1. Approximating the Minimum Degree Spanning Tree to within One from the Optimal Degree R96922115 陳建霖 R96922055 宋彥朋 B93902023 楊鈞羽 R96922074 郭慶徵 R96922142 林佳慶

2. Authors Martin Furer Department of Computer Science and Engineering The Pennsylvania State UniversityDepartment of Computer Science and Engineering The Pennsylvania State University Balaji Raghavachari Computer Science Department, EC 3.1 University of Texas at Dallas Computer Science Department, EC 3.1University of Texas at Dallas

3. Application Distribution of mail and news on the Internet Designing power grids

4. Similar approximation properties Edge coloring problem 3-colorability of planer graphs

5. Theorem 1 There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree

6. Theorem 2 There is a polynomial time approximation algorithm for the minimum degree Steiner tree problem which produces a Steiner tree of degree

7. Theorem 3 There is a polynomial time approximation algorithm for the directed version of the minimum degree spanning tree problem which produces a directed spanning tree of degree

8. Theorem 4 There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree at most

9. A Simple Approximation Algorithm

10. Improvement :the degree of vertex u in T If we now introduce an “improvement” in T by adding the edge and deleting one of the edge in C incident to

11. Example 3 <=2 1 1 2 GT

12. Locally Optimal Tree Def: A locally optimal tree (LOT) is a tree in which none of the non-tree edges produce any improvements. Its maximal degree denoted by k Def: contains those vertices of T which have degree at least i

13. Theorem For any b>1, the maximal degree k of a locally optimal tree T is less than Hence, k =

14. Proof Observe the ratio of can be larger than b at most times in a row. To be more precise, there is an i with k- +1 i k such that Otherwise, we have which lead to a contradiction

15. Proof Suppose we remove the vertices of from T. This split T into a forest F with t trees. And we can say that

16. Proof Now, we can say the average degree of vertices in in any spanning tree of G is at least

17. Proof

18. Algorithm (sketch) We start with an arbitrary spanning tree T of G We used the local optimality condition only on “ high”( ) degree vertices. The algorithm stops when no vertex in has a local improvement. Each phase of the algorithm can be implemented in polynomial time.

19. Theorem 1 There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree

20. Idea We just need to show that the number of phases is bounded by a polynomial in n.

21. Proof Recall :the degree of vertex u in T Define a potential function such that (c>2) Define total potential of the tree is the sum of the potentials of all vertices. We can say that where k is the maximal degree of T

22. Proof

23. For, where

24. Proof Observe we need phases,the potential reduces by a constant factor. On the other hand, k cannot decrease more than n times. Hence, bound on the number of phases.

25. The Steiner tree version

26. Steiner Problem Consider a graph G = (V,E) and a distinguished set of vertices D V. Steiner spanning tree: Find a tree which spans the vertices set D, paths in the tree may go through vertices of V － D. W1 W2 W3 W4 D1 D2 D3 D4 D5 D6 D7 D8

27. Steiner Problem (cont.) Minimum Degree Steiner (spanning) Tree –To find a tree of minimum degree, which spans the set D. –This is a generalization of the [Minimum Degree Spanning Tree] problem.

28. Steiner Problem (cont.) We start with an arbitrary tree T which spans D and retain only edges which separate the set D in T. In other words, there are no useless edges since every edge separates at least two distinguished vertices.

29. Let W be the set of vertices spanned by T. (W : all vertices in the tree T) Non-tree path: a path between two vertices in W which goes entirely through vertices of V － W except for the end point.

30. Some case of non-tree paths

31. Steiner Problem (cont.) Pseudo optimal Steiner tree (POST): –Every edge in the tree separates at least two distinguished vertices. –None of the non-tree paths produce any improvement for any vertex in S i for i = k- ⌈ log n ⌉, where k denotes the maximal degree of the tree. –If this property is true for all i, then we call it locally optimal Steiner tree (LOST).

32. Steiner Problem (cont.) Similar to the theorem 2.1, we have theorem 3.1: for any b>1, the maximal degree k of a LOST T is at most bΔ * + ⌈ log n ⌉. Hence k = O(Δ * + log n ). This idea can be converted into a polynomial time algorithm which produces a Steiner tree of degree O(Δ * + log n ).

33. Steiner Problem (cont.) Theorem 1.2. There is a polynomial time approximation algorithm for the minimum degree Steiner tree problem which produces a Steiner tree of degree O(Δ * + log n ).

34. Directed spanning trees version

35. Input : A directed graph G and anyone spanning tree together with r which is the root of the tree.

36. What is a directed spanning tree ? Def : A rooted spanning tree T of G is a subgraph of G with the following properties: –T does not contain any cycles. –The outdegree of every vertex except r is exactly one. –There is a path in T from every vertex to the root r.

37. An example

38. Recall The improvement in the undirected case 3 <=2 1 1 2

39. Where is the cycle in directed case ?

40. New “improvement” Idea : Consider a vertex v of indegree i. Try to reduce the indegree of the vertex by attaching one of the i subtrees of v to another vertex of smaller degree.

41. Step1 : Move the root of the subtree T’ that is being removed from v to a “convenient” vertex in that subtree. Step2 : Attach T’ to another vertex outside the tree to which the new root has a connection.

42. What is a convient vertex ? The set of convenient vertices are those in the strongly connected part of the root of T’ in the graph induced by the vertices of T’ with all non-tree edges removed from vertices of degree i – 1 or greater. We pick a convenient vertex from convenient set

43. T’

44. Algorithm in directed version The algorithm tries to decrease the degrees of vertices in S i for i =k – using the improvement step above.

45. A Lemma Let T be a directed spanning tree of degree k. Let S i consist of those vertices whose indegree is at least i. Suppose we remove the vertices of S i from T, breaking T into a forest F. Then there are at least t = IS i I× (i – 1) + 1 trees of F whose vertices do not have descendants of degree i or greater in T.

46. Let Y be “ a tree of F whose vertices do not have descendants of degree i or greater in T “ ? What is a Y ?

47. i descendent ancestor …

48. Why IS i I× (i – 1) + 1 ? We use induction on the size of |S i | –Basis : if |S i | =1, then there are (i-1) +1 = i subtrees like Y (o.k)  Note : Each addition of a vertex to the set S i can remove at most one tree from the set of candidate trees, but adds at least i more.

49. –suppose that |S i | = k holds, i.e : there are k ×(i-1)+1 subtrees like Y. –then if |S i | = k+1, we will have at least [k ×(i-1)+1]+i-1 = k×i-k+i = (k+1) ×(i-1)+1 By math induction, we complete the proof.

50. THEOREM 1.3 There is a polynomial time approximation algorithm for the directed version of the minimum degree spanning tree problem which produces a directed spanning tree of degree O(∆* + logn).

51. illustration As in the proof of the undirected case, we look for an i in the range k to k – where the set S i does not “expand” by more than a constant factor. We will have at least [ IS i I× (i – 1) + 1 ]+ IS i I−1 edges connecting these components and each one of these edges is incident to at least one vertex in S i-1

52. illustration Hence the average degree of vertices in S i-1 in any spanning tree of G is at least The maximal degree ∆* is at least the average degree of these vertices.

53. illustration Hence,

54. The △ * +1 algorithm

55. Definition k-1k k ≦ k-2 uv w k k-1 w uv If ρ(u) ≧ k – 1, we say that u blocks w from (u, v). If neither u nor v blocks w, then w benefits from (u, v). T T

56. The algorithm works in phases (Recall Def.) Suppose k were the maximal degree of a spanning tree T, Let S i be the set of vertices of degree at least i in T. In each phase we try to reduce the size of S k by one. If successful, we move on to the next phase. So we also update k after each phase. Even in a phase the number of vertices of degree at most k-1 increases so many, but it’s ok!

57. We use the local optimality property. many phases … try to improve Max degree = k but |S k | remains the same… Local Optimality Property Our solution

58. How many phases? As the size of S k reduces by one in each phase (except the last one), there are at most O (n / k) phases when the maximal degree is k. At most phases k=1 k=2 maximal degree from k to k-1 …

59. Theorem Let △ * be the degree of a MDST. Let S be the set S k. Let B be an arbitrary subset of vertices of degree k–1 in T. S U B suppose no such edges exist then k ≦△ *+1 F

60. Try to prove △ * ≧ something △ * ≧ average degree of any subset we want the vertices of high degree take S k SkSk |F| = t take S U B S U B S U B may be either connected or disconnected (Case 1) (Case 2)

61. S U B is connected Case 1 S U B is connected (a tree), then F contains exactly |S|k+|B|(k-1)-2(|S|+|B|-1) subtrees. Exactly |S|k+|B|(k-1)-2(|S|+|B|-1)+(|S|+|B|-1) edges connect to S U B. The average degree of vertices in S U B is exactly S U B

62. S U B is not connected Case 2 S U B is not connected (not a tree), then F contains more than |S|k+|B|(k-1)-2(|S|+|B|-1) subtrees. More than |S|k+|B|(k-1)-2(|S|+|B|-1)+(|S|+|B|-1) edges connect to S U B. S U B The average degree of vertices in S U B is more than

63. Finally we proved △ * ≧ k-1 △ * ≧ average degree of S U B = Therefore △ * ≧ k-1 k ≦ △ *+1 Note ： what if |B|=0?

64. Observation Let B = V k-1, apply Theorem 5.1, we are done. V k U V k-1 ≦ k-2 If no any S k on the cycle, just union and mark good k k k-1 k-2 kk-1 Case 1Case 2aCase 2b k-2 k-1… ≦ k-1

65. Algorithm Step 1. Given a SPT (the degree of all vertices is known), remove V k U V k-1 (marked as bad) from T and mark all the connected components as good. V k U V k-1 F ≦ k-2 good bad

66. Algorithm Step 2. Choose any edge between 2 components and find the cycle generated, also check all bad vertices on the cycle. If no such edges exist, stop. V k U V k-1 F u v ≦ k-2

67. Algorithm Step 3. (Case 1) At least a vertex in V k on the cycle, reduce the size of V k by one, go to Step 1. V k U V k-1 F u v ≦ k-2 kk-1

68. Algorithm Step 3. (Case 2) No any vertex in V k on the cycle, (imply) at least a bad vertex in V k-1 on the cycle, make a union of all components along with all bad vertices on the cycle, also mark the bad vertices as good, go to Step 2. V k U V k-1 F u v ≦ k-2 k-1 good

69. Illustration of example 1 ≦ k-2 k k-1 good k-2 ≦ k-1 k-1

70. Illustration of example 2 k or k-1 k-1 ≦ k-2 Must not be choosed

71. Illustration of example 3 k-1kk

72. Time analysis Step 1. Given a SPT (the degree of all vertices is known), remove V k U V k-1 (marked as bad) from T and mark all the connected components as good. Step 2. choose any edge between 2 components and find the cycle generated, also check all bad vertices on the cycle. If no such edges exist, stop. Step 3. (Case 1) At least a vertex in V k on the cycle, reduce the size of V k by one, go to Step 1. Step 3. (Case 2) No any vertex in V k on the cycle, (imply) at least a bad vertex in V k-1 on the cycle, make a union of all components along with all bad vertices on the cycle, also mark the bad vertices as good, go to Step 2.