Mazes
And Random Walks
Complexity
1
Can You Solve This Maze?
Complexity
2
The Solution
Complexity
3
What Will We Do?
• Our task is to show an algorithm for
general mazes.
• We have memory which is logarithmic
in the size of the maze.
Complexity
4
Introduction
• Objectives:
– To explore the undirected connectivity
problem
– To introduce randomized computations
• Overview:
– Undirected Connectivity
– Random Walks
Complexity
5
Undirected Connectivity
• Instance: An undirected graph
G=(V,E) and two vertices s,tV
• Problem: To decide if there is a path
from s to t in G
Complexity
6
What Do We Know?
Theorem:
Directed Connectivity is NL-Complete
Corollary:
Undirected Connectivity is in NL.
Complexity
7
Undirected Connectivity is in
NL: Revisit
Our non-deterministic algorithm:
At each node, nondeterministically
choose a neighbor
and jump to it
Complexity
8
What If We Don’t Have “Magic
Coins”?
• Non-deterministic “algorithms” use
“magic coins” to lead them to the
right solution if one exists.
• In real life, these are unavailable…
Complexity
9
Idea!
• What if we have plain coins?
• In other words, what if we randomly
choose a neighbor?
Complexity
10
Random Walks
Add a self loop to each vertex.
• Start at s.
• Let di be the degree of the current node.
• Jump to each of the neighbors with
probability 1/di.
• Stop if you get to t.
s
Complexity
t
11
Notations
• Let vt denote the node visited at time
t (v0=s).
• Let pt(i) = Pr[vt=i]
p0(s)=1
Complexity
s
a
p1(a)=0.5
t
12
Stationary Distribution
Lemma:
If G=(V,E) is a connected graph,
for any iV,
di
lim pt i 
t 
2E
Complexity
13
Weaker Claim
We’ll prove a weaker result:
Lemma: If for some t, for any iV,
di
pt i 
2E
then for any iV,
di
pt 1 i 
2E
Complexity
14
Proof
Proof: di=2|E|. If the ith node has weight di
at time t, then it retains this weight at
time t+1 (it’s reachable (only) from its di
neighbors). 
Complexity
15
Illustrated Proof
Complexity
16
Using the Asymptotic Estimate
Corollary: Starting from
some node i, we will revisit
i within expectedly 2|E|/di
steps.
Proof: Since the walk has no
“memory”, the expected
return time is the same as
the asymptotic estimate
Complexity
17
One-Sided Error
• Note that if the right
answer is ‘NO’, we
clearly answer ‘NO’.
• Thus, a random walk
algorithm has onesided error.
• Such algorithms are
called “Monte-Carlo”
algorithms.
Complexity
18
How Many Steps Are Needed?
If the right answer is ‘YES’, in how many
steps do we expect to discover that?
But every time we
get here, we get a
second chance!
s
The probability we head in the
right direction is 1/ds
t
. . .
Complexity
19
How Many Steps Are Needed?
• Since expectedly we return
to each vertex after
2|E|/di steps,
• We expect to head in the
right direction after |E|
steps (w.p. ½).
• By the linearity of the
expectation, we expect to
encounter t in
d(s,t)|E||V||E| steps.
Complexity
20
Randomized Algorithm for
Undirected Connectivity
1. Run the random walk from s for
2|V||E| steps.
2. If node t is ever visited, answer
“there is a path from s to t”.
3. Otherwise, reply “there is probably
no path from s to t”.
Complexity
21
PAP 401-404
Main Theorem
Theorem: The above algorithm
To maintain the current
position we only need
- uses logarithmic space
log|V| space
- always right for ‘NO’ instances.
- errs with probability at most ½ for
‘YES’ instances.
Markov: Pr(X>2E[X])<½
Complexity
22
Summary

• We explored the undirected connectivity
problem.
• We saw a log-space randomized algorithm
for this problem.
• We used an important technique called
random walks.
Complexity
23