Post on 06-Jul-2020
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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Example
Try to solve the TSP for the following weighted graph.
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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)!
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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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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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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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