Undirected ST-Connectivity In Log Space

Omer ReingoldSlides by Sharon Bruckner

Today

• Some history

• What are we adding to the mix?– Connectivity in expanders– Making expanders: powering it up and

cutting it down– Putting it together in Log-Space

• In conclusion

Some History

• What is ST-Connectivity?

• What do we know about it?

• What are we trying to accomplish?

What is ST-Connectivity?

• Given a graph G and two vertices s and t:– YES if there is a s-t path in G– NO otherwise

• Two flavors to the problem:– STCON: G is directed– USTCON G is undirected.

• Today we’ll talk about USTCON

What do we know about it?• Solved easily with BFS

– But – polynomial space!

• USTCON is in NL (=NSPACE(log))– Just guess the path

• USTCON is in DSPACE(log2n) (Savitch)– Guess a vertex on the path, recurse

• USTCON is in RL– Random walk– Randomness is a resource!

• USTCON is complete for SL

What are we trying to accomplish?

We would like an algorithm for USTCON which is deterministic

and works in space logarithmic in the size of

the graph

What are we adding to the mix?

Useful Notations

• The adjacency matrix of a graph G

• We’ll use D regular graphs and the normalized adjancecy matrix: (1/D)M ,

1 i, j EM i, j

0 otherwise

Useful Notations

• M is a stochastic matrix, and serves as the “random walk” matrix.

• Largest eigenvalue of M is 1, with eigenvector (1,1,…,1)єRn ,2nd largest eigenvalue is λ(G).

• A (N,D,λ) graph is a D regular graph over N vertices with λ(G) ≤ λ

Expanders – a Reminder

Two equivalent definitions of expander• (N,D, λ) is an expander iff the

spectral gap 1- λ > 0• (N,D, λ) is an expander if there

exists ε>0 such that for any set S of at most half the vertices in G, at least (1+ ε)|S| vertices of G are connected to some vertex in S

Note that the adjacency matrix in this case in normalized!

Expanders – a Reminder

2

2

21

2

G G

D D

,2

,min

V

S V S

E S SG

S D

Connectivity in Expanders

What would happen if each connected component of our graph was an expander?

We could decide USTCON in logspace!

Every expander has a O(log(N)) diameter

Theorem:For any s and t in an expander, there’s

a path from s to t of length O(log(N))

Proof

i

i

n(1 )

2n

log 1 log2

log n 1ilog 1

Logspace Algorithm for paths in an expander

For a (N,D, λ) expander there is a a space O(logD*logN) algorithm which decides USTCON for any s and t

Idea:•From any vertex s there are Dl=O(logN) different paths. •enumerate and check if any of them reach t.

How much does it cost?

•At each vertex we have a choice of D vertices, log(D) to represent 1…D.

•Each path is log(N) long.

• we need log(D) at each stage of a log(N) path, altogether O(logD*logN)

So far

We now know that if our graph G was an expander graph, USTCON can be solved in Log-Space.

How can we turn G into a graph of expanders?

Making expanders: powering it up and cutting it down

What do we want from the expander graph G’?

1. There is a path from s to t in G if and only if there is a path from s’ to t’ in G’.

2. Each connected component is an expander, with constant expansion

3. Construction in Log-Space.

The Plan

• New tools for our toolbox– Rotation Maps– Powering– Zig Zag products

• The actual construction• Why we got what we wanted• Why we got it in Log-Space

Overview• Series of powering by 8 and zig zag

products, one increases the spectral gap and the other one decreases it, but not by much.

8

O(Log(N))

Zig Zag Powering

Rotation Maps

• This is the notation we’ll use in this algorithm

GRot : N D N D Let G be a D regular undirected graph. The rotation map

GRot v,i w, j

If the edge (v,w) exists and is the ith edge coming out of v and the jth edge coming out of w

Rotation Map Example

1 23

4

17

…

…54

55

4

3 21

28

5554

…

…u

v

G

G

Rot u,17 v,28

Rot v,28 u,17

PoweringNo limitations on the degree → can expenderize with powering:

The k-th power of the D-regular graph G is Gk where there is an edge (u,v) iff there is a path of length ≤ k between u and v in G. There can be multiple edges!

≤k

Powering

What is it good for?• GN is an N-clique an expander• If G = (N,D, λ) then Gk is a (N,Dk, λk) λ(G)

The spectral gap• Why is this not enough?

Zig Zag Product

If G is a D-regular graph with N vertices and H is a d-regular graph with D vertices, we replace each v in G with a copy of H, Hv. Therefore, our new graph has [N]x[D] vertices.

H

Part of G

v is now Hv

v

Zig-Zag ProductThe resulting graph is a d2 regular

graph, with edges as follows:

zG H

z

G

v

w

a’

b’

Hv

Hw

a

b’

G Z H

a

b

i

j

a’ii’

bj

j’

Zig-Zag Product

From last week:

If G is an (N,D,λ) graph and H is a (D,d,) graph, then

The zig-zag product doesn’t hurt the spectral gap too badly!

211 (G z H ) 1 1

2

Satan’s Example

Technicalities

Rotation Map for Powering

u w

v

G

3

12 5

8

G

G

Rot u,3 v,12

Rot v,5 w,8

u v

G2

2GRot u, 3,12 w, 8,5

3,128,5

u w…a1 a2 ak

b1b(k-1)bk

G

G 1 1

G k k

Rot u,a v,b

...

Rot v ',a w,b

u va1,..,ak

bk,..,b1

Gk

2 1 2 k k k 1 1GRot u, a ,a ..., a w, b ,b ..., b

Rotation Map for Zig Zag

Hw

(w,b’) (w,b)

G

v

w

a’

b’

Hv

(v,a) (v,a’)i i’

j j’

Rot ((v,a), (i,j)) = zG H

H

G

H

w,b , j', i ' where

a ', i ' Rot a, i

w,b ' Rot v,a '

b, j' Rot b ', j

The Actual Construction

Show a transformation that turns every connected component of a graph into an expander.

But!Not any graph, but a D16 regular graph.

How to construct such a graph - later

Main Transformation

0

8

i i 1

G G

G G z H

Gi is a D16 regular graph, with N*(D16)i vertices

On input G and H, where G is a D16 regular graph on N vertices, and H is a D-regular graph on D16 vertices

ldefine G,H G , where l O log N

Details

•Set l to be the smallest integer s.t. If D is constant, then this number is O(logN).

•Gl has poly(N) vertices.•Every iteration can be done in logspace!

•Transforming the entire graph is equivalent to transforming each connected component separately

l2

2

1 11DN 2

More Details

S1

S2

S3

S1

S2

S3

Why ?

Show that this is true for a single iteration.

1

m

S

S

0

0

Adjacency matrix sorted by connected components

Both powering and zig zag maintain the block structure

Why is this an expander?

G and H be inputs of above. If and G is connected and non-bipartite then

Which means that the output of is indeed an expander.

H 1/ 2

G,H 1/ 2

Non bipartite means that G cannot be split into two sets of

vertices where all the edges of G are between those two sets

Proof

G0=G, connected and non-bipartite. Therefore, (inequality) Since

We’ll show and from the lemma we get

20G 1 1/ DN

l2 N2 21 1/ DN 1 1/ DN 1/ 2

2

i i 1i 0 G max G ,1/ 2

H 1/ 2

i 1 i 1 i 1G H 1 3/8 1 G 1 1/ 3 1 G

Proof Continued

We saw that if G = (N,D,λ) then Gt = (N,Dt,λt) =We can bound λ(Gi) by

For each i, one of two cases occurs:

Otherwise, it’s always true that

8

i 11 1/ 3 1 G

8

i 1 i

1 5 1G G

2 6 2

41 1/ 3 1 x x

42

i i 1 i 1 i 1G G because it 's the square of 1 1/ 3 1 G G

What do we want from the expander graph G’?

1. There is a path from s to t in G if and only if there is a path from s’ to t’ in G’.

2. Each connected component is an expander.

3. Construction in Log-Space.

Log Space Construction

Prove that for any D, can be calculated in space O(log(N)).

l16 16v N D , a D

RotG RotHInput Tape

Proof

Idea:Evaluating RotGi+1 for each pair takes just a constant additional amount of memory over evaluating RotGi. And all recursions use the same space!

Proof - Continued

A will allocate the variables:v in [N] (from the original G)a0, …, al in [D16] (vertex names in H)