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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1 MT6 Examination 16 Q1 (a) Part 1 Part1 Solutions (b) egree Sequence (,,,,, ) (c) Sum of the vertex degrees = = 18 Twice the number of edges = 9 = 18. Thus the Handshaking lemma... The sum of the vertex degrees = twice the number of edges holds. Q (a) (i) Three vertex disjoint st paths are :- (ii) Five edge disjoint st paths are :- sabct, shdt, sgfet sabct, sbt, shdt, sft, sgfet (b) κ(g) = a set of vertices disconnecting G are b, d, f λ(g) = a set of edges disconnecting G are sh, ha, hg, hd Q (a) (i) f = n + (ii) e = n (iii) v = n ( there are n triangles and n-sided polygons) ( there are n sides of polygons, 6n sides of triangles, giving 8n sides but all sides are shared so we divide by ) ( there are n vertices of n-sided polygons) (b) Euler's polyhedron formula is :- v e + f = n n + n + = n n + = Q (a) hoose E length (shortest edge) hoose length (next shortest edge) hoose (or E) length (next shortest edge) E an't choose E (or ) creates a cycle an't choose creates a cycle hoose length (next shortest edge) length = 1 Noel Eastham 17

2 MT6 Examination 16 Part 1 Part1 Solutions (b) Starting at, choose length (smallest weight edge incident to ) hoose E length (smallest weight edge joining or to a new vertex) hoose E length (smallest weight edge joining, or E to a new vertex) hoose length (smallest weight edge joining,, E or to a new vertex) (Note: at the second stage we could have chosen, the third and fourth choices would still have been the same so that we produce the same tree as for Kruskal's algorithm above.) E length = 1 Q (a) The potentials assigned to each vertex in the order in which they were assigned are S [], [6], [7], [8], F[11], [1], E[1], T[17]. (b) The shortest path from S to T is SET Q6 (a) (i) c 1 r 1 c r c r c r c r c 6 c 7 (ii) The framework is not rigid because the bipartite graph is not connected, there is no path from the bold section to the rest of the graph. (b) (i) bracing r c 6 would make the framework rigid. (ii) The resulting rigid framework would not be a minimum bracing since it contains a cycle r 1c 1r c r 1. Or the bipartite graph would not be a tree since it would contain 1 edges and 1 vertices. Noel Eastham 17

3 MT6 Examination 16 Part 1 Part1 Solutions Q7 Using the cycle agbfcedha. We list edges not in the cycle af, be, ch, dg. h a Put af inside the cycle, then be and dg are incompatible with af but are compatible with each other and so must be placed outside the cycle. Finally ch is incompatible with dg and be but compatible with af and so can be placed inside the cycle. The graph G is thus planar. d e c f g b Q8 (a) w x y z w 6 1 x y z (b) n optimum assignment is z, w, x, y Q9 (a) odewords of are, 1111, 1111, 1111 (b) The generator matrix of in standard form is (c) the parity check matrix of in standardform is Q1 (a) depth-first spanning tree for G is a e b f c d Noel Eastham 17

4 MT6 Examination 16 Part 1 Part1 Solutions (b) breadth-first spanning tree for G is a e b f d c (c) No, another depth-first spanning tree for G would be a e b f c d This has a vertex e of degree whereas the graph in (a) does not Q11 The parameters fit the requirements for onstruction with n = 1 and p = 7 v = p + 1 = 8 b = p = 1 r = p = 7 p + 1 k = = p 1 λ = = Take the non-zero squares modulo 7 together with an extra variety z and the set of non-squares modulo 7 together with and use the cyclic construction in each case; then put the two designs together. The non-zero squares modulo 7 are 1, and. The non-squares modulo 7 are, and 6 The required construction is E F G H I J K L M N z z z z z z z 6 1 Noel Eastham 17

5 MT6 Examination 16 Part Solutions Part Q1 (a) (i) T 1 is central since on first 6 8 removing leaf vertices, 1, 6, 7 and 8 and their associated edges gives (1) 7 1 emoving leaves 1 and and (1) () associated edges gives () (ii) T 1 has Prüfer sequence (,,,, 1, 1). Found by removing vertex joined to, then vertex joined to, then vertex joined to, then vertex 6 joined to, then vertex joined to 1, then vertex 7 joined to 1 leaving (1,8) connected (b) The labelled tree T with Prüfer sequence (1,,, 1,, ) is or redrawn as (c) (i) Two non-isomorphic unlabelled trees with five vertices each having five leaves (pick from the ) or or (ii) The degree sequence of a tree with 8 vertices and leaves must have the form (1, 1, 1, 1, 1, x, y, z). Since it is a tree with 8 vertices there must be 7 edges and so the sum of the vertex degrees is 1 by the handshaking lemma. Thus + x + y + z = 1 that is x + y + z = 9, where x, y, z. The only possible solutions are + + = 9, + + = 9 and + + = 9 correspondingly there are only possible degree sequences which are... (1, 1, 1, 1, 1,,, ), (1, 1, 1, 1, 1,,, ) and (1, 1, 1, 1, 1,,, ) (d) If T is a labelled tree with Prüfer sequence comprising distinct numbers each occuring twice then (i) There are 6 numbers in the Prüfer sequence and hence 8 vertices in the tree number appearing twice corresponds to a vertex of degree so there must be three vertices of degree. ut then since the sum of the vertex degrees is 1 the remaining vertices must all be of degree 1. That is T has leaves (ii) There is only one unlabelled tree with 8 vertices, leaves and three vertices of degree it is the one above with degree sequence (1, 1, 1, 1, 1,,, ) MT6 Exam 16 Solutions Page of 1

6 MT6 Examination 16 Part Solutions Q1 (a) (b) (i) Step urrent Vertex label Level Path Stack Top of Stack 1 root opposite to opposite to opposite to Level Level Level 1 Level Level 1 Level Level - empty empty - (ii) root This doesn't need to be redrawn, but it's easy to copy and paste on a computer ;-) (c) (i) and, and (ii) The lower vertices are and (d) is in error because is a lower vertex and should lead to the next stack as. is in error because for a 6-screen there can be at most vertices in a path stack. MT6 Exam 16 Solutions Page 6 of 1

7 MT6 Examination 16 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. Q1 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 Euler 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 Euler 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 Euler 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. MT6 Exam 16 Solutions Page 7 of 1

8 MT6 Examination 16 Part Solutions Q1 (a) The basic network conveying all the required information. W X S W Y 1 W 6 Z T (b) SW 1XT - send 6 - saturating arc W 1 SW 1W XT - send 1 - saturating arc W, X, XT (c) (i) The value of the corresponding flow in the basic network above is 1 as given below W 1 6,6,7 X 9,9 1,1,,,7,, 6,6 6,6 S T W Y,,,,1,, W,6,, Z There are : in stock in W 1, in stock in W and in stock in W There are : delivered to X, 6 delivered to Y and delivered to Z (ii) No, it is not a maximum flow. There is a flow augmenting path in the basic network SW XT which can carry a flow of 1 saturating. The maximum flow has value 16 with a minimum cut of {W 1, W, W,, Z} which has a capacity of = 16. (d) Yes, we need to send a further to output X. There is available in stock at W. We need to use SW X as a flow augmenting path to get the stock at W to X. This can be done by increasing the capacity of truck route W from 1 to which will then allow the supply to be made. MT6 Exam 16 Solutions Page 8 of 1

9 MT6 Examination 16 Part Solutions Q16 (a) The electrical network corresponding to the given oriented graph is V 1 L R (b) (i) Kirchhoff's current equation for is i 1 i = (ii) Kirchhoff's voltage equation for 1,,, is v 1 + v + v + v = (c) From the equation v 1 + v + v + v = we deduce one of 1,, or is a chord the other three are branches From the equation i + i i = we deduce one of, or is a branch the other two are chords If is a branch in the current equation then both and are chords, which is impossible for the voltage equation. If is a branch in the current equation then both and are chords, so 1, and are branches in the voltage equation. If is a branch in the current equation then both and are chords, so 1, and are branches in the voltage equation. There are just two spanning trees giving rise to these equations R 1 or 1 (d) (i) Fundamental cycle equations are v 1 + v + v + v =, v 1 + v + v + v = f= (ii) Fundamental cutset equations are i 1 i i =, i i i =, i i i = f= di (iii) omponent equations are v 1 = - V, v, v = i R, v = i R, L dt dv i dt = = ( e) The state variables are i and v MT6 Exam 16 Solutions Page 9 of 1

10 MT6 Examination 16 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. Q17. matching is a one-to-one association between the elements of two disjoint sets X and Y. It can be represented on a bipartite graph by a set of thick lines. The graph contains the elements of one set X as the vertices on one side of the graph and the elements of the other set Y on the other. The edges only join vertices in X to vertices in Y and represent the relationships between the elements of the two sets. y 1 y 1 y 1 x 1 x 1 x 1 y y y x y x y x y x y x y x y x y x y x y x y 6 x y 6 x y 6 matching is a subset of the edges, shown by the thick lines in the graphs above, with no vertices in common. Matching problems do not necessarily have equal numbers in the two sets, or a full complement of edges to form a complete bipartite graph. maximum matching is one in which the number of edges is as large as possible. The maximum matching algorithm labels a bipartite graph to improve the matching. In the first bipartite graph above since x is not in the matching we label x with (), y with (x ), x with (y ), y with (x ), x with (y ) and finally y 6 with (x ) to produce the alternating path x y x y x y 6 used to improve the matching as shown in the second graph above. There are now thick lines as opposed to. n alternating path alternates from from side to side in the graph and also in the nature of the edges which are alternately not in the matching and in the matching.there are thick lines in the last graph, so it is a maximum matching since the number of edges in a maximum matching can never be greater than the number of vertices in the smaller set. complete matching is a maximum matching in which the number of edges in the matching is the same as the number of vertices in each set. Matching problems are of three main types. The simplest is the marriage problem in which we have, for example, a set of applicants and a set of jobs. Each applicant is equally qualified but can only do certain jobs. We wish to match applicants to jobs so as to obtain a maximum matching. necessary condition for there to be a solution to this problem, that is for every applicant to be allocated a job they can do, is that 'for every subset of m applicants, they collectively must be able to do at least m jobs, for all values of m in the range 1 m n ' where n is the number of applicants. This condition is often called the Marriage Theorem, it is commonly phrased in terms of men marrying women they know. nother example is a room timetabling problem in which groups of students are to be placed in rooms according to the size of the group and the rooms into which they fit. The second type of matching problem is the assignment problem. This differs from the simple marriage problem in that applicants for jobs are not equally qualified. The suitability for a job is specified by a cost, lower cost means greater suitability for the job. The problem then is to find a best matching in the sense of minimising the costs involved. This can be done with the Hungarian algorithm. similar problem is giving gifts to people. If each gift is given a score representing its suitability for a particular person, we may wish to maximise suitability. Finally a modified form of assignment problem is the bottleneck problem in which we assign people to jobs on a production line. This can operate only as fast as the slowest worker, hence the bottleneck. We need to find a maximum matching in which the time taken by the slowest worker is as small as possible. For the gift example above the bottleneck problem is one of maximising the least suitable score of a gift in the matching. MT6 Exam 16 Solutions Page 1 of 1

11 MT6 Examination 16 Part Solutions Q18. (a) (i) Length n = 8, imension k =, Rate = (b) (ii) The number of codewords in = k = = 16 (iii) Since δ = the code can detect errors and correct 1 error. code is self-dual if the generator matrix of the code is also a parity check matrix of the code and k = n k. onsider which is the by zero matrix. Hence, since GG T = and = 8 then G = H and the code is self-dual (c) O = codeword for O = K = codeword for K (d) (i) G H T G = = The error syndromes indicate no errors in the first and third codewords an error in the third bit of the second codeword and the last bit of the fourth codeword. So the codewords are 1111, , 1111, Now 1111 is the sum of rows and of G and corresponds to 11 that is F and is row 1 of G and corresponds to 1 that is I and 1111 is the sum of rows 1, and of G and corresponds to 111 that is N and 1111 is row of G and corresponds to 1 that is E The intended message was hence FINE. k 1 = = n T r = = (e) If two bits are randomly in error then eg would become 11, there are three other codewords 1111, 1111, 1111, each of which are within distance of this error word Similarly is distance from , 1111, 1111 and y symmetry since δ =, this will be true in general and so codewords could have been sent. For use in the above parts (d) nand (e) the set of all codewords of G are as follows : MT6 Exam 16 Solutions Page 11 of 1

12 MT6 Examination 16 Part Solutions Q19 (a) (i) v = 9, b = 18, r = 8, k = so vr = 9 8 = 7 = 18 = bk and r(k 1) 8 λ= = = v 1 8 (ii) The 18 blocks, b, are the 18 rows of trees, the four trees per row is the block size k, giving 7 apple trees in total. There are 9 varieties of fertilizer to be tested on the trees and each fertilizer is to be tested against every other λ = times, by being used r = 8 times in total. (iii) No, the design is not resolvable since to be so would require each variety to occur once per replicate but the number of blocks b = 18 is not divisible by the replication r = 8. (b) Since it is required to test the fertilizer for time of application (early, usual or late) we require replicates, one corresponding to each time period within which each fertilizer will occur once. (c) (i) Using the extended rectangular lattice construction for varieties in a 6 6 array we have (i) The first two replicates are the rows and columns of the array E F G H I J K L (ii) The third replicate comes from the diagonals from bottom left to top right M N P Q R S (d) Yes, we can test more than fertilizers provided we have more than the 7 original trees in order to populate the extra space available in the blocks. It is possible to test 6 fertilizers in a -replicate design if we use blocks of size 6 and a 6 6 square lattice design. We would have 6 plots per block in 18 blocks giving 18 plots in total, there would be replicates of 6 varieties giving 6 = 18 plots. MT6 Exam 16 Solutions Page 1 of 1

13 MT6 Examination 16 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 polyhedron is a three dimensional solid whose faces are polygons in edge to edge contact. In general the polygons can be of any size and shape and the solid so formed can be convex or not. polygon is a plane figure whose edges are straight lines, it is called regular if all its edges are equal in length and consequently all of its internal angles are also equal. Of special interest are the convex polyhedra for which any line segment joining two points within or on the polyhedron itself lies within or on the polyhedron. The convex polyhedra for which the faces are congruent regular polygons are called the regular polyhedra or Platonic solids. ll vertices of such polyhedra have exactly the same arrangement of polygons around them and there are in fact just five examples of such polyhedra. They are the tetrahedron consisting of four equilateral triangles, the octahedron consisting of eight equilateral triangles, the cube consisting of six squares, the icosahedron consisting of twenty equilateral triangles and finally the dodecahedron consisting of twelve regular pentagons. The dual of a regular polyhedron is formed by placing a vertex at the centre of each face and joining these vertices by an edge whenever the corresponding faces are in edge to edge contact. In this way it is easily shown that the duals of Platonic solids are all Platonic solids. In fact the tetrahedron is self dual while the cube and octahedron are duals of each other, as are the icosahedron and dodecahedron. semi-regular polyhedron is one in which the polygonal faces, while they are all regular, are not congruent, however, each vertex has the same arrangement of polygons around it. There are an infinite number of semi-regular polyhedra consisting of two infinite sets plus a set of thirteen. The infinite sets are the prisms and antiprisms, these both have end faces as regular n-polygons but the prisms have n squares around the sides, while the antiprisms have offset end faces and n equilateral triangles around the sides. The thirteen individual semi-regular polyhedra are called the rchimedean solids, five have names which indicate the way they can be formed by truncation from the Platonic solids such as truncated tetrahedron, truncated cube etc. The cuboctahedron is a half way truncation between the cube and the octahedron and similarly the icosidodecahedron is a half way truncation between the icosahedron and the dodecahedron. Finally there are six which are derived from the cuboctahedron and icosidodecahedron either by inserting extra equilateral triangles producing snub versions or further truncation in two different ways. The five basic truncations, the cuboctahedron and icosidodecahedron and their snub versions and one of the further truncations of the cuboctahedron all have two types of faces. The other three have three types of faces. Euler s polyhedron formula is a simple formula which relates the number of vertices v, edges e and faces f of any convex polyhedron, v e + f =. In addition there is a simple result called the handshaking lemma for polyhedra which connects the degree of a face of a polyhedron to the number of edges. Since the degree of a face is defined to be the number of edges surrounding it, then it is obvious, since each edge is adjacent to two faces, that the sum of the face degrees must be equal to twice the number of edges. Euler s formula and the handshaking lemma have many important consequences not the least of which is the result given earlier that there are only five Platonic solids. Given any convex polyhedron we can imagine removing a face and then opening out the polyhedron and flattening it onto the plane. In this way we can obtain a planar graph with the same number of vertices and edges as the original polyhedron. If we count the region outside the graph as a face then there will also be the same number of faces as in the original polyhedron and so Euler s formula will apply to planar graphs as well as to polyhedra. There are many other types of polyhedra such as various types of pyramids which have different arrangements of polygons around different vertices or the stellated polyhedra which are non-convex. MT6 Exam 16 Solutions Page 1 of 1

14 MT6 Examination 16 Part Solutions Essay Outlines Q1. Introduction the definition of a planar graph. Euler 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. Q17. Introduction efinition of a Matching Maximum matching Labelling, alternating paths, complete matching. Marriage Problem The marriage Theorem, equal suitability ssignment Problem lower cost greater suitability, Hungarian algorithm. ottleneck ssignment Problem minimise the time taken by the slowest. Q. Introduction definition of polyhedron, polygon and convex Regular convex polyhedra Platonic Solids uals Semi-regular polyhedra rchimedean Solids Prisms and nti-prisms Euler s Polyhedron Formula Handshaking Lemma for Polyhedra Planar graphs Euler s formula the link with polyhedra. onclusion other polyhedra. MT6 Exam 16 Solutions Page 1 of 1