Solutions to Assignment 5

Transcription

1 Solutions to ssignment 5. We have captured several people whom we suspect are part of a spy ring. They are identified as,,,, E, F, and G. fter interrogation, admits to having met the other six. admits to having met five, to having met four, to having met three, E to having met two, F to having met two, and G 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 F 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? Solution: First 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() = 5, deg() = 4, deg() = 3, deg(e) = 2, deg(f ) = 2, deg(g) =. Therefore, not all of the spies are telling the truth. We know that F 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 G has met and, thus G would also be lying, but there is only one liar. If both E and G are telling the truth, then one of or is lying, and since deg(g) =, then has met all of,,, E, and F, and the situation (so far) is as shown in the diagram. If is lying, then deg() 5, and this leads to a contradiction. deg() 4, and this also leads to a contradiction. Therefore, the liar is either E or G. G Similarly, if is lying, then F E This much we did in class.

2 To solve the problem, we now consider now the two possible cases. ase. E is lying and G is telling the truth. In this case, G would have met only, and since E is lying, then E has met = 5 other spies, and since he cannot have met G, he must have met,,, and F. ut, who is telling the truth, has met 5 others, namely,,, E, and F, and has met all 6 others. Thus,, and E have all met F, but this would make F a liar, which is a contradiction. Thus, this case cannot occur. ase 2. E is telling the truth and G is lying. If G is lying, he must have met + 3 = 4 other spies, and the graph shows that this case is possible. G F Thus, the lying spy is G. E 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. Each 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,,,, E, F, and G. ll code words have different frequencies: occurs on the average of 0 times out of 00;, 20 out of 00;, 9 out of 00;, 3 out of 00; E, 7 out of 00; F, 4 out of 00; and G, 9 out of 00. 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 00 code words takes no more than 200 seconds. 2

3 Solution: Each code word is represented by a dash followed by 0 to 6 dots. frequencies have fewer dots, so the codes would be assigned as follows: Those with higher = = = = E = F = G = Since each dash signifies the beginning of a code word, there is no ambiguity. This code will take seconds to send the 00 code words = 9.5 Other more efficient solutions are possible. The record for the past 6 years is 85.5 seconds, and was attained by a student in this year s class. 4. 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 20 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 2 hours and 20 minutes. What is the maximum number of rooms? Solution: In 2 hours and 20 minutes, each keeper can search up to a maximum of 7 rooms, so the number of rooms cannot be greater than 4. 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 2 exactly, all the others have degree 3 or. There must be an even number of vertices with odd degree, so there can be a maximum of 2 vertices with odd degree. Thus there can be no more than 3 rooms in the temple. The diagram below shows that 3 rooms are possible. The two keepers can search the rooms in alphabetical order, and can find the tiger in 2 hours and 20 minutes. E G I K F H J L M 5. uring wartime, the queen and her prime minister live in fortified rooms. The queen has 5 rooms and the PM has 5, for a total of 20 rooms. Some of the rooms are connected by passageways. No two rooms are connected by more than one passageway. For each two of the queen s rooms, there is a unique path of passageways from one to the other. The combined total number of passageways for the queen s rooms and the PM s rooms is 25. Prove that there is a passageway between one of the queen s rooms and one of the prime minister s rooms. 3

4 Solution: Note first that we can represent the problem using a graph, with the vertices representing the rooms and the edges representing the passageways, since there is at most one edge between any two vertices. Since for each pair of vertices in the queen s complex there is a unique path from one to the other, the queen lives in a tree. The queen s tree has 5 vertices and hence 4 edges. Since there are a total of 25 edges for the queen s complex and the PM s complex, then there are edges connecting the( vertices ) in the PM s complex. However, the PM s complex has 5 vertices, and so 5 must have at most = 0 edges. The extra edge must therefore connect one of the PM s rooms 2 with one of the queen s rooms. 6. Let G = (V, E) be a graph with vertex set V and edge set E, and let p = V be the number of vertices in G, and q = E the number of edges in G. The average degree of the vertices in G is defined to be (G) = deg(v). p v V If G is a connected graph, what can you say about G if (a) (G) > 2? (b) (G) = 2? (c) (G) < 2? raw a few pictures before committing yourself!!! Solution: First note that (G) = p v V deg(v) = 2q p (a) (G) > 2 if and only if q > p. Suppose that G is a connected graph and q > p, then G is not a tree and hence contains a cycle. Remove an edge from the cycle, then the resulting graph H is still connected, has p vertices and q edges. Now, p < q = (q ) +, so that H is not a tree, so it also has a cycle. Therefore, G has at least two cycles. onversely, suppose that G is a connected graph which has at least two cycles, then we may remove an edge from two of the cycles and the resulting graph H is still connected, so that p (q 2) + = q < q, and (G) > 2. Thus, if G is a connected graph, then (G) > 2 if and only if G has at least two cycles. (b) (G) = 2 if and only if q = p. Suppose that G is a connected graph and q = p, then G is not a tree and hence contains a cycle. Remove an edge from the cycle, then the resulting graph H is still connected, has p vertices and q edges and p = (q ) +, so that H is a tree. Therefore, G has exactly one cycle. onversely, suppose that G is a connected graph with exactly one cycle, if we remove an edge from the cycle, the resulting graph H is still connected and has no cycles, hence is a tree. Therefore, since H has p vertices and q edges, we have p = (q ) + = q. Thus, if G is a connected graph, then (G) = 2 if and only if G has exactly one cycle. 4

5 (c) (G) < 2 if and only if q < p. Suppose that G is a connected graph and q < p, then since p and q are integers, we must have q + p. We showed in class that any connected graph with p vertices and q edges has p q +, therefore p = q + and G is a tree. onversely, suppose that G is a tree, then G is connected and p = q +, so that (G) = 2q p = p (2p 2) = 2 2 p < 2. Thus, if G is a connected graph, then (G) < 2 if and only if G is a tree. 7. The Floridian parliament has to choose its next prime minister. There are five candidates, namely: xel, oris, laudia, ieter, Erica The house is split into 3 different voting blocks, each with 30 voters. The preferences are shown below. (,,, E, ) (, E,,, ) (,,, E, ) Here (U, V, W, X, Y ) means that a voting block prefers U to V to W to X to Y. The Floridian system uses a sequence of one-on-one elections with the winner of one election being a candidate in the next election. The sequence is set by the chief justice. (a) raw the tournament which is generated by these preferences charts. (b) The chief justice wants laudia to win. What sequence should he use. Solution: The tournament generated by the preference charts is shown below. E Since the tournament contains the directed Hamiltonian path E the chief justice can arrange for laudia to win the election by holding 4 one-on-one elections as follows: Election : Election 2: Election 3: Election 4: vs E winner vs winner vs winner vs 5

6 8. Let G be a graph whose vertices correspond to the bit-strings of length n, a = a a 2 a n where a i = 0 or, and whose edges are formed by joining those bit-strings which differ in exactly two places. (a) Show that G is regular, that is, every vertex has the same degree, and find the degree of each vertex. (b) Find a necessary and sufficient condition that there exist a path joining two vertices a = a a 2 a n and b = b b 2 b n in G. (c) Find the number of connected components of G. Solution: (a) Each vertex has degree ( n 2). (There is an edge between a a 2 a n and b b 2 b n if and only if then the Hamming distance between them is two, so the degree of a a 2 a n is the number of words at distance two from a a 2 a n, or equivalently, the number of words that differ in exactly two places which is ( n 2).) (b) necessary and sufficient condition is that a a 2 a n and b b 2 b n have the same parity (ie, that both have an even number of ones or both have an odd number of ones). The reason is as follows: If the Hamming distance between a a 2 a n and b b 2 b n is an even number, then we can transform a a 2 a n into b b 2 b n by changing two bits at a time, so there is a path from a a 2 a n to b b 2 b n. onversely, if there is a path from a a 2 a n and b b 2 b n, then we must be able to transform a a 2 a n into b b 2 b n by changing two bits at a time. ut changing two bits of a a 2 a n preserves the parity. (c) It follows from (b) that there are exactly two components, one containing all bit strings of length n with an even number of s, the other containing those with an odd number of s. 9. Prove that if a tournament is not strongly connected, then there exists an arc such that if its orientation is reversed, the resulting tournament would be strongly connected. You may use Redei s theorem or the amion-moon theorem in your proof. Solution: y Redei s theorem, any tournament T has a Hamiltonian directed path, that is, a directed path starting at a vertex a and ending at a vertex b, and passing through every vertex in the tournament exactly once. If the tournament is not strongly connected, then a b, and the arc goes from a to b (otherwise, we would have a Hamiltonian cycle and the tournament would be strongly connected). Therefore, if we reverse the direction or orientation of the arc (a, b), then we have a Hamiltonian cycle and the resulting tournament is strongly connected. 0. You are given the weighted graph shown. 8 5 E F H G 2 6

7 (a) Use ijkstra s algorithm to construct a shortest-path spanning tree from the vertex in the graph above. You may do it graphically or by using a table. (b) For the same weighted graph as in part (a), use Kruskal s algorithm to find a minimal spanning tree. You may do it graphically or by using a table. (c) For each of the spanning trees found above, compare the weights of the paths from vertex to vertex G. lso, compare the total weights of the two spanning trees. What can you conclude? Solution: (a) Using ijkstra s algorithm, starting at, we find the shortest-path spanning tree for the graph as follows: Step Tree So Far Fringe Vertices andidate Edge (istance).,, E (8), (0), E(5) 2., E,, F, H (8), (0), EF (6), EH(8) 3., E, F,, G, H F (7), (0), F G(), EH(8) 4., E, F,,, G, H (), (0), F G(), EH(8) 5., E, F,, H,, G (), H(9), F G() 6., E, F,, H,, G (0), F G() 7., E, F,, H,, G F G() 8., E, F,, H,,, G The shortest-path spanning tree from the vertex, as well as the table, is shown below. E F G, 0,,,,,, 5 E F,8,,0,5,, E E, 8,,0 E,6, E,8 EF F F,7,,0 F, E,8 F 3 5,2,0 F, E,8 EH H,2 H,9 F, H H G,0 F, F, FG G (b) Using Kruskal s algorithm, the minimal spanning tree for the graph, as well as the table, is shown below. E F G H E F H G 2 F EF H G EH GH E FG H, (c) s is easily seen from the spanning tree, using ijkstra a algorithm we have d(, G) = and Total Weight = 7, while using Kruskal s algorithm we have d(, G) = 2 and Total Weight = 4. 7

8 We conclude that the shortest-path spanning tree generated by ijkstra s algorithm does not have to be a minimal spanning tree, and that the minimal spanning tree generated by Kruskal s algorithm does not have to be a shortest-path spanning tree. 8