Using latin squares Using latin squares to color split graphs to - PowerPoint PPT Presentation

Using latin squares Using latin squares to color split graphs to color split graphs Sheila Morais de Almeida (State University of Campinas) Sheila Morais de Almeida (State University of Campinas) Cé élia Picinin de Mello lia Picinin de Mello (State University of Campinas) C (State University of Campinas) Aurora Morgana (University of Rome Aurora Morgana (University of Rome “ “La Sapienza La Sapienza” ”) ) May - 2008

Outline Outline � Edge Edge- -Coloring Coloring � � The Classification Problem The Classification Problem � � Split Graphs Split Graphs � � Overfull Overfull Graphs Graphs � � The Edge The Edge- -Coloring Coloring Conjecture Conjecture for Split Graphs for Split Graphs � � Latin Squares Latin Squares � � The Edge The Edge- -Coloring of Split Graphs Using Latin Coloring of Split Graphs Using Latin � Squares Squares

Edge- -Coloring Coloring Edge A k k- -edge edge- -coloring coloring of a graph G is an assignment of k colors to the edges of G such that any two edges incident in a common vertex have distinct colors.

The Edge- -Coloring Problem Coloring Problem The Edge The minimum k required to perform a k-edge-coloring of a simple graph G is called chromatic chromatic index of G and is index denoted by χ ’(G) . χ ’(Bull)=3

The Edge- -Coloring Problem Coloring Problem The Edge χ ’ ≥ ∆ ∆ (G), χ ’(G) (G) ≥ (G), where ∆ (G) is the maximum degree of a graph G. χ ’(Bull)=3

The Edge- -Coloring Problem Coloring Problem The Edge There are graphs that have χ ’(G) > ∆ (G). In 1964, Vizing showed that every simple graph G has χ ’(G) ≤ ∆ (G)+1. ∆ (C 3 )=2 χ ’(C 3 ) = 3

The Edge- -Coloring Problem Coloring Problem The Edge A direct result of the Vizing’s Theorem is that any simple graph G has ∆ (G) ≤ χ ’(G) ≤ ∆ (G)+1 . Vizing restricted the Edge-Coloring Problem to the following problem: Given a simple graph G, is the chromatic index of G equal to ∆ (G) or ∆ (G)+1?

The Classification Problem The Classification Problem Given a simple graph G, is the chromatic index of G equal to ∆ (G) or ∆ (G)+1? This problem is known as the Classification Classification Problem . Problem

The Classification Problem The Classification Problem If χ ’(G) = ∆ (G), then G is Class 1. Otherwise, χ ’(G) = ∆ (G) + 1 and G is Class 2. Classe 1 Classe 2

The Classification Problem The Classification Problem In 1981, Holyer showed that the problem of deciding if a simple graph G is Class 1 is NP-Complete.

The Classification Problem The Classification Problem Even so, there are efficient algorithms to solve this problem when we are restricted to some classes of graphs. � bipartite graphs are Class 1. � a complete graph is Class 1 iff it has an even number of vertices. � a cycle (without chords) is Class 1 iff it has an even number of vertices. � split graphs with odd maximum degree are Class 1.

Split Graphs Split Graphs A graph G is a split graph if split graph the set of Q vertices of G admits a partition [Q, S], where Q is a clique and S is S a stable set.

Overfull Graphs Graphs Overfull Consider a graph G, with n vertices and m edges. The graph G is overfull if n is overfull odd and m > ∆ (G)  n/2  . m = 9 > ∆ (G)  n/2  = = 4  5/2  = 8

Subgraph- -Overfull Overfull Graphs Graphs Subgraph If G has a subgraph which is overfull and has maximum degree equal to ∆ (G), then G is subgraph- -overfull overfull . subgraph

Subgraph- -Overfull Overfull Graphs Graphs Subgraph If G has a subgraph which is overfull and has maximum degree equal to ∆ (G), then G is subgraph- -overfull overfull . subgraph m = 9 > ∆ (G)  n/2  = = 4  5/2  = 8

Neighborhood- -Overfull Overfull Neighborhood Graphs Graphs If G has an overfull subgraph induced by a vertex with degree ∆ (G) and its neighbors, then G is neighborhood neighborhood- - overfull . overfull m = 9 > ∆ (G)  n/2  = = 4  5/2  = 8

Subgraph- -Overfull Overfull Graphs Graphs Subgraph Overfull graphs graphs and and neighborhood neighborhood- -overfull overfull Overfull graphs are subgraph- -overfull graphs. overfull graphs. graphs are subgraph Every subgraph subgraph- -overfull overfull graph is graph is Class Class 2. 2. Every Class 2 Subgraph-overfull Neighborhood- overfull Overfull

Edge- -Coloring Coloring Conjecture Conjecture for Split for Split Edge Graphs Graphs Figueiredo, Meidanis Meidanis and and Mello show Mello show that that Figueiredo, every subgraph subgraph- -overfull overfull split graph is split graph is every neighborhood- -overfull overfull. . neighborhood They present present the the following following conjecture conjecture: : They A split graph G is Class Class 2 if, 2 if, and and only only if, G if, G A split graph G is is neighborhood neighborhood- -overfull overfull. . is

Graphs with with Universal Universal Vertices Vertices Graphs Planthold presents presents the the following following theorem theorem: : Planthold Every simple graph G simple graph G containing containing a a Every universal vertex vertex is is Class Class 2 2 iff iff G is G is universal subgraph- -overfull overfull. . subgraph K 5 minus one edge is overfull � K 5 is subgraph-overfull

Split Graphs Split Graphs In 1995, Chen, Fu, and Ko showed that split graphs with odd ∆ (G) are Class 1. ∆ (G)=7

Split Graphs Split Graphs Every split graph G with partition [Q, S] has a bipartite Q subgraph induced by the edges with a vertex in Q and another vertex S in S.

Split Graphs Split Graphs Considering the bipartite subgraph of G, we denote: Q d(Q) = max{d(v), v ∈ Q} and S d(S) = max{d(v), v ∈ S}. d(Q) = 2 d(S) = 3

Split Graphs Split Graphs Consider the partition [Q, S] of a split graph, where Q is a maximal clique. Q Chen, Fu and Ko also showed that every split graph with d(Q) ≥ d(S) is Class 1. S d(Q) = 3 d(S) = 1

Edge- -Coloring of Split Graphs Coloring of Split Graphs Edge � Split Graphs that are neighborhood-overfull are Class 2. � Split Graphs with odd maximum degree are Class 1. � Split Graphs with even maximum degree that are not neighborhood-overfull and contain a universal vertex are Class 1. � Split-Graphs with partition [Q, S], where Q is a maximal clique and such that d(Q) ≥ d(S) are Class 1. How about split graphs with even maximum degree and d(S) > d(Q), that are not neighborhood-overfull and do not contain universal vertices? Are these graphs Class 1?

Latin Square Square Latin A latin latin square square of of order order k k is is A 0 1 2 3 4 � a a k k x k- -matrix matrix x k � 1 2 3 4 0 � filled filled with with entries entries from from � {0, 1, ..., k- -1} 1} {0, 1, ..., k 2 3 4 0 1 � each each element element appears appears � 3 4 0 1 2 exactly once once in in each each row row exactly � each each element element appears appears 4 0 1 2 3 � exactly once once in in each each column column. . exactly

Commutative Latin Latin Square Square Commutative A latin square square M=[m M=[m i,j ] is commutative commutative if if A latin i,j ] is m i,j = m m j for 0 ≤ ≤ i,j i,j ≤ ≤ k k- -1. 1. m i,j = ,i for 0 j,i 0 1 2 3 4 0 1 2 3 4 1 2 3 4 0 4 0 1 2 3 2 3 4 0 1 3 4 0 1 2 3 4 0 1 2 2 3 4 0 1 4 0 1 2 3 1 2 3 4 0

Commutative Latin Latin Square Square Commutative A latin square square M=[m M=[m i,j ] is commutative commutative if if A latin i,j ] is m i,j = m m j for 0 ≤ ≤ i,j i,j ≤ ≤ k k- -1. 1. m i,j = ,i for 0 j,i 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 1 2 3 4 0 4 0 1 2 3 4 0 1 2 3 2 3 4 0 1 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 2 3 4 0 1 2 3 4 0 1 4 0 1 2 3 1 2 3 4 0 1 2 3 4 0

Commutative Latin Latin Square Square Commutative A latin square square M=[m M=[m i,j ] is commutative commutative if if A latin i,j ] is m i,j = m m j for 0 ≤ ≤ i,j i,j ≤ ≤ k k- -1. 1. m i,j = ,i for 0 j,i 0 1 2 3 4 0 1 2 3 4 1 2 3 4 0 4 0 1 2 3 2 3 4 0 1 3 4 0 1 2 3 4 0 1 2 2 3 4 0 1 4 0 1 2 3 1 2 3 4 0

Commutative Latin Latin Square Square Commutative A latin square square M=[m M=[m i,j ] is commutative commutative if if A latin i,j ] is m i,j = m m j for 0 ≤ ≤ i,j i,j ≤ ≤ k k- -1. 1. m i,j = ,i for 0 j,i 0 1 2 3 4 0 1 2 3 4 1 2 3 4 0 4 0 1 2 3 2 3 4 0 1 3 4 0 1 2 3 4 0 1 2 2 3 4 0 1 4 0 1 2 3 1 2 3 4 0

Idempotente Latin Square Square Idempotente Latin A latin square square M=[m M=[m i,j ] is idempotente idempotente if if A latin i,j ] is m i,i = i for 0 ≤ ≤ i i ≤ ≤ k k- -1. 1. m i,i = i for 0 0 3 1 4 2 0 1 2 3 4 3 1 4 2 0 1 2 3 4 0 1 4 2 0 3 2 3 4 0 1 4 2 0 3 1 3 4 0 1 2 2 0 3 1 4 4 0 1 2 3