3.3 the traveling-salesman problem

8/2/2019 3.3 the Traveling-salesman Problem

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Traveling Salesman Problem


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Traveling Salesman Problem (TSP)

Optimization problem:

Given a complete weighted graph, find a minimum weight

Hamiltonian cycle Decision Problem:

Given a complete weighted graph and an integerk, is there

a Hamiltonian cycle with total weight at most k?


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TSP as a Combinatorial

Optimization Problem

Ifn is the number of cities then (n-1)! is the total number ofpossible routes n=5 120 possible routes n=10 3,628,800 possible rotes n=20 2,432,902,008,176,640,000 possible routes

TSP is a NP-Complete combinatorial optimization problem =non-deterministic polynomial time

Scientific challenge for solving TSP

$1 million prize for solving the TSPas one representative problem of alarger class of NP-completecombinatorial optimization problemshas been offered by ClayMathematics Institute of Cambridge

(CMI)


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The traveling-salesman problem

The triangle inequality:if for all vertices u,v, wV, cost function w satisfies w(u, w)

w(u, v) + w(v, w).

The traveling-salesman problem with thetriangle inequality


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ApproxMSTTSP(G

)1 select a vertex rVto be a "root" vertex

2 compute a minimum spanning tree TforG from root r

using MSTPrim(G, w, r)

3 letLbe the list of vertices visited in a preorder tree walk

ofT

4 returnthe hamiltonian cycleHthat visits the vertices in

the orderL


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Even with a simple implementation ofMSTPrim, the running time of

ApproxMSTTSP(G) is O(|V|

2

). TheoremApproxMSTTSP(G) is a

polynomial-time 2-approximation

algorithm for the traveling salesmanproblem with the triangle inequality.


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Proof. LetH* denote an optimal tour for the given set ofvertices. Since we obtain a spanning tree by deleting anyedge from a tour, the weight of the minimum spanningtree Tis a lower bound on the cost of an optimal tour,

that is, w(T) w(H*)-w(e) w(H*)A full walkofTlists the vertices when they are firstvisited and also whenever they are

returned to after a visit to a subtree. Let us call this walkW

. Since the full walk traverses every edge ofTexactlytwice, we have

w(W)=2w(T)


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w(W)2w(H*)

and so the cost ofWis within a factor of 2 of the cost of

an optimal tour.

Unfortunately, Wis generally not a tour, since it visitssome vertices more than once. By the triangle inequality,

however, we can delete a visit to any vertex from Wand

the cost does not increase. By repeatedly applying this

operation, we can remove from Wall but the first visit to

each vertex.


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LetHbe the cycle corresponding to this preorder

walk. It is a hamiltonian cycle, since every vertex is

visited exactly once, and in fact it is the cycle

computed by ApproxMSTTSP(G). SinceHisobtained by deleting vertices from the full walkW,

we have

w(H)w(W)

So w(H)w(W) 2w(H*).


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Other heuristic strategies and Discussion

Nearest neighbor strategy

a b

d c

15

20

35

25

12

10

a b

d c

15

20

35

25

12

10


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NearestNeighbor(G)1 select an arbitrary vertexs to start the cycleH

2 us

3 whilethere are vertices not yet inHdo

4 select an edge (u,v) of minimum weight, where v isnot inH.

5 add edge (u,v) toH

6 u v7 Add the edge (u,s) toH

8 returnH


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a b

d c

15

20

35

25

12

10

a b

d c

15

20

35

25

12

10

a b

d c

15

20

35

25

12

10

a b

d c

15

20

35

25

12

10

H: (a, c), (c, b), (b, d), (d, a)

H=85

H*=72


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1. Like Prims MST algorithm, starts with an arbitraryvertex u and adds edge (u,v) of minimum weight such

that vertex v is not yet in the cycle.

2. However, it always adds edges from the endpoint of

the cycle constructed so far.

3. Worst case time of O(|V|2) for a graph with |V| vertices.


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Nearest Neighbor heuristic worst-case ratio:

R(|I|) = (1+lg|V|)

The ratio grows with |V|.


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For (a):

For(b):

.25.18

10

*)(

)()( ===

Hw

HwIR

.

8

4

*)(

)()(

w

Hw

HwIR

+==


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Shortest link strategy

ShortestLinkedHeuristic(G)

1H

2E' E

3 whileE' do

4 remove the lightest edge (u,v) fromE'

5 if(u,v) does not make a cycle with edges inH

and (u,v) would not be the third edge inHincident on u or

vthen

6 HH { u, v}

7 Add the edge connecting the endpoints of the path inH

8 returnH


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a b

d c

15

20

35

25

12

10

H: (b, c), (c, d), (b, a), (d, a)

H=77H*=72

a b

d c

15

35

25

12

10

20

a b

dc

15

35

25

12

10

20

a b

d c

15

35

25

12

10

20


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Shortest link strategy 1.Like Kruskal's MST algorithm, considers edges in order

of cost, from shortest to longest.

2.Adds edge (u, v) if it does not make a cycle with edgesin the path constructed so far and would not be the third

edge in the path incident on u orv.

3.When |E|-1 edges have been added, connect theendpoints of the path to form a cycle.

4.Running time O(|E|lg|E|) for a graph with |E| edges.

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Insertion heuristic

Nearest insertion(closest-point heuristic)

The vertex closest to any of the vertices already in the

tourHis inserted.

Farthest insertion

The vertex farthest to any of the vertices already in the

tourHis inserted.

Random insertionrandom insertion inserts a vertex uniformly at random.


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When inserting a vertex between twovertices vi and vk, we want to minimize the

added cost to the tour. Vertex vj

is added to

the tour by adding edges (vi, vj) and (vj, vk)

while removing (vi, vk). We choose i and k

as to minimize w(vi, vj) + w(vj, vk) w(vi,vk).


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NearestInsertion(G)

1 Choose an arbitrary node viand let the cycleH

consist of only vi

2 whilethere are vertices not yet inHdo

3 Find a node outsideHclosest to a node inH,

call it vj.

4 Find an edge (vi, vj)inHsuch that w(vi, vk) +

w(vk, vj) w(vi, vj)is minimal.

5 Construct a new cycleHby replacing (vi, vj)

with (vi, vk)and (vk, vj).

6 returnH


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a b

d c

15

20

35

25

12

10

H: (a, c), (c, d), (d, b), (b, a)

H=72H*=72

a b

d c

15

20

35

25

12

10

a b

d c

15

20

35

25

12

10

(a, c): 35+12-

15=32(b, c): 25+12-10=27

(a, b): 35+25-20=40

a b

d c

15

20

35

25

12

10


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Theorem. Any insertion heuristic order gives aR(|I|)=

. 1||lg +V


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Given a complete graph G=(V,E) with positive weights

(distances) on its edges,

The minimum traveling salesman problem consists ofminimizing the cost of a Hamiltonian cycle, the cost of

such a cycle being the sum of the weights on its edges.

The maximum traveling salesman problem consists of

maximizing the cost of a Hamiltonian cycle.


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Homework

Exercises: Page 94

1,2,4

Experiments:

Implement ApproxMSTTSP,NearestNeighbor,ShortestLinkedHeuristic, NearestInsertion and compare

these algorithms


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3.3 the traveling-salesman problem

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