Transcript Computational Geometry Seminar Lecture 4 More on straight-line embeddings Gennadiy Korol

Computational Geometry
Seminar Lecture 4
More on straight-line embeddings
Gennadiy Korol
Today
Restricted point set drawing.
Preliminaries.
Outerplanar graphs.
Restricted outerplanar graph drawing.
Angular resolution in graph drawing.
Lower bound on angular resolution for
general graphs.
 Lower bound on angular resolution for
general planar graphs.






Restricted point set drawing
 Until now we were allowed to define the position
of each point in order to create the proper
straight-line non crossing embedding of a graph.
 Even if G is planar, there exist point sets that do
not admit a straight-line embedding.
Input: K4
Input point set P:
Restricted point set drawing
 Interesting open question:
Given a planar graph G and an arbitrary point set
P in general position on the plane, does G have a
straight-line non crossing embedding in P?
 This problem for general planar graph is believed
to be NP-complete.
 Even then the progress have been made
when G is restricted to a subclass of planar
graphs.
Convex and non convex sets
 A set S in a vector space over R is called convex
if the line segment joining any pair of points of S
lies entirely in S.
Convex set:
Non convex set:
Convex hull
 Let P be the non-empty finite point set in the plane.
Convex hull of P is the boundary of the minimal
convex set that contains P.
 If points of P are not collinear convex hull of P is a
closed polygonal chain.
Rubber band analogy:
v1
v2
v5
v4
v3
Outerplanar graphs
 A graph is called outerplanar if it has an embedding
in the plane such that its vertices lie on a fixed circle
and the edges lie inside of the disk of the circle and
do not intersect (except for the end points).
All finite/countable trees are outerplanar:
Intuitively:
Outerplanar graphs
 Equivalently graph is outerplanar if there is some
face that includes every vertex.
 Every planar non crossing embedding of a tree will
produce exactly one face (otherwise there is a cycle
and graph is not a tree).
 Thus every tree is outerplanar.
All finite/countable trees are outerplanar:
1
2
Cycle!
Outerplanar graphs
 Obviously every outerplanar graph is planar.
 Not every planar graph is outerplanar, for instance
K4 is the smallest non-outerplanar graph known.
K4 is not outerplanar.
Reminder: minors
 Graph H is a minor of G if it can be obtained from G
by repeatedly contracting or removing edges, and by
deleting isolated nodes.
H is a minor of G:
Delete
Contracting
isolated
edges
nodes
Removing
edges
H:
G:321=H:
G
:
Outer-planarity criterion
 A beautiful analogue to Kuratowski's theorem
can be found for outer-planar graphs:
 A graph G is outer-planar if and only if it
contains neither K4 nor K2,3 as a minor.
K4
K2,3
Not outer-planar graph:
Maximal outerplanar graphs
 Outerplanar graph is called maximal if addition of a
new edge invalidates its outer-planarity.
 Can be looked upon as a triangulation of every
interior face of the graph.
G:32:is maximal outer-planar
G
Drawing outerplanar graphs
 Theorem:
Every outerplanar graph G on n vertices admits a
non crossing straight-line drawing on any set P of
n points in general position on the plane as its
vertex set.
 Fasten your seat belts, here comes the proof!
Proof of the theorem
Proof:
 We are going to prove a stronger statement.
 Given any two consecutive vertices v1, v2 on the
outer face of G and any two consecutive vertices
p1, p2 of the convex hull of P, there is a legal
drawing in which v1 is represented by p1 and v2
is represented by p2.
Proof of the theorem
Proof:
 Without loss of generality we can assume that
G is maximal outer-planar graph.
 Base case: n = 3, the statement is true as we
can always embed a triangle into any 3 points
on a plane such that v1 is represented with p1
and v2 is represented with p2.
 Assume that the assertion is true for all
outerplanar graphs with less than n vertices.
Proof continued
 Let v1, v2, … , vn denote the vertices of G along
the outer face in counter clockwise order.
 Let vi be the third vertex of the unique triangle
sitting on the edge {v1, v2} in E(G).
Input G:
v3
v4
v2
vi = v5
v1
v8
v6
v7
Proof continued
 Let p1, p2 be any 2 consecutive vertices of the
convex hull of P.
Input P:
w
p2
p1
Proof continued
 There is a point p in P - {p1, p2} such that:
a) Triangle <p1 p2 p> contains no point of P.
b) There is a line l through p and an interior point
of the segment p1p2 that has precisely i–2
points of P strictly on its right-hand (p2) side.
Input P:
p2
p1
p
l
Proof continued
 First select p’ such that p2p’ has exactly i-3
elements of P strictly on its right-hand side.
 If triangle p1 p2 p’ is empty, let p := p’
 Otherwise choose p to be the point in p1 p2 p’
such that the angle p2 p1 p is minimal.
Input P:
p2
p
p’
p1
p’
p’
p’
Proof continued
 Let G1 and G2 be the subgraphs of G induced by
{v2, v3, …, vi} and {vi, vi+1, ..., vn, v1} resp.
 Now let H1 and H2 denote the points of P on the
right and left half-planes of l respectively (incl p).
v3
Input G:
Input P:
H1
p2
v4
v2
G1
vi = v5
v1
G2
p1
p
l
v8
v6
v7
H2
Proof continued
 By induction hypothesis G1 has a non crossing
straight line drawing on the point set H1.
 So does G2 on H2.
 Add the edge {v1, v2} as p1p2 and we’re done.
Input G:
Input P:
H1
p2
G1
vi
G2
v2
p
v1
p1
H2
□
On constrained drawing
 Outerplanar graphs are known to be the largest
subclass of planar graphs that admits a straight
line drawing on a general set of points.
 Corollary:
Every tree with n vertices admits a non
crossing straight-line drawing on any set of n
points in general position in the plane.
Rooted tree drawing
 Theorem (Ikebe, Perles, Tamura and
Shinnichi)
 Every rooted tree with n vertices admits a non
crossing straight-line drawing on any set P of n
points in the plane in general position such that
the root is embedded in an arbitrarily specified
element of P.
 Proof is beyond the scope of our lecture.
Concluding
 Even when G is planar there are point sets that
forbid non-crossing straight line drawing.
 Outerplanar graphs have a drawing where all
vertices lie on some face.
 Graph is outerplanar iff it does not contain K4 or
K2,3 as a minor.
 Every outerplanar graph has non-crossing
straight-line embedding on an arbitrary set of
points in the plane.
 Every tree can be drawn on any set of points
while embedding the root at a specific point.
Moving on
 Here we leave our outer-planar graph study and
move on to general graph drawing questions.
 Good time for your questions.
Angular resolution
 Putting a graph on a small grid guarantees that
the picture does not get too “crowded”.
 Another parameter is the angular resolution of
the graph G, which is the smallest angle between
two adjacent edges in the best drawing of G.
Much
Poor angular
better layout:
resolution:
On angular resolution
 Clearly if the maximum degree of the vertices
in G is d, the angular resolution cannot
exceed 2 .
d
 This is not a tight bound. For instance K3 with
d=2 but with an angular resolution of π/3.
 This parameter is known to be NP-hard to
compute, even for d=4. But this problem is
not known to be in NP.
On angular resolution
 Theorem (Formann et al.)
Every graph with maximum degree d has a
straight-line drawing with resolution  c 
 Edges are allowed to cross in such
representation.
1
d
2
On angular resolution
 With a little care we can make sure that all
vertices are embedded into distinct points of a
polynomial size grid and the resolution remains
roughly the same.
 For planar graphs, the situation is better, a
resolution of at least c*(1/d) can be achieved.
 But again however, in such a representation
the edges may cross!
On angular resolution
 Theorem (Malitz, Papakostas)
Every planar graph with maximum degree d
admits a non-crossing straight-line drawing
with resolution of at least ad-2, where a > 0 is
an absolute constant.
 We are going to base the proof on 3 lemmas
and 1 important theorem.
Disk packing
 Disk packing P is a set {D1,..., Dn} of closed
disks of zero, finite or infinite radius on a
plane.
 Some disks maybe adjacent but none overlap
on interior points.
Disk packing P:
Disk packing
 Disk packing induces a planar graph G.
 Let F be an exterior face of G is this planar
drawing.
 We will say that P realizes the pair (G, F).
Disk packing P induces G:
F
Theorem 1
 Theorem 1 (Andrew and Thurston):
For every triangulated planar graph G with
exterior face F there exists a disk packing P
that realizes (G, F).
 No algorithm with polynomial time is known to
find such P.
Wheels
 Ordered disk packing P = (C, D0, …., Dt-1) is
called a wheel of length t with C as a hub if:
 All Di are adjacent to C.
 Each Di is adjacent to Di+1 mod t
P is a wheel of length 6:
C
Fans
 Ordered disk packing P = (C, D0, …., Dt) is called
a fan of length t with C as a hub if:
 All Di are adjacent to C.
 Each Di is adjacent to Di+1
P is a fan of length 4:
C
Lemma 1
 Lemma 1:
Let the ordered disk packing P = (C, A, D1,…, Dt, B)
be a fan of length t+2 with a hub C.
 Let rC, rA, r1,…, rt, rB, respectively denote the radii of
the discs in P. Assume that A and B are adjacent.
 Let r = min {rA, rB, rC}.
1
 0.15
 Then each Dj has radius rj >= at r, a 
32 3
P is a fan of length 4:
A
C
B
Lemma 2
 Lemma 2:
Take ordered disc packing P = (C, D0, …, Dt-1) that
is a wheel of length t with hub C.
 Let the discs in P have respective radii
rC, r0, …, rt-1.
 Then each disk Dj has radius rj >= at-3 rC.
P is a wheel of length 6:
D5
D4
D3
C
D2
D0
D1
Lemma 3
 Lemma 3:
Every triangulated planar graph G with exterior
face F is realized by a disk packing in which all
three exterior discs have the same radius.
Realizing K4:
Back to the theorem
 Theorem (Malitz, Papakostas)
Every planar graph with maximum degree d
admits a non-crossing straight-line drawing
with resolution of at least ad-2, where a > 0 is
an absolute constant.
 For proof for lemmas please refer to original
paper:
“On the Angular Resolution of Planar Graphs”
by Seth Malitz and Achilleas Papakostas.
Proving the theorem
 Proof:
 Given any triangulated planar graph G with
maximum degree d and exterior face F, we
know by the Theorem 1 and Lemma 3 that
there exists a disk packing P for (G, F) where
all three outer disks have the same radius, say
unit one.
 Consider any adjacent pair of discs C, D in P
and let D be the one with the smaller radius.
Proof, continued
 If C is not one of the three outer disks then C
and D are part of a wheel of length <= d with
C as a hub.
 By Lemma 2 the radius of D divided by that of
C is at least ad-3. (rD >= ad-3 rC)
C
Proof, continued
 Now suppose that C is an exterior disc and D is
an interior one.
 In this case C and D are part of a fan with hub C,
including 2 other exterior disks, named A and B.
 Using Lemma 1: rD/rC >= ad-2
C
A
B
Proof, continued
 Hence, given any two adjacent discs in P, the
radius of the smaller disk divided by that of the
larger is at least ad-2.
 Take any triangle in the straight-line drawing of
G induced by P as above.
 Call discs generating this triangle A, B and C
with radii rA <= rB <= rC respectively.
C
A
B
Proof, the end
 If we normalize the radii by taking rC = 1, by
the remarks above: ad-2 <= rA <= rB
 It is easy t see that the smallest angle L in the
triangle is achieved when rA = rB = ad-2
 Then sin(L/2) = ad-2 / (1 + ad-2)
 From basic calculus: L/2 > sin(L/2)
 Finally L > 2*ad-2 / (1 + ad-2) > ad-2
□
C
A
B
On angular resolution
 It turned out that by using disk packing
technique there is no lower bound decreasing
less than exponentially fast for the angular
resolution.
 Proved by Papacostas and Malitz in the same
paper.
Summing it up
 Angular resolution is an important measure in
graph drawing.
 A lower bound of order 1/d2 exists when edges
are allowed to cross.
 Can preserve the resolution while drawing on
the grid.
 A better lower bound exists for planar graphs
that is of order of 1/d.
 For non crossing drawing of planar graphs no
better bound than ad-2 is known (a ~ 0.15).