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 each
pair of them, the travelling salesman problem, or TSP for
short, is to find the cheapest way of visiting all of the cities
and 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
2
4
9
9
f
b
8
3
6
8
e
4
8
4
6
c
Weight Matrix
a b c d e
a 0 2 9 0 9
b 2 0 6 4 8
c 9 6 0 6 8
d 0 4 6 0 0
e 9 8 8 0 0
f 4 0 8 4 3
f
4
0
8
4
3
0
d
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The Hardness of TSP
Fact
If 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
5
12
8
2520
10
181,440
12
19,958,400
15
87,178,291,200
18
177,843,714,048,000
20
60,822,550,204,416,000
Time
12 microseconds
2.5 milliseconds
0.18 seconds
20 seconds
12.1 hours
5.64 years
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 a
spanning 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 in
G has to consist of two edges from v , say vu and vw , and a
path from u to w in the graph G \ {v } obtained from G by
removing 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 given
by distances satisfying the triangle inequality
d(x, z) ≤ d(x, y ) + d(y , z)
where d(x, y ) denotes the shortest distance along edges from
x to y .
Under these assumptions, we can construct a Hamilton cycle
of G by taking a minimum spanning tree and then replacing
edges with shortcuts, and obtain that
solution to TSP ≤ 2MST .
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Acknowledgements
Statements of results follow the notation and wording of
Anderson’s First Course in Discrete Mathematics.
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