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Running head: DISCREET MATHEMATICS1

DISCREET MATHEMATICS21.No, if you take relation R on a set 1. 2, 3WhereR={(1,1),(1,2),(2,1),(1,3),(3,1),(2,2),(2,3)}The function is symmetric since for allx,yεRthereis(y,x)εRAlso, it is transitive as for allx,y,zεA,(x,y)εR,(y,z)εRthereis(x,z)εRDespite being transitive and symmetric the relation is not reflexive as forxεA,(x,x)εRismissingi.e. for x=3,(3,3)≠R2.A functionf(x)isone to one iff an input yields a unique outputOn the other hand, a function is onto iff every element of the image has a preimageThe function f: Z6→Z6will be one to one and onto which is the same case forf: Z5→Z5A multiplicative inverse exists for the elements which are onto and one to one3.afbced

DISCREET MATHEMATICS3In the question the vertices a, c, e and f all have vertices with odd degree of 3. Since aEuclerian path graph cannot have more than 2 vertices of odd degree G have noEuclerian path.Adding one edge to the graph flips the parities of the vertices it connects that is odd toeven and even to odd. We will add edge between two vertices of odd degreeAdding ac givesfcdfaceabeThen addingefgivesafeabecdfcwhich now have a Euclerian path4.The vertices a, b, c, d, e, f is odd and of degree 3 and 1and g is an even vertex of 2Since a graph with a vertex of degree one cannot have a Hamilton circuit then G haveno Hamiltonian cycle5.A minimum spanning tree has the minimum weight than any other spanning tree of agraphhthe total weight is5+11+4=205a4f11cEdgeabacadaeafagahweight1843793284655

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