Computational Geometry Seminar Lecture

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Computational Geometry Seminar Lecture

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Title: Computational Geometry Seminar Lecture

1
Computational Geometry Seminar Lecture
103-quasi planar and 4-quasi planar graphs

David Malachi.

topological graph G is a graph drawn in the
plane.V(G) is a set of distinct points.E(G)
is a set of Jordan arcs. (E(G) are straight
lines ? geometric graph)
A topological graph is k-quasi-planar if no k of
its edges are pairwise crossing.(We shall refer
to 3-quasi-planar simply as quasi-planar)
It was conjectured that for every fixed k, the
maximum number of edges in a k-quasi-planar graph
on n vertices is O(n).
k 2 ? trivial (Planar graphs, for n2 exists
E(G) 3n 6).

Agarwal in 3-quasi-planar graphs E(G)
O(n).
Pach in 3-quasi-planar graphs E(G) 65n.
(Eyal Ackerman, Gabor Tardos)
Theorem 1
The maximum number of edges in a
quasi-planargraph on n 3 vertices is at most
8n - 20.
Theorem 2
For every positive integer n, there is a
quasi-planar graph on n vertices with 7n - O(1)
edges.

Proofs main concept.. (Theorem 1)
discharging method
Assign charges to elements of input.
Compute total charge.
discharging phase redistribute charges
between elements of input.
Compute total charge again.

5
Proof (Theorem 1)

Let G be a quasi-planar topological graph with n
vertices.
(WLOG with minimum intersections drawn as
possible)
Assume G is connected.G doesnt contain parallel
edges, self loops.
Let G new graph V(G) V(G) U X(G) X(G)
intersection points of edges of G (new
vertices) ?? u?X(G), d(u) 4 E(G)
subdivided edges of G with respect to X(G) F(G)
faces of G For a face f?F(G) f
number of edges comprising face f. v(f)
original vertices comprising f.

We assign charges to every face f in F(G).
For a face f, ch(f) is the charge of f.ch(f)
f v(f) - 4.
?G does not contain faces of size one or two.
? No 0-triangles (3 pairwise crossing).
? Every face in G has a non-negative charge.

Sum up all the charges, we get
Next, we redistribute the charges without
affecting the total chargefound in (1).
Our target ?
Modify ch(f) to a new charge ch (f) of a face f,
which satisfies ch (f) v(f)/5.
Note, The only faces that do not satisfy this
bound with the original charge ch are
1-triangles. (ch 31-4 0 , but v(f)/5 1/5
0)

8
Redistribution begins charge the 1-triangles

Let f be a 1-triangle.u is the original vertex
of f.e1 and e2 be the two sides of f incident
to u.e1 and e2 are edges of segments e1 and
e2.
We look for the first face along e1 (e2
symmetric) which is not a 0-quadrilateral,
denote fi .
?for sure exists, as the last face contains e1
endpoint.
?fi is not a triangle. (no 0-triangles)
We Shift 1/5 units of charge to from fi to f.
? fi lost charge through the edge shared with
fi-1

9
Current status of charges

For every 1-triangle f, ch(f) 1/5 v(f)/5.
For every 0-quadrilateral f, ch(f) 0 v(f)/5.
Let f be a face of G.
Assume f is not 0-quadrilateral, nor
1-triangle.? ch(f) (f v(f) 4) 1
ch(f) ch(f) - xf (xf is the total
charge f lost in the redistribution)
f lost at most 1/5 unit of charge through any one
of its edges only if both endpoints of the edge
are new.we have xf (f - v(f))/5
? For that f, ch (f) (2/5)v(f) (4/5)(ch(f)
- 1) 2v(f)/5
?ch (f) v(f)/5, for all faces f of G.

10
Second round of redistribution..

Now we collect all the extra charge at faces
incident to an original vertex, and place this
charge on the vertex.
For every face f with v(f) 0, we compute the
extra charge ch (f) - v(f)/5 ,
and distribute it evenly among the original
vertices v(f) of f.
Each original vertex of f receives (ch (f) -
v(f)/5) / v(f) units of charge from f.
ch(f) is the remaining charge at a face f
after this step, for any f.
ch(u) is the charge accumulated at an original
vertex u.

By the construction of ch we have ch (f)
v(f)/5 for every face f.
From equation (1) we have
Remains to prove a lower bound on ch (u) for an
original vertex u.
Claim For an original vertex u, ch (u) 4/5.

Suppose Claim is true,
Combining our bound for ch(f) for any face f
?F(G), and ch(u) for any u?V(G), we get
E(G) 8n 20. ? (Theorem)
Now on to the proof the Claim.

Proof ( Claim - ch(u)4/5 )
Let G the graph after the second round of
distribution, u? V(G).
Gu obtained from G by removing the vertex u and
all its incident edges.
Let G'u be the corresponding plane drawing. (With
respect to Gu)
- fu be the face of G'u containing u.- Let w be
a vertex of fu.? There is at least one face f of
G' that touches both u and w.
If f 5, then f alone contributes at least 4/5
unit of charge to u.Same holds for
4-quadrilateral, 3-quadrilateral, 2-quadrilateral
(with the two original vertices not being
neighbors).

As f cannot be a 1-triangle, the following cases
remained
- 2-triangle contributing 3/10.
- 3-triangle contributing 7/15 .
- 1-quadrilateral contributing 2/5 .
- 2-quadrilateral (with neighboring original
vertices) contributing at least 7/10 .

Note
Except for the 1-quadrilateral (and 1-triangle)
case we find an edge of G incident to u that is
not involved in any crossing, and therefore, the
faces on both sides of this edge contribute
charge to u.
There is at least one other vertex w' of fu
besides w. ( The minimal value of ch''(u) 4/5
is only possible if fu is subdivided into two
1-quadrilaterals and a few 1-triangles in G' .
? (claim).

16
Usage of theorem 1..

Let m0 be the largest value such that the
complete graph Km0 can be drawn as a
quasi-planar graph.? By Theorem 1 ,m0 14.
Claimm0 9 for a quasi-planar geometric
graph.
ProofOne can inspect and see for itself
By Aichholzer and Krasser K10 cannot be
drawn as a quasi-planar geometric graph.

A Reminder
Theorem 2 For every positive integer n, there is
a quasi-planar graph on n vertices with 7n -
O(1).
Proof
A constructive one
- Let be a hexagonal grid such that for each
hexagon we draw all the diagonals, but one, as
straight line edges.
- Add a curved edge for the missing diagonal.

Finally we add longer edges, two from every
vertex.
One can verify by inspection the
(3-)quasi-planarity of the obtained graph

- degree of vertices that are far enough from the
boundary is 14.
- degree of O(vn) vertices which are close to
the boundary is smaller than 14
?This quasi-planar graph already has 7n - O(vn)
edges.
To improve the error term
- wrap the graph around a cylinder so that we
have three hexagons around the cylinder.
- draw five more edges on each of the top and
bottom faces.
?now only O(1) vertices dont have degree 14.
? (Theorem 2)

Part 2
4-quasi planar graphs
By Eyal Ackerman

Theorem (4-quasi planar)
For any integer n 2, every topological graph on
n vertices with no four pairwise crossing edges
has at most 36(n - 2) edges.
Proof ( lets start)
Number of a 4-quasi planar graph on n3
vertices,
Is 3
? Assume n 4.

22
A few definitions

ep,q -
the segment of e between p and q.- e ? E(G)- p
and q are points on e.
lens
e1,e2 ? E(G) which intersect at least twice.
Region bounded by segments of e1 and e2 that
connect two consecutive intersection points is
called a lens.
A wedge is a triplet w (v, l, r) such that -
v?V (G).- l and r are edges emerging from v.- l
immediately follows r in a clockwise order.
(of the edges touching v).

23
Assumptions and markings

- G drawn with least possible number of
crossings. (with respect to quasi planarity)
- No three edges crossing at the same point.
- For every v?V(G), d(v)2.
G does not contain self loops/no parallel edges.
G does not contain Empty Lenses.(Lens which does
not contain a vertex of G).
G as beforeV(G) V(G) U X(G) (X(G) new
vertices)E(G) subdivided edges of G with
respect to X(G). (We will refer E(G) edges as
p-edges)
F(G) Faces of G. f number of p-edges
f. v(f) number of original vertices.

24
Proofs main idea is

Discharging method
Assign charges to faces of G.
Redistribute the charges, with certain logic.
No faces with negative charge.
All wedges have least of 1/18 unit of charge.
Total wedges is charge (2E(G) / 18 ) 4n
8.

25
Starting

As before, we assign charges to F(G)
ch(f) fv(f) 4
Total sum of charges is
Equality is due to Eulers formula (fe-v2)
the next equality

Minimum face size is 3.
Face of size one yields a self-intersecting edge.
Face of size two yields an empty lens or two
parallel edges
Only faces with negative charge are
0-triangles.(3 0 4 -1)
Solution
Charge the 0-triangles !

Let t be a 0-triangle - e1 be one of its
p-edges.- f1 be the other face incident to e1
f1 3
If v(f1) 0 or f1 4 (?f1s charge is
1)? move 1/3 units of charge from f1 to t.
( f1 contributed 1/3 units of charge to t
through p-edge e1 )
Otherwise, f1 must be a 0-quadrilateral.

We look at the face incident to the opposite edge
to e1,and continue on until we find a face which
is not a 0-quadrilateral, Denote - fi.
? fi will contribute 1/3 units of charge to t
through ei-1.
In a similar way t obtains 2/3 units of charge
from its other p-edges.

After re-distributing charges this way, the
charge of every 0-triangle is 0.
A face can contribute through each of its p-edges
at most once ?every face f such that fv(f)
6 still has a non-negative charge.
Remains to verify
1-quadrilaterals and 0-pentagons, which had only
one unit of charge to contribute, also have a
non-negative charge.
Observe
? 1-quadrilateral contributes to at most two
0-triangles (endpoints of a p-edge through which
it contributes must be vertices from X(G) )
? 0-pentagon, on the other hand, can contribute
to at most three 0-triangles by Observation 2.1

Observation 2.1
0-pentagon contributes charge to at most three
0-triangles. Moreover, if it contributes to
three 0-triangles, then the contribution must be
done through consecutive p-edges.
Proof (in waving hands)
One can easily inspect that a contribution to
three 0-triangles, through non-consecutive
p-edges implies four pairwise crossing edges.
?
At this stage, faces with positive charge are
called good faces, and faces with 0 charge are
called bad faces. (except for 0-triangles and
0-quadrilateral).0-pentagons that contributed
to three 0-triangles, and 0-hexagons that
contributed to six 0-triangle BAD

At this point, ALL faces have non-negative
charge.
Next stage is to put charge on the
wedges..First, a few definitions
A-crossing - Let w (v,l,r) be a wedge.
An A-crossing of w is a triplet cr (e,p,q)
such that - e is an edge crossing l at p, and
r at q- ep,q does not intersect l or r.- the
endpoints of e are not in we,p,q
cr is an uncut A-crossing of w, if ep,q is a
p-edge.
cr is farther than cr, for 2 A-crossings cr and
cr of w, if p?lv,p and q?rv,q.
cr is an empty A-crossing of w, if there are no
original vertices in wcr (wcr we,p,q )

X-crossing
Let w be a wedge, cr1 (e1,p1,q1) and cr2
(e2,p2,q2) be two A-crossings of w.
x (cr1,cr2) is an X-crossing of w if e1p1,q1
and e2p2,q2 intersect exactly once.
We say that x is empty if both cr1 and cr2 are
empty A-crossings.
wx represents the area wcr1 U wcr2 .
y is the intersection point of e1p1,q1 and
e2p2,q2.
Vis(x) is the curve (e1p1,y) U (e2y,q2).
Vis(x)l is e1p1,y.
Vis(x)r is e2y,q2.

Observation 2.4
x ( cr1 (e1, p1, q1) , cr2 (e2,p2, q2)) is
an empty X-crossing of a wedge w (v, l,
r),Vis(x)l part of e1 .an edge e that crosses
Vis(x)l (resp., Vis(x)r) must cross l (resp., r)
and must not cross e2 (resp., e1).
Let z be the crossing point of e and e1p1,y.
Since x is an empty X-crossing of w, e must
cross the boundary of wx at least one more time.
(no original vertex inside)
If it crosses e1p1,q1 at another point different
from z, then we have an empty lens.
if e crosses rv,q2 then it must also cross
e2p2,y ? four pairwise crossing edges e0,
e1,e2, r.
?e must cross l.
?e must not cross e2. (four pairwise crossing).
?

34
Charge the wedges

Let x and x be two X-crossings of a wedge w.
x is a farther X-crossing of w than x, if one
A-crossing of x is farther than one A-crossing of
x and the other A-crossing of x is not farther
than the other A-crossing of x.
An uncut A-crossing is farther (resp., closer)
than an X-crossing of the same wedge, if it is
farther (resp., closer) than both A-crossings of
the X-crossing.
Given a wedge w (v, l, r), we look for the
farthest empty uncut A-crossing or X-crossing
of w.

35
Start charging wedges

If there is no such uncut A-crossing or
X-crossing, then the face incident to w is not a
1-triangle.
?Its charge is at least 1/3 units, from which we
use 1/18 units to charge w.
Assume we find such A-crossing , cr (e, p, q).
Let f be the face incident to ep,q outside wcr.
f is not 0-triangle as this would yield an empty
lens or parallel edges.
f is not 0-quadrilateral since this would imply
an empty uncut A-crossing farther than cr.
- If f is a bad pentagon, then it follows from
Observation 2.1 that there is an empty
X-crossing, farther than cr.
?f must be a good face which will contribute 1/18
units of charge to w through ep,q.

36
2nd case

Assume we find such X-crossing x (cr1
(e1,p1,q1), cr2 (e2,p2,q2)). Vis(x)l
e1p1,y. y is the intersection point of e1p1,q1
and e2p2,q2 .
Let f1 be the face that is incident to y and
outside wx.
Several cases for f1
- Suppose f1 is a 0-triangle, e3 be the third
edge incident to it.
?By Observation 2.4 that e3 must cross l, thus
l, e1, e2, e3 are pairwise crossing.

Suppose f1 is a bad pentagon, then we consider
the possible casesnone, one, or both of e1p1,y
and e2y,q2 are p-edges
- In case both of them are p-edges ,then there is
an empty uncut A-crossing of w that is farther
than x.
In case one of them, WLOG e2y,q2 , is a p-edge,
then there is an empty X-crossing farther than
x.
In case none of them is a p-edge ,then there must
be four pairwise crossing edges.

In a similar way, if f1 is a bad hexagon then
there must be four pairwise crossing edges, or an
X-crossing of w farther than x.
So, if f1 is a bad face, it must be a
0-quadrilateral.
Let f2 be the face outside of wx that shares a
p-edge with f1 and is incident to Vis(x)l .
If there is no such face, or there is no face
outside wx that shares a p-edge with f1 and is
incident to Vis(x)r, then there is an empty
X-crossing farther than x.

f2 cannot be a 0-triangle . (four pairwise
crossing edges).
if f2 is a bad pentagon (or a bad hexagon), then
again (as before) there must be four pairwise
crossing edges or an empty X-crossing farther
than x.

So, lets assume f2 be a 0-quadrilateral,Lets
examine the next face, that is, the face f3 ! f1
such that f3 is outside wx, shares a p-edge with
f2, and is incident to Vis(x)l.
If there is no such a face, then again, we have
an empty X-crossing farther than x.
Else Apply the arguments for f2 and so on..
Until we find such a good face, which we must
encounter, otherwise we have an empty X-crossing
farther than x.
Denote fi -the first good face we encounter
along Vis(x)l, ei -be the p-edge of fi that is
contained in Vis(x)l.
?fi contributes 1/18 units of charge to w
through ei.

Left is to prove that after charging every wedge
of an original vertex, there are no faces with a
negative charge.
We show that by proofing a face cannot contribute
to too many" wedges.
Definition
For a (good) face f and one of its p-edges m, we
say that f is a possible X-contributor to a wedge
w through m, if there is an empty X-crossing of
w, x, such that f is outside wx and m?Vis(x)l.
Observation 2.5.Let f be a face and let m be one
of its p-edges. Then f is a possible
X-contributor through m to at most one wedge.

Proof (Observation 2.5)
Let f be a face.Suppose f is a possible
X-contributor through one of its p-edges m, to
two wedges, w1 (v1, l1, r1) and w2 (v2, l2,
r2).
Let e be the edge containing m.
There are four points p1, q1, p2, q2, such that
cr1 (e, p1, q1) is an A-crossing of w1.
cr2 (e, p2, q2) is an A-crossing of w2.
x1 ((e, p1, q1),(e1, p1, q1)) -
the empty X-crossing of w1, such that m
?Vis(x1)l.
x2 ((e, p2, q2),(e2, p2, q2)) -
the empty X-crossing of w2,such that m ?Vis(x2)l.

Lets we sort p1, q1, p2, q2 by the order in which
they appear when traversing e from one of its
endpoints to the other.(such that when
traversing m the face f is to our right.)
?pi must precede qi, for i 1,2 since f is
outside of wxi.
Assume, WLOG, that p1 precedes p2.
(Other case symmetric)
? Notice, By Observation 2.4 l2 must cross l1.
Since m ?( ep1,q1 U ep2,q2) , the order of
the four points is p1, p2, q1, q2 or p1, p2,
q2, q1

Suppose q1 precedes q2
w1x1 and w2x2 do not contain any original
vertex. We have 2 options
- l2 must cross l1 and r1.
- r2 must cross l1 and r1.
The first case yields an empty lens.
In the second case, note that e1 must cross
either l2 or r2 , yielding four pairwise
crossing edges.

Suppose q2 precedes q1
Then the edge e2 must cross e twice and
creating an empty lens, or cross l1 and yield
four pairwise crossing edges.

We conclude that f cannot be a possible
X-contributor through m to more than one wedge.
? (Observation 2.5)
?face f, such that f 7, ends up with a charge
of at least (when v(f)0)f - 4 - f( 1/3
1/18 ) 0.
?k-quadrilaterals, for k 0, and good pentagons
and hexagons, end up with a non-negative charge,
as their charge after charging the 0-triangles
was at least 1/3 .
Summing up the charge over all the wedges we have
2E(G)/18 4n 8 ? E(G) 36n
72
?