Solutions to Tutorial 1 (Week 8)

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1 The University of Syney School of Mathematics an Statistics Solutions to Tutorial 1 (Week 8) MATH2069/2969: Discrete Mathematics an Graph Theory Semester 1, In each part, etermine whether the two pictures represent the same isomorphism class of graphs. If the answer is yes, label the vertices so that they become two pictures of the same graph. If the answer is no, give a reason why. (a) (b) Solution: These two pictures represent the same isomorphism class of graphs, as the following labelling of their vertices makes clear. a a e c e c (c) Solution: These two pictures represent ifferent isomorphism classes. In the first graph (or, to say it properly, a graph in the isomorphism class represente by the first picture) there are two 4-cycles, namely the inner square an the outer square, whereas in the secon graph there are three 4- cycles; isomorphic graphs must contain the same number of 4-cycles. There are many other ifferences between the pictures which coul also be use to show that they cannot epict isomorphic graphs. Solution: In the first graph (known as the Petersen graph), there are no 4- cycles, whereas in the secon graph there are, so they cannot be isomorphic. One can also show this by consiering the 5-cycles: the secon graph has only the obvious two 5-cycles, namely the outer an inner rings, whereas in Copyright c 2018 The University of Syney 1

2 () the Petersen graph any ege forms part of four 5-cycles (an there are twelve 5-cycles in total). Solution: These are both pictures of the Petersen graph(or, more properly, the Petersen isomorphism class of graphs), as is shown by the following labelling: a a b f e g b g j e j f c i h c i h 2. Draw the complements of the following two graphs. Are these complements isomorphic to each other? a e c Solution: The complements of these graphs are: a e c These complements are not isomorphic to each other, because the first has two connecte components whereas the secon is connecte (it is a 6-cycle). This is one way to show that the original two graphs are not isomorphic to each other (because if they were isomorphic to each other, their complements woul have to be isomorphic to each other also). 3. Translate each of the following problems into a problem about a graph. You on t nee to solve the problems (in fact you can t, because the specific information require is lacking); just say how to efine the graph, an what the problem is asking in terms of that graph. (a) The communications between the computers in a network go through various ethernet cables, each of which joins two computers. How many ethernet cables have to be remove before the computers are no longer all able to communicate with each other? 2

3 Solution: Define a graph whose vertex set is the set of computers, where two vertices are ajacent if there is a cable joining those computers. The problem is asking how many eges have to be remove before the graph becomes isconnecte. (b) An airline flies various routes between cities aroun the worl. Is it possible to construct a roun-trip itinerary which visits every city that the airline flies to, without ever repeating a route? Solution: Define a graph whose vertex set is the set of cities, where two vertices are ajacent if there is a route between those two cities. The problem is asking whether there is a walk in the graph which uses every vertex, returns to its starting point, an never uses an ege more than once. (c) A omino is a rectangular tile isplaying i ots near one en an j ots near the other, where i,j {0,1,,6}. Suppose we have some ominoes which on t inclue any oubles (that is, i is always ifferent from j) an which are all ifferent (so no set {i,j} appears more than once). Is it possible to lay these ominoes out in a long line so that whenever two ens touch, they have the same number of ots? Solution: Define a graph whose vertex set is {0,1,,6}, where i an j are ajacent if the omino {i,j} is in our collection. The problem is asking whether there is a walk in the graph which uses every ege exactly once. *() Of the various animals in a zoo, some pairs are compatible with each other an some pairs aren t. What is the smallest number of cages require to house all the animals, if the only constraint is that any two animals in the same cage are compatible with each other? Solution: Define a graph whose vertex set is the set of animals, where two vertices are ajacent if they are not compatible. The problem is asking for the smallest number k such that there is a function f from the vertex set to {1,2,,k} with the property that f(v) f(w) whenever v an w are ajacent. 4. Give an example of a graph which is isomorphic to its own complement. Solution: The path graph P 4 has this property: its complement has vertex set {1,2,3,4} an eges {1,3}, {2,4}, {1,4}, which also form a path of length three with en-vertices 2 an Fix an integer n 2. Let i j be two vertices of the complete graph K n. (a) How many walks of length 2 are there in K n from i to j? Solution: A walk of length 2 from i to j has the form i,k,j, where {i,k} an {k,j} are eges. Since we are ealing with the complete graph K n, this just means that i k an k j. Since there are n 2 choices of k satisfying these inequalities, the answer is n 2. (b) How many walks of length 3 are there in K n from i to j? Solution: A walk of length 2 from i to j has the form i,k 1,k 2,j, where the only properties require of k 1,k 2 {1,2,,n} are k 1 i, k 2 j, an 3

4 k 1 k 2. If k 1 = j, there are n 1 ways to choose k 2. For any of the other n 2 possible values of k 1 (other than i an j), there are n 2 ways to choose k 2. So the answer is n 1+(n 2) 2 = n 2 3n+3. *(c) Let a l be the number of walks of length l in K n from i to j. Fin a close formula for a l in terms of n. (Hint: show that the sequence a 0,a 1,a 2, satisfies a recurrence relation.) Solution: Note first that a 0 = 0 an a 1 = 1. For l 2, a walk of length l from i to j has the form i,k 1,k 2,,k l 1,j, where k 1 i, k l 1 j, an k i k i+1 for i = 1,2,,l 2. As in the previous part, we consier the cases where k 1 = j an k 1 j separately. If k 1 j (an there are n 2 possible values of k 1 which satisfy this), then the number of ways to choose k 2,,k l 1 is a l 1, because k 1,k 2,,k l 1,j is a walk of length l 1 from k 1 to j. If k 1 = j, then there are n 1 ways to choose k 2, an whichever value of k 2 is chosen, the number of ways to choose k 3,,k l 1 is a l 2. (If l = 2 this oesn t quite make sense, but the argument gives the right answer of (n 1)a 0 = 0 anyway.) So we have the recurrence relation a l = (n 2)a l 1 +(n 1)a l 2, for all l 2. This is a homogenous linear recurrence relation (from the viewpoint of varying l, n is a constant). Its characteristic polynomial x 2 (n 2)x (n 1) has roots n 1 an 1, so the general solution is a l = C 1 (n 1) l +C 2 ( 1) l, for some constants C 1 an C 2. Imposing the initial conitions a 0 = 0 an a 1 = 1 gives C 1 = 1 = C n 2, so we obtain the close formula a l = (n 1)l ( 1) l. n 6. In this question, G enotes a connecte graph with at least 3 vertices. A vertex v of G is calle a cut vertex if the graph G v obtaine by removing it is not connecte. (a) If G is the cycle graph C n, how many cut vertices oes it have? Solution: If you remove any vertex from C n, you get a graph isomorphic to the path graph P n 1, which is connecte. So C n has no cut vertices. (b) If G is the path graph P n, how many cut vertices oes it have? Solution: If you remove one of the en-vertices from P n, you get a graph isomorphictothepathgraphp n 1, whichisconnecte; ifyouremoveanyone of the other vertices, i.e. 2,3,,n 1, you get a graph with two connecte components. So P n has n 2 cut vertices. (c) Prove that if G contains a brige, then it contains a cut vertex. Is the converse true? Solution: Supposethattheege{v,w}isabrigeofG. Theninthegraph G {v,w}, the vertices v an w belong to ifferent connecte components, say G 1 an G 2. Now the graph G v is a subgraph of G {v,w}, which contains all of G 2 as one connecte component, an all of G 1 v as another 4

5 collection of connecte components. Thus G v is not connecte, except in the case when G 1 v has no vertices, i.e. G 1 consists only of v. Similarly G w is not connecte, except in the case when G 2 consists only of w. Since G has at least 3 vertices, it is impossible for both these exceptions to hol, so either v or w must be a cut vertex of G. The converse is not true: the following connecte graph has a cut vertex v, but no briges. v *() Prove that v is a cut vertex of G if an only if there are two other vertices u an w of G such that v belongs to every path between u an w. Solution: If v is a cut vertex, then G v is not connecte, so there must be two vertices u an w which are not linke to each other in G v. So if we choose any path between u an w in G, then this path must not belong to G v, an the only way that can happen is if v belongs to the path. Conversely, if there are vertices u an w such that v belongs to every path in G between u an w, then there is no path in G v between u an w, so u an w are not linke in G v, which is therefore not connecte. *7. Prove that, for any graph G with at least one vertex, either G is connecte or its complement G is connecte. Solution: It suffices to prove that if G is not connecte, then G is. Suppose that G has connecte components G 1,G 2,,G s where s 2, an let V i enote the vertex set of G i. Then for i j, any vertices v i V i an v j V j are non-ajacent in G, so they are ajacent in G. If v i an v i are two vertices in the same V i, then for any j i an v j V j, the walk v i,v j,v i shows that v i is linke to v i in G. So G is connecte (an in fact any two vertices are joine by a walk of length 2). 5

Solutions to Tutorial 2 (Week 9)

Solutions to Tutorial 2 (Week 9) The University of Syney Shool of Mathematis an Statistis Solutions to Tutorial (Week 9) MATH09/99: Disrete Mathematis an Graph Theory Semester, 0. Determine whether eah of the following sequenes is the