Quadrangulations of polygonal regionsJurij Kovič

University of Primorska, Koper; Institute for Mathematics, Physics and Mechanics, Ljubljana

2015 International Conference on Graph TheoryKoper 27 May 2015

Purpose and structure of the talk

This talk is about:• CORRESPONDENCES, • CODES, • CLASSIFICATION & COMPUTATIONS.

The purpose of this talk is:• to present some CORRESPONDENCES between various mathematical structures;• to use these correspondences for the CODING of these structures;• to use this coding for the CLASSIFICATION of these objects and for various

COMPUTATIONS (operations, transformations) on them.

Content

1) Introduction 2) Correspondences 3) Codes 4) Classification and computations 5) Conclusion

We start with a few definitions, then follow the plan: CORRESPONDENCES → CODES → CLASSIFICATION & COMPUTATIONSand end with a few open problems.

1. Introduction

QUADRANGULATION → RHOMBIC QUADRANGULATION OF A POLYGONAL REGION → INTERSECTION GRAPH OF RHOMBIC PATHS → CIRCLE GRAPH →CYCLICAL CODE OF A CIRCLE GRAPH → CYCLICAL CODE OF ANY GRAPH → CLASSIFICATION OF GRAPHS → OPERATIONS ON CYCLICAL CODES → OPERATIONS ON ANY GRAPH

• First we define what we mean by a ‘‘quadrangulation‘‘, then we show how ‘‘rhombic quadrangulations‘‘ are related to graphs.

Quadrangulations of polygons

• Let Ω be a planar polygonal region (homeomorphic to a disc, with only one boundary component).• Definition. A quadrangulation Q(Ω) of a region Ω is a tiling of Ω with

quadrilaterals whose vertices may lie also in the interior of Ω.

Quadrangulations of surfaces

• Quadrangulations are studied also in 3D-space. • The shapes of the surfaces are often modeled by quadrangulations, using only

‘‘regular vertices‘‘ of valence 4 and ‘‘irregular vertices‘‘ of valence 3 and 5. Irregular vertices are eliminated, if possible‚ or ‘‘pushed‘‘ to the boundary.• Such ‘‘requadrangulations‘‘ (i.e. local changes of some small part of a

quadrangulation) are used in architecture to obtain more stable constructions (or in design, to obtain more beautiful regular patterns on shoes, clothes, etc.).

Opposite sides of quadrangulated regions • Observation. The number of boundary edges of a quadrangulated

region must be even.

• Proof. Just connect the midpoints of each quadrilateral with a curve. This automatically ‘‘connects‘‘ pairs of boundary edges. • Thus each side ai has its ‘‘opposite side‘‘ ai* and (ai*)* = ai.

Convex quadrangulations

• If all the quadrilaterals in a quadrangulation are convex, the quadrangulation is called a convex quadrangulation.• Question. Do all convex quadrangulations of a polygonal region Ω

define the same pairs (ai,ai*) of opposite sides of Ω?• No!

Rhombic quadrangulations

• Definition. If all the tiles are rhombs, then the quadrangulation is called a rhombic quadrangulation.

• Question. Do all rhombic quadrangulations of a polygonal region Ω define the same pairs (ai,ai*) of opposite sides of Ω?• Yes! And these opposite sides ai,ai* are always parallel!

Planar rhombic quadrangulations

• Planar rhombic quadrangulations have also other interesting properties. One of them is: • No two planar ‘‘rhombic paths‘‘ intersect more than once!

Planar rhombic paths ‘‘behave as lines‘‘• Claim. No two planar rhombic paths (in the tiling of Ω ) can have

more than one common rhomb.

• Thus two planar rhombic paths intersect once or not at all.• This is not true for the general planar quadrilateral paths, nor for the rhombic

paths in the 3D-space.

2. Corespondences

• Now we present a series of corespondences (‘‘iso-morphisms‘‘ and ‘‘homomorphisms‘‘) between various mathematical structures.• The first such result is:• Theorem. There is a corespondence between rhombic

quadrangulations Q(Ω) of planar regions Ω and graphs G(Q(Ω)).

• The region Ω may contain ‘‘polygonal holes‘‘.

Obvious differences

• At first glance, quadrangulations and graphs are two completely different things. In fact, ‘‘it is easy to see‘‘, ‘‘it is obvious‘‘, that they have nothing in common, nothing at all!

• And we can also ‘‘prove‘‘ this, since quadrangulations are, ‘‘geometrically speaking‘‘, obtained by glueing 2-dimensional objects (rhombs) together, while graphs are clearly 1-dimensional objects.

• So, why to search for similarities where there is no hope to find them?• Because ‘‘looking through the lens of algebra (or combinatorics)‘‘ these two

structures are the same!

Hidden correspondences

In mathematics (and perhaps elsewhere, too), it is always interesting (and useful) to find surprising correspondences (either 1-1 or only injective mapping A → B) between seemingly different, but in fact isomorphic structures A and B.

If B is somehow ‘‘more simple‘‘ or ‘‘easier to work with‘‘ than A, then we may ‘‘profit‘‘ from observing such a correspondence! In that case B can serve either as a ‘‘model‘‘ or as a ‘‘code‘‘ for A, and problems about A (e.g. the problem of the enumeration of its members) are translated into problems about B.

Intersection graphs

• To establish a correspondence between quadrangulations and graphs we use the well known concept of the intersection graph.• Definition. The intersection graph I(S) of the family of sets Si has the

vertices vi = Si and its edges e(i,j) are the intersections of sets Si and Sj.

• Examples. Triangle graphs, line graphs, etc.

Quadrangulations and graphs

• Correspondence. Each rhombic quadrangulation Q(Ω) of a polygonal region defines a graph G(Q) = I(P), obtained as the intersection graph of the family P = P(∂Ω) of rhombic paths Pi in the quadrangulation Q = Q(Ω)!• The vertices vi of G are rhombic paths Pi, i = 1,2,…,n. The edges e(i,j) of G are

intersection of rhombic paths – they correspond to rhombs R(i,j)!

Given Ω , is G(Q(Ω)) unique?

• Question. Is the graph G(Q(Ω)) independent of the quadrangulation Q? Is it uniquely determined by Ω, so that we can write G(Q(Ω)) = G(Ω) ?

Is it possible to reconstruct the graph G(Q(Ω)) only from the boundary ∂Ω (without knowing Q)?

• Question. Suppose somebody takes away (steals) all the rhombic tiles of the quadrangulation Q(Ω) and we are left only with the (drawing of the contour boundary of Ω. • Can we find out (deduce from the shape of Ω) if there was a rhomb

R(i,j) in the quadrangulation Q(Ω) (or, equivalently, if there is an edge e(i,j) in G(Q(Ω)))?• Yes! • This is easy to find out if Ω is homeomorphic to a disc.

• If Ω contains ‘‘polygonal holes‘‘, it can be proved (by induction) that G(Ω) is uniquely determined by Ω.

Cyclical sequences

• We prove the uniquenes of G(Q(Ω)) using the concept of a cyclical sequence (i.e. a sequence whose elements can be ‘‘shifted‘‘).

• Example. (a,b,c,d) = (b,c,d,a) = (c,d,a,b) = (d,a,b,c).• Example. (2,3,5) = (3,5,2) = (5,2,3).

Matching sequences

Definition. A cyclical sequence C = (a1,a2,…,a2n) is called a ‘‘matching sequence of length 2n‘‘, if each ai has its ‘‘match‘‘ ai*, and (ai*)* = ai.

Example. C = (a, b, c, a*, c*, b*)

The ‘‘matches‘‘ ai and ai* may be represented also with the identical symbols: C = (1, 2, 3, 1, 3, 2) or simply with the same colors: C = ( ).

Matching sequence C(Ω) of a polygonal region Ω with pairs (ai,ai*) of matching parallel edges

• Observation. To every polygonal region with pairs (ai,ai*) of matching parallel edges corresponds a matching sequence C(Ω) composed of these edges.

The ‘‘crossing condition‘‘ for the rhombic paths• Lemma. Two rhombic paths (ai,ai*) and (aj,aj*) in a rhombic quadrangulation Q(Ω) of a polygonal

region Ω intersect each other if and only if they form an ‘‘alternating pattern‘‘:

(…,ai,…,aj,…,ai*,...,aj*,…).So, the non-intersecting rhombic paths form the pattern:

(…,ai,…,ai*,…,aj,…,aj*,…).Proof. Use the ‘‘Jordan curve theorem‘‘, the propertythat two planar rhombic paths intersect at most once (andthe property that matching pairs are independent of the rhombic quadrangulation).Corollary. The existence of the rhomb R(i,j) in Q(Ω) depends

only on the matching sequence C(Ω)!

G(C) is also the ‘‘intersection graph‘‘ of the chords in a circle with endpoints ai,ai* arranged as in the cyclical sequence C

Obviously these 2n points can be placed in a ‘‘general position‘‘so that no three chords meet in the same point.

Such intersection graphs are known in the graph theory as ‘‘circle graphs‘‘.

G(Q(Ω)) = G(Ω)

• Theorem. Every rhombic quadrangulation Q(Ω) of a polygonal region defines the same graph G = G(Ω).• Proof (for the polygonal regions). Each two rhombic paths (representing

vertices of G(R(Ω)) intersect at most once. • And this depends only on which pairs of boundary edges are ‘‘alternating‘‘

and which are not! • Remark. General proof for the regions with polygonal holes rests on an

induction argument.

The graph G(Q(Ω)) depends only on the polygonal region Ω

Rhombic quadrangulations of planar polygonal regions correspond to quadrangulations of the projective plane

• Theorem. Every rhombic quadrangulation Q(Ω) of a planar polygonal region Ω coresponds to a quadrangulation of the projective plane.

• Proof. Identify (glue together) the opposite matching edges AB and A*B* (identify the endpoints A,A* and B,B*).• Corollary. The dual graph of this quadrangulation defines a map on the

projective plane whose 1-grid is a 4-valent graph G‘. Therefore to any circle graph G(C) corresponds a 4-valent graph G‘(C) whose adjacent vertices correspond to adjacent edges of G (although not all adjacent edges in G correspond to adjacent vertices in G‘).

3. Codes

• Observation. Every circle graph G(D) on n vertices can be ‘‘coded‘‘ by a cyclical matching sequence C of length 2n.

• Proof. The edges of G can be found algebraically (without looking at the circle diagram D) just by applying the ‘‘crossing condition‘‘ on the sequence C.

• Questions:• Can we use this fact on some well-known family of circle graphs?• Is every graph a circle graph?

Is every graph a circle graph? No!

It is a well known fact in graph theory, that: • circle graphs are a subclass of ‘‘string graphs‘‘ (i.e. intersection graphs

of curves on a plane) and

• a subdivision of K5 is not a string graph. Hence it is not true that every graph is a circle graph.

Can we find ‘‘some kind of a cyclical code‘‘ also for non-circle graphs? Maybe for the string graphs?

• What do we know about string graphs? • Every planar graph is a string graph (Schaefer and Štefankovič (2001)

credit this observation to Sinden (1966)• By Schneierman‘s conjecture (now proven) every planar graph may be

represented as the intersection graph of straight line segments.

Idea. If one cyclical sequence is not enough for the coding of a graph, use several sequences!

We are looking for something like this:

G = (a1,a2,…,ai) + (b1,b2,…,bj) +…+ (k1,k2,…,kl).

General idea → Technical details (‘‘paperwork‘‘)

Once you have this general idea in this ‘‘crystallized form‘‘ its ‘‘technical realization‘‘ is more or less obvious, although some time was needed to formulate it in this way!

In fact, to realize that just one cyclical sequence may not always be enough(and abandoning the naive hope: ‘‘One sequence suffices!‘‘) was a major step for me in formulating this idea.

m-cyclical graphs

We will see:

Some graphs may be coded with one cyclical sequence.Some graphs may be coded with two cyclical sequences, but no less.…Some graphs may be coded with m cyclical sequences, but no less.Such graphs will be called ‘‘m-cyclical graphs‘‘.

We will answer also the following question:Question. Is it true that every graph G is m-cyclical for some m = m(G)?

Maybe every string graph is m-cyclical for some m? But how to define cyclical sequences C1,…,Cm?

Step 1: Start drawing the ‘‘outer contour‘‘

This is ‘‘almost‘‘ a circle graph with the codeC1 = . But there are two more edges!

Step 2: Identify all the lines i containing crossings (i,j) that have not been used yet in and repeat Step 1 on them!

Now we have two cycles C1 or C2.Hence our graph is 2-cyclical, since G = C1 + C2.

Idea. In coding graphs with cyclical sequences we can use also ‘‘subtraction‘‘ of cyclical sequences!

Every string graph is ‘‘composed of‘‘ sums or differences of circle graphs. In this sense it is m-cyclical: G = C1 + C2 – C3 …+ Cm

• In other words: Every string graph can be coded by the respective cyclical codes Ci of these circle graphs!

What about general graphs?Is every graph m-cyclical for some m?

• Yes!

The coding of graphs with cyclical sequences• We know: Every graph is a disjoint union of planar graphs:

G = G1 + … + Gn

• It is known: Each planar graph is a string graph (Sinden, 1966).• We have shown: Each string graph Gi is m-cyclical:

Gi = Ci 1 + … - Ci m

Therefore: Theorem. Every graph is m-cyclical (it can be coded with the ‘‘sums‘‘ and ‘‘differences‘‘ of cyclical sequences: G = C1 + C2 +…- Cm )!

Remark. This works also for signed graphs! (with edges labeled 1,0 and -1).

Yet how to get (construct) these cycles (circle graphs)? How to get the code C1+…-Cm of G?

• If a graph is drawn on a torus or on some other surface, closed curves do not always correspond to the boundaries of discs!

To find a code C(G) of a graph G find the coresponding rhombic quadrangulation of some region Ω!

• To build a geometrical model of a graph whose cyclical code can be easily found it is enough to do the following:• Start with a regular 2n-gon and cover it with rhombs.• Label the edge vectors by 1,2,3,…,n,1*,2*,…,n*.• The rhombus (i,j) coresponds to the edge (i,j) in G.• Delete from the regular polygon all rhombi (i,j) that are not in G.• Try to reduce the number of ‘‘holes‘‘ by rearranging the rhombs (their

orientation must remain the same) ‘‘pushing‘‘ them towards the boundary.

Example: ‘‘Pushing‘‘ the rhomb (2,3) towards the boundary ‘‘using hexagons‘‘ before deleting it

Not every rhomb can be deleted without increasing the number of cyclical sequences in the code, or without increasing the number of chords! Some chords in the corresponding circle graph must then be colored with the same color.

Theorem. Every graph has a coding with cyclical sequences of the type G = C1 – C2 – … – Cm.Proof. If G is the graph on n vertices just delete from the tiling of the regular 2n-gon any rhomb R(i,j) such that the edge e(i,j) is not in G. This rhomb R(i,j) has a code (i,j,i,j).

All the deleted circle graphs (having no common chords passing through them) may be imagined as joined into one, since the corresponding cyclical matching sequences are just ‘‘concatenated‘‘ like this: (1,2,1,2)+(4,5,6,4,6,5) = (1,2,1,2,4,5,6,4,6,5).

Every (connected) graph has a coding: G = C or G = C1 – C2. • Theorem. Every graph is either a circle graph or a difference of two

circle graphs!

4. Classification and computations

Hence, there are only two classes of graphs:Class 1: Circle graphs.Class 2: Differences of two circle graphs.Classification completed!

This might have a practical application.

Imagine all graphs coded as G = C1 – C2

• Imagine that we have an effective algorithm for expressing every graph G in that form!• Then we could also make a list of codes of all circle graphs C1,C2,C3,…

• In fact any code of a given circle graph is equally good, since from any of them this graph may be easily reconstructed!

• Then we could calculate all possible differences Ci - Cj of these graphs.• And then any graph G(i,j) = Ci - Cj could be represented as a point (i,j) in a two-

dimensional grid! • This would surely simplify very much the making of any possible collection or

census of graphs!

Example. The code of a Petersen graph

Since the Petersen graph is the best example for almost everything in the graph theory, let us construct its code G = C1 – C2. It has three orbits of edges and a 5-fold rotational symmetry. The cyclical sequence C1 must contain 5 x 4 = 20 numbers.

Edges e(i,j) → Rhombs R(i,j)

• Each edge corresponds to a rhomb in the quadrangulation.

Since there are two vertex orbits in G,there are two orbits of rhombic paths in Q!

• The rombic path 1 contains rhombs 15,16,12.

We continue drawing the rhombic paths from the first orbit using the 5-fold rotational symmetry.

• Similarly we find the rhombic paths 2 and 3.

The same pattern must be very carefully exactly reproduced using the 5-fold symmetry.

• Similarly we find the rhombic paths 4 and 5.

We try to add a few details of the second orbit of rhombic paths.

We add the numbers of the rhombic paths and try to guess some more details. (A lot of trial and error!)

Now we forget geometry for a while! We have found a sequence 1569! Hence, because of the 5-fold rotational symmetry it must be:

C1 = (1, 5, 6, 9, 2, 1, 7, 10, 3, 2, 8, 6, 4, 3, 9, 7, 5, 4, 10, 8) !Now we can complete the diagram:

Let us draw this with the rhombs! 10 rhombs must be deleted! C1 = (1, 5, 6, 9, 2, 1, 7, 10, 3, 2, 8, 6, 4, 3, 9, 7, 5, 4, 10, 8) • C2 = (1, 9, 1, 6, 2, 10, 2, 7, 3, 6, 3, 8, 4, 7, 4, 9, 5, 8, 5,10) Hence the cyclical code of the Petersen graph is:

G = C1 – C2 = (1,5,9,6,2,1,10,7,3,2,6,8,4,3,7,9,5,4,8,10) – (1,9,1,6,2,10,2,7,3,6,3,8,4,7,4,9,5,8,5,10).

A method for representing a graph G in a form C or C1 – C2.

• Step 1. Start with a rhombic tiling of convex 2n-gon. • This polygon corresponds to the complete graph Kn with the cyclical code C(G) =

(1,2,…,n,1,2,…,n).

• Step 2. Delete the rhombs R(i,j), corresponding to the edges in Kn – G.• The remaining rhombs represent the edges of G.

• Step 3. By rearranging the remaining rhombs reduce the number of holes in the region covered by these rhombs as much as possible.• Step 4. You get a region with at most 2 boundary cycles. Write down the

corresponding cyclical sequences C1 and C2.

Step 1. Tiling of a convex 2n-gon with 1 + 2 + 3 + … + n – 1 = n(n-1)/2 rhombs

Step 2. G is the complement of Kn – G in Kn

Step 3: Rearranging the rhombs

• This can be done also in a form of a game, or of ‘‘solving a puzzle‘‘:• Given the rhombs (i,j) corresponding to the edges of G glue them

together in a plane so that:• Two adjacent rhombs share the same color along the common edge.• The number of ‘‘holes‘‘ in the covered region is minimal.

Algebraical formula A(G) of a graph G, defined as a formal sum of rhombs aiaj • To each graph G with vertices a1,a2,…,an (or just 1,2,…,n) corresponds

its algebraic formula A(G), obtained as follows:• Edge e(i,j) is represented as a formal product aiaj.• Think of ij as a vector product ai x aj whose absolute value represents the area

of the rhombus (i,j).

• We define: A(G) = Σ aiaj, where e(i,j) in E(G).

• Example. A(G) = ac + bc = (a + b)c.

The distributive law ac + bc = (a + b)c applies.

Algebraical formula for the complete graphA(Kn) = (1/2)(a1 + … + an)2, since ai

2 = 0.• Using this formula for the complete graph Kn we easily get the

algebraical formulas and also cyclical codes of other graphs.• Example. Delete a clique Km from a Kn. Suppose m = 3 and n = 5:• 2A(K5 – K3) = (a + b + c + d + e)2 – (a + b + c)2 = = (2(a + b + c) + (d + e)) (d + e) = = 2(a + b + c)(d + e) + (d + e)(d + e) = = 2(a + b + c)(d + e) + 2de = = 2((a + b + c)(d + e) + de)

Hence A(K5 – K3) = (a + b + c)(d + e) + de.

The formula A(G) may help us to construct the corresponding region Ω(G) covered by rhombs

• C(K5 – K3) = (a,b,c,d,e,a,b,c,d,e) – (a,b,c,a,b,c).

• A(K5 – K3) = (a + b + c)(d + e) + de• From the ‘‘form‘‘ of this formula we see that Ω(G) must be composed of the ‘‘direct product‘‘ of paths (a + b + c) and (d + e), and of an additional rhomb de.

From A(K5 – K3) to C(K5 – K3)So K5 – K3 is a circle graph!

General formula for A(Kn – Km)and the cyclical code C(Kn – Km)

• As before we get:

2A(Kn – Km) = (a1 + … + am + am+1 + … + an)2 – (a1 + … + am)2 =

= 2((a1 + … + am ) + am+1 + … + an)(am+1 + … + an) =

= 2((a1 + … + am )(am+1 + … + an) + (am+1 + … + an)(am+1 + … + an)

• The cyclical code C(Kn – Km) of the region Ω(Kn – Km) is easily

obtained from C(Kn) = (1,2,…,m, m+1,…,n, 1,2,…,m-1,m, m+1,…,n) by reversing the order in 1,2,…,m. Thus:

C(Kn – Km) = (1,2,…,m, m+1,…,n, m,m-1,…,2,1, m+1,…,n)

How to check if a given G on n vertices is 1-cyclical? How to construct C(G) from G when this is possible?

• A ‘‘brute force‘‘ approach with computers could be like this:

• Step 1. Generate all cyclical matching sequences C of length 2n.• Step 2. Calculate their graphs G(C).• Step 3. Make a list C → G(C) connecting each C to the graph G(C).• Step 4. Make a list G → C(G) connecting each graph G on n vertices to

its cyclical code C(G)!

So we can use our code G = C1 – C2 for the classification of graphs. What else can we do with it?

• Good codes should serve not only for the classification or enumeration purposes, but also for the computational purposes!• The Roman numerals I,II,III,IV,… were good only for counting, not for

calculations. The hindu-arabic numerals 0,1,2,3,4,5,6,7,8,9 and the position system of writing numbers changed the situation dramatically.• Similarly, using the code G = C1 – C2 we can define many

computations, transformations and operations on graphs.• Our first example of this kind is from mathematical chemistry!

Boundary codes of benzenoid systems (polyhexes)• In mathematical chemistry, a benzenoid system B is a mathematical model

of a hydro-carbon molecule, composed of hexagonal carbon chains and some hydrogen atoms (which are ignored in the model). • So, a benzenoid system is simply a planar polyhex. • Various ‘‘boundary codes‘‘ of benzenoids are known in the literature. From

them the symmetry properties of benzenoids can be deduced. • However, from none of them the number h(B) of hexagons in B can be

easily determined.

Cyclical matching code of benzenoid systems

• Observation. Since a benzenoid B admits a rhombic quadrangulation Q, it can be represented by a circle graph G, hence it can be coded by a cyclical matching sequence C!

The number h(B) of hexagons in a benzenoid B can be computed from its cyclical matching code

• We just determine the number of rhombs in the rhombic quadrangulation Ω(B) using the ‘‘crossing condition‘‘ and divide this number by 3 to get h(B).

Benzenoids B, their algebraical formulas A(G(B)) and their cyclical codes C(G(B))• 2A(G(B)) = (a1 + a2 + a5)2 + (a3 + a4 + a5)2 • A(G(B)) = a5(a1 + a2 + a3 + a4 )2 + a1a2 + a3a4

• C(B) = (1,2,3,4,5,3,4,1,2,5)

What information about the shape of Ω can be deduced from the cyclical code C(G(Ω))?

• Proposition. No two planar rhombic paths (in the tiling of Ω ) can have more than one common rhomb. • This is not true in the 3D-space, or for general quadrilateral tiles.

• Proof.

• Corollary. Suppose the adjacent boundary edges i and j of Ω form an obtuse exterior angle. Then there can be no rhomb (i,j) in Ω. • If i and j form an acute exterior angle, nothing can be said about the existence of rhomb (i,j) in

the tiling of Ω.

So in general the convex and concave boundary points of Ω are not deducible from C.

• However, for benzenoids, we have this characterization of convex and concave boundary points:• Claim. Suppose i,j are adjacent boundary edges in Ω(B). Then they

form an acute exterior angle (convex point) if and only if there is a rhomb (i,j) in G(B). And this depends only on the cyclical code C(B):

• The alternating pattern (…,i,j,…,i,…,j,…) corresponds to an edge (i,j) in G(B), hence to a convex boundary vertex between i and j (this is true also for polyominoes!); the non-alternating pattern (…,i,j,…,j,…,i,…) means there is no edge (i,j) in G(B), hence it corresponds to a concave boundary vertex between i and j.

The shape of a benzenoid is uniquely determined by its cyclical matching sequence.

• Hence the cyclical sequence of convex (1) and concave (2) boundary vertices of Ω(B) can be easily obtained from our cyclical code C(B).• This sequences is one of the well-known boundary codes of benzenoids.

• Corollary. The shape of a benzenoid B is uniquely determined by C(B)!

Operations on circle graphs

The cyclical code C, existing for all circle graphs, enables us to define new operations on all graphs.

• Idea. Suppose we have defined an operation on cyclical matching sequences, producing a cyclical matching sequence C1C2 from the cyclical matching sequences C1and C2.• Then, given two graphs G1 and G2 with cyclical matching sequences C1

and C2, the graph G = G1G2 is defined as the unique circle graph satisfying the relation C(G1G2 ) = C1C2.

How can circle graphs be multiplied?

• We can rephrase this question:• Question. How can two cyclical matching sequences be combined into a third

one?

• One possible answer is the ‘‘zip‘‘ operation, defined as follows:• Choose ai in the first sequence and bj in the second.

• Now use the ‘‘zip‘‘ formula: (a1,a2,…,a2m)(b1,b2,…,b2n) = Pi,j(ai,bj,ai+1,bj+1,…,).

• This product clearly depends on the choice of ai and bj.

5. Conclusion(open problems)

• There are many interesting open problems, regarding • rhombic quadrangulations • circle graphs • cyclical codes of graphs…

We mention a few of them.

‘‘Flip graphs‘‘ of quadrangulations

• Flip graphs are known in the theory of triangulations: Two triangulations of a given region Ω are called adjacent if they are obtained from each other by ‘‘flipping‘‘ an edge. • The graph whose vertices are triangulations Ti(Ω ) and whose edges

corresond to adjacent triangulations is called the flip graph of Ω.

• Observation. A similar trick can be used for quadrangulations. Here we ‘‘flip‘‘ three rhombs inside a hexagon!

Flip graph of P8

is C8

How to obtain the flip graph of all rhombic quadrangulations of a given polygonal region Ω?

• This question can be rephrased as follows:

• Find the algorithm for constructing the flip graph F(Ω) of all possible rhombic quadrangulations of a polygonal region Ω(C), associated to a given circle diagram D(C), associated to the given cyclical code C.

What is the number of rhombic quadrangulations of a given polygonal region Ω?

• Let r(Ω) denote the number of such quadrangulations. • Question. Given Ω, how to compute r(Ω)?

• Example. Let P2n denote the regular 2n-gon. Then:

r(P4) = 1 r(P6) = 2 r(P8) = 8

• Question.What is the general formula?

r(P2n) = ?

Some more open problems…

• What is the ratio between circle graphs on n vertices and all graphs on n vertices?

• Find an effective algorithm for writing a graph in the form G = C or G = C1 – C2!

Thank you!