Lemma 1 Let the components of, Suppose. Trees. A tree is a graph which is. (a) Connected and. (b) has no cycles (acyclic). (b)

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1 Trees Lemma Let the components of ppose "! be (a) $&%('*)+ - )+ / A tree is a graph which is (b) 0 %(')+ - 3)+ / 6 (a) (a) Connected and (b) has no cycles (acyclic) (b) roof Eery path 8 in which is not in mst contain Also )+ - )+ /: ppose ;<=?> is a path in G and F "!?@A3?@CB D Ë =F that ses Then clearly ; H (a) follows as now no new relations ;JI6F are added (b) Only possible new relations ;KI&F are for ; and F "! L in "M "! and so Bt LI becomes (only) new component Lemma Q O R <RT UV "! (a) $&%(' - contains a cycle is acyclic and has one less com- edges (b) X0 =%' ponent (c) has Y [Z is acyclic (forest) with (tree) W 3

2 Z (a) in implies there exists a path = > o Y contains the cycle > E > EH ppose contains the cycle a cycle of =?> Ë D?> Bt then contains the path > E to contradiction l- else is from The drop in the nmber of components follows from Lemma The rest follows from 6 (c) ppose for and Claim: has components Indction on $ : > has no edges : is acyclic and so is It follows from part (a) that joins ertices in distinct components of It follows from (b) that has one less component than End of proof of claim Ths 3 [Z (we assmed had Z components) Corollary If a tree has ertices then (a) It has 6 edges (b) It has at least ertices of degree ( ) roof (a) is part (c) of preios lemma since is connnected (b) Let be the nmber of ertices of degree in There are no ertices of degree 0 these wold form separate components Ths < o 8

3 ' ' ' Theorem ppose R JR 3 and R R = following three statements become eqialent (a) is connected (b) is acyclic (c) is a tree Let and for 6 The : > has components and has component Addition of each edge mst redce the nmber of components by Lemma (b) Ths acyclic implies is acyclic (b) follows as > is acyclic ince : We need to show that is connected is acyclic )+ )+ ( for : each Lemma (b) Ths )+ : triial 0 Ct edges Corollary If is a ertex of degree in a tree then is also a tree ct edge is a ct edge of Theorem cycle of if )+ 6 )+ is a ct edge iff is not on any roof ppose has ertices and edges Then has ertices and acyclic and so mst be a tree edges It roof ) increases iff there exist ;I F sch that all walks from ; to F se ppose there is a cycle D if O ; D F is a walk from ; to F sing ; F is a walk from ; to F that doesn t se Ths is not a ct edge containing Then

4 0 x Corollary Eery finite connected graph contains a spanning tree roof Consider the following process: starting with If there are no cycles stop If < is not a ct edge then contains a path from to ( I in and relations are maintained after deletion of ) o D is a cycle containing Corollary 3 A connected graph is a tree iff eery edge is a ct edge If there is a cycle delete an edge of a cycle Obsere that (i) the graph remains connected we delete edges of cycles (ii) the process mst terminate as the nmber of edges is assmed finite On termination there are no cycles and so we hae a connected acyclic spanning sbgraph ie we hae a spanning tree 3 Theorem 3 Let be a spanning tree of R <R 3 ppose (a) (b) Q contains a niqe cycle Q implies that tree of f is a spanning roof (a) Lemma (a) implies that has a cycle ppose that contains another cycle Let edge is connected has edges Bt contains a cycle contradictng Lemma (b) (b) is connected and has Therefore it is a tree edges 6

5 B Z Q For spanning tree Maximm weight trees is a connected graph is the weight of edge roblem: find a spanning tree of maximm weight ort edges so that Greedy Algorithm for $ if then Otpt do where does not contain a cycle Greedy always adds the maximm weight edge which does not make a cycle with preiosly chosen edges 8 Theorem The tree constrcted by GREEDY has maximm weight roof Let the edges of the greedy tree be in order of choice ote that Q Q since neither makes a cycle with Let where be the edges of any other tree Z ppose () is false There exists sch 6Z makes a cycle Each of Otherwise one of them wold hae been chosen in preference Let the components of the Q Each 3Z has both of its points in the same component We show that 6 () C I f J 0

6 B!! Z Let be the nmber of which hae both points in and let be the nmber of ertices of B (3) It follows from () and (3) that there exists sch B Z 6 D Bt () implies that the edges contain a cycle () ZA! sch that n= n= n=6 How many trees? Cayley s Formla rüfer s Correspondence There is a - correspondence between spanning trees of (the complete graph with ertex set ) and seqences Ths for = Cayley s Formla Assme some arbitrary ordering : for to do neighbor of least leaf of

7 0 Lemma 3 in roof Assme R : Assme : T appears indction appears indction appears exactly = empty string s :+G/ & where times R By indction on :G X8 l times times Constrction of Indctiely assme that for all R R there is an inerse fnction (Tre for ) ow define by pls edge where and Then C Ths has an inerse and the correspondence is established 6 mber of trees with a gien degree seqence A Corollary If then the nmber of spanning trees of qence with degree se- is "!!! roof From rüfer s correspondence and Lemma 3 this is the nmber of seqences of length in K K which appears times appears times and so on