MT365 Examination 2017 Part 1 Solutions Part 1


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1 MT xamination 0 Part Solutions Part Q. G (a) Number of vertices in G =. Number of edges in G = (i) The graph G is simple no loops or multiple edges (ii) The graph G is not regular it has vertices of deg., and vertices of deg. (iii) The graph G is bipartite the vertices can be split into two sets (circled and uncircled above) so that edges only join vertices in one set to those in the other. Q. S,9, T,,,,, S,,, 0,,,9 T S,, T (a) Q The network N changed to a basic network with flows, capacities and saturated arcs low ugmenting Path found alphabetically is SS T T. It will carry a flow of and will saturate. (Note the labels found are not asked for) G Noel astham 0
2 MT xamination 0 Part Solutions Part Q 9 (a) T is bicentral. Vertices removed first {,,,,, 9} second {, } icentre and Remove Sequence (, Remove Sequence (,, Remove Sequence (,,, Remove Sequence (,,,, Remove Sequence (,,,,, Remove Sequence (,,,,,, Remove Sequence (,,,,,,, Remove 9 Sequence (,,,,,,, ) inal Prufer Sequence (,,,,,,, ) Q S 0 T rcs and are redundant and so have been omitted Noel astham 0
3 MT xamination 0 Part Solutions Part Q. (a) c r c r c r r r c c c Note that: vertices and edges does not necessarily mean we have a tree! Since we have a cycle! r c c The framework is not rigid the section c r c r c is disconnected from the rest of the bipartite graph as indicated by the heavy lines in red above. (c) One bracing when added which will make the framework rigid is r c. Q. ( The graph is planar since f = and v e + f = + = ) (a) The face degrees are : face of degree and 9 faces of degree. The sum of the face degrees = deg f = x + 9 x = + = Twice the number of edges = e = x = deg f = e as required G or the graph G there is a four colouring of the edges as above so χ (G) y Vizing's Theorem χ (G) since the maximum vertex degree is Hence χ (G) = Noel astham 0
4 MT xamination 0 Part Solutions Part Q (a) w x y z w x y 0 0 z n optimum assignment is w, z, y, x. Q9 The given codewords are 0, 0, 0 (a) The remaining five coewords are 0000, 0, 00,, cycles to itself cycles to itself cycles 0 cycles The code is thus cyclic. (c) 0 0 G= thus H= [ ] Q M T H S R (a) H M S T R (i) S: HMTS (ii) S: HMST Q (a) v = v a, v = v b + v c, v = v c i = i a, i = i b  i c, (i = i b since i i = i b  i a ) Noel astham 0
5 MT xamination 0 Part Part Solutions Q (a) (i) = (ii) Outdegree sequence : (,,,, ) Indegree sequence : (,,,, ) (,) (,) (,) (,) (In,Out) (,) (i) cycle of length starting at vertex is (ii) closed trail in of length from vertex is (iii) closed walk in of length from vertex is (iv) walk in of length from vertex to vertex is (c) (i) = = There are walks from vertex to vertex in. (ii) The walks of length from vertex to vertex are and (d) Note as given in the question has element a incorrect but this does not affect the answer = = 0 y Theorem. of Networks : If = has all nondiagonal entries > 0 then the digraph is strongly connected, but + + has all nondiagonal entries > 0 except for the entry in position (,) and so will have the same. We note that there is a walk of length from vertex to vertex as in (iv) above. So is strongly connected. (e) (i) (ii) very strongly connected digraph is Hamiltonian is false. igraph above is a counterexample since it is strongly connected but there is no Hamiltonian cycle as we must go from to and each path out of omits a vertex before retuning to. very Hamiltonian digraph is strongly connected is true since the Hamiltonian cycle provides a path between each pair of vertices. MT xam 0 Solutions Page of
6 MT xamination 0 Part Part Solutions Q (a) hoose of length the shortest length hoose of length next shortest length hoose of length next shortest length hoose of length 9 next shortest length an't choose (), (), (), (), () all create cycles hoose of length next shortest length not creating a cycle Length of the minimum connector =. Remove vertex and start with vertex hoose edge of length smallest from hoose edge of length smallest from, hoose edge of length smallest from,, hoose edge of length smallest from,,, 9 Lower bound for TSP = + +! = + = (c) Remove vertex Lower bound for TSP = = 0 + = 9 0 (d) The better lower bound is since it is the larger of the two (e) If we remove S in calculating a lower bound for the new TSP then S Lower bound for new TSP = = + = 9 If the two vertices connecting Syd's home to the tree were and then there would be a Hamiltonian cycle of length this could be a solution to this TSP and so the best lower bound that would be possible. MT xam 0 Solutions Page of
7 MT xamination 0 Part Part Solutions This essay is just one version of what could be produced; it is probably more detailed than would be expected and is included to provide an overview of the topic as well as an example. Q graph is planar if it can be drawn on a flat piece of paper so that none of the edges intersect each other, consequently the edges will only meet at vertices. If this cannot be done then the graph is non planar. are must be taken, since it is quite possible to have a drawing of a graph in which some edges intersect but for which, on rearrangement of the edges while keeping them incident to the same vertices, the intersections can be removed; such a graph is planar. n important result that applies to all planar graphs is uler s formula which states that v e + f =, where v is the number of vertices, e is the number of edges and f is the number of faces. It is possible to deduce from uler s formula two inequalities which apply respectively to simple connected planar graphs with three or more vertices and to simple connected planar graphs with three or more vertices and no triangles. These inequalities put limits on the number of edges such graphs can have and allow us to deduce that the complete graph K and the complete bipartite graph K, are both non planar. The inequalities referred to above do not provide a complete characterisation of a planar graph since some graphs, such as the Petersen graph, satisfy both of the inequalities and yet are non planar. However, there is a criterion by which we can characterise all planar graphs and hence all non planar graphs, this is Kuratowski s theorem. This theorem states that a graph is planar if and only if it does not contain a subdivision of K or K,. subdivision of a graph is a graph that can be obtained from the given graph by the insertion of one or more vertices of degree two on any edges. learly the insertion (or removal) of such vertices will not change the planarity or otherwise of the given graph. Thus in a sense K and K, are basic non planar components for any non planar graph and a planar graph cannot contain either of them as a subgraph or as a subdivision. Kuratowski s theorem, although characterising planar graphs completely, does not provide an easy means of testing whether a given graph is planar. There are efficient algorithms for doing this but one simple test which can be applied to any graph, which is not too large and which contains a Hamiltonian cycle, is the cycle method. This algorithm proceeds by specifying a Hamiltonian cycle in the graph and then attempts to allocate all the edges of the graph, not in the cycle, to two sets of non intersecting edges. If this can be done, then one set can be drawn inside the cycle and the other outside the cycle and the graph is planar. If there is a conflict and the edges do not fall into two disjoint sets then the graph is non planar. Given any planar graph, it is possible to construct another graph called its dual by placing a vertex in each face and joining these vertices by edges so that one edge of the dual crosses each edge of the original. The dual of a planar graph is planar but is not necessarily unique; however, the dual of a dual is a graph isomorphic to the original graph. There is a one to one correspondence between edges in a graph and its dual, vertices of one correspond to faces of the other and cycles in one correspond to cutsets in the other, so uler s theorem still applies. Knowing whether a graph is planar or knowing how to decompose a given graph into two or more planar subgraphs is important in various applications. One such application is the printed circuit board problem in which electronic components must be connected by conducting strips which must not cross on a particular board. MT xam 0 Solutions Page of
8 MT xamination 0 Part Part Solutions Q. S H T J G (a) S G H J 0 * T * * (I) The next labels to be assigned are *, G, H, J (ii) The next potential to be assigned is G (iii) The remaining vertex potentials are, J 9, H 0, T. (iv) The shortest paths are of length, SHT (red), SGJT (green), SGT (blue) and SHT S G H J T S S S, S,,,,,,,, G, G, H, J 0,0 0,9 9,,,,, (i) The potentials assigned at the next step are 0, and G 9, (ii) The remaing potentials are H, J, T. (iii) The single longest path of length is SJT (c) If one arc is to be increased by (i) If the arc is S all the shortest paths are increased by to length, but there are more shortest paths since now S = S (SHT, SGJT, SGT) the longest path is unchanged (ii) If the arc is J all the shortest paths are unchanged the longest path is increased by to length 9 (iii) If the arc is JT the shortest path is still of length but SGJT is no longer one of them the longest path is increased by to length 9 MT xam 0 Solutions Page of
9 MT xamination 0 Q xtra Part Part Solutions or those who wish to see the full tables for both algorithms (a) The shortest path algorithm S G H J T 0 * * * The longest path algorithm S G H J T S S S, S,,,,,,,, G, G, H, J 0,,,,,,0,9 0 9,,,,, MT xam 0 Solutions Page 9 of
10 MT xamination 0 Part Part Solutions Q (a) (i) (ii) The set {,,, } can only cover the tutorials {,, } Since there is a set of tutors larger than the set of tutorials covered by those tutors, The Marriage Theorem states that a complete matching of tutors to tutorials is impossible. (iii) The maximum number of tutors that can be assigned to tutorials is. The allocation,,,, is such an allocation. (i) () () () () () () () () (G) breakthrough () G (G) () H (H) lternating path HG giving an improved matching as follows () () () () breakthrough () () () () () () () G () H (G) lternating path Maximum matching,,,,,, G, H (ii) No, the maximum matching is not unique there are the other has and G all others the same. MT xam 0 Solutions Page of
11 MT xamination 0 Part Part Solutions 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. It might seem that an electrical network should be modelled by a mathematical network, but this is not the case. When setting up the model at the outset, before any analysis has taken place, we do not have a predetermined direction associated with the current flow through, or the voltage across, any particular component. Thus the directions must be left to be determined later and a graph is the appropriate model, albeit with an arbitrary direction associated with each edge, so producing an oriented graph The two main aspects of an electrical network which require modelling are the components and their interconnections. The components can be of various kinds and are connected to other components through their terminals which can number two or more. Two terminal components are represented by a single edge, three terminal components by two edges and in general n terminal components by n edges as explained below. There are two laws used in the graphical representation of any physical network, they are the vertex law and the cycle law. The vertex law states that the algebraic sum of the through variables at any vertex is zero, while the cycle law states that the algebraic sum of the across variables around any cycle is zero. In the context of electrical networks the through variable is current, the across variable is voltage and the laws are generally known as Kirchoff s Laws. These laws allow the connectivity or topology of the network to be described in terms of a set of fundamental equations called the fundamental cycle and cutset equations. These equations are not unique for any network but are defined in terms of a spanning tree for the oriented graph representing the network. That these equations are only dependent on the interconnections of the components can be seen from the fact that they can be obtained directly from the incidence matrix for the oriented graph, which alone fully describes the connectivity. s mentioned above, a three terminal component is represented by two edges. It might seem obvious that we should have three edges connecting each terminal to every other so having six variables (three through and three across) to represent the component. Kirchoff s laws, however, generate two equations which connect these six variables so reducing the number of independent variables by two and hence the number of edges required by one. In a similar way every multiterminal component with n terminals has the number of edges in its graphical representation reduced by the equations generated by Kirchoff s laws so producing in every case a representation by a tree with n edges. These representations are by no means unique. The representation of electrical networks by graphs allows large and complex systems to be analysed through the simple ideas of graph theory and this facilitates the use of computers to solve large and complex electrical network problems. There are also other benefits from the use of graph theory. contrasting and alternative viewpoint allows ideas and concepts from graph theory to illuminate those in electrical network theory. simple example is the idea of the dual of a planar graph as defined in graph theory. When transferred to the study of electrical networks this introduces the idea of the dual of a planar electrical network. The benefit is that if we have solved an electrical network problem, we can immediately write down the solution of the corresponding dual network problem by just using the correspondences indicated by duality. MT xam 0 Solutions Page of
12 MT xamination 0 Part Part Solutions Q (a) (i) The system S has binary joints, ternary joints and binary links (ii) The direct graph representation of S is shown above next to the system (iii) M = g n = x = = This is the mobility criterion for graphs which applies to all systems with only binary links but with joints of any multiplicity. This expansion uses the links indicated by circled crosses (i) The interchange graph of the above system S is shown above next to the system (ii) This expansion uses the links indicated by circled crosses different expansion S of the system is given above with its interchange graph next to the system (iii) (c) These two interchange graphs are not isomorphic because the second contains a vertex of degree and no vertices of degree while the first contains two vertices of degree and none of degree There are four distinct nonisomorphic interchange graphs resulting from the expansions of the ternary joints of S. part from those above there are two more as follows Not isomorphic to that in This expansion because the two vertices uses the links of degree are joined by an indicated by edge while in the two circled crosses vertices of degree are not joined. Not isomorphic to that in This expansion because the two vertices uses the links of degree are joined to only indicated by one common vertex of degree circled crosses while in the two vertices of degree have two vertices of degree in common ue to the symmetry of the given system there are no other choices for the links to be used to perform the expansion possible see the diagrams with the circled crosses. MT xam 0 Solutions Page of
13 MT xamination 0 Part Part Solutions Q9 G H I J K L M (a) (i) v =, b =, r =, k = (ii) r(k ) ( ) λ= = = = v (iii) or a finite projective plane v = b = n + n + =, r = k = n + = so n = (i) No, Δ is not resolvable since for a replicate containing the block (0,,, 9) we cannot find a block containing since (0,,, ) contains 0 and (,,, ) contains while H (,,, ) contains and finally I(,,, 9) contains 9 (ii) Yes, Δ is a cyclic design as indicated below by rearranging the blocks H J I K L M G (c) If is the complement of Δ then (i) v= v=, b= b=, k= v k= = 9, r= b r= = 9 (ii) y Th. λ= b r+λ= + = the design is balanced (d) (i) r = the number of cards carrying a particular symbol k = the number of symbols on a particular card (ii) finite projective plane of order n = has + + = points and the same number of lines. There would be r = k = n + = different symbols and each would occur on cards in the full set giving x = symbols Since in a finite projective plane any two lines (cards) have only one common point (symbol) there will be one symbol which occurs on only cards and symbols which occur on cards while the remaining symbols occur on eight cards ( x + x + x = = 0 = x =  ) MT xam 0 Solutions Page of
14 MT xamination 0 Part Part Solutions This essay is just one version of what could be produced; it is probably more detailed than would be expected and is included to provide an overview of the topic as well as an example. Q0 linear binary code is one in which the sum x + y of any two codewords x and y is also a codeword. The operations are performed modulo since the codewords are strings of s and 0 s, elements from the finite field Z = ({0, }, +, ). If the codewords are of length n such as x = x x x n, where x i {0, } and there are k of them, then the code is called an (n, k) linear code. The k codewords form a subset of all possible binary words of length n, which themselves form an n dimensional vector space consisting of all n ntuples. Thus a linear code is a kdimensional subspace of an ndimensional vector space. very linear code contains the zero codeword 0 consisting of n zeros. Since codewords of an (n, k) linear code form a kdimensional subspace, it is possible to find a set of k of them which form a basis for this subspace. ll codewords can be written as linear combinations of the basic set. Such a set of k codewords is called a generator set for the code and can be written as a k n matrix G, the generator matrix for the code. very codeword is formed by a linear combination of the rows of G, there being k such combinations. ll codewords in a linear code have a weight defined as the number of s in the codeword. The smallest weight of any nonzero codeword is the minimum distance δ for the code and this indicates the error detecting and correcting properties of the code. So we use (n, k, δ) to specify the code. Related to this is a matrix H for which Hx = 0 for all codewords x. This matrix is called the parity check matrix because the n k rows of H specify combinations of the bits of the codewords which must always be zero for all codewords x. Parity check matrices are used in decoding received binary words. Multiplication of a received word by the parity check matrix produces a vector called an error syndrome which, if the code is a single error correcting code and the received word contains just one error, will be a column of the parity check matrix. The position of this column will indicate the position of the error in the received word. In a similar way, the error syndrome will be a linear combination of m columns from the parity check matrix if the code is an m error correcting code and the received word contains m errors. Two linear codes are equivalent if the bits in all the codewords of one can be rearranged by some fixed rule to produce all the codewords of the other. If this is the case, then the columns of the generator matrix of one code must be related by rearrangement to the columns of the generator matrix of the other by the same fixed rule. Hence every (n, k) linear code is equivalent to a code for which the generator matrix is in standard form with a unit sub matrix in the first k columns. If we take the rows of the parity check matrix of a code as the generator set of a new code we obtain the code words of a code called the dual code * of the code, so G* = H. lso the generator matrix of is the parity check matrix of * so H* = G. or any linear code we can produce another code called the xtended ode by adding an extra digit to make all codewords even parity. The Hamming codes are a set of single error correcting codes formed so the parity check matrix is a set of columns which are the m nonzero binary numbers of length m. Hamming odes are ( m, m m, ) codes. If we have a t error correcting code for which every binary word is within a distance t of a codeword the code is called perfect because there are no binary words that cannot be decoded unambiguously. The Hamming codes are all perfect codes. cyclic code is one in which, whenever x = x x x x n is a codeword then so is x = x x x n x. There may be several disjoint cycles in a cyclic code. Such codes are important since they can be practically implemented in a simple way. In conclusion many linear codes are of practical importance. The ReedMuller codes have been used for transmission from the Mariner 9 space probe because of their ability to correct many errors, while the ReedSolomon codes are used for digital music on s because of their ability to correct bursts of errors as produced by scratches. MT xam 0 Solutions Page of
15 MT xamination 0 Part Part Solutions ssay Outlines Q. Introduction the definition of a planar graph. uler s formula for planar graphs resulting inequalities. Kuratowski s Theorem subdivisions of a graph. Testing for planarity Hamiltonian cycle method. uality of planar graphs cycles and cutsets. onclusion importance of planar graphs for printed circuit boards. Q. Introduction  the use of oriented graphs to model networks The vertex and cycle laws for through and across variables respectively undamental cycles and cutsets Incidence matrices The representation of nterminal components by trees onclusion dual graphs and dual networks Q0. Introduction definition of a linear binary code Generator set of codewords Generator matrices rror orrection and etection Parity heck matrices quivalent odes, ual odes and xtended odes Hamming odes, Perfect odes and yclic codes onclusion Reed  Muller and Reed  Solomon odes MT xam 0 Solutions Page of
Part 1. Twice the number of edges = 2 9 = 18. Thus the Handshaking lemma... The sum of the vertex degrees = twice the number of edges holds.
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