{ 1} Definitions. 10. Extremal graph theory. Problem definition Paths and cycles Complete subgraphs

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1 Problem definition Paths and cycles Complete subgraphs 10. Extremal graph theory Definitions Let us examine the following forbidden subgraph problems: At most how many edges are in a graph of order n which does not contain a path of length n? a cycle of length at least (at most) l? a complete graph K r? k+1 independent edges? (*) a Kuratowsky graph? etc Why are these problems extremal? Graph Theory 8 1 Graph Theory 8 If for a given class of graphs a certain graph parameter, say the number of edges or the minimum degree, is at most some number f, then the graphs for which equality holds are extremal graphs of the inequality. Usually,, let ex(n;f) denote the extremal graph of order n which does not contain a given subgraph F,, then the graph has ex(n;f) edges. Examples: An acyclic graph of order n has at most n 1 edges, and the extrem- al graphs are trees of order n. If we are looking for the extremal graph which does contain k+1 independent edges, then in ex(n;f F )we) have F=(k+1)K. Let C denote an arbitrary cycle of a graph G with order n.. Then ex(n;c) = n 1,, since a graph order n contains at most n 1 edges to remain acyclic. Graph Theory Paths and Cycles For a given graph G denotes the minimal degree of vertices and g is the girth and denotes the length of a shortest cycle. Theorem 10.1.: For g 3 and 3 put { 1} δ ( g 1 ) 1 + ( δ 1 ) δ n0 ( g, δ ) = g {( δ 1 ) 1} δ Then a graph G with minimum degree n 0 (g, ) vertices. if g is odd if g is even. and girth g has at least We will examine separately the two cases if g is even or odd. Graph Theory 8 4 1

2 Suppose that g = d+1, d G: x 1. Let x There is no two x y paths of length at most d. V(G) and y V(G). Suppose the opposite.. Then the length of both paths (l( 1 and l ) d. So, l 1 + l d and so we have a cycle in G with length at most d, which contradicts that g = d+1. Graph Theory 8 5 y In a connected graph G the distance d(u,v) between two vertices u and v is the minimum of the lengths of the u v paths of G. We can classify the vertices into different classes accordind to their distance from the vertex x: So Since deg(x) ), so there are at least vertices at distance 1 from x. We can reach the vertices are distance from x through vertices which are at distance 1 from x.. So we have at least ( -1) vertices at distance. The number of vertices at distance 3 is at least ( -1). The number of vertices at distance d is at least ( -1). d-1 n 1 + δ + δ ( δ 1 ) δ ( δ 1 ) δ (g = 1+ ( 1) 1) δ Graph Theory 8 6 Suppose that g = d, d Pick two adjacent vertices, say x and y. 1. Then there are at least ( -1) vertices at distance 1 from {x,y}, Let us looking for an extremal graph G 0 with parameters The above theorem shows that G 0 is regular of degree. Furthermore, and g. and at least ( -1) vertices at distance,, and so on,, at least ( -1) d-1 vertices at distance d-1 from {x,y}. So n = + ( δ 1) ( δ 1) = δ g ( 1) 1) d-1 if g = d+1 then G 0 has diameter d and if g = d then every vertex is within distance d 1 of each pair of adjacent vertices. We call G 0 a Moore graph of degree and girth g, or,, if g = d+1,, a Moore graph of degree and diameter d. This completes the proof. Graph Theory 8 7 Graph Theory 8 8

3 A Moore graph with =3 and g=7 Graph Theory 8 9 A Moore graph with =4 and g=6 Graph Theory 8 10 Let us see now what we can say about long cycles and paths in a graph. The length of a longest cycle in a graph G is called as circumference. Example: : the circumference of a Hamiltonian graph is n, while the longest path is n-1. Is there any connection between the length of a circumference and the length of a longest path? Theorem 10.3.: Every non-hamiltonian connected graph contains at least as long paths as the circumference of the graph. Let C = x 1 x x l be a longest cycle. l < n since the graph is not Hamiltonian. G y x 1 x x l x 3 x l- x 4 x 6 x 5 x l-1 Since l < n then there exists a vertex y V(G) which is not on C and is adjacent with a vertex of C, say x 1, and then yx 1 x x l is a path of length l. Graph Theory 8 11 Graph Theory 8 1 3

4 Lemma 10.4.: Let G be a connected graph of order n 3 such that for any two non-adjacent vertices x and y we have d(x)+ d(y) k. If k = n then G is Hamiltonian. If k < n then G contains a path of length k,, and a cycle of length at least (k+)/. Suppose that G is not Hamiltonian.. Let P = x 1 x x l be a longest path in G. Since P is maximal, so all the neigbours of x 1 and x l are in P. The path P cannot contain vertices x i and x i+1 such that x 1 is adjacent to x i+1 and x l is adjacent to x i. Since, otherwise x 1 x x i x l x l-1 x i+1 is a cycle of length l. x i+1 Graph Theory 8 13 P l =x 1 x x i x l x l-1 x x i+1 x 1 x x 3 x i x i+1 x l-1 x l We define the following two sets: (x 1 )={x i : x 1 x i E(G)} (x l )={x i+1 : x i x l E(G)} These two sets are disjoint subsets of {x,x 3,,x,x l } and so k d(x 1 ) + d(x l ) l 1 n 1. The first two assertions of the theorem follow from this inequality: if k = n this is impossible,, so G is Hamiltonian, if k < n then P has length l 1 k. Graph Theory 8 14 It remains to prove the statement about cycles. Assume that d(x 1 ) d(x l ) so d(x 1 ) least integer not less than z. Put t = max {i: x 1 x i E(G)}. Then k/, where x 1 x x 3 x 4 x t x l-1 x l k k + t d( x1 ) and so G contains a cycle of length t: C = x 1 x x 3 x t. z denotes the The Lemma has the following consequence: Theorem 10.5.: Let G be a graph of order n without a path of length k 1.. Then k 1 e( G ) n. A graph is an extremal graph (that is equality holds for it) iff all its components are complete graphs of order k. We fix k and apply induction on n. If n k then the statement is trivial. Let n > k and suppose that the statetemnt is true for all n' < n. If G is disconnected,, the induction hypothesis implies the result. Graph Theory 8 15 Graph Theory

5 So we can suppose that G is connected. If G contains K k (a complete k order graph) then it has a path of length k. So we can suppose that G does not contain K k. So by the Lemma there exist two non-adjacent vertices x and y,, for which d(x) ) + d(y) ) < k. So, at least one of them, say x k 1 d( x ). Since G x is not an extremal graph. So k 1 k 1 k 1 e(g ) d( x ) + e(g x ) < + ( n 1) = n which completes the proof. Graph Theory Complete Subgraphs What is the maximal number of edges in a graph of order n not containing a K r (a complete graph of order n)? ex(n;k r )? If G is (r( 1)-partite then it does not contain a K r since every vertex class of G contains at most one vertex of a complete subgraph. Consequence: ex(n;k r ) the maximal size of an (r( 1)-partite graph of order n. Is there a unique (r 1)-partite graph of order n that has maximal size? How looks like this graph? Let T r-1 (n) be the (r( 1)-partite complete graph which has in k. class n + k 1 r 1 vertices. Graph Theory 8 18 Let we realize that T r-1 (n) is the complete (r 1)-partite graph of order n whose classes are as equal as possible: : there are n k vertices in the kth class and n 1 n n r-1 n Theorem 10.6.: If G is an (r( 1)-partite graph of order n and maxim- al size then G = T r-1 (n). n-(m i +m j ) further vertices ith Class jth Class Let G be a graph with maximal size and suppose the classes of G are not as equal as possible. Say,, there are m i vertices in the ith class and m j in the jth class and m j m i +. Transfer one vertex from the jth class to the ith class. Graph Theory 8 19 The number of ignored edges: n (m i +m j ) + m i = n m j The number of new edges: (m j 1) + (n (m i + m j ), So the difference is: m j 1+n m i m j n+m j = m j 1 m i m i + 1 m i = 1. So we would increase the number of edges. Graph Theory 8 0 5

6 n = 7 r = ,1 3 1, 1,1,1 4, 1,1, 1,1,1,1 5,3 1,, 1,1,1, 1,1,1,1,1 6 3,3,, 1,1,, 1,1,1,1, 7 3,4,,3 1,,, 1,1,1,, If k = 1 then n 1 = If k = then n = If k = 3 then n 3 = 3 T r-1 (n)) vertices Graph Theory 8 1 The Turan graph T 3 (7). Graph Theory 8 Denote the number of edges in T r-1 (n) by t r-1 (n). Example: n t( n ) = 4 Consequence: ex(n;k r ) t r-1 (n). The fundamental theorem of Turán states that this trivial inequality, in fact,, an equality for every n and r. First, we shall show that the degree sequence of a graph without a K r is dominated by the degree sequence of an (r-1)-partite( graph. In view of the remarks above this will imply Turán's theorem. We say that a sequence a 1,a, a n b 1,b,,b n if a i b i, 1 i n. dominates the sequence Theorem 10.7.: Let G be a graph with vertex set V that does not contain K r. Then there is an (r( 1)-partite graph H with vertex set V such that for every vertex z V we have d G (z) d H (z). If G is not a complete (r 1)-partite graph then there is at least one vertex z for which the inequality above is strict. We shall apply induction on r. For r = there is nothing to prove since G is the empty graph, which is 1-partite. Assume now that r values of r. 3 and the assertion holds for smaller Graph Theory 8 3 Graph Theory 8 4 6

7 Pick a vertex x V for which d G (x) is maximal. Let G(x) ) = W, and let G 0 = G(W). G 0 does not contain a K r-1 otherwise with x it would form a K r. By the induction hypothesis we can replace G 0 by an (r( )-partite graph H 0 with vertex set W in such a way that d G0 (y) d H0 (y) for every y W. Add H 0 the vertices in V W and join each vertex in V W to each vertex in W. Denote this graph by H. We will prove that H has the required properties. Let us examine the degrees: If z If z V W then d H (z) ) = d H (x) ) = d G (x) d G (z). W then d (z) = d H0 H(z) ) + n - W d G (z) What can we say about the extremal case e(g) ) = e(h)? Since e(g) ) = e(h) then e(g 0 ) = e(h 0 ), so by the induction hypothesis G 0 is a complete (r )-partite graph, which implies that G is a complete (r 1)-partite graph. H does not contain K r, since a complete subgraph of H has at most one vertex in V W and at most r vertices in W. Graph Theory 8 5 Graph Theory 8 6 Now,, as a consequence,, we can state the Turán's theorem: Theorem (Turán( Turán): ex(n;k r ) = t r-1 (n) and T r-1 (n) is the unique graph of order n and size t r-1 (n) that does not contain a complete graph of order r. Since T r-1 (n) is the unique (r-1)-partite graph of order n and maximum size,, so both assertions follow from the Theorem Graph Theory 8 7 Problem of Zarankiewicz: : If G (m,n) is a bipartite graph with m vertices in the first class and n vertices in the second one. What is the maximum size of a graph G (m,n) if it does not contain a complete bipartite graph G (s,t)? The maximum usually denoted by z(m,, n ; s, t). There are not known exact values for z(m,, n ; s, t) but upper bounds can be given. The following lemma seems ato imply a very good upper bound for the function z. Lemma 10.9.: Let m, n, s, t, r, k be integers, s m, m t n, 0 k, 0 r < m, and let G = G (m,n) ) be a graph of size z = my = km + r without a K(s,t) subgraph having s vertices in the first class and t in the second. Then y k k + 1 n m ( m r ) + r ( s 1 ). t t t t Graph Theory 8 8 7

8 Denote by V 1 and V the the vertex classes of G. We say that a subset T of t elements of V belongs to a vertex x V 1 if x is joined to every vertex in T.. We call this set as t-set. d( x ) The number of t-sets belonging to a vertex x is. t Since the assumption on G is exactly that each t-set in V belongs to at most s 1 vertices of V 1, we find that x V 1 d( x ) n ( s 1) t t (*) As x V 1 u for u d(x) = z = my = km+ r, and f(u)= u is a convex function of t t, the inequality (*) implies the statement of the theorem. Theorem : z( m,n; s,t ) ( s 1 ) 1 t ( n t + 1 )m 1 1 t + ( t 1 )m. Let G=G (m,n) be an extremal graph for the function z(m,n;s,t), that is let G be a bipartite graph of size z(m,n;s,t) ) = my without a K(s,t) subgraph. As y n,, the inequality of the Lemma implies t t ( y ( t 1 )) ( s 1 )( n ( t 1 ) m 1. Graph Theory 8 9 Graph Theory