Edge Tree Spanners

Abstract

A tree t-spanner of a graph G is a spanning tree T of G in which any two adjacent vertices of G have distance at most t in T. The line graph L(G) of a graph G is the intersection graph of the edges of G. We define the edge tree t-spanner of a graph G as a spanning tree T of L(G) in which any two edges that share an endpoint in G have distance at most t in T. Although determining if G has a tree 3-spanner is an open problem for more than 20 years, we settle that deciding if a graph G has an edge tree 3-spanner is polynomial-time solvable. As a consequence, we present polynomial time algorithms for the edge tree t-spanner problem for several graph classes such as trees, join of graphs, split graphs, P ₄-tidy, and (1, 2)-graphs. Moreover, we establish that deciding whether a graph G has an edge tree 8-spanner is NP-complete, even if G is bipartite.

1 Introduction

The problem of looking for a spanning tree with constraints on the vertices’ or edges’ distances is a combinatorial challenge with many applications and approaches [1, 11]. A tree t-spanner of a graph G is a spanning tree T of G in which any two adjacent vertices of G have distance at most t in T. A graph G having a tree t-spanner is called a t-admissible graph. The smallest t for which a graph G is t-admissible is the stretch index of G and is denoted by σ _T(G) (or simply σ(G)). The t-admissibility problem aims to decide whether a given graph G has σ(G) ≤ t. The problem of determining the tree stretch index, i.e. the minimum stretch spanning tree problem (MSST) has been studied by establishing bounds on σ(G) or developing the computational complexity of the decision version of MSST for several graph classes [2,3,4]. Cai and Corneil [2] proved that t-admissibility is NP-complete, for t ≥ 4, whereas 2-admissible graphs can be recognized in polynomial-time. However, the characterization of 3-admissible graphs is still an open problem.

The characterization for 2-admissible graphs [2], stated in Theorem 1, deals with triconnected components of a connected graph, defined as any maximal subgraph that does not contain two vertices whose removal disconnects the graph (the authors also consider K ₂ and K ₃ as triconnected components). A nonseparable graph is a graph without a cut vertex, i.e., a vertex whose removal disconnects the graph. A star with n + 1 vertices is the complete bipartite graph K _1,n. A v-centered star is a star centered on v, that is a universal vertex. Similarly, a bi-star is a graph such that there is an edge uv and every edge of E shares an endpoint with uv. Hence, uv is a universal edge of the bi-star. A uv-centered bi-star is a bi-star centered on a universal edge uv.

Theorem 1 ([2])

A nonseparable graph G is 2-admissible if and only if G contains a spanning tree T such that for each triconnected component H of G, T ∩ H is a spanning star of H.

Given a graph G, its line graph L(G) is obtained as follows: V (L(G)) = E(G); E(L(G)) = {{uv, uw}|uv, uw ∈ E(G)}. I.e., each edge of G is a vertex of L(G) and if two edges share an endpoint, then their corresponding vertices are adjacent in L(G). The distance between two edges e ₁ and e ₂ of G, for e ₁, e ₂ ∈ E(G) is the distance between their corresponding vertices in L(G).

We define the edge tree t-spanner of a graph G as a spanning tree T of L(G) such that, for any two adjacent edges of G, their distance is at most t in T. Therefore, an edge tree t-spanner of G is a tree t-spanner of L(G).

A graph G that has an edge tree t-spanner is called edge t-admissible. The smallest t for which G is an edge t-admissible graph is the edge stretch index of G, and is denoted by $\sigma ^{\prime }_T(G)$ (or simply σ′(G)). The edge t-admissibility problem aims to decide whether a given graph G has σ′(G) ≤ t. Figure 1 depicts the relation between the edge tree spanner of a graph and the tree spanner of its line graph.

Fig. 1

A graph G, a tree 3-spanner of L(G) in red, and G with the related edge 3-spanner in red

An immediate consequence of MSST is that the property of being t-admissible graph is not hereditary, i.e., if G is t-admissible then there may exist a subgraph H of G that is not t-admissible. Indeed, the addition of a universal vertex u to any t-admissible graph results in a 2-admissible graph by a u-centered star.

On the other hand, regarding the edge tree t-spanner, in Sect. 3 we prove that being an edge 3-admissible graph is a hereditary property, and based on that, we are able to decide whether G is edge 3-admissible in polynomial time. Moreover, in Sect. 4 we determine polynomial time algorithms to obtain the edge stretch index for some edge 4-admissible and edge 5-admissible classes, such as split graphs, join graphs, P ₄-tidy graphs and (1, 2)-graphs. In Sect. 5, we prove that edge 8-admissibility is NP-complete for (2, 0)-graphs, i.e. bipartite graphs. In Sect. 6, we present concluding remarks. Next (Sect. 2), we relate admissibility and edge admissibility problems, presenting immediate consequences and preliminary results.

2 Admissibility Versus Edge Admissibility for Graph Classes

Since induced cycles in a graph G correspond to cycles of the same length in L(G), we have that σ′(C _n) = σ(C _n) = n − 1. Although cycle graphs satisfy σ′ = σ, for several other classes the stretch index is different of the edge stretch index.

For instance, trees are 1-admissible and the unique edge 1-admissible graphs are the ones such that their line graphs are trees. Since line graphs are claw-free, then path graphs are the unique edge 1-admissible graphs. In Proposition 1 we determine the edge stretch index of trees.

Proposition 1

Let G be a tree. If G is a path graph then σ′(G) = 1, otherwise σ′(G) = 2.

Proof

Note that if G is a path, then L(G) is a path and σ′(G) = 1. For any other tree there is a vertex of degree at least 3, implying a complete subgraph of length at least 3 in L(G). Each internal node u of G correspond to a maximal complete subgraph of L(G) of size d _G(u) and two of such maximal complete subgraphs share at most a vertex in L(G). Hence, any triconnected component of L(G) is a complete subgraph and satisfies Theorem 1. □

Since the study of edge tree spanners is equivalent to the study of tree spanners of line graphs, and deciding whether a graph is 2-admissible is polynomial-time solvable, Theorem 1 implies Corollary 1.

Corollary 1

Edge 2-admissibility is polynomial-time solvable.

The edge stretch index of cycle graphs and complete graphs are useful to characterize edge 3-admissible graphs, as discussed in Sect. 3.

Complete graphs are 2-admissible, however their line graphs are not. In order to prove that σ′(K _n) = 4, from Lemma 1 we have that σ′(K ₅) ≤ 4, and it is possible to prove that K ₅ is not edge 3-admissible, as highlighted below.

To prove that K ₅ is not edge 3-admissible, one can verify by a case analysis that it is not possible obtain a spanning tree T such that T ∩ L(K ₅) has at least 3 internal nodes. Clearly, T ∩ L(K ₅) cannot have more than 3 internal nodes, because otherwise the edge factor of such a tree would be at least 4. Moreover, it is not possible obtain a spanning tree T such that T ∩ L(K ₅) is a bi-star or it is a tree with three internal nodes whose leaves at distance 4 in T are not adjacent in L(K ₅).

In Sect. 3 we prove that being edge 3-admissible is a hereditary property for induced subgraphs (Lemma 2), then Corollary 3 states that σ′(K _n) = 4, for n ≥ 5.

A graph G has a distance two dominating edge uv if every edge of E(G) has a vertex in N[u] ∪ N[v] as one of its endpoints, where N[x] is the closed neighborhood of x, i.e. N[x] = N(x) ∪{x}. Moreover, G has two adjacent distance two dominating edges uv and vw if every edge of E(G) has a vertex in N[u] ∪ N[v] ∪ N[w] as one of its endpoints.

Lemma 1

A graph G with a distance two dominating edge uv has σ′(G) ≤ 4.

Proof

Since G has a distance two dominating edge uv, there is a spanning tree with diameter at most four of L(G) with the vertex uv as its root, the vertices {ux | ux ∈ E(G)}∪{vy | vy ∈ E(G)} adjacent to uv, and the remaining vertices of L(G) adjacent to some vertex in {ux | ux ∈ E(G)}∪{vy | vy ∈ E(G)}. □

Figure 2 depicts graphs with distance two dominating edges and their edge tree 4-spanners, as the proof of Lemma 1. A graph is split if its vertex set can be partitioned into a stable set and a clique. The join between two graphs G ₁ and G ₂ results in the graph G such that V (G) = V (G ₁) ∪ V (G ₂) and E(G) = E(G ₁) ∪ E(G ₂) ∪{uv | u ∈ V (G ₁) and v ∈ V (G ₂)}.

Fig. 2

A split graph and a join graph with their edge tree 4-spanners

Several graph classes can be constructed by join and complement of join operations, i.e. union operations. Cographs are the P ₄-free graphs, i.e. graphs without a P ₄ as an induced subgraph, and G is a cograph iff it has the following recursive definition: (i) G is a K ₁; (ii) G is a join of cographs; (iii) G is a union of cographs. A generalization of cographs are the graphs with few P ₄’s, such as P ₄-sparse and P ₄-tidy [7].

A graph is P ₄-sparse if for each set of 5 vertices, there is at most one induced P ₄. A graph is P ₄-tidy if for each induced P ₄ of G, say P, there is at most one vertex v ∈ V (G) ∖ V (P) such that V (P) ∪{v} induces at most two P ₄’s in G. P ₄-tidy generalizes P ₄-sparse graphs, and G is a P ₄-tidy graph iff it has the following recursive definition: (i) G is P ₅, C ₅, $\overline {P_5}$, or K ₁; (ii) G is a join of P ₄-tidy graphs; (iii) G is a union of P ₄-tidy graphs; (iv) G is a spider; (v) G is an almost spider. A graph is a spider graph if its vertex set can be partitioned into $\mathscr {S},\mathscr {K}$ and $\mathscr {R}$ such that (i) $\mathscr {K}$ is a clique ($\mathscr {K}$ is called body), $\mathscr {S}$ is a stable set and $|\mathscr {S}|=|\mathscr {K}| \geq 2$; (ii) each vertex of $\mathscr {R}$ ($\mathscr {R}$ is called head) is adjacent to all vertices of $\mathscr {K}$ and is non-adjacent to any vertex of $\mathscr {S}$; (iii) There is a bijection $f: \mathscr {S} \mapsto \mathscr {K}$ such that, for all $x \in \mathscr {S}$, either N(x) = {f(x)}, or $N(x) = \mathscr {K} - \{f(x)\}$. A graph is an almost-spider graph if it can be constructed from a spider graph $G = (\mathscr {S}, \mathscr {K}, \mathscr {R})$ by adding a vertex v′ which is either a false twin of v or a true twin of v, such that $v \in \mathscr {S} \cup \mathscr {K}$ [10].

Split graphs, join graphs and P ₄-tidy graphs are 3-admissible [3, 4]. Corollary 2 follows from Lemma 1 and: for split graphs, any clique’s edge is distance two dominating; for join graphs between G ₁ and G ₂, any uv such that u ∈ V (G ₁) and v ∈ V (G ₂) is distance two dominating; for P ₄-tidy graphs, any edge between the head and the body is distance two dominating.

Corollary 2

Split graphs, join graphs and P ₄ -tidy graphs are edge 4-admissible.

Since 3-admissibility is still open and t-admissibility is NP-complete, for t ≥ 4, we are interested to establish the computational complexity of determining the edge stretch index. In Sect. 3, we prove that edge 3-admissibility is polynomial-time solvable, and as an immediate consequence, we are able to determine in polynomial time the edge stretch index for any edge 4-admissible graph, such as split graphs, join graphs and P ₄-tidy graphs (Corollary 6).

3 Edge 3-Admissibility Is Polynomial-Time Solvable

Lemma 2

Edge 3-admissibility is a hereditary property for induced subgraphs.

Proof

Assume that there is an edge 3-admissible graph G with an induced subgraph H such that H is not edge 3-admissible. W.l.o.g. let G′ be an induced subgraph of G such that: |V (G′)| = |V (H)| + 1, u ∈ V (G′) ∩ V (H); G′ is edge 3-admissible; H is edge k-admissible for k ≥ 4; T′ is an edge tree 3-spanner of G′; and T is an edge k-tree spanner of H with k ≥ 4. In any edge tree k-spanner T of H there is a path P with k + 1 vertices using edges of T and an edge of G′ not in T between the two endpoints of this path (see Fig. 3a that considers k = 5). Since G′ is edge 3-admissible, the addition of the vertex u must remove a part of that path P from T. For the sake of contradiction, assume T″ is a tree that contains at least three internal nodes among the edges incident to u. Since these edges have u as endpoint, then the leaves that are at distance 4 in T″ correspond to adjacent edges in G′, a contradiction. Therefore, the edges incident to u must be a bi-star in T′ (see Fig. 3b).

Fig. 3

(a) V (H) = {v, w, x, y, z, t} and a path P in red. (b) In red a bi-star satisfying Case 1. (c) In red a bi-star satisfying Case 2

W.l.o.g. assume that u is adjacent to all vertices of G related to the path P of T. The edges of the bi-stars cover at most four vertices of P. We have two cases: Case 1: the bi-star connects consecutive vertices of P. In this case it does not reduce the distance between the vertices of P in T′ (e.g. see Fig. 3b, the distance between vw and vt is 5 in T′) and T′ is not an edge tree 3-spanner, a contradiction; Case 2: the bi-star connects non-consecutive vertices of P. In this case it does reduce the distance between vertices of P, however, the vertex xy between this non consecutive vertex of P is connected to leaves of the two centers of the bi-star in L(G), which implies that T′ is not edge 3-admissible, a contradiction (Fig. 3c). □

Corollary 3

Any complete graph K _n has σ′(K _n) = 4, for n ≥ 5.

Proof

Since σ′(K ₅) = 4 (Sect. 2) and for n ≥ 5, K _n has a K ₅ as an induced subgraph, then, by Lemma 2, we have that K _n are not edge 3-admissible, for n  ≥ 5. Furthermore, complete graphs have a distance two dominating edge, hence by Lemma 1, σ′(K _n) ≤ 4, for n ≥ 5, and the result follows. □

Line graphs of K _n are complement of Kneser graph KG _n,2 [8], then $\sigma (\overline {KG_{n,2}})=~4$.

Note that C _k and K _k, for k ≥ 5 are not subgraphs of edge 3-admissible graphs. See Fig. 4 for examples of C ₄ and K ₄ where all vertices have degree at least 3 and 4 in G, resp. Suppose H is an induced C ₄ (or K ₄) in G. In L(G[H]) there must be a path through all L(C ₄)’s vertices (or through four L(K ₄)’s vertices) and one more vertex corresponding to an edge that does not belong to the C ₄ (to the K ₄) in H. Hence, it implies that σ′(H) ≥ 4, and Corollary 4 follows.

Fig. 4

C ₄ and K ₄ whose vertices have degree at least 3 and 4 in G, resp. Note that d _T(e ₁, e ₂) = 4

Corollary 4

Let G be an edge 3-admissible graph. If X ∈{C ₄, K ₄} is an induced subgraph of G, then there is a vertex v ∈ V (X) such that N _G(v) ⊆ V (X).

By Corollary 4, any edge 3-admissible graph has vertices of degree 2 and 3 in each induced C ₄’s and K ₄, resp. Hence, Construction 2 presents a way to break C ₄’s and K ₄’s into P ₅’s and K ₃’s, resp., in order to present a stronger necessary condition in Lemma 4.

Construction 2

Let G be a graph that satisfies: G does not have induced C _k nor K _k , for k ≥ 5, as induced subgraphs; for each induced C ₄ there is a vertex of degree two in G; and for each induced K ₄ there is a vertex of degree three in G. We construct a graph H from G as follows:

• 1. each induced C ₄ = a, b, c, d, a, for d _G(a) = 2, is transformed into a P ₅ = a, b, c, d, a′ by adding a new vertex a′ and the edge da′, and removing the edge da;

• 2. each induced K ₄ = {a, b, c, d}, for d _G(a) = 3, is transformed into three complete graphs K ₃ by adding a new vertex a′ and: removing edge ba; adding edges ba′ and ca′.

Lemma 3

A graph G is edge 3-admissible if and only if the graph H from Construction 2 is edge 3-admissible.

Proof

If G is edge 3-admissible, then all edges of an edge tree 3-spanner of G are used to obtain a spanning tree of H and we do not increase the edge stretch index from G to H, because, by construction, we are not increasing a maximum path between any two adjacent vertices of G in H. If H is edge 3-admissible, then all edges of an edge tree 3-spanner of H are used for a spanning tree of G and, since we are identifying vertices that belong only to C ₄’s or K ₄’s in G, such identification does not affect cycles that give the edge tree 3-spanner of H and does not increase such index of G by the used edges of H. □

A k-tree is a graph obtained from a K _k+1 by repeatedly adding vertices in such a way that each added vertex v has exactly k neighbors defining a clique of size k + 1. A partial k-tree is a subgraph of a k-tree [9].

Lemma 4

Let G be an edge 3-admissible graph. If H is the graph obtained from G in Construction 2 , then H is a chordal partial 2-tree graph.

Proof

If G is edge 3-admissible with X ∈{C ₄, K ₄} as an induced subgraph, then, by Corollary 4, X must have at least one vertex a such that N(a) ⊆ X. Based on that, in Construction 2 we obtain a graph without C ₄’s nor K ₄’s. Since, by Lemma 3, the transformed graph H from an edge 3-admissible graph G is also edge 3-admissible, we have that the length of any clique is at most 3 and it does not have C _k, for k ≥ 4. Since chordal graphs with maximum clique of length 3 are partial 2-tree [9], we have that H is a chordal partial 2-tree graph. □

By Lemma 4, edge tree 3-spanner graphs are formed by 2-trees where either an edge or a vertex connects two 2-trees. Hence, for the former case such edge is a bridge and for the later case it is a cut vertex of the graph. Lemmas 5 and 6 present conditions that force spanning trees correspond to edge 3-admissible graphs.

Lemma 5

Given an edge 3-admissible graph G and two 2-trees A ₁ and A ₂ connected by a bridge uv, such that |V (A _i)| > 3 for i ∈{1, 2}, then for any edge 3-spanner T, uv is a pendant vertex in T[A ₁ ∪{u, v}], i.e. $d_{T[A_1 \cup \{u,v\}]}(uv)=1$.

Proof

Assume u ∈ A ₁, u, x, y is a triangle and v ∈ A ₂. Suppose $d_{T[A_1 \cup \{u,v\}]}(uv)\geq 2$, hence xy must be adjacent to either ux or to uy in T. W.l.o.g., let xy be adjacent to uy, then, there is an edge wx in A ₁ which implies the distance between wx and xy to be equal to 4 by a path through uv, a contradiction. □

Each bridge forces a unique way to obtain an edge tree 3-spanner of G. Hence, by Lemma 5, assume G is 2-edge connected, i.e. there is not a bridge in G. Otherwise, we consider each connected component separately after the bridges removal of G.

Now, consider the case that G has a cut vertex. Let a windmill graph Wd(3, n) be the graph constructed for n ≥ 2 by identifying n copies of K ₃ at a universal vertex. Since an edge 3-admissible graph is partial 2-tree, we have that if there is a cut vertex u in G, then G[N _G[u]] contains a windmill graph Wd(3, d), for $2\leq d \leq \frac {d_{G}(u)}{2}$. Let a diamond graph be a K ₄ minus an edge. Each K ₃ of a windmill centered in u has two vertices of degree 2, or it has a cut vertex of G distinct of u, or it belongs to a diamond graph of G.

Lemma 6

Let G be 2-edge connected graph with a cut vertex u and edge 3-admissible. If the associated windmill graph Wd(3, n) centered in u satisfies n ≥ 3, then u belongs to at most 2 diamonds in G.

Proof

Assume that u is center of the windmill graph Wd(3, 3) and it belongs to 3 diamonds D ₁, D ₂ and D ₃ in G. We prove that G is not edge 3-admissible, and then it implies that if G is edge 3-admissible, then u does not belong to more than 3 diamonds for every n ≥ 3, either, because the hereditary property proved in Lemma 2.

Note that L(H), for H = Wd(3, 3) ∪ D ₁ ∪ D ₂ ∪ D ₃, is composed by a K ₆ and the addition of three other subgraphs, named B ₁, B ₂ and B ₃, constructed by a join between a vertex and a C ₄. Moreover, each edge of a perfect matching of the K ₆, {e ₁, e ₂, e ₃}, is identified to an edge of B ₁, B ₂ and B ₃ that belongs to the C ₄s, resp. Suppose that L(H) is 3-admissible, hence for any tree 3-spanner T of L(H) we have that T ∩ L(H) is a fl-centered bi-star, for f and l being any two K ₆’s vertices. Since any vertex of the K ₆ belongs to exactly one of the other three subgraphs added to it, i.e. each K ₆’s vertex belongs to either B ₁, B ₂ or B ₃, then at least two adjacent vertices of L(H) are adjacent to leaves of the fl-centered bi-star, implying σ′(H) = 4. □

If there is a vertex u that belongs to Wd(3, 2) then there are two solutions in T ∩ Wd(3, 2), less than isomorphism. Consider a Wd(3, 2) such that V (Wd(3, 2)) = {u, v, w, v′, w′} such that u, v, w and u, v′, w′ induce K ₃’s. Note that an edge tree 3-spanner T ∩ Wd(3, 2) can be formed as follows: Case 1: {uv, uw}, {uv, vw}, {uv, uv′}, {uv′, uw′}, {uv′, v′w′}; Case 2: {uv, uw}, {uv, vw}, {uv, uv′}, {uv, uw′}, {uv′, v′w′}. Any other edge tree spanner of Wd(3, 2) is not edge tree 3-spanner.

Although a Wd(3, 2) graph centered in u may have two spanning trees, if each triangle also belongs to a diamond, let D ₁ and D ₂ be such diamonds with vertices V (D ₁) = {u, v, w, x} and V (D ₁) = {u, v′, w′, x′}, then the previous Case 1 is the unique edge tree 3-spanner for T ∩ Wd(3, 2), less than isomorphism.

Furthermore, let H = Wd(3, 2) ∪ D ₁ be formed by a Wd(3, 2) centered in u with vertices V (Wd(3, 2)) = {u, v, w, v′, w′} such that vw belongs to the diamond D ₁ with vertices V (D ₁) = {v, w, s, t}, then we have that H is not edge 3-admissible, which can be verified by conditions above and a simple case analyses.

Hence, we have presented necessary conditions of a 2-edge connected graph G satisfying Construction 2 to be edge 3-admissible when it has a cut vertex.

Now, consider G a biconnected graph. Theorem 2 characterizes such graphs. The diameter of a graph G is the greatest distance between any pair of vertices, and is denoted by D(G).

Theorem 2

Given G a biconnected graph with D(G) ≤ 3. We have that σ′(G) ≤ 3 if and only if either there is distance two dominating edge e ₁ = uv or for any edges e ₁ = uv, e ₂ = uw, and e ₃∉N(u) ∪ N(v) ∪ N(w), e ₃ is adjacent to edges only of N(v) (or equivalently, only of N(w)).

Proof

If G has a dominating edge, for D(G) ≤ 3, then σ′(G) ≤ 3 by a uv centered bi-star. Or, if any edge is not dominated by e ₁ but it is adjacent to edges only of N(v), then in the solution spanning tree such vertex is adjacent to a leaf of v and it does not turn σ′(G) ≥ 4 because it is not adjacent to leaves of u. Assume that G is edge 3-admissible, there is not a distance two dominating edge and there is an edge e ₃, such that e ₃∉N(u) ∪ N(v) ∪ N(w) that is adjacent to edges of N(v) and N(w). In this case e ₃ is connected to leaves of the two centers of the bi-star in L(G), which implies that T′ is not edge 3-admissible, a contradiction. □

Note that Theorem 2 gives another argument on the lower bound of Corollary 3, since a K _n does not satisfy conditions of Theorem 2.

Corollary 5

Edge 3-admissibility is polynomial-time solvable.

4 Edge Stretch Index for Split and Generalized Split Graphs

Since σ′(G) ≤ 4 for graphs with a distance two dominating edge (Theorem 1), the polynomial time algorithm for edge 3-admissible of Corollary 5 also works for these graphs and their subclasses, such as split graphs, join graphs and P ₄-tidy graphs. I.e., we know whether these graphs have σ′(G) = 2, σ′(G) = 3 or σ′(G) = 4.

Corollary 6

Edge t-admissibility is polynomial-time solvable for split graphs, join graphs and P ₄ -tidy graphs.

As presented in Corollary 6, we are able to determine the edge stretch index for split graphs. Split graphs can be generalized as the (k, ℓ)-graphs, which are the graphs that the vertex set can be partitioned into k stable sets and ℓ cliques. The (k, ℓ)-graphs are also denoted as the generalized split graphs [5].

In [4], the dichotomy P versus NP-complete on deciding the stretch index for (k, ℓ)-graphs was partially classified. One of the open problems regarding MSST is to establish the computational complexity for (1, 2)-graphs. Next, we prove that the edge stretch index for (1, 2)-graphs can be determined in polynomial time.

We denote a (1, 2)-graph as a graph G = (V, E) where V  is partitioned into $V=\mathscr {K}_1 \cup \mathscr {K}_2 \cup S$, such that each $\mathscr {K}_i$ induces a clique and S is a stable set.

Lemma 7

If G is a (1, 2)-graph, then G is edge 5-admissible.

Proof

Since G is connected, there is a path between a vertex $u \in \mathscr {K}_1$ and $v\in \mathscr {K}_2$ by an edge uv or by a P ₃ = u, w, v. Figure 5 depicts the cases of (1, 2)-graphs and their edge 5-tree spanners. In Fig. 5a there is an induced C ₆ by two vertices of each clique and two vertices of S, implying a non-edge in any tree, hence σ′(G) ≤ 5. □

Fig. 5

Cases of (1, 2)-graphs and the corresponding edge tree spanners. (a) an edge 5-admissible graph. (b) and (c) are edge 4-admissible graphs

Theorem 3

A (1, 2)-graph $G=(\mathscr {K}_1 \cup \mathscr {K}_2 \cup S, E)$ has σ′(G) ≤ 4 if and only if G has a distance two dominating edge or two adjacent distance two dominating edges that are adjacent to at least one edge of each pair of edges incident to a vertex of S such that one endpoint of an edge of this pair is in $\mathscr {K}_1$ and another one in $\mathscr {K}_2$.

Proof

From Lemma 1, if G has a distance two dominating edge, then G is edge 4-admissible. Moreover, if G has two distance two dominating edges e ₁ and e ₂ adjacent to at least one edge of each pair of edges incident to a vertex of S such that one endpoint of an edge of this pair is in $\mathscr {K}_1$ and an endpoint of the other edge is in $\mathscr {K}_2$, one obtain an edge tree 4-spanner T of G by selecting any spanning tree of L(G) that maximizes the degrees of these two distance two dominating edges in T.

Conversely, for the sake of contradiction assume that G does not have such distance two dominating edges and T is an edge tree 4-spanner of G. Since G is connected, there is a vertex of S adjacent to both $\mathscr {K}_1$ and $\mathscr {K}_2$ and we can select these two edges of S to be two distance two dominating edges of G. Therefore, for all distance two dominating edges e ₁ and e ₂ of G we have two edges e _i and e _f incident to a vertex of S such that these edges are both not adjacent to e ₁ and e ₂. Therefore, in the best case scenario these two edges are adjacent to edges $e^{\prime }_1$ and $e^{\prime }_2$ adjacent to e ₁ and e ₂. However, we have a path in T e _i $e^{\prime }_1$ e ₁ e ₂ $e^{\prime }_2$ e _f with these two edges e _i and e _f sharing an endpoint, which implies that T is not an edge 4-tree spanner of G. □

Corollary 7

Edge t-admissibility is polynomial-time solvable for (1, 2)-graphs.

5 Edge 8-Admissibility Is NP-Complete for Bipartite Graphs

Next, we present a polynomial time transformation from 3-sat [6] to edge 8-admissibility for (2, 0)-graphs, i.e. bipartite graphs.

Construction 3

Given an instance I = (U, C) of 3-sat we construct a graph G as follows. We add a P ₂ with labels x and x′ to G. For each variable u ∈ U we add a C ₈ to G with three consecutive vertices labeled as u, m _u , and $\overline {u}$ and the other five consecutive vertices labeled as u ₁ to u ₅ . For each u _i, i = 1, …, 5, u and $\overline {u}$ we add a pendant vertex. For each variable u ∈ U we add the edge xm _u to G. For each clause c ₁ = (u, v, w) ∈ C, we add two vertices vertex c ₁ and $c_1^{\prime }$ to G and the edges $c_1c_1^{\prime }$ , c ₁ u, c ₁ v, and c ₁ w. For each variable u ∈ U we add a P ₄ to G with endpoints labeled p _u1 and p _u4 and the edges p _u1 x and p _u4 m _u.

Figure 6 depicts an example of a graph obtained from a 3-sat instance.

Fig. 6

Graph obtained from Construction 3 on the instance $I=(\{u,v,w\},\{(u,v,w),(\overline {u},v,\overline {w}\})$ and an edge tree 8-spanner of it in red

The key idea of the proof of Theorem 4 is that, for each variable u ∈ U, we have exactly one edge in the edge tree 8-spanner T which is near to x and u or $\overline {u}$. We relate this proximity to a true assignment of that literal. Next, we require that at least one edge incident to each clause to be connected to a true literal. Otherwise, if they are all false literals, we end up with two of the edges incident to that clause being vertices of L(G) with distance at least 9 in T.

Theorem 4

Edge 8-admissibility is NP -complete for bipartite graphs.

Proof

By construction, G is bipartite. Moreover, not only the problem is in NP, but also the size of the graph G, obtained from Construction 3 on an instance I = (U, C) of 3-sat, is polynomially bounded by the size of I. We prove that G is edge 8-admissible if and only if there is a truth assignment to I. Consider a truth assignment of I = (U, C). We obtain an edge tree 8-spanner T of G as follows (see Fig. 6).

Add to T the edges: {x′x, xm _u | u ∈ U}; {xm _u, m _u u | u ∈ U and u is true} or

$\{xm_u, m_u \overline {u}\ |\ u \in U \mbox{ and } \overline {u} \mbox{ is true}\}$; $\{um_u, \overline {u}m_u\ |\ u \in U\}$; For each clause select a true literal and add to T: {c′c, uc | c is a clause with the selected true literal u};

{uc, um _u | c is a clause with the selected true literal u};

$\{\overline {u}c,\overline {u}m_u\ | \ c\mbox{ is a clause } \mbox{with the selected true literal } \overline {u}\}$;

{uc, vc | c is a clause with the selected trueliteral u and v is other literal of c};

For each variable u ∈ U add to T the edges: $\{m_up_{u_4},p_{u_4}p_{u_3}\}$; $\{p_{u_4}p_{u_3},p_{u_3}p_{u_2}\}$; $\{p_{u_3}p_{u_2},p_{u_2}p_{u_1}\}$; $\{p_{u_2}p_{u_1},p_{u_1}x\}$; $\{p_{u_1}x,xm_u\}$; {um _u, uu ₁}; $\{\overline {u}m_u,\overline {u}u_5\}$; {uu ₁, u ₁ u ₂}; {u ₃ u ₄, u ₄ u ₅}; $\{u_4u_5,\overline {u}u_5\}$; and each pendant G is added to a solution tree as Fig. 6

Consider an edge tree 8-spanner T of G (resp. tree 8-spanner of L(G)), we present a truth assignment of I = (U, C). First we claim that for each variable u ∈ U, there is exactly one of these two edges in T: {xm _u, um _u} and $\{xm_u, \overline {u}m_u\}$. Assume that both edges are in T. There are in L(G) two adjacent vertices u _i u _i+1 and u _i+1 u _i+2 of the cycle C ₉ of variable u with distance 9 in T, a contradiction. Now, assume that both edges are not in T. We consider two cases. If there are no edges $p_{u_4}m_u, um_u$ or $p_{u_4}m_u, \overline {u}m_u$, then there are in L(G) two adjacent vertices $p_{u_4}m_u$ and um _u (or $\overline {u}m_u$) with distance at least 9 in T, since it is necessary to make a path passing through xx′, a contradiction. Otherwise, there is an edge $p_{u_4}m_u,um_u$ or $p_{u_4}m_u, \overline {u}m_u$. In both cases, let c ₁ = (u, v, w) be a clause that contains u, there are in L(G) two adjacent vertices c ₁ v, vv ₁ that have distance at least 9 in T, a contradiction.

Hence, relate the edge {xm _u, um _u} or $\{xm_u, m_u\overline {u}\}$ in T for each variable u ∈ U to a true assignment to the literal u or $\overline {u}$. Assume that there is a clause with three false literals c ₃ = (x, y, z). No matter how we connect the vertices $c_3^{\prime }c_3$, c ₃ x, c ₃ y and c ₃ z in T, two of them have distance at least 9 in T, a contradiction. Therefore, each clause has at least one true literal, and this is a truth assignment of I. □

Construction 3 can be adapted in order to prove that edge 2k-admissibility is NP-complete, for k ≥ 5. It can be obtained by subdividing the edge m _u x and the cycles corresponding to each variable u.

6 Concluding Remarks

We have obtained the edge stretch index of some graph classes, or equivalently, the stretch index of line graphs, such as gridline graphs (line graphs of bipartite graphs); complement of Kneser graphs KG _n,2 (line graphs of complete graphs); and line graphs of (k, ℓ)-graphs. Although deciding the 3-admissibility is open for more than 20 years, we characterize the edge 3-admissible graphs in polynomial time, and we also prove that edge 8-admissibility is NP-complete, even for bipartite graphs. Hence, some open questions arise, such as determine the computational complexity of edge t-admissibility for 4 ≤ t ≤ 7, and t = 2k + 1, k ≥ 4.