Hamilton Paths A school bus picks up children along a planned route. (credit: “Kids at School Bus Stop” by Ty Hatch/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Describe and identify Hamilton paths. Evaluate Hamilton paths in real-world applications. Distinguish between Hamilton paths and Euler trails. In the United States, school buses carry 25 million children between school and home every day. The total distance they travel is around 6 billion kilometers per year. In the city of Boston, Massachusetts, the 2016 budget for running those buses was $120 million dollars. In 2017, the city held a competition to find ways to cut costs and the Quantum Team from the MIT Operations Research Center came to the rescue, using a computer algorithm to identify the most efficient and least costly routes, which saved the city of Boston $5 million each year and even reduced daily CO 2 emissions by 9,000 kilograms! ( This U.S. city put an algorithm in charge of its school bus routes and saved $5 million , Sean Fleming, World Economic Forum) The problem the Quantum Team tackled involves graph theory. Imagine a graph in which vertices are the bus depot, the school, and the bus stops along a particular route. The bus must start at the depot, visit every stop exactly once, and end at the school. The route is a special kind of path that visits every vertex exactly once. Can you guess what those paths are called? Hamilton Paths Just as circuits that visit each vertex in a graph exactly once are called Hamilton cycles (or Hamilton circuits), paths that visit each vertex on a graph exactly once are called Hamilton paths . As we explore Hamilton paths, you might find it helpful to refresh your memory about the relationships between walks, trails, and paths by looking at . We know that paths are walks that don’t repeat any vertices or edges. So, a Hamilton path visits every vertex without repeating any vertices or edges. shows a path from vertex A to vertex E and a Hamilton path from vertex A to vertex E . Walks, Trails, and Paths Path or Hamilton Path? Identifying Hamilton Paths Which of the following sequences of vertices is a Hamilton path for Graph Q in ? Graph Q a → d → b → c → e → g → f c → b → e → h → g → f → d → a h → e → g → d → b → e → g → f → d → a → b → c Sequence 1 is a path, because it is a walk that doesn’t repeat any vertices or edges, but not a Hamilton path because it skips vertex h . Sequence 2 is a path that visits each vertex exactly once; so, it is a Hamilton path. Sequence 3 is a walk, but it is not a path because it visits vertices g , e , and b each more than once; so, it cannot be a Hamilton path. So, we can see that only sequence 2 is a Hamilton path. TIP! Since a Hamilton path visits each vertex exactly once, it must have the same number of vertices listed as appear in the graph. Finding Hamilton Paths Suppose you were visiting an aquarium with some friends. The map of the aquarium is given in . The letters represent the exhibits. Map of Aquarium Exhibits shows a graph of the aquarium in which each vertex represents an exhibit and each edge is a route between the pair of exhibits that doesn’t bypass another exhibit. Graph of Aquarium Let’s see if we can plan a tour of the exhibits that visits each exhibit exactly once, beginning at exhibit O and ending at exhibit C . Suppose that, after exhibit O , we plan to visit exhibit Q and then exhibit M . After M , should we plan to visit N , L , or R ? Take a look at . If R is not chosen next, that will cause a problem later on. Do you see what it is? Choosing Vertex L , N , or R If L or N is chosen next, the only way to get to R later will be to go from S to R , and then we will not be able to continue without repeating a vertex. So, we will pick R next, and then the only option is S . After S we have another choice to make. As shown in , the next choice is between B and E . Keeping in mind that the goal is to end at C , which would be the better choice? Choosing Vertex B or E If you said vertex B , you are right! Otherwise, we will not be able to visit B later. After B , the only option is E . Then we can choose either D or G . Either will work fine. Let’s choose G as shown in . After G , you must visit H , but should you visit K or L after that? Choose Vertex L or K If you said to go to vertex L next, you are right! Otherwise, it will be impossible to visit N without repeating a vertex. So, next is L , then N , then K , and then at J you have another decision to make we can see in . Should you choose F , I , or P next? Choose Vertex F , I , or P If you said P , you are right! If you choose either of the other two vertices, you will not be able to visit P later without passing through another vertex twice. Once P is chosen, vertex I must be next followed by F . Then you have to choose between A and D as shown in . Choose Vertex A or D In this case, we must go to D then to A so that we can visit C without backtracking. The complete Hamilton path is shown in . Complete Hamilton Path from O to C So, one Hamilton path that begins at O and ends at C is O → Q → M → R → S → B → E → G → H → L → N → K → J → P → I → F → D → A → C . There is no set sequence of steps that can be used to find a Hamilton path if it exists, but it does help to keep in mind where we are headed and avoid choices that will make returning to a particular vertex impossible without repeating vertices. Let’s practice finding Hamilton paths. Finding a Hamilton Path Use to find a Hamilton path between vertices C and D . Graph G If we start at vertex C , A must be next. Then we must choose between B and F . If we choose F , we will have to backtrack to get to include B ; so, we must choose B . Once we choose B , we must choose F next. After F , we choose E , because we want to end at D . So, a Hamilton path between C and D is C → A → B → F → E → D . Existence of a Hamilton Path It turns out that there is no Hamilton path between vertices A and E in Graph G in . To understand why, let’s imagine there is a red apple tree on one side of a bridge and a green apple tree on the other side of the bridge. Now suppose someone asked you to pick up all the fallen apples under each tree without crossing the bridge more than once, and making sure that the first apple you pick up and the last apple you pick up are both red. You would say, that is impossible! To have the first and last apple be red would either require leaving the green apples on the ground or crossing the bridge twice. Let’s see how this relates to finding a Hamilton path between A and E in Graph G . The edge AC is a bridge because, if it were removed, the graph would become disconnected with two components, the component { C } and the component { A , B , D , E , F }. So, we can think of the vertices A , B , D , E , and F as the red apples, vertex C as the green apple, and the edge AC is the bridge between them as in . Bridge between Red and Green Apples The creation of a Hamilton path requires a visit to each vertex, just as picking up all the apples requires a visit to each apple. A and E are both red apples; so, a path from A to E would both start and end at a red apple, just as you were asked to do. And you wouldn’t be able to cross the bridge twice because that would mean visiting A twice, which is not allowed in a Hamilton path. So, it is impossible to find a Hamilton path from A to E just as it was impossible to pick up all the apples without crossing the bridge more than once. By the same reasoning, if a graph has a bridge, there will never be a Hamilton path that begins and ends on the same side of that bridge, meaning beginning and ending at vertices that would be in the same component if the bridge were removed from the graph. Finding a Hamilton Path If One Exists Find a Hamilton path from vertex s to vertex v for each graph in or indicate that there is none. Graphs A , B , C , and D Graph A : Edge uw is a bridge connecting component { s , t , u , v } to the component { w , x , y , z }. There is no Hamilton path from vertex s to vertex v because they would be part of the same component if the bridge uw were removed. Graph B : There are no bridges in Graph B . The only method we have to determine if a Hamilton path from vertex s to vertex v exists is to try every possibility. From vertex s , we can visit either vertex y or vertex t . We will try vertex y first and then come back to see what happens with vertex t . After visiting y , we must visit z and then u , but then we have to decide between vertices r , t , and v next as shown in . Choose Vertex r , t , or v Vertex v is not an option since we want to end at v . Vertex t is not an option since that would force us to go to visit s a second time. So, we must go to vertex r next. After vertex r , we must visit x , then w , then v , but we missed vertex t as shown in . Missed Vertex t Let’s go back to the beginning and choose t instead of y . After t , we must go to u and then we have a choice to make between r , v , and z as shown in . Choose Vertex v , r , or z Vertex v is not an option since we want to end at v . Vertex z is not an option since that would force us to go y and then to visit s a second time. So, we must go to vertex r next. After r , we must go to x then w then v , where we have to stop even though we have missed vertices y and z , as shown in . Missed Vertices y and z So, we have tried every possible route and there are no Hamilton paths between s and v in Graph B . Graph C : In Graph C , there is a Hamilton path, s → t → u → x → w → v . Graph D : In Graph D , there is a bridge, tx , which would form components { r , s , t , u , q } and { v , w , x , y , z } if it were removed. Since s and v would be in different components, it is possible there is a Hamilton path between them. The only way to know is to try all possibilities. If we begin at s , we can go to r then t , or we can go directly to t , either way, we have a problem as you can see in . Vertices Visited Twice or Skipped If we visit all the vertices in the component { r , s , t , u , q }, we will have to visit t a second time in order to cross the bridge. If we visit t only once, we have to skip some of the vertices. So, there is no Hamilton path between s and v . There is not a short way to determine if there is a Hamilton path between two vertices on a graph that works in every situation. However, there are a few common situations that can help us to quickly determine that there is no Hamilton path. Some of these are listed in . Scenario Diagram Scenario 1 If an edge ab is a bridge, then there is no Hamilton path between a pair of vertices that are on the same side of edge ab . We saw this in Graph A of . No Hamilton path between any two vertices in the component { a , c , d , f }. No Hamilton path between any two vertices in { b , e , h , g , i }. Scenario 2 If an edge ab is a bridge with at least three components on each side, then there is no Hamilton path beginning or ending at a or b . We saw this in Graph D of . No Hamilton path beginning or ending at a or b . Scenario 3 If a graph is composed of two cycles joined only at a single vertex p , and v is any vertex that is NOT adjacent to p , then there are no Hamilton paths beginning or ending at either p or v . We saw this in Graph B of . No Hamilton path can be formed starting or ending at vertices, r , v , or u because they are not adjacent to p . Some Impossible Hamilton Paths There are also many other situations in which a Hamilton path is not possible. These are just a few that you will encounter. Hamilton Path or Euler Trail? We learned in Euler Trails that an Euler trail visits each edge exactly once, whereas a Hamilton path visits each vertex exactly once. Let’s practice distinguishing between the two. Distinguishing between Hamilton Path and Euler Trail Use to determine if the given sequence of vertices is a Hamilton path, an Euler trail, both, or neither. Graphs A , F , and K Graph A , e → b → a → e → d → c → b Graph F , f → g → j → h → i Graph K , k → l → m → n → o Since the sequence covers every edge once but visits vertices more than once, it is only an Euler trail. Since the sequence visits every vertex exactly once but skips some edges, it is only a Hamilton path. Since the sequence visits each edge and each vertex exactly once, it is both an Euler trail and a Hamilton path. As we saw in , the Emerald City lies at the center of the Magical Land of Oz, with Gillikin Country to the north, Winkie Country to the east, Munchkin Country to the west, and Quadling Country to the south. Munchkin Country and Winkie Country each shares a border with Gillikin Country and Quadling Country. Let’s apply graph theory to Dorothy’s famous journey through Oz one more time! Draw a graph in which each vertex is one of the regions of Oz. Is there a Hamilton path that Dorothy could follow, instead of the yellow brick road, to lead her from the land of the Munchkins, through all the regions of Oz exactly once, and end in the Emerald City? If so, what might it be? Compare your results with those of a classmate. Check Your Understanding Key Terms Hamilton path Key Concepts A Hamilton path visits every vertex exactly once. Some Hamilton paths are also Euler trails, but some are not.

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Traveling Salesperson Problem Each door on the route of a traveling salesperson can be represented as a vertex on a graph. (credit: "Three in a row, Heriot Row" by Jason Mason/Flickr, CC BY 2.1) Learning Objectives After completing this section, you should be able to: Distinguish between brute force algorithms and greedy algorithms. List all distinct Hamilton cycles of a complete graph. Apply brute force method to solve traveling salesperson applications. Apply nearest neighbor method to solve traveling salesperson applications. We looked at Hamilton cycles and paths in the previous sections Hamilton Cycles and Hamilton Paths . In this section, we will analyze Hamilton cycles in complete weighted graphs to find the shortest route to visit a number of locations and return to the starting point. Besides the many routing applications in which we need the shortest distance, there are also applications in which we search for the route that is least expensive or takes the least time. Here are a few less common applications that you can read about on a website set up by the mathematics department at the University of Waterloo in Ontario, Canada: Design of fiber optic networks Minimizing fuel expenses for repositioning satellites Development of semi-conductors for microchips A technique for mapping mammalian chromosomes in genome sequencing Before we look at approaches to solving applications like these, let's discuss the two types of algorithms we will use. Brute Force and Greedy Algorithms An algorithm is a sequence of steps that can be used to solve a particular problem. We have solved many problems in this chapter, and the procedures that we used were different types of algorithms. In this section, we will use two common types of algorithms, a brute force algorithm and a greedy algorithm . A brute force algorithm begins by listing every possible solution and applying each one until the best solution is found. A greedy algorithm approaches a problem in stages, making the apparent best choice at each stage, then linking the choices together into an overall solution which may or may not be the best solution. To understand the difference between these two algorithms, consider the tree diagram in . Suppose we want to find the path from left to right with the largest total sum. For example, branch A in the tree diagram has a sum of 10 + 2 + 11 + 13 = 36 . Points Along Different Paths To be certain that you pick the branch with greatest sum, you could list each sum from each of the different branches: A : 10 + 2 + 11 + 13 = 36 B : 10 + 2 + 11 + 8 = 31 C : 10 + 2 + 15 + 1 = 28 D : 10 + 2 + 15 + 6 = 33 E : 10 + 7 + 3 + 20 = 40 F : 10 + 7 + 3 + 14 = 34 G : 10 + 7 + 4 + 11 = 32 H : 10 + 7 + 4 + 5 = 26 Then we know with certainty that branch E has the greatest sum. Branch E Now suppose that you wanted to find the branch with the highest value, but you only were shown the tree diagram in phases, one step at a time. Tree Diagram Phase 1 After phase 1, you would have chosen the branch with 10 and 7. So far, you are following the same branch. Let’s look at the next phase. Tree Diagram Phase 2 After phase 2, based on the information you have, you will choose the branch with 10, 7 and 4. Now, you are following a different branch than before, but it is the best choice based on the information you have. Let’s look at the last phase. Tree Diagram Phase 3 After phase 3, you will choose branch G which has a sum of 32. The process of adding the values on each branch and selecting the highest sum is an example of a brute force algorithm because all options were explored in detail. The process of choosing the branch in phases, based on the best choice at each phase is a greedy algorithm. Although a brute force algorithm gives us the ideal solution, it can take a very long time to implement. Imagine a tree diagram with thousands or even millions of branches. It might not be possible to check all the sums. A greedy algorithm, on the other hand, can be completed in a relatively short time, and generally leads to good solutions, but not necessarily the ideal solution. Distinguishing between Brute Force and Greedy Algorithms A cashier rings up a sale for $4.63 cents in U.S. currency. The customer pays with a $5 bill. The cashier would like to give the customer $0.37 in change using the fewest coins possible. The coins that can be used are quarters ($0.25), dimes ($0.10), nickels ($0.05), and pennies ($0.01). The cashier starts by selecting the coin of highest value less than or equal to $0.37, which is a quarter. This leaves $ 0.37 − $ 0.25 = $ 0.12 . The cashier selects the coin of highest value less than or equal to $0.12, which is a dime. This leaves $ 0.12 − $ 0.10 = $ 0.02 . The cashier selects the coin of highest value less than or equal to $0.02, which is a penny. This leaves $ 0.02 − $ 0.01 = $ 0.01 . The cashier selects the coin of highest value less than or equal to $0.01, which is a penny. This leaves no remainder. The cashier used one quarter, one dime, and two pennies, which is four coins. Use this information to answer the following questions. Is the cashier’s approach an example of a greedy algorithm or a brute force algorithm? Explain how you know. The cashier’s solution is the best solution. In other words, four is the fewest number of coins possible. Is this consistent with the results of an algorithm of this kind? Explain your reasoning. The approach the cashier used is an example of a greedy algorithm, because the problem was approached in phases and the best choice was made at each phase. Also, it is not a brute force algorithm, because the cashier did not attempt to list out all possible combinations of coins to reach this conclusion. Yes, it is consistent. A greedy algorithm does not always yield the best result, but sometimes it does. The Traveling Salesperson Problem Now let’s focus our attention on the graph theory application known as the traveling salesperson problem (TSP) in which we must find the shortest route to visit a number of locations and return to the starting point. Recall from Hamilton Cycles , the officer in the U.S. Air Force who is stationed at Vandenberg Air Force base and must drive to visit three other California Air Force bases before returning to Vandenberg. The officer needed to visit each base once. We looked at the weighted graph in representing the four U.S. Air Force bases: Vandenberg, Edwards, Los Angeles, and Beal and the distances between them. Graph of Four California Air Force Bases Any route that visits each base and returns to the start would be a Hamilton cycle on the graph. If the officer wants to travel the shortest distance, this will correspond to a Hamilton cycle of lowest weight. We saw in that there are six distinct Hamilton cycles (directed cycles) in a complete graph with four vertices, but some lie on the same cycle (undirected cycle) in the graph. Complete Graph Cycle Cycle Cycle Clockwise Hamilton Cycle a → b → c → d → a a → b → d → c → a a → c → b → d → a Counterclockwise Hamilton Cycle a → d → c → b → a a → c → d → b → a a → d → b → c → a Hamilton Cycles in a Complete Graph with Four Vertices Since the distance between bases is the same in either direction, it does not matter if the officer travels clockwise or counterclockwise. So, there are really only three possible distances as shown in . Three Possible Distances The possible distances are: 396 + 410 + 106 + 159 = 1071 207 + 410 + 439 + 159 = 1215 396 + 439 + 106 + 207 = 1148 So, a Hamilton cycle of least weight is V → B → E → L → V (or the reverse direction). The officer should travel from Vandenberg to Beal to Edwards, to Los Angeles, and back to Vandenberg. Finding Weights of All Hamilton Cycles in Complete Graphs Notice that we listed all of the Hamilton cycles and found their weights when we solved the TSP about the officer from Vandenberg. This is a skill you will need to practice. To make sure you don't miss any, you can calculate the number of possible Hamilton cycles in a complete graph. It is also helpful to know that half of the directed cycles in a complete graph are the same cycle in reverse direction, so, you only have to calculate half the number of possible weights, and the rest are duplicates. In a complete graph with n vertices, The number of distinct Hamilton cycles is ( n − 1 ) ! . There are at most ( n − 1 ) ! 2 different weights of Hamilton cycles. TIP! When listing all the distinct Hamilton cycles in a complete graph, you can start them all at any vertex you choose. Remember, the cycle a → b → c → a is the same cycle as b → c → a → b so there is no need to list both. Calculating Possible Weights of Hamilton Cycles Suppose you have a complete weighted graph with vertices N, M, O , and P . Use the formula ( n − 1 ) ! to calculate the number of distinct Hamilton cycles in the graph. Use the formula ( n − 1 ) ! 2 to calculate the greatest number of different weights possible for the Hamilton cycles. Are all of the distinct Hamilton cycles listed here? How do you know? Cycle 1: N → M → O → P → N Cycle 2: N → M → P → O → N Cycle 3: N → O → M → P → N Cycle 4: N → O → P → M → N Cycle 5: N → P → M → O → N Cycle 6: N → P → O → M → N Which pairs of cycles must have the same weights? How do you know? There are 4 vertices; so, n = 4 . This means there are ( n − 1 ) ! = ( 4 − 1 ) ! = 3 ⋅ 2 ⋅ 1 = 6 distinct Hamilton cycles beginning at any given vertex. Since n = 4 , there are ( n − 1 ) ! 2 = ( 4 − 1 ) ! 2 = 6 2 = 3 possible weights. Yes, they are all distinct cycles and there are 6 of them. Cycles 1 and 6 have the same weight, Cycles 2 and 4 have the same weight, and Cycles 3 and 5 have the same weight, because these pairs follow the same route through the graph but in reverse. TIP! When listing the possible cycles, ignore the vertex where the cycle begins and ends and focus on the ways to arrange the letters that represent the vertices in the middle. Using a systematic approach is best; for example, if you must arrange the letters M, O, and P, first list all those arrangements beginning with M, then beginning with O, and then beginning with P, as we did in Example 12.42. The Brute Force Method The method we have been using to find a Hamilton cycle of least weight in a complete graph is a brute force algorithm, so it is called the brute force method . The steps in the brute force method are: Step 1: Calculate the number of distinct Hamilton cycles and the number of possible weights. Step 2: List all possible Hamilton cycles. Step 3: Find the weight of each cycle. Step 4: Identify the Hamilton cycle of lowest weight. Applying the Brute Force Method On the next assignment, the air force officer must leave from Travis Air Force base, visit Beal, Edwards, and Vandenberg Air Force bases each exactly once and return to Travis Air Force base. There is no need to visit Los Angeles Air Force base. Use to find the shortest route. Distances between Five California Air Force Bases Step 1: Since there are 4 vertices, there will be ( 4 − 1 ) ! = 3 ! = 6 cycles, but half of them will be the reverse of the others; so, there will be ( 4 − 1 ) ! 2 = 6 2 = 3 possible distances. Step 2: List all the Hamilton cycles in the subgraph of the graph in . Subgraph with Cities B, E, T , and V To find the 6 cycles, focus on the three vertices in the middle, B, E, and V . The arrangements of these vertices are BEV, BVE, EBV, EVB, VBE , and VEB . These would correspond to the 6 cycles: 1: T → B → E → V → T 2: T → B → V → E → T 3: T → E → B → V → T 4: T → E → V → B → T 5: T → V → B → E → T 6: T → V → E → B → T Step 3: Find the weight of each path. You can reduce your work by observing the cycles that are reverses of each other. 1: 84 + 410 + 207 + 396 = 1097 2: 84 + 396 + 207 + 370 = 1071 3: 370 + 410 + 396 + 396 = 1572 4: Reverse of cycle 2, 1071 5: Reverse of cycle 3, 1572 6: Reverse of cycle 1, 1097 Step 4: Identify a Hamilton cycle of least weight. The second path, T → B → V → E → T , and its reverse, T → E → V → B → T , have the least weight. The solution is that the officer should travel from Travis Air Force base to Beal Air Force Base, to Vandenberg Air Force base, to Edwards Air Force base, and return to Travis Air Force base, or the same route in reverse. Now suppose that the officer needed a cycle that visited all 5 of the Air Force bases in . There would be ( 5 − 1 ) ! = 4 ! = 4 × 3 × 2 × 1 = 24 different arrangements of vertices and ( 5 − 1 ) ! 2 = 4 ! 2 = 24 2 = 12 distances to compare using the brute force method. If you consider 10 Air Force bases, there would be ( 10 − 1 ) ! = 9 ! = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 362 , 880 different arrangements and ( 10 − 1 ) ! 2 = 9 ! 2 = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 2 = 181 , 440 distances to consider. There must be another way! The Nearest Neighbor Method When the brute force method is impractical for solving a traveling salesperson problem, an alternative is a greedy algorithm known as the nearest neighbor method , which always visit the closest or least costly place first. This method finds a Hamilton cycle of relatively low weight in a complete graph in which, at each phase, the next vertex is chosen by comparing the edges between the current vertex and the remaining vertices to find the lowest weight. Since the nearest neighbor method is a greedy algorithm, it usually doesn’t give the best solution, but it usually gives a solution that is "good enough." Most importantly, the number of steps will be the number of vertices. That’s right! A problem with 10 vertices requires 10 steps, not 362,880. Let’s look at an example to see how it works. Suppose that a candidate for governor wants to hold rallies around the state. They plan to leave their home in city A , visit cities B, C, D, E , and F each once, and return home. The airfare between cities is indicated in the graph in . Airfares between Cities A, B, C, D, E , and F Let’s help the candidate keep costs of travel down by applying the nearest neighbor method to find a Hamilton cycle that has a reasonably low weight. Begin by marking starting vertex as V 1 for "visited 1st." Then to compare the weights of the edges between A and vertices adjacent to A : $250, $210, $300, $200, and $100 as shown in . The lowest of these is $100, which is the edge between A and F . Finding the Second Vertex Mark F as V 2 for "visited 2nd" then compare the weights of the edges between F and the remaining vertices adjacent to F : $170, $330, $150 and $350 as shown in . The lowest of these is $150, which is the edge between F and D . Finding the Third Vertex Mark D as V 3 for "visited 3rd." Next, compare the weights of the edges between D and the remaining vertices adjacent to D : $120, $310, and $270 as shown in . The lowest of these is $120, which is the edge between D and B . Finding the Fourth Vertex So, mark B as V 4 for "visited 4th." Finally, compare the weights of the edges between B and the remaining vertices adjacent to B : $160 and $220 as shown in . The lower amount is $160, which is the edge between B and E . Finding the Fifth Vertex Now you can mark E as V 5 and mark the only remaining vertex, which is C , as V 6 . This is shown in . Make a note of the weight of the edge from E to C , which is $180, and from C back to A , which is $210. Finding the Sixth Vertex The Hamilton cycle we found is A → F → D → B → E → C → A . The weight of the circuit is $ 100 + $ 150 + $ 120 + $ 160 + $ 180 + $ 210 = $ 920 . This may or may not be the route with the lowest cost, but there is a good chance it is very close since the weights are most of the lowest weights on the graph and we found it in six steps instead of finding 120 different Hamilton cycles and calculating 60 weights. Let’s summarize the procedure that we used. Step 1: Select the starting vertex and label V 1 for "visited 1st." Identify the edge of lowest weight between V 1 and the remaining vertices. Step 2: Label the vertex at the end of the edge of lowest weight that you found in previous step as V n where the subscript n indicates the order the vertex is visited. Identify the edge of lowest weight between V n and the vertices that remain to be visited. Step 3: If vertices remain that have not been visited, repeat Step 2. Otherwise, a Hamilton cycle of low weight is V 1 → V 2 → ⋯ → V n → V 1 . Using the Nearest Neighbor Method Suppose that the candidate for governor wants to hold rallies around the state but time before the election is very limited. They would like to leave their home in city A , visit cities B , C , D , E , and F each once, and return home. The airfare between cities is not as important as the time of travel, which is indicated in . Use the nearest neighbor method to find a route with relatively low travel time. What is the total travel time of the route that you found? Travel Times between Cities A, B, C, D, E and F Step 1: Label vertex A as V 1 . The edge of lowest weight between A and the remaining vertices is 85 min between A and D . Step 2: Label vertex D as V 2 . The edge of lowest weight between D and the vertices that remain to be visited, B, C, E , and F , is 70 min between D and F . Repeat Step 2: Label vertex F as V 3 . The edge of lowest weight between F and the vertices that remain to be visited, B, C, and E , is 75 min between F and C . Repeat Step 2: Label vertex C as V 4 . The edge of lowest weight between C and the vertices that remain to be visited, B and E , is 100 min between C and B . Repeat Step 2: Label vertex B as V 5 . The only vertex that remains to be visited is E . The weight of the edge between B and E is 95 min. Step 3: A Hamilton cycle of low weight is A → D → F → C → B → E → A . So, a route of relatively low travel time is A to D to F to C to B to E and back to A . The total travel time of this route is: 85 min + 70 min + 75 min + 100 min + 95 min + 90 min = 515 min or 8 hrs 35 min Check Your Understanding Key Terms brute force algorithm greedy algorithm traveling salesperson problem (TSP) brute force method nearest neighbor method Key Concepts A brute force algorithm always finds the ideal solution but can be impractical whereas a greedy algorithm is efficient but usually does not lead to the ideal solution. A Hamilton cycle of lowest weight is a solution to the traveling salesperson problem. The brute force method finds a Hamilton cycle of lowest weight in a complete graph. The nearest neighbor method is a greedy algorithm that finds a Hamilton cycle of relatively low weight in a complete graph. Formulas In a complete graph with n vertices, the number of distinct Hamilton cycles is ( n − 1 ) ! . In a complete graph with n vertices, there are at most ( n − 1 ) ! 2 different weights of Hamilton cycles.

Math


Trees In graph theory, graphs known as trees have structures in common with live trees. (credit: “Row of trees in Roslev” by AKA CJ/Flickr, Public Domain) Learning Objectives After completing this section, you should be able to: Describe and identify trees. Determine a spanning tree for a connected graph. Find the minimum spanning tree for a weighted graph. Solve application problems involving trees. We saved the best for last! In this last section, we will discuss arguably the most fun kinds of graphs, trees. Have you every researched your family tree? Family trees are a perfect example of the kind of trees we study in graph theory. One of the characteristics of a family tree graph is that it never loops back around, because no one is their own grandparent! What Is A Tree? Whether we are talking about a family tree or a tree in a forest, none of the branches ever loops back around and rejoins the trunk. This means that a tree has no cyclic subgraphs, or is acyclic . A tree also has only one component. So, a tree is a connected acyclic graph. Here are some graphs that have the same characteristic. Each of the graphs in is a tree. Graphs T , P , and S Let’s practice determining whether a graph is a tree. To do this, check if a graph is connected and has no cycles. Identifying Trees Identify any trees in . If a graph is not a tree, explain how you know. Graphs M , N , and P Graph M is not a tree because it contains the cycle ( b , c , f ). Graph N is not a tree because it is not connected. It has two components, one with vertices h, i, j , and another with vertices k, l, m . Graph P is a tree. It has no cycles and it is connected. Types of Trees Mathematicians have had a lot of fun naming graphs that are trees or that contain trees. For example, the graph in is not a tree, but it contains two components, one containing vertices a through d , and the other containing vertices e through g , each of which would be a tree on its own. This type of structure is called a forest . There are also interesting names for trees with certain characteristics. A path graph or linear graph is a tree graph that has exactly two vertices of degree 1 such that the only other vertices form a single path between them, which means that it can be drawn as a straight line. A star tree is a tree that has exactly one vertex of degree greater than 1 called a root, and all other vertices are adjacent to it. A starlike tree is a tree that has a single root and several paths attached to it. A caterpillar tree is a tree that has a central path that can have vertices of any degree, with each vertex not on the central path being adjacent to a vertex on the central path and having a degree of one. A lobster tree is a tree that has a central path that can have vertices of any degree, with paths consisting of either one or two edges attached to the central path. Examples of each of these types of structures are given in . Forest Graph F Six Types of Trees Identifying Types of Trees Each graph in is one of the special types of trees we have been discussing. Identify the type of tree. Graphs U and V Graph U has a central path a → b → d → f → i → l → o → q . Each vertex that is not on the path has degree 1 and is adjacent to a vertex that is on the path. So, U is a caterpillar tree. Graph V is a path graph because it is a single path connecting exactly two vertices of degree one, r → s → u → v → w . Characteristics of Trees As we study trees, it is helpful to be familiar with some of their characteristics. For example, if you add an edge to a tree graph between any two existing vertices, you will create a cycle, and the resulting graph is no longer a tree. Some examples are shown in . Adding edge bj to Graph T creates cycle ( b , c , i , j ). Adding edge rt to Graph P creates cycle ( r , s , t ). Adding edge tv to Graph S creates cycle ( t , u , v ). Adding Edges to Trees It is also true that removing an edge from a tree graph will increase the number of components and the graph will no longer be connected. In fact, you can see in that removing one or more edges can create a forest. Removing edge qr from Graph P creates a graph with two components, one with vertices o, p and q , and the other with vertices r, s , and t . Removing edge uw from Graph S creates two components, one with just vertex w and the other with the rest of the vertices. When two edges were removed from Graph T , edge bf and edge cd , creates a graph with three components as shown in . Removing Edges from Trees A very useful characteristic of tree graphs is that the number of edges is always one less than the number of vertices. In fact, any connected graph in which the number of edges is one less than the number of vertices is guaranteed to be a tree. Some examples are given in . Number of Vertices and Edges in Trees vs. Other Graphs The number of edges in a tree graph with n vertices is n − 1 . A connected graph with n vertices and n − 1 edges is a tree graph. Exploring Characteristics of Trees Use Graphs I and J in to answer each question. Graphs I and J Which vertices are in each of the components that remain when edge be is removed from Graph I ? Determine the number of edges and the number of vertices in Graph J . Explain how this confirms that Graph J is a tree. What kind of cycle is created if edge im is added to Graph J ? When edge be is removed, there are two components that remain. One component includes vertices a, b , and c . The other component includes vertices d, e , and f . There are seven vertices and six edges in Graph J . This confirms that Graph J is a tree because the number of edges is one less than the number of vertices. The pentagon ( i , h , j , l , m ) is created when edge im is added to Graph J . Graph Theory in the Movies In the 1997 film Good Will Hunting , the main character, Will, played by Matt Damon, solves what is supposed to be an exceptionally difficult graph theory problem, “Draw all the homeomorphically irreducible trees of size n = 10 .” That sounds terrifying! But don’t panic. Watch this great Numberphile video to see why this is actually a problem you can do at home! The Problem in Good Will Hunting by Numberphile Spanning Trees Suppose that you planned to set up your own computer network with four devices. One option is to use a “mesh topology” like the one in , in which each device is connected directly to every other device in the network. Common Network Configurations The mesh topology for four devices could be represented by the complete Graph A 1 in where the vertices represent the devices, and the edges represent network connections. However, the devices could be networked using fewer connections. Graphs A 2 , A 3, and A 4 of show configurations in which three of the six edges have been removed. Each of the Graphs A 2 , A 3 and A 4 in is a tree because it is connected and contains no cycles. Since Graphs A 2 , A 3 and A 4 are also subgraphs of Graph A 1 that include every vertex of the original graph, they are also known as spanning trees . Network Configurations for Four Devices By definition, spanning trees must span the whole graph by visiting all the vertices. Since spanning trees are subgraphs, they may only have edges between vertices that were adjacent in the original graph. Since spanning trees are trees, they are connected and they are acyclic. So, when deciding whether a graph is a spanning tree, check the following characteristics: All vertices are included. No vertices are adjacent that were not adjacent in the original graph. The graph is connected. There are no cycles. Identifying Spanning Trees Use to determine which of graphs M 1 , M 2 , M 3 , and M 4 , are spanning trees of Q . Graphs Q , M 1 , M 2 , M 3, and M 4 Graph M 1 is not a spanning tree of Graph Q because it has a cycle ( c , d , f , e ). Graph M 2 is a spanning tree of Graph Q because it has all the original vertices, no vertices are adjacent in M 2 that weren’t adjacent in Graph Q , Graph M 2 is connected and it contains no cycles. Graph M 3 is not a spanning tree of Graph Q because vertices a and f are adjacent in Graph M 3 but not in Graph Q . Graph M 4 is not a spanning tree of Graph Q because it is not connected. So, only graph M 2 is a spanning tree of Graph Q . Constructing a Spanning Tree Using Paths Suppose that you wanted to find a spanning tree within a graph. One approach is to find paths within the graph. You can start at any vertex, go any direction, and create a path through the graph stopping only when you can’t continue without backtracking as shown in . First Phase to Construct a Spanning Tree Once you have stopped, pick a vertex along the path you drew as a starting point for another path. Make sure to visit only vertices you have not visited before as shown in . Intermediate Phase to Construct a Spanning Tree Repeat this process until all vertices have been visited as shown in . Final Phase to Construct a Spanning Tree The end result is a tree that spans the entire graph as shown in . The Resulting Spanning Tree Notice that this subgraph is a tree because it is connected and acyclic. It also visits every vertex of the original graph, so it is a spanning tree. However, it is not the only spanning tree for this graph. By making different turns, we could create any number of distinct spanning trees. Constructing Spanning Trees Construct two distinct spanning trees for the graph in . Graph L Two possible solutions are given in and . First Spanning Tree for Graph L Second Spanning Tree for Graph L Revealing Spanning Trees Another approach to finding a spanning tree in a connected graph involves removing unwanted edges to reveal a spanning tree. Consider Graph D in . Graph D Graph D has 10 vertices. A spanning tree of Graph D must have 9 edges, because the number of edges is one less than the number of vertices in any tree. Graph D has 13 edges so 4 need to be removed. To determine which 4 edges to remove, remember that trees do not have cycles. There are four triangles in Graph D that we need to break up. We can accomplish this by removing 1 edge from each of the triangles. There are many ways this can be done. Two of these ways are shown in . Removing Four Edges from Graph D Spanning Trees in Graph Theory Removing Edges to Find Spanning Trees Use the graph in to answer each question. Graph V Determine the number of edges that must be removed to reveal a spanning tree. Name all the undirected cycles in Graph V . Find two distinct spanning trees of Graph V . Graph V has nine vertices so a spanning tree for the graph must have 8 edges. Since Graph V has 11 edges, 3 edges must be removed to reveal a spanning tree. ( a , c , d ), ( a , c , f ), ( a , d , c , f ), and ( b , e , h , i , g ) To find the first spanning tree, remove edge ac , which will break up both of the triangles, remove edge cf , which will break up the quadrilateral, and remove be , which will break up the pentagon, to give us the spanning tree shown in . Spanning Tree Formed Removing ac, cf , and be To find another spanning tree, remove ad , which will break up ( a , c , d ) and ( a , d , c , f ), remove af to break up ( a , c , f ), and remove hi to break up ( b , e , h , i , g ). This will give us the spanning tree in . Spanning Tree Formed Removing ad , af , and hi Chains of Affection Here is a strange question to ask in a math class: Have you ever dated your ex’s new partner’s ex? Research suggests that your answer is probably no. When researchers Peter S. Bearman, James Moody, and Katherine Stovel attempted to compare the structure of heterosexual romantic networks at a typical midwestern high school to simulated networks, they found something surprising. The actual social networks were more like spanning trees than other possible models because there were very few short cycles. In particular, there were almost no four-cycles. Chains of Affection “…the prohibition against dating (from a female perspective) one’s old boyfriend’s current girlfriend’s old boyfriend – accounts for the structure of the romantic network at [the highschool].” In their article “Chains of Affection: The Structure of Adolescent Romantic and Sexual Networks,” the researchers went on to explain the implications for the transmission of sexually transmitted diseases. In particular, social structures based on tree graphs are less dense and more likely to fragment. This information can impact social policies on disease prevention. (Peter S. Bearman, James Moody, and Katherine Stovel, “Chains of Affection: The Structure of Adolescent Romantic and Sexual Networks,” American Journal of Sociology Volume 110, Number 1, pp. 44-91, 2004) Kruskal’s Algorithm In many applications of spanning trees, the graphs are weighted and we want to find the spanning tree of least possible weight. For example, the graph might represent a computer network, and the weights might represent the cost involved in connecting two devices. So, finding a spanning tree with the lowest possible total weight, or minimum spanning tree , means saving money! The method that we will use to find a minimum spanning tree of a weighted graph is called Kruskal’s algorithm . The steps for Kruskal’s algorithm are: Step 1: Choose any edge with the minimum weight of all edges. Step 2: Choose another edge of minimum weight from the remaining edges. The second edge does not have to be connected to the first edge. Step 3: Choose another edge of minimum weight from the remaining edges, but do not select any edge that creates a cycle in the subgraph you are creating. Step 4: Repeat step 3 until all the vertices of the original graph are included and you have a spanning tree. Use Kruskal's Algorithm to Find Minimum Spanning Trees in Graph Theory Using Kruskal’s Algorithm A computer network will be set up with six devices. The vertices in the graph in represent the devices, and the edges represent the cost of a connection. Find the network configuration that will cost the least. What is the total cost? Graph of Network Connection Costs A minimum spanning tree will correspond to the network configuration of least cost. We will use Kruskal’s algorithm to find one. Since the graph has six vertices, the spanning tree will have six vertices and five edges. Step 1: Choose an edge of least weight. We have sorted the weights into numerical order. The least is $100. The only edge of this weight is edge AF as shown in . Step 1 Select Edge AF Step 2: Choose the edge of least weight of the remaining edges, which is BD with $120. Notice that the two selected edges do not need to be adjacent to each other as shown in . Step 2 Select Edge BD Step 3: Select the lowest weight edge of the remaining edges, as long as it does not result in a cycle. We select DF with $150 since it does not form a cycle as shown in . Step 3 Select Edge DF Repeat Step 3: Select the lowest weight edge of the remaining edges, which is BE with $160 and it does not form a cycle as shown in . This gives us four edges so we only need to repeat step 3 once more to get the fifth edge. Repeat Step 3 Select Edge DF Repeat Step 3: The lowest weight of the remaining edges is $170. Both BF and CE have a weight of $170, but BF would create cycle ( b , d , f ) and there cannot be a cycle in a spanning tree as shown in . Repeat Step 3 Do Not Select Edge BF So, we will select CE , which will complete the spanning tree as shown in . Repeat Step 3 Select Edge CE The minimum spanning tree is shown in . This is the configuration of the network of least cost. The spanning tree has a total weight of $ 100 + $ 120 + $ 150 + $ 160 + $ 170 = $ 700 , which is the total cost of this network configuration. Final Minimum Spanning Tree Check Your Understanding Key Terms acyclic tree forest path graph or linear graph star tree root starlike tree caterpillar tree lobster tree spanning tree minimum spanning tree Key Concepts A brute force algorithm always finds the ideal solution but can be impractical whereas a greedy algorithm is efficient but usually does not lead to the ideal solution. A Hamilton cycle of lowest weight is a solution to the traveling salesperson problem. The brute force method finds a Hamilton cycle of lowest weight in a complete graph. The nearest neighbor method is a greedy algorithm that finds a Hamilton cycle of relatively low weight in a complete graph. Videos The Problem in Good Will Hunting by Numberphile Spanning Trees in Graph Theory Use Kruskal's Algorithm to Find Minimum Spanning Trees in Graph Theory Formulas The number of edges in a tree graph with n vertices is n − 1 . A connected graph with n vertices and n − 1 edges is a tree graph. Projects Everyone Gets a Turn! – Graph Colorings Let’s put your knowledge of graph colorings to work! Your task is to plan a field day following these steps. Select between seven and ten activities for your field day. You can look online for ideas. Create a survey asking for the participants to select the three to five events in which they would most like to participate. Survey between seven and ten people. Use the results of your survey to create a graph in which each vertex represents one of the events. A pair of vertices will be adjacent if there is at least one participant who would like to participate in both events. Find a minimum coloring for the graph. Explain how you found it and how you know the chromatic number of the graph. Use your solution to part d to determine the minimum number of timeslots you must use to ensure that everyone has the opportunity to participate in their top three events. Find the complement of the graph you created. Explain what the edges in this graph represent. A Beautiful Day in the Neighborhood – Euler Circuits Let’s apply what you have learned to the community in which you live. Using resources such as your county’s property appraiser’s website, create a detailed graph of your neighborhood in which vertices represent turns and intersections. Represent a large enough part of your community to include no fewer than 10 intersections or turns. Then use your graph to answer the following questions. Label the edges of your graph. Determine if your graph is Eulerian. Explain how you know. If it is not, eulerize it. Find an Euler circuit for your graph. Give the sequence of vertices that you found. What does the Euler circuit you found in part c represent for your community? Describe an application for which this Euler circuit might be used. Dream Vacation – Hamilton Cycles and Paths Where in the world would you like to travel most: the Eiffel Tower in Paris, a Broadway musical in New York city, a bike tour of Amsterdam, the Tenerife whale and dolphin cruises in the Canary Islands, the Giza Pyramid in Cairo, or maybe the Jokhang Temple in Tibet? Let's plan your dream vacation! Which four destinations are at the top of your bucket list? Draw a complete weighted graph with five vertices representing the four destinations and your home city, and the weights representing the cost of travel between cities. Use a website (such as Travelocity ) to find the best airfare between each pair of cities. List the airlines and flight numbers along with the prices. Include cost for ground transportation from the nearest airport if there is no airport at the destination you want to visit. Use the nearest neighbor algorithm to find a Hamilton cycle of low weight beginning and ending in your hometown. What is the weight of this circuit and what does it represent? Use the brute force method to find a Hamilton cycle of lowest weight beginning and ending in your hometown. What is the weight of this circuit? Is it the same or different from the weight of the Hamilton cycle you found in Exercise 4? Suppose that instead of returning home, you planned to move to your favorite location on the list, but you wanted to stop at the other three destinations once along the way. Where would you move? List all Hamilton paths between your hometown and your favorite location. Find the weights of all the Hamilton paths you found in Exercise 6. Chapter Review Graph Basics Graph Structures Comparing Graphs Navigating Graphs Euler Circuits Euler Trails Hamilton Cycles Hamilton Paths Traveling Salesperson Problem Trees Chapter Test

Math


Introduction Math can be found in many different areas and subjects. (credit: modification of work "Chemistry/Physics Library" by University Libraries/Flickr, CC BY 2.0) Where do we find math around us? Math can be found in areas that are expected and sometimes in areas that are surprising. There are many ways that mathematical concepts, such as those in this text, are infused in the world around us. In this chapter, we will explore a sampling of five distinct areas from everyday life where math’s impact plays a meaningful role. Math's impact on art can be found in numerical relationships that are known to create or enhance beauty. The Fibonacci numbers are one mathematical example that can be found in nature such as in petal count of a rose. On a different note, a mathematical exploration can aid in making a convincing argument on how we can positively impact our environment. Whether looking at the choices of a single individual or the larger impact offered from a collaborative effort, there are measurable responses to positively address climate change. Turning to medicine, which has been a topic of global importance in recent years, we will explore how math is used to determine drug dosage rates and test the validity of a vaccine. Switching back to items of aesthetic nature, we will examine some foundational components of music which, like art, brings beauty and joy to our lives. Finally, we will explore some ways that math is used in sports to predict future performance and analyze tournaments styles.

Math


Math and Art Sunflower seeds appear in a pattern that involves the Fibonacci sequence. (credit: “Sunflower Surprise” by frankieleon/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify and describe the golden ratio. Identify and describe the Fibonacci sequence and its application to nature. Apply the golden ratio and the Fibonacci sequence relationship. Identify and compute golden rectangles. Art is the expression or application of human creative skill and imagination, typically in a visual form such as painting or sculpture, producing works to be appreciated primarily for their beauty or emotional power. Oxford Dictionary Art, like other disciplines, is an area that combines talent and experience with education. While not everyone considers themself skilled at creating art, there are mathematical relationships commonly found in artistic masterpieces that drive what is considered attractive to the eye. Nature is full of examples of these mathematical relationships. Enroll in a cake decorating class and, when you learn how to create flowers out of icing, you will likely be directed as to the number of petals to use. Depending on the desired size of a rose flower, the recommendation for the number of petals to use is commonly 5, 8, or 13 petals. If learning to draw portraits, you may be surprised to learn that eyes are approximately halfway between the top of a person’s head and their chin. Studying architecture, we find examples of buildings that contain golden rectangles and ratios that add to the beautifying of the design. The Parthenon ( ), which was built around 400 BC, as well as modern-day structures such the Washington Monument are two examples containing these relationships. These seemingly unrelated examples and many more highlight mathematical relationships that we associate with beauty in artistic form. The Parthenon in Greece demonstrates the golden ratio. (credit: “Parthénon” by Julien Maury/Flickr, Public Domain Mark 1.0) Golden Ratio The golden ratio , also known as the golden proportion, is a ratio aspect that can be found in beauty from nature to human anatomy as well as in golden rectangles that are commonly found in building structures. The golden ratio is expressed in nature from plants to creatures such as the starfish, honeybees, seashells, and more. It is commonly noted by the Greek letter ϕ (pronounced “fee”). ϕ = 1 + 5 2 , which has a decimal value approximately equal to 1.618. Consider : Note how the building is balanced in dimension and has a natural shape. The overall structure does not appear as if it is too wide or too tall in comparison to the other dimensions. Vitruvian Man by Leonardo da Vinci (credit: "Vitruvian Man" by Leonardo da Vinci/Wikimedia Commons, Public Domain) The golden ratio has been used by artists through the years and can be found in art dating back to 3000 BC. Leonardo da Vinci is considered one of the artists who mastered the mathematics of the golden ratio, which is prevalent in his artwork such as Virtuvian Man ( ). This famous masterpiece highlights the golden ratio in the proportions of an ideal body shape. The golden ratio is approximated in several physical measurements of the human body and parts exhibiting the golden ratio are simply called golden. The ratio of a person’s height to the length from their belly button to the floor is ϕ or approximately 1.618. The bones in our fingers (excluding the thumb), are golden as they form a ratio that approximates ϕ . The human face also includes several ratios and those faces that are considered attractive commonly exhibit golden ratios. Using Golden Ratio and a Person’s Height If a person’s height is 5 ft 6 in, what is the approximate length from their belly button to the floor rounded to the nearest inch, assuming the ratio is golden? Step 1: Convert the height to inches 5 ft 6 in = 66 in Step 2: Calculate the length from the belly button to the floor, L . 66 / L = 1.618 L = 40.8 in The length from the person’s belly button to the floor would be approximately 41 in. Fibonacci Sequence and Application to Nature Rose petals appear in a Fibonacci spiral. (credit: “rilke4” by monchoohcnom/Flickr, Public Domain Mark 1.0) The Fibonacci sequence can be found occurring naturally in a wide array of elements in our environment from the number of petals on a rose flower to the spirals on a pine cone to the spines on a head of lettuce and more. The Fibonacci sequence can be found in artistic renderings of nature to develop aesthetically pleasing and realistic artistic creations such as in sculptures, paintings, landscape, building design, and more. It is the sequence of numbers beginning with 1, 1, and each subsequent term is the sum of the previous two terms in the sequence (1, 1, 2, 3, 5, 8, 13, …). The petal counts on some flowers are represented in the Fibonacci sequence. A daisy is sometimes associated with plucking petals to answer the question “They love me, they love me not.” Interestingly, a daisy found growing wild typically contains 13, 21, or 34 petals and it is noted that these numbers are part of the Fibonacci sequence. The number of petals aligns with the spirals in the flower family. Applying the Fibonacci Sequence to Rose Petals Suppose you were creating a rose out of icing, assuming a Fibonacci sequence in the petals, how many petals would be in the row following a row containing 13 petals? The number of petals on a rose is often modeled with the numbers in the Fibonacci sequence, which is 1, 1, 2, 3, 5, 8, 13,…, where the next number in the sequence is the sum of 8 + 13 = 21 . There would be 21 petals on the next row of the icing rose. Golden Ratio and the Fibonacci Sequence Relationship Mathematicians for years have explored patterns and applications to the world around us and continue to do so today. One such pattern can be found in ratios of two adjacent terms of the Fibonacci sequence. Recall that the Fibonacci sequence = 1, 1, 3, 5, 8, 13,… with 5 and 8 being one example of adjacent terms. When computing the ratio of the larger number to the preceding number such as 8/5 or 13/8, it is fascinating to find the golden ratio emerge. As larger numbers from the Fibonacci sequence are utilized in the ratio, the value more closely approaches ϕ , the golden ratio. Finding Golden Ratio in Adjacent Fibonacci Terms The 24th Fibonacci number is 46,368 and the 25th is 75,025. Show that the ratio of the 25th and 24th Fibonacci numbers is approximately ϕ . Round your answer to the nearest thousandth. 75,025 / 46,368 = 1.618 ; The ratio of the 25th and 24th term is approximately equal to the value of ϕ rounded to the nearest thousandth, 1.618. The pyramids of Giza in Egypt (credit: “Giza Pyramids” by Vincent Brown/Flickr, CC BY 2.0) Golden Rectangles Turning our attention to man-made elements, the golden ratio can be found in architecture and artwork dating back to the ancient pyramids in Egypt ( ) to modern-day buildings such as the UN headquarters. The ancient Greeks used golden rectangles —any rectangles where the ratio of the length to the width is the golden ratio—to create aesthetically pleasing as well as solid structures, with examples of the golden rectangle often being used multiple times in the same building such as the Parthenon, which is shown in . Golden rectangles can be found in twentieth-century buildings as well, such as the Washington Monument. Looking at another man-made element, artists paintings often contain golden rectangles. Well-known paintings such as Leonardo da Vinci’s The Last Supper and the Vitruvian Man contain multiple golden rectangles as do many of da Vinci’s masterpieces. Whether framing a painting or designing a building, the golden rectangle has been widely utilized by artists and are considered to be the most visually pleasing rectangles. Finding Golden Rectangle in Frames A frame has dimensions of 8 in by 6 in. Calculate the ratio of the sides rounded to the nearest thousandth and determine if the size approximates a golden rectangle. 8/6 = 1.333; A golden rectangle’s ratio is approximately 1.618. The frame dimensions are close to a golden rectangle. M.C. Escher M.C. Escher (credit: "M.C. Escher" by Hans Peters (ANEFO)/Dutch National Archives, CC0 1.0 Public Domain) Mauritis Cornelis Escher was a Dutch-born world-famous graphic artist and his work can be found in murals, stamps, wallpaper designs, illustrations in books, and even carpets. Over his lifetime, M.C. Escher created hundreds of lithographs and wood engravings as well as more than 2,000 sketches. Escher’s work is characterized with the infusion of geometric designs that obey most of the mathematical rules. If you study his work closely, you can see where he breaks a mathematical relationship to create famous illusions such as soldiers marching around the top of a square turret where the soldiers appear to be always going uphill but are contained on a single set of stairs in a square. Look closely and the golden ratio as well as golden rectangles abound in Escher’s work. Like many famous people, M.C. Escher did not find success in his early school years. Before finding success, Escher failed his final school exam and quit a short stint in architecture. Finding a graphic arts teacher who recognized Escher’s talent, Escher completed art school and enjoyed traveling through Italy, where he found much of his inspiration for his work. Check Your Understanding Key Terms golden ratio ϕ Fibonacci sequence golden rectangle Key Concepts The golden ratio, ϕ , can be found in nature, and the relationship is often associated with beauty and balance. The Fibonacci sequence reflects a pattern of numbers that can be found in various places in nature. The sequence can be used to predict other values that follow the Fibonacci pattern. State some naturally occurring applications of the Fibonacci sequence. State some naturally occurring applications of the golden ratio. Determine if a rectangle is golden. State some artistic applications of the golden rectangle.

Math


Math and the Environment Solar panels harness the sun's energy to power homes, businesses, and various methods of transportation. (credit: modification of work “Craters of the Moon solar array” by NPS Climate Change Response/Flickr, Public Domain Mark 1.0) Learning Objectives After completing this section, you should be able to: Compute how conserving water can positively impact climate change. Discuss the history of solar energy. Compute power needs for common devices in a home. Explore advantages of solar power as it applies to home use. Climate change and emissions management have been debated topics in recent years. However, more and more people are recognizing the impacts that have resulted in temperature changes and are seeking timely and effective action. The World Meteorological Organization shared in a June 2021 publication that “2021 is a make-or-break year for climate action, with the window to prevent the worst impacts of climate change—which include ever more frequent more intense droughts, floods and storms—closing rapidly.” The problem no longer belongs to a few countries or regions but rather is a worldwide concern measured with increasing temperatures leading to decreased glacier coverage and resulting rise in sea levels. The good news is, there are small steps that each of us can do that collectively can positively impact climate change. Making a Positive Impact on Climate Change—Water Usage Our use of water is one element that impacts climate change. Having access to clean, potable water is critical for not only our health but also for the health of our ecosystem. About 1 out of 10 people on our planet do not have easy access to clean water to drink. As each of us conserves water, we prolong the life span of fresh water from our lakes and rivers and also reduce the impact on sewer systems and drainage in our communities. Additionally, as we conserve water, we also conserve electricity that is used to bring water to and in our homes. So, what can we do to help conserve water? Brushing Your Teeth (One Person’s Contribution) Brushing your teeth with the water running continually uses about 4 gal of water. Turning the faucet off when you are not rinsing uses less than one-fourth of a gallon of water. Considering the recommendation to brush your teeth twice a day, how much water would be saved in a week if the faucet was off when not rinsing? Leaving the water running continually: Step 1: Calculate gallons used not with water running continually: Brushing twice a day for 7 days using 4 gal of water for each brushing ( 2 times a day ) ( 7 days ) ( 4 gal ) = 56 gal Step 2: Calculate gallons used turning the faucet off when you are not rinsing: Brushing twice a day for 7 days using 0.25 gal of water for each brushing ( 2 times a day ) ( 7 days ) ( 0.25 gal ) = 3.5 gal Step 3: Calculate savings: Savings = 56 gal − 3.5 gal = 52.5 gal During one week, 52.5 gal of water would be saved if one person turned the faucet off except when rinsing when brushing your teeth. Brushing Your Teeth (Multiple People’s Contribution – Town) Using the data in , how much water would be saved in a month if one-fifth of a town’s population of 15,000 turned the faucet off when brushing their teeth except when rinsing? Step 1: From , we found that 1 person saves 52.5 gal per week. Step 2: Calculate the population to save water: One-fifth of 15,000 people = 3,000 people Step 3: One-fifth of a town’s population turning off the faucet when brushing their teeth for a month: ( 3,000 ) ( 52.5 gal per week ) ( 4 weeks ) = 630,000 gal During one month, 630,000 gallons of water would be saved if one-fifth of a town of 15,000 people turned the faucet off except when rinsing when brushing their teeth. Brushing Your Teeth (Multiple People’s Contribution – State) Using the data in , how much water would be saved in a year if one-fourth of the population of the state of Minnesota, which is approximately 5.6 million people, turned the faucet off when brushing their teeth except when rinsing for a year (52 weeks)? Step 1: From , we found that 1 person saves 52.5 gal per week. Step 2: Calculate the population to save water: One-fourth of 5.6 million people = 1.4 million people Step 3: One-fourth of a town’s population turning off the faucet when brushing their teeth for a month: ( 1.4 million ) ( 52.5 gal per week ) ( 52 weeks ) = 3,822 million gal During one year, 3,822 million gal of water would be saved if one-fourth of the state of Minnesota turned the faucet off except when rinsing when brushing their teeth. History of Solar Energy In the mid-1800s, Willoughby Smith discovered photoconductive responsiveness in selenium. Shortly thereafter, William Grylls Adams and Richard Evans Day discovery that selenium can produce electricity if exposed to the sun was a major breakthrough. Less than 10 years later, Charles Fritts invented the first solar cells using selenium. Jumping a mere 100 years later, Bell Labs in the United States produced the first practical photovoltaic cells in the mid-1950s and developed versions used to power satellites in the same decade. Solar panel use has exploded in recent decades and is now used by residences, organizations, businesses, and government buildings such as the White House, space to power satellites, and various methods of transportation. One reason for the expansion is a continuing drop in cost combined with an increase in performance and durability. In the mid-1950s, the cost of a solar panel was around $300 per watt capability. Twenty years later, the cost was a third of the 1950s’ cost. Currently, solar panel cost has dropped to less than $1 per watt while decreasing in size as well as increasing in longevity. The dropping price and improved performance has moved solar to a modest investment that can pay for itself in less than half the time of systems from 15 years ago. Solar Power’s Age The sun has been harnessed by humans for centuries. The earliest recorded use of tapping the sun’s energy for power dates back to the seventh century BC when man focused the sun’s rays through a magnifying glass to create fire. Four thousand years later, we find historical record of using mirrors to focus the sun and light torches, often for ceremonial proceedings. Use of the sun to light torches continued through the centuries and has been recorded by various cultures including the Chinese civilization in 20 AD and beyond. In more recent years, the sun was harnessed to power ovens on ships traversing to oceans in the 1700s. At the same time, the power of the sun was utilized to power steamboats through the 1800s. Mária Telkes, a Hungarian-born American scientist, invented a widely deployed solar seawater distiller used on World War II life rafts. Soon after, she partnered with architect Eleanor Raymond to design the first modern home to be completely heated by solar power. Air warmed on rooftop collectors transferred heat to salts, which stored the heat for later use. Although solar panels as we know them today are relatively new in history, use of the sun to harness power is much older. Compute Power Needs for Common Home Devices A kilowatt (kW) is 1,000 watts (W). A kilowatt-hour (kWh) is a measurement of energy use, which is the amount of energy used by a 1,000-watt device to run for an hour. Using the definition of a kilowatt-hour, to calculate how long it would take to consume 1 kWh of power, we divide 1,000 by the watts use of a device. 1,000 / watts = time needed to use 1 kW For example, a 75 W bulb would take 1,000 ÷ 75 = 13.3 hours to use 1 kW of power. watts / 1 , 000 = kilowatt hours Calculating the Kilowatt-Hours Needed to Run a Television A 48 in plasma television uses about 200 W. How many kilowatt-hours are needed to run the television in a month if the television is one for an average of 2.5 hours a day? Step 1: 1,000 / ( 200 watts ) = 5 hours to use 1 kW Step 2: ( 2.5 hours a day ) ( 30 days ) = 75 hours of use Step 3: 75 hour / 5 hours per kW = 15 kW The television will consume about 15 kW in a month. Calculating the Cost to Run a Refrigerator A medium-sized Energy Star–rated refrigerator uses about 575 W and runs for about 8 hours per day. What is the monthly (30 days) cost of running the refrigerator if the electric rate is 12 cents per kilowatt-hour? Step 1: Calculate the watts per day: ( 575 W ) ( 8 hours ) = 4,600 W per day Step 2: Calculate the kilowatt-hours. ( 4,600 ) / ( 1,000 ) = 4.6 kWh Step 3: Calculate the daily cost. ( 4.6 kWh ) ( 12 cents ) = 55 cents = $ 0.55 Step 4: Calculate the monthly cost. ( $ 0.55 ) ( 30 ) = $ 16.50 It would cost about $16.50 to run the refrigerator for a month. Calculating the Kilowatt-Hours to Run an Oven An electric oven is labeled as 4,000 W. How much would it cost to bake a cake for 30 minutes if the electric rate is 14 cents per kilowatt-hour? Step 1: Determine the time it takes to use 1 kW of power: 1,000 / ( 4,000 watts ) = 0.25 hours to use 1 kW For every 15 minutes, the oven uses 1 kW of power. Step 2: Determine how many kilowatt-hours are needed to bake the cake for 30 minutes: ( 30 minutes ) / ( 15 minutes per kW ) = 2 kW Step 3: Calculate the cost of the oven usage: ( 2 kW ) ( 14 cents per kWh ) = 28 cents It would cost about 28 cents to bake the cake. Solar Advantages There are multiple advantages that solar power can offer us today including reducing greenhouse gas and CO 2 emissions, powering vehicles, reducing water pollution, reducing strain on limited supply of other power options such as fossil fuels. We will look further at reducing greenhouse gas and CO 2 emissions. Any gas that prevents infrared radiation from escaping Earth's atmosphere is a greenhouse gas. There are 24 currently identified greenhouse gases of which carbon dioxide is one. When measuring the impact of any of the greenhouse gases, the measurements are given in units of carbon dioxide emissions. For this reason, greenhouse gas and carbon dioxide have become interchangeable in discussions. Charles Fritts and Mohammad M. Atalla Charles Fritts, a New York inventor, is credited with creating the first solar cell, which he installed on a rooftop in New York City in 1884. While the solar cell was not very efficient, having a rate of conversion between 1 to 2%, this was a major step early in solar power energy. Today’s solar cells have an efficiency on average of 15 to 20%, which yields a notably higher impact. Nonetheless, the work that Fritts successfully completed marked the start of solar energy through the use of photovoltaic solar panels in the United States. Mohamed M. Atalla was an Egyptian-born scientist who moved to the United States to complete his studies, and undertook research and development at Bell Laboratories in New Jersey. Many of the early efficiency gains in solar cells were due to his development of processes for using silicon within electronic devices. Atalla's work led to the invention of silicon transistors and microchips (including his own invention of the MOSFET, the most widely used transistor in the world), and quickly increased the efficiency of solar cells. Calculating the Solar Power for Average Home Use in Kilowatts If a home uses approximately 30 kW of electricity per day, what size solar system would be needed to fuel 80% of a home’s needs for a month (30 days)? ( 30 kW hours ) ( 30 days ) ( 0.80 ) = 720 kW A solar system capable of producing 720 kW a month would be needed. Check Your Understanding Key Terms greenhouse gas CO 2 emissions watt kilowatt (kW) Key Concepts Recognize how water conservation by one person, family, community, or nation can positively impact the world’s freshwater supply. Recall components from the history of solar use by mankind. Calculate electrical demand given watts. Recognize advantages of residential solar power. Formulas 1,000 / watts = time needed to use 1 kWh watts / 1 , 000 = kilowatt hours

Math


Math and Medicine Shoppers wear masks during the Covid-19 pandemic. (credit: "True Covid Scene - Mask Buying" by Joey Zanotti/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Compute the mathematical factors utilized in concentrations/dosages of drugs. Describe the history of validating effectiveness of a new drug. Describe how mathematical modeling is used to track the spread of a virus. The pandemic that rocked the world starting in 2020 turned attention to finding a cure for the Covid-19 strain into a world race and dominated conversations from major news channels to households around the globe. News reports decreeing the number of new cases and deaths locally as well as around the world were part of the daily news for over a year and progress on vaccines soon followed. How was a vaccine able to be found so quickly? Is the vaccine safe? Is the vaccine effective? These and other questions have been raised through communities near and far and some remain debatable. However, we can educate ourselves on the foundations of these discussions and be more equipped to analyze new information related to these questions as it becomes available. Concentrations and Dosages of Drugs Consider any drug and the recommended dosage varies based on several factors such as age, weight, and degree of illness of a person. Hospitals and medical dispensaries do not stock every possible needed concentration of medicines. Drugs that are delivered in liquid form for intravenous (IV) methods in particular can be easily adjusted to meet the needs of a patient. Whether administering anesthesia prior to an operation or administering a vaccine, calculation of the concentration of a drug is needed to ensure the desired amount of medicine is delivered. The formula to determine the volume needed of a drug in liquid form is a relatively simple formula. The volume needed is calculated based on the required dosage of the drug with respect to the concentration of the drug. For drugs in liquid form, the concentration is noted as the amount of the drug per the volume of the solution that the drug is suspended in which is commonly measured in g/mL or mg/mL. Suppose a doctor writes a prescription for 6 mg of a drug, which a nurse calculates when retrieving the needed prescription from their secure pharmaceutical storage space. On the shelves, the drug is available in liquid form as 2 mg per mL. This means that 1 mg of the drug is found in 0.5 mL of the solution. Multiplying 6 mg by 0.5 mL yields 3 mL, which is the volume of the prescription per single dose. Volume needed = ( medicine dosage required ) ( weight of drug by volume ) . A common calculation for the weight of a liquid drug is measured in grams of a drug per 100 mL of solution and is also called the percentage weight by volume measurement and labeled as % w/v or simply w/v. Note that the units for a desired dose of a drug and the units for a solution containing the drug or pill form of the drug must be the same. If they are not the same, the units must first be converted to be measured in the same units. Suppose you visit your doctor with symptoms of an upset stomach and unrelenting heartburn. One possible recourse is sodium bicarbonate, which aids in reducing stomach acid. Calculating the Quantity in a Mixture How much sodium bicarbonate is there in a 250 mL solution of 1.58% w/v sodium bicarbonate? 1.58 % w / v = 1.58 g sodium bicarbonate in 100 mL. If there is 250 mL of the solution, we have 2.5 times as much sodium bicarbonate as in 100 mL. Thus, we multiply 1.58 by 2.5 to yield 3.95 g sodium bicarbonate in 250 mL solution. Calculating the Quantity of Pills Needed A doctor prescribes 25.5 mg of a drug to take orally per day and pills are available in 8.5 mg. How many pills will be needed each day? The prescription and the pills are in the same units which means no conversions are needed. We can divide the units of the drug prescribed by the units in each pill: 25.5 / 8.5 = 3 . So, 3 pills will be needed each day. Calculating the Drug Dose in Milligrams, Based on Patient Weight A patient is prescribed 2 mg/kg of a drug to be delivered intramuscularly, divided into 3 doses per day. If the patient weighs 45 kg, how many milligrams of the drug should be given per dose? Step 1: Calculate the total daily dose of the drug based on the patient’s weight (measured in kilograms): ( 2 mg / kg ) ( 45 kg ) = 90 mg Step 2: Divide the total daily dose by the number of doses per day: 90 mg / 3 = 30 mg The patient should receive 30 mg of the drug in each dose. Note that the units for a patient’s weight must be compatible with the units used in the medicine measurement. If they are not the same, the units must first be converted to be measured in the same units. Calculating the Drug Dose in Milliliters, Based on Patient Weight A patient is prescribed 2 mg/kg of a drug to be delivered intramuscularly, divided into 3 doses per day. If the drug is available in 20 mg/mL and the patient weighs 60 kg, how many milliliters of the drug should be given per dose? Step 1: Calculate the total daily dose of the drug (measured in milligrams) based on the patient’s weight (measured in kilograms): ( 2 mg / kg ) ( 60 kg ) = 120 mg Step 2: Calculate the volume in each dose: ( 120 mg daily total ) / ( 3 doses a day ) = 40 mg per dose Step 3: Calculate the volume based on the strength of the stock: ( prescribed dose needed ) / ( stock dose ) = volume ( 40 mg per dose ) / ( 20 mg / mL ) = 2 mL The patient should receive 2 mL of the stock drug in each dose. Math Statistics from the CDC The Centers for Disease Control and Prevention (CDC) states that about half the U.S. population in 2019 used at least one prescription drug each month, and about 25% of people used three or more prescription drugs in a month. The resulting overall collective impact of the pharmaceutical industry in the United States exceeded $1.3 trillion a year prior to the 2020 pandemic. Validating Effectiveness of a New Vaccine The process to develop a new vaccine and be able to offer it to the public typically takes 10 to 15 years. In the United States, the system typically involves both public and private participation in a process. During the 1900s, several vaccines were successfully developed, including the following: polio vaccine in the 1950s and chickenpox vaccine in the 1990s. Both of these vaccines took years to be developed, tested, and available to the public. Knowing the typical timeline for a vaccine to move from development to administration, it is not surprising that some people wondered how a vaccine for Covid-19 was released in less than a year’s time. Lesser known is that research on coronavirus vaccines has been in process for approximately 10 years. Back in 2012, concern over the Middle Eastern respiratory syndrome (MERS) broke out and scientists from all over the world began working on researching coronaviruses and how to combat them. It was discovered that the foundation for the virus is a spike protein, which, when delivered as part of a vaccine, causes the human body to generate antibodies and is the platform for coronavirus vaccines. When the Covid-19 pandemic broke out, Operation Warp Speed, fueled by the U.S. federal government and private sector, poured unprecedented human resources into applying the previous 10 years of research and development into targeting a specific vaccine for the Covid-19 strain. Shibo Jiang Dr. Shibo Jiang, MD, PhD, is co-director the Center for Vaccine Development at the Texas Children’s Hospital and head of a virology group at the New York Blood Center. Together with his colleagues, Jiang has been working on vaccines and treatments for a range of viruses and infections including influenzas, HIV, Sars, HPV and more recently Covid-19. His work has been recognized around the world and is marked with receiving grants amounting to over $20 million from U.S. sources as well as the same from foundations in China, producing patents in the United States and China for his antiviral products to combat world concerns. Jiang has been a voice for caution in the search for a vaccine for Covid-19, emphasizing the need for caution to ensure safety in the development and deployment of a vaccine. His work and that of his colleagues for over 10 years on other coronaviruses paved the way for the vaccines that have been shared to combat the Covid-19 pandemic. Mathematical Modeling to Track the Spread of a Vaccine With a large number of people receiving a Covid-19 vaccine, the concern at this time is how to create an affordable vaccine to reach people all over the world. If a world solution is not found, those without access to a vaccine will serve as incubators to variants that might be resistant to the existing vaccines. As we work to vaccinate the world, attention continues with tracking the spread of the Covid-19 and its multiple variants. Mathematical modeling is the process of creating a representation of the behavior of a system using mathematical language. Digital mathematical modeling plays a key role in analyzing the vast amounts of data reported from a variety of sources such as hospitals and apps on cell phones. When attempting to represent an observed quantitative data set, mathematical models can aid in finding patterns and concentrations as well as aid in predicting growth or decline of the system. Mathematical models can also be useful to determine strengths and vulnerabilities of a system, which can be helpful in arresting the spread of a virus. The chapter on Graph Theory explores one such method of mathematical modeling using paths and circuits. Cell phones have been helpful in tracking the spread of the Covid-19 virus using apps regulated by regional government public health authorities to collect data on the network of people exposed to an individual who tests positive for the Covid-19 virus. Gladys West Gladys West (credit: "Dr. Gladys West Hall" by The US Air Force/Wikimedia Commons, Public Domain) Dr. Gladys West is a mathematician and hidden figure with a rich résumé of accomplishments spanning Air Force applications and work at NASA. Born in 1930, West rose and excelled both academically and in her professional life at a time when Black women were not embraced in STEM positions. One of her many accomplishments is the Global Positioning System (GPS) used on cell phones for driving directions. West began work as a human computer, someone who computes mathematical computations by hand. Considering the time and complexity of some calculations, she became involved in programming computers to crunch computations. Eventually, West created a mathematical model of Earth with detail and precision that made GPS possible, which is utilized in an array of devices from satellites to cell phones. The next time you tag a photo or obtain driving directions, you are tapping into the mathematical modeling of Earth that West developed. Consider the following graph ( ): Contact Tracing for Math 111 Section 1 At the center of the graph, we find Alyssa, whom we will consider positive for a virus. Utilizing the technology of phone apps voluntarily installed on each phone of the individuals in the graph, tracking of the spread of the virus among the 6 individuals that Alyssa had direct contact with can be implemented, namely Suad, Rocio, Braeden, Soren, and Sandra. Let’s look at José’s exposure risk as it relates to Alyssa. There are multiple paths connecting José with Alyssa. One path includes the following individuals: José to Mikaela to Nate to Sandra to Alyssa. This path contains a length of 4 units, or people, in the contact tracing line. There are 2 more paths connecting José to Alyssa. A second path of the same length consists of José to Lucia to Rocio to Braeden to Alyssa. Path 3 is the shortest and consists of José to Lucia to Rocio to Alyssa. Tracking the spread of positive cases in the line between Alyssa and José aids in monitoring the spread of the infection. Now consider the complexity of tracking a pandemic across the nation. Graphs such as the one above are not practical to be drawn on paper but can be managed by computer programs capable of computing large volumes of data. In fact, a computer-generated mathematical model of contact tracing would look more like a sphere with paths on the exterior as well as on the interior. Mathematical modeling of contact tracing is complex and feasible through the use of technology. Using Mathematical Modeling For the following exercises, use the sample contact tracing graph to identify paths ( ). Contact Tracing for ECON 250 Section 1 How many people have a path of length 2 from Jeffrey? Find 2 paths between Kayla and Rohan. Find the shortest path between Yara and Kalani. State the length and people in the path. 5 (Lura, Naomi, Kalani, Vega, Yara) Answers will vary. Two possible answers are as follows: Kayla, Jeffrey, Rohan Kayla, Lura, Yara, Lev, Vega, Uma, Kalani, Rohan Length is 4. People in path = Yara, Lev, Vega, Uma, Kalani Check Your Understanding Key Terms concentrations/dosages of drugs mathematical modeling Key Concepts Compute volumes of prescription drugs in liquid and pill form. Validate the effectiveness of a new drug. Mathematical modeling can be used to describe and track the spread of a virus. Formula Volume needed = ( medicine dosage required ) ( weight of drug by volume ) .

Math