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

Getting Started – Background and Definitions

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

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

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

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

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

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

The Proof

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!

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

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) …

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

(Planar graph used in the paper)

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

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”

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

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

Example:

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

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: ->

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

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?

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

Tiling

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

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

Parity Enforcement (Propagation)

Flipping Parity: NOT Tiles A horizontal NOT tile with and

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 20+20 = 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.

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.

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

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

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

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

Questions / Comments?

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 

Have a nice weekend!