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 selfloops 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.
2 80 Trees Proof Let G be a graph ad let there be exactly oe path betwee every pair of vertices i G. So G is coected. Now G has o cycles, because if G cotais a cycle, say betwee vertices u ad v, the there are two distict paths betwee u ad v, which is a cotradictio. Thus G is coected ad is without cycles, therefore it is a tree. Coversely, let G be a tree. Sice G is coected, there is at least oe path betwee every pair of vertices i G. Let there be two distict paths betwee two vertices u ad v of G. The uio of these two paths cotais a cycle which cotradicts the fact that G is a tree. Hece there is exactly oe path betwee every pair of vertices of a tree. The ext two results give alterative methods for defiig trees. Theorem 4. A tree with vertices has 1 edges. Proof We prove the result by usig iductio o, the umber of vertices. The result is obviously true for = 1, ad 3. Let the result be true for all trees with fewer tha vertices. Let T be a tree with vertices ad let e be a edge with ed vertices u ad v. So the oly path betwee u ad v is e. Therefore deletio of e from T discoects T. Now, T e cosists of exactly two compoets T 1 ad T say, ad as there were o cycles to begi with, each compoet is a tree. Let 1 ad be the umber of vertices i T 1 ad T respectively, so that 1 + =. Also, 1 < ad <. Thus, by iductio hypothesis, umber of edges i T 1 ad T are respectively 1 1 ad 1. Hece the umber of edges i T = = = 1. Theorem 4.3 Ay coected graph with vertices ad 1 edges is a tree. Proof Let G be a coected graph with vertices ad 1 edges. We show that G cotais o cycles. Assume to the cotrary that G cotais cycles. Remove a edge from a cycle so that the resultig graph is agai coected. Cotiue this process of removig oe edge from oe cycle at a time till the resultig graph H is a tree. As H has vertices, so umber of edges i H is 1. Now, the umber of edges i G is greater tha the umber of edges i H. So 1 > 1, which is ot possible. Hece, G has o cycles ad therefore is a tree. Defiitio: A graph is said to be miimally coected if removal of ay oe edge from it discoects the graph. Clearly, a miimally coected graph has o cycles. Here is the ext characterisatio of trees. Theorem 4.4 A graph is a tree if ad oly if it is miimally coected. Proof tree. Let the graph G be miimally coected. The G has o cycles ad therefore is a Coversely, let G be a tree. The G cotais o cycles ad deletio of ay edge from G discoects the graph. Hece G is miimally coected.
3 Graph Theory 81 The followig results give some more properties of trees. Theorem 4.5 A graph G with vertices, 1 edges ad o cycles is coected. Proof Let G be a graph without cycles with vertices ad 1 edges. We have to prove that G is coected. Assume that G is discoected. So G cosists of two or more compoets ad each compoet is also without cycles. We assume without loss of geerality that G has two compoets, say G 1 ad G (Fig. 4.1(b)). Add a edge e betwee a vertex u i G 1 ad a vertex v i G. Sice there is o path betwee u ad v i G, addig e did ot create a cycle. Thus G e is a coected graph (tree) of vertices, havig edges ad o cycles. This cotradicts the fact that a tree with vertices has 1 edges. Hece G is coected. Theorem 4.6 Fig. 4.1(b) Ay tree with at least two vertices has at least two pedat vertices. Proof Let the umber of vertices i a give tree T be ( > 1). So the umber of edges i T is 1. Therefore the degree sum of the tree is ( 1). This degree sum is to be divided amog the vertices. Sice a tree is coected it caot have a vertex of 0 degree. Each vertex cotributes at least 1 to the above sum. Thus there must be at least two vertices of degree exactly 1. Secod proof We use iductio o. The result is obviously true for all trees havig fewer tha vertices. We kow that T has 1 edges, ad if every edge of T is icidet with a pedat vertex, the T has at least two pedat vertices, ad the proof is complete. So let there be some edge of T that is ot icidet with a pedat vertex ad let this edge be e = uv (Fig. 4.). Removig the edge e, we see that the graph T e cosists of a pair of trees say T 1 ad T with each havig fewer tha vertices. Let u V(T 1 ), v V(T ), ad V(T 1 ) = 1, V(T ) =. Applyig iductio hypothesis o both T 1 ad T, we observe that each of T 1 ad T has two pedat vertices. This shows that each of T 1 ad T has at least oe pedat vertex that is ot icidet with the edge e. Thus the graph T e+e = T has at least two pedat vertices. Fig. 4.
4 8 Trees Third proof Let T be a tree with ( > 1) vertices. The umber of edges i T is 1 ad the sum of degrees i T is ( 1), that is, d i = ( 1). Assume T has exactly oe vertex v 1 of degree oe, while all the other 1 vertices have degree. The sum of degrees is d(v 1 )+d(v )+...+ d(v ) = 1 + ( 1). So, ( 1) 1 + ( 1), implyig 0 1, which is absurd. Hece T has at least two vertices of degree oe. The followig result characterises tree degree sequeces. Theorem 4.7 oly if Proof Necessity The sequece [d i ] 1 of positive itegers is a degree sequece of a tree if ad (i) d i 1 for all i, 1 i ad (ii) d i =. Sice a tree has o isolated vertex, therefore d i 1 for all i. Also, ( 1), as a tree with vertices has 1 edges. d i = Sufficiecy We use iductio o. For =, the sequece is [1, 1] ad is obviously the degree sequece of K. Suppose the claim is true for all positive sequeces of legth less tha. Let [d i ] 1 be the odecreasig positive sequece of terms, satisfyig coditios (i) ad (ii). The d 1 = 1 ad d > 1 (by Theorem 4.5). Now, cosider the sequece D = [d, d 3,..., d 1, d 1], which is a sequece of legth 1. Obviously i D, d i 1 ad d i = d + d d 1 + d 1 = d 1 + d + d d 1 + d 1 1 = = ( 1) (because d 1 = 1). So D satisfies coditios (i) ad (ii), ad by iductio hypothesis there is a tree T realisig D. I T, add a ew vertex ad joi it to the vertex havig degree d 1 to get a tree T. Therefore the degree sequece of T is [d 1, d,..., d ]. Theorem 4.8 A forest of k trees which have a total of vertices has k edges. Proof Let G be a forest ad T 1, T,..., T k be the k trees of G. Let G have vertices ad T 1, T,..., T k have respectively 1,,..., k vertices. The k =. Also, the umber of edges i T 1, T,..., T k are respectively 1 1, 1,..., k 1. Thus umber of edges i G = k 1 = k k = k. The followig result characterises trees as subgraphs of a graph. Theorem 4.9 Let T be a tree with k edges. If G is a graph whose miimum degree satisfies δ(g) k, the G cotais T as a subgraph. Alteratively, G cotais every tree of order atmost δ(g)+1 as a subgraph. Proof We use iductio o k. If k = 0, the T = K 1 ad it is clear that K 1 is a subgraph of ay graph. Further, if k = 1, the T = K ad K is a subgraph of ay graph whose miimum
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
7 Graph Theory 85 are defied oly with respect to a give spaig tree. A edge that is a brach of oe spaig tree T 1 (i a graph G) may be a chord with respect to aother spaig tree T. I Figure 4.5, u 1 u u 3 u 4 u 5 u 6 is a spaig tree, u u 4 ad u 4 u 6 are chords. Fig. 4.5 A coected graph G ca be cosidered as a uio of two subgraphs T ad T, that is G = T T, where T is a spaig tree, T is the complemet of T i G. T beig the set of chords is called the co tree, or chord set. The followig result provides the umber of chords i ay graph with a spaig tree. Theorem 4.13 With respect to ay of its spaig trees, a coected graph of vertices ad m edges has 1 tree braches ad m +1 chords. Proof Let G be a coected graph with vertices ad m edges. Let T be the spaig tree. Sice T cotais all vertices of G, T has 1 edges ad thus the umber of chords i G is equal to m ( 1) = m +1. Defiitio: Let G be a graph with vertices, m edges ad k compoets. The rak r ad ullity µ of G are defied as r = k ad µ = m +k. Clearly, the rak of a coected graph is 1 ad the ullity is m + 1. It ca be see that rak of G = umber of braches i ay spaig tree (or forest) of G. Also, ullity of G = umber of chords i G. So, rak + ullity = umber of edges i G. The ullity of a graph is also called its cyclomatic umber, or first Betti umber. Theorem 4.14 If T is a tree with k 0 vertices of odd degree, the E(T) is the uio of k pairwise edgedisjoit paths. Proof We prove the result for every forest G, usig iductio o k. If k = 0, the G has o pedat vertex ad therefore o edge. Let k > 0 ad let each forest with k vertices of odd degree has decompositio ito k 1 paths. Sice k > 0, some compoet of G is a tree with at least two vertices. This compoet has at least two pedat vertices. Let P be the path coectig two pedat vertices. Deletig E(P) chages the parity of the vertex degree oly for the ed vertices of P ad it makes them eve. Thus G E(P) is a forest with k vertices of odd degree. So by the iductio hypothesis, G E(P) is the uio of
8 86 Trees k 1 pair wise edgedisjoit paths. These k 1 edgedisjoit paths together with P partitio E(G) ito k pair wise edgedisjoit paths (Fig. 4.6). Fig. 4.6 Theorem 4.15 Let T be a otrivial tree with the vertex set S ad S = k, k 1. The there exists a set of k pairwise edgedisjoit paths whose ed vertices are all the vertices of S. Proof Obviously, there exists a set of k paths i T whose ed vertices are all the vertices of S. Let P = {P 1, P,..., P k } be such a set of k paths ad let the sum of their legths be the miimum. We show that the paths of P are pairwise edgedisjoit. Assume to the cotrary, ad let P i ad P j, i j, be paths havig a edge i commo. The P i ad P j have path P i j of legth 1 i commo. Therefore, P i P j the symmetric differece of P i ad P j is a disjoit uio of two paths, say Q i ad Q j, with their ed vertices beig disjoit pairs of vertices belogig to S (Fig. 4.7). If P i ad P j are replaced by Q i ad Q j i P, the the resultig set of paths has the property that their ed vertices are all the vertices of S ad that the sum of their legths is less tha the sum of the legths of the paths i P. This is a cotradictio to the choice of P. Fig. 4.7
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.
11 Graph Theory 89 Remarks The above result implies that for ay vertex v of a coected graph G, there exists a image Φ v (G) which is a spaig tree of G preservig distaces from v. This is called a isometric tree of G at v. If there is oly oe such tree (upto isomorphism) at v, we say that G has a uique isometric tree at v. If G has the same uique isometric tree at each vertex v, the G is said to have a uique isometric tree (or uique distace tree). K, ad the Peterso graph are examples of graphs havig uique isometric trees, while K 3, 3 does ot have a uique isometric tree at ay vertex. Every tree has a uique isometric tree. The ext result due to Chartrad ad Stewart [5] gives the ecessary coditio for a graph to have a uique isometric tree. Theorem 4.18 Let G be a coected graph with d = r, which has a uique isometric tree. The the ed vertices of every diametral path of G has degree 1. Proof Let G be a coected graph with d = r ad let P be a diametral path with ed vertices u ad v. If possible let d(u G) > 1. Let T u be the isometric tree at u. It is easy to see that T u ca be chose to cotai P. Sice u has degree at least i G, there is a vertex u i adjacet to u ad ot lyig i P. Clearly, d Tu (u i, v) = 1+d. Let c be a cetral vertex of G. The for ay two vertices w 1 ad w of G, we have d G (w 1, c) r = 1 d ad d G(w, c) r = 1 d. Therefore, d G (w 1, w ) d(w 1, c)+d(c, w ) d. Sice T c is isometric with G at c, we also have d T c (w 1, w ) d. Thus o path T c has legth greater tha d, whereas there is a path i T u of legth 1 + d. Therefore T c T u ad G does ot have a uique isometric tree. This cotradicts the hypothesis. Hece the result follows. Remark The above coditio is ecessary but ot sufficiet. To see this, cosider the graph give i Figure 4.9. Fig. 4.9 Graph without a uique isometric tree Chartrad ad Schuster [54], ad Kudu [14] have give some more results o the graphs with uique isometric trees. Defiitio: of G. The complexity τ(g) of a graph G is the umber of differet spaig trees
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 oeoe correspodece betwee the elemets of S 1 ad the spaig trees of G e. Also, there is a oeoe 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
13 Graph Theory 91 Label the edges of G arbitrarily. Choose e 1 as the first cyclic edge. The τ(g) is the sum of the complexities of the graphs give i Figure 4.10(b) ad (c). Now, choose e 4 i both G 1 ad G as the ext cyclic edge. The τ(g) is the sum of the complexities of the graphs i Figure 4.10(d) ad (e). Sice there are o more cyclic edges, the algorithm termiates, ad we have τ(g) = = Number of Labelled Trees Let us cosider the problem of costructig all simple graphs with vertices ad m edges. There are ( 1)/ uordered pairs of vertices. If the vertices are distiguishable from each other ( (i.e., labelled ) graphs), the the umber of ways of selectig m edges to form the ( 1) graph is. m Thus the umber of simple labelled graphs with vertices ad m edges is ( ) ( 1). (A) m Clearly, may of these graphs ca be isomorphic (that is they are same except for the labels of their vertices). Thus the umber of simple, ulabelled graphs of vertices ad m edges is much smaller tha that give by (A) above. Theorem 4.0 The umber of simple, labelled graphs of vertices is ( 1). Proof The umber of simple graphs of vertices ad 0, 1,,..., ( 1)/ edges are obtaied by substitutig 0, 1,,..., ( 1)/ for m i (A). The sum of all such umbers is the umber of all simple graphs with vertices. Therefore the total umber of simple, labelled graphs of vertices is ( ( 1) 0 ) + by usig the idetity ( ( 1) 1 ) + ( ) k + 0 ( ( 1) ( ) k + 1 ) ( ) k ( ( 1) ) ( 1) = ( 1), ( ) k = k. k The followig result was proved idepedetly by Tutte [5] ad NashWilliams [167]. We prove the ecessity ad for sufficiecy the reader is referred to the origial papers of Tutte ad NashWilliams. Theorem 4.1 A simple coected graph G cotais k pairwise edgedisjoit spaig trees if ad oly if, for each partitio π of V(G) ito p parts, the umber m (π) of edges of G joiig distict parts is at least k(p 1).
14 9 Trees Proof Necessity Let G has k pairwise edgedisjoit spaig trees. If T is oe of them, ad if π = {V 1, V,..., V p } is a partitio of V(G) ito p parts, the idetificatio of each part V i ito a sigle vertex v i, 1 i p, results i a coected graph G 0 (possibly with multiple edges) o {V 1, V,..., V p }. Clearly, G 0 cotais a spaig tree with p 1 edges, ad each such edge belogs to T, ad jois distict partite sets of π. Sice this is true for each of the k edge disjoit spaig trees of G, the umber of edges joiig distict parts of π is at least k(p 1). Cayley [46] i 1889 discovered the formula τ(k ) =. Clearly, the umber of spaig trees of K is same as the umber of olabelisomorphic trees o vertices. Several proofs of this result have appeared sice Cayley s discovery. Moo [164] has outlied te such proofs, ad a complete presetatio of some of these ca also be foud i Lovasz [15]. Here we give two proofs, ad the first is due to Prufer [1]. Theorem 4. (Cayley) There are labelled trees with vertices,. Proof Let T be a tree with vertices ad let the vertices be labelled 1,,...,. Remove the pedat vertex (ad the edge icidet to it) havig the smallest label, say u 1. Let v 1 be the vertex adjacet to u 1. From the remaiig 1 vertices, let u be the pedat vertex with the smallest label ad let v be the vertex adjacet to u. We remove u ad the edge icidet o it. We repeat this operatio o the remaiig vertices, the o 3 vertices, ad so o. This process completes after steps, whe oly two vertices are left. Let the vertices after each removal have labels v 1, v,..., v. Clearly, the tree T uiquely defies the sequece (v 1, v,..., v ). (4..1) Coversely, give a sequece of labels, a vertex tree is costructed uiquely as follows. Determie the first umber i the sequece 1,, 3,...,, (4..) that does ot appear i (4..1). Let this umber be u 1. Thus the edge (u 1, v 1 ) is defied. Remove v 1 from sequece (4..1) ad u 1 from (4..). I the remaiig sequece of (4..), fid the first umber which does ot appear i the remaiig sequece of (4..1). Let this be u ad thus the edge (u, v ) is defied. The costructio is cotiued till the sequece (4..1) has o elemet left. Fially, the last two vertices remaiig i (4..) are joied. For each of the elemets i sequece (4..1), we choose ay oe of the umbers, thus formig ( )tuples, each defiig a distict labelled tree of vertices. Sice each tree defies oe of these sequeces uiquely, there is a oe oe correspodece betwee the trees ad the sequeces.
15 Graph Theory 93 Example Cosider the tree show i Figure Pedat vertex with smallest label is u 1. Remove u 1. Let v 1 be adjacet to u 1 (label of v 1 is 1). Pedat vertex with smallest label is 4. Remove 4. Here 4 is adjacet to 1. Pedat vertex with smallest label is 1. Remove 1. Here 1 is adjacet to 3. Remove 3. The 3 is adjacet to 5. Remove 6. So 6 is adjacet to 5. Remove 5. Remove 7. 7 is adjacet to 5. So 5 is adjacet to 9. Sequece (v 1, v,..., v ) is (1, 1, 3, 5, 5, 5, 9). Fig Theorem 4.3 If D = [ d i ] 1 is the degree sequece of a tree, the the umber of labelled trees with this degree sequece is ( )! (d 1 1)!(d 1)!...(d 1)!. Proof We first observe that, whe askig for all possible trees with the vertex label set V = {v 1, v,..., v } with degree sequece D = [ d i ] 1, it is ot ecessary that d i = d (v i ) ad it is ot ecessary that the sequece be mootoic odecreasig. Therefore we assume that D = [ d i ] 1 is a iteger sequece satisfyig the coditios d i = ( 1) ad d i 1. We use iductio o. The result is obvious for = 1,. For =, the sequece is [ d 1, d ] ad the oly degree sequece i this case is [ 1, 1]. Clearly, there is oly oe labelled tree with this degree sequece. Also, ( )! (d 1 1)!...(d 1) = ( )! (1 1)!(1 1)! = 1. Now, assume that the result is true for all sequeces of legth 1. Let D = [ d i ] 1 be a legth sequece. By assumptio there is a d i = 1 ad let it be d = 1. Let T be a tree realisig D = [ d i ] 1. Now, removig v, we get a tree T 1 o the vertex set {v 1, v,..., v 1 } with degrees d 1,..., d j 1, d j 1, d j+1,..., d 1, where v j is the vertex to which v is adjacet i
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.
18 96 Trees Defiitio: Let G be a coected graph with vertices ad m edges. The umber of chords of G relative to a spaig tree T of G is m +1 = µ. The µ distict cycles of a coected graph G correspodig to the distict chords of G relative to a spaig tree T of G are said to form a set of fudametal cycles of G. If G is a discoected graph with k compoets G 1, G,..., G k ad T i, 1 i k, are a set of k spaig trees of G i, the the uio of the set of fudametal cycles of G i with respect to T i is a set of fudametal cycles for G. It is to be oted that differet spaig trees give differet sets of fudametal cycles. The followig result characterises cycles i terms of the set of all spaig trees. Theorem 4.5 Ay cycle of a coected graph G cotais at least oe chord of every spaig tree of G. Proof Let C be a cycle ad assume the result is ot true. So there exists a spaig tree T of G such that C is cotaied i the edge set E(G) E(T), where T is the cotree of G correspodig to T. This meas that the tree T cotais the cycle C, which is a cotradictio. Theorem 4.6 A set of edges C of a coected graph G is a cycle of G if ad oly if it is a miimal set of edges cotaiig at least oe chord of every spaig tree of G. Proof Let C be a cycle of G. The it cotais at least oe chord of every spaig tree of G. If C is ay proper subset of C, the C does ot cotai a cycle ad is a forest. A spaig tree T of G ca therefore be costructed cotaiig C. Clearly, C does ot cotai ay chord of T. Thus o proper subset of C has the stated property, provig that C is miimal with respect to the property. To prove sufficiecy, let C be miimal set with the stated property. The C is ot acyclic. Therefore C cotais at least a cycle C. But by the ecessary part, C is miimal with respect to the property ad hece C = C, that is, C is a cycle. 4.5 Geeratio of Trees Defiitio: Let T 1 ad T be two spaig trees of a coected graph G ad let there be edges e 1 T 1 ad e T such that T 1 e 1 + e = T (ad hece T e + e 1 = T 1 ). The trasformatio T 1 T is called a elemetary tree trasformatio (ETT), or a fudametal exchage. If e 1 ad e are adjacet i G, the the ETT is called a eighbour trasformatio (NT). If e is a pedat edge of T 1 (ad hece e is a pedat edge of T ) the ETT is called a pedatedge trasformatio (PET) or a edlie trasformatio. Defiitio: Let I be the collectio of all spaig trees of a coected graph G. Let Tr(G) be the graph whose vertices t i correspod to the elemets T i of I, ad i which t i ad t j are adjacet if ad oly if there is a ETT betwee T i ad T j, that is, if ad oly if E(T i ) E(T j ) = {e i, e j }. The Tr(G) is called the tree graph of G. The distace d(t i, T j )
19 Graph Theory 97 betwee the spaig trees T i ad T j of G is defied to be the distace betwee t i ad t j i Tr(G). Theorem 4.7 The tree graph Tr(G) of a coected graph is coected. Proof Let G be a coected graph with vertices ad let Tr(G) be its tree graph. To prove that Tr(G) is coected, it is eough to show that ay two spaig trees of G ca be obtaied from each other by a fiite sequece of ETT s. Let T ad T be two distict spaig trees of G. The there is a set S = {e 1, e,..., e k } of some k edges of T which are ot i T. Sice a spaig tree has 1 edges, there is a correspodig set S = {e 1, e,..., e k } of edges of T which are ot i T. Thus T + e 1 cotais a uique fudametal cycle T e 1. As T is a tree, at least oe edge of T e 1 (which is a brach of T) will ot be i T ad thus is a member of S. Without loss of geerality, let this edge be e 1. Defie T 1 = T e 1 + e 1. The T 1 ca be obtaied from T by a ETT ad therefore T 1 ad T have oe more edge i commo. Repeatig this process k 1 more times, we get a sequece of spaig trees T 0 = T, T 1, T,..., T k 1, T k = T such that there is a ETT T i T i+1, 0 i k 1. Theorem 4.8 A elemetary tree trasformatio ca be obtaied by a sequece of eighbour trasformatios. Proof Let T ad T = T x + y be spaig trees of the graph G, where x ad y are oadjacet edges of G. The we ca choose a set of edges e 1, e,..., e k such that x, e 1, e,..., e k, y is a path i T + y. Defie T 1 = T x+e 1 ad T i = T i 1 + e i 1 + e i, i k ad T k+1 = T k e k + y. The T k+1 = T, ad is obtaied from T by a sequece of k + 1 eighbour trasformatios through the itermediate trees T i, 1 i k. Defiitio: A spaig tree of a graph G correspodig to a cetral vertex of the tree Tr(G) is called a cetral tree. The set of diameters of the spaig trees of a coected graph G is the tree diameter set of G. A set of positive itegers is a feasible tree diameter set if it is the tree diameter set of some graph. For example, the graph i Figure 4.1 has oe spaig tree of diameter seve ad all others of diameter five. Fig. 4.1 The girth g(g) of a graph G is the legth of a smallest cycle of G. A cycle of smallest legth is called a girdle of G. The circumferece c(g) of a graph G is the legth of the logest cycle of G. A cycle of maximum legth is called a hem of G.
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 hcycle 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 hcycle 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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