Durham Goode dmg53
Myle Ott mao37
Random Walks on Undirected Graphs (3/14/07)
Discovery Time – huv
(a.k.a. hitting time)
- the expected time to reach v starting from u
Lemma:
- If u and v are connected by an edge then huv + hvu ≤ 2m
- m is the number of edges in the graph
Proof:
- The probability of traversing any edge is equally likely.
-
1
.
2m
Therefore, the frequency of seeing a given edge in a random walk is 2m
The probability of traversing an edge in a given direction is therefore
u
-
v
Expected length 2m
Based on the condition that the walk traversed edge uv, the expected time until
the next traversal of edge uv is 2m.
But since random walks are memoryless, we can drop the condition.
Expected length 2m
u
-
-
v
The path taken by the random walk could have touched u and gone back to v
multiple times before actually taking edge uv. Therefore, since the total path
length is expected to be 2m, the path may go to u and come back to v in less than
2m steps.
Therefore huv + hvu ≤ 2m
Circular Graph:
- Given a circular graph with n vertices, how long does it take to go from vertex u
to vertex v?
u
v
-
huv + hvu ≤ 2m
-
Due to symmetry: huv = hvu
-
2huv ≤ 2m
-
huv ≤ n
Line Graph:
- Erase the edge between u and v in the circle graph and you get a line.
n
u
-
v
huv = n2
Nonsymmetrical Discovery Time:
n
2
u
v
n
2
-
huv = O(n2)
When starting from v it will return to v about n times. If it goes in the “lollipop”
side it will take about n steps to get out of there, and if it takes the u side it will
take about n steps to get to u.
Therefore: hvu = O(n3)
Definitions
-
commute_time(u, v): time a random walk takes to get from u to v, and back to u.
-
return_time(u): time it takes to leave and return to u
1
return_time(u):
, where frequ is the stationary percentage of time you are at
freq u
u on a long random walk.
cover_time(u): time to reach all vertices, starting at u
cover_time(G): maxu cover_time(u)
Adding edges may increase or decrease cover time:
u
-
v
maxu cover_time(u) is u to v: n2
n
2
u
v
n
2
-
maxu cover_time(u) is v to u: n3 (u to v would be n2)
-
cover_time for clique of size n: n log n
Theorem
-
commute_time(u, v) = 2 m ruv, where m is the number of edges, and r is the
resistance when you put 1 amp in at u, and take it out at v.
Proof next class
Claim
-
For any n-vertex graph, commute_time(u, v) ≤ n3
Proof
-
Path from u to v is at most length n
Therefore, ruv ≤ n
#edges ≤ n2
So, commute_time ≤ n3
Theorem
-
Time for a random walk to cover all vertices is bounded by 2m(n-1)
Proof
-
Consider a spanning tree
v
huv + hvu ! 2m
-
u
Time to go u to v, and v to u is 2m, so we are bounded by 2m(n-1) edges