Isometric path complexity of graphs

A set $S$ of isometric paths of a graph $G$ is"$v$-rooted", where $v$ is a vertex of $G$, if $v$ is one of the end-vertices of all the isometric paths in $S$. The isometric path complexity of a graph $G$, denoted by $ipco(G)$, is the minimum integer $k$ such that there exists a vertex $v\in V(G)$ satisfying the following property: the vertices of any isometric path $P$ of $G$ can be covered by $k$ many $v$-rooted isometric paths. First, we provide an $O(n^2 m)$-time algorithm to compute the isometric path complexity of a graph with $n$ vertices and $m$ edges. Then we show that the isometric path complexity remains bounded for graphs in three seemingly unrelated graph classes, namely, hyperbolic graphs, (theta, prism, pyramid)-free graphs, and outerstring graphs. Hyperbolic graphs are extensively studied in Metric Graph Theory. The class of (theta, prism, pyramid)-free graphs are extensively studied in Structural Graph Theory, e.g. in the context of the Strong Perfect Graph Theorem. The class of outerstring graphs is studied in Geometric Graph Theory and Computational Geometry. Our results also show that the distance functions of these (structurally) different graph classes are more similar than previously thought. There is a direct algorithmic consequence of having small isometric path complexity. Specifically, we show that if the isometric path complexity of a graph $G$ is bounded by a constant, then there exists a polynomial-time constant-factor approximation algorithm for ISOMETRIC PATH COVER, whose objective is to cover all vertices of a graph with a minimum number of isometric paths. This applies to all the above graph classes.


Introduction
A path is isometric if it is a shortest path between its endpoints.An isometric path cover of a graph G is a set of isometric paths such that each vertex of G belongs to at least one of the paths.The isometric path number of G is the smallest size of an isometric path cover of G. Given a graph G and an integer k, the objective of the algorithmic problem Isometric Path Cover is to decide if there exists an isometric path cover of cardinality at most k.Isometric Path Cover has been introduced and studied in the context of pursuit-evasion games [2,3].However, until recently the algorithmic aspects of Isometric Path Cover remained unexplored.After proving that Isometric Path Cover remains NP-hard on chordal graphs (graphs without any induced cycle of length at least 4), Chakraborty et al. [8] provided constant-factor approximation algorithms for many graph classes, including interval graphs, chordal graphs, and more generally, graphs with bounded treelength.To prove the approximation ratio of their algorithm, the authors introduced a parameter called isometric path antichain cover number of a graph G, denoted as ipacc (G) (see Definition 6), and proved (i) when ipacc (G) is bounded by a constant, Isometric Path Cover admits a constant-factor approximation algorithm on G; and (ii) the isometric path antichain cover number of graphs with bounded treelength is bounded.
The objectives of this paper are three fold: (A) provide a more intuitive definition of isometric path antichain cover number; (B) provide a polynomial-time algorithm to compute ipacc (G); and (C) prove that it remains bounded for seemingly unrelated graph classes.Along the way, we also extend the horizon of approximability of Isometric Path Cover.To achieve (A) we introduce the following new metric graph parameter, that we will show to be always equal to the isometric path antichain cover number, and whose definition is simpler.Definition 1.Given a graph G and a vertex v of G, a set S of isometric paths of G is v-rooted if v is one of the end-vertices of all the isometric paths in S. The isometric path complexity of a graph G, denoted by ipco (G), is the minimum integer k such that there exists a vertex v ∈ V (G) satisfying the following property: the vertices of any isometric path P of G can be covered by k many v-rooted isometric paths.
A consequence of Dilworth's theorem is that for any graph G, ipacc (G) = ipco (G) (see Lemma 7).We will give a polynomial-time algorithm to compute ipco (G), and therefore ipacc (G) for an arbitrary undirected graph G.This achieves (B).Finally, to achieve (C), we consider the following three seemingly unrelated graph classes, namely, δ-hyperbolic graphs, (theta, prism, pyramid)-free graphs and outerstring graphs, and show that their isometric path complexity is bounded by a constant.δ-hyperbolic graphs: A graph G is said to be δ-hyperbolic [21] if for any four vertices u, v, x, y, the two larger of the three distance sums d (u, v) + d (x, y), d (u, x) + d (v, y) and d (u, y)+ d (v, x) differ by at most 2δ.A graph class G is hyperbolic if there exists a constant δ such that every graph G ∈ G is δ-hyperbolic.This parameter comes from geometric group theory and was first introduced by Gromov [21] in order to study groups via their Cayley graphs.The hyperbolicity of a tree is 0, and in general, the hyperbolicity measures how much the distance function of a graph deviates from a tree metric.Many structurally defined graph classes like chordal graphs, cocomparability graphs [14], asteroidal-triple free graphs [15], graphs with bounded chordality or treelength are hyperbolic [9,23].Moreover, hyperbolicity has been found to capture important properties of several large practical graphs such as the Internet graph [26] or database relation graphs [31].Due to its importance in discrete mathematics, algorithms, metric graph theory, researchers have studied various algorithmic aspects of hyperbolic graphs [9,16,11,17].Note that graphs with diameter 2 are hyperbolic, which may contain any graph as an induced subgraph.
(theta, prism, pyramid)-free graphs: A theta is a graph made of three vertex-disjoint induced paths P 1 = a . . .b, P 2 = a . . .b, P 3 = a . . .b of lengths at least 2, and such that no edges exist between the paths except the three edges incident to a and the three edges incident to b.A pyramid is a graph made of three induced paths P 1 = a . . .b 1 , P 2 = a . . .b 2 , P 3 = a . . .b 3 , two of which have lengths at least 2, vertex-disjoint except at a, and such that b 1 b 2 b 3 is a triangle and no edges exist between the paths except those of the triangle and the three edges incident to a.A prism is a graph made of three vertex-disjoint induced paths P 1 = a 1 . . .b 1 , P 2 = a 2 . . .b 2 , P 3 = a 3 . . .b 3 of lengths at least 1, such that a 1 a 2 a 3 and b 1 b 2 b 3 are triangles and no edges exist between the paths except those of the two triangles.A graph G is (theta, pyramid, prism)-free if G does not contain any induced subgraph isomorphic to a theta, pyramid or prism.A graph is a 3-path configuration if it is a theta, pyramid or prism.The study of 3-path configurations dates back to the works of Watkins and Meisner [32] in 1967 and plays "special roles" in the proof of the celebrated Strong Perfect Graph Theorem [12,19,28,30].Important graph classes like chordal graphs, circular arc graphs, universally-signable graphs [13] exclude all 3-path configurations.Popular graph classes like perfect graphs, even hole-free graphs exclude some of the 3-path configurations.Note that, (theta, prism, pyramid)-free graphs are not hyperbolic.To see this, consider a cycle C of order n.Clearly, C excludes all 3-path configurations and has hyperbolicity Ω(n).
Outerstring graphs: A set S of simple curves on the plane is grounded if there exists a horizontal line containing one endpoint of each of the curves in S. A graph G is an outerstring graph if there is a collection C of grounded simple curves and a bijection between V (G) and C such that two curves in S intersect if and only if the corresponding vertices are adjacent in G.The term "outerstring graph" was first used in the early 90's [24] in the context of studying intersection graphs of simple curves on the plane.Many well-known graph classes like chordal graphs, circular arc graphs [20], circle graphs (intersection graphs of chords of a circle [18]), or cocomparability graphs [14] are also outerstring graphs and thus, motivated researchers from the geometric graph theory and computational geometry communities to study algorithmic and structural aspects of outerstring graphs and its subclasses [4,5,6,22,25].Note that, in general, outerstring graphs may contain a prism, pyramid or theta as an induced subgraph.Moreover, cycles of arbitrary order are outerstring graphs, implying that outerstring graphs are not hyperbolic.
It is clear from the above discussion that the classes of hyperbolic graphs, (theta, prism, pyramid)-free graphs, and outerstring graphs are pairwise incomparable (with respect to the containment relationship).We show that the isometric path complexities of all the above graph classes are small.

Our contributions
The main technical contribution of this paper are as follows.First we prove that the isometric path complexity can be computed in polynomial time.
Theorem 2. Given a graph G with n vertices and m edges, it is possible to compute ipco (G) in O(n 2 m) time.
Recall that, the above theorem and Lemma 7 imply that for any undirected graph G, ipacc (G) can be computed in polynomial time.Then we show that the isometric path complexity remains bounded on hyperbolic graphs, (theta, pyramid, prism)-free graphs, and outerstring graphs.Specifically, we prove the following theorem.
To the best of our knowledge, the isometric path complexity being bounded (by constant(s)) is the only known non-trivial property shared by any two or all three of these graph classes.Theorem 3 shows that isometric path complexity (equivalently isometric path antichain cover number), as recently introduced graph parameters, are general enough to unite these three graph classes by their metric properties.We hope that this definition will be useful for the field of metric graph theory, for example by enabling us to study (theta,prism,pyramid)-free graphs and outerstring graphs from the perspective of metric graph theory.
We provide a unified proof for Theorem 3(b) and 3(c) by proving that the isometric path complexity of (t-theta, t-pyramid, t-prism)-free graphs [29] (see Section 4 for a definition) is bounded by a linear function of t.Due to the above theorems, we also have as corollaries that there is a polynomial-time approximation algorithm for Isometric Path Cover with approximation ratio (a) 4δ+3 on δ-hyperbolic graphs, (b) 73 on (theta, prism, pyramid)-free graphs, (c) 95 on outerstring graphs, and (d) 8t + 63 on (t-theta, t-pyramid, t-prism)-free graphs.
To contrast with Theorem 3, we construct highly structured graphs with small treewidth and large isometric path complexity.A wheel consists of an induced cycle C of order at least 4 and a vertex w / ∈ V (C) adjacent to at least three vertices of C. The three path configurations introduced earlier and the wheel together are called Truemper configurations [30] and they are important objects of study in structural and algorithmic graph theory [1,19] Organisation.In Section 2, we recall some definitions and some results.In Section 3, we present an algorithm to compute the isometric path complexity of a graph and prove Theorem 2. In Section 4, we prove Theorem 3. In Section 5, we prove Theorem 4. We conclude in Section 6.

Definitions and preliminary observations
In this section, we recall some definitions and some related observations.A sequence of distinct vertices forms a path P if any two consecutive vertices are adjacent.Whenever we fix a path P of G, we shall refer to the subgraph formed by the edges between the consecutive vertices of P .The length of a path P , denoted by |P |, is the number of its vertices minus one.A path is induced if there are no graph edges joining non-consecutive vertices.A path is isometric if it is a shortest path between its endpoints.For two vertices u, v of a graph G, d (u, v) denotes the length of an isometric path between u and v.
In a directed graph, a directed path is a path in which all arcs are oriented in the same direction.For a path P of a graph G between two vertices u and v, the vertices V (P ) \ {u, v} are internal vertices of P .A path between two vertices u and v is called a (u, v)-path.Similarly, we have the notions of isometric (u, v)-path and induced (u, v)-path.The interval I(u, v) between two vertices u and v consists of all vertices that belong to an isometric (u, v)-path.For a vertex r of G and a set S of vertices of G, the distance of S from r, denoted as d (r, S), is the minimum of the distance between any vertex of S and r.For a subgraph For a graph G and a vertex r ∈ V (G), consider the following operations on G. First, remove all edges xy from G such that d (r, x) = d (r, y).Let G ′ r be the resulting graph.Then, for each edge e = xy ∈ E(G ′ r ) with d (r, x) = d (r, y) − 1, orient e from y to x.Let − → G r be the directed acyclic graph formed after applying the above operation on G ′ .Note that this digraph can easily be computed in linear time using a Breadth-First Search (BFS) traversal with starting vertex r.
The known approximation algorithm for Isometric Path Cover from [8] can now be stated as follows: (i) For each vertex r ∈ V (G), compute − → G r and find a minimum path cover C r of − → G r , and then (ii) report a C r with minimum cardinality.The following definition is inspired by the terminology of posets (as the graph − → G r can be seen as the Hasse diagram of a poset) and will be useful to analyze the above algorithm.Definition 5.For a graph G and a vertex r ∈ V (G), two vertices x, y ∈ V (G) are antichain vertices if there are no directed paths from x to y or from y to x in − → G r .A set X of vertices of G is an antichain set if any two vertices in X are antichain vertices.

Definition 6 ([8]). Let r be a vertex of a graph G. For a subgraph H, A r (H) shall denote the maximum antichain set of
The isometric path antichain cover number of graph G, denoted as ipacc (G), is defined as the minimum over all possible antichain covers of its associated directed acyclic graphs: For technical purposes, we also introduce the following definition.For a graph G and a vertex r of G, let ipco − → G r denote the minimum integer k such that any isometric path P of G can be covered by k r-rooted isometric paths (The notation reflects that it is a dual notion of ipacc − → G r ).Using Dilworth's Theorem we prove the following important lemma.

Lemma 7. For any graph G and vertex r, ipco
Proof.Let r be a vertex of G such that any isometric path of G can be covered by ipco − → G r r-rooted isometric paths.Let P be an arbitrary isometric path of G. Since two vertices of an antichain of − → G r cannot be covered by a single r-rooted path and P is covered by ipco − → G r r-rooted path, we deduce Conversely, consider a vertex r ∈ V (G).By definition of ipco − → G r , there is an isometric path P that cannot be covered by (ipco − → G r − 1) r-rooted isometric paths.By Dilworth theorem, P contains an The second part of the lemma follows immediately.
We also recall the following theorem and proposition from [8].

Proposition 9 ([8]
).Let G be a graph and r, an arbitrary vertex of G. Consider the directed acyclic graph − → G r , and let P be an isometric path between two vertices x and y in G.
Proof.Orient the edges of P from y to x in G. First, observe that P must contain a set E 1 of oriented edges such that Let the vertices of the largest antichain set of P in − → G r , i.e., A r (P ), be ordered as a 1 , a 2 , . . ., a t according to their occurrence while traversing P from y to x.For i ∈ [2, t], let P i be the subpath of P between a i−1 and a i .Observe that for any i ∈ [2, t], since a i and a i−1 are antichain vertices, there must exist an oriented edge

Proof of Theorem 2
In this section we provide a polynomial-time algorithm to compute the isometric path complexity of a graph.Let G be a graph.In the following lemma, we provide a necessary and sufficient condition for two vertices of an isometric path to be covered by the same isometric r-rooted path in − → G r for some vertex r ∈ V (G).
Lemma 10.Let r be vertex of G.
r) + 1 for every i ∈ {0, . . ., k − 1} then the path obtained by concatenating an isometric (r, u)-path and the path P is an isometric r-rooted (r, v)-path containing u, v in − → G r (P ).Now suppose that there exists an isometric r-rooted path containing u, v in − → G r (P ), i.e., d (r, v) − d (r, u) = d (u, v) .Then, along any path from u to v, we need to traverse at least d (u, v) edges increasing the distance to r.Since P is an isometric (u, v)-path, it contains exactly d (u, v) edges.Hence, d (r, v i+1 ) = d (r, v i ) + 1 for every i ∈ {0, . . ., k − 1}.

Notations and preliminary observations
We now introduce some notations that will be used to describe the algorithm and prove its correctness.Consider three vertices r, x, v of G such that x = v.Let P r ց (x, v) denote the set of all isometric (x, v)paths P containing a vertex u that is adjacent to v and satisfies d (r, u) = d (r, v) − 1. Analogously, let P r → (x, v) denote the set of all isometric (x, v)-paths P containing a vertex u that is adjacent to v and satisfies d (r, u) = d (r, v) and let P r ր (x, v) denote the set of all isometric (x, v)-paths P containing a vertex u that is adjacent to v and satisfies d (r, u) = d (r, v) + 1. Observe that the set of isometric (x, v)-paths is precisely P r ց (x, v) ∪ P r → (x, v) ∪ P r ր (x, v) and that some of these sets may be empty.Given a path P , we denote by |S r (P ) | the minimum size of a set of isometric r-rooted paths covering the vertices of P .We denote by γ r ց (x, v) and β r ց (x, v) respectively the minimum of |S r (P ) | and |S r (P − {v}) | over all paths P ∈ P r ց (x, v).More formally, Finally, let γ r (x, v) = max γ r ց (x, v), γ r → (x, v), γ r ր (x, v) be the maximum of |S r (P )| over all isometric (x, v)-paths P .In our algorithm, we will need also to consider the case where v = x as an initial case.For practical reasons, we let γ r (x, Based on the above notations and Lemma 7, we have the following observation.
Observation 11.For any graph G and any vertex r of G, we have ipco Observation 11 implies that to compute the isometric path complexity of a graph it is enough to compute the parameter γ r (x, v) for all r, x, v ∈ V (G) in polynomial time.In the next section, we focus on achieving this goal without computing explicitly any of the sets P r ց (x, v), P r → (x, v) or P r ր (x, v).(Note that the size of these sets could be exponential in the number of vertices of the graph).

An algorithm to compute γ r (x, v)
Throughout this section, let r and x be two fixed vertices of G.We shall call r as the "root" and x as the "source" vertex.The objective of this section is to compute the parameter γ r (x, v) for all vertices v ∈ V (G).
In the sequel, since we always refer to a fixed root r and source x, we omit r and x and use the shorthand γ(v) for γ r (x, v).We do the same with the notations γ ր (v), γ → (v), γ ց (v), β ր (v), and β ց (v) that also refer to fixed vertices r and x In the following lemmas, we shall provide explicit (recursive) formulas to compute γ ր (v), γ → (v), γ ց (v), β ր (v), and β ց (v).Using these formulas, we will show how to compute γ(v) for all v ∈ V (G) in a total of O(|E(G)|)-time.
Observation 12.If r is the root vertex, x the source vertex, and v is distinct from x, then Lemma 13.If r is the root vertex, x the source vertex, and Proof.Observe that P r → (x, v) is empty if and only if there is no vertex ) is empty, then γ → (v) = 0 and we are done.Suppose now that P r → (x, v) = ∅.
and consider a set S of isometric r-rooted paths covering the vertices of Q of size |S r (Q) | and a (r, v i )-shortest path P i .Observe that S ∪ {P i } is a set of isometric r-rooted paths covering the vertices of Consider now an isometric (x, Let P ′ be the isometric (x, v i )-path obtained by appending v i to Q ′ .Consider a set S ′ of isometric r-rooted paths covering the vertices of P ′ of size |S r (P ′ ) | and let P ′ i be a path of S ′ covering v i .By Lemma 10, no vertex of Q ′ is covered by ) is empty, then γ ց (v) = 0 and we are done.Assume now that P r ց (x, v) = ∅.If v is adjacent to x, then P = (x, v) is the unique isometric (x, v)-path, and since P r ց (x, v) = ∅, we have d (r, x) = d (r, v) − 1.Then P can be covered by any isometric (r, v)-path containing x, and thus Assume now that v is not adjacent to x.Let P = (x = v 0 , . . ., v i−1 , v i = v) be a path such that |S r (P ) | = γ ց (v), let Q = (v 0 , . . ., v i−1 ), and let R = (v 0 , . . ., v i−2 ).Note that d (r, First suppose that d (r, v i−2 ) = d (r, v i−1 ) − 1.We claim that |S r (P ) | ≤ |S r (Q) |.Indeed, consider a set S of isometric r-rooted paths covering the vertices of Q of size |S r (Q) |.Let P i−1 ∈ S be a path covering v i−1 .By Lemma 10 and since d (r, v i−2 ) = d (r, v i−1 )−1, we can assume that P i−1 is an isometric (r, v i−1 )-path.Consider the path P i obtained by appending v i at the end of P i−1 and observe that P i is an isometric (r, v i )-path covering the same vertices as P i−1 as well as v i .Consequently, replacing P i−1 by P i in S, we obtain a set of isometric r-rooted paths of size . As in the previous case, we show that |S r (P ) | ≤ |S r (Q) |.Indeed, consider a set S of isometric r-rooted paths covering the vertices of Q of size |S r (Q) |.Let P i−1 ∈ S be a path covering v i−1 .By Lemma 10 and since d (r, Consider a set S of isometric r-rooted paths covering the vertices of R of size |S r (R) | and a (r, v i )-shortest path P i containing v i−1 .Observe that S ∪ {P i } is a set of isometric r-rooted paths covering the vertices of P .Consequently, . Let P ′ be the isometric (x, v i )-path obtained by appending v i to Q ′ .Consider a set S ′ of isometric r-rooted paths covering the vertices of P ′ of size |S r (P ′ ) | and let P ′ i be the path of S ′ covering v i .By Lemma 10, the only vertex of Q ′ that can be covered by Since the formula for computing γ ց (v) (given in the statement of the lemma) takes into account these three exclusive alternatives, it computes γ ց (v) correctly.Lemma 15.If r is the root vertex, x the source vertex, and v is a vertex distinct from x, then γ ր Proof.The proof is similar to the the proof of Lemma 14.Now we provide a BFS based algorithm to compute the above parameters.Let r and x be fixed root and source vertices of G, respectively.For a vertex u ∈ V (G), let D(u) = {γ(u), γ ր (u), γ → (u), γ ց (u), β ր (u), β ց (u)}.Clearly, the set D(x) can be computed in constant time.Now let X i be the set of vertices at distance i from x. Clearly, the sets X i can be computed in O(|E(G)|)-time (using a BFS) and X 0 = {x}.Let i ≥ 1 be an integer and assume that for all vertices u ∈ i−1 j=0 X j , the set D(u) is already computed.Let v ∈ X i be a vertex.Then due to the formulas given in Observation 12 and Lemmas 13-15, the set D(v) can be computed by observing only the sets Hence, for all vertices v ∈ V (G), the sets D(v) can be computed in a total of O(|E(G)|) time.Hence, we have the following lemma.
Lemma 16.For a root vertex r and source vertex x, for each vertex v ∈ V (G), the value γ r (x, v) can be computed in O(|E(G)|) time.
We can now finish the proof of Theorem 2. Let G be a graph with n vertices and m edges.For a root vertex r, by applying Lemma 16, for every source By repeating this for every root r ∈ V (G), it is possible to compute

Proof of Theorem 3
First we prove Theorem 3(a).We recall the definition of Gromov products [21] and its relation with hyperbolicity.For three vertices r, x, y of a graph G, the Gromov product of x, y with respect to r is defined as Then, a graph G is δ-hyperbolic [10,21] if and only if for any four vertices x, y, z, r, we have (x|y) r ≥ min {(x|z) r , (y|z) r } − δ.
Let G be a graph with hyperbolicity at most δ.Due to Lemma 7, in order to prove Theorem 3(a), it is enough to show that ipacc (G) ≤ 4δ + 3. Aiming for a contradiction, let r be a vertex of G and P be an isometric path such that |A r (P ) | ≥ 4δ + 4. Let a 1 , a 2 , . . ., a 2δ+2 , . . ., a 4δ+4 be the vertices of A r (P ) ordered as they are encountered while traversing P from one end-vertex to the other.Let x = a 1 , z = a 2δ+2 , y = a 4δ+4 .Let Q denote the (y, z)-subpath of P .Observe that, |A r (Q) | ≥ 2δ + 2. Then we have (x|y) r ≥ min {(x|z) r , (y|z) r } − δ.Without loss of generality, assume that (x|z) r ≤ (y|z) r .Hence, But this directly contradicts Proposition 9, which implies that This completes the proof of Theorem 3(a).Now, we shall prove Theorems 3(b) and 3(c).First, we shall define the notions of t-theta, t-prism, and t-pyramid [29].For an integer t ≥ 1, a t-prism is a graph made of three vertex-disjoint induced paths P 1 = a 1 . . .b 1 , P 2 = a 2 . . .b 2 , P 3 = a 3 . . .b 3 of lengths at least t, such that a 1 a 2 a 3 and b 1 b 2 b 3 are triangles and no edges exist between the paths except those of the two triangles.For an integer t ≥ 1, a t-pyramid is a graph made of three induced paths P 1 = a . . .b 1 , P 2 = a . . .b 2 , P 3 = a . . .b 3 of lengths at least t, two of which have lengths at least t + 1, they are pairwise vertex-disjoint except at a, such that b 1 b 2 b 3 is a triangle and no edges exist between the paths except those of the triangle and the three edges incident to a.For an integer t ≥ 1, a t-theta is a graph made of three internally vertex-disjoint induced paths P 1 = a . . .b, P 2 = a . . .b, P 3 = a . . .b of lengths at least t + 1, and such that no edges exist between the paths except the three edges incident to a and the three edges incident to b.A graph G is (t-theta, t-pyramid, t-prism)-free if G does not contain any induced subgraph isomorphic to a t-theta, t-pyramid or t-prism.When t = 1, (t-theta, t-pyramid, t-prism)-free graphs are exactly (theta, prism, pyramid)-free graphs.Now, we shall show that the isometric path antichain cover number of (t-theta, t-pyramid, t-prism)free graphs are bounded above by a linear function on t.We shall show that, when the isometric path antichain cover number of a graph is large, the existence of a structure called "t-fat turtle" (defined below) as an induced subgraph is forced, which, cannot be present in a ((t − 1)-theta, (t − 1)-pyramid, (t − 1)-prism)-free graph.
Definition 17.For an integer t ≥ 1, a "t-fat turtle" consists of a cycle C and an induced (u, v)-path P of length at least t such that all of the following hold: (a) V (P ) ∩ V (C) = ∅, (b) For any vertex w ∈ (V (P ) \ {u, v}), N (w) ∩ V (C) = ∅ and both u and v have at least one neighbour in C, (c) For any vertex w ∈ N (u) ∩ V (C) and w ′ ∈ N (v) ∩ V (C), the distance between w and w ′ in C is at least t, (d) There exist two vertices {c, The tuple (t, C, P, c, c ′ ) defines the t-fat turtle.See Figure 2 for an example.
In the following observation, we show that any (t-theta, t-pyramid,t-prism)-free graph cannot contain a (t + 1)-fat turtle as an induced subgraph.
Lemma 18.For some integer t ≥ 1, let G be a graph containing a (t + 1)-fat turtle as an induced subgraph.Then G is not (t-theta, t-pyramid, t-prism)-free.
Proof.Let (t+1, C, P, c, c ′ ) be a (t+1)-fat turtle in G. Let the vertices of C be named c = a 0 , a 1 , . . ., a k = c ′ , a k+1 , . . ., a |V (C)| as they are encountered while traversing C starting from c in a counter-clockwise manner.Denote by u, v the end-vertices of P .By definition, there exist two distinct components Let i − and i + be the minimum and maximum indices such that a i − and a i + are adjacent to u.Let j − and j + be the minimum and maximum indices such that a j − and a j + are adjacent to v. By definition, i − ≤ i + < j − ≤ j + .Let P 1 be the (a i − , a j + )-subpath of C containing c. Let P 2 be the (a i + , a j − )-subpath of C that contains c ′ .Observe that P 1 and P 2 have length at least t (by definition).Now we show that P, P 1 , P 2 together form one of theta, pyramid or prism.If a i − = a i + and a j − = a j + , then P, P 1 , P 2 form a t-theta.If i − ≤ i + − 2 and j − ≤ j + − 2, then also P, P 1 , P 2 form a t-theta.If j − = j + − 1 and i − = i + − 1, then P, P 1 , P 2 form a t-prism.In any other case, P, P 1 , P 2 form a t-pyramid.
In the remainder of this section, we shall prove that there exists a linear function f (t) such that if the isometric path antichain cover number of a graph is more than f (t), then G is forced to contain a (t + 1)-fat turtle as an induced subgraph, and therefore is not (t-theta, t-pyramid,t-prism)-free.We shall use the following observation.
Observation 19.Let G be a graph, r be an arbitrary vertex, P be an isometric (u, v)-path in G and Q be a subpath of an isometric (v, r)-path in G such that one endpoint of Q is v.Let P ′ be the maximum (u, w)-subpath of P such that no internal vertex of P ′ is a neighbour of some vertex of Q.We have that |A   Let Q (r, u) be any isometric (u, r)-path of G. Observe that w is not adjacent to any vertex of Q (r, u).Otherwise, d (u, w) ≤ 2+d (r, u)−d (r, w), which contradicts Proposition 9. Let P (z, w) be the maximum (z, w)-subpath of P (u, w) such that no internal vertex of P (z, w) has a neighbour in Q (r, u).Observe that P (z, w) is an isometric path, and z has a neighbour in Q (r, u).Again applying Observation 19, we have the following: Let a 1 , a 2 , . . ., a k be the vertices of A r (P (z, w)) ordered according to their appearance while traversing P (z, w) from z to w. Due to Claim 20.2, we have that k ≥ 4t + 26.Let c = a 2t+13 and Q (r, c) denote an isometric (c, r)-path.Let T (r, c 1 ) be the maximum subpath of Q (r, c) such that no internal vertex of T (r, c 1 ) is adjacent to any vertex of P (z, w).Observe that neither z nor w can be adjacent to c 1 (due to Proposition 9).Morevoer, if c 1 is a vertex of P (z, w) then we must have c 1 = c.Proof.Let P (c, w) denote the (c, w)-subpath of P (z, w).Observe that |A r (P (c, w)) | ≥ 2t + 14.First, consider the case when x lies in the (z, c)-subpath of P (z, w).In this case, P (c, w) is a subpath of X and therefore |A r (X) | ≥ 2t + 14.Now consider the case when x lies in P (c, w).In this case, applying Observation 19, we have that |A • For a vertex e ′ 1 ∈ N (e 1 ) ∩ V (Z) and e ′ 2 ∈ N (e 2 ) ∩ V (Z), the distance between e ′ 1 and e ′ 2 is at least t + 1.This statement follows from Claim 20.Proof.To prove the lemma, we shall need to recall a few definitions and results from the literature.A graph G is a string graph if there is a collection S of simple curves on the plane and a bijection between V (G) and S such that two curves in S intersect if and only if the corresponding vertices are adjacent in G. Let G be a graph with an edge e.The graph G/e is obtained by contracting the edge e into a single vertex.Observe that string graphs are closed under edge contraction [24].We shall use the following result.
Proposition 22 ([24]).Let G be an outerstring graph with an edge e.Then G/e is an outerstring graph.
A full subdivision of a graph is a graph obtained by replacing each edge of G with a new path of length at least 2. We shall use the following result implied from Theorem 1 of [24].
Proposition 23 ([24]).Let G be a string graph.Then G does not contain a full subdivision of K 3,3 as an induced subgraph.
For a graph G, the graph G + is constructed by introducing a new apex vertex a and connecting a with all vertices of G by new copies of paths of length at least 2. We shall use the following result of Biedl et al. [4].
Proposition 24 (Lemma 1, [4]).A graph G is an outerstring graph if and only if G + is a string graph.Now we are ready to prove the lemma.Let G be an outerstring graph.Assume for the sake of contradiction that G contains an induced subgraph H which is a 4-theta, 4-pyramid, or a 4-prism.Since every induced subgraph of an outerstring graph is also an outerstring graph, we have that H is an outerstring graph.Let E be the set of edges of H whose both endpoints are part of some triangle.Now consider the graph H 1 = H/E which is obtained by contracting all edges in E. By Proposition 22, H 1 is an outerstring graph and it is easy to check that H 1 is a 3-theta.Let u and v be the vertices of H 1 with degree 3 and w 1 , w 2 , w 3 be the set of mutually non-adjacent vertices such that for each i ∈ {1, 2, 3} d (u, w i ) = 2 and d (v, w i ) ≥ 2. Since H 1 is a 3-theta, w 1 , w 2 , w 3 exist.Now consider the graph H + 1 and a be the new apex vertex.Due to Proposition 24, we have that H + 1 is a string graph.But notice that, for each pair of vertices in {x, y} ⊂ {w 1 , w 2 , w 3 , u, v, a}, there exists a unique path of length at least 2 connecting x, y.This implies that H + 1 (which is a string graph) contains a full subdivision of K 3,3 , which contradicts Proposition 23.

Proof of Theorem 4
We shall provide a construction for every k ≥ 4, this implies the statement of Theorem 4 for any k ≥ 1.First we shall prove Theorem 4(a).For a fixed integer k ≥ 4, first we describe the construction of a graph X k as follows.Consider k + 1 paths P 1 , P 2 , . . ., P k+1 each of length k and having a common endvertex a.For i ∈ [k + 1], let the other endvertex of P i be denoted as b i .Moreover, for i ∈ [k + 1], let the neighbours of a and b i in P i be denoted as a ′ i and b ′ i , respectively.For i ∈ [k], introduce an edge between b i and b i+1 .The resulting graph is denoted X k and the special vertex a is the apex of X k .See Figure 4 We shall use the following result relating hyperbolicity and isometric cycles.An induced cycle C of a graph G is isometric if for any two vertices u, v of C, the distance between u, v in C is the same as that in G.

Proposition 25 ([33]
).Let G be a graph containing an isometric cycle of order k with k ≡ c (mod 4).Then the hyperbolicity of G is at least ⌈ k 4 ⌉ − 1 2 if c = 1 and ⌈ k 4 ⌉, otherwise.We now prove the following lemmas.Proof.Since removing the special vertex a from G results in a graph with treewidth 2, it follows that G has treewidth at most 3. Observe that G has an isometric cycle of length at least 2k.Therefore, due to Proposition 25, G has hyperbolicity at least ⌈ k 2 ⌉ − 1.Let H and H ′ denote the two copies of Y k used to construct G. First we shall show that H does not contain a theta or a prism.Consider the graph H 1 obtained by removing the apex of H. Observe that H 1 does not contain a vertex v such that the vertices in N [v] induce a K 1,3 .Hence H does not contain a theta.It also can be verified that H 1 does not contain a prism.Since the neighbourhood of a is triangle-free, it follows that H does not contain a prism.Similarly, H ′ does not contain a theta or a prism.Now, from our construction, it follows that G does not contain a theta or a prism.Moreover, for any induced cycle C of G, and any vertex w / ∈ C, observe that w has at most two neighbours in C. Therefore, G is wheel-free.Using arguments similar to the ones used in the proof of Lemma 26, we have that ipacc (G) ≥ k.Observe that H does not contain a vertex v such that the vertices in N [v] induce a K 1,3 .Therefore, H does not contain a theta or a pyramid.Similarly, H ′ does not contain a theta or a pyramid.Due to our construction, it follows that G does not contain a theta or a pyramid.Moreover, for any induced cycle C of G, and any vertex w / ∈ C, observe that w has at most two neighbours in C. Therefore, G is wheel-free.Using arguments similar to the ones used in the proof of Lemma 26, we have that ipacc (G) ≥ k.
An isometric path cover C of a graph G is rooted if there exists a vertex v such that all paths in C are v-rooted isometric paths.
Lemma 29.For k ≥ 4, let G be the graph constructed by taking two distinct copies of W k and identifying the two apex vertices.Then G is a (prism, pyramid, wheel)-free planar graph such that any rooted isometric path cover of G has cardinality at least k 2 but there is an isometric path cover of G of cardinality 3k + 1.
Proof.The construction ensures that G is a (prism, pyramid, wheel)-free planar graph.Let H and H ′ denote the two copies of W k used to construct G and a denote the apex vertex.Observe that there are k 2 vertices at maximum distance from the apex vertex a in H and a a-rooted isometric path can only cover one of them.Therefore, at least k 2 many a-rooted isometric paths are needed to cover the graph H.As H ′ is isomorphic to H, it has the above properties.Since a is a cut-vertex in G, it is easy to verify that for any vertex v ∈ V (G), any v-rooted isometric path cover of G requires k 2 many paths.On the other hand, it is easy to check that G has an isometric path cover of cardinality 3k + 1. Indeed k + 1 geodesics are sufficient to cover the vertices of the maximal isometric paths containing a, 2k geodesics are sufficient to cover the remaining vertices of G.

Conclusion
In this paper, we have introduced the new graph parameter isometric path complexity.We have shown that the isometric path complexity of a graph with n vertices and m edges can be computed in O(n 2 m)time.It would be interesting to provide a faster algorithm to compute the isometric path complexity of a graph.We have derived upper bounds on the isometric path complexity of three seemingly (structurally) different classes of graphs, namely hyperbolic graphs, (theta,pyramid,prism)-free graphs and outerstring graphs.An interesting direction of research is to generalise the properties of hyperbolic graphs or (theta,pyramid,prism)-free graphs to graphs with bounded isometric path complexity.
Note that, in our proofs we essentially show that, for any graph G that belongs to one of the above graph classes, any vertex v of G, and any isometric path P of G, the path P can be covered by a small number of v-rooted isometric paths.This implies our "choice of the root" is arbitrary.This motivates the following definition.The strong isometric path complexity of a graph Gis the minimum integer k such that for each vertex v ∈ V (G) we have that the vertices of any isometric path P of G can be covered by k many v-rooted isometric paths.Our proofs imply that the strong isometric path complexity of graphs from all the graph classes addressed in this paper are bounded.We also wonder whether one can find other interesting graph classes with small (strong) isometric path complexity.
Our results imply a constant-factor approximation algorithm for Isometric Path Cover on hyperbolic graphs, (theta, pyramid, prism)-free graphs and outerstring graphs.However, the existence of a constant-factor approximation algorithm for Isometric Path Cover on general graphs is not known (an O(log n)-factor approximation algorithm is designed in [27]).

Figure 1 :
Figure 1: Inclusion diagram for graph classes.If a class A has an upward path to class B, then A is included in B. Constant bounds for the isometric path complexity on graph classes marked with * are contributions of this paper.

Figure 2 :
Figure 2: An example of a 4-fat turtle.Let C be the cycle induced by the black vertices, P be the path induced by the white vertices.Then the tuple (4, C, P, c, c ′ ) defines a 4-fat turtle.

Figure 3 :
Figure 3: Illustration of the notations used in the proof of Lemma 20.
(a).For a fixed integer k ≥ 4, consider the graph X k and for each i ∈ [k], introduce an edge between b i and b ′ i+1 .Let Y k denote the resulting graph and the special vertex a is the apex of Y k .See Figure 4(b).For a fixed integer k ≥ 4, consider the graph Y k and for each {i, j} ⊆ [k], introduce an edge between a ′ i and a ′ j .Let Z k denote the resulting graph and the special vertex a is the apex of Z k .See Figure 4(c).For a fixed integer k ≥ 4, consider the graph X k .For each i ∈ [k], delete the edge b i b i+1 and introduce k new vertices, each of which is adjacent to only b i and b i+1 .Call the resulting graph W k .See Figure 4(d).

Lemma 26 . 2 ⌉ − 1 .
For k ≥ 4, let G be the graph constructed by taking two distinct copies of X k and identifying the two apex vertices.Then G is a (pyramid, prism)-free graph with treewidth 2, hyperbolicity at least⌈ k 2 ⌉ − 1 and ipacc (G) ≥ k.Proof.Since G is triangle-free, clearly G is (pyramid, prism)-free.Moreover, for any induced cycle C of G, and any vertex w / ∈ C, observe that w has only one neighbour in C. Therefore, G is also wheel-free.Observe that G has an isometric cycle of length at least 2k.Therefore, due to Proposition 25, G has hyperbolicity at least ⌈ k Since removing the vertex a from G makes it acyclic, the treewidth of G is two.Let H and H ′ denote the two copies of X k used to construct G. Let r be any vertex of G and, without loss of generality, assume that r is a vertex of H ′ .Consider the graph − → G r .Now recall the construction of H (which is isomorphic to X k ) and consider the path Q = b 1 b 2 . . .b k .Observe that Q is an isometric path and for any two vertices u, v ∈ V (Q) we have d (r, u) = d (r, v).Therefore, A r (Q) ≥ k.Hence, ipacc (G) ≥ k.Lemma 27.For k ≥ 4, let G be the graph constructed by taking two distinct copies of Y k and identifying the two apex vertices.Then G is a (theta, prism)-free graph with treewidth 3, hyperbolicity at least ⌈ k 2 ⌉−1, and ipacc (G) ≥ k.

Lemma 28 .
For k ≥ 4, let G be the graph constructed by taking two distinct copies of Z k and identifying the two apex vertices.Then G is a (theta, pyramid)-free graph with hyperbolicity at least ⌈ k 2 ⌉ − 1 and ipacc (G) ≥ k.Proof.Observe that G has an isometric cycle of length at least 2k.Therefore, due to Proposition 25, G has hyperbolicity at least ⌈ k 2 ⌉ − 1.Let H and H ′ denote the two copies of Y k used to construct G.