# 4. Trees. 4.1 Basics. Definition: A graph having no cycles is said to be acyclic. A forest is an acyclic graph.

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1 4. Trees Oe of the importat classes of graphs is the trees. The importace of trees is evidet from their applicatios i various areas, especially theoretical computer sciece ad molecular evolutio. 4.1 Basics Defiitio: A graph havig o cycles is said to be acyclic. A forest is a acyclic graph. Defiitio: A tree is a coected graph without ay cycles, or a tree is a coected acyclic graph. The edges of a tree are called braches. It follows immediately from the defiitio that a tree has to be a simple graph (because self-loops ad parallel edges both form cycles). Figure 4.1(a) displays all trees with fewer tha six vertices. Fig. 4.1(a) The followig result characterises trees. Theorem 4.1 A graph is a tree if ad oly if there is exactly oe path betwee every pair of its vertices.

5 Graph Theory 83 degree is oe. Assume the result is true for all trees with k 1 edges (k ) ad cosider a tree T with exactly k edges. We kow that T cotais at least two pedat vertices. Let v be oe of them ad let w be the vertex that is adjacet to v. Cosider the graph T v. Sice T v has k 1 edges, the iductio hypothesis applies, so T v is a subgraph of G. We ca thik of T v as actually sittig iside G (meaig w is a vertex of G, too). Sice G cotais at least k + 1 vertices, ad T v cotais k vertices, there exist vertices of G that are ot a part of the subgraph T v. Further, sice the degree of w i G is at least k, there must be a vertex u ot i T v that is adjacet to w. The subgraph T v together with u forms the tree T as a subgraph of G (Fig. 4.3). Fig Rooted ad Biary Trees A tree i which oe vertex (called the root) is distiguished from all the others is called a rooted tree. A biary tree is defied as a tree i which there is exactly oe vertex of degree two ad each of the remaiig vertices is of degree oe or three. Obviously, a biary tree has three or more vertices. Sice the vertex of degree two is distict from all other vertices, it serves as a root, ad so every biary tree is a rooted tree. Below are give some properties of biary trees. Theorem 4.10 Every biary tree has a odd umber of vertices. Proof Apart from the root, every vertex i a biary tree is of odd degree. We kow that there are eve umber of such odd vertices. Therefore whe the root (which is of eve degree) is added to this umber, the total umber of vertices is odd. Corollary 4.1 There are 1 (+1) pedat vertices i ay biary tree with vertices. Proof Let T be a biary tree with vertices. Let q be the umber of pedat vertices i T. Therefore there are q iteral vertices i T ad so q 1 vertices of degree 3. Thus the umber of edges i T = 1 [3( q 1)++q]. But the umber of edges i T is 1.

6 84 Trees Hece, 1 [3( q 1)++q]= 1, so that q = 1 (+1). The followig result is due to Jorda [1]. Theorem 4.11 (Jorda) Every tree has either oe or two ceters. Proof The maximum distace, max d(v, v i ) from a give vertex v to ay other vertex occurs oly whe v i is a pedat vertex. With this observatio, let T be a tree havig more tha two vertices. Tree T has two or more pedat vertices. Deletig all the pedat vertices from T, the resultig graph T is agai a tree. The removal of all pedat vertices from T uiformly reduces the eccetricities of the remaiig vertices (vertices i T ) by oe. Therefore the ceters of T are also the ceters of T. From T we remove all pedat vertices ad get aother tree T. Cotiuig this process, we either get a vertex, which is a ceter of T, or a edge whose ed vertices are the two ceters of T. Defiitio: Trees with ceter K 1 are called uicetral ad trees with ceter K are called bicetral trees. Spaig trees A tree is said to be a spaig tree of a coected graph G, if T is a subgraph of G ad T cotais all vertices of G. Example Cosider the graph of Fig. 4.4, where the bold lies represet a spaig tree. Fig. 4.4 The followig result shows the existece of spaig trees i coected graphs. Theorem 4.1 Every coected graph has at least oe spaig tree. Proof Let G be a coected graph. If G has o cycles, the it is its ow spaig tree. If G has cycles, the o deletig oe edge from each of the cycles, the graph remais coected ad cycle free cotaiig all the vertices of G. Defiitio: A edge i a spaig tree T is called a brach of T. A edge of G that is ot i a give spaig tree T is called a chord. It may be oted that braches ad chords

9 Graph Theory 87 Theorem 4.16 ( If u is a vertex of a -vertex tree T, the d(u, v). ) v V (T) Proof Let T(V, E) be a tree with V =. Let u be ay vertex of T. We use iductio o. If =, the result is trivial. Let >. The graph T u is a forest ad let the compoets of T u be T 1, T,..., T k, where k 1. Sice T is coected, u has a eighbour i each T i. Also, sice T has o cycles, u has exactly oe eighbour v i i each T i. For ay v V(T i ), the uique u v path i T passes through v i ad we have d T (u, v) = 1+d Ti (v i, v). Let i = (T i ) (Fig. 4.8). The we have d T (u, v) = i + d Ti (v i, v). (4.16.1) v V(T i ) v V(T i ) By the iductio hypothesis, we have d Ti (v i, v) v V(T i ) ( i ). Fig. 4.8 We ow sum the formula (4.16.1) for distaces from u over all the compoets of T u ad we get d T (u, v) ( 1)+ v V(T i ) i ( i ). ( Now, we have i = 1. Clearly, i ) ( ) i, because the right side couts the i i edges i K i or K 1, ad the left side couts the edges i a subgraph of K i, the subgraph beig uio of disjoit cliques K 1 (, K 1, )..., K k. 1 Thus, d T (u, v) ( 1)+ = v V(T) ( ).

10 88 Trees Corollary 4. The sum of the distaces from a pedat vertex of the path P to all other vertices is 1 ( i =. ) i=0 Corollary 4.3 If H is a subgraph of a graph G, the d G (u, v) d H (u, v). Proof Every u v path i H appears also i G, ad G may have additioal u v paths that are shorter tha ay u v path i H. Corollary 4.4 Proof If u is a vertex of a coected graph G, the ( ) (G) d(u, v). v V(G) Let T be a spaig tree of G. The d G (u, v) d T (u, v), so that d G (u, v) vεv(g) ( ) (G) d T (u, v). vεv(g) The sum of the distaces over all pairs of distict vertices i a graph G is the Wieer idex W(G) = d(u, v). O assigig vertices for the atoms ad edges for the atomic u, vεv(g) bods, we ca use graphs to study molecules. Wieer [68] origially used this to study the boilig poit of paraffi. Theorem 4.17 Let v be ay vertex of a coected graph G. The G has a spaig tree preservig the distaces from v. Proof Let G be a coected graph. We fid a spaig tree T of G such that for each u V = V(G) = V(T), d G (v, u) = d T (v, u). Cosider the eighbourhoods of v, N i (v) = {u V : d G (v, u) = i}, 1 i e, where e = e(v). Let H be the graph obtaied from G by removig all edges i each < N i (v) >. Clearly, H is coected. Let < B i (v) > H deote the iduced subgraph of H, iduced by the ball B i (v). Clearly, < B 1 (v) > H does ot cotai ay cycle. If < B (v) > H cotais cycles, remove edges from [ N 1 (v), N (v)] sequetially, oe edge from each cycle, till it becomes acyclic. Proceedig successively by removig edges from [ N i (v), N i+1 (v)] to make < B i+1 (v) > H acyclic for 1 i e 1, we get a spaig tree of H ad hece of G. Sice i this procedure oe distace path from v to each of the other vertices remais itact, we have d G (v, u) = d T (v, u) for each u V.

12 90 Trees The followig result gives a recursive formula for τ(g). Theorem 4.19 For ay cyclic edge e of a graph G, τ(g) = τ(g e)+τ(g pe). Proof Let S be the set of spaig trees of G ad let S be partitioed as S 1 S, where S 1 is the set of spaig trees of G ot cotaiig e ad S is the set of the spaig trees of G cotaiig e. Sice e is a cyclic edge, G e is coected ad there is a oe-oe correspodece betwee the elemets of S 1 ad the spaig trees of G e. Also, there is a oe-oe correspodece betwee the spaig trees of G þe ad the elemets of S. Thus, τ(g) = S 1 + S = τ(g e)+τ(g þe). Remarks 1. The above recurrece relatio is valid eve if e is a cut edge. This is because τ(g e) = 0 ad every spaig tree of G cotais every cut edge.. The recurrece relatio is valid eve if G is a geeral graph ad e is a multiple edge, but ot whe e is a loop. 3. The complexity of ay graph G is computed by repeatedly applyig the above recurrece. We observe that o applyig the elemetary cotractio to a multiple edge, the resultig graph ca have a loop ad by remark () the procedure ca be still cotiued. At each stage of the algorithm, oly a edge belogig to the proper cycle is chose. The algorithm starts with a give graph ad produces two graphs (possibly geeral) at the ed of the first stage. At each subsequet stage oe proper cyclic edge from each graph is chose (if it exists) for applyig the recurrece. O termiatio of the algorithm, we get a set of graphs (or geeral graphs) oe of which have a proper cycle. The τ(g) is the sum of the umber of these graphs. If H is ay of these graphs, the τ(h) is the product of its edges, igorig the loops. Example Cosider the graph G give i Figure Fig. 4.10

16 94 Trees T. Clearly, the coverse is also true. Therefore, by iductio hypothesis, the umber of trees T 1 is = = ( 3)! (d 1 1)!...(d j 1 1)!(d j 1 1)!(d j+1 1)!...(d 1 1)! ( 3)!(d j 1) (d 1 1)!...(d j 1 1)![(d j 1)(d j )!](d j+1 1)!...(d 1 1)! ( 3)!(d j 1) (d 1 1)!...(d j 1)!...(d 1 1)!(d 1)! = ( 3)!(d j 1). (d j 1)! j=1 Sice v j is ay oe of the vertices v 1,..., v 1, the umber of trees T is 1 j=1 ( 3)!(d j 1) = ( 3)! (d j 1)! (d j 1)! j=1 j=1 (d j 1), as d = 1 ad d 1 = 1 1 = 0 j=1 = ( 3)! (d j 1)! j=1 ( ), sice (d j 1) = ( 1) = j=1 = ( )!. (d j 1)! j=1 Now, we use Theorem 4. to obtai τ(k ) =, which forms the secod proof of Cayley s Theorem. Secod Proof of Theorem 4. degree sequece [d i ] 1 is We kow the umber of labelled trees with a give ( )!. (d j 1)! j=1 The total umber of labelled trees with vertices is obtaied by addig the umber of labelled trees with all possible degree sequeces.

17 Graph Theory 95 Therefore, τ(k ) = d i 1 d i = ( )! (d j 1)! j=1 Let d i 1 = k i. So d i 1 gives d i 1 0, or k i 0. Also, k i = (d i 1) = d i = =. Thus, τ(k ) = k i 0 1 k i =. ( )! k 1!k!...k! = ( )! k 1!k!k! 1k1 1 k...1 k k i 0 1 k i = = ( ), by multiomial theorem. Hece, τ(k ) ==. Note The multiomial distributio is give by! x 1!x!...x k! px 1 1 px...px k k = (p 1 + p p k ), where x i =. 4.4 The Fudametal Cycles Defiitio: Let T be a spaig tree of a coected graph G. Let T be the spaig subgraph of G cotaiig oly the edges of G which are ot i T (i.e., T is the relative complemet of T i G). The T is called the co tree of T i G. The edges of T are called braches ad the edges of T are called chords of G relative to the spaig tree T. Theorem 4.4 If T is a spaig tree of a coected graph G ad f is a chord of G relative to T, the T + f cotais a uique cycle of G. Proof Let f = uv. The there is a uique u v path P i T. Clearly, P+ f is a cycle of G, sice T is acyclic, ay cycle C of T + e should cotai e, ad C e is a u v path i T. Sice there is a uique path i T, T + e cotais a uique cycle of G. Remarks 1. If f 1 ad f are two distict chords of the coected graph G relative to a spaig tree T, the there are two uique distict cycles C 1 ad C of G cotaiig respectively f 1 ad f.. If e E(G) ad T is a spaig tree of G, the T + e cotais a uique cycle of K.

20 98 Trees Let (δ, g) deote the miimum order (miimum vertices) of a graph with miimum degree at least δ ( 3) ad girth at least g ( ). Let (, g) deote the maximum order of a graph with degree at most ad girth at most g. The followig upper boud for (δ, g) ca be foud i Bollobas [9]. Theorem 4.9 (Bollobas) (δ, g) (δ) g. Proof Clearly, (δ, g) deotes the miimum order of a graph with miimum degree at least δ ( 3) ad girth at least g ( ). Therefore we costruct a graph with atmost (δ) g vertices with these properties. Let = (δ) g. Cosider all graphs with vertex set V = {1,,..., } ad havig exactly δ edges. Sice there are ( ) possible positios to accommodate these δ edges, the umber of such graphs ) (( = δ ). Amog the available vertices, the umber of ways a h-cycle ca be formed is = 1 Obviously, 1 ( h) (h 1)! ( h) (h 1)! < 1 h h. The umber of graphs i the set which cotai a give h-cycle is ( ( ) ) = h. δ h Hece the average umber of cycles of legth at most g 1 i these graphs g 1 (( 1 < h=3 h h ) h δ h < g 1 (δ) h < (δ) g =. h=3 )/(( ) ) δ Sice the average is less tha, there is a elemet i the set with value less tha or equal to 1. Thus there is a graph G o vertices with δ edges ad at most 1 cycles of legth at most g 1. Removig oe edge from each of these cycles, we get a graph G 0 with girth at least g. The umber of edges removed is atmost 1, so that m(g o ) δ ( 1) (δ 1)+1 ad (G 0 ) =. Thus G 0 G δ 1, ad hece G 0 cotais

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