On a property of the $n$-dimensional cube

arXiv:1110.6010v1 [cs.DM] 27 Oct 2011 On a property of the n-dimensional cube Rafayel Kamalian a ∗and Arpine Khachatryan b † aInstitute for Informatics and Automation Problems, National Academy of Sciences of Republic of Armenia, 0014, Armenia bIjevan Branch of Yerevan State University, 4001, Armenia We show that in any subset of the vertices of n-dimensional cube that contains at least 2n−1 + 1 vertices (n ≥4), there are four vertices that induce a claw, or there are eight vertices that induce the cycle of length eight. 1. Introduction and definitions We consider finite graphs G = (V, E) with vertex set V and edge set E. The graphs contain no multiple edges or loops. The n-dimensional cube will be denoted by Qn, and claw is the complete bipartite graph K1,3. Moreover, the vertex of degree three in the claw is called a claw-center. Non-defined terms and concepts can be found in [ 1]. The main result of the paper is the following: Theorem 1 Let n ≥4 and let V ′ ⊆V (Qn). If |V ′| ≥2n−1 + 1, then at least one of the following two conditions holds: (a) there are four vertices in V ′ that induce a claw; (b) there are eight vertices that induce a simple cycle. Proof. Our proof is by induction on n. Suppose that n = 4. Clearly, without loss of generality, we can assume that |V ′| = 9. Consider the following partition of the vertices of Q4: V1 = {(0, α2, α3, α4) : αi ∈{0, 1}, 2 ≤i ≤4}, V2 = {(1, α2, α3, α4) : αi ∈{0, 1}, 2 ≤i ≤4}. Clearly, the subgraphs of Q4 induced by V1 and V2 are isomorphic to Q3. Define: V ′ 1 = V1 ∩V ′, V ′ 2 = V2 ∩V ′. We will assume that |V ′ 1| ≥|V ′ 2|. We will complete the proof of the base of induction, by considering the following cases: Case 1: |V ′ 1| = 8 and |V ′ 2| = 1. Clearly, any vertex from V ′ 1 is a claw-center. ∗email: rrkamalian@yahoo.com †email: khachatryanarpine@gmail.com 2 Rafayel Kamalian, Arpine Khachatryan Case 2: |V ′ 1| = 7 and |V ′ 2| = 2. It is not hard to see that V ′ 1 contains a claw-center. Case 3: |V ′ 1| = 6 and |V ′ 2| = 3. Again, it is a matter of direct verification that V ′ 1 contains a claw-center. Case 4: |V ′ 1| = 5 and |V ′ 2| = 4. Consider the subgraph G1 of Q4 induced by V ′ 1. Clearly, if G1 contains a vertex of degree three, then this vertex is a claw-center. Thus, any vertex in G1 has degree at most two. It is not hard to see that this implies that G1 contains no isolated vertex. Moreover, since |V ′ 1| = 5, we have that G1 is a connected graph, and therefore it is the path of length four. Now, let a1, a2, a3 be the internal vertices of G1, and let b1, b2 be the end-vertices of G1. Clearly, we can assume that neither of a1, a2, a3 has a neighbour in V ′ 2. Since |V2| = 8 and |V ′ 2| = 4, we have that there are five possibilities for V ′ 2. We invite the reader to check that in four of these cases one can find a claw-center in V ′ 2, and in the final case V ′ has a vertex z such that V ′\{z} induces a simple cycle. Now, let us assume that the statement is true for n−1, and let V ′ ⊆V (Qn) be a subset with |V ′| ≥2n−1 + 1. Consider the following partition of the vertices of Qn: V1 = {(0, α2, ..., αn) : αi ∈{0, 1}, 2 ≤i ≤n}, V2 = {(1, α2, ..., αn) : αi ∈{0, 1}, 2 ≤i ≤n}. Clearly, the subgraphs of Qn induced by V1 and V2 are isomorphic to Qn−1. Moreover, it is not hard to see that at least one of the following two inequalities is true: |V1∩V ′| ≥2n−2+1 and |V2 ∩V ′| ≥2n−2 + 1. Thus the proof follows from the induction hypothesis. □ For the case of n = 3 we have: Proposition 1 Let V ′ ⊆V (Q3) and let |V ′| ≥6. Then at least one of the following two conditions holds: • there are four vertices in V ′ that induce a claw; • there are six vertices in V ′ that induce a simple cycle. Acknowledgement We would like to thank Zhora Nikoghosyan and Vahan Mkrtchyan for their attention to this work. REFERENCES 1. F. Harary, Graph Theory, Addison-Wesley, Reading, MA, 1969.

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