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 cj for each clause
Insert OR vertex c’j to 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 tiles Quadrupeds 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
Quadruped = 4 black points and 4 white points.
To connect q4, we need a path of at least alpha (same for q1 – even by the path of q4). Similarly beta for each q2 and q3. Thus lower bound is 2(alpha + beta)
Also, the above is the best we can do.
To achieve the lower bound, each point must fall on this lowest possible path of 2(alpha + beta). Starting with q2, we can connect it with either a horizontal or vertical edge. If we use vertical solid edge, then we need to connect q3 with a horizontal solid edge, then q4 with a vertical edge and q1 with a horizontal edge, giving a total length of 2(alpha + beta). Similarly for the other edges.
Forbidden region – because if there is any point inside it (say triangle b2, b3, q4) then you could make a tree less than alpha to connect to q4. But otherwise, if the point is outside the dashed diagonal b2, b3, the minimum distance with which you could connect it to q4 would be greater than alpha/2 + alpha/2. So you can’t go below the lower bound if you are outside this region!
What do the points mean in the graph transformation in terms of the original SAT variables?

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
2 ways to connect q5 and q6, either through dashed edges q2-q5 and q3-q6 giving a total of = 40 or through the Steiner point giving 24+4.
So to connect q5 and q6 by solid edges, it can be done only if either q2 and q3 (and thus q4 and q1) are connected by a solid edge. Thus the parity of left forest should be 1 and right should be 0.

33. OR tile Total width 192 Always tries to achieve parity = 1
The right side of this tile is always a CLAUSE tile. So it will try to reduce the cost of that tile to min cost of 108 on the right. So it will try to have the 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 together Basic 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!