Bipartite powers of k-chordal graphs

Let k be an integer and k \geq 3. A graph G is k-chordal if G does not have an induced cycle of length greater than k. From the definition it is clear that 3-chordal graphs are precisely the class of chordal graphs. Duchet proved that, for every positive integer m, if G^m is chordal then so is G^{m+2}. Brandst\"adt et al. in [Andreas Brandst\"adt, Van Bang Le, and Thomas Szymczak. Duchet-type theorems for powers of HHD-free graphs. Discrete Mathematics, 177(1-3):9-16, 1997.] showed that if G^m is k-chordal, then so is G^{m+2}. Powering a bipartite graph does not preserve its bipartitedness. In order to preserve the bipartitedness of a bipartite graph while powering Chandran et al. introduced the notion of bipartite powering. This notion was introduced to aid their study of boxicity of chordal bipartite graphs. Given a bipartite graph G and an odd positive integer m, we define the graph G^{[m]} to be a bipartite graph with V(G^{[m]})=V(G) and E(G^{[m]})={(u,v) | u,v \in V(G), d_G(u,v) is odd, and d_G(u,v) \leq m}. The graph G^{[m]} is called the m-th bipartite power of G. In this paper we show that, given a bipartite graph G, if G is k-chordal then so is G^{[m]}, where k, m are positive integers such that k \geq 4 and m is odd.


Introduction
A hole is a chordless (or an induced) cycle in a graph.The chordality of a graph G, denoted by C(G), is defined to be the size of a largest hole in G, if there exists a cycle in G.If G is acyclic, then its chordality is taken as 0. A graph G is k-chordal if C(G) ≤ k.In other words, a graph is k-chordal if it has no holes with more than k vertices in it.Chordal graphs are exactly the class of 3-chordal graphs and chordal bipartite graphs are bipartite, 4-chordal graphs.k-chordal graphs have been studied in the literature in [2], [5], [6], [8], [9] and [16].For example, Chandran and Ram [5] proved that the number of minimum cuts in a k-chordal graph is at most (k+1)n Paulraja [1] proved that odd powers of chordal graphs are chordal.Chang and Nemhauser [7] showed that if G and G 2 are chordal then so are all powers of G. Duchet [10] proved a stronger result which says that if G m is chordal then so is G m+2 .Brandstädt et al. in [3] showed that if G m is k-chordal then so is G m+2 , where k ≥ 3 is an integer.Studies on families of graphs that are closed under powering can also be seen in the literature.For instance, it is known that interval graphs, proper interval graphs [14], strongly chordal graphs [13], circular-arc graphs [15] [12], cocomparability graphs [11] etc. are closed under taking powers.
Subclasses of bipartite graphs, like chordal bipartite graphs, are not closed under powering since the m-th power of a bipartite graph need not be even bipartite.Chandran et al. in [4] introduced the notion of bipartite powering to retain the bipartitedness of a bipartite graph while taking power.The m-th bipartite power G [m] of a bipartite graph G is the bipartite graph obtained from G by adding edges (u, v) where d G (u, v) is odd and less than or equal to m.Note that G [m] = G [m+1] for each odd m.It was shown in [4] that the m-th bipartite power of a tree is chordal bipartite.The intention there was to construct chordal bipartite graphs of high boxicity.The fact that the chordal bipartite graph under consideration was obtained as a bipartite power of a tree was crucial for proving that its boxicity was high.Since trees are a subclass of chordal bipartite graphs, a natural question that came up was the following: is it true that the m-th bipartite power of every chordal bipartite graph is chordal bipartite?In this paper we answer this question in the affirmative.In fact, we prove a more general result.

Our Result
Let m, k be positive integers with k ≥ 4. Let G be a bipartite graph.If G is k-chordal, then so is G [m] .Note that the special case when k = 4 gives us the following result: chordal bipartite graphs are closed under bipartite powering.

Graph Preliminaries
Throughout this paper we consider only finite, simple, undirected graphs.For a graph G, we use V (G) to denote the set of vertices of G. Let E(G) denote its edge set.For every x, y ∈ V (G), d G (x, y) represents the distance between x and y in G.For every The length of a path P is the number of edges in it and is denoted by ||P ||.A cycle C with vertex set V (C) = {v 1 , v 2 , . . ., v n }, and edge set

Holes in Bipartite Powers
Let H be a bipartite graph.Let B(H) be a family of graphs constructed from H in the following manner: Proof: Any hole of size greater than 4 in H ′ cannot have more than one vertex from the same bag, say B v , as such vertices have the same neighborhood.Hence, the vertices of a hole (of size greater than 4) in H ′ belong to different bags and thus there is a corresponding hole of the same size in H.
Proof: We prove this by contradiction.Let p denote the size of a largest induced cycle, say is bipartite).Between each u i−1 and u i , where i ∈ {0, . . ., p − 1}, there exists a shortest path of length not more than m in G (i) .Let P i be one such shortest path between u i−1 and u i in G.
Let H be the subgraph induced on the vertex set As mentioned in the beginning of this section, construct a graph H ′ from H, where H ′ ∈ B(H), in the following manner: for each , let B v have as many vertices as the number of paths in {P 0 . . .
that no two paths Q i and Q j (where i = j) share a vertex (i)  .From our construction of H ′ from H it is easy to see that such paths exist.Let The reader may also note that the cycle C (= u 0 u 1 . . .u p−1 u 0 ) which is present in G [m] will be present in H [m] and thereby in H ′[m] too.
In order to prove the theorem, it is enough to show that there exists an induced cycle of size at least p in H ′ .Then by combining Observation 3 and the fact that H is an induced subgraph of G, we get k ≥ C(G) ≥ C(H) ≥ C(H ′ ) ≥ p contradicting our assumption that p > k.Hence, in the rest of the proof we show that C(H ′ ) ≥ p.
Consider the following drawing of the graph H ′ .Arrange the vertices u 0 , u 1 , . . ., u p−1 in that order on a circle in clockwise order.Between each u i−1 and u i on the circle arrange the vertices v i,1 , v i,2 , . . ., v i,ri in that order in clockwise order.Recall that these vertices are the internal vertices of path Q ′ i .Claim 4.1.In this circular arrangement of vertices of H ′ , each vertex has an edge (in H ′ ) with both its left neighbor and right neighbor in the arrangement. Let We define the clockwise distance from x 1 to x 2 , denoted by clock dist(x 1 , x 2 ), as the minimum non-negative integer s such that j = i + s.Similarly, the clockwise distance from x 2 to x 1 , denoted by clock dist(x 2 , x 1 ), is the minimum non-negative integer s ′ such that i = j + s ′ .Let x, y, z ∈ V (H ′ ).We say y < x z if scanning the vertices of H ′ in clockwise direction along the circle starting from x, vertex y is encountered before z.
Claim 4.2.There always exists a vertex which is the farthest neighbor of x before z, unless (i) throughout this proof expressions involving subscripts of u, P , Q, and Q ′ are to be taken modulo p.Every such expression should be evaluated to a value in {0, . . ., p − 1}.For example, consider a vertex u i , where i < p Then, p + i = i.
Let {A, B} be the bipartition of the bipartite graph H ′ .We categorize the edges of H ′ as follows: an edge (x, y) ∈ E(H ′ ) is called an l-edge, if l = min(clock dist(x, y), clock dist(y, x)).Proof: Suppose H ′ has an l-edge, where l > 2, between x ∈ Q i and y ∈ Q i+l (see Fig. 1).Let We consider the following two cases: Case 1: l is even In this case u i−1 and u i+l−1 will be on the same side of the bipartite graph H ′ .Without loss of generality, let u i−1 , u i+l−1 ∈ A. Then, u i , u i+l ∈ B. We know that, for every Summing up the two inequalities we get, (a 1 + b 1 ) + (a 2 + b 2 ) ≥ 2m + 2. This implies that either ||Q ′ i || or ||Q ′ i+l || is greater than m which is a contradiction.Case 2: l is odd (proof is similar to the above case and hence omitted).
Hence we prove the claim.
We find a cycle C ′ = z 0 z 1 . . .z q z 0 in H ′ using Algorithm 3.1 (i) .Please read the algorithm before proceeding further. . (i) throughout this proof expressions involving subscripts of z are to be taken modulo q + 1.Every such expression should be evaluated to a value in {0, . . ., q}.For example, consider a vertex za, where a < q + 1.Then, q + 1 + a = a.(z a , z b ) is an l-edge.We know that (z 0 , z 1 ) is also an l-edge with Step 2 of the algorithm we know that z 0 is the first vertex (in a clockwise scan) in Q 0 which has an l-edge to a vertex in Q l .This implies that, since z 0 < za z b , z a = z 0 which is a contradiction.Hence we prove the claim.
What is left now is to show that q + 1 ≥ p, i.e., ||C ′ || ≥ ||C||, where C ′ = z 0 . . .z q z 0 and C = u 0 . . .u p−1 u 0 .In order to show this, we state and prove the following claims.Claim 4.7.For every j ∈ {0, . . ., p − 1}, (V (Q Proof: Suppose the claim is not true.Find the minimum j that violates the claim.Clearly, j = 0 as z 0 ∈ V (Q 0 ).We claim that z q ∈ V (Q j−1 ).Suppose z q / ∈ V (Q j−1 ).Let a = max{i | z i ∈ V (Q j−1 )} (note that, since j = 0, by the minimality of j, (V (Q j−1 )∪V (Q j ))∩V (C ′ ) = ∅ and therefore V (Q j−1 ) ∩ V (C ′ ) = ∅).Since z a = z q , by the maximality of a, we have ).Thus z a = z q and z a+1 is not the farthest neighbor of z a before z 0 .This is a contradiction to the way z a+1 is chosen by Algorithm 3.1.Hence, z q ∈ V (Q j−1 ).We know that (z q , z 0 ) ∈ E(H ′ ) with . Thus, we have z 1 , z q ∈ V (Q l ) and z 1 < z0 z q .But this contradicts the fact that z 1 is the last vertex in Q l encountered in a clockwise scan that has z 0 as its neighbor.
Claim 4.8.Let (z a , z a+1 ), (z b , z b+1 ) ∈ E(C ′ ) be two 2-edges, where a < b.Let P , P ′ denote the clockwise z a+1 − z b , z b+1 − z a paths respectively in C ′ .Both P and P ′ contain at least one 0-edge.Proof: Consider the path P (proof is similar in the case of path P ′ ).Path P is a non-trivial path only if z a+1 = z b .Suppose z a+1 = z b (see Fig. 2).Let z a ∈ V (Q f ).For the sake of ease of notation, assume f = 1 (the same proof works for any value of f ).Let We know that, for every ), we have, Summing up the two inequalities, we get Since b = a + s + 1 and c b+1 + d b+1 ≤ m, we get Substituting for s = 1 in Inequality Claim 4.9.For every j, j ′ ∈ {0, . . ., p − 1}, where j < j ′ and ( Proof: By Claim 4.7, (i) j ′ = j + 1 or j ′ = j − 1, and (ii) there exist r, r ′ ∈ {0, . . ., q} such that (z r , z r+1 ) is a 2-edge with its endpoints on Q j−1 and Q j+1 and (z r ′ , z r ′ +1 ) is a 2-edge with its endpoints on Q j ′ −1 and Q j ′ +1 .By Claim 4.8, we know that if P , P ′ denote the clockwise z r+1 − z r ′ , z r ′ +1 − z r paths respectively in C ′ , then both P and P ′ contains at least one 0-edge.This proves the claim.
In order to show that the size of cycle C ′ (= z 0 . . .z q z 0 ) is at least p, we consider the following three cases:- In this case, for every j ∈ {0, . . .p − 1}, Q j contributes to V (C ′ ) and therefore ||C ′ || ≥ p = ||C||.
Case |{Q j ∈ {Q 0 . . .Q p−1 } | V (Q j ) ∩ V (C ′ ) = ∅}| = 1: Let Q j be that only path (among Q 0 . . .Q p−1 ) that does not contribute to V (C ′ ).Then we claim that there exists a Q j ′ , where j ′ = j, such that V (C ′ ) ∩ V (Q j ′ ) ≥ 2. Suppose the claim is not true then it is easy to see that ||C ′ || = p − 1 which is an odd number thus contradicting the bipartitedness of H ′ .Hence the claim is true.Now, by applying the claim it is easy to see that Scan vertices of H ′ starting from any vertex in clockwise direction.Claim 4.9 ensures that between every Q j and Q j ′ , which do not contribute to V (C ′ ), encountered there exists a Q i which compensates by contributing at least two vertices to V (C ′ ).Therefore, ||C ′ || ≥ p = ||C||.

Discussion
An interesting open question that naturally follows from our result is the following: given a graph G and positive integers k, m where k ≥ 4, if G [m] is k-chordal, then is G [m+2] also k-chordal?As mentioned earlier, Brandstädt et al. in [3] showed a similar result in the context of ordinary graph powering.They showed that, for every graph G, if G m is k-chordal, then so is G m+2 , where k, m are positive integers with k ≥ 3. A straightforward extension of their proof technique doesn't seem to work in our context due to the bipartite nature of the powering that we consider.

Fig. 1 :
Fig.1:x ∈ V (Qi), y ∈ V (Q i+l ) and let (x, y) ∈ E(H ′ ) be an l-edge, where l > 2. The dotted line between ui−1 and ui indicate the path Qi.Similarly, the dotted line between u i+l−1 and u i+l indicate the path Q i+l .

Fig. 2 :
Fig. 2: Figure illustrates the case when path P defined in Claim 4.8 is a trivial path.The dotted lines between each ui−1 and ui indicate the path Q ′ i .Each continuous arc corresponds to an edge in the cycle C ′ = z0 . . .zqz0.

Fig. 3 :
Fig. 3: Figure illustrates the case when path P defined in Claim 4.8 is P = za+1za+2 . . .za+1+s, where s ≥ 1 and za+1+s = z b .The dotted lines between each ui−1 and ui indicate the path Q ′ i .Each continuous arc corresponds to an edge in the cycle C ′ = z0 . . .zqz0.