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Isometric embeddings of subdivided complete graphs in the hypercube
Electronic Notes in Discrete Mathematics 29 (2007) 277–281 www.elsevier.com/locate/endm
Isometric embeddings of subdivided complete graphs in the hypercube Laurent Beaudou 1 Sylvain Gravier 2 Institut Fourier UMR 5582 CNRS – Universit´e Joseph Fourier Grenoble, France
Kahina Meslem 3 Laboratoir LAID3 Facult´e des Math´ematiques, U.S.T.H.B. Algiers, Algeria
Abstract Isometric subgraphs of hypercubes are known as partial cubes. These graphs have first been investigated by Graham and Pollack [3], and Djokovi`c [2]. Several papers followed with various characterizations of partial cubes. In this paper, we prove that a subdivision of a complete graph of order n (n ≥ 4), is a partial cube if and only if this one is isomorphic to S(Kn ), or there exist n − 1 non-subdivided edges of Kn adjacent to a common vertex in the subdivision and the other edges of Kn are subdivided an odd number of times. Keywords: partial cubes, subdivision of a graph, isometric embeddings.
1 2 3
Email: [email protected] Email: [email protected] Email: kahina [email protected]
1571-0653/$ – see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2007.07.048
278
L. Beaudou et al. / Electronic Notes in Discrete Mathematics 29 (2007) 277–281
Introduction A subgraph H of G is called isometric if and only if dG (u, v) = dH (u, v) for all u, v ∈ V (H). Isometric subgraphs of hypercubes are called partial cubes. Partial cubes have first been investigated by Graham and Pollak [3], and Djokovi`c [2]. Later, several characterizations were shown using a relation defined on the edges set or constructive operations. Partial cubes have found different applications, for instance, in [4], interesting applications in chemical graph theory were established. Clearly, partial cubes are bipartite. The simple way to obtain a bipartite graph is to subdivide every edge of G by a single vertex. Such a graph is a subdivision of G and denoted S(G). However, the main question is how to determine which subdivision is a partial cube. In this paper we prove a conjecture due to A¨ıder, Gravier and Meslem [1] which characterizes all the subdivisions of a clique that are a partial cube. Either it is S(Kn ), or one of the vertices has no incident subdivided edge and all other edges are subdivided an odd number of times.
1
Preliminary definitions and main result
We only consider finite, simple, loopless, connected and undirected graphs. Given a subset S of V , the induced subgraph S of G is the maximal subgraph of G with vertex set S. For a graph G, the distance dG (u, v) between vertices u and v is defined as the number of edges on a shortest uv-path (or uv-geodesic). A subgraph H of G is called isometric if dG (u, v) = dH (u, v) for all distinct vertices u and v in V (H). An isometric subgraph of hypercube Qn is called partial cube. A graph G is an isometric embedding in the hypercube if it is isomorphic to a partial cube. A subdivision of a graph G, noted sub(G), is a graph obtained from G by adding vertices to the edges of G. A vertex v in G which is adjacent to all its neighbors of G in sub(G) is said universal in sub(G) . That means that all the edges of G incident to v, are not subdivided. S(G) is the subdivision of G where each edge of G contains exactly one added vertex. Our proposal is to demonstrate the following theorem conjectured in [1] Theorem 1.1 Let G be a subdivision of a complete graph Kn (n ≥ 4). G is a partial cube if and only if, either G is isomorphic to S(Kn ) or G contains a universal vertex u and the number of added vertices to each edge not incident to u in Kn is odd.
L. Beaudou et al. / Electronic Notes in Discrete Mathematics 29 (2007) 277–281
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279
Sketch of the proof
G is a subdivision of Kn . A vertex u in G is said principal in G if u belongs to Kn . We will be interested about paths that join principal vertices in G. Thus, the subdivided edge between x and y, principal vertices of G, is denoted P (x, y). It is a path. An edge that joins two principal vertices in G, x and y is said plain. For each x and y principal vertices in G, we say that x sees y if the path joining these vertices in G is geodesic. Theorem 2.1 [5] Let k ≥ 3. Then a subdivided k-wheel W is a partial cube if and only if W is isomorphic to Wk (m1 , ..., mk ; 0, ..., 0), where mi is odd for i = 1, ..., k, or W = W3 (1, 1, 1; 1, 1, 1). Lemma 2.2 [1] Let G be a subdivision of Kn (n ≥ 4) where each edge in Kn is an isometric path in G. G is a partial cube if and only if G contains a universal vertex and the other edges of Kn have exactly one added vertex or G is isomorphic to S(Kn ). It is easy to see that the sufficient condition is satisfied. Thus, we just prove the necessary condition by enouncing the useful propositions and main intermediate results of the proof. Proposition 2.3 Let x, y be principal vertices of G, then a xy-geodesic is either isomorphic to P2 or P3 (considering only principal vertices). From now on, for any x and y principal vertices of G, we will note x → y v if P (x, y) is geodesic and x − → y if the path along x, v and y is a xy-geodesic. Remark 2.4 If G is a partial cube, then G is bipartite, and all its cycles are even. Lemma 2.5 If P (x, y) is plain and y → v then a xv-geodesic is contained in x, y, v. For any vertex u of G we define two sets Ku and Lu that split the remaining vertices of G : Ku contains principal vertices x such that P (u, x) is isometric and Lu contains principal vertices a such that P (u, a) is not isometric. We will denote them K and L when there is no doubt. Definition 2.6 We define some types associated to u for the vertices x of Ku as follows : x
•
x has type L if there exists y in Ku such that u − → y.
•
x has type I if there exists y in Lu such that u − → y.
x
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L. Beaudou et al. / Electronic Notes in Discrete Mathematics 29 (2007) 277–281 x
•
x has type C if there exist y, z in Ku such that y − → z.
•
x has type Λ if there exist y, z in Lu such that y − → z.
•
x has type R if there exist y in Ku and z in Lu such that y − → z.
x
x
Remark 2.7 If there is no vertex v in G such that G\v is isometric, then for any u in G, every vertex of Ku has at least one of these types. Moreover, there x is no vertex x with only type L (u − → y) because P (u, y) is also geodesic, and G\x would be isometric. Lemma 2.8 If there is no vertex v in G such that G\v is isometric, then for any u in G, there is no vertex with type C in Ku . We now prove the induction. Proposition 2.9 [1] Let G be a subdivision of K5 . G is a partial cube if and only if G is isomorphic to S(K5 ) or G contains a universal vertex u and the number of the added vertices to each edge not incident to u in K5 is odd. With this proposition and the Theorem 2.1, our result is proven for n = 4, 5. Let us suppose that there exists n ≥ 6 such that the partial cube G is a subdivision of Kn . By Lemma 2.2 we can assume that G is not isomorphic to S(Kn ), and that the theorem is proven for any m < n. Case 1 : there exists u principal in G such that G\u is isometric. As G\u is isometric in G which is a partial cube, G\u is also a partial cube. With the induction hypothesis, there exists x ∈ G\u a universal vertex in G\u or G\u is isomorphic to S(Kn−1 ). Quite easily, we prove that both cases verify our Theorem. Case 2 : for all u principal in G, G\u is not isometric. In this case, we know that there exists three principal vertices u, vK , vL vk such that u −→ vL . We will now denote K for Ku and L for Lu . These are clearly two non-empty sets because vK ∈ K and vL ∈ L. By Lemma 2.8 there is no vertex with type C in K. Lemma 2.10 There are no vertices x and y in K such that P (x, u) and P (x, y) are plain. Proposition 2.11 Each vertex a of L sees exactly one vertex of K. We will now prove that |K| = 1.
L. Beaudou et al. / Electronic Notes in Discrete Mathematics 29 (2007) 277–281
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For this, we will proceed by contradiction. Thus, let us suppose that there exist x and y distinct vertices of K. Both must have a type R,I or Λ (by Remark 2.7 and Lemma 2.8). Each one of these types, implies that x and y sees at least one vertex in L. Let a be a vertex such that x sees a. Then we consider a shortest path from a to y. It cannot be direct because of Proposition 2.11. It cannot go through x, else it would induce an isometric subdivision of K4 and x would be universal : P (x, u) and P (x, y) would then be plain which b is forbidden by Lemma 2.10. So there exists b in L such that a − → y. As b sees y y we can assume that u − → b (by Proposition 2.11). We may then consider y x → a, u − →b that (x, a, y, b) are the vertices of that kind (x, y ∈ K, a, b ∈ L, u − b → y) that minimize the quantity dG (x, a) + dG (a, b) + dG (b, y). and a − Claim 2.12 u, x, y, a, b, is isometric. Then, this subdivision of K5 is a partial cube and by the induction hypothesis, it is either isomorphic to S(K5 ) or has a universal vertex. It cannot be isomorphic to S(K5 ) because P (u, a) would then be a ua-geodesic and it is not. By consequence it must have a universal vertex which can neither be u (it does not see a), nor a or b (they do not see u), nor x or y (then we would have P (x, u) and P (x, y) plain which is forbidden by Lemma 2.10). Thus, this subgraph is not a partial cube. As a conclusion, we can assume |K| = 1 and then, u sees only one vertex in G. It implies that G\u is isometric. This contradicts our hypothesis. Consequently, the Theorem 1.1 is proven. 2
References [1] M. A¨ıder, S. Gravier, K. Meslem, Isometric embeddings of subdivided connected graphs in the hypercube, Submitted (2005). [2] D. Djokovi`c, Distance preserving subgraphs of the hypercubes, Journal of Combinatorial Theory, Ser B41 (1973), 263–267. [3] R.L Graham, H.Pollack On the addressing problem for loop switching, Bell System Technol., J.50 (1971) 2495–2519. [4] W.Imrich, S. Klavˇzar, Product Graphs: structure and recognition, Wiley, New York, (2000). [5] S. Gravier, S. Klavˇzar, M. Mollard Isometric embeddings of subdivided wheels in hypercubes, Discrete Mathematics, 269 (2003) 287–293.
Isometric embeddings of subdivided complete graphs in the hypercube Laurent Beaudou 1 Sylvain Gravier 2 Institut Fourier UMR 5582 CNRS – Universit´e Joseph Fourier Grenoble, France
Kahina Meslem 3 Laboratoir LAID3 Facult´e des Math´ematiques, U.S.T.H.B. Algiers, Algeria
Abstract Isometric subgraphs of hypercubes are known as partial cubes. These graphs have first been investigated by Graham and Pollack [3], and Djokovi`c [2]. Several papers followed with various characterizations of partial cubes. In this paper, we prove that a subdivision of a complete graph of order n (n ≥ 4), is a partial cube if and only if this one is isomorphic to S(Kn ), or there exist n − 1 non-subdivided edges of Kn adjacent to a common vertex in the subdivision and the other edges of Kn are subdivided an odd number of times. Keywords: partial cubes, subdivision of a graph, isometric embeddings.
1 2 3
Email: [email protected] Email: [email protected] Email: kahina [email protected]
1571-0653/$ – see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2007.07.048
278
L. Beaudou et al. / Electronic Notes in Discrete Mathematics 29 (2007) 277–281
Introduction A subgraph H of G is called isometric if and only if dG (u, v) = dH (u, v) for all u, v ∈ V (H). Isometric subgraphs of hypercubes are called partial cubes. Partial cubes have first been investigated by Graham and Pollak [3], and Djokovi`c [2]. Later, several characterizations were shown using a relation defined on the edges set or constructive operations. Partial cubes have found different applications, for instance, in [4], interesting applications in chemical graph theory were established. Clearly, partial cubes are bipartite. The simple way to obtain a bipartite graph is to subdivide every edge of G by a single vertex. Such a graph is a subdivision of G and denoted S(G). However, the main question is how to determine which subdivision is a partial cube. In this paper we prove a conjecture due to A¨ıder, Gravier and Meslem [1] which characterizes all the subdivisions of a clique that are a partial cube. Either it is S(Kn ), or one of the vertices has no incident subdivided edge and all other edges are subdivided an odd number of times.
1
Preliminary definitions and main result
We only consider finite, simple, loopless, connected and undirected graphs. Given a subset S of V , the induced subgraph S of G is the maximal subgraph of G with vertex set S. For a graph G, the distance dG (u, v) between vertices u and v is defined as the number of edges on a shortest uv-path (or uv-geodesic). A subgraph H of G is called isometric if dG (u, v) = dH (u, v) for all distinct vertices u and v in V (H). An isometric subgraph of hypercube Qn is called partial cube. A graph G is an isometric embedding in the hypercube if it is isomorphic to a partial cube. A subdivision of a graph G, noted sub(G), is a graph obtained from G by adding vertices to the edges of G. A vertex v in G which is adjacent to all its neighbors of G in sub(G) is said universal in sub(G) . That means that all the edges of G incident to v, are not subdivided. S(G) is the subdivision of G where each edge of G contains exactly one added vertex. Our proposal is to demonstrate the following theorem conjectured in [1] Theorem 1.1 Let G be a subdivision of a complete graph Kn (n ≥ 4). G is a partial cube if and only if, either G is isomorphic to S(Kn ) or G contains a universal vertex u and the number of added vertices to each edge not incident to u in Kn is odd.
L. Beaudou et al. / Electronic Notes in Discrete Mathematics 29 (2007) 277–281
2
279
Sketch of the proof
G is a subdivision of Kn . A vertex u in G is said principal in G if u belongs to Kn . We will be interested about paths that join principal vertices in G. Thus, the subdivided edge between x and y, principal vertices of G, is denoted P (x, y). It is a path. An edge that joins two principal vertices in G, x and y is said plain. For each x and y principal vertices in G, we say that x sees y if the path joining these vertices in G is geodesic. Theorem 2.1 [5] Let k ≥ 3. Then a subdivided k-wheel W is a partial cube if and only if W is isomorphic to Wk (m1 , ..., mk ; 0, ..., 0), where mi is odd for i = 1, ..., k, or W = W3 (1, 1, 1; 1, 1, 1). Lemma 2.2 [1] Let G be a subdivision of Kn (n ≥ 4) where each edge in Kn is an isometric path in G. G is a partial cube if and only if G contains a universal vertex and the other edges of Kn have exactly one added vertex or G is isomorphic to S(Kn ). It is easy to see that the sufficient condition is satisfied. Thus, we just prove the necessary condition by enouncing the useful propositions and main intermediate results of the proof. Proposition 2.3 Let x, y be principal vertices of G, then a xy-geodesic is either isomorphic to P2 or P3 (considering only principal vertices). From now on, for any x and y principal vertices of G, we will note x → y v if P (x, y) is geodesic and x − → y if the path along x, v and y is a xy-geodesic. Remark 2.4 If G is a partial cube, then G is bipartite, and all its cycles are even. Lemma 2.5 If P (x, y) is plain and y → v then a xv-geodesic is contained in x, y, v. For any vertex u of G we define two sets Ku and Lu that split the remaining vertices of G : Ku contains principal vertices x such that P (u, x) is isometric and Lu contains principal vertices a such that P (u, a) is not isometric. We will denote them K and L when there is no doubt. Definition 2.6 We define some types associated to u for the vertices x of Ku as follows : x
•
x has type L if there exists y in Ku such that u − → y.
•
x has type I if there exists y in Lu such that u − → y.
x
280
L. Beaudou et al. / Electronic Notes in Discrete Mathematics 29 (2007) 277–281 x
•
x has type C if there exist y, z in Ku such that y − → z.
•
x has type Λ if there exist y, z in Lu such that y − → z.
•
x has type R if there exist y in Ku and z in Lu such that y − → z.
x
x
Remark 2.7 If there is no vertex v in G such that G\v is isometric, then for any u in G, every vertex of Ku has at least one of these types. Moreover, there x is no vertex x with only type L (u − → y) because P (u, y) is also geodesic, and G\x would be isometric. Lemma 2.8 If there is no vertex v in G such that G\v is isometric, then for any u in G, there is no vertex with type C in Ku . We now prove the induction. Proposition 2.9 [1] Let G be a subdivision of K5 . G is a partial cube if and only if G is isomorphic to S(K5 ) or G contains a universal vertex u and the number of the added vertices to each edge not incident to u in K5 is odd. With this proposition and the Theorem 2.1, our result is proven for n = 4, 5. Let us suppose that there exists n ≥ 6 such that the partial cube G is a subdivision of Kn . By Lemma 2.2 we can assume that G is not isomorphic to S(Kn ), and that the theorem is proven for any m < n. Case 1 : there exists u principal in G such that G\u is isometric. As G\u is isometric in G which is a partial cube, G\u is also a partial cube. With the induction hypothesis, there exists x ∈ G\u a universal vertex in G\u or G\u is isomorphic to S(Kn−1 ). Quite easily, we prove that both cases verify our Theorem. Case 2 : for all u principal in G, G\u is not isometric. In this case, we know that there exists three principal vertices u, vK , vL vk such that u −→ vL . We will now denote K for Ku and L for Lu . These are clearly two non-empty sets because vK ∈ K and vL ∈ L. By Lemma 2.8 there is no vertex with type C in K. Lemma 2.10 There are no vertices x and y in K such that P (x, u) and P (x, y) are plain. Proposition 2.11 Each vertex a of L sees exactly one vertex of K. We will now prove that |K| = 1.
L. Beaudou et al. / Electronic Notes in Discrete Mathematics 29 (2007) 277–281
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For this, we will proceed by contradiction. Thus, let us suppose that there exist x and y distinct vertices of K. Both must have a type R,I or Λ (by Remark 2.7 and Lemma 2.8). Each one of these types, implies that x and y sees at least one vertex in L. Let a be a vertex such that x sees a. Then we consider a shortest path from a to y. It cannot be direct because of Proposition 2.11. It cannot go through x, else it would induce an isometric subdivision of K4 and x would be universal : P (x, u) and P (x, y) would then be plain which b is forbidden by Lemma 2.10. So there exists b in L such that a − → y. As b sees y y we can assume that u − → b (by Proposition 2.11). We may then consider y x → a, u − →b that (x, a, y, b) are the vertices of that kind (x, y ∈ K, a, b ∈ L, u − b → y) that minimize the quantity dG (x, a) + dG (a, b) + dG (b, y). and a − Claim 2.12 u, x, y, a, b, is isometric. Then, this subdivision of K5 is a partial cube and by the induction hypothesis, it is either isomorphic to S(K5 ) or has a universal vertex. It cannot be isomorphic to S(K5 ) because P (u, a) would then be a ua-geodesic and it is not. By consequence it must have a universal vertex which can neither be u (it does not see a), nor a or b (they do not see u), nor x or y (then we would have P (x, u) and P (x, y) plain which is forbidden by Lemma 2.10). Thus, this subgraph is not a partial cube. As a conclusion, we can assume |K| = 1 and then, u sees only one vertex in G. It implies that G\u is isometric. This contradicts our hypothesis. Consequently, the Theorem 1.1 is proven. 2
References [1] M. A¨ıder, S. Gravier, K. Meslem, Isometric embeddings of subdivided connected graphs in the hypercube, Submitted (2005). [2] D. Djokovi`c, Distance preserving subgraphs of the hypercubes, Journal of Combinatorial Theory, Ser B41 (1973), 263–267. [3] R.L Graham, H.Pollack On the addressing problem for loop switching, Bell System Technol., J.50 (1971) 2495–2519. [4] W.Imrich, S. Klavˇzar, Product Graphs: structure and recognition, Wiley, New York, (2000). [5] S. Gravier, S. Klavˇzar, M. Mollard Isometric embeddings of subdivided wheels in hypercubes, Discrete Mathematics, 269 (2003) 287–293.