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


 Edwin Maxwell
 2 years ago
 Views:
Transcription
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
1.3. VERTEX DEGREES & COUNTING
35 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 36 its eighbor o P. Note that P has at least three vertices. If G x v is coected, let y = v. Otherwise, a compoet cut off from P x v
More information2. Degree Sequences. 2.1 Degree Sequences
2. Degree Sequeces The cocept of degrees i graphs has provided a framewor for the study of various structural properties of graphs ad has therefore attracted the attetio of may graph theorists. Here we
More information{{1}, {2, 4}, {3}} {{1, 3, 4}, {2}} {{1}, {2}, {3, 4}} 5.4 Stirling Numbers
. Stirlig Numbers Whe coutig various types of fuctios from., we quicly discovered that eumeratig the umber of oto fuctios was a difficult problem. For a domai of five elemets ad a rage of four elemets,
More informationx(x 1)(x 2)... (x k + 1) = [x] k n+m 1
1 Coutig mappigs For every real x ad positive iteger k, let [x] k deote the fallig factorial ad x(x 1)(x 2)... (x k + 1) ( ) x = [x] k k k!, ( ) k = 1. 0 I the sequel, X = {x 1,..., x m }, Y = {y 1,...,
More information5 Boolean Decision Trees (February 11)
5 Boolea Decisio Trees (February 11) 5.1 Graph Coectivity Suppose we are give a udirected graph G, represeted as a boolea adjacecy matrix = (a ij ), where a ij = 1 if ad oly if vertices i ad j are coected
More informationMath Discrete Math Combinatorics MULTIPLICATION PRINCIPLE:
Math 355  Discrete Math 4.14.4 Combiatorics Notes MULTIPLICATION PRINCIPLE: If there m ways to do somethig ad ways to do aother thig the there are m ways to do both. I the laguage of set theory: Let
More informationSection IV.5: Recurrence Relations from Algorithms
Sectio IV.5: Recurrece Relatios from Algorithms Give a recursive algorithm with iput size, we wish to fid a Θ (best big O) estimate for its ru time T() either by obtaiig a explicit formula for T() or by
More information1. MATHEMATICAL INDUCTION
1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: 1 + 2 + 3 +... + ( + 1 2 (1.1 STEP 1: For 1 (1.1 is true, sice 1 1(1 + 1. 2 STEP 2: Suppose (1.1 is true for some k 1, that is 1
More informationInfinite Sequences and Series
CHAPTER 4 Ifiite Sequeces ad Series 4.1. Sequeces A sequece is a ifiite ordered list of umbers, for example the sequece of odd positive itegers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...
More informationIn nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008
I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces
More informationDepartment of Computer Science, University of Otago
Departmet of Computer Sciece, Uiversity of Otago Techical Report OUCS200609 Permutatios Cotaiig May Patters Authors: M.H. Albert Departmet of Computer Sciece, Uiversity of Otago Micah Colema, Rya Fly
More informationLecture 7: Borel Sets and Lebesgue Measure
EE50: Probability Foudatios for Electrical Egieers JulyNovember 205 Lecture 7: Borel Sets ad Lebesgue Measure Lecturer: Dr. Krisha Jagaatha Scribes: Ravi Kolla, Aseem Sharma, Vishakh Hegde I this lecture,
More informationrepresented by 4! different arrangements of boxes, divide by 4! to get ways
Problem Set #6 solutios A juggler colors idetical jugglig balls red, white, ad blue (a I how may ways ca this be doe if each color is used at least oce? Let us preemptively color oe ball i each color,
More informationMath 475, Problem Set #6: Solutions
Math 475, Problem Set #6: Solutios A (a) For each poit (a, b) with a, b oegative itegers satisfyig ab 8, cout the paths from (0,0) to (a, b) where the legal steps from (i, j) are to (i 2, j), (i, j 2),
More informationChapter 5 O A Cojecture Of Erdíos Proceedigs NCUR VIII è1994è, Vol II, pp 794í798 Jeærey F Gold Departmet of Mathematics, Departmet of Physics Uiversity of Utah Do H Tucker Departmet of Mathematics Uiversity
More informationif A S, then X \ A S, and if (A n ) n is a sequence of sets in S, then n A n S,
Lecture 5: Borel Sets Topologically, the Borel sets i a topological space are the σalgebra geerated by the ope sets. Oe ca build up the Borel sets from the ope sets by iteratig the operatios of complemetatio
More informationORDERS OF GROWTH KEITH CONRAD
ORDERS OF GROWTH KEITH CONRAD Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really wat to uderstad their behavior It also helps you better grasp topics i calculus
More informationLecture 13. Lecturer: Jonathan Kelner Scribe: Jonathan Pines (2009)
18.409 A Algorithmist s Toolkit October 27, 2009 Lecture 13 Lecturer: Joatha Keler Scribe: Joatha Pies (2009) 1 Outlie Last time, we proved the BruMikowski iequality for boxes. Today we ll go over the
More information0,1 is an accumulation
Sectio 5.4 1 Accumulatio Poits Sectio 5.4 BolzaoWeierstrass ad HeieBorel Theorems Purpose of Sectio: To itroduce the cocept of a accumulatio poit of a set, ad state ad prove two major theorems of real
More informationMATH 361 Homework 9. Royden Royden Royden
MATH 61 Homework 9 Royde..9 First, we show that for ay subset E of the real umbers, E c + y = E + y) c traslatig the complemet is equivalet to the complemet of the traslated set). Without loss of geerality,
More informationModule 4: Mathematical Induction
Module 4: Mathematical Iductio Theme 1: Priciple of Mathematical Iductio Mathematical iductio is used to prove statemets about atural umbers. As studets may remember, we ca write such a statemet as a predicate
More informationSAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx
SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL 006 3 4 Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x 3 3 + 3 of the iterval
More informationAsymptotic Growth of Functions
CMPS Itroductio to Aalysis of Algorithms Fall 3 Asymptotic Growth of Fuctios We itroduce several types of asymptotic otatio which are used to compare the performace ad efficiecy of algorithms As we ll
More information1 The Binomial Theorem: Another Approach
The Biomial Theorem: Aother Approach Pascal s Triagle I class (ad i our text we saw that, for iteger, the biomial theorem ca be stated (a + b = c a + c a b + c a b + + c ab + c b, where the coefficiets
More informationTHE ARITHMETIC OF INTEGERS.  multiplication, exponentiation, division, addition, and subtraction
THE ARITHMETIC OF INTEGERS  multiplicatio, expoetiatio, divisio, additio, ad subtractio What to do ad what ot to do. THE INTEGERS Recall that a iteger is oe of the whole umbers, which may be either positive,
More informationReview for College Algebra Final Exam
Review for College Algebra Fial Exam (Please remember that half of the fial exam will cover chapters 14. This review sheet covers oly the ew material, from chapters 5 ad 7.) 5.1 Systems of equatios i
More informationDiscrete Mathematics and Probability Theory Spring 2014 Anant Sahai Note 13
EECS 70 Discrete Mathematics ad Probability Theory Sprig 2014 Aat Sahai Note 13 Itroductio At this poit, we have see eough examples that it is worth just takig stock of our model of probability ad may
More informationThe Field of Complex Numbers
The Field of Complex Numbers S. F. Ellermeyer The costructio of the system of complex umbers begis by appedig to the system of real umbers a umber which we call i with the property that i = 1. (Note that
More informationChapter 6: Variance, the law of large numbers and the MonteCarlo method
Chapter 6: Variace, the law of large umbers ad the MoteCarlo method Expected value, variace, ad Chebyshev iequality. If X is a radom variable recall that the expected value of X, E[X] is the average value
More informationSolutions to Exercises Chapter 4: Recurrence relations and generating functions
Solutios to Exercises Chapter 4: Recurrece relatios ad geeratig fuctios 1 (a) There are seatig positios arraged i a lie. Prove that the umber of ways of choosig a subset of these positios, with o two chose
More informationConvexity, Inequalities, and Norms
Covexity, Iequalities, ad Norms Covex Fuctios You are probably familiar with the otio of cocavity of fuctios. Give a twicedifferetiable fuctio ϕ: R R, We say that ϕ is covex (or cocave up) if ϕ (x) 0 for
More informationCS103X: Discrete Structures Homework 4 Solutions
CS103X: Discrete Structures Homewor 4 Solutios Due February 22, 2008 Exercise 1 10 poits. Silico Valley questios: a How may possible sixfigure salaries i whole dollar amouts are there that cotai at least
More informationThe Euler Totient, the Möbius and the Divisor Functions
The Euler Totiet, the Möbius ad the Divisor Fuctios Rosica Dieva July 29, 2005 Mout Holyoke College South Hadley, MA 01075 1 Ackowledgemets This work was supported by the Mout Holyoke College fellowship
More informationPage 2 of 14 = T(2) + 2 = [ T(3)+1 ] + 2 Substitute T(3)+1 for T(2) = T(3) + 3 = [ T(4)+1 ] + 3 Substitute T(4)+1 for T(3) = T(4) + 4 After i
Page 1 of 14 Search C455 Chapter 4  Recursio Tree Documet last modified: 02/09/2012 18:42:34 Uses: Use recursio tree to determie a good asymptotic boud o the recurrece T() = Sum the costs withi each level
More information4.1 Sigma Notation and Riemann Sums
0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas
More informationπ d i (b i z) (n 1)π )... sin(θ + )
SOME TRIGONOMETRIC IDENTITIES RELATED TO EXACT COVERS Joh Beebee Uiversity of Alaska, Achorage Jauary 18, 1990 Sherma K Stei proves that if si π = k si π b where i the b i are itegers, the are positive
More informationSequences and Series
CHAPTER 9 Sequeces ad Series 9.. Covergece: Defiitio ad Examples Sequeces The purpose of this chapter is to itroduce a particular way of geeratig algorithms for fidig the values of fuctios defied by their
More informationSequences II. Chapter 3. 3.1 Convergent Sequences
Chapter 3 Sequeces II 3. Coverget Sequeces Plot a graph of the sequece a ) = 2, 3 2, 4 3, 5 + 4,...,,... To what limit do you thik this sequece teds? What ca you say about the sequece a )? For ǫ = 0.,
More informationARITHMETIC AND GEOMETRIC PROGRESSIONS
Arithmetic Ad Geometric Progressios Sequeces Ad ARITHMETIC AND GEOMETRIC PROGRESSIONS Successio of umbers of which oe umber is desigated as the first, other as the secod, aother as the third ad so o gives
More information6 Algorithm analysis
6 Algorithm aalysis Geerally, a algorithm has three cases Best case Average case Worse case. To demostrate, let us cosider the a really simple search algorithm which searches for k i the set A{a 1 a...
More informationRiemann Sums y = f (x)
Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, oegative fuctio o the closed iterval [a, b] Fid
More informationVladimir N. Burkov, Dmitri A. Novikov MODELS AND METHODS OF MULTIPROJECTS MANAGEMENT
Keywords: project maagemet, resource allocatio, etwork plaig Vladimir N Burkov, Dmitri A Novikov MODELS AND METHODS OF MULTIPROJECTS MANAGEMENT The paper deals with the problems of resource allocatio betwee
More informationNotes on exponential generating functions and structures.
Notes o expoetial geeratig fuctios ad structures. 1. The cocept of a structure. Cosider the followig coutig problems: (1) to fid for each the umber of partitios of a elemet set, (2) to fid for each the
More informationWeek 3 Conditional probabilities, Bayes formula, WEEK 3 page 1 Expected value of a random variable
Week 3 Coditioal probabilities, Bayes formula, WEEK 3 page 1 Expected value of a radom variable We recall our discussio of 5 card poker hads. Example 13 : a) What is the probability of evet A that a 5
More informationThe Field Q of Rational Numbers
Chapter 3 The Field Q of Ratioal Numbers I this chapter we are goig to costruct the ratioal umber from the itegers. Historically, the positive ratioal umbers came first: the Babyloias, Egyptias ad Grees
More informationLecture 4: Cauchy sequences, BolzanoWeierstrass, and the Squeeze theorem
Lecture 4: Cauchy sequeces, BolzaoWeierstrass, ad the Squeeze theorem The purpose of this lecture is more modest tha the previous oes. It is to state certai coditios uder which we are guarateed that limits
More informationPermutations, the Parity Theorem, and Determinants
1 Permutatios, the Parity Theorem, ad Determiats Joh A. Guber Departmet of Electrical ad Computer Egieerig Uiversity of Wiscosi Madiso Cotets 1 What is a Permutatio 1 2 Cycles 2 2.1 Traspositios 4 3 Orbits
More information8.3 POLAR FORM AND DEMOIVRE S THEOREM
SECTION 8. POLAR FORM AND DEMOIVRE S THEOREM 48 8. POLAR FORM AND DEMOIVRE S THEOREM Figure 8.6 (a, b) b r a 0 θ Complex Number: a + bi Rectagular Form: (a, b) Polar Form: (r, θ) At this poit you ca add,
More informationSoving Recurrence Relations
Sovig Recurrece Relatios Part 1. Homogeeous liear 2d degree relatios with costat coefficiets. Cosider the recurrece relatio ( ) T () + at ( 1) + bt ( 2) = 0 This is called a homogeeous liear 2d degree
More information2.7 Sequences, Sequences of Sets
2.7. SEQUENCES, SEQUENCES OF SETS 67 2.7 Sequeces, Sequeces of Sets 2.7.1 Sequeces Defiitio 190 (sequece Let S be some set. 1. A sequece i S is a fuctio f : K S where K = { N : 0 for some 0 N}. 2. For
More informationChapter 7 Methods of Finding Estimators
Chapter 7 for BST 695: Special Topics i Statistical Theory. Kui Zhag, 011 Chapter 7 Methods of Fidig Estimators Sectio 7.1 Itroductio Defiitio 7.1.1 A poit estimator is ay fuctio W( X) W( X1, X,, X ) of
More informationNPTEL STRUCTURAL RELIABILITY
NPTEL Course O STRUCTURAL RELIABILITY Module # 0 Lecture 1 Course Format: Web Istructor: Dr. Aruasis Chakraborty Departmet of Civil Egieerig Idia Istitute of Techology Guwahati 1. Lecture 01: Basic Statistics
More informationDivide and Conquer, Solving Recurrences, Integer Multiplication Scribe: Juliana Cook (2015), V. Williams Date: April 6, 2016
CS 6, Lecture 3 Divide ad Coquer, Solvig Recurreces, Iteger Multiplicatio Scribe: Juliaa Cook (05, V Williams Date: April 6, 06 Itroductio Today we will cotiue to talk about divide ad coquer, ad go ito
More informationA probabilistic proof of a binomial identity
A probabilistic proof of a biomial idetity Joatho Peterso Abstract We give a elemetary probabilistic proof of a biomial idetity. The proof is obtaied by computig the probability of a certai evet i two
More informationCS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations
CS3A Hadout 3 Witer 00 February, 00 Solvig Recurrece Relatios Itroductio A wide variety of recurrece problems occur i models. Some of these recurrece relatios ca be solved usig iteratio or some other ad
More informationRecursion and Recurrences
Chapter 5 Recursio ad Recurreces 5.1 Growth Rates of Solutios to Recurreces Divide ad Coquer Algorithms Oe of the most basic ad powerful algorithmic techiques is divide ad coquer. Cosider, for example,
More informationSection 11.3: The Integral Test
Sectio.3: The Itegral Test Most of the series we have looked at have either diverged or have coverged ad we have bee able to fid what they coverge to. I geeral however, the problem is much more difficult
More informationLecture Notes CMSC 251
We have this messy summatio to solve though First observe that the value remais costat throughout the sum, ad so we ca pull it out frot Also ote that we ca write 3 i / i ad (3/) i T () = log 3 (log ) 1
More information1.3 Binomial Coefficients
18 CHAPTER 1. COUNTING 1. Biomial Coefficiets I this sectio, we will explore various properties of biomial coefficiets. Pascal s Triagle Table 1 cotais the values of the biomial coefficiets ( ) for 0to
More informationAlternatives To Pearson s and Spearman s Correlation Coefficients
Alteratives To Pearso s ad Spearma s Correlatio Coefficiets Floreti Smaradache Chair of Math & Scieces Departmet Uiversity of New Mexico Gallup, NM 8730, USA Abstract. This article presets several alteratives
More informationTaking DCOP to the Real World: Efficient Complete Solutions for Distributed MultiEvent Scheduling
Taig DCOP to the Real World: Efficiet Complete Solutios for Distributed MultiEvet Schedulig Rajiv T. Maheswara, Milid Tambe, Emma Bowrig, Joatha P. Pearce, ad Pradeep araatham Uiversity of Souther Califoria
More informationCOMP 251 Assignment 2 Solutions
COMP 251 Assigmet 2 Solutios Questio 1 Exercise 8.34 Treat the umbers as 2digit umbers i radix. Each digit rages from 0 to 1. Sort these 2digit umbers ith the RADIXSORT algorithm preseted i Sectio
More informationLecture 2: Karger s Min Cut Algorithm
priceto uiv. F 3 cos 5: Advaced Algorithm Desig Lecture : Karger s Mi Cut Algorithm Lecturer: Sajeev Arora Scribe:Sajeev Today s topic is simple but gorgeous: Karger s mi cut algorithm ad its extesio.
More information+ 1= x + 1. These 4 elements form a field.
Itroductio to fiite fields II Fiite field of p elemets F Because we are iterested i doig computer thigs it would be useful for us to costruct fields havig elemets. Let s costruct a field of elemets; we
More informationFactors of sums of powers of binomial coefficients
ACTA ARITHMETICA LXXXVI.1 (1998) Factors of sums of powers of biomial coefficiets by Neil J. Cali (Clemso, S.C.) Dedicated to the memory of Paul Erdős 1. Itroductio. It is well ow that if ( ) a f,a = the
More information.04. This means $1000 is multiplied by 1.02 five times, once for each of the remaining sixmonth
Questio 1: What is a ordiary auity? Let s look at a ordiary auity that is certai ad simple. By this, we mea a auity over a fixed term whose paymet period matches the iterest coversio period. Additioally,
More informationChapter 5: Inner Product Spaces
Chapter 5: Ier Product Spaces Chapter 5: Ier Product Spaces SECION A Itroductio to Ier Product Spaces By the ed of this sectio you will be able to uderstad what is meat by a ier product space give examples
More informationTHE ABRACADABRA PROBLEM
THE ABRACADABRA PROBLEM FRANCESCO CARAVENNA Abstract. We preset a detailed solutio of Exercise E0.6 i [Wil9]: i a radom sequece of letters, draw idepedetly ad uiformly from the Eglish alphabet, the expected
More informationMeasurable Functions
Measurable Fuctios Dug Le 1 1 Defiitio It is ecessary to determie the class of fuctios that will be cosidered for the Lebesgue itegratio. We wat to guaratee that the sets which arise whe workig with these
More informationHandout: How to calculate time complexity? CSE 101 Winter 2014
Hadout: How to calculate time complexity? CSE 101 Witer 014 Recipe (a) Kow algorithm If you are usig a modied versio of a kow algorithm, you ca piggyback your aalysis o the complexity of the origial algorithm
More informationNUMBERS COMMON TO TWO POLYGONAL SEQUENCES
NUMBERS COMMON TO TWO POLYGONAL SEQUENCES DIANNE SMITH LUCAS Chia Lake, Califoria a iteger, The polygoal sequece (or sequeces of polygoal umbers) of order r (where r is r > 3) may be defied recursively
More informationCME 302: NUMERICAL LINEAR ALGEBRA FALL 2005/06 LECTURE 8
CME 30: NUMERICAL LINEAR ALGEBRA FALL 005/06 LECTURE 8 GENE H GOLUB 1 Positive Defiite Matrices A matrix A is positive defiite if x Ax > 0 for all ozero x A positive defiite matrix has real ad positive
More informationA CHARACTERIZATION OF MINIMAL ZEROSEQUENCES OF INDEX ONE IN FINITE CYCLIC GROUPS
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5(1) (2005), #A27 A CHARACTERIZATION OF MINIMAL ZEROSEQUENCES OF INDEX ONE IN FINITE CYCLIC GROUPS Scott T. Chapma 1 Triity Uiversity, Departmet
More informationI. Chisquared Distributions
1 M 358K Supplemet to Chapter 23: CHISQUARED DISTRIBUTIONS, TDISTRIBUTIONS, AND DEGREES OF FREEDOM To uderstad tdistributios, we first eed to look at aother family of distributios, the chisquared distributios.
More informationRamseytype theorems with forbidden subgraphs
Ramseytype theorems with forbidde subgraphs Noga Alo Jáos Pach József Solymosi Abstract A graph is called Hfree if it cotais o iduced copy of H. We discuss the followig questio raised by Erdős ad Hajal.
More informationTrigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE 6.44. The absolute value of the complex number z a bi is
0_0605.qxd /5/05 0:45 AM Page 470 470 Chapter 6 Additioal Topics i Trigoometry 6.5 Trigoometric Form of a Complex Number What you should lear Plot complex umbers i the complex plae ad fid absolute values
More informationChapter Gaussian Elimination
Chapter 04.06 Gaussia Elimiatio After readig this chapter, you should be able to:. solve a set of simultaeous liear equatios usig Naïve Gauss elimiatio,. lear the pitfalls of the Naïve Gauss elimiatio
More informationApproximating Area under a curve with rectangles. To find the area under a curve we approximate the area using rectangles and then use limits to find
1.8 Approximatig Area uder a curve with rectagles 1.6 To fid the area uder a curve we approximate the area usig rectagles ad the use limits to fid 1.4 the area. Example 1 Suppose we wat to estimate 1.
More information6 Borel sets in the light of analytic sets
Tel Aviv Uiversity, 2012 Measurability ad cotiuity 86 6 Borel sets i the light of aalytic sets 6a Separatio theorem................. 86 6b Borel bijectios.................... 87 6c A oborel aalytic set
More informationSection 1.6: Proof by Mathematical Induction
Sectio.6 Proof by Iductio Sectio.6: Proof by Mathematical Iductio Purpose of Sectio: To itroduce the Priciple of Mathematical Iductio, both weak ad the strog versios, ad show how certai types of theorems
More informationLecture 5: Span, linear independence, bases, and dimension
Lecture 5: Spa, liear idepedece, bases, ad dimesio Travis Schedler Thurs, Sep 23, 2010 (versio: 9/21 9:55 PM) 1 Motivatio Motivatio To uderstad what it meas that R has dimesio oe, R 2 dimesio 2, etc.;
More informationAdvanced Probability Theory
Advaced Probability Theory Math5411 HKUST Kai Che (Istructor) Chapter 1. Law of Large Numbers 1.1. σalgebra, measure, probability space ad radom variables. This sectio lays the ecessary rigorous foudatio
More informationOverview of some probability distributions.
Lecture Overview of some probability distributios. I this lecture we will review several commo distributios that will be used ofte throughtout the class. Each distributio is usually described by its probability
More informationIncremental calculation of weighted mean and variance
Icremetal calculatio of weighted mea ad variace Toy Fich faf@cam.ac.uk dot@dotat.at Uiversity of Cambridge Computig Service February 009 Abstract I these otes I eplai how to derive formulae for umerically
More information2.3. GEOMETRIC SERIES
6 CHAPTER INFINITE SERIES GEOMETRIC SERIES Oe of the most importat types of ifiite series are geometric series A geometric series is simply the sum of a geometric sequece, Fortuately, geometric series
More informationYour organization has a Class B IP address of 166.144.0.0 Before you implement subnetting, the Network ID and Host ID are divided as follows:
Subettig Subettig is used to subdivide a sigle class of etwork i to multiple smaller etworks. Example: Your orgaizatio has a Class B IP address of 166.144.0.0 Before you implemet subettig, the Network
More informationN04/5/MATHL/HP2/ENG/TZ0/XX MATHEMATICS HIGHER LEVEL PAPER 2. Thursday 4 November 2004 (morning) 3 hours INSTRUCTIONS TO CANDIDATES
c IB MATHEMATICS HIGHER LEVEL PAPER DIPLOMA PROGRAMME PROGRAMME DU DIPLÔME DU BI PROGRAMA DEL DIPLOMA DEL BI N/5/MATHL/HP/ENG/TZ/XX 887 Thursday November (morig) hours INSTRUCTIONS TO CANDIDATES! Do ot
More informationSECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES
SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES Read Sectio 1.5 (pages 5 9) Overview I Sectio 1.5 we lear to work with summatio otatio ad formulas. We will also itroduce a brief overview of sequeces,
More informationON THE EDGEBANDWIDTH OF GRAPH PRODUCTS
ON THE EDGEBANDWIDTH OF GRAPH PRODUCTS JÓZSEF BALOGH, DHRUV MUBAYI, AND ANDRÁS PLUHÁR Abstract The edgebadwidth of a graph G is the badwidth of the lie graph of G We show asymptotically tight bouds o
More informationTHE UNLIKELY UNION OF PARTITIONS AND DIVISORS
THE UNLIKELY UNION OF PARTITIONS AND DIVISORS Abdulkadir Hasse, Thomas J. Osler, Mathematics Departmet ad Tirupathi R. Chadrupatla, Mechaical Egieerig Rowa Uiversity Glassboro, NJ 828 I the multiplicative
More information23 The Remainder and Factor Theorems
 The Remaider ad Factor Theorems Factor each polyomial completely usig the give factor ad log divisio 1 x + x x 60; x + So, x + x x 60 = (x + )(x x 15) Factorig the quadratic expressio yields x + x x
More informationwhere: T = number of years of cash flow in investment's life n = the year in which the cash flow X n i = IRR = the internal rate of return
EVALUATING ALTERNATIVE CAPITAL INVESTMENT PROGRAMS By Ke D. Duft, Extesio Ecoomist I the March 98 issue of this publicatio we reviewed the procedure by which a capital ivestmet project was assessed. The
More information7. Sample Covariance and Correlation
1 of 8 7/16/2009 6:06 AM Virtual Laboratories > 6. Radom Samples > 1 2 3 4 5 6 7 7. Sample Covariace ad Correlatio The Bivariate Model Suppose agai that we have a basic radom experimet, ad that X ad Y
More information1. C. The formula for the confidence interval for a population mean is: x t, which was
s 1. C. The formula for the cofidece iterval for a populatio mea is: x t, which was based o the sample Mea. So, x is guarateed to be i the iterval you form.. D. Use the rule : pvalue
More informationThe following example will help us understand The Sampling Distribution of the Mean. C1 C2 C3 C4 C5 50 miles 84 miles 38 miles 120 miles 48 miles
The followig eample will help us uderstad The Samplig Distributio of the Mea Review: The populatio is the etire collectio of all idividuals or objects of iterest The sample is the portio of the populatio
More informationLecture 4: Cheeger s Inequality
Spectral Graph Theory ad Applicatios WS 0/0 Lecture 4: Cheeger s Iequality Lecturer: Thomas Sauerwald & He Su Statemet of Cheeger s Iequality I this lecture we assume for simplicity that G is a dregular
More informationwhen n = 1, 2, 3, 4, 5, 6, This list represents the amount of dollars you have after n days. Note: The use of is read as and so on.
Geometric eries Before we defie what is meat by a series, we eed to itroduce a related topic, that of sequeces. Formally, a sequece is a fuctio that computes a ordered list. uppose that o day 1, you have
More information. P. 4.3 Basic feasible solutions and vertices of polyhedra. x 1. x 2
4. Basic feasible solutios ad vertices of polyhedra Due to the fudametal theorem of Liear Programmig, to solve ay LP it suffices to cosider the vertices (fiitely may) of the polyhedro P of the feasible
More informationSection 9.2 Series and Convergence
Sectio 9. Series ad Covergece Goals of Chapter 9 Approximate Pi Prove ifiite series are aother importat applicatio of limits, derivatives, approximatio, slope, ad cocavity of fuctios. Fid challegig atiderivatives
More informationJournal of Combinatorial Theory, Series A
Joural of Combiatorial Theory, Series A 118 011 319 345 Cotets lists available at ScieceDirect Joural of Combiatorial Theory, Series A www.elsevier.com/locate/jcta Geeratig all subsets of a fiite set with
More information