that |vi –vj|2· and
| h vi –vj, u i | ¸ 0.01
d
d
<d
Check if Su and Tu have size (n)
u
Tu
Su
If Su, Tu still have size (n), output them
Main difficulty: Show that whp only o(n) points get deleted
Obs: Deleted pairs are stretched and they form a matching.
“Stretched pair”: vi, vj such that |vi –vj|2· and
| h vi –vj, u i | ¸ 0.01
d
log n
O( 1 ) £ standard deviation
) PrU [ vi, vj get stretched] = exp( - 1 )
= exp( - )
E[# of stretched pairs] = O( n2 ) £ exp(- )
logn
Vi
0.01
d
Suppose with probability (1) there is a matching of (n) stretched pairs
Vj
u
Ball (vi , )
|vfinal - vi| < r
| <vfinal – vi, u>| ¸ 0.01r
0.01
0.01
0.01
r
d
d
d
d
= O( r ) x standard dev.
Vj
vfinal
Vi
r steps
u
Contradiction!!
Reason: Isoperimetric inequality for spheres
<d
A : measurable set with (A) ¸ 1/4
A
A : points with distance · to A
(A) ¸ 1 – exp(-2 d)
A
log n
S
Our Thm: If G has expansion , then a d-regular expander flow can be routed in it where d=
“Expander”
G = (V, E)
Idea: Embed a d-regular (weighted) graph such that 8 S w(S, Sc) = (d |S|)
(*)
(certifies expansion = (d) )
S
Graph w satisfies (*) iff L(w) = (1) [Cheeger]
Cf. Jerrum-Sinclair, Leighton-Rao(embed a complete graph)
n-cycle
Take any 3-regular expander on n nodes
Put a weight of 1/3n on each edge
Embed this into the n-cycle
Routing of edges does not exceed any capacity ) expansion =(1/n)
log n
Pij = paths whose endpoints are i, j
8i jp 2 Pij fp = d (degree)
8e 2 E p 3 e fp· 1 (capacity)
8S µ V i 2 S j 2 Scp 2 Pij fp¸0 d |S| (demand graph is an expander)
fp¸ 0 8 paths p in G
)
(d) ·(G) · O(d )
log n
log n
O(n2) time algorithm that given any graph G finds for some d >0
Ingredients: Approximate eigenvalue computations; Approximate flow computations (Garg-Konemann; Fleischer)
Random sampling (Benczur-Karger + some more)
Idea: Define a zero-sum game whose optimum solution is an expander
flow; solve approximately using Freund-Schapire approximate solver.
G = (V, E); = (G)
For every distribution on n/3 –balanced cuts {zS} (i.e., SzS =1)
there exist (n) disjoint pairs (i1, j1), (i2, j2), ….. such that for each k,
Conjecture ) existence of d-regular expander flows for d =
log n