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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.
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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.
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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|.
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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
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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
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World Wide Web
Web graph.
Node: web page.
Edge: hyperlink from one page to another.
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cnn.com
netscape.com
novell.com
cnnsi.com
timewarner.com
hbo.com
sorpranos.com
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9-11 Terrorist Network
Social network graph.
Node: people.
Edge: relationship between two people.
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Reference: Valdis Krebs, http://www.firstmonday.org/issues/issue7_4/krebs
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Ecological Food Web
Food web graph.
Node = species.
Edge = from prey to predator.
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Reference: http://www.twingroves.district96.k12.il.us/Wetlands/Salamander/SalGraphics/salfoodweb.giff
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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.
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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.
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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.
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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
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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.
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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
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Phylogeny Trees
Phylogeny trees. Describe evolutionary history of species.
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GUI Containment Hierarchy
GUI containment hierarchy. Describe organization of GUI widgets.
Reference: http://java.sun.com/docs/books/tutorial/uiswing/overview/anatomy.html
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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.
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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.
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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.
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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.
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Implementing BFS
Q: What’s a good way to implement the above algorithm?
A: Use a queue for the “frontier”
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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
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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.
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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
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each edge (u, v) is counted exactly twice
in sum: once in deg(u) and once in deg(v)
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Connected Component
Connected component. Find all nodes reachable from s.
Connected component containing node 1 = { 1, 2, 3, 4, 5, 6, 7, 8 }.
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Q1: Finding connected components
Give an algorithm to find the set of all connected components of an
undirected graph.
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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.
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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
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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
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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.
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recolor lime green blob to blue
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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)
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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.
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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.
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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.
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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.
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