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PPT – Approximation Algorithms for the Traveling Salesperson Problem PowerPoint presentation | free to view - id: 164ebe-NjdkO

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Approximation Algorithms for the Traveling

Salesperson Problem

Approximation Algorithm

- Up to now, the best algorithm for solving an

NP-complete problem requires exponential time in

the worst case. It is too time-consuming. - To reduce the time required for solving a

problem, we can relax the problem, and obtain a

feasible solution close to an optimal solution

Approximation Algorithm

Error Ratio Bound

(1?)-approximation algorithm

An approximation algorithm for Euclidean

traveling salesperson problem (ETSP)

- The ETSP is to find a shortest closed path

through a set S of n points in the plane. - The ETSP is NP-hard.

Approximation Algorithm for ETSP

- Input A set S of n points in the plane.
- OutputAn approximate traveling salesperson tour

of S. - Step 1 Find a minimal spanning tree T of S.
- Step 2 Find a minimal Euclidean weighted

matching M on the set of vertices of odd degrees

in T. Let GM?T. - Step 3 Find an Eulerian cycle of G and then

traverse it to find a Hamiltonian cycle as an

approximate tour of ETSP by bypassing all

previously visited vertices.

Eulerian Cycle

- An Eulerian path (Eulerian trail, Euler walk) in

a graph is a path that uses each edge precisely

once. If such a path exists, the graph is called

traversable. - An Eulerian cycle (Eulerian circuit, Euler tour)

in a graph is a cycle with uses each edge

precisely once. If such a cycle exists, the graph

is called Eulerian.

- L. Euler showed that an Eulerian cycle exists if

and only if all vertices in the graph are of even

degree and all edges are contained in the same

component. - L. Euler also showed an Eulerian path exists, if

and only if at most two vertices in the graph are

of odd degree and all edges are contained in the

same component.

- Leonhard Euler (April 15, 1707 - September 18,

1783) (pronounced "oiler") was a Swiss

mathematician and physicist. He is considered

(together with Gauss) to be the greatest

mathematician ever. - Leonhard Euler stated and solved the problem of

Seven Bridges of Königsberg in 1736, which is the

first formally discussed problem in graph theory.

Eulerian Cycle exists becausedegree(V1)

4 degree(V2) 2 degree(V3) 4 degree(V4) 4.

This is the Eulerian Cycle.

- A Hamiltonian path (also called traceable path)

is a path that visits each vertex exactly once. - A Hamiltonian cycle (also called Hamiltonian

circuit, vertex tour or graph cycle) is a cycle

that visits each vertex exactly once, except for

the starting vertex.

Minimal Euclidean Weighted Matching Problem

- Given a set of points in the plane, the minimal

Euclidean weighted matching problem is to join

the points in pairs by line segments such that

the total length is minimum.

Approximation Algorithm for ETSP

- E.g.
- Step1

Approximation Algorithm for ETSP

- Step2The number of points with odd degrees must

be even. , which is even

One edge contributes 2 degrees

Approximation Algorithm for ETSP

- Step3

P3 and P4 are visited twice. By bypassing P3 and

P4 and connecting P6 to P1 directly, we obtain a

Hamiltonian cycle.

Approximation Algorithm for ETSP

- Time complexity O(n3)
- Step 1 O(nlogn)
- Step 2 O(n3)
- Step 3 O(n)
- How close the approximate solution to an optimal

solution?

How good is the solution ?

- The approximate tour is within 3/2 of the optimal

one. - Reasoning
- L optimal ETSP tour, T MST, Lp a path

derived by removing one edge from L (Lp is also a

spanning tree) - ? length(T)?length(Lp)?length(L)
- Let Lpj1i1j2i2j3i2m, where i1, i2,, i2m is

the set of odd-degree vertices in T where indices

of vertices in the set are arranged in the same

order as they are in the optimal ETSP tour L.

How good is the solution ?

- Let M be the minimal Euclidean weighted matching.
- Consider the two matchingsM1i1,i2,i3,i4,,

i2m-1,i2m and M2i2,i3,i4,i5,,i2m,i1.

We have - length(L)? length(M1) length(M2) (triangular

inequality) ? 2 length(M ) - ? length(M)? 1/2 length(L )
- G T?M
- ? length(T) length(M) ? length(L) 1/2

length(L) - 3/2 length(L)

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The Bottleneck Traveling Salesperson Problem

(BTSP)

- Minimize the longest edge of a tour.
- This is a mini-max problem.
- This problem is NP-hard.
- The input data for this problem fulfill the

following assumptions - The graph is a complete graph.
- All edges obey the triangular inequality rule.

An Algorithm for Finding an Optimal Solution

- Step1 Sort all edges in G (V,E) into a

nondecreasing sequence e1?e2??em. Let

G(ei) denote the subgraph obtained from G by

deleting all edges longer than ei. - Step2 i?1
- Step3 If there exists a Hamiltonian cycle in

G(ei), then this cycle is the solution hence,

stop. - Step4 i?i1. Go to Step 3.

An Example for BTSP Algorithm

- e.g.

- There is a Hamiltonian cycle, A-B-D-C-E-F-G-A, in

G(BD). - The optimal solution is 13.

Time complexity of the algorithm

- The Hamiltonian cycle problem is NP-hard.
- The algorithm cannot be a polynomial one.

Theorem for Hamiltonian Cycles

- Def The t-th power of G(V,E), denoted as

Gt(V,Et), is a graph that an edge (u,v)?Et if

there is a path from u to v with at most t edges

in G. - Theorem If a graph G is bi-connected, then G2

has a Hamiltonian cycle.

An Example for the Theorem

G2

A Hamiltonian cycle A-B-C-D-E-F-G-A

An Approximation Algorithm for BTSP

- Input A complete graph G(V,E) where all edges

satisfy triangular inequality. - Output A tour in G whose longest edge is not

greater than twice of the value of an optimal

solution to the special bottleneck traveling

salesperson problem of G. - Step 1 Sort the edges into e1?e2??em.
- Step 2 i 1.
- Step 3 If G(ei) is bi-connected, construct

G(ei)2, find a Hamiltonian cycle in G(ei)2 and

return this as the output. - Step 4 i i 1. Go to Step 3.

An Example

Add some more edges. Then it becomes bi-connected.

- A Hamiltonian cycle A-G-F-E-D-C-B-A.
- The longest edge 16

Time complexity

- The algorithm is of polynomial time
- The determination of bi-connectedness can be

solved by polynomial algorithms. - If G is bi-connected, then there exists a

polynomial algorithm to solve the Hamiltonian

cycle problem of G2.

How Good is the Solution ?

- The value of an approximate solution is bounded

by two times that of an optimal solution. - Reasoning
- A Hamiltonian cycle is bi-connected.
- eop the longest edge of an optimal solution
- G(ei) the first bi-connected graph
- ei?eop
- The length of the longest edge in G(ei)2?2ei
- (triangular inequality)

?2eop