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Branch and bounding : Data structures
Branch and bound is a state space search method that generates all children of a node before expanding any children. It associates a cost or profit with each node and uses a min or max heap to select the next node to expand. For the travelling salesman problem, it constructs a permutation tree representing all possible routes and uses lower bounds and reduced cost matrices at each node to prune the search space and find an optimal solution.
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Explains the concept of Branch and Bound applied to the Travelling Salesman Problem.
Describes Branch and Bound as a state space search method generating all children of a node before expanding.
Outlines how least cost or maximum profit is achieved via min/max heaps for live nodes.
Details ways to select E-nodes using FIFO (BFS) and LIFO (D-Search) approaches in Branch and Bound.
Lists key requirements: branching and lower bounding necessary for the algorithm's operation.
Explains least-cost search with an equation associating costs to nodes in the search tree.
Illustrates cost function in the 8-puzzle problem using g(x) and h(x) metrics.
Details calculations of cost function values based on various states in the 8-puzzle setup.
Introduces the Travelling Salesman Problem (TSP) as finding minimum cost cycles in a network.
Visual representation of a four-vertex network with associated costs among vertices.
Shows state space diagrams to visualize possible paths through the Travelling Salesman Problem.
Presents mathematical formulas describing the structure of cost matrices involved in TSP.
Gives an example cost matrix for a TSP setup with n=5, ready for analysis.
Visualizes the construction of a state space tree highlighting vertex visits and costs.
Describes methods for reducing cost matrices by row and column to assist in TSP calculations.
Explains the computation of reduced costs and identifying lower bounds through matrix reductions.
Details the costs associated with node calculations based on previous nodes and cost matrix.
Continues with cost calculations for the next node, highlighting cumulative cost computations.
Presents the formula and results for calculating the cost of node 4, tying to reduced costs.
Summarizes current live nodes, their associated costs, and the exploration considerations.
Details the exploration from node 4 to node 2 with re-evaluated cost computations.
Summarizes the updated state of live nodes and their costs after transitioning to node 6.
Explains the new paths and costs as nodes are explored further in the state space tree.
Presents further explorations in the state space with newly derived costs for node transitions.
Reiterates the costs for all live nodes, indicating future paths available from node 10.
Concludes with the selection of routes from node 10 to node 11, detailing costs and paths.
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Branch and bounding : Data structures
- 1. Branch and Bound Travelling Salesman Problem
- 2. Branch and Boundis a state space search method in which all the children of a node are generated before expanding any of its children.
- 3. Least Cost ormax 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.
- 4. 1 Ways toselect E-Node 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 8 9 6 7 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.
- 5. 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.
- 6. Searching: Least-costsearch (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)
- 7. Example: 8-puzzle ^ = g(x) +h(x) Cost function: C Where, h(x)=the number of misplaced tiles g(x)=the number of moves so far Assumption: move one tile in any direction cost 1, Initial State Final State 1 2 3 5 6 7 8 4 1 2 3 5 8 6 7 4
- 8. 1 2 3 5 6 7 8 4 ^ C 1 2 3 5 6 4 7 8 ^ C 1 2 3 5 6 7 8 4 ^ C 1 2 5 6 3 7 8 4 1 4 5 1 2 3 1 4 5 ^ C 1 2 3 5 8 6 7 4 ^ C 1 2 3 5 6 7 8 4 ^ C 1 3 5 2 6 7 8 4 2 1 3 2 3 5 2 3 5 ^ C 1 2 3 5 8 6 7 4 ^ C 1 2 3 5 8 6 7 4 3 2 5 3 0 3
- 9. 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.
- 10. Four-Vertex network 12 3 30 10 20 6 5 4 4
- 11. State space diagramor permutation tree A B 1 2 3 4 C D E 3 4 2 4 2 3 F G H I J K 4 3 4 2 3 2 L M N O P Q
- 12. formulas • S={1,p,1|p=Permutationof (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
- 13. EXAMPLE Let thefollowing cost matrix be for n=5 in graph G(V,E):
- 14. State space tree 1 25 Vertex = 2 Vertex = 5 35 53 4 25 5 2 31 Vertex = 2 Vertex = 5 6 7 8 Vertex = 3 Vertex = 5 10 Vertex = 3 3 Vertex = 3 Vertex = 4 28 50 36 52 28 9 Vertex = 3 11 28
- 15. Reduction by row: 10 2 2 3 4 Reduction by column: 1 3
- 16. The reduced cost=(10+2+2+3+4)+(1+3) =25 Cost of node1=Lower Bound=25 Reduced Matrix:
- 17. Cost (2): •Theresulting cost matrix is: T(2)=T(1)+A(1,2)+r =25+10+0 =35
- 18. Cost (3): •Theresulting cost matrix is: T(3)=T(1)+A(1,3)+r =25+17+11 =53
- 19. Cost (4): •Theresulting cost matrix is: T(4)=T(1)+A(1,4)+r =25+0+0 =25
- 20. • In summary: So the live nodes we have so far are: o2: cost(2) = 35, path: 1->2 o3: cost(3) = 53, path: 1->3 o4: cost(4) = 25, path: 1->4 o5: cost(5) = 31, path: 1->5 Explore the node with the lowest cost: Node 4 has a cost of 25 Vertices to be explored from node 4: 2, 3, and 5 Now we are starting from the cost matrix at node 4 is: Cost(4):
- 21. •Choose to goto 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
- 22. •In summary: Sothe live nodes we have so far are: o2: cost(2) = 35, path: 1->2 o3: cost(3) = 53, path: 1->3 o5: cost(5) = 31, path: 1->5 o6: cost(6) = 28, path: 1->4->2 o7: cost(7) = 50, path: 1->4->3 o8: 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
- 23. •Choose to goto 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
- 24. •Choose to goto 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
- 25. • In summary: So the live nodes we have so far are: o2: cost(2) = 35, path: 1->2 o3: cost(3) = 53, path: 1->3 o5: cost(5) = 31, path: 1->5 o7: cost(7) = 50, path: 1->4->3 o8: cost(8) = 36, path: 1->4->5 o9: cost(9) = 52, path: 1->4->2->3 o10: 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:
- 26. •Choose to goto 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