On the metric subgraphs of a graph

The three subgraphs of a connected graph induced by the center, annulus and periphery are called its metric subgraphs. The main results are as follows. (1) There exists a graph of order $n$ whose metric subgraphs are all paths if and only if $n\ge 13$ and the smallest size of such a graph of order $13$ is $22;$ (2) there exists a graph of order $n$ whose metric subgraphs are all cycles if and only if $n\ge 15,$ and there are exactly three such graphs of order $15;$ (3) for every integer $k\ge 3,$ we determine the possible orders for the existence of a graph whose metric subgraphs are all connected $k$-regular graphs; (4) there exists a graph of order $n$ whose metric subgraphs are connected and pairwise isomorphic if and only if $n\ge 24$ and $n$ is divisible by $3.$ An unsolved problem is posed.


Introduction
We consider finite simple graphs.For terminology and notations we follow the books [3] and [11].The order of a graph G, denoted |G|, is its number of vertices, and the size is its number of edges.We denote by V (G) and E(G) the vertex set and edge set of a graph G respectively.Denote by d G (u, v) the distance between two vertices u and v in G.If the graph G is clear from the context, we simply write d(u, v).The eccentricity, denoted by e(v), of a vertex v in a graph G is the distance to a vertex farthest from v. Thus e(v) = max{d(v, u)|u ∈ V (G)}.If e(v) = d(v, x), then the vertex x is called an eccentric vertex of v.The radius of a graph G, denoted rad(G), is the minimum eccentricity of all the vertices in V (G), whereas the diameter of G, denoted diam(G), is the maximum eccentricity.For graphs we will use equality up to isomorphism, so G 1 = G 2 means that G 1 and G 2 are isomorphic.A graph is called null if it has order 0; otherwise it is non-null.
Let G be a connected graph.A vertex u is a central vertex of G if e(u) = rad(G).
The center of G, denoted C(G), is the set of all central vertices of G.A vertex v is a peripheral vertex of G if e(v) = diam(G).The periphery of G, denoted P (G), is the set of all peripheral vertices.A vertex w is an annular vertex of G if rad(G) < e(w) < diam(G).
The annulus of G, denoted A(G), is the set of all annular vertices.Definition 1.Let G be a connected graph.The subgraph of G induced by its center is called the central subgraph of G; the subgraph of G induced by its annulus is called the annular subgraph of G; the subgraph of G induced by its periphery is called the peripheral subgraph of G.These three subgraphs are called the metric subgraphs of G.
We will determine possible orders for the existence of a graph whose metric subgraphs are either all paths, or all cycles, or all connected k-regular graphs with k ≥ 3, or connected and pairwise isomorphic graphs, and we also consider the smallest size problem.At the end we pose an unsolved problem.

Main results
For two graphs G and H, G ∨ H denotes the join of G and H, which is obtained from the disjoint union G + H by adding edges joining every vertex of G to every vertex of H.
Given two vertex subsets S and T of a graph, we denote by [S, T ] the set of edges having one endpoint in S and the other in T. P n , C n and K n denote the path of order n, the cycle of order n and the complete graph of order n respectively.As usual, qK 2 denotes the graph consisting of q pairwise vertex-disjoint edges, and deg(v) denotes the degree of a vertex v.We denote by G the complement of a graph G. G and H by x 1 , . . ., x s and y 1 , . . ., y t respectively, the nice connection of G and H with respect to this vertex labeling is the graph obtained from the disjoint union G + H by adding the edges x i y i , i = 1, 2, . . ., s and the edges x 1 y j , j = s + 1, . . ., t.
Given two graphs, there are possibly many nice connections of them, depending on the vertex labeling.For our purposes below, any nice connection works.
We will need the following lemmas.
Lemma 1 (Lesniak [10]).Let G be a connected graph of order n.Then for every integer k with rad(G) < k ≤ diam(G), there exist at least two vertices in G of eccentricity k.Proof.Let v be a central vertex of H and let x be an eccentric vertex of v in G. Then showing that rad(H) ≥ diam(G).Combining this inequality with the fact that rad(H) ≤ |H|/2 we obtain |H| ≥ 2 diam(G). 2 For a graph G and S ⊆ V (G), the neighborhood of S is defined to be N (S) = {x ∈ V (G) \ S| x has a neighbor in S}.We will repeatedly use the fact that the eccentricities of two adjacent vertices differ by at most 1.
Since a connected graph of order at most 3 has radius at most 1, we obtain |A(G)| ≥ 4. Note that every non-null graph has at least one central vertex.We obtain This completes the proof. 2 Theorem 6.There exists a connected graph of order n whose metric subgraphs are all paths if and only if n ≥ 13, and the smallest size of such a graph of order 13 is 22.
Proof.If G is a connected graph of order n whose metric subgraphs are all paths, then by Lemma 5, n ≥ 13.
Conversely, for every n ≥ 13 we will construct such a graph of order n.First, the graph G 1 in Figure 1 is a graph of order 13 whose metric subgraphs are all paths.
The central subgraph, the annular subgraph and the peripheral subgraph of G 1 are the paths u, Q = v 1 v 2 v 3 v 4 , and w 1 w 2 ...w 8 respectively.Next for a given integer n ≥ 14, in G 1 replacing the vertex u by the path P n−12 and then taking the join of P n−12 and the path Q we obtain a graph.This is a graph of order n whose metric subgraphs are all paths.
Finally we show that the smallest size of such a graph of order 13 is 22.Let H be a graph of order 13 whose metric subgraphs are all paths.Denote the center, annulus and We conclude that the size of H is at least 3 + 7 + 8 + 4 = 22.Conversely, the graph G 1 in Figure 1 is a graph of order 13 and size 22 whose metric subgraphs are all paths.This completes the proof. 2 Remark 1.By the proof of Theorem 6, it is not difficult to check that there are exactly 64 connected graphs of order 13 and size 22 whose metric subgraphs are all paths.
We will need the following two results.
Lemma 7 (Kim, Rho, Song and Hwang [9]) Let G be a graph of order n with radius r and minimum degree k where r ≥ 3 and k ≥ 2. Then n ≥ 2r(k + 1)/3.Lemma 8. Let k and n be integers with 1 ≤ k ≤ n − 1.Then there exists a k-regular graph of order n if and only if kn is even.If kn is even and k ≥ 2, then there exists a hamiltonian k-regular graph of order n.
Lemma 8 can be found in [6, pp.12-13].Its first part is well-known, but its hamiltonian part is usually not stated.
Next we determine the possible orders for the existence of a graph whose metric subgraphs are all connected and k-regular.The answer depends on the nature of k and there are six cases.Theorem 9. Let k ≥ 2 be an integer and denote q = k/3 .There exists a connected graph of order n whose metric subgraphs are all connected and k-regular if and only if (1) n ≥ 14q + 6 when k ≡ 0 mod 3 and q is even; (2) n is even and n ≥ 14q + 8 when k ≡ 0 mod 3 and q is odd; (3) n is even and n ≥ 14q + 12 when k ≡ 1 mod 3 and q is even; (4) n ≥ 14q + 11 when k ≡ 1 mod 3 and q is odd; (5) n ≥ 14q + 15 when k ≡ 2 mod 3 and q is even; (6) n is even and n ≥ 14q + 16 when k ≡ 2 mod 3 and q is odd.
Proof.Let G be a connected graph of order n whose metric subgraphs are all connected and k-regular.Denote the center, annulus and periphery of G by C, A and P respectively.

Since the central subgraph
We then estimate the cardinality of A. Since the annular subgraph W = G[A] is kregular, we have |A| ≥ k + 1 where equality holds if and only if W is complete.By Lemma 4, rad(W ) ≥ 2, implying that W is incomplete.Hence |A| ≥ k + 2. When k is odd, the order of W must be even since W is k-regular, and consequently |A| ≥ k + 3.
Conversely, for every order n in the range as stated in the theorem, we construct a connected graph of order n whose metric subgraphs are all connected and k-regular.
Notation.For a positive even integer f, we denote by K f − P M the graph obtained from K f by deleting a perfect matching; i.e., K f − P M = (f /2)K 2 .For an integer g ≥ 3, we denote by K g − HC the graph C g .For an integer s ≥ 2, we denote by K s − HP the graph P s .

Note that every vertex in H
(1) i has a unique neighbor in R (1) i for i = 1, . . ., 8. It is easy to verify that rad(M 1 ) = 2 and diam(M 1 ) = 4, the three graphs C (1) , A (1) and P (1)   are connected and k-regular, and they are the central subgraph, annular subgraph and peripheral subgraph of M 1 which has order n.
where u 3q+4 = u 1 .Partition the vertex set of A (2) into eight subsets (four pairs) = {u 4+2j | j = 0, 1, . . ., (q − 3)/2} = {u q+1+2j | j = 1, . . ., (q + 1)/2} = {u 2q+2+2j | j = 1, . . ., (q + 1)/2} such that A (2) [V Remark 2. In Theorem 9, the condition of being connected on metric subgraphs is essential.For example, Theorem 9 asserts that the smallest order of a connected graph whose metric subgraphs are all cubic is 22.Let Q be the graph obtained from the disconnected graph in Figure 2 by adding all the edges x i y j for i = 1, . . ., 4 and j = 1, . . ., 6. Then Q is a graph of order 18 whose metric subgraphs are all cubic.The peripheral subgraph of Q is disconnected.Now we determine the smallest graphs whose metric subgraphs are all cycles.Lemma 5 shows that connectedness conditions on the metric subgraphs of a graph yields some restriction on the order of the graph.Finally we pose the following question.
Question.Let k ≥ 2 be an integer.What is the smallest order of a graph whose metric subgraphs are all k-connected?
Using the information obtained in this paper, it is easy to show that the answer is 15 when k = 2.

Lemma 2 .
If G is a graph with a nonempty annulus, then rad(G) ≥ 2, diam(G) ≥ 4, and |A(G)| ≥ 2. Proof.Since diam(G) ≤ 2 rad(G) [11, p.78], if rad(G) = 1 then diam(G) ≤ 2, implying that the annulus of G is empty, a contradiction.Hence rad(G) ≥ 2. Consequently diam(G) ≥ rad(G) + 2 ≥ 4. The assertion |A(G)| ≥ 2 follows from Lemma 1 and the condition that G has a nonempty annulus. 2 We denote by e G (v) the eccentricity of a vertex v in G. Recall that |G| denotes the order of a graph G.The following lemma is of independent interest.Lemma 3. Let H be the peripheral subgraph of a connected graph G.If H is connected, then rad(H) ≥ diam(G) and |H| ≥ 2 diam(G).

Lemma 4 .
Let W be the annular subgraph of a connected graph G.If W is non-null and connected, then rad(W ) ≥ 2 and consequently |A(G)| ≥ 4.

2 Lemma 5 .
If a connected graph has a connected peripheral subgraph and a non-null connected annular subgraph, then its order is at least 13.Proof.Suppose G is a connected graph whose peripheral subgraph is connected and whose annular subgraph is non-null and connected.Consider the diameter.By Lemma 2, diam(G) ≥ 4. Then by Lemma 3, |P (G)| ≥ 2 diam(G) ≥ 8. Lemma 4 gives |A(G)| ≥ 4.
periphery of H by C, A and P respectively.By the proof of Lemma 5, we have |C| = 1, |A| = 4, |P | = 8, diam(H) = 4 and hence rad(H) = 2. Let C = {u}, let the annular subgraph of H be the path v 1 v 2 v 3 v 4 , and let the peripheral subgraph of H be the path w 1 w 2 ...w 8 .Then e(u) = 2, e(v i ) = 3, i = 1, . . ., 4, and e(w j ) = 4, j = 1, . . ., 8. Since the eccentricities of two adjacent vertices differ by at most 1, N (u) ⊆ A. Now the condition e(u) = 2 implies that each vertex in P has a neighbor in A. Hence |[P, A]| ≥ 8. Clearly, any eccentric vertex of a vertex in P lies in P. Since e(v 2 ) = 3, every eccentric vertex of v 2 lies in P. Let w k be an eccentric vertex of v 2 .Then v 4 is the unique neighbor of w k in A. Since d(u, w k ) ≤ 2, u and v 4 are adjacent.Similarly, considering the vertex v 3 we deduce that u and v 1 are adjacent.We claim that N (u) = A. Otherwise deg(u) ≤ 3. Without loss of generality, suppose v 3 and u are nonadjacent.Denote S = {v 1 , v 2 } and T = {v 4 }.Then each vertex in P has a neighbor in S ∪ T. We will repeatedly use this fact.Since w 1 is the only possible eccentric vertex of w 5 , we have d(w 1 , w 5 ) = 4. Let x ∈ S ∪ T be a neighbor of w 1 and let y ∈ S ∪ T be a neighbor of w 5 .Then we have the following two possible cases.Case 1. x ∈ S and y ∈ T. Since w 8 is the unique eccentric vertex of w 4 , w 8 and v 4 are nonadjacent.Hence w 8 has a neighbor in S. Consequently, the neighbor of w 4 in S ∪ T must be v 4 .To keep d(w 4 , w 8 ) = 4, w 7 and v 4 cannot be adjacent.Thus w 7 has a neighbor in S. Similarly, to keep d(w 5 , w 1 ) = 4, w 2 and v 4 cannot be adjacent.Thus w 2 has a neighbor in S. Now w 6 is the only eccentric vertex of w 2 .Hence w 6 has no neighbor in S, implying that v 4 is the only neighbor of w 6 in S ∪ T. Since w 3 has a neighbor in S ∪ T, we deduce that e(w 7 ) ≤ 3, a contradiction.Case 2. x ∈ T and y ∈ S. The condition e(w 5 ) = 4 implies that the two vertices w 4 and w 6 are nonadjacent to v 4 .Thus, both w 4 and w 6 have a neighbor in S. Since w 8 is the unique eccentric vertex of w 4 , w 8 and v 4 are adjacent and w 3 has a neighbor in S. If w 7 is adjacent to v 4 , using the fact that w 2 has a neighbor in S ∪ T we deduce that e(w 6 ) ≤ 3, a contradiction.Hence w 7 has a neighbor in S.But then e(w 7 ) ≤ 3, a contradiction again.This shows that deg(u) = 4.

Theorem 10 .
The minimum order of a connected graph whose metric subgraphs are all cycles is 15 and there are exactly three such graphs of order 15, all of which have size 35 and are depicted in Figure3.Proof.The case k = 2 of Theorem 9 asserts that the minimum order of a connected graph whose metric subgraphs are all cycles is 15.Let G be a connected graph of order 15 whose metric subgraphs are all cycles, and let C, A and P be the center, annulus and periphery of G respectively.Then |C| + |A| + |P | = 15.By the first four paragraphs of the proof of Theorem 9 we have |C| ≥ 3, |A| ≥ 4 and |P | ≥ 8. Hence |C| = 3, |A| = 4 and |P | = 8.Denote by H the peripheral subgraph of G. Then H = C 8 and rad(H) = 4.By Lemma 2, rad(G) ≥ 2, diam(G) ≥ 4 and by Lemma 3, we have 4 = rad(H) ≥ diam(G).Thus diam(G) = 4 and consequently rad(G) = 2, since A is nonempty.Combining the fact that the eccentricities of two adjacent vertices differ by at most 1 and the condition that rad(G) = 2, we deduce that for any vertex x ∈ C and any vertex y ∈ P, d(x, y) = 2. Let x, z, y be a path.Then z ∈ A. Thus every vertex in P has a neighbor in A. Note that the three metric subgraphs of G are the cycles C 3 , C 4 and C 8 .Let y be the antipodal vertex of y on the even cycle G[P ].Then clearly y is the unique eccentric rad(H) ≥ 4.Recall that the eccentricity of any vertex in a disconnected graph is infinity.We will construct a graph Q whose metric subgraphs are all isomorphic to H. Let H c , H a and H p be three pairwise vertex-disjoint copies of H, let V (H a ) = {x 1 , . . ., x k } and let f be an isomorphism from H a to H p .Then the graph Q is obtained from H c + H a + H p by first taking the join of H c and H a and then adding the edges x i f (x i ) for i = 1, . . ., k.It is easy to verify thatQ has radius 2 and diameter 4, and the central subgraph, annular subgraph and peripheral subgraph of Q are H c , H a and H p respectively.Next we prove the second assertion of Theorem 11.Suppose R is a graph of order n whose metric subgraphs are connected and pairwise isomorphic.Let W be the peripheral subgraph of R. By Lemma 2, diam(R) ≥ 4 and by Lemma 3, |W | ≥ 2 diam(R).Hence |W | ≥ 2 × 4 = 8.It follows that n = 3|W | ≥ 24 and n is divisible by 3. Conversely, suppose n ≥ 24 is an integer that is divisible by 3. Let H be the path P n/3 .By the first part of Theorem 11 proved above, there exists a graph whose metric subgraphs are all isomorphic to H.This completes the proof. 2 Definition 2. Given a sequence of graphs H 1 , H 2 , . . ., H p , their circular join is defined to be the graph obtained from the disjoint union H 1 + H 2 + . . .+ H p by adding edges joining each vertex of H i to each vertex of H i+1 for every i = 1, 2, . . ., p where H p+1 means H 1 .Definition 3. Let G and H be two graphs with |G| ≤ |H|.Labeling the vertices of