On X Free Covers

On X Free Covers

Khurram H.Sha?que and Ronald D.Dutton

School of Computer Science

University of Central Florida

Orlando,FL USA32816

khurram@https://www.doczj.com/doc/6c17739745.html,,dutton@https://www.doczj.com/doc/6c17739745.html,

Abstract

For an arbitrary graph G=(V,E),let X be a graphical property that can be possessed,or satis?ed by the subsets of V.For example,being a clique (maximal complete subgraph),a maximal independent set,an edge,a closed neighborhood,a minimal dominating set,etc.Let C X={A|A?V,|A|>1, and A possesses or satis?es property X}.A set S is an X cover(or X free)if A∩S(or A∩(V?S))is not empty for every A∈C X.Further,S is an X free cover when S is both X free and an X cover.

Many vertex-partitioning problems can be viewed as that of?nding an X free cover.In this paper,we present properties of X free covers and investi-gate the conditions for their existence in some special cases.This provides an underlying uni?cation theory for these and other similar properties. Keywords:vertex cover,edge cover,independent set,vertex-partition.

1Introduction

Vertex partitioning problems are widely studied in discrete optimization. Mainly,these problems involve splitting the vertex set into two or more sets such that each set satis?es a given criteria.Many of these problems are special cases of the classical Set Splitting problem.Given a?nite set T and a collection C of subsets of T,the problem of?nding a partition of T into two subsets T1and T2,such that no subset in C is entirely contained in either T1 or T2is called Set Splitting[5].

Let X denote an arbitrary property of a set of vertices in a graph G=(V,E). Examples of X include:is a clique(maximal complete subgraph),is a maximal independent set,is an edge,is a closed neighborhood,is a minimal dominating set,is acyclic,is a tree,is planar,etc.Further,assume that C X={A|A?V, |A|>1,and A possesses or satis?es property X}.Typical questions that arise

are:What are the minimum and maximum sets in C X for a given graph?, What are the upper and lower bounds on the cardinalities of these sets?,Can the vertex set of a graph be partitioned into two ore more sets in C X?,etc.

In this paper,we consider a special case of Set Splitting problem,where the ?nite set T is the vertex set V of a simple graph G=(V,E),and the collection C is de?ned by the subsets of V that satisfy some property X,i.e.,C X.The problem is then to partition the vertex set of a graph into two sets,such that, none of these sets contains a subset that satis?es property X.Such a partition is called an X free partition.A bipartition of the vertex set of a graph is an X partition if both of the sets in the partition satisfy property X.

A set S?V is an X free set,if for all A∈C X,A?S=?,i.e.,S does not contain any set having property X as a subset.For example,if X is the property of being an edge then an X free set(or edge free set)is an independent set.Similarly,if X is the property of being independent then an X free set is a clique.An X free set S is maximal if?v/∈S,?A?S such that A∪{v}∈C X.

A maximum X free set is a maximal X free set of largest cardinality.

A set S?V is an X cover if for all A∈C X,A∩S=?,i.e.,S contains at least one vertex from each set of G that possesses property X.For example,if X is the property of being an edge then an X cover set(or edge cover set)S is a set that contains at least one end vertex of every edge of G.Note that such a set is called vertex cover set or hitting set in the literature.An X cover set S is minimal if no proper subset of S has property X.A minimum X cover set is a minimal X cover of smallest cardinality.

The following theorem shows the duality between X covers and X free sets. Theorem1.S?V is an X cover if and only if V?S is X free.

Proof.A set S is an X free set if and only if,for every set A∈C X,A?S=?if and only if,for every set A∈C X,A∩(V?S)=?if and only if V?S is an X cover.

A set S is an X free cover if S is both X free and an X cover.Equivalently, S is an X free cover if for all A∈C X,A∩S=?and A∩(V?S)=?.Thus, we have the following:

Theorem2.A set S is an X free cover if and only if V?S is an X free cover.

Corollary3.A graph G has an X free partition if and only if G has an X free cover.

In general,the decision version of Splitting Set problem is NP-Complete. The case where all subsets in the collection C are of the same size k is also NP-Complete,for all k>2.The problem is equivalent to a variant of Satis?ability problem called Monotone NAESAT problem,where the clauses contain only

unnegated literals and the goal is to decide whether a truth assignment exists such that all clauses have at least one true literal and one false literal.

The case,where the collection C is de?ned by some graph property X has also been investigated for a number of cases of X.The case where X is the property of having a path of length greater than some?xed k is studied in[3]. In[6],it is shown that the problem of determining whether an arbitrary graph has a’cycle’free cover is NP Complete.Achlioptas[1]proved that if X is any ?xed graph G of order greater than2then the problem of deciding whether an arbitrary graph has an X free cover,is NP Complete.An additive induced hereditary graph family is a set of graphs that is closed under taking vertex disjoint unions and taking induced subgraphs.In a related work,Farrugia[4] showed that vertex partitioning into?xed additive induced hereditary graph families is NP Hard.

This work provides a uni?ed setting in which these and other similar con-cepts can be studied.We begin by exploring the existence and complexity of computing X free covers for some special cases of property X.

2Basic Properties

Let X and Y be any two properties that can be satis?ed or possessed by the subsets of V.

Proposition4.If X=?Y,i.e.,C X?C Y then every Y free cover is also an X free cover.

A property X is called additive if C X is closed under unions,whereas it is called hereditary if C X is closed under taking subsets.For example,the property‘is acyclic’is both an additive and a hereditary property.A property X is k?ary if it is only de?ned on the subsets of cardinality greater than or equal to k.For example,the property‘is a cycle’is a3-ary property in a simple graph.A k?ary property is hereditary if C X is closed under taking subsets of cardinality greater than or equal to k.

Theorem5.For an additive hereditary k?ary property X,G has an X free cover if and only if for all A∈C X,|A|<2k?1.

Proof.Let X be an additive hereditary k?ary property and let V=S∪T be an X free partition in a graph G.Assume to the contrary that there exists a set A∈C X such that|A|≥2k?1.Without loss of generality,it may be assumed that A∩S≥k.Since X is hereditary,A∩S∈C X,which contradicts S being X free.

Assume now that for all A∈C X,|A|<2k?1and that G does not have an X free cover.Let S be a minimal X cover.Since G does not have an X free cover,there must exist a set A?S such that A∈C X.By the minimality of

set S,for every vertex v∈S,there is a set B∈C X such that S∩B={v}. Then,by additivity of X,A∪B∈C X.Since|B∩A|≤1and|B|≥k and |A|≥k,we have that|A∪B|≥2k?1,a contradiction.

Let X be a k?ary property.A property X is the complement of X,if for

every set A?V of cardinality greater than or equal to k,A∈C X⊕C

X ,where

P⊕Q=(P\Q)∪(Q\P).

Proposition6.If a k?ary property X is hereditary then X is additive. Proposition7.If X is hereditary then G has an X free cover if and only if one of the following is true:

(i)|V|<2k?1,

(ii)there exists a set A∈C X,such that|A|≥|V|?k+1,or

(iii)the graph G has an X partition.

Theorem8.If X is a hereditary k?ary property and|V|>4k?3then G has an X free cover if and only if G does not have an X free cover.

Proof.Let X be a hereditary k?ary property and let G be a graph with |V|>4k?3.

=?Let A?V be an X free cover in the graph G.From Theorem2,it may

be assumed that|A|≥2k?1and thus A∈C

X .Assume to the contrary that

G has an X free cover P.Once again,from Theorem2,V?P is also an X

free cover.Since A∈C

X ,A∩P=?and A∩(V?P)=?.Assume,without

loss of generality,that|A∩P|≥k.Thus,P∈C X.Since X is a hereditary property,A∩P∈C X,which contradicts A being an X free cover.

?=Let A?V be an X free cover in the graph G.Once again,it may be assumed that|A|≥2k?1and thus A∈C X.Assume to the contrary that G has an X free cover S.Without loss of generality,it may be assumed that S∩A≥k.Since X is hereditary,S∩A∈C X,which contradicts S being X free.

A property X is superhereditary if whenever a set S has property X,so does every proper superset S ?S.For example,the property‘contains a triangle’is superhereditary.

Proposition9.A property X is superhereditary if and only if its complement X is hereditary.

Let Z be the property of being an X cover for some property X,i.e., C Z={A|A?V and A is an X cover}.By the de?nition of X free cover, we have that G has an X free cover if and only if G has a Z partition.

Theorem10.If Z is the property of being an X cover for some superheredi-tary property X then G has a Z free cover if and only if G has an X partition. Proof.Let X be a superhereditary property and Z be the property of being an X cover.Consider an X partition V=A∪B.By de?nition of X cover, every X cover S∈C Z must have nonempty intersections with both sets A and B,which implies that both A and B are Z free covers.

Now,assume that G has a Z free cover T.By Theorem2,U=V?T is also a Z free cover.We claim that{T,U}?C X.Assume to the contrary that

T/∈C X,i.e.,T∈C

X .Since X is superhereditary,X is hereditary,and hence,

T is X free.Thus,U is an X cover,i.e.,U∈C Z,which contradicts U being Z free.Hence T∈C

X

and similarly U∈C X.

3Special Cases

3.1When X=‘is an Edge’

If X is the property of being an edge then an edge free set is an independent set,and an edge cover is a set that contains at least one end vertex of every edge of the graph.The following well known result is immediate within this framework.

Proposition11.A graph has an edge free cover if and only if it is bipartite.

3.2When X=‘is Independent’

If X is the property of being independent then an independent free set is a clique.

Proposition12.A graph G has an’independent set’free cover if and only if its complement G is bipartite.

3.3When X=‘is Neighborhood’

The open neighborhood of a vertex v is the set N(v)={u|uv∈E},whereas closed neighborhood of v is the set N[v]=N(v)∪{v}.

If X is the property of being a closed neighborhood,then an X cover is a set S such that every vertex v∈V is either an element of S or is adjacent to an element of S,i.e.,S is a dominating set.By Theorem1,an X free cover is a dominating set S,such that V?S is also a dominating set.The following is a direct consequence of a theorem in[8].

Proposition13.If X is the property of being a closed neighborhood,then a graph G has an X free cover if and only if G does not have any isolated vertex.

If X is the property of being an open neighborhood,then an X cover is a set S such that every vertex v∈V is adjacent to an element of S,i.e.,S is a total dominating set[7].Once again,an X free cover is a total dominating set S,such that V?S is also a total dominating set.

De?ne a problem ONFC(OPEN NEIGHBORHOOD FREE COVER)to be the problem of deciding whether a given graph has an‘open neighborhood’free cover.We show that ONFC is NP-Complete by giving a polynomial transfor-mation from NAE3SAT problem,which is de?ned as follows:

INSTANCE NAE3SAT:a set U={u1,u2,...,u n}of variables and a col-lection C={C1,C2,...,C m}of clauses over U,where each clause contains exactly three literals(variables or their complements),with no literal appear-ing more than once in any given clause.

QUESTION:Is there a truth assignment that makes one or two(but not all three)literals true in each clause?

It may be assumed that each literal appears in at least one of the clauses, otherwise,for each literal u j that does not appear in any of the clauses,one can add another variable y and two clauses C 1={u j,u j,y}and C 2={u j,u j,y}. These two clauses are satis?ed by any truth assignment and do not a?ect the truth assignment of the original problem.

Theorem14.Given a graph G,the problem of deciding whether the graph G has an‘open neighborhood’free cover is NP-Complete.

Proof.Given an instance of NAE3SAT with n variables and k clauses,we transform it into an instance of ONFC by constructing a graph G=(V,E)as follows:

Let V=R∪S∪S∪T,where R={r i,1≤i≤m},S={s i,1≤i≤n}, S={s i,1≤i≤n},and T={t i,1≤i≤n}.For each variable u i∈U,create a P3=s i t i s i.For each clause C i∈C,create a vertex r i in V such that N(r i)={s j|u j∈C i}∪{s k|u k∈C i}.

The order of the constructed graph,|V|=3n+m and the size of the graph, |E|=3n+3m,which is polynomially related to the size of the NAE3SAT problem.

We claim that the constructed graph G has an‘open neighborhood’free cover if and only if the given instance of NAE3SAT has a satisfying truth assignment.The proof of the claim is as follows:

=?Suppose that the given instance of NAE3SAT has a satisfying truth as-signment f:U?→{0,1}.De?ne a partition of vertex set V=A∪B as follows:A=R∪{s i|f(u i)=1}∪{s i|f(u i)=0}and B=V?A.To show that ?v∈V,N(v)∩A=?and N(v)∩B=?,i.e.,A is an’open neighborhood’free cover,we consider three exhaustive cases.Case1:v∈R.Since f is a satisfying assignment,every clause C i contains a literal that is assigned the value1and a literal that is assigned the value0.Hence,for all v∈R,v is adjacent to at least one vertex in the set A and at least one vertex in the set

B.Case2:v∈S∪S.By assumption,each literal appears in at least one of the clauses.Hence,each vertex in set S∪S is adjacent to at least one vertex in R?A.Also,by construction,each vertex in set S∪S is adjacent to one vertex in T?B.Case3:v∈T.By construction,each v∈T is adjacent to a vertex s i∈S and s i∈S and thus has a neighbor in both sets A and B.

?=Suppose now that the constructed graph G has an’open neighborhood’free cover A,and let B=V?A.De?ne a truth assignment f:U?→{0,1}, such that f(u i)=1if and only if s i∈A.Since each vertex t i∈T is adjacent to only two vertices,s i∈S and s i∈S,exactly one of these vertices must be in set A.Thus,for each literal u i,f(u i)=1if and only if f(u i)=0,i.e.,f is a legal assignment.Also,each vertex r i∈R has at least one vertex in A and one vertex in B and hence each clause C i has at least one true literal and at least one false literal.Thus,f is a satisfying assignment.

3.4When X=‘is a dominating set’

Given a set S?V,a vertex v is called k?dominated by S,if it is dominated by(adjacent to)at least k vertices in S,i.e.,|N(v)∩S|≥k.If every vertex in V?S is k?dominated by S,then S is called a k?dominating set.A set T is called total k?dominating set if every vertex in V is k?dominated by T[7]. Note that,a1?dominating set is simply a dominating set,whereas a total 1?dominating set is a total dominating set.

If X is the property of being a k?dominating set,then a set S is X free if and only if there is a vertex v∈V?S,such that,|N(v)∩S|

Theorem15.If X is the property of being a k?dominating set then a graph G has an X free cover if and only if there exists{u,v}?V,such that, |N[u]∩N[v]|≤2k?2.

Proof.Let X be the property of being a k?dominating set.Set A is an X free cover if and only if both A and V?A are X free if and only if there is a vertex u∈V?A such that|N(u)∩A|

Corollary16.If X is the property of being a dominating set then a graph G has an X free cover if and only if the diameter of G is greater than2.

If X is the property of being a total k?dominating set,then a set S is X free if and only if there is a vertex v∈V such that|N(v)∩S|

Proof.Let X the property of being a total k?dominating set.Set A is an X free cover if and only if there are vertices u and v such that|N(u)∩A|

A set S is called distance k?dominating set if every vertex in V?S is within distance k of at least one vertex in S,i.e.,for all v∈V?S,d(v,S)≤k.

Theorem18.If X is the property of being a distance k?dominating set then a graph G has an X free cover if and only if the diameter of G is greater than 2k.

Proof.Let X be the property of being a total k?dominating set.

=?Consider a graph G and let u and v be any two vertices with distance d(u,v)>2k.Let A={w|w∈V∧d(u,w)≤k}.By construction,u∈A and v∈V?A.Since d(u,V?A)>k,V?A is X free.Also,since d(u,v)>2k, the distance of v from every vertex in A is at least k+1,i.e.,d(v,A)>k. Hence,A is also X free.

?=Let A be an X free cover in graph G.Then by the de?nitions of X and X free cover,there exist vertices u∈A and v∈V?A,such that d(u,V?A)>k and d(v,A)>k,which is possible only if d(u,v)>2k.

3.5When X=‘is an alliance’

A vertex v in a set A?V is said to be k?satis?ed with respect to A if |N(v)∩A|≥|N(v)∩(V?A)|+k,where?δ

B is a vertex partition such that each vertex has as many or more neighbors outside the set in which it occurs than inside it,then for all k>0,both A and B are defensive k?alliance free.Such a partition is called an unfriendly partition and one exists for every?nite graph[2].In addition,there exists a polynomial algorithm for?nding an unfriendly partition of a given graph.Thus,for all k>0,a defensive k?alliance free cover can be found in polynomial time.For k=0,we have shown the following in[10].

Theorem19.A connected graph G has an alliance free cover if and only if G has a block that is other than an odd clique or an odd cycle. References

[1]D.Achlioptas,“The complexity of G-free colourability,”Discrete Mathe-

matics,165-166,(1997),pp.2130.

[2]R.Aharoni, https://www.doczj.com/doc/6c17739745.html,ner,and K.Prikry,“Unfriendly partitions of a

graph,”Journal of Combinatorial Theory Series B,50(1990),pp1-10. [3]G.Chartrand,D.P.Geller,and S.Hedetniemi,“A generalization of the

chromatic number,”in Proc.Cambridge Philosophical Society,64(1968), pp.265-271.

[4]A.Farrugia,“Vertex partitioning into?xed additive induced hereditary

properties,”preprint,2004.

[5]M.R.Gary and D.S.Johnson,Computers and intractability,Freeman,

NY,1979.

[6]S.L.Hakimi and E.F.Schmeichel,“A note on the vertex arboricity of a

graph,”SIAM J.Discrete Mathematics,1(1989),pp.6467.

[7]T.W.Haynes,S.T.Hedetniemi,and P.J.Slater,Fundamentals of dom-

ination in graphs,Marcel Dekker,NY,1998.

[8]O.Ore,Theory of graphs,AMS Colloquium Publications38,American

Mathematical Society,Providence,RI,1962.

[9]K.H.Sha?que and R.D.Dutton,“Maximum alliance-free and minimum

alliance-cover sets,”Congressus Numerantium162(2003),pp.139-146.

[10]K.H.Sha?que and R.D.Dutton,“Partitioning a graph into alliance free

sets,”preprint,2004.

小学五年级英语听力和答案

小学五年级英语听力及答案 练习一 一、听录音，完成下列句子中所缺的单词或短语，每空一词。 1、A: _______do you ________ _________? B: _______ 7:20. 2、A: What do you do on the __________? B: I often _________ _________. _______I _______mountains. 3、A: Do you _______do ________ _________in the morning? B: Yes, I do. 4、A: _______ _______ do you like ______? B: ________. 5、A: ______ do you like _______? B: ______ I can ________. 6、A: What can you do in _______? B: I can ______ _______. 7、A: What’s your ________ ________? B: I like _________. 8、A: Is your ________ birthday in ________? B: Yes, ______ ______. A: ________ the _______? B: It’s __________ _______. 9、A: _______is Christmas _______? B: It’s ___________ ________. 10、A: How many __________ are there in _________? B: There are _________. 二、听问句，选答语。 ( ) 1、A. At 6:30 B. In 6:30 C. On 6:30 ( ) 2、A. I like Monday. B. I like spring. C. I like January. ( ) 3、A. I often climb mountains. B. I am a student. C. I can plant trees. ( ) 4、A. It’s warm. B. It’s cold. C. It’s hot. ( ) 5、A. I can skate. B. I often skate. C. I often swim. ( ) 6、A. Yes, it is. B. No, she isn’t. C. Yes, she is. ( ) 7、A. Because I can swim. B. Because I can plant flowers. C.Because I can skate. ( ) 8、A. There are twelve. B. There are seven. C. There are three. ( ) 9、A. I go to school at 7:00. B. I go home at 5:00. C. I go to work at 8:00. ( ) 10、A. It’s in spring. B. It’s in January. C. It’s June. 练习二 一、你将听到一个句子，根据听到的内容选择符合的一项。

(完整word版)多功能办公楼智能照明控制系统方案

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