COMP 482: Design and
Analysis of Algorithms
Spring 2013
Lecture 4
Prof. Swarat Chaudhuri
Q3: Analyzing an algorithm
You have an array A with integer entries A[1],…, A[n]
Output a 2-D array B such that B[i,j] contains the sum A[i] + A[i+1] + …
+ A[j]
Here is an algorithm:
For i=1, 2, …, n
For j = i+1, 2, …, n {
Add up entries A[i] through A[j]
Store result in B[i,j]
}
}
Obtain an upper bound and a lower bound for the algorithm.
2
The lower bound is the more interesting one
Consider the times during the execution of the algorithm when i ≤ n/4
and j ≥ 3n/4.
In these cases, j − i + 1 ≥ 3n/4 − n/4 + 1 > n/2. Therefore, adding up the
array entries A[i] through A[j] would require at least n/2
operations, since there are more then n/2 terms to add up.
How many times during the execution of the given algorithm do we
encounter such cases? There are (n/4)2 pairs (i, j) with i ≤ n/4 and j
≥ 3n/4. the given algorithm enumerates over all of them, and as
shown above, it must perform at least n/2 operations for each such
pair. Therefore, the algorithm must perform at least n/2.(n/4)2 =
n3/32 operations. This is Ω(n3), as desired.
3
Graphs
3.1 Basic Definitions and Applications
Undirected Graphs
Undirected graph. G = (V, E)
V = nodes.
E = edges between pairs of nodes.
Captures pairwise relationship between objects.
Graph size parameters: n = |V|, m = |E|.




V = { 1, 2, 3, 4, 5, 6, 7, 8 }
E = { 1-2, 1-3, 2-3, 2-4, 2-5, 3-5, 3-7, 3-8, 4-5, 5-6 }
n=8
m = 11
5
Some Graph Applications
Graph
Nodes
Edges
transportation
street intersections
highways
communication
computers
fiber optic cables
World Wide Web
web pages
hyperlinks
social
people
relationships
food web
species
predator-prey
functions
function calls
scheduling
tasks
precedence constraints
circuits
gates
wires
software systems
6
World Wide Web
Web graph.
Node: web page.
Edge: hyperlink from one page to another.


cnn.com
netscape.com
novell.com
cnnsi.com
timewarner.com
hbo.com
sorpranos.com
7
9-11 Terrorist Network
Social network graph.
Node: people.
Edge: relationship between two people.


Reference: Valdis Krebs, http://www.firstmonday.org/issues/issue7_4/krebs
8
Ecological Food Web
Food web graph.
Node = species.
Edge = from prey to predator.


Reference: http://www.twingroves.district96.k12.il.us/Wetlands/Salamander/SalGraphics/salfoodweb.giff
9
Graph Representation: Adjacency Matrix
Adjacency matrix. n-by-n matrix with Auv = 1 if (u, v) is an edge.
Two representations of each edge.
Space proportional to n2.
Checking if (u, v) is an edge takes (1) time.
Identifying all edges takes (n2) time.




1
2
3
4
5
6
7
8
1
0
1
1
0
0
0
0
0
2
1
0
1
1
1
0
0
0
3
1
1
0
0
1
0
1
1
4
0
1
0
1
1
0
0
0
5
0
1
1
1
0
1
0
0
6
0
0
0
0
1
0
0
0
7
0
0
1
0
0
0
0
1
8
0
0
1
0
0
0
1
0
10
Graph Representation: Adjacency List
Adjacency list. Node indexed array of lists.
Two representations of each edge.
degree = number of neighbors of u
Space proportional to m + n.
Checking if (u, v) is an edge takes O(deg(u)) time.
Identifying all edges takes (m + n) time.




1
2
3
2
1
3
4
5
3
1
2
5
7
4
2
5
5
2
3
4
6
6
5
7
3
8
8
3
7
8
11
Paths and Connectivity
Def. A path in an undirected graph G = (V, E) is a sequence P of nodes
v1, v2, …, vk-1, vk with the property that each consecutive pair vi, vi+1 is
joined by an edge in E.
Def. A path is simple if all nodes are distinct.
Def. An undirected graph is connected if for every pair of nodes u and
v, there is a path between u and v.
12
Cycles
Def. A cycle is a path v1, v2, …, vk-1, vk in which v1 = vk, k > 2, and the
first k-1 nodes are all distinct.
cycle C = 1-2-4-5-3-1
13
Trees
Def. An undirected graph is a tree if it is connected and does not
contain a cycle.
Theorem. Let G be an undirected graph on n nodes. Any two of the
following statements imply the third.
G is connected.
G does not contain a cycle.
G has n-1 edges.



14
Rooted Trees
Rooted tree. Given a tree T, choose a root node r and orient each edge
away from r.
Importance. Models hierarchical structure.
root r
parent of v
v
child of v
a tree
the same tree, rooted at 1
15
Phylogeny Trees
Phylogeny trees. Describe evolutionary history of species.
16
GUI Containment Hierarchy
GUI containment hierarchy. Describe organization of GUI widgets.
Reference: http://java.sun.com/docs/books/tutorial/uiswing/overview/anatomy.html
17
3.2 Graph Traversal
Connectivity
s-t connectivity problem. Given two nodes s and t, is there a path
between s and t?
s-t shortest path problem. Given two nodes s and t, what is the length
of the shortest path between s and t?
Applications.
Facebook.
Maze traversal.
Erdos number.
Kevin Bacon number.
Fewest number of hops in a communication network.





19
Breadth First Search
BFS intuition. Explore outward from s in all possible directions, adding
nodes one "layer" at a time. Effect: find “shallow” paths to nodes.
s
L1
L
L
2
n-1
BFS algorithm.
L0 = { s }.
L1 = all neighbors of L0.
L2 = all nodes that do not belong to L0 or L1, and that have an edge
to a node in L1.
Li+1 = all nodes that do not belong to an earlier layer, and that have
an edge to a node in Li.




Theorem. For each i, Li consists of all nodes at distance exactly i
from s. There is a path from s to t iff t appears in some layer.
20
Implementing BFS
Q: What’s a good way to implement the above algorithm?
A: Use a queue for the “frontier”
21
Breadth First Search
Property. Let T be a BFS tree of G = (V, E), and let (x, y) be an edge of
G. Then the level of x and y differ by at most 1.
L0
L1
L2
L3
22
Breadth First Search: Analysis
Theorem. The above implementation of BFS runs in O(m + n) time if
the graph is given by its adjacency list representation.
Pf.


Easy to prove O(n2) running time:
– at most n lists Li
– each node occurs on at most one list; for loop runs  n times
– when we consider node u, there are  n incident edges (u, v),
and we spend O(1) processing each edge
Actually runs in O(m + n) time:
– when we consider node u, there are deg(u) incident edges (u, v)
– total time processing edges is uV deg(u) = 2m
▪
each edge (u, v) is counted exactly twice
in sum: once in deg(u) and once in deg(v)
23
Connected Component
Connected component. Find all nodes reachable from s.
Connected component containing node 1 = { 1, 2, 3, 4, 5, 6, 7, 8 }.
24
Q1: Finding connected components
Give an algorithm to find the set of all connected components of an
undirected graph.
25
Connected Component
Connected component. Find all nodes reachable from s.
s
R
u
v
it's safe to add v
Theorem. Upon termination, R is the connected component containing s.
BFS = explore in order of distance from s.

26
Q2: Flood Fill
Flood fill. Given lime green pixel in an image, change color of entire
blob of neighboring lime pixels to blue.
recolor lime green blob to blue
27
Flood Fill
Flood fill. Given lime green pixel in an image, change color of entire
blob of neighboring lime pixels to blue.
recolor lime green blob to blue
28
Flood Fill
Flood fill. Given lime green pixel in an image, change color of entire
blob of neighboring lime pixels to blue.
Node: pixel.
Edge: two neighboring lime pixels.
Blob: connected component of lime pixels.



recolor lime green blob to blue
29
Depth-first search
Use recursion
DFS intuition. Explore outward from s along one path as far as
possible, and backtrack when you cannot progress. Effect: find
faraway nodes.
DFS(u):
Mark u as “Explored” and add u to R
For each edge (u,v) incident to u
If v is not marked “Explored” then
Recursively call DFS(v)
30
Depth-first search
Property. For a given recursive call DFS(u), all nodes marked
“Explored” between the beginning and end of this recursive call are
descendants of u in T.
Theorem. Let T be a depth-first search tree, let x and y be nodes in T,
and let (x,y) be an edge of G that is not an edge of T. Then one of x
or y is an ancestor of the other.
31
Q3: BFS and DFS trees
We have a connected graph G = (V, E) and a specific vertex u. Suppose
we compute a DFS tree rooted at u, and obtain a tree T that
includes all nodes of G. Suppose we then compute a BFS tree rooted
at u, and obtain the same tree T.
Prove that G = T.
32
Answer
Suppose G has an edge e = {a, b} that does not belong to T.
As T is a DFS tree, one of the two ends must be an ancestor of the
other—say a is an ancestor of b.
(*) Since T is a BFS tree, the distance of the two nodes from u in T can
differ at most by one.
But if a is an ancestor of b, and (*) holds, then a must be the direct
parent of b. This means that {a, b} is an edge in T. Contradiction.
33
Q4: Finding a cycle
Given a graph G, determine if it has a cycle. If so, the algorithm should
output this cycle.
Answer: Assume that G is connected; otherwise work on the connected
components.
Run BFS from an arbitrary node s, and obtain a BFS tree T. If every
edge of G appears in the tree, then G = T and there is no cycle.
Otherwise, there is an edge e = (v, w) that is in G but not in T. Consider
the least common ancestor u of v and w in T. We get a cycle from
edge e and paths u-v and u-w in T.
34