January –February 2020 ISSN: 0193-4120 Page No. 2189-2191 2189 Published by: The Mattingley Publishing Co., Inc. Complement Connected Majority Neighborhood Number 1 I. Paulraj Jayasimman, 2 J. Joseline Manora 1 Academy of Maritime Education and Training Deemed to be University, kanathur, Chennai 2 TBML College, Porayar, Nagai Article Info Volume 82 Page Number: 2189 - 2191 Publication Issue: January-February 2020 Article History Article Received: 14 March 2019 Revised: 27 May 2019 Accepted: 16 October 2019 Publication: 12 January 2020 Abstract A set of vertices in a graph is a majority neighborhood set of if the subgraph [] ∈ contains atleast 2 vertices and 2 edges, where [] is the subgraph of . In this article majority neighborhood connected complement set of are introduced and this number determined () for various graph structures. Keywords: Connected Majority neighborhood number, Complement connected Majority neighborhood. 1. Introduction In this article the author use the simple and undirected graph. The parameters 0 , and () was studied in the following articles [1],[5],[6]-[8]. 2. Majority neighborhood connected complement number of a graph Definition 2.1 Let be a graph with vertices and edges. A complement connected majority neighborhood set such that is a connected majority neighborhood set in and majority neighborhood set in . The complement majority neighborhood number is the minimum size of connected complement majority neighborhood set of and is denoted by (). Theorem If = with ≥ 4 then = −3 2 . Proof If = , = = (), the set = { 1 , 2 , 3 ,.. −3 2 } be the CMN- set . In covers −3 2 vertices and deg = − 3 ∈ and( ) = (− 3), each = (− 3) ≥ (−3) 2 ∈ and each ≥ −3 2 . Hence = −3 2 . Theorem For the path graph , = −3 2 , ≥ 4. Theorem For the = 2 +1 with ≥ 1 then =2 Proof: Let = 2 +1 with =2 +1 and = 3in . Let = , 1 be the connected neighborhood setin . Since () =2 + 1, = 3 and by the super hereditary property () ≥ (). Therefore is connected majority neighborhood set. In complement of , =2 +1and () = 2− 2. = 0, =2− 2 ∈ . Therefore, =2− 1, ∈ and = − 22− 2⇒ =2− 1 ≥ 2 +1 2 = 2 ∈ , = − 22− 2≥ (2−2) 2 = 2 ∈ . Therefore, is majority neighborhood set.Hence =2 . Theorem If = 1, with ≥ 2then =2. Theorem For the wheel graph with ≥ 3 = 2 = 3,4 ≥ 8 3 = 5,6,7 Proof. Let = with ≥ 3 and() = , () = 2.