© J. Christopher Beck 2008 1

Lecture 5: Graph Theory andLongest Paths

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Outline What is Graph Theory? Graph Theory Basics Finding the Longest Path

Why? (modeling) How

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Readings

Gibbons,Algorithmic Graph Theory, Cambridge University Press, 1985

Sedgewick, Algorithms in C++, 3rd Edition, 2002

Wikipedia (“Graph Theory”), accessed August 12, 2008

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The Bridges of Konigsberg

Can you walk a route that crosses each bridge exactly once?

Euler, 1736Islands Bridges

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The Bridges of Konigsberg

Node or vertex

Edge or arc

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The Bridges of Konigsberg

So, now the problem is: can you traverse every edge exactly once? Any ideas?

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Graph Theory

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 Just a number associated with each

edge

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Graph Theory

Graph theory is the secret to theuniverse

A tremendous number of problems can be modeled and solved using graph theory OR, epidemiology, geometry, IT (the web

is a graph), social networks, biological pathways, chemical reactions, …

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CPM

Can we pose this as agraph theory problem? Why would we want to?

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CPM Model

Job Node Precedence Directed Edge Processing time Edge Weight

(This is the job-on-arc format: P p. 52)

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Our Small Example

Job p(j) Predecessors1 2 -2 3 -3 1 -4 4 1,25 2 26 1 4

1

2

3

64

5

1

2

3

64

5

2

3

3

4

What are we missing?

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Dummy Nodes1

2

3

64

5

0

0

0

0

1

2 64

5

2

3

3

4

7

1

2

1

Two dummy nodes(0 and n+1)

3

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Makespan?

0

0

0

0

1

2 64

5

2

3

3

4

7

1

2

1

3

How do we find the makespan?

Critical Path?

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Initialize Nodes & Queueclass 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) ;

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Process Each Nodewhile(!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);

}After executing this, where is the makespan?

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Critical Path

How would you modify this algorithm to also allow you to find a critical path?

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Add a Fieldclass Node {

public int label;

public int numPredecessors;

public int head;

public Node criticalPredecessor;

};

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Add a Linewhile(!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);

}

Then follow thecriticalPredecessor links

from nodeLast backto node0.

Does this form of algorithm (i.e., use ofcriticalPredecessor) remind you of anything?

© J. Christopher Beck 2008 19

Example 4.2.3

Jobs 1 2 3 4 5 6 7 8 9 10

11

12

13

14

pj 5 6 9 12

7 12

10

6 10

9 7 8 7 5

Model and solve using a graph

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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.