Leave No Stone Unturned: Improved Approximation Algorithm for Degree-Bounded MSTs

1. Leave No Stone Unturned: Improved Approximation Algorithm for Degree-Bounded MSTs Raja Jothi University of Texas at Dallas raja@utdallas.edu Joint work with Balaji Raghavachari

2. Menu • Appetizer • On-tray • Dessert

3. Degree-Bounded MSTs Problem Definition Given n points in the Euclidean plane, the degree-δ MST asks for a minimum weight spanning tree in which the degree of each vertex is at most δ. Complexity • δ=5: Polynomial time solvable [Monma & Suri ’92] • NP-hard for δ=3 [Papadimitriou & Vazirani ’84] • NP-hard for δ=4?

4. Previous Results • Degree-2 MST • PTASs [Arora ’96 & Mitchell ‘97] • Degree-3 MST • 1.5-approximation [Khuller, Raghavachari & Young ‘96] • 1.402-approximation [Chan ‘03] • Degree-4 MST • 1.25-approximation [Khuller, Raghavachari & Young ‘96] • 1.143-approximation [Chan ‘03]

5. Previous Results…(continued) • Degree-3 MST in Rd, d > 2 • 5/3-approximation [Khuller, Raghavachari & Young ‘96] • 1.633-approximation [Chan ‘03] • Degree-δ MST in Rd • QPTAS – (nO(log n)) [Arora & Chang ‘03]

6. Our Result For any arbitrary collections of points in the plane, there always exists a degree-4 MST of weight at most (√2 + 2 )/3 < 1.1381 times the weight of an MST.

7. Preliminaries • Input: Degree-5 MST T • Root T at an arbitrary vertex • Every vertex, but root, has at most 4 children

8. Khuller, Raghavachari & Young’s recipe Degree-3 MST

9. Khuller, Raghavachari & Young’s recipe Degree-3 MST

10. Khuller, Raghavachari & Young’s recipe Degree-3 MST

11. Khuller, Raghavachari & Young’s recipe Degree-3 MST

12. Khuller, Raghavachari & Young’s recipe Degree-4 MST

13. Khuller, Raghavachari & Young’s recipe Degree-4 MST

14. Khuller, Raghavachari & Young’s recipe Degree-4 MST

15. Khuller, Raghavachari & Young’s recipe Degree-4 MST

16. Menu • Appetizer • On-tray • Dessert

17. Ingredients Strengthened triangle inequality If |AB| ≤ |BC|, then |AC| ≤ F(θ)|AB| + |BC| Where F(θ) = sqrt(2(1-cosθ)) – 1 C B’ θ B A

18. Ingredients…(continued) Bounds on edge weights of an MST Let AB and BC be two edges that intersect at point B in an MST. Let θ= ABC ≤ 90o. Then 2|BC|cosθ≤ |AB| ≤ A |BC| 2cosθ θ B C

19. Ingredients…(continued) Charging Scheme Let |av| ≤ |bv|. Then, cost of the new tree formed by replacing bv by ab is at most (|av| + |bv| + |cv| + |dv|) * F(θ) / k p a d v θ c b

20. Ingredients…(continued) Charging Scheme Let |av| ≤ |bv|. Then, cost of the new tree formed by replacing bv by ab is at most (|av| + |bv| + |cv| + |dv|) * F(θ) / k p a d v θ c b

21. Ingredients…(continued) p a d v c b

22. Ingredients…(continued) p a d v c b

23. Ingredients…(continued) v p b a d a v c b

24. Ingredients…(continued) v p b a d a v c b

25. Chan’s degree-4 recipe 2 biological child & 1 foster child v

26. Chan’s degree-4 recipe 2 biological child & 1 foster child v recurse recurse recurse

27. Chan’s degree-4 recipe 3 biological child & 1 foster child v

28. Chan’s degree-4 recipe 3 biological child & 1 foster child v

29. Chan’s degree-4 recipe 3 biological child & 1 foster child v

30. Chan’s degree-4 recipe 3 biological child & 1 foster child v recurse recurse recurse

31. Chan’s degree-4 recipe 4 biological child & 1 foster child v

32. Chan’s degree-4 recipe 4 biological child & 1 foster child v recurse recurse recurse

33. Snippets of our recipe vvertex under consideration v1-v4v’sbiological children v’ v’s foster child Obj. Reduce the degree of v to 3 Note max{θ1,θ2,θ3,θ4+θ5} ≥ 120o Solve it by case-by-case analysis. v’ θ5 θ4 v1 v v4 θ1 θ3 θ2 v2 v3

34. Secrets behind our recipe’s success • Let θ5≤ 60o. • Then |vv’| ≥ |vv1|. • Otherwise |v1v’| ≥ |vv1|, which contradicts the fact that vv1was chosen over v1v’ to be the MST edge. • Removal of vv’ and adding v1v’ results in savings, which is used for future extra charge accounting. • Smarter charging scheme. v’ θ5 θ4 v1 v v4 θ1 θ3 θ2 v2 v3

35. A stone we turned Case θ2 ≥ 120o, θ4 ≥ 60o,θ5 ≤ 60o ,θ3 > 90o v’ ≤ 60o ≥ 60o θ5 θ4 v1 v v4 θ1 θ3 > 90o θ2 ≥ 120o v2 v3

36. A stone we turned Case θ2 ≥ 120o, θ4 ≥ 60o,θ5 ≤ 60o ,θ3 > 90o For simplicity, let |vv1| = x1; … ; |vv4| = x4 and |vv’| = x5 x5 ≤ 60o ≥ 60o θ5 θ4 x1 v x4 θ1 θ3 > 90o θ2 ≥ 120o x2 x3

37. A stone we turned Case θ2 ≥ 120o, θ4 ≥ 60o,θ5 ≤ 60o ,θ3 > 90o For simplicity, let |vv1| = x1; … ; |vv4| = x4 and |vv’| = x5 Subcase x3or x4is the 2nd smallest among {x1, x2, x3, x4} AND x1≤x2, x3, x4 x5 ≤ 60o ≥ 60o θ5 θ4 x1 v x4 θ1 θ3 > 90o θ2 ≥ 120o x2 x3

38. A stone we turned Case θ2 ≥ 120o, θ4 ≥ 60o,θ5 ≤ 60o ,θ3 > 90o For simplicity, let |vv1| = x1; … ; |vv4| = x4 and |vv’| = x5 Subcase x3or x4is the 2nd smallest among {x1, x2, x3, x4} AND x1≤x2, x3, x4 Savings = 2(sin(90-θ5)-1) x1 x5 ≤ 60o ≥ 60o θ5 θ4 x1 v x4 θ1 θ3 > 90o θ2 ≥ 120o x2 x3

39. A stone we turned Case θ2 ≥ 120o, θ4 ≥ 60o,θ5 ≤ 60o ,θ3 > 90o For simplicity, let |vv1| = x1; … ; |vv4| = x4 and |vv’| = x5 Subcase x3or x4is the 2nd smallest among {x1, x2, x3, x4} AND x1≤x2, x3, x4 Savings = 2(sin(90-θ5)-1) x1 It is as if we have at least an additional 2(sin(90-θ5)-1) x1/0.1381 to charge. x5 ≤ 60o ≥ 60o θ5 θ4 x1 v x4 θ1 θ3 > 90o θ2 ≥ 120o x2 x3

40. A stone we turned • W.l.o.g., let x3≤ x4 • Savings = 2(sin(90-θ5)-1) x1/0.1381) • Additional cost due to transformations is ≤ F(θ3) * (x1 + x2 + x3 + x4 + Savings) 3 + 2cos(120o–θ5/2) * (1+Savings) which is bounded by 0.0709(x1 + x2 + x3 + x4 + Savings) x5 ≤ 60o ≥ 60o θ5 θ4 x1 v x4 θ1 θ3 > 90o θ2 ≥ 120o x2 x3

41. A stone we turned • W.l.o.g., let x3≤ x4 • Savings = 2(sin(90-θ5)-1) x1/0.1381) • Additional cost due to transformations is ≤ F(θ3) * (x1 + x2 + x3 + x4 + Savings) 3 + 2cos(120o–θ5/2) * (1+Savings) which is bounded by 0.0709(x1 + x2 + x3 + x4 + Savings) x5 ≤ 60o ≥ 60o θ5 θ4 x1 v x4 θ1 θ3 > 90o θ2 ≥ 120o x2 x3

42. A stone we turned • There are cases where the ratio is bounded by (√2 – 1 )/3 < 0.1381 • That makes the approximation ratio of our algorithm to be (√2 + 2 )/3 < 1.1381.

43. Dessert • Our ratio of (√2 + 2 )/3 < 1.1381 CANNOT be improved by using just local changes. • Improvement of the ratio requires a global approach. • There exists a degree-4 tree whose weight is (2sin36o + 4)/5 < 1.0352 times the weight of MST.

44. Future Problems Points in the plane • Is degree-4 MST problem NP-hard? • Improve the 1.1381 ratio for degree-4 trees. • Improve the 1.402 ratio for degree-3 trees. • Is there a PTAS for degree-δ MST problem? Points in higher dimensions • Improve the 1.633 ratio for degree-3 trees. • Approximation of degree-δ trees in general metric spaces (no triangle inequality), within ratios better than 2.