Graph Theory(Due with the Final Exam)

Transcription

1 Graph Theory(ue with the Final Exam) Possible Walking Tour.. Is it possible to start someplace(either in a room or outside) and walk through every doorway once and only once? Explain. If it is possible, show how such a walk can be made. E O The ridges Of Madison ounty.. Here is a diagram of Madison county. The county has 3 islands and bridges. Is it possible to walk through Madison ounty and cross each bridge once and only once? Explain. If it is possible, show how such a walk can be made.

2 Knight s Tour. 3. Following the rules of hess, is it possible for a knight to tour the chess board, visiting every square once and only once, and return to its initial square? Such a path would be a Hamiltonian circuit treating the squares of the chess board as the vertices of a graph. Here is one of several solutions of the problem: a) For a 3 3 chess board, etermine if a knight s tour is possible. b) For a chess board, We can represent the graph as lthough the graph is not complete, see if you can find a Hamiltonian circuit.

3 id You Hear The One bout The Traveling Salesman?. onsider the traveling salesman problem for the following weighted graph: a) Find the Optimal circuit using the rute Force method: ircuit from to E E E E E E E E E E E E E E E E E E E E E E E E Weight E b) Show that the Nearest Neighbor method starting at any of the vertices will not find the Optimal circuit: Vertex Nearest Neighbor Weight E

4 Trace own n nswer. 5. an the following figure be traced with a single, continuous, non-overlapping stroke of a pen? In other words, is there an Euler path or circuit? on t Miss The Forest For The Trees.. Use Kruskal s lgorithm to find the minimum spanning tree for the following weighted graph: 0 0 E F 9 G H I 9 J

5 F G H J K L M N Traces Of Love?. Is it possible to trace every edge exactly once without lifting your pencil and return to where you started (an Euler ircuit) in the following graphs? a) b) c) P Q R S T U V L O V E R

6 Keep It In The Family.. family with seven members in different parts of the country has a relative working overseas. The family wants to set up a telephone-calling network so everyone will know the latest news about the overseas relative, for the least total cost. The relative overseas will call Felix, and then Felix will start the message through the network. The table below shows the cost for a 5-minute phone call between each pair of family members. my Felix Hillary Kit Owen Pearl Robin my - $3.50 $.5 $3.0 $.0 $.5 $5.0 Felix $ $3.5 $.50 $.50 $.0 $3.0 Hillary $.5 $3.5 - $.95 $3.5 $.0 $3.50 Kit $3.0 $.50 $.95 - $.5 $3.30 $3.0 Owen $.0 $.50 $3.5 $.5 - $.95 $3.5 Pearl $.5 $.0 $.0 $3.30 $.95 - $3.0 Robin $5.0 $3.0 $3.50 $3.0 $3.5 $3.0 - a) What is the total cost of the least expensive calling network they can set up?(minimal spanning tree) b) escribe the least expensive calling network.(minimal spanning tree) Table For Two, Nonsmoking, nd Plenty Of Light. 9. restaurant has opened an outdoor patio for evening dining. The owner wants to hang nine decorative light fixtures at designated locations on the overhead latticework. ecause of the layout of the patio and the latticework, it is not possible to install wiring between every pair of lights. The table below shows the distances in feet between lights that can be linked directly. The main power supply from the restaurant building is at location X. The owner wants to use the minimum amount of wire to get all nine lights connected. X E F G H I X E F G H I a) What is the minimum amount of wire needed to connect all nine lights and the power supply?(minimal spanning tree) b) escribe the connections which use the minimal amount of wire.?(minimal spanning tree)

7 Pentagonal Hamiltonian. 0. Find a Hamiltonian circuit for the graph: Information concerning which vertices of a graph are connected by an edge can be stored in a matrix called the adjacency matrix. For example: has the adjacency matrix E F G H I J K L M N O P Q R S T

8 while has and has Let s get back to the first example,

9 If you multiply the adjacency matrix with itself(square it), you get a matrix which tells you the number of different paths of length edges that connect the indicated vertices If you cube it, you get a matrix which tells you the number of different paths of length 3 edges that connect the indicated vertices Now let s add some restrictions to the second graph by specifying the allowable direction of travel along the edges to get a directed graph. To From If you multiply the adjacency matrix with itself(square it), you get a matrix which tells you the number of different directed paths of length edges that connect the indicated vertices

10 The Matrix.. onsider the following directed graph: E Here is the associated adjacency matrix: From E a) How many edges leave from vertex to adjacent vertices? How many edges enter vertex from adjacent vertices? b) How many different ways can you travel from vertex to vertex in a path consisting of directed edges? c) How many different ways can you travel from vertex to vertex in a path consisting of 3 directed edges? To E The Matrix, Part Two.. onsider the given directed graph: a) Find the associated adjacency matrix. b) Square the adjacency matrix. c) How many different ways are there to go from to using two directed edges? d) How many different ways are there to go from to using two directed edges? e) How many different ways are there to go from to using three directed edges? f) How many different ways are there to go from to using three directed edges?

11 onsider the following problem: We look at a diagram of four city blocks, where the lines represent streets. North East Suppose that the density of traffic on these streets and hence the times required to travel each block differ. The travel times in minutes are listed below: North 3 East 5 We want to determine the fastest way of going from point to point subject to the following: t any corner we must go either East or North. So for example, we might take the path indicated in the following diagram:

12 North 3 East 5 It takes minute to travel this particular path. We will designate this path as ENEN using E to designate an East step and N to designate a North step. nother possible path is shown below: North 3 East 5 This path requires minutes, and we designate it as NNEE. Observe that any allowable path has steps: two East steps and two North steps. We may represent the steps as follows: Step Step Step 3 Step

13 Once we have chosen the two steps that will be East steps, the remaining two must be North! steps. So this means that there will be allowable paths from to. They!! are listed in the following table along with the travel times: Path Travel Time EENN 9 ENEN 3 ENNE NNEE 5 NENE 5 NEEN We would like to select the path which requires the smallest number of minutes of travel. In the above list, minutes is the minimum travel time, so we should choose the path NEEN as illustrated in the following diagram: North 3 East So we see that we can solve such problems, and that the time spent is not unreasonable. s a matter of fact, in order to solve the above problem, we performed additions(3 per path) and 5 comparisons. One method for making such comparisons is the following: ompare the time for Path to the time for Path ; take the smaller of these two times and compare with the time for Path 3; and so on. Notice that the number of additions is one less than the number of steps, and the number of comparisons is less than the total number of allowable paths. Now suppose that we have a diagram of 00(0 by 0) city blocks. Then each path would consist of 0 North steps and 0 East steps. The number of allowable paths is 5

14 0!,5 0! 0 0!. In order to solve this larger problem the same way we solved 0 0 the smaller problem, we would need to perform 9,5 3,50,3 additions. We would also have to make,55 comparisons. dedicated student(with very little sleep) could solve this problem in a month. computer performing additions and comparisons at the rate of,000,000 per second would require approximately 3.5 seconds of computing time to solve the problem. Now let s consider a diagram of 900(30 by 30) city blocks. Each allowable path would consist of 30 East steps and 30 North steps. The number of allowable paths would be 0! 0 30,,5,5,, 30! 0 30!. The number of additions per path is 59, so to determine the path of least time, we would need to perform 59,,5,5,,,9,0,3,3,,0 additions and,,5,5,,3 comparisons. This totals,095,,93,9,5,39. computer performing additions and comparisons at the rate of,000,000 per second would take more than 5,000 years to find the best path. So we have the following table: Number of locks Number of Operations Human omputer 3 0 minutes ,95,9 month 3.5 seconds 900,095,,93,9,5,39-5,000 years The solution procedure that we have been using is called irect Search. The irect Search procedure involves finding a value associated with each alternative(in our case the travel time associated with each path). There are many real-life problems in the physical, social, and management sciences that present more alternatives than our 900 block problem. Therefore, we need a more efficient method for solving problems of this type. The method that we now illustrate is the basic idea underlying a relatively new(invented by Richard ellman around 950) branch of Mathematics known as ynamic Programming. The ynamic Programming method will solve a 900 block problem using fewer than 3,000 operations, an almost unbelievable reduction from the,095,,93,9,5,39 operations required by irect Search.

15 Let s go back to the by block problem and introduce a coordinate system with the origin at the point so that we can label the street intersections. 0, North, 3, 0,,, East 0,0,0 5,0 Let s start at and work backwards as follows: If we arrive at corner,, we have no choice but to go East, which require 3 minutes. If we arrive at corner,, we have no choice but to go North, which will require minutes. Similarly, if we arrive at corner,0, we have no choice but to go North, which require 3 minutes to reach. If we should arrive at corner 0,, we North have no choice but to 0 3 go East, 3 which require 0 minutes to get to. We indicate these numbers and directions in the following East figure: 3

16 If we arrive at corner,, we do have a choice of East or North. Going East would require 0 minutes, while going North would require minutes, so we ll go East. This is indicated in the following figure: North East 3 If we arrive at corner,0, again, we have a choice of East or North. North would yield a trip of minutes to, while East would take minutes, so we should go North. North East 5 3

17 If we complete this process, we end up with the following diagram: North East 3 5 So from the diagram, we can see the fastest path from to. It will take minutes and is indicated in the following figure: North 3 0 East 5 You can see it s the path NEEN, just like we found using the direct search. The new method is more efficient.

18 e ynamic, nd Get With The Program. 3. onsider the following diagram: There are 0 different North/East paths from to. Find the path of least travel time using ynamic Programming.

19 . onsider the following diagram: on t Stop Now There are 0 different North/East paths from to. Find the path of least travel time using ynamic Programming. Even I Think It s Odd, Or Oddly, I Think It s Even. 5. If a graph has no edges, then the degree of each vertex is zero, and therefore the sum of the degrees of the vertices is also zero. For each edge that is added to the graph, it adds to the total of the degrees of the vertices, so this means that the sum of the degrees of the vertices in a graph must be an even number(twice the number of edges). Using the previous conclusion, what can you say about the number of odd degree vertices in a graph?

20 Short ircuit In Ferris Euler s ay Off.. In this problem, you will show that a graph that contains an odd vertex, cannot have an Euler ircuit. Suppose that vertex has odd degree, i.e., 3, 5,. If there is an Euler ircuit for the graph, then the circuit can be assumed to start and end at vertex. ase I: has degree. If the circuit starts at vertex, then there is no way to get back to. ase II: has degree 3. 3 If the circuit starts at vertex, then it must come back to, and then leave again. s in ase I, there is no way for it to get back to. a) Reason out ase III: has degree 5 in a similar way as ases I and II. b) Try to find an argument that works for all odd degrees of vertex. {Hint: Edges away from and towards can be grouped in pairs.}