Lecture 5:
Graph Theory and
Longest Paths
© J. Christopher Beck 2008
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Outline
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What is Graph Theory?
Graph Theory Basics
Finding the Longest Path
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Why? (modeling)
How
© J. Christopher Beck 2008
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Readings
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Gibbons,
Algorithmic Graph Theory, Cambridge
University Press, 1985
Sedgewick, Algorithms in C++, 3rd
Edition, 2002
Wikipedia (“Graph Theory”), accessed
August 12, 2008
© J. Christopher Beck 2008
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The Bridges of Konigsberg
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Can you walk a
route that crosses
each bridge
exactly once?
Euler, 1736
Islands
© J. Christopher Beck 2008
Bridges
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The Bridges of Konigsberg
Edge or arc
Node or vertex
© J. Christopher Beck 2008
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The Bridges of Konigsberg

So, now the
problem is: can
you traverse every
edge exactly once?

Any ideas?
© J. Christopher Beck 2008
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Graph Theory
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Nodes and edges form a graph, G(V,E),
with V being the set of nodes, E being
the set of edges
Edges may be directed or un-directed
Edges may have a weight
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Just a number associated with each edge
© J. Christopher Beck 2008
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Graph Theory
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Graph theory is
the secret to the
universe
A tremendous number of problems can
be modeled and solved using graph
theory
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OR, epidemiology, geometry, IT (the web is
a graph), social networks, biological
pathways, chemical reactions, …
© J. Christopher Beck 2008
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CPM
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Can we pose this as a
graph theory problem?
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Why would we want to?
© J. Christopher Beck 2008
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CPM Model
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Job  Node
Precedence  Directed Edge
Processing time  Edge Weight
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(This is the job-on-arc format: P p. 52)
© J. Christopher Beck 2008
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Our Small Example
1
Job p(j) Predecessors
1 2
2 3
3 1
4 4
1,2
5 2
2
6 1
4
2
4
3
5
1
2
3
© J. Christopher Beck 2008
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2
3
3
4
4
6
5
What are we missing?
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Dummy Nodes
1
2
3
4
6
5
0
0
0
0
1
2
3
2
3
3
4
4
6
2
5
1
7
1
© J. Christopher Beck 2008
Two dummy nodes
(0 and n+1)
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Makespan?
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How do we find the
makespan?
Critical Path?
0
0
0
0
1
2
3
2
3
3
4
4
6
2
5
1
7
1
© J. Christopher Beck 2008
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Initialize Nodes & Queue
class Node {
public int label;
public int numPredecessors;
public int head;
};
// initialize Nodes.label and numPredecessors
// from input data. head = 0
Node node0 = initial dummy node
Node nodeLast = final dummy node
queue.push(node0) ;
© J. Christopher Beck 2008
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Process Each Node
while(!queue.empty()) {
Node n = queue.pop();
for each e = out-going edge of n
Node next = e.getOtherNode(n);
if (n.head + e.weight > next.head)
next.head = n.head + e.weight;
--next.numPredecessors;
if (next.numPredecessors == 0)
queue.push(next);
}
© J. Christopher Beck 2008
After executing this, where is the makespan?
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Critical Path
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How would you modify
this algorithm to also
allow you to find a
critical path?
© J. Christopher Beck 2008
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Add a Field
class Node {
public int label;
public int numPredecessors;
public int head;
public Node criticalPredecessor;
};
© J. Christopher Beck 2008
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Add a Line
Then follow the
criticalPredecessor links
from nodeLast back
to node0.
while(!queue.empty()) {
Node n = queue.pop();
for each e = out-going edge of n
Node next = e.getOtherNode(n);
if (n.head + e.weight > next.head)
next.head = n.head + e.weight;
next.criticalPredecessor = n;
--next.numPredecessors;
if (next.numPredecessors == 0)
queue.push(next);
Does this form of algorithm (i.e., use of
© J. Christopher Beck 2008
criticalPredecessor) remind you of anything?
}
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Example 4.2.3
Jobs 1 2 3 4
pj
5
6
7
8
9 10 11 12 13 14
5 6 9 12 7 12 10 6 10 9
7
8
7
5
Model and solve using a graph
© J. Christopher Beck 2008
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The Bridges of Konigsberg

So, now the
problem is: can
you traverse every
edge exactly once?

Any ideas?
Hint: The degree of a node is the number of its edges.
Think about the degrees of the nodes in any path.
© J. Christopher Beck 2008
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