The Rectilinear Steiner Arborescence Problem is NP-Complete

1. The Rectilinear Steiner Arborescence Problem is NP-Complete Weiping Shi (Dept. of Electrical Engg. TAMU) Chen Su (Inet Technologies, TX) SIAM Journal of Computing 2006 Presenter: Vishal Kapoor

2. Getting Started –Background and Definitions

3. Steiner Tree • A minimal length tree connecting a set of points in a plane • May have Steiner Points

4. Rectilinear Steiner Tree • A Steiner tree in which every path is made up of straight line segments

5. Rectilinear Steiner Arborescence (RSA) • Directed Steiner tree rooted at origin • Points are in the first quadrant of the plane • Every segment in the tree is directed left to right or bottom to top

6. Difference? • RSA is a shortest distance tree with respect to the origin

7. Rectilinear Steiner Minimum Arborescence (RSMA) • A minimum length RSA • Difference between RSA and RSMA:

8. Applications • Useful in VLSI routing • It has been shown that routing trees based on such structures may have much lesser delays than those based on traditional Steiner trees

9. The Proof

10. My thoughts about the proof • Very “artistic” (geometric) proof • Cited in literature as “pretty” • Uses graph drawing techniques • One of the clearest papers that I have read – provides details very clearly!

11. The Decision Problem • A Set of points P = {p1, p2 … pn} in the plane and a positive number k • Is there an RSA of total edge length k or less? • Proof: Reduction from 3-SAT

12. 3-SAT • A set of variables V = {v1, v2 … vn} and clauses C = {c1, c2 … cn} • Each clause has at most 3 literals. • Is there an assignment of variables so that all clauses are satisfied? • Satisfying the 3-CNF form is NP-Complete • Eg: (v1 OR v2 OR v6) AND (v1 OR v8) …

13. 3-SAT problem as a planar graph (Planar 3-SAT)

14. (Planar graph used in the paper)

15. A Quick Recap • Steiner Tree / Points • Rectilinear Steiner Tree • RSA, RSMA + differences • The Decision Problem • 3-SAT, Planar 3-SAT

16. Reduction Technique • Based on Component Design • Steps: • 3-SAT graph is reduced to another planar graph H • H is embedded into a grid (as graph R) • Each vertex of R is replaced by a “tile”

17. First Step – Transforming G to H • H is planar with maximum degree = 3 • Each vertex of G contribute deg(v) vertices in H • Eg:

18. Properties of H:New Vertices – OR, NOT • “Nice” form • If not in Nice form: • Insert NOT vertex wij • Suppose the clause looks like below: • Insert CLAUSE vertex cjfor each clause • Insert OR vertex c’jto connect big clauses

19. Example:

20. Second Step: Embedding H as graph R in a grid • Can be done in polynomial time O(V2) - Valiant

21. Properties of R:Nice “drawing” techniques • Vertices only unit distance apart are connected • For each CLAUSE vertex: • Positive variable enters from left • Negative variable enters from below Eg: ->

22. … cont’d 3. Each OR vertex occupies 2 horizontal vertices in R

23. Final Step: Covering R with tilesQuadrupeds and forbidden regions • These are a set of RSAs that connect the white points; each RSA is rooted at a black point • Has 2 min forests, both of length • Black points are interface points, white points are “inside” points

24. Another Recap • 3 steps for reduction • 1. Transforming G into H • Always looking out for "nice" clauses • Adding OR, NOT, CLAUSE vertices to make "nice" form • 2. Embedding H into a grid as R • Adding OR vertices having 2-horizontal-vertices • Quadrupeds, Forbidden regions, minimum forests made of RSAs • Now we have to see what is tiling

25. Tiling

26. A basic tile • Represents a single variable • Made up of overlapping quadrupeds • Has height and width = 96 units • Only OR tile has height = 96 and width = 2x96 • Has

28. Parity of a tile • 1 if the rightmost vertex connected by horizontal edge • else 0 • Tiles enforce (propagate) parities on neighboring tiles

29. Parity Enforcement (Propagation)

30. Flipping Parity: NOT Tiles • A horizontal NOT tile with and

32. CLAUSE tiles • Have the same interface to the left and to the bottom (Why?) • Min forest (black) has length 108 achieved iff tile to left has parity 1 and tile below has 0

33. OR tile • Total width 192 • Always tries to achieve parity = 1

34. Another Recap • A basic tile that represents ordinary variables • Definition of Parity and Parity enforcement (propagation) • NOT, CLAUSE and OR tiles and their properties and dimensions

35. Let’s go to the Final Result • We are done – we have all the structures that we need • All structures till were made / manipulated locally and can be constructed in polynomial time • Thus the final reduction should be polynomial time

36. Connecting tiles togetherBasic Tiles and Trials • Intuitively, trials “close” the basic tiles • Are not in any forbidden region, so don’t affect how white points are connected

38. RSA problem is NP-Complete • L = sum of min edge lengths to connect all black points + min forest length of each type of tile • A planar 3-SAT having m clauses has a satisfying assignment iff the set of points has an RSA of length L+108m

39. Questions / Comments?

40. Homework –A Descriptive Question • RSA is NP-Complete by reduction to 3-SAT • 3-SAT is “Strongly NP-Complete” • What does “Strongly NPC” mean in a general sense? Give a precise mathematical definition and 1 example explaining the idea • What types of problems are not Strongly NP-complete? What are they called? • Is there a complexity class/classes for such languages? • What does this mean in terms of circuit complexity? A PTIME algorithm for solving the homework: 1: Choose any 1 color and answer questions of that color only 2: Google for a column (and a small paper) on NP-hardness by Garey & Johnson for checking this out Evaluation: I will grade based on the quality of your answer – not quantity 

41. Have a nice weekend!