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CS 312: Algorithm Analysis

This work is licensed under a Creative Commons Attribution-Share Alike 3.0 Unported License . CS 312: Algorithm Analysis. Lecture #34: Branch and Bound Design Options for Solving the Traveling Salesman Problem: Tight Bounds.

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CS 312: Algorithm Analysis

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  1. This work is licensed under a Creative Commons Attribution-Share Alike 3.0 Unported License. CS 312: Algorithm Analysis Lecture #34: Branch and Bound Design Options for Solving the Traveling Salesman Problem: Tight Bounds Slides by: Eric Ringger, with contributions from Mike Jones, Eric Mercer, and Sean Warnick

  2. Announcements • Homework #24 due now • Homework #25 due Friday • Project #7: TSP • ASAP: Read the helpful “B&B for TSP Notes” linked from the schedule • Read Project Instructions • Today: We continue discussing main ideas • Next Wednesday: Early day • Week from Friday: due

  3. Objectives • Review the Traveling Salesman Problem (TSP) • Develop a good bound function for the TSP • Reason about Tight Bounds • Augment general B&B algorithm

  4. 1 2 5 3 4 Traveling Salesman (Optimization) Problem • Rudrata or Hamiltonian Cycle • Cycle in the graph that passes through each vertex exactly once + • Find the least cost or “shortest”cycle 9 1 8 2 10 6 5 3 12 7 Distinguish from theTSP search problem and theTSP decision problem 4

  5. How to solve? • If with B&B, what do we need?

  6. How to solve? • If with B&B, what do we need?

  7. 1 2 5 3 4 Initial BSSF How to compute? Should be quick. What if you have a complete graph? What if you don’t? 9 1 8 2 10 6 5 3 12 7 4

  8. 1 2 5 3 4 Simple-Minded Initial BSSF Cost of BSSF = 9+5+4+12+1 = 31 9 1 8 2 10 6 5 3 12 7 4

  9. 1 2 5 3 4 A Bound on Possible TSP Tours We need a bound function. Lower or Upper? How to compute? 9 1 8 2 10 6 5 3 12 7 4

  10. 1 2 5 3 4 A Bound on Possible TSP Tours We need a bound function. Lower or Upper? How to compute? 9 1 8 2 10 6 5 3 12 7 4

  11. 1 2 5 3 4 A Bound on Possible TSP Tours 9 1 8 2 10 6 5 3 12 7 4 What’s the cheapest way to leave each vertex?

  12. 1 2 5 3 4 Bound on Possible TSP Tours Rough draft bound= 8+6+3+2+1 = 20 9 1 8 2 10 6 5 3 12 7 4 Save the sum of those costs in the bound (as a rough draft).

  13. 1 2 5 3 4 Bound on Possible TSP Tours Rough draft bound = 20 9-8=1 1 8-8=0 2 10 6 4 3 12 7 4 For a given vertex, subtract the least cost departure from each edge leaving that vertex.

  14. 1 2 5 3 4 Bound on Possible TSP Tours Rough draft bound = 20 1 0 0 0 9 0 2 0 6 1 1 Repeat for the other vertices. What do the numbers on the edges mean now?

  15. 1 2 5 3 4 Bound on Possible TSP Tours Rough draft bound = 20 1 0 0 0 9 0 2 0 6 1 1 Now, can we find a tighter lower bound?

  16. 1 2 5 3 4 Bound on Possible TSP Tours Rough draft bound = 20 1 0 0 0 9 0 2 0 6 1 1 Does that set of edges now having 0 residual cost arrive at every vertex?

  17. 1 2 5 3 4 Bound on Possible TSP Tours Rough draft bound = 20 1 0 0 0 9 0 2 0 6 1 1 In this case, those edges never arrive at vertex #3.

  18. 1 2 5 3 4 Bound on Possible TSP Tours Rough draft bound = 20 1 0 0 0 9 0 2 0 6 1 1 We have to take an edge to vertex 3 from somewhere. Assume we take the cheapest.

  19. 1 2 5 3 4 Bound on Possible TSP Tours Bound = 21 1 0 0 0 9 0 1 0 6 0 1 Subtract its cost from other edges entering vertex 3 and add the cost to the bound. We have just tightened the bound.

  20. This Bound • It will cost at least this much to visit all the vertices in the graph. • There’s no cheaper way to get in and out of each vertex. • Each edge is now labeled with the extra cost of choosing that edge. • The bound is not a solution; it’s a bound! • Why are tight bounds desirable?

  21. 9 1 1 8 2 2 5 10 6 3 4 12 7 3 4 4 Bound on Possible TSP Tours To: 1 2 3 4 5 From: 1 2 3 4 5 Our algorithm can do this reasoning using a cost matrix.

  22. 1 0 1 0 0 2 5 9 0 0 2 6 1 3 4 1 Bound on Possible TSP Tours To: 1 2 3 4 5 From: 1 2 3 4 5 Reduce all rows.

  23. 1 0 1 0 0 2 5 9 0 0 2 6 1 3 4 1 Bound on Possible TSP Tours To: 1 2 3 4 5 From: 1 2 3 4 5 Then reduce column #3. Now we have a tighter bound.

  24. Search • Let’s start the search • Arbitrarily start at vertex 1 • Why is this OK? • Focus on: • the bound function and • the reduced cost matrix representation of states

  25. Using this bound for TSP in B&B Start at vertex 1 in graph (arbitrary) BSSF=31 bound = 21 What should our state expansion strategy be?

  26. Using this bound for TSP in B&B Start at vertex 1 in graph (arbitrary) BSSF=31 bound = 21 1-2 1-5 1-3 1-4 bound = 21 bound = 21+1

  27. 1 1 2 2 5 5 3 3 4 4 Focus: going from 1 to 2 BSSF=31 bound = 21 Before After 1 0 1 0 1-2 0 0 0 9 9 0 1 0 1 6 6 0 0 1 1 bound = 22 Add extra cost from 1 to 2, exclude edges from 1 or into 2.

  28. 1 1 2 2 5 5 3 3 4 4 Focus: going from 1 to 2 BSSF=31 bound = 21 Before After 1 0 1 0 1-2 0 0 0 9 9 0 1 0 1 6 6 0 0 1 1 bound = 22+1 No edges into vertex 4 w/ 0 reduced cost.

  29. 1 2 5 3 4 Focus: going from 1 to 2 BSSF=31 bound = 21 1 0 1-2 0 8 1 6 0 0 bound = 21+1+1 Add cost of reducing edge into vertex 4.

  30. Bounds for other choices Agenda: BSSF=31 bound = 21 1-2(23),1-4(21) 1-2 1-5 1-3 1-4 bound = 23 bound = 21

  31. 1 2 5 3 4 Leaving Vertex 4 BSSF=31 1-4 0 0 bound = 21 0 0 1 0 6 0 1-4-2 1-4-3 1-4-5 bound = 22 bound = 21 bound = 28

  32. 1 2 5 3 4 Leaving Vertex 4 BSSF=31 1-4 Agenda: 0 0 1-4-2(22), 1-4-3(21) 1-4-5(28),1-2(23) bound = 21 0 0 1 0 6 0 1-4-2 1-4-3 1-4-5 bound = 22 bound = 21 bound = 28

  33. 1 2 5 3 4 Leaving Vertex 3 BSSF=31 1-4-3 0 bound = 21 0 0 0 0 1-4-3-2 bound = 21

  34. 1 2 5 3 4 Leaving Vertex 3 BSSF=31 1-4-3 Agenda: 0 bound = 21 0 4-2(22), 3-2(21) 4-5(28), 1-2(23), 0 0 0 1-4-3-2 bound = 21

  35. Search Tree for This Problem b=21 1-to-2 1-to-4 b=23 b=21 4-to-2 4-to-3 4-to-5 b=22 b=21 b=28 3-to-2 b=21 2-to-5

  36. Termination Criteria for a B&B Algorithm • Repeat until • Agenda is empty • Or time is up • Or BSSF cost is equal to original LB

  37. Update: Branch and Bound functionBandB(v) BSSF  quick-solution(v) // BSSF.cost holds cost Agenda.clear() v.b  bound(v) Agenda.add(v, v.b) while !Agenda.empty() andtime remains and BSSF.cost != v.bdo u  Agenda.first() Agenda.remove_first() children = generate_children_ascending(u) for each w in childrendo if ! time remains then break w.b  bound(w) if (w.b < BSSF.cost) then if criterion(w) then BSSF  w • Agenda.prune(BSSF.cost) else ifpartial_criterion(w) then Agenda.add(w, w.b) return BSSF

  38. Assignment • HW #25: • Compute bound for TSP instance using today’s method • Reason about search for TSP solution

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