Turan’s theorem and extremal graphs Question: How many edges a simple graph must have to guarantee that the graph contains a triangle? Since K m,m and K m, m+1 do not contain triangles, we see that if the graph has n vertices, then  n 2 /4  edges are not enough. Mantel (1907) proved that if there are more edges then the graph contains a triangle.

p2. K n/2, n/2 has n 2 /4 edge but without a triangle.

p3. Proof of Mantel’s : Let G have vertex set {1,...,n} and no triangle. Give vertex i a weight z i  0 and  z i =1 and we wish to maximize S=  z i z j, where the sum is over all edges {i, j}. Suppose k and l are not joined. Let the neighbors of k have total weight x, and those of l total weight y, where x  y. Since (z k +e)x+(z l -e)y  z k x+z l y, so we can shift the weight from l to k to make S larger. Thus S is maximized if all the weight concentrates on an edge. Therefore S  1/4. Why? On the other hand let all z i =1/n, then S=|E|/n 2. So |E|  n 2 /4. 

p4.

p5. M(n,p)=(p-2)n 2 /2(p-1)-r(p-1-r)/2(p-1). n=t(p-1)+r Thm 4.1. (Turan 1941) If a simple graph on n vertices has more than M(n,p) edges, then it contains a K p. Proof: By induction on t: It is obvious, if t=0. Consider a graph G with n vertices with maximum edges but without K p. Clearly G must have K p-1, say H. Each of the remaining vertices is joined to at most p-2 vertices of H. The remaining n-p+1 vertices do not contain K p. Since n-p+1=(t-1)(p-1)+r, we can apply the induction hypothesis to these points. So the the number of edges of G is at most: M(n-p+1, p)+(n-p+1)(p-2)+C(p-1, 2), which is M(n,p).

p6. The girth of a graph G is the size of a smallest polygon (cycle) in G. Thm If a graph G on n vertices has more than edges, then G has girth at most 4. Pf: By contradiction, suppose G has girth at least 5. Let y 1,...,y d be x’s neighbors and d=deg(x). No two them are adjacent. Why? Also no vertex can be adjacent to more than one of y 1,...,y d. Why? Thus (deg(y 1 )-1)+...+ (deg(y d )-1)+(d+1)  n. Why?

Thm 4.3. If a simple graph G on n vertices has all vertices of degree at least n/2, then it contains a Hamiltonian circuit. Pf: By contradiction, suppose it is not true. Let G be such a counterexample with the maximum number of edges. Let y and z be two non-adjacent vertices. Since adding {y, z} creates a Hamiltonian circuit, there exists a simple path from y to z with vertices y=x 1, x 2,...,x n =z, say. {i: y is adjacent to x i+1 } and {i: z is adjacent to x i } each have at least n/2 neighbors and are in {1,2,...,n-1}, so they must intersect at some element {i0}. Then y=x 1, x 2,...,x i0, z=x n, x n-1,...,x i0+1, x 1 =y is Ham circuit. Contradiction! y=x 1 x2x2 xi0xi0 x i0+1 x n =z