Das Fraktionale Prize-Collecting Steiner Tree Problem auf Baumgraphen

Transcription

1 Das Fraktionale Prize-Collecting Steiner Tree Problem auf Baumgraphen Gunnar W. Klau (TU Wien) Ivana Ljubić (TU Wien) Petra Mutzel (Uni Dortmund) Ulrich Pferschy (Uni Graz) René Weiskircher (TU Wien)

2 Motivation

3 Motivation

4 Modell

5 Kosten und Profite Kosten einer Kante: Kosten für Aufgraben, Verlegen der Rohre, Wiederherstellen der Strasse Profit eines Knoten: Kunden-Knoten: Erwartete Zahlungen für Wärmelieferung bei Anschluß an das Netzwerk Kreuzungsknoten: Profit=0

6 Prize-Collecting Steiner Tree Problem (PCSTP) Input: G = (V,E) p: V N 0 (Profit der Knoten) c: E N 0 (Kosten der Kanten) Status: NP-hard Output: Zusammenhängender Teilgraph G =(V,E ),

7 Fraktionale Zielfunktion: Problem mit der linearen Zielfunktion: Kosten Profite Höhere Gewinnsumme Netzwerk 1 5 Mill. 5.2 Mill. Netzwerk 2 220, ,000 Höhere Rentabilität Alternative Zielfunktion: Maximiere Rentabilität

8 Fraktionales PCST auf Bäumen Im weiteren wird der interessante Spezialfall G ist ein Baum behandelt. Lineare Zielfunktion in O(n) Zeit einfach lösbar Fraktionale Zielfunktion: Vergleichsstudie für verschiedene Suchstrategien Komplexitätsverbesserung bei Megiddo-Methode erschienen in: Proceedings of the 11. ESA European Symposium on Algorithms, 2003, Springer Lecture Notes in Computer Science 2832.

9 Subdividing the Problem Blocks Trees Makes sense for village topologies

10 Trees: The Linear Case Assume: optimal rooted sub-tree of T i has value P (T i ) e 1 v e n Optimal sub-tree with root v contains edge e i if: T 1 T n

11 Example

12 Example

13 Fractional Objective Function: Parametric Search Assumption: Goal: We have algorithm A for solving linear version of Problem P Solving the fractional version P F of P where the objective function is the quotient of profits and costs

14 Fractional Objective Function: Parametric Search Multiply all costs with a factor t Q + new linear problem P(t): There is a t* with the properties: 1 P(t*) has same optimal sub-trees as P F 2 Objective value o(t*) = 0 3 The optimal solution for P F is t* Apply linear algorithm to P(t)

15 Fractional Objective Function: Geometric Interpretation linear function for each subtreet T o(t): piecewise linear, convex, monotonically decreasing t* t

16 Fractional Objective Function: How to find t*? (1) Straightforward Approach: fi Binary Search + easy to handle upper bound for t*? no polynomial running time bound stopping criterion

17 Example: Binary Search i 1 t i 27/2 2t i r 4t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i o(t i ) -54 z frac 0

18 Example: Binary Search i 2 t i 27/4 2t i r 4t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i o(t i ) -27 z frac 0

19 Example: Binary Search i 3 t i 27/8 2t i r 4t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i o(t i ) -27/2 z frac 0

20 Example: Binary Search i 4 t i 27/16 2t i r 4t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i o(t i ) -41/8 z frac 5/6

21 Example: Binary Search i 5 t i 27/32 2t i r 4t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i o(t i ) 353/32 z frac 49/45

22 Example: Binary Search i 6 t i 81/64 2t i r 4t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i o(t i ) -83/32 z frac 5/6

23 Example: Binary Search i 7 t i 135/128 2t i r 4t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i o(t i ) 13/4 z frac 37/32

24 Example: Binary Search i 8 t i 297/256 2t i r 4t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i o(t i ) -1/256 z frac 29/25

25 Example: Binary Search i 9 t i 567/512 2t i r 4t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i o(t i ) 25/16 z frac 37/32

26 Example: Binary Search i 10 t i 1161/1024 2t i r 4t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i o(t i ) 23/32 z frac 37/32

27 Example: Binary Search i 11 t i 2349/2048 2t i r 4t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i o(t i ) 113/257 z frac 37/32

28 Example: Binary Search i 12 t i 4725/4096 2t i r 4t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i o(t i ) 659/4096 z frac 29/25

29 Fractional Objective Function: How to find t*? (2) More advanced: fi Newton s Method + still quite easy to handle + exploits the geometric properties of o(t) + polynomial running time bounds

30 Fractional Objective Function: Newton s Method o(t) t 1 t 2 t* t

31 Example: Newton i t i o(t i ) t i+1 27/28

32 Example: Newton 4t i 2t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i i 2 t i 27/28 o(t i ) 10 t i+1 37/32

33 Example: Newton 4t i 2t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i i 3 t i 37/32 o(t i ) 151/32 t i+1 29/25

34 Example: Newton 4t i 2t i 9t i 6t i 5 9 5t i 9t i 7t i t i 2t i 4t i t i i 4 t i 29/25 o(t i ) 0 t i+1 =t* 29/25

35 Newton s Method Running time bounds: Radzik: iterations Theorem: iterations running time Worst case example:...

36 Fractional Objective Function: How to find t*? (3) Different approach: fi Megiddo s Method Instead of searching for t*, the linear problem is solved with fixed costs resp. profits replaced by linear functions.

37 Fractional Objective Function: Megiddo s Method Simulate run of linear algorithm on P(t*) Consider all leafs of the tree Edge e is assigned label true if: Solving linear problem P(t) decides label of e and identifies t as upper or lower bound for t*

38 Megiddo s Method Interval for t*: Root of a-bt

39 Megiddo s Method Interval for t*: a-bt Root of

40 Megiddo s Method Interval for t*: Root of a-bt Solve P (t 0 ) t 0 will be new t l or t h Interval becomes smaller Decision like before

41 Speed-Up Collect roots t of the decisions for all leafs of the tree and sort them Binary search on sorted sequence executing A until all decisions are fixed Delete all current leafs from tree and add their linear function to the function of their parent if edge label is true Start next iteration

42 Example: Megiddo r

43 Example: Megiddo r t l 0 t h

44 Example: Megiddo r t l 0 t h calling linear alg. witht= 8/7... o(t) = 3/7 > 0 t l = 8/7

45 Example: Megiddo r t l 8/7 t h calling linear alg. witht= 3/2... o(t) = -4 < 0 t h = 3/2

46 Example: Megiddo r t l 8/7 t h 3/2

47 Example: Megiddo r t l 8/7 t h 3/2

48 Example: Megiddo r t l 8/7 t h 3/2

49 Example: Megiddo r t l 8/7 t h 3/2

50 Example: Megiddo r t l 8/7 t h 3/2

51 Example: Megiddo r t l 8/7 t h 3/2 calling linear alg. witht= 24/19... o(t) = -49/19 < 0 t h = 24/19

52 Example: Megiddo r t l 8/7 t h 24/19

53 Example: Megiddo r t l 8/7 t h 24/19

54 Megiddo: Analysis No degree 2 vertices (i.e. no paths): Every iteration deletes at least half of all remaining vertices O(nlog logn) total running time General trees (with vertices of degree 2): path contraction

55 Megiddo: Path Contraction Case 1: optimal solution contains subtree T i Contains cost of the complete path T i

56 Megiddo: Path Contraction Case 2: optimal solution cuts off subtree T i Contains cost of the best part of the path T i

57 Megiddo: Path Contraction Combination: put artificial vertex between root and T i Contains cost of the best part of the path Edge contains cost of the remaining path T i

58 Megiddo: Path Contraction How to find the best vertex on the path? m vertices on the path m candidates each corresponds to a linear cost function find minimum of m linear functions m breakpoints in O(mlog logm) time collect them and perform binary search together with the roots from the leafs

59 Megiddo: Path Contraction Analysis: If path contraction is performed in every iteration at least one third of all remaining vertices are deleted in every iteration Recursion for overall running time T(n): T(n) n logn + T(2/3 n)... O(nlog logn) total running time for general trees

60 Computational Experiments Computations on randomly generated trees with 1,000-10,000 vertices 100 graphs for each size Two different tree sets: Maximum 2 children per vertex Maximum 10 children per vertex Two different sets of random profits/costs Range 1-10,000 Range 1-5

61 Number of Calls to Linear Algorithm by Megiddo Degree 2 c/p Degree 10 c/p Degree 2 c/p 1-5 Degree 10 c/p Number of Vertices

62 Running Times Degree 2 c/p Degree 10 c/p Degree 2 c/p 1-5 Degree 10 c/p Number of Vertices Seconds

63 Comparison: Number of Calls Megiddo D2 Megiddo D10 Binary Search D2 Binary Search D10 Newton D2 Newton D Number of vertices