TSP in line graphs

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TSP in line graphs

• 2-optimal Euler path problem

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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.

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Figure 1
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Theorem 2

• If all the vertices of a connected graph have
  equal in-degrees and out-degrees, then the graph
  has an Euler circuit.

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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.

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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.

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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.

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Figure 2
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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.

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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)

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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.

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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.

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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.

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Figure 3
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Step1we get 5 circles shown below
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Step2the undirected graph constructed is
e,f,g,h
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i,j,k,l
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a,b,c,d
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m,n,o,p
q,r,s,t
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Step3the minimum spanning tree for the above
tree is as follows
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Step4The 2-optimal Euler path is shown below
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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.

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