Branch and BoundTravelling Salesman Problem

Branch and Bound is a state space search method in which all the children of a node are generated before expanding any of its children.

Least Cost or max profit

This scheme associates a cost or profit with each node. If we are searching for a solution with least cost, then the list of live nodes can be set up as a min heap. The next E-node is the live node with least cost. If we want a solution with maximum profit, the live node list can be set up as a max heap. The next E-node is the live node with maximum profit.

1

2 3 4 5

Live Node: 2, 3, 4, and 5

1

2 3 4 5

6 7 8 9

1

2 3 4 5

6 78 9

FIFO Branch & Bound (BFS)

Children of E-node are inserted in a queue.

LIFO Branch & Bound (D-Search)

Children of E-node are inserted in a stack.

Ways to select E-Node

Requirements

• Branching: A set of solutions, which is represented by a node, can be partitioned into mutually exclusive sets. Each subset in the partition is represented by a child of the original node.

• Lower bounding: An algorithm is available for calculating a lower bound on the cost of any solution in a given subset.

Searching: Least-cost search (LC)•Cost and approximation•Each node, X, in the search tree

is associated with a cost: C(X)•C(X) = cost of reaching the

current node, X (E-node), from the root + the cost of reaching an answer node from X.

C(X) = g(X) + h(X)

Example: 8-puzzle Cost function: C

^

= g(x) +h(x)

Where,h(x)=the number of misplaced tilesg(x)=the number of moves so far

Assumption: move one tile in any direction cost 1,

Initial State Final State1 2 3

5 6

7 8 4

1 2 3

5 8 6

7 4

1 2 3

5 6

7 8 4

1 2 3

5 6 4

7 8

1 2 3

5 6

7 8 4

1 2

5 6 3

7 8 4

541^C

321^C

541^C

1 2 3

5 8 6

7 41 2 3

5 6

7 8 4

1 3

5 2 6

7 8 4

312^

C

532^C

532^C

1 2 3

5 8 6

7 4

1 2 3

5 8 6

7 4

523^C

303^C

Travelling salesman problem

In this problem we are given an n vertex network (either directed or undirected) and are to find a cycle of minimum cost that includes all n vertices. Any cycle that includes all n vertices of a network is called a tour. In the traveling-salesperson problem, we are to find a least-cost tour.

Four-Vertex network

3

30

10

20

6

5

4

1 2

4

State space diagram or permutation tree

A

B

DC E

GF H I J K

L M N O P Q

1

23

4

3 4 2 4 2 3

4 3 4 2 3 2

formulas• S={1,p,1|p=Permutation of (2,3,…,n)}

Size of S=(n-1)!

oT(S)=T(R)+A(i,j)+r

where,

R=parent node

A=reduced cost matrix for node R

S=child of R such that <R,S>=<i,j> belongs to E

r=reduced constant

EXAMPLELet the following cost matrix be for n=5 in graph G(V,E):

State space tree

Vertex = 3 Vertex = 5

6 7 8

10

4 535 53 25

Vertex = 2 Vertex = 5Vertex = 3

3

Vertex = 2 Vertex = 5Vertex = 4Vertex = 3

28 50 36

52 28

251

2 31

9

11 28

Vertex = 3

Reduction by row: 1

02

23

4

Reduction by column:

1 3

The reduced cost=(10+2+2+3+4)+(1+3) =25Cost of node1=Lower Bound=25

Reduced Matrix:

Cost (2):• The resulting cost matrix is:

T(2)=T(1)+A(1,2)+r =25+10+0 =35

Cost (3):

• The resulting cost matrix is:

T(3)=T(1)+A(1,3)+r =25+17+11 =53

Cost (4):

• The resulting cost matrix is:

T(4)=T(1)+A(1,4)+r =25+0+0 =25

• In summary:So the live nodes we have so far are:

o2: cost(2) = 35, path: 1->2o3: cost(3) = 53, path: 1->3o4: cost(4) = 25, path: 1->4o5: cost(5) = 31, path: 1->5

Explore the node with the lowest cost: Node 4 has a cost of 25Vertices to be explored from node 4: 2, 3, and 5Now we are starting from the cost matrix at node 4 is:

Cost(4):

•Choose to go to vertex 2: Node 6 (path is 1->4->2)

The resulting cost matrix is:

T(6)=T(4)+A(4,2)+r =25+3+0 =28

•In summary: So the live nodes we have so far are:

o 2: cost(2) = 35, path: 1->2o 3: cost(3) = 53, path: 1->3o 5: cost(5) = 31, path: 1->5o 6: cost(6) = 28, path: 1->4->2o 7: cost(7) = 50, path: 1->4->3o 8: cost(8) = 36, path: 1->4->5

Explore the node with the lowest cost: Node 6 has a cost of 28

Vertices to be explored from node 6: 3 and 5

Now we are starting from the cost matrix at node 6 is:

Cost(6)=28

•Choose to go to vertex 3: Node 9 ( path is 1->4->2->3 )

The resulting cost matrix is:

T(9)=T(6)+A(2,3)+r =28+11+13 =52

•Choose to go to vertex 3: Node 10 ( path is 1->4->2->5)

The resulting cost matrix is:

T(10)=T(6)+A(2,5)+r =28+0+0 =28

•In summary: So the live nodes we have so far are:

o 2: cost(2) = 35, path: 1->2o 3: cost(3) = 53, path: 1->3o 5: cost(5) = 31, path: 1->5o 7: cost(7) = 50, path: 1->4->3o 8: cost(8) = 36, path: 1->4->5o 9: cost(9) = 52, path: 1->4->2->3o 10: cost(2) = 28, path: 1->4->2->5

Explore the node with the lowest cost: Node 10 has a cost of 28

Vertices to be explored from node 10: 3

Now we are starting from the cost matrix at node 10 is:

•Choose to go to vertex 3: Node 11( path is 1->4->2->5->3 )

The resulting cost matrix is:

T(11)=T(10)+A(5,3)+r =28+0+0 =28