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COMBINATORIAL CONFIGURATIONS. Incidence structure. An incidence structure C is a triple C = (P, L ,I) where P is the set of points, L is the set of blocks or lines I P L is an inciden ce relation . Elements from I are called flags .
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Incidence structure • An incidence structureCis a triple • C = (P,L,I) where • P is the set ofpoints, • Lis the set of blocks or lines • I P Lis anincidence relation. • Elements from I are calledflags. • The bipartite incidence graph G(C) with black vertices P, white verticesLand edges I is known as theLevi graph of the structure C.
(Combinatorial) Configuration • A (vr,bk) configuration is an incidence structure C = (P,L,I) of points and lines,such that • v = |P| • b = |L| • Each point lies on r lines. • Each line contains k points. • Two lines intersect in at most one point. • Warning: Levi graph is semiregular of girth 6
Symmetric configurations • A (vr,bk) configuration is symmetric, if • v = b (this is equivalent to r = k). • A (vk,vk) configuration is usually denoted by (vk) configuration.
Motivation 5 Example 1: Fano configuration (smallest projective plane). 6 7 3 1 4 2
Exercises • N1: Determina all lines passing through the point 3 of Fano plane. • N2: How many lines pass through 4 and 6? • N3: How many lines pass through 1, 2, 3? • N4*: Is it possible to draw Fano plane with straight lines? • N5*: Describe various properties of Fano plane. For example: Each number from 1 to 7 appears exactly three times in the table.
Another Example • The figure on the left has 4 points 1,2,3,4 and four circles a,b,c,d. Combinatorial description is given by the following configuration table. 1 2 4 3
Another Example - Continuation • Configuration table admits a “graphical” description.
Another Example - Exercises B • N1: Determine the configuration table for the vertices and faces of the tetrahedron. • N2: Determine the configuration table for the vertices and edges of the tetrahedron. • N3: Explore the relationship between these structures.. 4 3 1 A D 2 C
Small Configurations • Triangle is the only (32)configuration. • Pasch configuration (62,43) and its dualcomplete quadrangle (43,62) share the same Levi graph S(K4).
Miquel “Configuration” • Configuration constists of 4 points and 6 circles (see figure on the left.) It is called the Miquel configuration. [Note that two lines are considered as “degenerate circles”] • Configuration results from Miquel theorem.
Miquel Configuration - Exercises • N1: Show that the configuration on the left is combinatorially equivalent of the Miquel configuration. • N2: How many points and curves does the configuration on the left have?
Examples • 1. Each graph G = (V,E) is an incidence structure: P = V, L = E, (x,e) 2 I if and only if x is an endvertex of e. • 2. Any family of sets FµP(X) is an incidence structure. P = X, L = F, I = 2. • 3. A line arrangement L = {l1, l2, ..., ln} consisting of a finite number of n distinct lines in Euclidean plane E2 defines an incidence structure. Let V denote the set of points from E2 that are contained in at least two lines from L. Then: P = V, L = L and I is the point-line incidence in E2.
Exercises • N1. Draw the Levi graph of the incidence structure defined by the complete bipartite graph K3,3. • N2. Draw the Levi graph of the incidence structure defined by the powerset P({a,b,c}). • N3. Determine the Levi graph of the incidence structure, defined by an arrangemnet of three lines forming a triangle in E2.
Example of Rank 2 Geometry • Graph H on the left is known as the Heawood graph. • H is connected • H is trivalent: d(H) = D(H) = 3. • H je bipartite. • H is a Pasini geometry.
Another View • Geometry of the Heawood graph H has another interpretation. • Rank = 2. There are two types of objects in Euclidean plane, say, points and curves. • There are 7 points, 7 curves, 3 points on a curve, 3 curves through a points. • The corresponding Levi graph is H!
In other words ... • The Heawood graph (with a given black and white coloring) is the same thing as the Fano plane (73), the smallest finite projective plane. • Any incidence geometry can be interpeted in terms of abstract points, lines. • If we want to distinguish geometry (geometric interpretation) from the associated graph (combinatorial description) we refer to the latter the Levi graph of the corresponding geometry.
Predsedniške volitve 1997 in Möbius-Kantorjeva konfiguracija • 8 kandidatov: Bernik, Cerar, Kovač, Kučan, Miklavčič, Peršak, Podobnik, Poljšak • Soočenja na TV Slovenija: 8 večerov, vsak večer po trije kandidati
Predsedniške volitve 1997 Nekega večera v studiu Čez dolgih sedem dni
Vprašanje • Ali lahko razvrstimo kandidate v 8 trojic, tako da se nobena 2 ne bosta srečala več kot enkrat?
Möbius - Kantorjeva konfiguracija 124 edina (83) konfiguracija 235 6 346 457 568 7 3 671 782 8 5 813 1 2 4 2 3 8 9 6 4 7 5 1
Levijev graf Möbius - Kantorjeve konfiguracije drugačen pogled na rešitev 813 1 8 782 124 7 2 671 235 6 3 568 346 4 5 457
Nekoliko spremenjeni pogoji • Denimo, da bi se na predsedniške volitve letos prijavilo 9 kandidatov. • Ali jih lahko razvrstimo v 9 trojic, tako da se nobena 2 ne srečata več kot enkrat?
Vse (93) konfiguracije 2 3 8 9 6 4 7 5 1 2 3 8 9 6 4 7 5 1
Problem sadovnjaka Na vrtu v vrstah po tri devet se dreves zasadi. Glavni problem je le v tem, da naj se deset vrst dobi!
Pappusova konfiguracija rešitev problema sadovnjaka
1 2 3 4 5 6 7 8 9 Pappusova konfiguracija se edina lahko vloži v afino ravnino reda 3 Omogoča razvrstitev 12 kandidatov v 9 večerov
Pappusova konfiguracija • Pappusova (93) konfiguracija sestoji iz devetih točk in devetih premic. Točkam lahko pripišemo homogene koordinate (a,b,c), pa tudi premicam lahko pripišemo homogene koordinate [p,q,r] pri čemer incidenco določa zveza ap+bq+cr=0. • Ta primer lahko interpretiramo kot zgled ortogonalne reprezentacije grafov, pri katerih u~v implicira r(u) ^r(v).
Pappus Configuration 1 2 3 2’’ 3’’ 1’’ 2’ 3’ 1’
Exercises • N1: How many points and lines has Pappus configuration? • N2: How many lines pass through the same point of Pappus configuration? • N3: What is the maximum number of lines common to the same pair of points? • N4: Write down the configuration table for Pappus configuration. • N5*: Plant 9 in 10 lines, in such a way that each line contains exactly 3 trees!
Problem: Find a Model! • We know that G(10,3) embeds in double torus. • Is there a nice model out there? • Here are the faces: • 7, 18, 19, 10, 11, 12, 13, 8 • 6, 15, 14, 13, 12, 17, 18, 7 • 5, 6, 7, 8, 9, 10, 19, 20 • 2, 3, 14, 15, 16, 17, 12, 11 • 1, 2, 11, 10, 9, 4, 5, 20 • 1, 16, 15, 6, 5, 4, 3, 2 • 1, 20, 19, 18, 17, 16 • 3, 4, 9, 8, 13, 14
Heawood Graph in Torus • On the left there is a hexagonal embedding of the Heawood graph in torus.
Möbius-Kantor Graph in Double Torus • Möbius-Kantor graph in double torus
Möbius-Kantor Graph in Double Torus • Möbius-Kantor graph in double torus . • The embedding is octagonal. • The map is regular.
Almost Finished Medium Model • Vertices: black • Edges: red • Faces: yellow.