Connected Graphs and Distance - BUKU GRAPHS & DIGRAPH PDF

30 CHAPTER 1. INTRODUCTION TO GRAPHS 10. For which integersx(0≤x≤7), if any, is the sequence 7,6,5,4,3,2,1, x

graphical?

11. Use Theorem 1.13 to determine whether the sequences : 6,6,5,4,3,2,2 is graphical.

12. We have seen that there is only one graphical sequence d₁, d₂, d₃, d₄, d₅ with 4 =d₁≥d₂≥d₃≥d₄≥d₅= 1 such that at least one term is 3 and at least one term is 2. How many graphical sequencesd₁, d₂, d₃, d₄, d₅are there with 4 =d₁ ≥d₂≥d₃≥d₄ ≥d₅ = 2 such that at least one term is 3?

13. Show that for every finite setSof positive integers, there exists a positive integerk such that the sequence obtained by listing each element ofS a total ofktimes is graphical. Find the minimum such kforS={2,6,7}. 14. According to Theorem 1.15, for each integern≥2, there exist exactly two distinct graphical sequences of length nhaving exactly two equal terms.

What terms are equal for these two sequences?

15. Two finite sequencess₁ands₂of nonnegative integers are calledbigraph- ical if there exists a bipartite graphGwith partite sets V₁ andV₂ such that si lists the degrees of the vertices ofGinVi fori= 1,2. Prove that the sequencess₁:a₁, a₂, . . . , ar ands₂:b₁, b₂, . . . , btof nonnegative integers with r≥2,a₁ ≥a₂ ≥ · · · ≥ar, b₁≥b₂ ≥ · · · ≥bt, 0< a₁ ≤t and 0 < b₁≤rare bigraphical if and only if the sequences s^′₁ :a₂, a₃,· · ·, ar

ands^′₂:b₁−1, b₂−1, . . . , ba1−1, ba1+1, . . . , bt are bigraphical.

Walks, Trails and Paths

For two (not necessarily distinct) vertices u and v in a graph G, a u−v walk W in G is a sequence of vertices in G, beginning with uand ending at v such that consecutive vertices inW are adjacent inG. Such a walkW inG can be expressed as

W = (u=v₀, v₁, . . . , vk=v), (1.2) where vivi+1 ∈ E(G) for 0 ≤ i ≤ k−1. (The walk W is also commonly denoted by W : u= v₀, v₁, . . . , vk = v.) Nonconsecutive vertices in W need not be distinct. The walk W is said to contain each vertex vi (0 ≤ i ≤ k) and each edge vivi+1 (0 ≤i≤k−1). The walk W can therefore be thought of as beginning at the vertex u = v₀, proceeding along the edge v₀v₁ to the vertex v₁, then along the edge v₁v₂ to the vertexv₂, and so forth, until finally arriving at the vertexv=vk. The number of edges encountered inW (including multiplicities) is the lengthofW. Hence the length of the walkW in (1.2) is k. In the graphGof Figure 1.34,

W₁= (x, w, y, w, v, u, w) (1.3) is an x−wwalk of length 6. This walk encounters the vertex w three times and the edgewy twice.

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u v

x y

G: w

Figure 1.34: Walks in a graph

A walk whose initial and terminal vertices are distinct is an open walk;

otherwise, it is a closed walk. Thus the walk W₁ in (1.3) in the graph G of Figure 1.34 is an open walk. It is possible for a walk to consist of a single vertex, in which case it is atrivial walk. A trivial walk is therefore a closed walk.

A walk in a graph G in which no edge is repeated is a trail in G. For example, in the graph G of Figure 1.34, T = (u, v, y, w, v) is a u−v trail of length 4. While no edge of T is repeated, the vertex v is repeated, which is allowed. On the other hand, a walk in a graphGin which no vertex is repeated is called a path. Every nontrivial path is necessarily an open walk. Thus P^′= (u, v, w, y) isu−ypath of length 3 in the graphGof Figure 1.34. Many proofs in graph theory make use of u−v walks or u−v paths of minimum length (or of maximum length) for some pair u, v of vertices of a graph. The proof of the following theorem illustrates this.

32 CHAPTER 1. INTRODUCTION TO GRAPHS Theorem 1.16 Letuandvbe two vertices of a graphG. For everyu−vwalk W inG, there exists au−v pathP such that every edge of P belongs toW. Proof. Among allu−v walks inG, let

P = (u=u₀, u₁, . . . , uk=v)

be au−v walk of minimum length for which every edge of P belongs to W. Thus the length of P is k. We claim that P is a u−v path. Assume, to the contrary, that this is not the case. Then some vertex of G must be repeated in P, sayui=uj for somei andj with 0≤i < j ≤k. If we then delete the verticesui+1, ui+2, . . . , uj from P, we arrive at the u−vwalk

W^′= (u=u₀, u₁, . . . , ui−1, ui=uj, uj+1, . . . , uk=v)

whose length is less thankand such that every edge ofW^′ belongs toW, which is impossible.

The Adjacency Matrix of a Graph

We have seen that a graph can be defined or described by means of sets (the definition) or diagrams. As Example 1.3 suggests, there are also matrix representations of graphs. Suppose thatGis a graph of ordern, whereV(G) = {v₁, v₂, . . . , vn}. The adjacency matrix of G is the n×n zero-one matrix A(G) = [aij], or simply A= [aij], where

aij =

1 ifvivj∈E(G) 0 ifvivj∈/E(G).

Figure 1.35 shows the adjacency matrix of a graphG.

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...

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0 1 1

0 0

v₃ v₄

v₅ G:

0 0

v₂ v₁

0 1

1 1 0

1 0 1 0

Figure 1.35: A graph and its adjacency matrix

There are several observations that can be made about the adjacency matrix Aof a graphGof ordern. First, all entries along the main diagonal ofAare 0 since no vertex ofGis adjacent to itself. Second,Ais a symmetric matrix, that is, rowi ofA is identical to columni of Afor every integeri with 1≤i ≤n. Also, if we were to add the entries in rowi(or in columni), then we obtain the degree ofvi.

Wheneveraij = 1, this means thatGcontains the edge vivj and therefore avi−vj path of length 1 and, of course, avi−vj walk of length 1 as well. Not

only can the adjacency matrix ofG be used to identify whetherGcontains a vi−vjwalk of length 1, it can be used to determine whetherGcontains avi−vj

walk of lengthkfor an arbitrary positive integerkand, in fact, the number of such walks. Before stating a theorem that provides us with this information, we need to know when twou−v walks in a graph Gare considered to be the same.

Twou−vwalksW = (u=u₀, u₁, . . . , uk =v) andW^′= (u=v₀, v₁, . . . , vℓ= v) in a graph areequalifk=ℓ andui=vi for alliwith 0≤i≤k.

Theorem 1.17 Let Gbe a graph with vertex set V(G) ={v₁, v₂, . . . , vn}and adjacency matrixA. For each positive integerk, the number of differentvi−vj

walks of lengthk inGis the (i, j)-entry in the matrixA^k.

Proof. Leta⁽ij^k⁾denote the (i, j)-entry in the matrixA^k for a positive integer k. Thus, A¹ =Aand a⁽¹⁾_ij =aij. We proceed by induction onk. For vertices vi and vj of G, there can be only one vi−vj walk of length 1 or no vi−vj

walks of length 1, and this occurs ifaij= 1 oraij = 0, respectively. Therefore, the (i, j)-entry of the matrixAis the number ofvi−vj walks of length 1 inG.

Thus the basis step of the induction is established.

We now verify the inductive step. Assume, for a positive integerk, thata⁽ij^k⁾

is the number of different vi−vj walks of length k in G. We show that the (i, j)-entrya⁽_ij^k⁺¹⁾ inA^k⁺¹ gives the number of differentvi−vj walks of length k+ 1 inG. First, observe that everyvi−vj walk of lengthk+ 1 inGis obtained from avi−vtwalk of lengthk for some vertexvt inGthat is adjacent tovj.

SinceA^k⁺¹ =A^k·A, it follows that the (i, j)-entrya⁽ij^k⁺¹⁾ in A^k⁺¹ can be obtained by taking the inner product of rowiofA^k and columnj ofA. That is,

a⁽ij^k⁺¹⁾=a⁽i^k1⁾a₁j+a⁽i^k2⁾a₂j+. . .+a⁽in^k⁾anj =

t=1

a⁽it^k⁾atj. (1.4) By the induction hypothesis, for each integertwith 1≤t≤n, the integera⁽_it^k⁾ is the number of different vi−vt walks of length k in G. Ifatj = 1, then vt

is adjacent to vj and so there are a⁽it^k⁾ different vi−vj walks of length k+ 1 in Gwhose next-to-last vertex isvt. On the other hand, ifatj = 0, thenvt is not adjacent to vj and there are no vi−vj walks of length k+ 1 in Gwhose next-to-last vertex is vt. In any case, a⁽it^k⁾·atj gives the number of different vi−vjwalks of lengthk+ 1 inGwhose next-to-last vertex isvt. Consequently, the total number of different vi−vj walks of length k+ 1 inGis the sum in (1.4), which isa⁽_ij^k⁺¹⁾.

By the Principle of Mathematical Induction,a⁽_ij^k⁾is the number of different vi−vj walks of length kinGfor every positive integerk.

Before giving an example to illustrate Theorem 1.17, we make a few observations. Let G be a graph of order n with V(G) = {v₁, v₂, . . . , vn} and

34 CHAPTER 1. INTRODUCTION TO GRAPHS A^k = h

a⁽_ij^k⁾i

, where A is the adjacency matrix ofG. By Theorem 1.17, a⁽²⁾_ii gives the number of different vi−vi walks of length 2 in G. Since a vi−vi

walk of length 2 is (vⁱ, vt, vi) for some vertex vt adjacent tovi, it follows that a⁽²⁾_ii = degvi for every vertexvi ofG. Fori6=j,a⁽²⁾_ij is the number of different vi−vj pathsof length 2 inG. Again, by Theorem 1.17,a⁽³⁾ii gives the number of differentvi−vi walks of length 3 in G. Since a vi−vi walk of length 3 is (vⁱ, vs, vt, vi) for adjacent vertices vsand vt, each of which is adjacent to vi, it follows that vi must belong to a triangle. Not only is (vⁱ, vs, vt, vi) a vi−vi

walk of length 3, so too is (vⁱ, vt, vs, vi) a (different)vi−vi walk of length 3 in G. Therefore,a⁽³⁾_ii is twice the number of triangles inGthat containvi.

As an illustration, consider the graphGof Figure 1.36 having the adjacency matrixA. We can computeA²without matrix multiplication by observing that the (i, i) entry of A², 1≤i≤4, is degvi, and the (i, j) entry ofA², i 6=j, is the number of differentvi−vj paths of length 2. We now turn to A³. Since the differentv₁−v₃ walks of length 3 inGare

W₁= (v₁, v₃, v₁, v₃),W₂= (v₁, v₂, v₁, v₃), W₃= (v₁, v₃, v₂, v₃),W₄= (v₁, v₃, v₄, v₃),

the (1,3) entry of A³ is 4. The entire matrix A³ can be computed in this manner.

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.v3

v1 v2







0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 0





 A²=







2 1 1 1 1 2 1 1 1 1 3 0 1 1 0 1





 A³=







2 3 4 1 3 2 4 1 4 4 2 3 1 1 3 0







Figure 1.36: A graphGand powers of its adjacency matrix

Circuits and Cycles

A nontrivial closed walk in a graph G in which no edge is repeated is a circuitin G. For example,

C= (u, w, x, y, w, v, u)

is a circuit in the graphGof Figure 1.34. In addition to the required repetition of uin this circuit,w is repeated as well. This is acceptable provided no edge is repeated. A circuit

C= (v=v₀, v₁, . . . , vk =v),

k ≥ 2, for which the vertices vi, 0≤ i ≤k−1, are distinct is a cycle in G. Therefore,

C^′= (u, v, y, x, w, u)

is a cycle of length 5 in the graphGof Figure 1.34. As in the case of the graph Ck, a cycle of lengthk≥3 is called ak-cycle. A 3-cycle is also referred to as a triangle. A cycle of even length is aneven cycle, while a cycle of odd length is anodd cycle.

The subgraph induced by the edges in a path (v₁, v₂,· · ·, vk) or a cycle (v₁, v₂,· · ·, vk, v₁),k≥3, is itself called apathorcycle. Consequently, paths and cycles have two interpretations in graphs – as sequences of vertices and as subgraphs. This is the case with trails and circuits as well. The path P^′ and the cycleC^′ described earlier in the graphGof Figure 1.34 correspond to the subgraphs shown in Figure 1.37. We have seen that a graph of ordern that is a path or a cycle is denoted byPn andCn, respectively.

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x G: w

u v

y w

x y

v u

P^′ : C^′ : w

Figure 1.37: A path and cycle in a graph

The length of a smallest cycle in a graphG(containing cycles) is the girth ofG, denoted byg(G). Thus g(Kⁿ) = 3 forn≥3 andg(K^s,t) = 4 fors, t≥2, while the girth of the Petersen graph is 5.

Connected Graphs

Two verticesu and v in a graphG are connected if G contains a u−v path. The graphGitself isconnectedif every two vertices ofGare connected.

By Theorem 1.16, a graphGis connected if Gcontains au−v walk for every two verticesuandv ofG. A graphGthat is not connected is adisconnected graph. The graphF of Figure 1.38 is connected sinceF contains au−vpath (and au−v walk) for every two verticesuandv inF. On the other hand, the graphH is disconnected since, for example,H contains no y₄−y₅ path.

A connected subgraphH of a graphG is a component of Gif H is not a proper subgraph of any connected subgraph of G. Thus every component of G is an induced subgraph ofG. The number of components in a graphG is denoted by k(G). Therefore, G is connected if and only if k(G) = 1. For

36 CHAPTER 1. INTRODUCTION TO GRAPHS

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x₂

x₄ x₆

x₅

y₅ y₆ y₇

x₁ y₁ y₂

y₃ y₄ x₃

F : H :

Figure 1.38: A connected graph and disconnected graph

the setsS₁={y₁, y₂, y₃, y₄} andS₂={y₅, y₆, y₇} of vertices of the graphH of Figure 1.38, the induced subgraphsH[S₁] andH[S₂] are (the only) components ofH. Therefore,k(H) = 2.

Ifuandvare distinct vertices in a connected graphG, then there is au−v path in G. In fact, there may very well be several u−v paths in G, possibly of varying lengths. This information can be used to provide a measure of how closeuandv are to each other or how far from each other they are. The most common definition of distance between two vertices in a connected graph is the following.

Distance in Graphs

ThedistancedG(u, v) from a vertexuto a vertexv in a connected graph G, or simplyd(u, v) if the graphGis clear, is the minimum of the lengths of the u−v paths inG. Au−vpath of lengthd(u, v) is called au−vgeodesic. In the graphGof Figure 1.39, the pathP = (v₁, v₅, v₆, v₁₀) is a v₁−v₁₀geodesic and sod(v₁, v₁₀) = 3. Furthermore,

d(v₁, v₁) = 0,d(v₁, v₂) = 1,d(v₁, v₆) = 2,d(v₁, v₇) = 3 andd(v₁, v₈) = 4.

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v₁ v₂ v₃

v₈

v₆ v₇

G: v₅

v₉ v₁₀

v₄

Figure 1.39: Distances in a graph

The distanced defined above satisfies each of the following properties in a connected graphG:

(1) d(u, v)≥0 for every two verticesuand vofG;

(2) d(u, v) = 0 if and only ifu=v;

(3) d(u, v) =d(v, u) for allu, v∈V(G) (thesymmetric property);

(4) d(u, w)≤d(u, v) +d(v, w) for allu, v, w∈V(G) (the triangle inequality).

Sincedsatisfies the four properties (1)-(4),dis ametriconV(G) and (V(G), d) is ametric space. Sincedis symmetric, we can speak of the distance between two verticesuandv rather than the distance fromuto v.

We are now prepared to present a useful characterization of bipartite graphs.

Theorem 1.18 A nontrivial graph G is a bipartite graph if and only if G contains no odd cycles.

Proof.Suppose first that Gis bipartite. Then V(G) can be partitioned into partite setsU andW (and so every edge ofGjoins a vertex ofU and a vertex of W). Let C = (v₁, v₂, . . . , vk, v₁) be a k-cycle of G. We may assume that v₁∈U. Thusv₂∈W,v₃∈U, and so forth. In particular,vi∈U for every odd integeri with 1≤i≤kand vj ∈W for every even integerj with 2≤j ≤k. Sincev₁∈U, it follows thatvk∈W and sokis even.

For the converse, let Gbe a nontrivial graph containing no odd cycles. If G is empty, then G is clearly bipartite. Hence it suffices to show that every nontrivial component of Gis bipartite and so we may assume that Gitself is connected. Letube a vertex ofGand let

U ={x∈V(G) : d(u, x) is even} andW ={x∈V(G) : d(u, x) is odd} where u ∈ U then. We show that G is bipartite with partite sets U and W. It remains to show that no two vertices of U are adjacent and no two vertices ofW are adjacent. Suppose thatW contains two adjacent verticesw₁ andw₂. LetP₁ be au−w₁ geodesic andP₂au−w₂ geodesic. Letzbe the last vertex thatP₁andP₂have in common (possiblyz=u). Then the length of thez−w₁ subpathP₁^′ ofP₁and the length of thez−w₂subpathP₂^′ ofP₂are of the same parity. Thus the pathsP₁^′ andP₂^′ together with the edgew₁w₂produce an odd cycle. This is a contradiction. The argument that no two vertices of U are adjacent is similar.

Theeccentricitye(v) of a vertexv in a connected graphGis the distance betweenv and a vertex farthest fromv in G. Thediameterdiam(G) of Gis the greatest eccentricity among the vertices ofG, while theradiusrad(G) is the smallest eccentricity among the vertices ofG. The diameter ofGis therefore the greatest distance between any two vertices ofG. A vertexvwithe(v) = rad(G) is called acentral vertexof Gand a vertexv with e(v) = diam(G) is called aperipheral vertex ofG. Two verticesuandv ofGwithd(u, v) = diam(G) areantipodal verticesofG. Necessarily, ifuand vare antipodal vertices in G, then each ofuandvis a peripheral vertex. For the graphGof Figure 1.39,

38 CHAPTER 1. INTRODUCTION TO GRAPHS e(v₆) = 2, e(v₂) =e(v₃) =e(v₄) =e(v₅) =e(v₇) =e(v₉) =e(v₁₀) = 3,

e(v₁) =e(v₈) = 4

and so diam(G) = 4 and rad(G) = 2. In particular,v₆is the only central vertex ofGandv₁ andv₈are the only peripheral vertices ofG. Sinced(v₁, v₈) = 4 = diam(G), it follows thatv₁andv₈are antipodal vertices ofG. It is certainly not always the case that diam(G) = 2 rad(G) as, for example, diam(P₄) = 3 and rad(P₄) = 2. Indeed, the following can be said about the radius and diameter of a connected graph.

Theorem 1.19 For every nontrivial connected graphG, rad(G)≤diam(G)≤2 rad(G).

Proof. The inequality rad(G)≤diam(G) is immediate from the definitions.

Letuandwbe two vertices such thatd(u, w) = diam(G) and letvbe a central vertex of G. Therefore,e(v) = rad(G). By the triangle inequality,

diam(G) =d(u, w)≤d(u, v) +d(v, w)≤2e(v) = 2 rad(G), as desired.

Theorem 1.19 gives a lower bound (namely, rad(G)) for the diameter of a connected graphGas well as an upper bound (namely, 2 rad(G)). This is one of many results for which a question of “sharpness” is involved. These involve the question: Just how good is this result? Ordinarily, there are many interpretations of such a question. We shall consider some possible interpretations in the case of the upper bound.

Certainly, the upper bound in Theorem 1.19 would not be considered sharp if diam(G) < 2 rad(G) for every graph G; however, it would be considered sharp indeed if diam(G) = 2 rad(G) for every graphG. In the latter case, we would have a formula or an identity, not just a bound. Actually, there are graphs G for which diam(G) < 2 rad(G) and there are graphs H for which diam(H) = 2 rad(H). This alone may be a satisfactory definition of sharpness.

A more likely interpretation is the existence of an infinite class H of graphs such that diam(H) = 2 rad(H) for each graph H belonging to H. Such a class exists; for example, let H consist of the graphs of the type Kt∨K₂. One disadvantage of this example is that for each H ∈ H, diam(H) = 2 and rad(H) = 1. Perhaps a more satisfactory class (which fills a more satisfactory requirement for sharpness) is the class of paths P₂k+1, k ≥ 1. In this case, diam(P₂k+1) = 2k and rad(P₂k+1) = k; that is, for each positive integer k, there exists a connected graphGsuch that diam(G) = 2 rad(G) = 2k.

Center and Periphery

The subgraph induced by the central vertices of a connected graph G is the center of Gand is denoted by Cen(G). If every vertex of Gis a central

vertex, then Cen(G) = Gand Gis self-centered. The subgraph induced by the peripheral vertices of a connected graph Gis the periphery ofG and is denoted by Per(G).

For the graphGof Figure 1.39, the center ofGconsists of the isolated vertex v₆ and the periphery consists of the two isolated verticesv₁ andv₈. The graph H of Figure 1.40 has radius 2 and diameter 3. Therefore, every vertex ofH is either a central vertex or a peripheral vertex. Indeed, the center of H is the triangle induced by the three “exterior” vertices of H, while the periphery of H is the 6-cycle induced by the six “interior” vertices ofH.

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u₆ v₃ v₂

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u₅ u₆ u₁ u₂

Per(H) : v₁

v₃ v₂

H :

Cen(H) :

Figure 1.40: The center and periphery of a graph

It is not difficult to see that Cen(P₂k+1) =K₁ and Cen(P₂k) =K₂ for all k ≥1. Also Cen(Cn) =Cn for all n≥3. Hence there are many graphs that are centers of graphs. In an observation first made by Stephen Hedetniemi (see [31]), there is no restriction on which graphs can be the center of some graph.

Theorem 1.20 Every graph is the center of some graph.

Proof. LetG be a graph. We construct a graphH from G by first adding two new verticesuandvtoGand joining them to every vertex ofGbut not to each other. The construction of H is completed by adding two other vertices u₁ and v₁, where u₁ is joined to u and v₁ is joined to v. (This construction is illustrated in Figure 1.41.) Since e(u₁) =e(v₁) = 4, e(u) =e(v) = 3 and eH(x) = 2 for every vertex xin G, it follows that V(G) is the set of central vertices ofH and so Cen(H) =H[V(G)] =G.

40 CHAPTER 1. INTRODUCTION TO GRAPHS

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Figure 1.41: A graph with a given center

Example 1.21

Figure 1.42 shows a graphGthat represents the street system of a community, where each edge is a street and each vertex is an intersection denoted by si

(1 ≤i≤20). The community wants to build an emergency facility at one of the intersections so that the number of blocks needed to drive from the facility to the intersection farthest from it will be as small as possible. What are the possible locations for the emergency facility?

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Figure 1.42: A graph representing a street system in Example 1.21 To answer this question, we need to place an emergency facility at a central vertex of the graph G. Consequently, the eccentricity of each vertex must be computed. These are shown in Figure 1.42. Since the minimum eccentricity (radius) ofGis 4, the emergency facility should be placed ats₄, s₇, s₁₀ors₁₄.

Not every graph is the periphery of some graph however, as Halina Bielak and Maciej Syslo [19] showed.

Theorem 1.22 A nontrivial graph G is the periphery of some graph if and only if every vertex ofGhas eccentricity1or no vertex ofGhas eccentricity1.

Proof. If every vertex ofGhas eccentricity 1, thenGis complete and Per(G) = G; while if no vertex ofGhas eccentricity 1, then let F be the graph obtained from G by adding a new vertex w and joining w to each vertex of G. Since eF(w) = 1 andeF(x) = 2 for every vertexxofG, it follows that every vertex ofGis a peripheral vertex ofF and so Per(F) =F[V(G)] =G.

Dalam dokumen BUKU GRAPHS & DIGRAPH PDF (Halaman 43-58)