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NADAR SARASWATHI COLLEGE OF ARTS
AND SCIENCE,THENI.
DEPARTMENT OF COMPUTER
SCIENCE
TRAVELING
SALESPERSON
BY
B.POORANI
TRAVELING SALESPERSON
 Let G=(V,E) be a directed graph defining an instance of the
traveling salesperson problem.
 Let Cij be the cost of edges (i,j),Cij=infinity if (i,j) and let
V=N.
 Each leaf node L is a solution node and represents the tour
defined by the path from the root to L.
The progress of the LCBB algorithm on the
problem instance.
The initial reduced matrix is that upper=∞.The
portion of the state space tree that gets generated.
Starting with the root node as the E-node ,nodes 2,
3,4 and 5 are generated.
 The matrix of is obtained from the by (1) setting all entries
in row 1 and column 3 to ∞.
 (2) setting the element at position (3,1) to ∞,and (3)
reducing column 1 by subtracting by 11.
 The value of upper is unchanged and node 4 becomes the
next E-node . Its children 6,7 and 8 are generated.
 The live nodes at this time are node 2,3,5,6,7 and 8 .
 Node 6 has least c value and become the next E-node.
 Nodes 9 and 10 are generated.
 Node 10 is the next E-node . The solution node , node 11 , is
generated.
 The tour length for this node is c(11)=28 and upper is
updated to 28.
 For the next E-node , node 5,c(5)=31>upper.
 To use LCBB to search the traveling salesperson state
space tree , we need to define a cost function c(.) and two
other function c^(.) and u(.).
 c(A)={length of tour defined by the path from the root to A
, if A is a leaf .cost of a minimum-cost leaf in the subtree A ,
if A is not a leaf.
 A different LCBB algorithm can be arrived at by
considering a different tree organization for the solution
space.
 This organization is reached by regarding a tour as a
collection of n edges. If G=(V,E) has e edges , then every
tour contains exactly n of the e edges.
 A possible organization for the state space is a binary tree
in which a left branch represents the inclusion of a
particular edges while the right branch represents the
exclusion of the edges.
 An example of low LCBB would work on the dynamic
binary tree formulation . consider the cost matrix .
 We must decide which edges to use to partition the solution
space into two subsets . if edges (i,j) is used , then the left
subtree of the root represents all tour including edges (i,j)
and the right subtree represents all tour that do not
include edges (i,j).
Thank you

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traveling salesperson.pptx

  • 1. NADAR SARASWATHI COLLEGE OF ARTS AND SCIENCE,THENI. DEPARTMENT OF COMPUTER SCIENCE TRAVELING SALESPERSON BY B.POORANI
  • 2. TRAVELING SALESPERSON  Let G=(V,E) be a directed graph defining an instance of the traveling salesperson problem.  Let Cij be the cost of edges (i,j),Cij=infinity if (i,j) and let V=N.  Each leaf node L is a solution node and represents the tour defined by the path from the root to L.
  • 3. The progress of the LCBB algorithm on the problem instance. The initial reduced matrix is that upper=∞.The portion of the state space tree that gets generated. Starting with the root node as the E-node ,nodes 2, 3,4 and 5 are generated.
  • 4.  The matrix of is obtained from the by (1) setting all entries in row 1 and column 3 to ∞.  (2) setting the element at position (3,1) to ∞,and (3) reducing column 1 by subtracting by 11.  The value of upper is unchanged and node 4 becomes the next E-node . Its children 6,7 and 8 are generated.  The live nodes at this time are node 2,3,5,6,7 and 8 .
  • 5.  Node 6 has least c value and become the next E-node.  Nodes 9 and 10 are generated.  Node 10 is the next E-node . The solution node , node 11 , is generated.  The tour length for this node is c(11)=28 and upper is updated to 28.  For the next E-node , node 5,c(5)=31>upper.
  • 6.  To use LCBB to search the traveling salesperson state space tree , we need to define a cost function c(.) and two other function c^(.) and u(.).  c(A)={length of tour defined by the path from the root to A , if A is a leaf .cost of a minimum-cost leaf in the subtree A , if A is not a leaf.
  • 7.  A different LCBB algorithm can be arrived at by considering a different tree organization for the solution space.  This organization is reached by regarding a tour as a collection of n edges. If G=(V,E) has e edges , then every tour contains exactly n of the e edges.  A possible organization for the state space is a binary tree in which a left branch represents the inclusion of a particular edges while the right branch represents the exclusion of the edges.
  • 8.  An example of low LCBB would work on the dynamic binary tree formulation . consider the cost matrix .  We must decide which edges to use to partition the solution space into two subsets . if edges (i,j) is used , then the left subtree of the root represents all tour including edges (i,j) and the right subtree represents all tour that do not include edges (i,j).