Graph_Theory_Notes2.pdf

# Theorem 6 erd os a r enyi and v t s os 1966 in any

• Notes
• 10

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Theorem 6. (Erd¨ os, A. R´ enyi and V. T. S´ os, 1966) In any friendship graph of order n , there exists a vertex of degree n - 1 (i.e. Δ( G ) = n - 1). Proof. Suppose otherwise. Then there is at least one friendship graph such that the result fails. Let G be a friendship graph of order n and maximum degree Δ( G ) < n - 1. Step 1: We first show that any two nonadjacent vertices of G have the same degree. Let x and y be nonadjacent vertices of G . For each v N ( x ), let f ( v ) be the unique common neighbour of v and y . Since x and f ( v ) have a unique common neighbour, namely v , the mapping f : N ( x ) N ( y ) is one-to-one (that is, for distinct v, v 0 N ( x ) we have f ( v ) 6 = f ( v 0 )). Hence deg( x ) = | N ( x ) | ≤ | N ( y ) | = deg( y ). Similarly, deg( y ) deg( x ). Therefore, deg( x ) = deg( y ). Hence any two nonadjacent vertices of G have the same degree. In other words, if two vertices are adjacent in G , then they have the same degree in G . Proof (continued). Step 2: We now prove that G is connected. In fact, since Δ( G ) < n - 1 by our assumption, we have δ ( G ) = n - 1 - Δ( G ) > 0. Hence G has no isolated vertices. In other words, every connected component of G contains at least two vertices. Suppose to the contrary that G is disconnected. Let H and K be two components of G . By what we just proved above, each of H and K has order at least two. So we can take two distinct vertices of H , say u and v , and two distinct vertices of K , say x and y . Then there is no edge of G joining one of u and v and one of x and y . So ux, uy, vx and vy are all edges of G . But this means that u and v have at least two common neighbours in G , namely x and y , which is a contradiction. Therefore, G must be connected. Proof (continued). Step 3: We now prove that G is regular, and hence G is k -regular for some k 1. In fact, since G is connected by Step 2, for any two vertices x, y of G , there exists an ( x, y )-path P : x = x 0 , x 1 , . . . , x k = y in G . Since x and x 1 are nonadjacent in G , by Step 1, we have deg G ( x ) = deg G ( x 1 ). Similarly, deg G ( x 1 ) = deg G ( x 2 ) , deg G ( x 2 ) = deg G ( x 3 ) , . . . , deg G ( x k - 1 ) = deg G ( y ).

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Hence deg G ( x ) = deg G ( y ). Since this is true for any two vertices x, y of G , we conclude that all vertices of G have the same degree in G . In other words, G is k -regular for some k 1. (Note that k cannot be 0 for otherwise G has no edge at all and so cannot be a friendship graph.) Proof (continued). Step 4: We now prove that n = k 2 - k + 1. This is achieved by counting the number of 2-paths of G in two ways, where a 2-path is a path of length two. Note first that any 2-path of G uses exactly one vertex as its middle vertex. Since G is k -regular, for any fixed vertex v , there are precisely ( k 2 ) 2-paths of G with middle vertex v , namely the paths x, v, y for the ( k 2 ) pairs of distinct neighbours x, y of v in G . Since G has order n , the number of 2-paths of G is equal to n ( k 2 ) . On the other hand, each pair of distinct vertices x, y of G gives rise to exactly one 2-path of G , namely the path x, z, y , where z is the unique common neighbour of x and y . Hence the total number of 2-paths of G is also equal to ( n 2 ) .
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• One '14
• Graph Theory, Shortest path problem, vertices, • Bridges

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