computational geometry seminar lecture #10 3-quasi planar and 4-quasi planar graphs david malachi

Computational Geometry Seminar

Lecture #10

3-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 n>2 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-planar

graph on n ≥ 3 vertices is at most 8n − 20. Theorem 2For every positive integer n, there is a quasi-planar graph

on n vertices with 7n − O(1) edges.

Proof`s 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.

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 doesn’t 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)

uX(G), d(u) = 4

E(G’) = subdivided edges of G with respect to X(G) F(G’) = faces of G’

For a face fF(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 = 3+1-4 = 0 , but v(f)/5 = 1/5 > 0)

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 ”

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‘.

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 uV(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).

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.

Claim:

m0 ≥ 9 for a quasi-planar geometric graph.Proof:

One 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).

ProofA 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(√n) vertices which are close to the boundary is smaller

than 14

This quasi-planar graph already has 7n − O(√n) 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 don’t 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 n=3 vertices,Is 3 < 36(3 - 2) = 36.

Assume n ≥ 4.

A few definitions

e|p,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 - vV (G).- l and r are edges emerging from v.- l immediately follows r in a clockwise order. (of the edges touching v).

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 vV(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 before…V(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.

Proof`s 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 (2|E(G)| / 18 ) ≤ 4n – 8.

Starting…

As before, we assign charges to F(G’):ch(f) = |f|+v(f) – 4

Total sum of charges is :

Equality is due to Euler`s formula (f=e-v+2) + 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 (f1`s 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 |f|+v(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 2.10-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- e|p,q does not intersect l or r.- the endpoints of e are not in w|e,p,q

• cr is an uncut A-crossing of w, if e|p,q is a p-edge.• cr is farther than cr`, for 2 A-crossings cr and cr` of w, if p`l|v,p

and q`r|v,q.• cr is an empty A-crossing of w, if there are no original vertices in w|

cr (w|cr = w|e,p,q )

X-crossingLet 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 e1|p1,q1 and e2|p2,q2 intersect

exactly once.We say that x is empty if both cr1 and cr2 are empty A-crossings.

w|x represents the area w|cr1 U w|cr2 .y is the intersection point of e1|p1,q1 and e2|p2,q2.Vis(x) is the curve (e1|p1,y) U (e2|y,q2).Vis(x)l is e1|p1,y.Vis(x)r is e2|y,q2.

Observation 2.4x = ( 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 e1|p1,y.Since x is an empty X-crossing of w, e’ must cross the boundary of w|x

at least one more time. (no original vertex inside)If it crosses e1|p1,q1 at another point different from z, then we have

an empty lens.

if e’ crosses r|v,q2 then it must also cross e2|p2,y four pairwise crossing edges: e0, e1,e2, r.

e’ must cross l.e’ must not cross e2. (four pairwise crossing).

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.

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 e|p,q outside w|cr.- 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 e|p,q.

2nd case

Assume we find such X-crossing x = (cr1 = (e1,p1,q1), cr2 = (e2,p2,q2)). Vis(x)l = e1|p1,y. y is the intersection point of e1|p1,q1 and e2|p2,q2 .

Let f1 be the face that is incident to y and outside w|x.

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 cases:none, one, or both of e1|p1,y and e2|y,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 e2|y,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 w|x 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 w|x 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 w|x, 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.

DefinitionFor 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 w|x and mVis(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 w|xi.

Assume, WLOG, that p1 precedes p2. (Other case – symmetric)

Notice, By Observation 2.4 l2 must cross l1.

Since m ( e|p1,q1 U e|p2,q2) , the order of the four points is p1, p2, q1, q2 or p1, p2, q2, q1

Suppose q1 precedes q2 :

w1|x1 and w2|x2 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

2|E(G)|/18 ≤ 4n – 8 |E(G)| ≤ 36n – 72

The End….

Thank you !