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TSP in line graphs

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Contract each circle Ci in H into a single node ni .There is an ... Euler circuit for G by merging circles based on the minimum spanning tree obtained in Step3. ... – PowerPoint PPT presentation

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Title: TSP in line graphs


1
TSP in line graphs
  • 2-optimal Euler path problem

2
Euler circuit in a directed graph
  • An Euler circuit in a directed graph is a
    directed circuit that visits every edge in G
    exactly once.

b
d
e
g
a
c
f
Figure 1
3
Theorem 2
  • If all the vertices of a connected graph have
    equal in-degrees and out-degrees, then the graph
    has an Euler circuit.

4
2-path
  • A 2-path is a directed subgraph consisting of two
    consecutive edges. For example
  • v is called the midpoint.
  • Every 2-path is given a cost (positive number)
  • A Euler circuit of m edges contains m 2-paths,and
    the cost of the Euler circuit is the total weight
    of the m 2-paths.

u
v
w
5
2-optimal Euler circuit problem
  • Instance A directed graph, each of its vertices
    has in-degree 2 and out-degree 2 and 2-path has a
    cost.
  • QuestionFind an Euler circuit with the smallest
    cost among all possible Euler circuits.

6
Line graph
  • If we view each edge in the above graph G(Figure
    1) as a vertex,and each 2-path in the above graph
    as an edge,we get another graph L(G), called line
    graph.

d
b
e
a
c
g
f
Figure 2
7
Continue
  • Observation An Euler circuit in graph G
    corresponds to a Hamilton circuit in graph L(G).
  • 2-optimal Euler path problem becomes a TSP
    problem in line graph.

8
Theorem 3
  • TSP in line graph with in-degree 2 and out-degree
    2 can be solved in polynomial time. (TSP in
    general is NP-complete)

9
Facts
  • For each node in G,there are four 2-paths that
    form 2 pairs of 2-paths.
  • In an Euler circuit,one of the pairs is used.
  • A pair of 2-path for v is good if the total cost
    of this pair of 2-paths is not less than that of
    the other pair.

10
Algorithm
  • For each node v in G, fixed the good pair of
    2-path for v. (This leads to a set of edge
    disjoint circles H.)
  • Contract each circle Ci in H into a single node
    ni .There is an edge(undirected) between ni and
    nj if Ci and Cj have a common vertex,say,v. The
    weight on edge (ni,nj) is w(e1)w(e2)-w(e3)-w(e4)

    where e1 and e2 forms a bad pair for v,and e3 and
    e4 forms a good pair for v.Call H be the
    resulting undirected graph.
  • Construct a minimum spanning tree H
  • Construct an Euler circuit for G by merging
    circles based on the minimum spanning tree
    obtained in Step3.

11
Example
  • Suppose that the directed Euler graph G is Given
    in Figure 3(next page).There are 5 cycles.The
    2-paths in circles have cost 1 and the costs on
    other 2-paths are listed below
    w(a,e)2,w(h,b)2 w(f,i)3,w(l,g)3
    w(g,t)2,w(s,h)11
    w(k,m)1,w(p,l)2
    w(t,o)2,w(n,q)3.

12
j
f
i
k
e
l
a
d
m
p
g
b
h
c
o
t
n
s
r
q
Figure 3
13
Step1we get 5 circles shown below
14
Step2the undirected graph constructed is
e,f,g,h
4
i,j,k,l
2
1
11
a,b,c,d
3
m,n,o,p
q,r,s,t
15
Step3the minimum spanning tree for the above
tree is as follows
4
2
1
3
16
Step4The 2-optimal Euler path is shown below
17
Theorem 4
  • The above algorithm is correct.
  • Proof
  • The total cost of the 2-paths obtained in Step1
    is not greater than that of the optimal solution.
  • To obtain an Euler circuit, we have to merge the
    k circles obtained in Step1.(needs (k-1) merge
    operations)
  • These (k-1) merge operations correspond to (k-1)
    edges in the undirected graph.(In fact, a
    spanning tree)
  • The final cost is the cost of all the 2-paths
    obtained in Step1 the cost of the (minimum)
    spanning tree.
  • Since we use a minimum spanning tree, the cost of
    our solution is the smallest.

18
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