The Travelling Salesman Problem

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Overview

The Travelling Salesman Problem

Example

Lower Bounds for TSP

An Upper Bound for TSP

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Outline

The Travelling Salesman Problem

Example

Lower Bounds for TSP

An Upper Bound for TSP

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The Travelling Salesman Problem

Given a collection of cities and the cost of travel between eachpair of them, the travelling salesman problem, or TSP forshort, is to find the cheapest way of visiting all of the citiesand returning to your starting point.

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A Tour of 13509 U.S. Cities

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Outline

The Travelling Salesman Problem

Example

Lower Bounds for TSP

An Upper Bound for TSP

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Example

Try to solve the TSP for the following weighted graph.

a

b

c

d

e

f

2

99

4

6

4

8

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8

8

4

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Weight Matrix

a b c d e fa 0 2 9 0 9 4b 2 0 6 4 8 0c 9 6 0 6 8 8d 0 4 6 0 0 4e 9 8 8 0 0 3f 4 0 8 4 3 0

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The Hardness of TSP

FactIf there are n cities with complete set of routes between them,then the number of possible tours is (n−1)!

2.

Number of Cities Number of Tours Time5 12 12 microseconds8 2520 2.5 milliseconds

10 181,440 0.18 seconds12 19,958,400 20 seconds15 87,178,291,200 12.1 hours18 177,843,714,048,000 5.64 years20 60,822,550,204,416,000 1927 years

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Outline

The Travelling Salesman Problem

Example

Lower Bounds for TSP

An Upper Bound for TSP

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A Basic Lower Bound for TSP

Lower bounds can be found by using spanning trees.

Since removal of one edge from any Hamilton cycle yields aspanning tree, we see that

Solution to TSP > minimum length of a spanning tree (MST).

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A Better Lower Bound for TSP

Consider any vertex v in the graph G . Any Hamilton cycle inG has to consist of two edges from v , say vu and vw , and apath from u to w in the graph G \ {v} obtained from G byremoving v and its incident edges.

Since this path is a spanning tree of G \ {v}, we have

Solution to TSP ≥ (sum of lengths of two shortest edges from v)

+ (MST of G \ {v}).

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Outline

The Travelling Salesman Problem

Example

Lower Bounds for TSP

An Upper Bound for TSP

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An Upper Bound for TSP

Suppose that G is a complete graph with edge weights givenby distances satisfying the triangle inequality

d(x , z) ≤ d(x , y) + d(y , z)

where d(x , y) denotes the shortest distance along edges fromx to y .

Under these assumptions, we can construct a Hamilton cycleof G by taking a minimum spanning tree and then replacingedges with shortcuts, and obtain that

solution to TSP ≤ 2MST .

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Acknowledgements

Statements of results follow the notation and wording ofAnderson’s First Course in Discrete Mathematics.

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