Discrete Structures

# Lecture Notes, 6 Dec 2010

 Last Update: 6 December 2010 Note: or material is highlighted

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### §9.1; §9.5 to p. 640;§9.6, esp. pp. 653–655: Graphs (cont'd)

1. Graphs can be used to model:

1. computer networks

1. including the Internet:

• nodes = computers
arcs = network connections

or

• nodes = local networks
arcs = network connections to the Internet

2. and the World Wide Web:

3. and social networks

2. tournaments:

• a•→•b for: a beats b

3. road maps

4. words and their meanings

• Let each vertex represent a word,
and let each edge connect a word with the words in its definition

• See Massé 2008

5. the human brain (or the mind (via a "semantic network"))

• Let vertices represent neurons (or else ideas)
• Let edges represent synapses (or else associations between ideas)

6. anything where ∃ binary relation:

2. Euler Paths & Circuits:

• pronounced "OILer"

1. Read about the 7 bridges of Königsberg:

• The K-Bridge graph:

• Let nodes represent land, arcs represent bridges:

2. Defs:

3. More Defs:

1. A connected graph =def
a graph such that ¬∃ 2 subgraphs with no common edge.

2. A multigraph =def a graph with duplicate edges

• i.e.) some pairs of vertices are connected by ≥2 edges

3. Let G=(V,E) be a graph.
Let v ∈ V.
Then degree(v) =def
the number of edges with v as an endpoint

• i.e.) degree(v) = |{e ∈ E | ∃v′[e = {v,v&prime}]}|

4. Thm 1:

A connected multigraph has an Euler circuit
each vertex has even degree.

1. e.g.)

2. ∴ the Königsberg Bridge graph has no Euler circuit!

3. Neither does:

5. Thm 2:

A connected multigraph has an Euler path
(but no Euler circuit)
it has exactly 2 vertices of odd degree

3. Traveling Salesman Problem:

1. Salesman wants to visit n cities exactly once each
& return to starting point by shortest path
(e.g., to save gas or time)
(possibly traveling over same road >1 times)

1. This is called a Hamiltonian circuit:

• travel through every vertex exactly once
& return to starting point

• inspired by a wooden puzzle devised by Hamilton

2. Compare Euler circuit:

• travel through every edge exactly once, & return

3. See articles by Brian Hayes, especially Hayes 2008.

2. Fact 1:

There are no known necessary & sufficient conditions for Hamiltonian circuits
(as there are for Euler circuits)

3. Fact 2:

Let graph G have n vertices.
Let v0 be the starting vertex.
∴ ∃ ≤ (n–1)! different H-circuits to examine.

• ∃ ≤ n–1 choices for v1
• ∃ ≤ n–2 choices for v2, etc.

But can travel H-circuit in reverse order:

• same trip, only backwards

∴ "only" need to examine ≤ (n–1)!/2 H-circuits to find shortest.

But n=25 → need to examine 24!/2 ≈ 3.1*1023 circuits (in the worst case)!

At 1 nanosecond/circuit (= 10–9sec/circuit), need ∼10,000,000 years!!

• (actually, 9,837,144 years, 321 days, 4 hours, 48 minutes,
but who's counting?)

4. This is computable in theory

• You could easily write a program to find all Hamiltonian circuits and then find the shortest.

but not computable in practice

• What does that mean exactly? Next time!

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