Solutions to Tutorial 2 (Week 9)

Transcription

1 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 egree sequene of a graph. If so, raw a piture of suh a graph. (a) (,,) Solution: Yes, this is the egree sequene of the yle graph C (or any graph isomorphi to it): (b) (0,,,) Solution: This sequene is not graphi. If there were a graph G with egree sequene (0,,,), then G woul have verties, one of whih (the one with egree 0) is not ajaent to any other vertex, an one of whih (the one with egree ) is ajaent to every other vertex. This is a ontraition. () (,,,,,) Solution: Sine the sum = is o, this annot be the egree sequene of a graph. (The Han-shaking Lemma implies that the sum of the egrees is always even.) () (,,,,) Solution: This is not graphi, beause in a graph with verties, the egree of any vertex is at most. (e) (,,,,) Solution: This is the egree sequene of a graph with the following piture (this is the isomorphism lass of the omplete bipartite graph K, ): (f) (,,,) Solution: If this were the egree sequene of a graph G, then G woul have verties, three of whih were ajaent to all other verties; it is then impossible for the fourth vertex not to be ajaent to the other three also. (g) (,,,,,) Solution: Aoring to the reursive rule given in letures, (,,,,,) is graphi if an only if (,,,,) is graphi, whih in turn hols if an only if (0,0,,) is graphi. But (0,0,,) is the egree sequene of a graph with verties an one ege, so the original sequene is graphi also. To Copyright 0 The University of Syney

2 onstrut a graph with egree sequene (,,,,,), we an start with a graph with egree sequene (0,0,,): then a another vertex an join it to the appropriate existing verties to reate a graph with egree sequene (,,,,): then a another vertex an join it to the appropriate existing verties to reate the esire graph: (h) (,,,,,,) Solution: Aoring to the reursive rule, (,,,,,,) is graphi if an only if (0,,,,,) is graphi; but the latter sequene is not graphi, beause the vertex of egree nees other verties of nonzero egree to be ajaent to.. Suppose that G has verties, an every vertex has egree at least. (a) Is it possible that G is regular of egree? Solution: No, beause thenthe sumof theegrees woul be =, whereas we know by the Han-shaking Lemma that the sum of the egrees is always even. (b) Prove that G is onnete. Solution: If G were not onnete, then G woul have onnete omponents G,G,,G s for some s. There must be one of these onnete omponents, say G i, whih has at most verties, beause if every G i ha at least verties, then the total number of verties woul be at least s =, ontrary to assumption. But if G i has at most verties, then every vertex of G i has egree at most 0, again ontrary to assumption. This ontraition proves that G must be onnete.. For eah of the following graphs, either fin an Eulerian iruit, fin an Eulerian trail, or explain why neither exists. (a) b e f a Solution: Sine every vertex has even egree, we know that there must be an Eulerian iruit. One of many suh iruits is the following walk: a,b,,a,e,f,,e,b,f,,,a.

3 (b) () e j i f a g h Solution: Sine there are more than two verties of o egree (in fat, every vertex has o egree), there annot be either an Eulerian iruit or an Eulerian trail in the Petersen graph. a b b e Solution: Sine there are exatly two verties of o egree, namely b an, there is no Eulerian iruit but there is an Eulerian trail whih starts at b an finishes at. An example of suh a trail is the following: f b,,a,b,,e,f,,. *() a b e f g h Solution: Sine there are exatly two verties of o egree, namely f an k, there is no Eulerian iruit but there is an Eulerian trail whih starts at f an finishes at k. To fin suh a trail, we an a the ege {f,k} to make the graph Eulerian, an then fin an Eulerian iruit in the moifie graph by the metho of removing the eges of yles. In the following iagram, the yles with verties f,,a,e an f,g,i,l,k are otte to iniate that they have been remove. a b i f g h e i l It is easy to fin Eulerian iruits in the omponents of the subgraph whih remains: for example,,,e,b, an g,h,j,k,i,h,k,g. We an then inorporate these Eulerian iruits into the remove yles to obtain one big Eulerian iruit: f,,,e,b,,a,e,f,g,h,j,k,i,h,k,g,i,l,k,f. Removing the extraneous ege {f, k} (whih is the last one use in this iruit) we obtain an Eulerian trail in the original graph: f,,,e,b,,a,e,f,g,h,j,k,i,h,k,g,i,l,k. j k l j k

4 . Fin the number of verties, the number of eges, an the egrees of: (a) the omplete bipartite graph K p,q ; Solution: The number of verties is p+q, an the number of eges is pq. Theverties,,,pallhaveegreeq, antheverties p+,p+,,p+q all have egree p. (b) the ube graph Q m. Solution: Remember the efinition of Q m : the verties are the m-tuples (x,x,,x m ) where every x i is either 0 or, an two suh m-tuples are ajaent if an only if they iffer in exatly one oorinate. The number of verties is learly m. Every vertex has egree m, beause there are m hoies for whih oorinate to hange. Hene by the Han-shaking Lemma the number of eges is mm = m m. *. What is the smallest number n for whih there exist two non-isomorphi graphs with n verties having the same egree sequene? Solution: The answer is. From the lassifiation of graphs with at most verties isusse in letures, it an be seen that for n the egree sequene etermines the isomorphism lass of the graph uniquely. However, there are two nonisomorphi graphs with verties whih both have egree sequene (,,,,): namely, the path graph P an the non-onnete graph whose two omponents are a path P an a yle C.. Determine whether the graphs represente by the following pitures are Hamiltonian or not. Explain your answers. (a) (b) Solution: This graph is Hamiltonian; a spanning yle is obtaine by eleting the top ege. Solution: This graph (the omplete bipartite graph K, ) is not Hamiltonian. Sine it has verties an eges, any spanning yle woul have to be obtaine by eleting a single ege; but it is lear that eleting any ege of this graph will not result in a yle. Alternatively, if you imagine the top three verties as oloure white an the bottom two as oloure blak, then every ege joins a white vertex to a blak vertex, so a spanning yle woul have to alternate between white an blak verties, whih is impossible as there are not the same number of eah.

5 () *() Solution: This graph is not Hamiltonian. If there was a spanning yle, it woul have to ontain 0 eges (i.e. all but two of the eges of the graph). Moreover, it woul have to ontain two of the eges ening at eah vertex. Consiering all the verties of egree, we see that this fores the yle to ontain all the eges in the outer an inner ring; this alreay onstitutes 0 eges, but they o not form a single yle. Solution: This graph is Hamiltonian. To fin a spanning yle, we an make an arbitrary hoie of whih two eges ening at the entral vertex are inlue, an then buil the rest of the yle by ontinuing to ensure that at eah vertex, two eges ening there are inlue. Here is one example of a spanning yle: *(e) Solution: This graph is not Hamiltonian. It is a bipartite graph; in other wors, the verties an be oloure white an blak in suh a way that every ege joins a white vertex to a blak vertex: Any spanning yle woul have to alternate between white an blak verties. But there are white verties an blak verties, so this is impossible.

6 *(f) Solution: This graph is not Hamiltonian. If there were a spanning yle, then for every one of the verties of egree in the mile ring, the spanning yle woul have to use both of the eges ening at that vertex. This means that it woul have to use every ege of the mile ring, an then oul not use any of the eges joining the mile ring to the outer ring an the inner ring. So it woul be impossible for it to ontain all the verties.. Show that a Hamiltonian graph annot ontain a brige. Solution: Suppose for a ontraition that G was a Hamiltonian graph ontaining a brige {v,w}. Let C be a spanning yle of G. Sine {v,w} is a brige, it is not ontaine in any yle, so it is not an ege of C. Thus C is a spanning yle of the graph G {v,w}. But the latter graph is not onnete, so we have our ontraition.. A lub plans a series of inners in whih the members will be sitting aroun a roun table. They want the seating plans for these inners to have the property that every member sits next to two people they have never sat next to before. For instane, if there are five members, they an hol two inners with the following seating plans: After these two inners ever member has sat next to the other four, so there annot be any further inners. (a) If there are members, fin the maximum possible number of inners the lub oul hol. Solution: At every inner, member sits next to two new members; there are other members in total, so there oul not be any more than inners. To see that the answer is, we nee to atually fin seating plans with the

7 require property. Here is one solution: **(b) Repeat the previous part where is replae by any o number n. Solution: As in the previous part, we see that there annot be more than n inners; the question is whether it is possible to hol this many. Eah inner orrespons to a spanning yle of K n, so we an rephrase the question ( in graph-theoreti terms as follows: is it possible to partition the n ) eges of Kn into n subsets, eah of whih forms a spanning yle? The answer is yes, an one suh partition is efine as follows. It is slightly easier to work with K m where m = n is even: it is enough to show that we an fin m spanning paths of K m whih have no eges in ommon, an no en-verties in ommon. (Then we an join the extra vertex n to the en-verties of eah path to get our spanning yles of K n.) We an imagine the m verties of K m evenly istribute aroun the irumferene of a irle, an numbere in anti-lokwise orer. For example, here is a piture of the ase m =, with a partiular spanning path shown: If we rotate the eges of this spanning path aroun by angles of π, π, an, we get three other spanning paths: π These four spanning paths o not share any eges or en-verties, beause the smallest rotationneee to take anege of the original pathinto another ege of that path is a half-turn, i.e. rotation by π. The same onstrution works for a general even number m. The spanning path with verties orere,,m,,m,,m,, m, m+, m+ has the property that its eges stritly inrease in length up to the mile, an then

8 the eges after the mile are the rotations by π of the eges before the mile, in reverse orer. So the smallest rotation neee to take an ege into another ege is rotation by π, whih means that the spanning paths obtaine from this one by rotation by kπ for 0 k < m have no eges or m en-verties in ommon.