The Rectilinear Steiner
Arborescence Problem is NPComplete
Weiping Shi (Dept. of Electrical Engg. TAMU)
Chen Su (Inet Technologies, TX)
SIAM Journal of Computing 2006
Presenter: Vishal Kapoor
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Getting Started –
Background and Definitions
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Steiner Tree
• A minimal length tree connecting a set of
points in a plane
• May have Steiner Points
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Rectilinear Steiner Tree
• A Steiner tree in which every path is made
up of straight line segments
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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
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Difference?
• RSA is a shortest distance tree with
respect to the origin
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Rectilinear Steiner Minimum
Arborescence (RSMA)
• A minimum length RSA
• Difference between RSA and RSMA:
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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
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The Proof
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My thoughts about the proof
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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!
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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
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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) …
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3-SAT problem as a planar graph
(Planar 3-SAT)
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(Planar graph used in the paper)
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A Quick Recap
• Steiner Tree / Points
• Rectilinear Steiner Tree
• RSA, RSMA + differences
• The Decision Problem
• 3-SAT, Planar 3-SAT
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Reduction Technique
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Based on Component Design
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Steps:
1. 3-SAT graph is reduced to another planar
graph H
2. H is embedded into a grid (as graph R)
3. Each vertex of R is replaced by a “tile”
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First Step – Transforming G to H
• H is planar with maximum degree = 3
• Each vertex of G contribute deg(v) vertices
in H
• Eg:
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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
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Example:
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Second Step: Embedding H as
graph R in a grid
• Can be done in polynomial time O(V2) Valiant
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Properties of R:
Nice “drawing” techniques
1. Vertices only unit distance
apart are connected
2. For each CLAUSE vertex:
-
Positive variable enters from
left
- Negative variable enters from
below
Eg: ->
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… cont’d
3. Each OR vertex
horizontal vertices in R
occupies 2
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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
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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
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Tiling
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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
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Parity of a tile
• 1 if the rightmost vertex connected by
horizontal edge
• else 0
• Tiles enforce (propagate) parities on
neighboring tiles
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Parity Enforcement (Propagation)
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Flipping Parity: NOT Tiles
• A horizontal NOT tile with
and
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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
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OR tile
• Total width 192
• Always tries to achieve parity = 1
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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
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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
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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
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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
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Questions / Comments?
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Homework –
A Descriptive Question
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RSA is NP-Complete by reduction to 3-SAT
3-SAT is “Strongly NP-Complete”
1.
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?
2.
3.
4.
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 
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Have a nice weekend!
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