Math 222 (A1) Solutions to Assignment 4

Transcription

1 Math (1) Solutions to ssignment 1. We have captured several people whom we suspect are part of a spy ring. They are identified as,,,,,, and. fter interrogation, admits to having met the other six. admits to having met five, to having met four, to having met three, to having met two, to having met two, and to having met one. None of them would identify whom they knew, and no spy would claim to have met more people than he has actually met. ssume that is telling the truth and there is only one liar. Who is the lying spy, if it is known in addition that the number of acquaintances he gives is 3 less than the true value? Why? irst recall the following from class: In the acquaintanceship graph, with vertices being the spies and with an edge between two vertices if and only if the corresponding spies have met, if all of the spies were telling the truth, there would be an even number of vertices with odd degree, but the degree sequence is deg() = 6, deg() =, deg() =, deg() = 3, deg() =, deg( ) =, deg() = 1. Therefore, not all of the spies are telling the truth. We know that is telling the truth and there is only one liar. ny spy who lies will claim to know less than he actually does, so must be telling the truth, since there are only 6 other spies. lso, is telling the truth, since if not, then has met 6 or more spies, which implies that has met and, thus would also be lying, but there is only one liar. If both and are telling the truth, then one of or is lying, and since deg() = 1, then has met all of,,,, and, and the situation (so far) is as shown in the diagram. If is lying, then deg(), and this leads to a contradiction. Similarly, if is lying, then deg(), and this also leads to a contradiction. Therefore, the liar is either or. This much we did in class. Solution: To solve the problem, we now consider now the two possible cases. ase 1. is lying and is telling the truth. In this case, would have met only, and since is lying, then has met + 3 = other spies, and since he cannot have met, he must have met,,, and. ut, who is telling the truth, has met others, namely,,,, and, and has met all 6 others. Thus,, and have all met, but this would make a liar, which is a contradiction. Thus, this case cannot occur. ase. is telling the truth and is lying. If is lying, he must have met = other spies, and the graph shows that this case is possible. Thus, the lying spy is. 1

2 . substantial collection of jewels is in a chest in one of two underground labyrinths, and they are in a room with an odd number of doors. ach door connects two different rooms. The first labyrinth has two entrance doors and the other has three, and only one of these labyrinths has any rooms with an odd number of doors. ssume that the labyrinth with two entrance doors does not contain any room with an odd number of doors. Prove that it is possible to enter this labyrinth by one entrance door and exit by the other. Solution: We know from the hypotheses that the labyrinth with two entrance doors has only rooms with an even number of doors. Suppose it is not possible to enter through one entrance door and exit by the other. Remove all of the labyrinth except for the part accessible from one of the entrance doors. In other words, look at the component that contains one of the entrance doors. onsider the graph that represents this component, where the area outside the labyrinth is represented by one vertex and the rooms are represented by other vertices. oors are represented by edges between the vertices. The graph is connected (all rooms belong to the same component) and the outside vertex has degree exactly one. y the parity theorem, there must be an even number of vertices with odd degree, which means that at least one other door has odd degree. ut this vertex would then represent a room with an odd number of doors. However, this room is part of the original labyrinth. Thus, we have a contradiction. 3. There are seven code words, denoted by,,,,,, and. ll code words have different frequencies: occurs on the average of 10 times out of 100;, 0 out of 100;, 9 out of 100;, 31 out of 100;, out of 100;, out of 100; and, 19 out of 100. ll our messages are to be sent in dots and dashes, but unlike Morse code, which has pauses between letters, we want the code words to go out without pauses. Trained people can send dots accurately at the rate of two per second, including the silence before the next dot or dash. ashes are slower, however, achieving rates of only one per second. Suppose that we are not allowed to use dot dot as a code word. esign an unambiguous code such that an average message of 100 code words takes no more than 00 seconds. Solution: ach code word is represented by a dash followed by 0 to 6 dots. Those with higher frequencies have fewer dots, so the codes would be assigned as follows: = = = = = = = Since each dash signifies the beginning of a code word, there is no ambiguity. This code will take = 191. seconds to send the 100 code words. Other more efficient solutions are possible. I believe the record for the past years is 186. seconds.

3 . tiger of the fiercest species has escaped from a zoo and is hiding in an abandoned temple. The zoo has keepers who are trained to catch tigers. The tiger may be in any room. The tiger may run from one room to another while they are looking for it, although it won t run into a room that has a keeper in it. The temple has no windows and only one entrance. Of its rooms, all but one connect to one other room or to three others. The last room connects to two other rooms. There are no doors of any kind between rooms. The rooms are dark and the tiger may find many hiding places, even though none of the rooms is very large. There is only one way to walk from any room to any other in the temple. We do not want to lay traps or put barriers between rooms. The temple is small enough that moving from room to room takes almost no time. oing a thorough search of a room takes a keeper 0 minutes. If he finds the tiger he can use his stun gun. Time is of the essence, because the tiger can scatch its way through various parts of the temple wall in under three hours. It is imperative that at no time should the tiger have a free path to the entrance. esign the layout of the temple with the properties above as stated by hief Inspector Singh, so that an escaped tiger inside the temple can be trapped by two keepers in hours and 0 minutes. What is the maximum number of rooms? Solution: In hours and 0 minutes, each keeper can search up to a maximum of rooms, so the number of rooms cannot be greater than 1. onsider the rooms as being the vertices of a graph and the connections between the rooms as being the edges (we are ignoring the entrance door). One vertex has degree exactly, all the others have degree 3 or 1. There must be an even number of vertices with odd degree, so there can be a maximum of 1 vertices with odd degree. Thus there can be no more than 13 rooms in the temple. The diagram below shows that 13 rooms are possible. The two keepers can search the rooms in alphabetical order, and can find the tiger in hours and 0 minutes. I K H J L M. graph has 11 vertices labeled to K, and the following 18 edges with their lengths given: (8), (), (), (6), (), (), (), (), () H(), H(), I(), J(), I(), J(), HK(), IK(), JK() pply Kruskal s algorithm to construct a minimum-length spanning tree. Solution: We apply Kruskal s algorithm as summarized in the following chart. I I H H J J HK JK H I J K

4 The graphical solution below is also acceptable. The edges were selected using the order of the edges as given in the previous table. K 8 J 6 I H 6. pply ijkstra s algorithm to construct a shortest-path spanning tree from the vertex in the graph in the preceding problem. Solution: We apply ijkstra s algorithm as summarized in the following chart: H I J K, 0,,,,,,,,,,, 8,,,,,,,,,, 8,, 9,10,,,,,, 8, 9,10, 9,,,,, 9,10, 9,,,,,10, 9,13,,,,10,13,11,13,,13,11,13,,13,13 I,16,13 I,16 I,16 I H J IK

5 graphical solution is given below, and is also acceptable. Using ijkstra s algorithm, starting at, the shortest-path spanning tree for the graph is shown below: K 8 J 6 I H. graph consists of vertices V 1, V, V 3, V, V, V 6, V. Two vertices V i and V j are joined by an edge if and only if the absolute difference between i and j is neither 3 nor. Is this graph planar? If so, draw it without crossing edges. If not, explain why not. Solution: The graph is nonplanar since it contains a subdivision of K 3,3 (see the diagram below). V 1 V 1 V V V V V 6 V 6 V 3 V 3 V V V V (original graph) (subdivision of K 3,3 ) elete the edges V V 3, V V, V 6 V, and V V 1, from the original graph, and the resulting subgraph has V, V 3, and V 6 in one bipartition set and V 1, V, and V in the other bipartition set, with V subdividing the edge V V.

6 8. labyrinth contains 1 rooms. ach room is connected to at least 6 others by a direct tunnel that does not pass through any other room. ach of nine of the rooms are directly connected to at least others. Show that you can pass from one room to any other room by a sequence of tunnels. Solution: onsider the graph whose vertices are the rooms, where there is an edge joining two vertices if and only if there is at least one tunnel directly connecting the corresponding rooms. Supposing that the graph is not connected, then it has at least two connected components, and each component must contain at least vertices, since each room is connected to at least 6 others. Since 1 = + 8, the graph can only have two connected components, one containing vertices and the other containing 8 vertices. Now, there are nine rooms each connected to at least others. Since the component with eight vertices cannot account for all of these, at least one of these nine rooms must belong to the other component. That is, in the component with rooms, one of the rooms is adjacent to others. ut this is a contradiction, since in a -vertex component, each vertex is adjacent to at most 6 others. Therefore the graph must be connected. Note: With only 8 rooms connected to others (instead of 9) it is possible to have components, namely K + K 8. 6