bounding the mixing time via spectral gap graph random walk seminar fall 2009
DESCRIPTION
Bounding the mixing time via Spectral Gap Graph Random Walk Seminar Fall 2009. Ilan Ben-Bassat Omri Weinstein. Outline Why spectral gap? Undirected Regular Graphs. Directed Reversible Graphs. Example (Unit hypercube). Conductance. Reversible Chains. - PowerPoint PPT PresentationTRANSCRIPT
Ilan Ben-Bassat Omri Weinstein
Outline
Why spectral gap? Undirected Regular Graphs. Directed Reversible Graphs. Example (Unit hypercube). Conductance. Reversible Chains.
P- Transition matrix (ergodic) of anundirected regular graph.
P is real, stochastic and symmetric, thus: All eigenvalues are real. P has n real (orthogonal) eigenvectors.
P’s eigenvalues satisfy:
Why do we have an eigenvalue 1? Why do all eigenvalues satisfy ? Why are there no more 1’s? Why are there no (-1)’s?
1...1 1210 N
1||
Laplacian MatrixL = I – P
So, L is symmetric and positive semi definite.
L has eigenvalue 0 with eigenvector 1v.
n
ji
n
jijiij
n
jj
n
jiijji
n
iiijji
n
ii
ttt yyPyPyyyPyyyPyyIyyLyy1, 1,
2
1
2
1,1
2
1
2 0)(2
1)2(
2
1
Claim: The multiplicity of 0 is 1.2
1,
)(2
10 ji
n
jiij
t vvPLvv
So?...
vvPIvLv )(
vPvv vvvPv )1(
IntuitionHow could eigenvalues and mixing time be connected?
111100111100 ...)...( Nt
Ntt
NNtt vPavPavPavavavaPvP
1111110111111000 ...... Nt
NNt
Nt
NNtt vavaavavava
t1
Spectral Gap and Mixing Time
The spectral gap determines the mixing rate:
Larger Spectral Gap = Rapid Mixing
max1
As for P’s spectral decomposition, P has anorthonormal basis of eigenvectors.
We can bound by .
So, the mixing time is bounded by:
|| , jtjiP t
max
maxlog
log)(
Assume directed reversible graph (or general undirected graph).
We have no direct spectral analysis.
But P is similar to a symmetric matrix!
ProofLet be a matrix with diagonal entries .
Claim: is a symmetric matrix.
2/1D )(w
2/12/1 PDDS
j
jiijjiijjijcolumnilineji PDPDDPDDPDs
1)()()( ,
2/1,
2/12/1,
2/1)(
2/1)(
2/1,
i
ijjij Ps
1)( ,,
ijjjii PP ,, From reversibility:
vDvDS
vvDSDvSDD
vvP
T
TT
T
2/12/1
2/12/12/12/1 )(
S and P have the same eigenvalues.What about eigenvectors?
Still:
Why do all eigenvalues satisfy ? (same)
Why do we have an eigenvalue 1? (same)
Why is it unique?same for (-1).
As for 1: Omri will prove:
1210 ...1 N
1||
iii
jijiiji
y y
yyP
T 2
2
01
)(
min1
According to spectral decomposition ofsymmetric matrices:
Note:
1
0
)(1
0
)()(N
i
ii
N
i
iii EeeS
T
2/1)0()()()()( ;;02 DeEEEE Tiiji
Main Lemma:For every , we define:
So, for every we get:
U
})(
|)(),(|{max
, j
jjiP t
UjiU
0t
)(minmax
iIi
t
U
So, we can get
1
0
)(N
i
iti
t ES
min
maxmaxmax
|),(|
,
11max)(
tt
ji
t
ji
jjiP
UjiU j
jt
t
max
min
log
)log()(
Now, we can bound the mixing time:
SummaryWe have bounded the mixing time forirreducible, a-periodic reversible
graphs.
Note:Reducible graphs have no unique
eigenvalue.Periodic graphs – the same (bipartite
graph).
Graph ProductLet . The product
Is defined by and
2,1),,( iEVG iii
),(21 EVGGG
21 VVV ]}),([]),([|)),(),,{(( 12121221212121 EvvandwworEwwandvvwwvvE
0
1
2K(0,0)
(0,1) (1,1)
(0,1)
22 KK
222 ... KKKQn
..0.
....
....
1..0
1GA
Entry (i,j)
...),(
....
...),(
),(.),(
1
11
in
i
jj
vw
vw
vwvw
..0.
....
....
..
||
||2
2
2
V
VG IAGA
So, only the permutations that were counted for the determinant of AG1, will be counted here. We instead of we getSo,
)(k )(22 G
AI
The eigenvectors of Qn are We now re-compute every eigenvalue by: Adding n (self loops) Dividing by 2n (to get a transition matrix).
Now we get
And the mixing time satisfies:
},...2,1,0{ n
n2
11max
)()log()( 21 nOn