MT365 Examination 2007 Part 1. Q1 (a) (b) (c) A

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1 MT6 Examination Part Solutions Q (a) (b) (c) F F F E E E G G G G is both Eulerian and Hamiltonian EF is both an Eulerian trail and a Hamiltonian cycle. G is Hamiltonian but not Eulerian and EF is a Hamiltonian cycle. G is Eulerian but not Hamiltonian and EF is an Eulerian trail. Q (a) The number of edges in G =. The sum of the vertex degrees in G = = 6 Since the sum of the vertex degrees (6) is equal to twice the number of edges ( ) the handshaking lemma is verified (b) simple connected graph with the same degree sequence as G is Q basic network equivalent to the given network is S 9 S S 6 T T T Noel Eastham

2 Q (a) (i) {,,,, E, F} is a triangle (ii) {,,, E, F} is a pentagon (b) (i) {,,, F} is a set of four points whose convex hull is a triangle (ii) {,,. E} is a set of four points whose convex hull is a quadrilateral Q (a) Using Kruskal s algorithm and selecting edges in the order (), E(), E(), E() gives the minimum connector E (b) Using Prim s algorithm and selecting edges in the order E(), (), (), E() gives the minimum connector E or E Q6 (a) The first fit packing method gives an answer of workers as follows F E 9 G 6 H W W W W W (b) The first fit decreasing packing method gives an answer of workers as follows E 9 G F 6 H W W W W MT6 Exam Solutions Noel Eastham Page of

3 Q (a) The interchange graph for the system is 6 (b) The mobility of the system considered as a planar kinematic system containing only revolute pairs is determined by noting the number of links n = and the number of joints j = M = (n ) j = ( ) = = = Q (a) Since there are x vertices of degree, y vertices of degree 6 and z vertices of degree The total number of vertices v is given by v = x + y + z The total number of edges e is given by e = x + 6y + z so e = x + y + z The total number of faces f is given by f = e so f x y z (b) Using Euler s formula v e + f = we obtain x y z x y z x y z x + y + z 6x 9y z + x + 6y + z = 6 x z = 6 and so x = 6 + z as required Q9 a w w x y z 6 a b x 6 b 6 c y c d 6 d z First revised cost matrix First Partial graph MT6 Exam Solutions Noel Eastham Page of

4 Q(a) The five remaining codewords are found by forming all possible sums of the three given,,,, (b) The minimum distance = since this is the smallest weight of a non-zero codeword. Hence the code can detect two errors and correct a single error. Q(a) a (b) a b f b f c e c e d depth first search spanning tree d breadth first search spanning tree Q(a) t c (b) t so c f I t c Q(a) v = 9, b =, k =, r = (b) is a resolvable design with replicates 6 9 G F E E F F E E H H I G I G H G I H F I (c) is a balanced design since (v ) = r (k ) gives = and so = MT6 Exam Solutions Noel Eastham Page of

5 Part Q(a) The chromatic number (G) since the maximum vertex degree is by rooks Theorem. The chromatic number (G) since G contains a cycle of length. So (G) = n explicit colouring of G with three colours is or (b) plane drawing of G is (c) The chromatic index '(G) = since a -edge colouring is given below and an edge colouring with fewer colours is impossible since the degrees of all of the vertices is. or The bounds given by Vizing s theorem are d '(G) d + where d is the maximum vertex degree, which in this case is and so '(G) and the chromatic index is equal to the lower bound. (d) The Independence number of G is 6. We obtain a lower bound for ind(g) from the condition (G) ind(g) n where n is the number of vertices. In this case it gives ind(g) 6, so ind(g) 6/ or ind(g) 6 since ind(g) is an integer. n upper bound for ind(g) comes from the outer cycle of length in G. Four alternate vertices in this cycle, such as those coloured red or () above, are adjacent to all others in the cycle and to four vertices in one of the inner squares, but are not adjacent to any of the vertices in the other inner square. Two vertices from opposite corners of the other inner square, coloured red or () again, are independent. Together these 6 vertices form a maximal set of independent vertices in G and so ind(g) 6. MT6 Exam Solutions Noel Eastham Page of

6 Q(a) pruned quad tree that stores the given image is (b) path arising from the steps of the north neighbour algorithm starting at the pixel in rc6 is North Neighbour rc6 Start Vertex rc6 (c) (d) The columns,,, are shifted right to become columns, 6,, and vice versa. In the pruned quad tree above, at the level directly below the root, we label the vertices as,, and. In order to produce the quad tree for the new image the subtrees rooted at and are interchanged and the subtrees rooted at and are interchanged. No changes at all are made within any of the subtrees themselves. If we count the root as level then the screen image above has a pruned quad tree with levels and we are interchanging subtrees rooted at level. In screen image if we exchange columns,,, with columns, 6,, and also columns 9,,, with columns,,, 6 then the associated quad tree has levels counting the root as level. In this case we leave level alone but we perform the same interchange of subtrees rooted at with and with at level as in the diagram below. MT6 Exam Solutions Noel Eastham Page 6 of

7 This essay is just one version of what could be produced; it is more detailed than would be expected and is included to provide an overview of the topic as well as an example. Q6 tree is a connected graph with no cycles. Trees can be labelled or unlabelled, possibly rooted and if so possibly binary. rooted tree has one vertex designated as the root vertex, with branches fanning out from it; if at most two branches arise from a vertex in a rooted tree then the tree is referred to as a binary tree. ounting graphs with particular properties can often be difficult but since trees are one of the simplest types of graphs, many tree counting problems can be solved. These solutions have important uses since trees often model real practical situations. Unlabelled trees can model organic molecules; counting the number of unlabelled trees on six vertices allows the enumeration of the number of organic molecules with the formula 6 H. Labelled trees can model the expansion of a multiple kinematic joint; the number of labelled trees on four vertices enumerates the number of ways of expanding a quaternary joint into binary joints. Labelled trees with n vertices can be counted by constructing a one one correspondence with a Prüfer Sequence. Prüfer Sequence is a list of n numbers, any one of which may take n values. The one-one correspondence is made in two stages. Firstly a given labelled tree with n vertices can be converted to a Prüfer Sequence by finding the vertex of degree one with the smallest vertex label, writing down the label of the vertex to which this is joined and then removing this vertex of degree one and its corresponding branch. This process is repeated until a single branch joining two vertices remains. This produces a list of n numbers, the Prüfer Sequence. onversely, a Prüfer Sequence can be converted to a labelled tree by finding the smallest number in the list of numbers to n not in the sequence, labelling a vertex with that number and joining it by an edge to a vertex with a label given by the first number in the Prüfer Sequence. These numbers are then removed from the list and the sequence and the process repeated until just two numbers remain in the list, when vertices with these labels are joined by an edge. Since there are n numbers in a Prüfer Sequence each of which can take n values, there are n n different Prüfer Sequences and so there are therefore n n labelled trees, this result is called ayley s Theorem. inary trees are rooted trees in which each vertex is of degree at most two. These are examples of unlabelled trees whose number can be found by the use of a recurrence relation. enote by u n the number of binary trees with n vertices. Such a binary tree either has a single edge emerging to the left from the root, a single edge emerging to the right from the root or two edges emerging from the root. If the numbers of these three types of binary tree are represented by a n, b n and c n then u n = a n + b n + c n. We can, by considering how such binary trees are formed from binary trees with fewer than n vertices build a recurrence relation for u n in terms of u n, u n,... u, u. Such a recurrence relation can be used to determine the number of unlabelled binary trees with n vertices. The method of deriving a recurrence relation can also be used to find the number of alkanes, a type of organic molecule, with the chemical formula n H n+. This was done for the first time by rthur ayley in the s. No unlabelled tree representing an alkane can have a vertex of degree more than four since this is the number of chemical bonds linking an individual carbon atom to its neighbours. In order to derive the recurrence relation ayley considered these trees as being of two types, either central or bicentral. central tree is one for which removal of all vertices of degree one and their associated branches in successive stages leaves at the end a single vertex. If there is a pair of vertices joined by a single edge left at the final stage the tree is called bicentral. ayley considered each tree as a rooted tree with the centre or bicentre as the root and built his recurrence relation by considering the rooted trees formed by removing the central or bicentral root. result first found by Georg Pólya in the 9s can be used to enumerate many types of graph including unlabelled trees with a particular number of vertices. Pólya s result is associated with Group Theory and uses a polynomial called the cycle index in order to complete the enumeration. MT6 Exam Solutions Noel Eastham Page of

8 Q For the following weighted digraph S 6 E F G H T (a) Each row of the following table corresponds to an iteration of the shortest path algorithm S E F G H T * The potentials are boxed above and were allocated in the order S(), (), (), (), E(), (), F(), H(), G(), T(). There are two shortest paths SEHT and SEHT both of length (b) Each row of the following table corresponds to an iteration of the longest path algorithm. The second row of the table indicates the vertices from which the vertex above is reachable. S E F G H T S S, S,E,,, F,E F,E G,H,,,9,,, 9,, The potentials are boxed above and were allocated in the order S(), (), (), (), F(), E(), (), H(), G(), T(). There is a single longest path SFEGT of length MT6 Exam Solutions Noel Eastham Page of

9 Q(a) Kirchhoff s current equation for vertex is i i i = or i + i i = Kirchhoff s voltage equation for cycle involving components, and is v + v + v = or v v v = (b) Each fundamental cycle contains a unique chord, the rest are branches. Each fundamental cutset contains a unique branch, the rest are chords. Thus two of the three edges,, are branches and one of the edges, is a branch. There are 6 possible spanning trees arising from choosing any two of the three edges,, in three ways with any one of the two edges, in two ways. The following are the six possibilities : {,, } {,, } {,, } {,, } {,, } {,, } (c) (i) Fundamental ycles are {,,, } or v + v + v + v = Fundamental utsets are {,, } or v + v + v = {, } or {, } i i = {,, } or {,, } i i i = {,, } or {,, } i i i = (ii) The five component equations are di v = V, v L dt, v = R i, v = R i, dv i dt (iii) The two state variables are the variables which are differentiated in the component equations. They are thus i and v and the state equations obtained by eliminating variables between the omponent, Fundamental ycle and Fundamental utset equations will involve these variables and their derivatives only. MT6 Exam Solutions Noel Eastham Page 9 of

10 This essay is just one version of what could be produced; it is more detailed than would be expected and is included to provide an overview of the topic as well as an example. Q9 Menger s theorem, in its various forms, is concerned with the number of different paths joining two vertices s and t in a connected graph or digraph. The paths can be edge or arc disjoint meaning they have no edges or arcs in common, or they can be vertex disjoint in which case they have no vertices in common with the exception of s and t. set of edges (or arcs) separating s from t is a set of edges (or arcs) whose removal from the graph destroys all paths from s to t. Similarly a set of vertices separating s from t is a set of vertices whose removal from the graph again destroys all paths from s to t. The edge form of Menger s Theorem for graphs states that the maximum number of edge disjoint st paths equals the minimum number of edges separating s from t. In a similar way, in a digraph, the maximum number of arc disjoint st paths equals the minimum number of arcs separating s from t. The result just quoted for digraphs can be proved as a consequence of the max flow min cut theorem. This is done by constructing a basic network from the digraph, by giving each arc a capacity of. Now a maximum flow from s to t must consist of a collection of flows of value along a set of arc disjoint paths and so the value of the maximum flow from s to t must equal the number of such arc disjoint paths from s to t. minimum cut in the basic network will be a minimum set of arcs separating s from t and the capacity of this cut will equal the number of arcs in the cut. Since the max flow min cut theorem states that the value of a maximum flow is equal to the capacity of a minimum cut it must therefore be the case that the number of arc disjoint st paths is equal to the number of arcs separating s from t, which is Menger s theorem for digraphs in its arc form. Menger s theorem for graphs in its edge form can be proved similarly by first introducing two arcs for each edge in the graph and applying Menger s theorem for digraphs. The vertex form of Menger s theorem states that the maximum number of vertex disjoint st paths is equal to the minimum number of vertices separating s from t. This is identical for graphs and digraphs and is proved for digraphs by introducing an arc of capacity replacing each vertex. rcs into the replaced vertex are directed into the start vertex of the replacement arc and arcs out of the replaced vertex are directed out of the finish vertex of the replacement arc. pplying the arc form of Menger s theorem for digraphs then enables us to prove the vertex form of Menger s theorem for digraphs. Finally the vertex form of Menger s theorem for graphs can be proved by performing both the transformations mentioned above, that is the replacement of each edge by two arcs and each vertex by an arc and then an application of the arc form of Menger s theorem. The edge connectivity of a connected graph is defined to be the smallest number of edges whose removal disconnects the graph. On applying Menger s theorem, this means there must be at least edge-disjoint paths joining any pair of vertices. This leads to a corollary of Menger s theorem which states that a connected graph has edge connectivity if and only if every pair of vertices are joined by or more edge-disjoint paths, and at least one pair of vertices is joined by exactly edge-disjoint paths. There is a corresponding result for the vertex connectivity This states that any connected graph, other than a complete graph, has vertex connectivity if and only if every pair of non-adjacent vertices are joined by or more vertex-disjoint paths and at least one non-adjacent pair of vertices is joined by exactly vertex-disjoint paths. These ideas are important in the design of reliable telecommunications networks. The edges of a graph represent the communication links while the vertices represent the exchanges or subscribers. telecommunication network will be reliable if, when a link or exchange is broken or fully occupied, there are alternative paths for communication between subscribers. The existence of alternative paths depends on the maximum number of edge disjoint paths between vertices and the maximum number of vertex disjoint paths between non adjacent vertices and so is related to the edge and vertex connectivity of the network. In practice to fully analyse the reliability of a telecommunication network we need information on all the cutsets as well as the edge and vertex connectivity. MT6 Exam Solutions Noel Eastham Page of

11 Q(a) onsider the octahedron Q E G 6 F H (i) The Incidence matrix of the incidence structure S whose points are the vertices and whose lines are the faces of Q is E F G H 6 (ii) The block table for S is E F G H (iii) The values of d p and d l are d p =, d l = (iv) The ube and Octahedron are dual polyhedra, vertices of the octahedron correspond to faces of the cube and faces of the octahedron correspond to vertices of the cube. The Incidence structure S* of the cube is the dual of the incidence structure S of the octahedron. S* is obtained by interchanging the roles of points and lines in S. So S (P, L) becomes S*(L, P) which is the incidence structure of the cube. The incidence matrix of the incidence structure S* of the cube is the transpose T of the incidence matrix of the incidence structure of the octahedron (b) (i) The completed table is n vn en dn 6 6 (ii) The handshaking lemma states the sum of the vertex degrees equals twice the number of edges so vndn e n hence en vndn n (n ) n(n ) MT6 Exam Solutions Noel Eastham Page of

12 Q(a) (i) Length n =, imension k =, Rate = k/n = / = / Minimum distance = (smallest weight of a non-zero codeword) so the code can correct error and detect errors. G 6 G 6 H 6 H 6 (b) (i) T I M The codewords for T, I and M are the first, second and last rows of the final matrix (ii) J O E This calculation can be understood as follows. The product of the Parity heck matrix H with the three received words produces a set of three error syndromes in the columns of the central matrix. These indicate errors in the rd bit of the first word, the sixth bit of the second word and no error in the last word. These errors are corrected (in red) to produce a set of three correct codewords. The Generator matrix G has the columns of the unit sub matrix in its rd, th, 6th and th columns so taking the rd, th, 6th and th bits of the codewords (as underlined) generates the message bits of the symbols of the message. These have been written in columns in the final matrix so enabling decoding of the message words to proceed easily and produce the message JOE. MT6 Exam Solutions Noel Eastham Page of

13 This essay is just one version of what could be produced; it is more detailed than would be expected and is included to provide an overview of the topic as well as an example. Q Latin square is an n n square array of n symbols in which each symbol occurs once in each row and column. Some examples of size, and are There are various methods of constructing Latin squares; the simplest methods are the cyclic construction and a construction using Steiner triples. The cyclic construction is seen in the Latin squares above where the symbols are shifted one place to the right in each successive row. This construction can be extended by moving the symbols r places to the right and the last r symbols to the front, r and n must have no common factor but if n is prime then r can have any value from to n. Latin squares can also be constructed by using a Steiner triple system. Given a Steiner triple system with n varieties label the rows and columns of the Latin square with these n varieties and put each of them in the corresponding diagonal position. Then for every position in the Latin square labelled by X and Y find the triple containing X and Y in the Steiner triple system and place the other symbol Z in that triple in the position labelled by X and Y in the Latin square. Latin squares are useful in the design of experiments particularly for situations in which there are two types of block which cross each other. Such a situation occurs in the testing of say four drugs,, and which are to be given to four patients over four three month periods. The Latin square above could be used for this experiment where the rows correspond to patients and the columns to each of the three month periods. Latin squares have properties which make them suitable for situations in which each block is complete (i.e. contains all varieties), each block of one type has exactly one plot in common with each block of the other type and each block of each type has the same number of plots as there are varieties. Two Latin squares are orthogonal if when one is placed directly over the other each of the n ordered pairs of symbols occur once and once only. Thus the two Latin squares are orthogonal since contains all n pairs once Such a set of orthogonal Latin squares are very useful in the construction of resolvable designs. resolvable design is one in which the blocks of the design can be grouped into replicates, a replicate being a set of plots containing each variety exactly once. Thus given a set of 6 varieties arranged in a square array as follows we can construct two replicates by using the rows and columns of the square array of numbers and two more by choosing those numbers in the same position as a particular letter in each of the Latin squares above. Finally Latin squares also have applications in the theory of error correcting codes and in edge colourings of complete bipartite graphs K n,n. These arise from an alternative way of describing a Latin square as a list of n triples (I, j, k) where i, j, and k are the row, column and symbol numbers associated with the Latin square. MT6 Exam Solutions Noel Eastham Page of

14 Essay Outlines Q6. Introduction definition of a tree types of trees. pplications of tree counting problems labelled and unlabelled trees. ounting labelled trees Prüfer Sequences. ounting binary trees by use of recurrence relations. ounting chemical trees central and bicentral trees. onclusion Pólya s counting theorem. Q9. Introduction edge disjoint and vertex disjoint paths separating s from t. Menger s Theorem for Graphs and igraphs edge and arc form. Proof of Menger s Theorem for igraphs Max-flow min-cut theorem. Menger s Theorem for Graphs and igraphs vertex form. orollaries of Menger s Theorem edge and vertex connectivity. onclusion application to telecommunication networks. Q. Introduction definition of a Latin Square onstruction of Latin Squares cyclic and Steiner triple constructions Uses of Latin Squares Orthogonal Latin Squares Uses of orthogonal Latin Squares. onclusion applications of Latin Square to odes and colourings. MT6 Exam Solutions Noel Eastham Page of