Apportionment Methods Schoolchildren depend on apportionment of resources like laptops among schools and classroom. (credit: “Richmond Public Schools” by Virginia Department of Education/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Describe and interpret the apportionment problem. Apply Hamilton’s Method. Describe and interpret the quota rule. Apply Jefferson’s Method. Apply Adams’s Method. Apply Webster’s Method. Compare and contrast apportionment methods. Identify and contrast flaws in various apportionment methods. A Closer Look at the Apportionment Problem In Standard Divisors, Standard Quotas, and the Apportionment Problem we calculated the standard divisor and the standard quotas in various apportionment scenarios. The results of those calculations routinely led to fractions and decimals of units. However, the seats in the House of Representatives, laptops in a classroom, or a variety of other resources, are indivisible, meaning they cannot be divided up into fractional parts. This leaves a decision to be made. For example, if the standard quota for the number of laptops to be distributed to a classroom is 12.44 units, how do we deal with the fractional part of 0.44? It is unclear if the classroom should receive 12 units, 13 units, or some other value. Let’s try traditional rounding to the nearest whole number value. Installing Emergency Lights The board of trustees of a college has recently approved the installation of 70 new emergency blue lights in three parking lots. The number of lights in each lot will be proportionate to the size of the parking lot, which is to be measured in acres. The total number of acres is 34; so the standard divisor is 34 70 ≈ 0.4857 . The standard quota for each lot is listed in the table below. Use this information to answer each question. Lot Acres Lot’s Standard Quota A 15 15 ÷ 0.4857 ≈ 30.88 emergency blue lights B 9 9 ÷ 0.4857 ≈ 18.53 emergency blue lights C 10 10 ÷ 0.4857 = 20.59 emergency blue lights Use traditional rounding to determine the number of lights assigned to each lot. Find the sum of the values from part 1. Does the sum found in part 2 equal the number of lights available? If traditional rounding is used, there will be 31, 19, and 21 lights distributed to each lot, respectively. The total of these values is 71. No, the total from part 2 is one more than the number of lights available. In other words, one of the parking lots must get 1 fewer light than apportioned. demonstrates that we cannot successfully apportion indivisible resources by rounding off each standard quota using traditional rounding. This leaves us with a problem. What is a fair way to distribute the fractional parts of the standard quotas? We will refer to this as the apportionment problem . Several methods for making this decision will be discussed. A-10C Thunderbolt II Aircraft In 2015, the U.S. Air Force had a fleet of approximately 281 A-10C Thunderbolt II aircraft. Suppose that the Air Force administration wanted to distribute 27 aircrafts across six bases based on the number of qualified pilots stationed at those bases. Use the information in the table below to answer each question. Base Pilots (A) Alpha 13 (B) Bravo 12 (C) Charlie 5 (D) Delta 16 (E) Echo 7 (F) Foxtrot 9 Identify the states, the seats, and the state population (the basis for the apportionment) in this scenario. Find the standard divisor for the apportionment of the aircraft. Round to four decimal places as needed. Include the units. Find each Air Force base’s standard quota for the apportionment of the aircraft. Round to the nearest hundredth as needed. What are the units? How does this example demonstrate the apportionment problem? Will traditional rounding solve the problem? The states are the bases, the seats are the aircraft, and the state populations are the pilots at a given base. Standard Divisor = Total Population House Size = 13 + 12 + 5 + 16 + 7 + 9 27 = 62 27 ≈ 2.2963 pilots per aircraft. A 13 2.2963 ≈ 5.66 , B 12 2.2963 ≈ 5.23 , C 5 2.2963 ≈ 2.18 , D 16 2.2963 ≈ 6.97 , E 7 2.2963 ≈ 3.05 , F 9 2.2963 ≈ 3.92 . The units are aircraft. This example demonstrates the apportionment problem because it is not possible to send a fractional number of aircraft to an Air Force base. On the other hand, if we use traditional rounding methods to get whole numbers, the results are 5 + 5 + 2 + 7 + 3 + 4 = 26 aircraft will be apportioned, which is one less than the number of aircraft that were supposed to be apportioned. Hamilton's Method of Apportionment One of the problems encountered when standard quotas are transformed into whole numbers using traditional rounding is that it is possible for the sum of the values to be greater than the number of seats available. A reasonable way to avoid this is to always round down, even when the first decimal place is five or greater. For example, a standard quota of 12.33 and a standard quota of 12.99 would both round down to 12. This is called the lower quota . Lower Quota for Apportionment of Aircraft The Air Force administration wants to distribute 27 aircrafts across six bases based on the number of qualified pilots stationed at those bases. The standard quotas for each base are listed in the table below. Use this information to answer the questions. Base Standard Quota (A) Alpha 13 2.2963 ≈ 5.66 aircraft (B) Bravo 12 2.2963 ≈ 5.23 aircraft (C) Charlie 5 2.2963 ≈ 2.18 aircraft (D) Delta 16 2.2963 ≈ 6.97 aircraft (E) Echo 7 2.2963 ≈ 3.05 aircraft (F) Foxtrot 9 2.2963 ≈ 3.92 aircraft Give the lower quota for each Air Force base. Find the sum of the lower quotas. By how much does this sum fall short of the actual number of aircraft? Round down. The lower quota for each Air Force base is 5, 5, 2, 6, 3, 3, respectively. The sum is 24. This is 3 fewer than the actual number of aircraft. If the standard quotas are all rounded down, their sum will always be less than or equal to the house size. Then, it would only remain to find a fair way to distribute any remaining seats. Alexander Hamilton, who was a general in the American Revolution, author of the Federalist Papers, and the first secretary of the treasury, took this approach to apportionment. Steps for Hamilton’s Method of Apportionment There are five steps we follow when applying Hamilton’s Method of apportionment: Find the standard divisor. Find each state’s standard quota. Give each state the state’s lower quota (with each state receiving at least 1 seat). Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain. Check the solution by confirming that the sum of the modified quotas equals the house size. Hamilton Method of Apportionment Hawaiian School Districts Suppose that the Hawaii State Department of Education has a budget for 616 schools and is doing a research study to determine the equitable number of schools to have in each of the five counties based on the residents under 19 years old, This data is provided in the table below. Using the Hamilton method, calculate how many schools would be funded in each state. Hawaii Honolulu Kalawao Kauai Maui Total Residents under age 19 46,310 224,230 20 16,560 38,450 325,570 Step 1: Calculate the standard divisor. Divide the total population, 325,570, by the house size, 616 seats. The standard divisor is 528.52. Step 2: Find each state’s standard quota: Hawaii Honolulu Kalawao Kauai Maui Total Standard Quota 46 , 310 528.52 ≈ 87.62 224 , 230 528.52 ≈ 424.26 20 528.52 ≈ 0.04 16 , 560 528.52 ≈ 31.33 38 , 450 528.52 ≈ 72.75 616 Step 3: Find each state’s lower quota and their sum: Hawaii Honolulu Kalawao Kauai Maui Total Lower Quota 87 424 1 31 72 615 Step 4: Compare the sum of the states’ lower quotas, 615, to the house size, 616. One seat remains to be apportioned and must be given to the state with the largest fractional part: Maui with 0.75. So, the final Hamilton quotas are as follows: Hawaii 87, Honolulu 424, Kalawao 1, Kauai 31, and Maui 73. Step 5: Find the total to confirm the sum of the quotas equals the house size, 616. Then 87 + 424 + 1 + 31 + 73 = 616 . The apportionment is complete. Apportionment Calculators Check out websites such as Ms. Hearn Math for a free Hamilton apportionment calculator. This can be a useful tool to confirm your results! The Quota Rule A characteristic of an apportionment that is considered favorable is when the final quota values all either result from rounding down or rounding up from the standard quotas. The value that results from rounding down is called the lower quota, and the value that results from rounding up is called the upper quota . As we explore more methods of apportionment, we will consider whether they satisfy the quota rule. If a scenario exists in which a particular apportionment allocates a value greater than the upper quota or less than the lower quota, then that apportionment violates the quota rule and the apportionment method that was used violates the quota rule. Which Apportionment Method Satisfies the Quota Rule? Several apportionment methods have been used to allocate 125 seats to ten states and the results are shown in the table below. Determine which apportionments do not satisfy the quota rule and justify your answer. State A State B State C State D State E State F State G Standard Quota 41.26 16.00 5.77 2.64 7.82 10.47 0.21 Lower Quota 41 16 5 2 7 10 0 Upper Quota 42 17 6 3 8 11 1 Method X 43 16 5 2 7 10 1 Method Y 41 16 6 2 8 10 1 Method Z 42 16 7 3 7 9 1 Look for states such that the number of seats allocated differs from the lower or upper quota. Method X violates the quota rule because State A receives 43 seats instead of 41 or 42. Method Z violates the quota rule because State C receives 7 seats instead of 5 or 6 and State F receives 9 instead of 10 or 11. It is possible for an apportionment method to satisfy the quota rule in some scenarios but violate it in others. However, because the Hamilton method always begins with the lower quota and either adds one to it or keeps it the same, the final Hamilton quota will always consist of values that are either lower quota values or upper quota values. When an apportionment method has this characteristic, it is said to satisfy the quota rule. So, we can say: The Hamilton method of apportionment satisfies the quota rule. Although the Hamilton method of apportionment satisfies the quota rule, it can result in some unexpected outcomes, which has caused it to pass in and out of favor of the U.S. government over the years. There are several apportionment methods that have been popular alternatives, such as Jefferson’s method of apportionment that the founders of Imaginaria should consider. Jefferson’s Method of Apportionment Another approach to dealing with the fractional parts of the standard quotas is to modify the standard divisor so that the total of the resulting modified lower quotas is the necessary number of seats. This is the approach used by Jefferson. In Jefferson’s method, the change to the standard divisor is made so that the total of the modified lower quotas equals the house size. The change in the standard divisor to get the modified divisor is relatively small. There is not a formula for this. The modified divisor is found by “guess and check.” It is important to remember that increasing the divisor decreases the quotas, but decreasing the divisor increases the quotas. So, if you need a larger quota, try reducing the divisor, and if you need a smaller quota, try increasing the divisor. Modifying a Standard Divisor Suppose the population of a state is 50 and the standard divisor is 12.5. Find the state’s standard quota. Increase the standard divisor by 2 units and use the modified divisor to determine the modified quota for the state. Decrease the modified divisor from part 2 by 1.5 units and use the new modified divisor to determine the modified quota for the state. Choose any value of divisor between the value of the modified divisor from part 2 and the value of the modified divisor from part 3 and use it to determine the modified quota for the state. Which modified quota was the largest, the modified quota from part 2, from part 3, or from part 4? Explain why. The state’s standard quota is 50 12.5 = 4 . The modified divisor is 14.5. The modified quota is 50 14.5 ≈ 3.45 . The modified divisor is 13. The modified quota is 50 13 ≈ 3.85 . One value between 13 and 14.5 is 13.5. With a modified divisor of 13.5, the modified quota is 50 13.5 ≈ 3.70 . The modified quota from part 3 was the largest because the divisor was the smallest of the three. Dividing the same number by a smaller value gives a larger result. When you use Jefferson’s method, you might have to adjust the divisor several times find modified lower quotas that sum to the house size. First, guess what the divisor should be based on the sum of the lower quotas and then increase or decrease it from there based on whether the sum needs to be smaller or larger respectively. If the result still does not produce lower quotas that sum to the house size, adjust again. Keep a record of the values that didn't work to help you narrow your search. Steps for Jefferson’s Method of Apportionment We take four steps to apply Jefferson’s Method of apportionment: Step 1: Find the standard divisor. Step 2: Find each state’s quota. This will be the standard quota the first time Step 2 is completed and the standard divisor is used, but Step 2 may be repeated as needed using a modified divisor and resulting in modified quotas. Step 3: Find the states’ lower quotas (with each state receiving at least one seat), and their sum. Step 4: If the sum from Step 3 equals the number of seats, the apportionment is complete. If the sum of the lower quotas is less than the number of seats, reduce the standard divisor. If the sum of the lower quotas is greater than the number of seats, increase the standard divisor. Return to Step 2 using the modified divisor. Hawaiian State Representative Districts Suppose that the Hawaii State Department of Education has a budget for 616 schools and is doing a research study to determine the equitable number of schools to have in each of five counties based on the residents under the age of 19. With the data in the table below, apply Jefferson’s method to apportion the schools to the counties. Hawaii Honolulu Kalawao Kauai Maui Total Residents under Age 19 46,310 224,230 20 16,560 38,450 325,570 Step 1: The process for finding the standard divisor, standard quotas, and lower quotas is the same in the Hamilton and Jefferson methods of apportionment. We walked through the Hamilton Method in , and following these steps resulted in lower quotas as shown in the table below. Hawaii Honolulu Kalawao Kauai Maui Total Standard Quota 46 , 310 528.52 ≈ 87.62 224 , 230 528.52 ≈ 424.26 20 528.52 ≈ 0.04 16 , 560 528.52 ≈ 31.33 38 , 450 528.52 ≈ 72.75 616 Lower Quota 87 424 1 31 72 615 Step 2: Compare the sum of the states’ lower quotas, 615, to the house size, 616. Since 615 is less than 616, use a modified divisor that is less than the standard divisor of 528.52. Try 526.00. Step 3: Find each state’s modified quota, lower quota, and the sum of the lower quotas based on the modified divisor of 526: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 526.00 ≈ 88.04 224 , 230 526.00 ≈ 426.29 20 526.00 ≈ 0.04 16 , 560 526.00 ≈ 31.48 38 , 450 526.00 ≈ 72.75 616 Lower Quota 88 426 1 31 72 618 Step 4: The new sum of the lower quotas is 2 units greater than 616. We have overshot the goal. So, increase the divisor to a value between 526.00 and 528.52. Try 527.00. Step 5: Repeat the process of finding the quotas. Find each state’s modified quota, lower quota, and the sum of the lower quotas based on the modified divisor of 526.00: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 527.00 ≈ 87.87 224 , 230 527.00 ≈ 425.48 20 527.00 ≈ 0.04 16 , 560 527.00 ≈ 31.42 38 , 450 527.00 ≈ 72.96 616 Lower Quota 87 425 1 31 72 616 Step 6: The new sum of the lower quotas equals the house size. The apportionment is complete. The apportionment is: Hawaii County 87, Honolulu County 425, Kalawao County 1, Kauai 31, and Maui 72 schools. When using Jefferson’s method, the modified divisors you use may be different from what another person chooses, but final apportionment values will be the same. Notice that, in this apportionment, Mythbury received more than the upper quota. Since this apportionment of representatives to Imaginarian states by Jefferson’s method does not satisfy the quota rule, we say that: Jefferson’s method violates the quota rule. We have discussed two apportionment methods: one that satisfies the quota rule and one that does not. Before you decide which method to use in Imaginaria, there are a couple more options to consider. Jefferson Apportionment Method Apportionment Calculators It is possible to create Excel spreadsheets that complete the calculations necessary to complete a Jefferson Apportionment. In some cases, this work has already been done and posted online. Check out websites such Ms. Hearn Math for a free Jefferson apportionment calculator. This can be a useful tool to confirm your results! Adams’s Method of Apportionment Adams’s method of apportionment is another method of apportionment that is based on a modified divisor. However, instead of basing the changes on the sum of the lower quotas, as Jefferson did, Adams used the upper quotas. To apply Adams’s Method of apportionment, there are four steps we follow: Find the standard divisor. Find each state’s quota. This will be the standard quota the first time Step 2 is completed, and the standard divisor is used, but Step 2 may be repeated as needed using a modified divisor and resulting in modified quotas. Find the states’ upper quotas and their sum. If the sum from Step 3 equals the number of seats, the apportionment is complete. If the sum of the upper quotas is less than the number of seats, reduce the standard divisor. If the sum of the upper quotas is greater than the number of seats, increase the standard divisor. Return to Step 2 using the modified divisor. Hawaiian School Districts As in earlier examples, suppose that the Hawaii State Department of Education has a budget for 616 schools and is doing a research study to determine the equitable number of schools to have in each of the five counties based on the residents under the age of 19. Use the data in the following table and the Adams method to apportion the schools to the counties. Hawaii Honolulu Kalawao Kauai Maui Total Residents under Age 19 46,310 224,230 20 16,560 38,450 325,570 Step 1: The steps of finding the standard divisor and each state’s quota are the same in the Jefferson and Adams methods. As in , the standard divisor is 528.52. Step 2: Find each state’s upper quota and their sum: Hawaii Honolulu Kalawao Kauai Maui Total Standard Quota 46 , 310 528.52 ≈ 87.62 224 , 230 528.52 ≈ 424.26 20 528.52 ≈ 0.04 16 , 560 528.52 ≈ 31.33 38 , 450 528.52 ≈ 72.75 616 Upper Quota 88 425 1 32 73 619 Step 3: Compare the sum of the states’ upper quotas, 619, to the house size, 616. Since 619 is greater than 616, we need to reduce the size of the quotas. Use a modified divisor that is greater than the standard divisor of 528.52. Try 534.00. Step 4: Find each state’s modified quota, upper quota, and the sum of the upper quotas based on the modified divisor of 534: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 534.00 ≈ 86.72 224 , 230 534.00 ≈ 419.91 20 534.00 ≈ 0.04 16 , 560 534.00 ≈ 31.01 38 , 450 534.00 ≈ 72.00 616 Upper Quota 88 420 1 32 72 613 Step 5: The new sum of the upper quotas is 3 units less than 616. Larger quotas are needed. So, decrease the divisor to a value between 534.00 and 528.52. Try 532.00. Step 6: Find each state’s modified quota, upper quota, and the sum of the upper quotas based on the modified divisor of 532.00: County Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 532.00 ≈ 87.05 224 , 230 532.00 ≈ 421.48 20 532.00 ≈ 0.04 16 , 560 532.00 ≈ 31.13 38 , 450 532.00 ≈ 72.27 616 Upper Quota 88 422 1 32 73 616 Step 7: The new sum of the upper quotas equals the house size. The apportionment is complete. The apportionment is Hawaii County 88, Honolulu County 422, Kalawao County 1, Kauai 32, and Maui 73 schools. When using Adams’s method, just as with Jefferson’s method, the modified divisors you use may be different from what another person chooses, but final apportionment values will be the same. In this apportionment, Mythbury received less than the state’s lower quota. So, this apportionment is an example of a scenario in which the Adams’s method violates the quota rule. Adams’s method of apportionment violates the quota rule. So far, only Hamilton’s method satisfies the quota rule, but there is one more apportionment method you should consider for Imaginaria. Adams Method Apportionment Calculator Apportionment Calculators Check out websites such as Ms. Hearn Math for a free Adams Method apportionment calculator. This can be a useful tool to confirm your results! Webster’s Method of Apportionment Webster’s method of apportionment is another method of apportionment that is based on a modified divisor. However, instead of basing the changes on the sum of the lower quotas, as Jefferson did or the sum of the upper quotas as Adams did, Webster used traditional rounding. To apply Webster’s method of apportionment, there are four steps we take: Find the standard divisor. Find each state’s quota. This will be the standard quota the first time Step 2 is completed, and the standard divisor is used, but Step 2 may be repeated as needed using a modified divisor and resulting in modified quotas. Round each state’s quota to the nearest whole number and find the sum of these values. If the sum of the rounded quotas equals the number of seats, the apportionment is complete. If the sum of the rounded quotas is less than the number of seats, reduce the divisor. If the sum of the rounded quotas is greater than the number of seats, increase the divisor. Return to Step 2 using the modified divisor. When using Webster’s method, just as with Jefferson’s method, the modified divisors you use may be different from what another person chooses, but final apportionment values will be the same. Hawaiian School Districts Use the data in the table below to apportion 616 schools to Hawaiian counties. This time, use Webster’s method. Hawaii Honolulu Kalawao Kauai Maui Total Residents under Age 19 46,310 224,230 20 16,560 38,450 325,570 To apply Webster’s method of apportionment, there are four steps we take: Step 1: The processes of finding the standard divisor and standard quota are the same in the Jefferson, Adams, and Webster’s methods. As in the previous examples, the standard divisor is 528.52. Step 2: Find each state’s rounded quota and their sum: Hawaii Honolulu Kalawao Kauai Maui Total Standard Quota 46 , 310 528.52 ≈ 87.62 224 , 230 528.52 ≈ 424.26 20 528.52 ≈ 0.04 16 , 560 528.52 ≈ 31.33 38 , 450 528.52 ≈ 72.75 616 Rounded Quota 88 424 1 31 73 617 Step 3: Compare the sum of the states’ rounded quotas, 617, to the house size, 616. Since 617 is greater than 616, we need to reduce the size of the quotas. Use a modified divisor that is greater than the standard divisor of 528.52. Try 534.00. Step 4: Find each state’s modified quota, rounded quota, and the sum of the rounded quotas based on the modified divisor of 534: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 534.00 ≈ 86.72 224 , 230 534.00 ≈ 419.91 20 534.00 ≈ 0.04 16 , 560 534.00 ≈ 31.01 38 , 450 534.00 ≈ 72.00 616 Upper Quota 87 420 1 31 72 612 Step 5: The new sum of the rounded quotas is 4 units less than 616. Larger quotas are needed. So, decrease the divisor to a value between 534.00 and 528.52. Try 530.00. Step 6: Find each state’s modified quota, rounded quota, and the sum of the rounded quotas based on the modified divisor of 530.00: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 530.00 ≈ 87.38 224 , 230 530.00 ≈ 423.08 20 530.00 ≈ 0.04 16 , 560 530.00 ≈ 31.25 38 , 450 530.00 ≈ 72.55 616 Upper Quota 87 423 1 31 73 615 Step 7: The new sum of the rounded quotas is 1 unit less than 616. Larger quotas are needed. So, decrease the divisor to a value between 528.52 and 530.00. Try 529.50. Step 8: Find each state’s modified quota, rounded quota, and the sum of the rounded quotas based on the modified divisor of 529.50: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 529.50 ≈ 87.46 224 , 230 529.50 ≈ 423.48 20 529.50 ≈ 0.04 16 , 560 529.50 ≈ 31.27 38 , 450 529.50 ≈ 72.62 616 Upper Quota 87 423 1 31 73 615 Step 9: The new sum is still only 1 unit less than 616. Larger quotas are needed, but not much larger. So, decrease the divisor to a value between 528.52 and 529.50. Try 529.30. Step 10: Find each state’s modified quota, rounded quota, and the sum of the rounded quotas based on the modified divisor of 529.30: Hawaii Honolulu Kalawao Kauai Maui Total Modified Quota 46 , 310 529.30 ≈ 87.49 224 , 230 529.30 ≈ 423.63 20 529.30 ≈ 0.04 16 , 560 529.30 ≈ 31.29 38 , 450 529.30 ≈ 72.64 616 Upper Quota 87 424 1 31 73 616 Step 11: The new sum of the rounded quotas equals the house size. The apportionment is complete. The apportionment is Hawaii County 87, Honolulu County 424, Kalawao County 1, Kauai 31, and Maui 73 schools. So far, we know that the Hamilton method satisfies the quota rule, while the Jefferson and Adams methods do not. The apportionments in the Example and Your Turn above are both scenarios in which the Webster method satisfies the quota rule. Does it always? We have a little more work to do to find out. However, one thing is clear. Not all apportionment methods have the same results. Before you make such an important decision for Imaginaria, it’s important to think about the differences in the apportionments that result from these four methods. How will the differences affect the citizens of Imaginaria? Apportionment Calculators Check out websites such as Ms. Hearn Math for a free Webster Method apportionment calculator. This can be a useful tool to confirm your results! Comparing Apportionment Methods Recall that the four apportionment methods discussed in this chapter differ in two main ways: Whether or not a modified divisor is used The type of rounding of the quotas that is used How might these differences affect Imaginarians? In the next two examples, we will compare the results when different apportionment methods are applied to the same scenario. Hawaiian School Districts with Different Apportionment Methods Let’s use the results from , , , and to compare the four apportionment methods we have discussed. The following table summarizes the results of the results of the Hamilton, Jefferson, Adams and Webster methods when applied to the apportionment of 616 schools to Hawaiian counties. Hawaii Honolulu Kalawao Kauai Maui Under 19 years old 46,310 224,230 20 16,560 38,450 Hamilton 87 424 1 31 73 Jefferson 87 425 1 31 72 Adams 88 422 1 32 73 Webster 87 424 1 31 73 Do any of the apportionment methods result in the same apportionment? If so, which ones? Which apportionment method would the citizens of the largest county likely favor most and least? Justify your answer. As a group, which apportionment method would the citizens of the other four counties likely favor most and least? Justify your answer. Yes, the Hamilton and Webster methods result in the same apportionment. The largest county is Honolulu. The citizens would likely favor the Jefferson method of apportionment most since they received the most seats by that method. They would likely favor the Adams method of apportionment least because they received the least number of seats by that method. As a group, the other four counties received 192 seats by either the Hamilton or Webster method, 194 seats by the Adams method, and 191 seats by the Jefferson method. They would likely favor the Adams method the most and favor the Jefferson methods the least. The Adams method favored the smaller states and the Jefferson method favored the larger states in the previous example, but is this the case in general? Since the Jefferson method begins with the lower quotas, any adjustment to the quotas will be an increase. As you have seen, this is accomplished by using a modified divisor that is smaller than the standard divisor. The next example compares the impact of a decreasing divisor on the modified quotas of large states to the impact of the same size decrease on small states. Effect of Decreasing Divisors on Modified Quotas The following table displays the effect of reducing the size of the divisor. Observe the effect this has on the modified quotas of smaller states versus larger states and use the table answer each question. Modified Quotas State Population Divisor: 10,500 Divisor: 10,000 Divisor: 9,500 A 10,000 0.95 1 1.05 B 100,000 9.52 10 10.53 C 1,000,000 95.24 100 105.26 When the divisor decreases from 10,500 to 10,000, how many representatives are gained by each state based on the lower quota? When the divisor decreases from 10,000 to 9,500, how many representatives are gained by each state based on the lower quota? Which state gains the most representatives each time the divisor is decreased? Since a state must have at least one seat, State A begins with 1 seat and still has one seat. State B begins with 9 seats and increases to 10 seats. State C begins with 95 seats and increases to 100 seats. So, State A gains 0, B gains 1, and C gains 5 seats. State A begins with 1 and still has 1. State B begins with 10 and still has 10. State C begins with 100 and increases to 105. So, State A gains 0, State B gains 0, and State C gains 5. State C, the largest state, gains the most representatives each time the divisor is decreased. This example demonstrates that the Jefferson method is biased toward states with larger populations because the modified divisor is smaller than the standard divisor. On the other hand, the Adams’s method, which begins with the upper quotas, must increase the standard divisor in order to reduce the quotas. Once again, the effect on the number of seats is greater for the larger states, but this time they are decreased. This means that the Adams’s method favors states with smaller populations. Flaws in Apportionment Methods As we have seen, different apportionment methods can have the same results in some scenarios but different results in others. Citizens of states which receive fewer seats with a particular apportionment method will view the apportionment method as flawed and argue in favor of a different method. This inevitably creates debates regarding the use of one method over another. Methods that favor larger states are likely to be challenged by smaller states, methods that favor smaller states are likely to be challenged by larger states, and methods that violate the quota rule are likely to be challenged by states of any size depending on the circumstances. Suppose that the State of Hawaii House of Representatives had 51 representatives, each with their own district. Imagine that redistricting were underway, and the representative districts were to be apportioned to each of five counties based on population. The following table shows the apportionment that would result from the use of the Jefferson, Adams, and Webster methods of apportionment. Hawaii Honolulu Kalawao Kauai Maui Population 201,500 974,600 100 72,300 167,400 Lower Quota 7 35 0 2 6 Upper Quota 8 36 1 3 7 Jefferson 7 35 1 2 6 Adams 7 34 1 3 6 Webster 7 34 1 3 6 From the table, you can see that Hawaii, Kalawao, and Maui receive the same number of seats regardless of the method used. However, citizens of Honolulu would likely reject the Adams and Webster methods arguing that they violate the quota rule. Similarly, citizens of Kauai would probably reject the Jefferson method based on the argument that it unfairly favors the larger states. This scenario demonstrates that the Adams and Webster methods violate the quota rule, but the Jefferson method also violates the quota rule at times. The Hamilton method is the only method that satisfies the quota rule in all scenarios. It also consistently favors neither larger nor smaller states. Unfortunately, it can have some strange and results in certain circumstances, which you will see in the next section. Gerrymandering: A Subtle Way to Impact Apportionment In addition to your choice of voting method and your choice of apportionment method, there is another important decision to make which could potentially have a huge impact on the fairness of elections in Imaginaria—the creation of electoral districts. In example above, we imagined that there were 51 state legislators in Hawaii, each representing their own district. But how did the legislators decide on the boundaries for these districts? Typically, boundaries are drawn so that each district has approximately the same number of residents, but the percentage of residents in each district with a particular political affiliation can swing the power from one group to another. When the districts are drawn to impact the power of a political party, ethnic or racial group, or other group, this is called gerrymandering. For example, districts can be drawn so that a particular group is spread thinly across districts, increasing the likelihood that they will not have strong representation. This cartoon map conveys the idea that the drawing of the map may impact election outcomes. (credit: “The Gerry Mander”/Wikimedia Commons, Public Domain) "The Gerry-mander" first appeared in this cartoon-map in the Boston Gazette, March 26, 1812, and was soon reproduced in several other Massachusetts newspapers in response to election district changes initiated by governor Eldridge Gerry. Note that while the practice is named after him, Gerry was not the first to employ it. Jonathan Mattingly Jonathan Mattingly is a mathematician who was featured in a Nature article titled “The Mathematicians Who Want to Save Democracy” . Mattingly is a mathematician at Duke University in North Carolina and he runs election simulations based on alternate versions of electoral districts in order to analyze the effects of gerrymandering. He has even been asked to testify as an expert witness in court. Mattingly and other mathematicians who are working on the problem will potentially have an impact on the redistricting that will occur as a result of the 2020 census. (Carrie Arnold, “The Mathematicians Who Want to Save Democracy,” Nature 546 , 200–202, 2017.) Check Your Understanding Key Terms apportionment problem lower quota upper quota Key Concepts Hamilton’s method of apportionment uses the standard divisor and standard lower quotas, and it distributes any remaining seats based on the size of the fractional parts of the standard lower quota. Hamilton’s method satisfies the quota rule and favors neither larger nor smaller states. Jefferson’s method of apportionment uses a modified divisor that is adjusted so that the modified lower quotas sum to the house size. Jefferson’s method violates the quota rule and favors larger states. Adams’s method of apportionment uses a modified divisor that is adjusted so that the modified upper quotas, sum to the house size. Adams’s method violates the quota rule and favors smaller states. Webster’s method of apportionment uses a modified divisor that is adjusted so that the modified state quotas, rounded using traditional rounding, sum to the house size. Webster’s method violates the quota rule but favors neither larger nor smaller states. Videos Hamilton Method of Apportionment Jefferson Apportionment Method Adams Method Apportionment Calculator

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Fairness in Apportionment Methods In this seating chart for the House of Representatives, each color indicates representatives from a particular state. Learning Objectives After completing this section, you should be able to: Describe and illustrate the Alabama paradox. Describe and illustrate the population paradox. Describe and illustrate the new-states paradox. Identify ways to promote fairness in apportionment methods. Apportionment Paradoxes The citizens of Imaginaria will want the apportionment method to be as fair as possible. There are certain characteristics that they would reasonably expect from a fair apportionment. If the house size is increased, the state quotas should all increase or remain the same but never decrease. If one state’s population is growing more rapidly than another state’s population, the faster growing state should not lose a seat while a slower growing state maintains or gains a seat. If there is a fixed number of seats, adding a new state should not cause an existing state to gain seats while others lose them. However, apportionment methods are known to contradict these expectations. Before you decide on the right apportionment for Imaginarians, let’s explore the apportionment paradox , a situation that occurs when an apportionment method produces results that seem to contradict reasonable expectations of fairness. There is a lot that the founders of Imaginaria can learn from U.S. history. The constitution of the United States requires that the seats in the House of Representatives be apportioned according to the results of the census that occurs every decade, but the number of seats and the apportionment method is not stipulated. Over the years, several different apportionment methods and house sizes have been used and scrutinized for fairness. This scrutiny has led to the discovery of several of these apportionment paradoxes. The Alabama Paradox At the time of the 1880 U.S. Census, the Hamilton method of apportionment had replaced the Jefferson method. The number of seats in the House of Representatives was not fixed. To achieve the fairest apportionment possible, the house sizes were chosen so that the Hamilton and Webster methods would result in the same apportionment. The chief clerk of the Census Bureau calculated the apportionments for house sizes between 275 and 350. There was a surprising result that became known as the Alabama paradox , which is said to occur when an increase in house size reduces a state’s quota. Alabama would receive eight seats with a house size of 299, but only receive seven seats if the house size increased to 300. (Michael J. Caulfield (Gannon University), "Apportioning Representatives in the United States Congress - Paradoxes of Apportionment," Convergence (November 2010), DOI:10.4169/loci003163) The 1880 Alabama Quota The 1880 census recorded the population of Alabama as 1,513,401 and that of the U.S. as 62,979,766. Calculate the standard divisor and standard quota for the State of Alabama based on a house size of 299. Calculate the standard divisor and standard quota for the State of Alabama based on a house size of 300. Did the standard quota increase or decrease when the house size increased? Consider the Hamilton method of apportionment. Explain how Alabama’s final quota could be smaller with a larger standard quota. The standard divisor is 62,979,766 ÷ 299 = 210 , resulting in 634.6689 citizens per seat. The standard quota for Alabama is 1,513,401 ÷ 210,634.7 = 7.1850 seats. The standard divisor is 62,979,766 ÷ 300 = 209 , resulting in 932.5533 citizens per seat. The standard quota for Alabama is 1,513,401 ÷ 209,932.6 = 7.2090 seats. The standard quota increased. In each case, the state would receive the lower quota of 7 and then be awarded one more seat if the fractional part of the standard quota were high enough relative to the fractional parts of the other states’ standard quotas. When the house size was 299, Alabama received one of the remaining seats after the lower quotas were distributed. When the house size was 300, Alabama did not receive one of the remaining seats after the lower quotas were distributed. It must have been the case that either the fractional part 0.2090 ranked lower amongst the other fractional parts of the state quotas than the fractional part 0.1850 did, or there were fewer remaining seats, or both. After the 1900 census, the Census Bureau again calculated the apportionment based on various house sizes. It was determined that Colorado would receive three seats with a house size of 356, but only two seats with a house size of 357. Hamilton’s Method and the Alabama Paradox Suppose that States A and B each have a population of 6, while State C has a population of 2. Use the Hamilton method to apportion 10 seats. Use the Hamilton method to apportion 11 seats. Does this example demonstrate the Alabama paradox? If so, how? Step 1: The total population is 14. The standard divisor is 14 ÷ 10 = 1.4 individuals per seat. Step 2: The states’ standard quotas are as follows: A 6 ÷ 1.4 ≈ 4.29 , B 6 ÷ 1.4 ≈ 4.29 , and C 2 ÷ 1.4 ≈ 1.43. Step 3: The states’ lower quotas are as follows: A 4, B 4, and C 1. Step 4: The sum of the lower quotas is 9, which means there is one seat remaining to be apportioned. State C has the highest fractional part and receives the additional seat. Step 5: The final apportionment is as follows: A 4, B 4, and C 2, which sums to 10. Step 1: The total population is 14. The standard divisor is 14 ÷ 11 ≈ 1.2727 individuals per seat. Step 2: The states’ standard quotas are: A 6 ÷ 1.2727 ≈ 4.71 , B 6 ÷ 1.2727 ≈ 4.71 , and C 2 ÷ 1.2727 ≈ 1.57 . Step 3: The states’ lower quotas are: A 4, B 4, and C 1. Step 4: The sum of the lower quotas is 9, which means there are two seats remaining to be apportioned. A and B have the highest fractional parts and receive the additional seats. Step 5: The final apportionment is: A 5, B 5, and C 1. Yes, this demonstrates the Alabama paradox because State C receives two seats if the house size is 10, but only one seat if the house size is 11. The Population Paradox It is important for the founders of Imaginaria to keep in mind that the populations of states change as time passes. Some populations grow and some shrink. Some populations increase by a large amount while others increase by a small amount. These changes may necessitate a reapportionment of seats, or the recalculation of state quotas due to a change in population. It would be reasonable for Imaginarians to expect that the state with a population that has grown more than others will gain a seat before the other states. Once again, this is not always the case with the Hamilton method of apportionment. The population growth rate of a state is the ratio of the change in the population to the original size of the population, often expressed as a percentage. This value is positive if the population is increasing and negative if the population is decreasing. The population paradox occurs when a state with an increasing population loses a seat while a state with a decreasing population either retains or gains seats. More generally, the population paradox occurs when a state with a higher population growth rate loses seats while a state with a lower population growth rate retains or gains seats. Notice that the population paradox definitions has two parts. If either part applies, then the population paradox has occurred. The first part of the definition only applies when one state has a decreasing population. The second part of the definition applies in all situations, whether there is a state with a decreasing population or not. It will be easier to identify situations that involve a decreasing population. The other situations requires the calculation of a growth rate. The reason that we don't have to calculate a growth rate when one state has a decreasing population and the other has an increasing population is that increasing population has a positive growth rate which is always greater than the negative growth rate of a decreasing population. A state must lose a seat in order for the population paradox to apply. It is not enough for a state with a lower growth rate to gain a seat while a state with a higher growth rate retains the same number of seats. Apportionment of Respirators to Hospitals Suppose that 18 respirators are to be apportioned to three hospitals based on their capacities. The Hamilton method is used to allocate the respirators in 2020, then to reallocate based on new capacities in 2021. The results are shown in the table below. How does this demonstrate the population paradox? Hospital Capacity in 2020 Respirators in 2020 Change in Capacity Growth Rate = Change in Capacity Capacity in 2020 Capacity in 2021 Respirators in 2021 A 825 9 57 57 825 ≈ 0.0691 = 6.91 % 882 9 B 613 7 13 13 613 ≈ 0.0212 = 2.12 % 626 6 C 239 2 3 3 239 = 0.0126 = 1.26 % 242 3 Hospital B lost a respirator while hospital C gained one, even though hospital B had a higher growth rate than hospital C. The growth rate of a population can be calculated by subtracting the previous population size from the current population size, and then dividing the difference by the previous population size. population growth rate = current population size − previous population size previous population size Make sure to calculate the subtraction before the division. If you are entering the values in a calculator, it is helpful to put parentheses around the subtracted terms. The Congress of Costaguana The country of Costaguana has three states: Azuera with a population of 894,000; Punta Mala with a population of 696,000; and Esmeralda with a population of 215,000. There are 38 seats in the Congress of Costaguana. The apportionment of the seats is determined by Hamilton’s method to be: 19 for Azuera, 15 for Punta Mala, and 4 for Esmerelda. A census reveals that the population has grown and the seats must be reapportioned. If Azuera now has 953,000 residents, Punta Mala now has 706,000 residents, and Esmerelda now has 218,000 residents, how many seats will each state receive upon reapportionment? How is this an example of the population paradox? The Hamilton reapportionment is: 19 for Azuera, 14 for Punta Mala, and 5 for Esmerelda. This is an example of the population paradox because Punta Mala lost a seat to Esmerelda, even though Punta Mala’s population grew by 1.44 percent while Esmerelda’s only grew by 1.40 percent. The New-States Paradox As a founder of Imaginaria, you might also consider the possibility that Imaginaria could annex nearby lands and increase the number of states. This occurred several times in the United States such as when Oklahoma became a state in 1907. The House size was increased from 386 to 391 to accommodate Oklahoma’s quota of five seats. When the seats were reapportioned using Hamilton’s method, New York lost a seat to Maine despite the fact that their populations had not changed. This is an example of the new-state paradox , which occurs when the addition of a new state is accompanied by an increase in seats to maintain the standard ratio of population to seats, but one of the existing states loses a seat in the resulting reapportionment. New State of Oscuridad The country of San Lorenzo has grown from two states to three. The house size of the congress has been increased by eight and the seats have been reapportioned to accommodate the new state of Oscuridad. The constitution mandates the use of the Hamilton method of apportionment. Use this information and the following table to answer the questions. State Population (in hundreds) Original Apportionment Reapportionment Clara 7,100 39 40 Velasco 9,080 51 50 Oscuridad 1,500 Not Applicable 8 What was the original house size? What is the new house size? How is this reapportionment an example of the new-states paradox? 39 + 51 = 90 40 + 50 + 8 = 98 The original state of Velasco lost a seat to the original state of Clara when the new state of Oscuridad was added. The Growing Country of Gulliversia The country of Gulliversia has two states: Lilliput with a population of 700,000 and Brobdingnag with a population of 937,000. The constitution of Gulliversia requires that the 90 congressional seats be apportioned by Hamilton’s method. Lilliput has received 38 seats while Brobdingnag has received 52 seats. Recently, the island of Houyhnhnmsland with a population of 170,000 has joined the union, becoming a state of Gulliversia. When Houyhnhnmsland is included, nine additional seats must be apportioned to maintain the same ratio of seats to citizens. Use Hamilton’s method to reapportion the 99 seats to the three states. How is the resulting apportionment an example of the new-states paradox? The reapportionment gives 39 seats to Lilliput, 51 seats to Brobdingnag, and 9 seats to Houyhnhnmsland. This is an example of the new-states paradox because the original state of Brobdingnag lost a seat to the original state of Lilliput when the new state was added to the union. When a new state is added, it is necessary to determine the amount that the house size must be increased to retain the original ratio of population to seats, in other words to keep the original standard divisor. To calculate the new house size, divide the new population by the original standard divisor, and round to the nearest whole number. New House Size = New Population Original Standard Divisor rounded to the nearest whole number. Oklahoma Joins the Union Oklahoma was admitted as the 46th state on November 16, 1907. Before Oklahoma joined the union, the U.S. population was approximately 75,030,000 and the House of Representatives had 386 seats. The new state had a population of approximately 970,000. Use this information to estimate the original standard divisor to the nearest hundred, the new population, the new house size, and the number of seats Oklahoma should receive. Step 1: Original Standard Divisor ≈ 75 , 030 , 000 386 ≈ 194 , 400 Step 2: New Population ≈ 75 , 030 , 000 + 970 , 000 = 76 , 000 , 000 Step 3: New House Size ≈ 76,000,000 194 , 400 ≈ 391 Step 4: There are 391 - 386 = 5 new seats to be apportioned to Oklahoma. The Search for the Perfect Apportionment Method The ideal apportionment method would simultaneously satisfy the following four fairness criteria. Satisfy the quota rule Avoid the Alabama paradox Avoid the population paradox Avoid the new-states paradox We have seen that the Hamilton method allows the Alabama paradox, the population paradox, and the new-states paradox in some apportionment scenarios. Let’s explore the results of the other methods of apportionment we have discussed in some of the same scenarios. Orange Grove and the New-States Paradox The incorporated town of Orange Grove consists of two subdivisions: The Oaks with 1,254 residents and The Villages with 10,746 residents. A council with 100 members supervises the municipality’s operations. The council votes to annex an unincorporated subdivision called The Lakes with a population of 630. They plan to increase the size of the council to maintain the ratio of seats to residents such that the new council will have 100 seats plus the number of seats given to The Lakes. Use each of the following apportionment methods and indicated number of additional seats to find the original and new apportionment and determine whether the new-state paradox occurs. Jefferson’s method with five additional seats. Adams’s method with six additional seats. Webster’s method with five additional seats Using a modified divisor of 119, the original apportionment would have been: The Oaks 10 and The Villages 90. Using a modified divisor of 119, the new apportionment would be: The Oaks 10, The Villages 90, and The Lakes 5. The new-state paradox does not occur. Using a modified divisor of 121, the original apportionment would have been: The Oaks 11 and The Villages 89. Using a modified divisor of 121, the new apportionment would be: The Oaks 11, The Villages 89, and The Lakes 6. The new-state paradox does not occur. Using the standard divisor of 120, the original apportionment would have been: The Oaks 10 and The Villages 90. Using a modified divisor of 119.5, the new apportionment would be: The Oaks 10, The Villages 90, and The Lakes 5. The new-state paradox does not occur. We have seen in our examples that neither the population paradox nor the new-states paradox occurred when using the Jefferson, Adams, and Webster methods. It turns out that, although all three of these divisor methods violate the quota rule, none of them ever causes the population paradox, new-states paradox, or even the Alabama paradox. On the other hand, the Hamilton method satisfies the quota rule, but will cause the population paradox, the new-states paradox, and the Alabama paradox in some scenarios. In 1983, mathematicians Michel Balinski and Peyton Young proved that no method of apportionment can simultaneously satisfy all four fairness criteria. There are other apportionment methods that satisfy different subsets of these fairness criteria. For example, the mathematicians, Balinski and Young who proved the Balinski-Young Impossibility Theorem created a method that both satisfies the quota rule and is free of the Alabama paradox. (Balinski, Michel L.; Young, H. Peyton (November 1974). “A New Method for Congressional Apportionment.” Proceedings of the National Academy of Sciences . 71 (11): 4602–4606.) However, no method may always follow the quota rule and simultaneously be free of the population paradox. (Balinski, Michel L.; Young, H. Peyton (September 1980). "The Theory of Apportionment" (PDF). Working Papers. International Institute for Applied Systems Analysis. WP-80-131.) So, as you and your fellow founders of Imaginaria make the important decision about the right apportionment method for Imaginaria, do not look for a perfect apportionment method. Instead, look for an apportionment method that best meets the needs and concerns of Imaginarians. Check Your Understanding Key Terms apportionment paradox Alabama paradox reapportionment population growth rate population paradox new-state paradox Key Concepts Several surprising outcomes can occur when apportioning seats that voters may find unfair: Alabama paradox, population paradox, and new-state paradox. Apportionment methods are susceptible to apportionment paradoxes and may violate the quota rule. The Balinsky-Young Impossibility Theorem indicates that no apportionment can satisfy all fairness criteria. Formulas population growth rate = current population size - previous population size previous population size New House Size = New Population Original Standard Divisor rounded to the nearest whole number. Projects The First Census: Challenges of Collecting Population Data As you and your fellow founders write the Imaginarian constitution, you must also design systems to collect accurate information about the population of Imaginaria. Why is this important and what are the challenges you will face? Let’s find out! Research and answer each question. (Questions adapted from “Authorizing the First Census–The Significance of Population Data,” Census.gov, United States Census Bureau) The First Congress laid out a plan for collecting this data in Chapter II, An Act Providing for the Enumeration of the Inhabitants of the United States , which was approved March 1, 1790. What group did Congress select to carry out the first enumeration? Why did they choose this group? What might be the advantages and disadvantages to this approach? Choose at least two challenges faced by the U.S. government during the first enumeration and explain how the information gathered might have helped address them. President James Madison was a U.S. Representative who participated in the census debate in the first congress. What were the main points of his remarks and how were they relevant to the overall debate about the first enumeration? What issues do you think the U.S. Census Bureau encounters today as it continues to collect and process data about the U.S. population that might be significant to you and the other founders of Imaginaria? The Party System: How Many Political Parties Are Enough? The electoral system of Imaginaria will likely involve multiple political parties. The way these parties interact with the system may be determined by the founders of the new democracy. Let’s explore the ways in which political parties interact with the governments of various democracies by researching and answering each of the following questions. What concerns did the founders of the United States have about political parties? Have any of their concerns become a reality? Were political parties addressed in the U.S. Constitution? How did they become such a large part of politics in the U.S.? As a founder of Imaginaria, would you address political parties in your constitution? What is frontloading? Does our current system of frontloading impact fair representation? Why do two small, racially homogeneous states hold their primaries first? Do you think this impacts the final results? How does the interaction political parties and the electoral system in the United States differ from that of other countries? Give at least three specific examples. Of these three, which would you be most likely to use as a model for Imaginaria? Technology and Voting: How Has the Digital Age Impacted Elections? Unlike most other countries, Imaginaria will be founded in the digital age. Let’s explore the impact this might have on how you choose to set up your electoral system. Research and answer each question. Participation in an electoral system if very important. In what ways does the Internet positively affect participation? In what ways does it negatively affect participation. What roll, if any, should government of Imaginaria play in tempering or promoting the Impact of the Internet? Find at least three examples of governments that utilize Internet voting around the world. What concerns have slowed the spread of this technology? What are the advantages and disadvantages of Internet voting? Would you be in favor of Internet voting for Imaginaria? Why or why not? Chapter Review Voting Methods Fairness in Voting Methods Standard Divisors, Standard Quotas, and the Apportionment Problem Apportionment Methods Fairness in Apportionment Methods Chapter Test

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Introduction Networks connect cities around the globe. (credit: "Globalization" by faith.e.murphy/Flickr, Public Domain) In this chapter, you will learn the fundamental skills needed to work with graphs used in an area of mathematics known as graph theory. You can think of these graphs as a kind of map. Maps have served many purposes over the course of history. You probably use GPS maps to navigate to various destinations. A scientist from ancient Greece named Ptolemy wanted an accurate map of the world to make more accurate astrological predications. In recent years, neurobiologists have mapped the cerebral cortex to better understand the human brain. Social network analysts map online interactions to assist advertisers in reaching target audiences. Like other maps, the graphs you will study in this chapter can serve many purposes, but they do not have a lot of the details you might expect in a map such as size, shape, and distance between objects. All of that is stripped away so that we can focus on one element of maps, the connections between objects.

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Graph Basics Cell phone networks connect individuals. (credit: "Business people using their phones" by Rawpixel Ltd./Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify parts of a graph. Model applications of graph basics. When you hear the word, graph , what comes to mind? You might think of the xy -coordinate system you learned about earlier in this course, or you might think of the line graphs and bar charts that are used to display data in news reports. The graphs we discuss in this chapter are probably very different from what you think of as a graph. They look like a bunch of dots connected by short line segments. The dots represent a group of objects and the line segments represent the connections, or relationships, between them. The objects might be bus stops, computers, Snapchat accounts, family members, or any other objects that have direct connections to each other. The connections can be physical or virtual, formal or casual, scientific or social. Regardless of the kind of connections, the purpose of the graph is to help us visualize the structure of the entire network to better understand the interactions of the objects within it. Parts of a Graph In a graph , the objects are represented with dots and their connections are represented with lines like those in . displays a simple graph labeled G and a multigraph labeled H . The dots are called vertices ; an individual dot is a vertex , which is one object of a set of objects, some of which may be connected. We often label vertices with letters. For example, Graph G has vertices a, b, c , and d , and Multigraph H has vertices, e, f, g , and h . Each line segment or connection joining two vertices is referred to as an edge . H is considered a multigraph because it has a double edge between f and h , and a double edge between h and g . Another reason H is called a multigraph is that it has a loop connecting vertex e to itself; a loop is an edge that joins a vertex to itself. Loops and double edges are not allowed in a simple graph. To sum up, a simple graph is a collection of vertices and any edges that may connect them, such that every edge connects two vertices with no loops and no two vertices are joined by more than one edge. A multigraph is a graph in which there may be loops or pairs of vertices that are joined by more than one edge. In this chapter, most of our work will be with simple graphs, which we will call graphs for convenience. A Graph and a Multigraph It is not necessary for the edges in a graph to be straight. In fact, you can draw an edge any way you want. In graph theory, the focus is on which vertices are connected, not how the connections are drawn (see ). In a graph, each edge can be named by the two letters of the associated vertices. The four edges in Graph X in are ab, ac, ad , and ae . The order of the letters is not important when you name the edge of a graph. For example, ab refers to the same edge as ba . A graph may have vertices that are not joined to other vertices by edges, such as vertex f in Graph X in , but any edge must have a vertex at each end. Different Representations of the Same Graph Identifying Edges and Vertices Name all the vertices and edges of graph F in . Graph F The vertices are v, w, x, y , and z . The edges are vw, vx, wx, wz, xy , and xz . When listing the vertices and edges in a graph, work in alphabetical order to avoid accidentally listing the same item twice. When you are finished, count the number of vertices or edges you listed and compare that to the number of vertices or edges on the graph to ensure you didn’t miss any. Since the purpose of a graph is to represent the connections between objects, it is very important to know if two vertices share a common edge. The two vertices at either end of a given edge are referred to as neighboring , or adjacent . For example, in , vertices x and w are adjacent, but vertices y and w are not. Identifying Vertices That Are Not Adjacent Name all the pairs of vertices of graph F in that are not adjacent. The pairs of vertices that are not adjacent in graph F are v and y, v and z, w and y , and y and z . Sergey Brin and Laurence Page The “Google boys,” Sergey Brin and Laurence Page, transformed the World Wide Web in 1998 when they used the mathematics of graph theory to create an algorithm called Page Rank, which is known as the Google Search Engine today. The two computer scientists identified webpages as vertices and hyperlinks on those pages as edges because hyperlinks connect one website to the next. The number of edges influences the ranking of a website on the Google Search Engine because the websites with more links to “credible sources” are ranked higher. ("Page Rank: The Graph Theory-based Backbone of Google," September 20, 2011, Cornell University, Networks Blog. Analyzing Geographical Maps with Graphs Commercial airlines' route systems create a global network. When graphs are used to model and analyze real-world applications, the number of edges that meet at a particular vertex is important. For example, a graph may represent the direct flight connections for a particular airport as in . Representing the connections with a graph rather than a map shifts the focus away from the relative positions and toward which airports are connected. In , the vertices are the airports, and the edges are the direct flight paths. The number of flight connections between a particular airport and other South Florida airports is the number of edges meeting at a particular vertex. For example, Key West has direct flights to three of the five airports on the graph. In graph theory terms, we would say that vertex FYW has degree 3. The degree of a vertex is the number of edges that connect to that vertex. Direct Flights between South Florida Airports Determining the Degree of a Vertex Determine the degree of each vertex of Graph J in . If graph J represents direct flights between a set of airports, do any of the airports have direct flights to two or more of the other cities on the graph? Graph J For each vertex, count the number of edges that meet at that vertex. This value is the degree of the vertex. In , the dashed edges indicate the edges that meet at the marked vertex. Degrees of Vertices of Graph J Vertex a has degree 3, vertex b has degree 1, vertices c and d each have degree 2, and vertex e has degree 0. Airports a, c, and d have direct flights to two or more of the other airports. Graphs are also used to analyze regional boundaries. The states of Utah, Colorado, Arizona, and New Mexico all meet at a single point known as the “Four Corners,” which is shown in the map in . Map of the Four Corners In , each vertex represents one of these states, and each edge represents a shared border. States like Utah and New Mexico that meet at only a single point are not considered to have a shared border. By representing this map as a graph, where the connections are shared borders, we shift our perspective from physical attributes such as shape, size and distance, toward the existence of the relationship of having a shared boundary. Graph of the Shared Boundaries of Four Corners States Graphing the Midwestern States A map of the Midwest is given in . Create a graph of the region in which each vertex represents a state and each edge represents a shared border. Map of Midwestern States Step 1: For each state, draw and label a vertex as in . Vertex Assigned to Each Midwestern State Step 2: Draw edges between any two states that share a common land border as in . Edge Assigned to Each Pair of Midwestern States with Common Border The graph is given in . Final Graph Representing Common Boundaries between Midwestern States Graph Theory: Create a Graph to Represent Common Boundaries on a Map Graphs of Social Interactions Geographical maps are just one of many real-world scenarios which graphs can depict. Any scenario in which objects are connected to each other can be represented with a graph, and the connections don’t have to be physical. Just think about all the connections you have to people around the world through social media! Who is in your network of Twitter followers? Whose Snapchat network are you connected to? Graphing Chloe’s Roblox Friends Roblox is an online gaming platform. Chloe is interested to know how many people in her network of Roblox friends are also friends with each other so she polls them. Explain how a graph or multigraph might be drawn to model this scenario by identifying the objects that could be represented by vertices and the connections that could be represented by edges. Indicate whether a graph or a multigraph would be a better model. The objects that are represented with vertices are Roblox friends. A Roblox friendship between two friends will be represented as an edge between a pair of vertices. There will be no double edges because it is not possible for two friends to be linked twice in Roblox ; they are either friends or they are not. Also, a player cannot be a friend to themself, so there is no need for a loop. Since there are no double edges or loops, this is best represented as a graph. Using Graph Theory to Reduce Internet Fraud Could graphs be used to reduce Internet fraud? At least one researcher thinks so. Graph theory is used every day to analyze our behavior, particularly on social network sites. Alex Buetel, a computer scientist from Carnegie Mellon University in Pittsburgh, Pennsylvania, published a research paper in 2016 that discussed the possibilities of distinguishing the normal interactions from those that might be fraudulent using graph theory. Buetel wrote, “To more effectively model and detect abnormal behavior, we model how fraudsters work, catching previously undetected fraud on Facebook, Twitter, and Tencent Weibo and improving classification accuracy by up to 68%.” In the same paper, the researcher discusses how similar techniques can be used to model many other applications and even, “predict why you like a particular movie.” (Alex Beutel, "User Behavior Modeling with Large-Scale Graph Analysis," http://reports-archive.adm.cs.cmu.edu/anon/2016/CMU-CS-16-105.pdf, May 2016, CMU-CS-16-105, Computer Science Department, School of Computer Science, Carnegie Mellon University, Pittsburgh, PA) Check Your Understanding Key Terms vertex edge loop graph (simple graph) multigraph adjacent (neighboring) degree Key Concepts Graphs and multigraphs represent objects as vertices and the relationships between the objects as edges. The degree of a vertex is the number of edges that meet it and the degree can be zero. An edge must have a vertex at each end. Multigraphs may contain loops and double edges, but simple graphs may not. Videos Graph Theory: Create a Graph to Represent Common Boundaries on a Map

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Graph Structures Neuroimaging shows brain activity. (credit: "MRI Scan" by NIH Image Gallery/Flickr, Public Domain) Learning Objectives After completing this section, you should be able to: Describe and interpret relationships in graphs. Model relationships with graphs. Graph theory is used in neuroscience to study how different parts of the brain connect. Neurobiologists use functional magnetic resonance imaging (fMRI) to measure levels of blood in different parts of the brain, called nodes. When nodes are active at the same time, it suggests there is a functional connection between them so they form a network. This network can be represented as a graph where the vertices are the nodes and the functional connections are the edges between them. (Mikey Taylor, "Graph Theory & Machine Learning in Neuroscience," Medium.com, June 24, 2020. Importance of the Degrees of Vertices One reason scientists study these networks is to determine how successful the communication within a network continues to be when it experiences failures in nodes and connections. Graphs can be used to study the resilience of these networks. (Mikey Taylor, "Graph Theory & Machine Learning in Neuroscience," Medium.com, June 24, 2020) Using Graphs to Understand Relationships The graphs in represent neural networks, where the vertices are the nodes, and the edges represent functional connections between them. Which graph do you think would represent a network with the highest resistance to failure? Which graph would probably be the most vulnerable to failure? How might this relate to the degrees of the vertices? Models of Neural Networks Graph B represents a network that would have the highest resistance to failure because there are more edges connecting the nodes indicating there are more connections between nodes than on either of the other graphs. This would make the network more resistant to failure because there are more options to work around a faulty edge or node. Graph A would probably be the most susceptible to failure because the network depends on a few particular edges and vertices for connections. There are more vertices of higher degree in Graph B than in Graph A because there are more edges connecting the nodes. Relating the Number of Edges to the Degrees of Vertices In the applications of graph theory to neuroscience and sociology in and YOUR TURN 12.6 , there was a correlation between the degrees of vertices and the resilience of a network. Researchers in many fields have also observed a direct relationship between the number of edges in a graph and the degrees of the vertices. To begin to understand this relationship, consider a graph with five vertices and zero edges as in . Graph with Five Vertices and zero Edges Instead of being marked with a name, each vertex in is marked with its degree. In this case, all of the degrees are 0 so the sum of the degrees is also zero. Suppose that we add an edge between any two existing vertices and indicate the degrees of the vertices. This gives us a graph with five vertices and one edge like the graph in . Graph with Five Vertices and One Edge Note that the degrees of two vertices increased, each by 1. So, the sum of the degrees is now 2. Suppose that we continue in this way, adding one edge at a time and making note of the number of edges and the sum of the degrees of the vertices as in . Comparing Number of Edges to Sum of Degrees demonstrates a characteristic that is true of all graphs of any shape or size. When the number of edges is increased by one, the sum of the degrees increases by two. This happens because each edge has two ends and each end increases the degree of one vertex by one unit. As a result, the sum of the degrees of the vertices on any graph is always twice the number of edges. This relationship is known as the Sum of Degrees Theorem . For the Sum of Degrees Theorem, sum of the degrees = 2 × number of edges or number of edges = sum of degrees 2 Finding the Sum of Degrees Suppose that a graph has five edges. Find the sum of the degrees of the vertices. Draw two different graphs that demonstrate this conclusion. The sum of the degrees is twice the number of edges: 2 × 5=10. The sum of the degrees is 10. See . Completeness Suppose that there were five strangers in a room, A , B , C , D , and E , and each one would be introduced to each of the others. How many introductions are necessary? One way to begin to answer this question is to draw a graph with each vertex representing an individual in the room and each edge representing an introduction as in . Model of Introductions between Five Strangers Let’s approach the problem by thinking about how many new people Person A would meet, then Person B , and so on, making sure not to repeat any introductions. The first graph in shows Person A meeting Persons B , C , D , and E , for a total of 4 introductions. The next graph shows that Person B still has to meet Persons C , D , and E , for a total of 3 more introductions. The next graph shows that Person C still has to meet Persons D and E , which is 2 more introductions. The next graph shows that Person D only remains to meet Person E , which is 1 more introduction. The final graph has 4 + 3 + 2 + 1 = 10 edges representing 10 introductions. The last graph is an example of a complete graph because each pair of vertices is joined by an edge. Another way of saying this is that the graph is complete because each vertex is adjacent to every other vertex. shows complete graphs with three, four, five, and six vertices. Complete Graphs with Up to Six Vertices Suppose we want to know the number of introductions necessary in a room with six people. This would be represented by a complete graph with six vertices, and the total number of introductions would be 5 + 4 + 3 + 2 + 1 = 15 , the number of edges in the graph. In fact, you can always find the number of introductions in a room with n people by adding all the whole numbers from 1 to n − 1 . The number of edges in a complete graph with n vertices is the sum of the whole numbers from 1 to n − 1 , 1 + 2 + 3 + ⋯ + ( n − 1 ) . Suppose that we want to determine how many introductions are necessary in a room with 500 strangers. In other words, suppose that we want to determine the number of edges in a complete graph with 500 vertices. Adding up all the numbers from 1 to 499 could take a long time! In the next example, we use the Sum of Degrees Theorem to make the problem more manageable. Using the Sum of Degrees Theorem Use the Sum of Degrees Theorem to determine the number of introductions required in a room with 6 strangers. 10 strangers. n strangers. Since there are 6 strangers, there are 6 vertices. Since each individual must meet 5 other individuals, there are 5 edges meeting at each vertex which means each vertex has degree 5. Since there are 6 vertices of degree 5, the sum of degrees is 6 ⋅ 5 = 30 . By the Sum of Degrees Theorem, the number of edges is half the sum of the degrees, which is 30 2 = 15 . So, the total number of introductions is 15. Since there are 10 strangers, there are 10 vertices. Since each individual must meet 9 other individuals, there are 9 edges meeting at each vertex which means each vertex has degree 9. Since there are 10 vertices of degree 9, the sum of degrees is 10 ⋅ 9 = 90 . By the Sum of Degrees Theorem, the number of edges is half the sum of the degrees, which is 90 2 = 45 . So, the total number of introductions is 45. Since there are n strangers, there are n vertices. Since each individual must meet n − 1 other individuals, there are n − 1 edges meeting at each vertex which means each vertex has degree n − 1 . Since there are n vertices of degree n − 1 , the sum of degrees is n ( n − 1 ) . By the Sum of Degrees Theorem, the number of edges is half the sum of the degrees, which is n ( n − 1 ) 2 . So, the total number of introductions is n ( n − 1 ) 2 . Now we have a shorter way to calculate the number of introductions in a room with n strangers, and the number of edges on a complete graph with n vertices. Let’s update our formula. The number of edges in a complete graph with n vertices is 1 + 2 + 3 + ⋯ + ( n − 1 ) = n ( n − 1 ) 2 . Subgraphs Sometimes a graph is a part of a larger graph. For example, the graph of South Florida Airports from is part of a larger graph that includes Orlando International Airport in Central Florida, which is shown in Orlando and South Florida Airports The graph in includes an additional vertex, MCO, and additional edges shown with dashed lines. The graph of direct flights between South Florida airports from is called a subgraph of the graph that also includes direct flights between Orlando and the same South Florida airports in . In general terms, if Graph B consists entirely of a set of edges and vertices from a larger Graph A , then B is called a subgraph of A . Identifying a Subgraph In , Graph G is given, along with four diagrams. Determine whether each diagram is or is not a subgraph of Graph G and explain why. Graph G and Potential Subgraphs Diagram J is not a subgraph of Graph G because edge ec is not in Graph G . Diagram K is a subgraph of Graph G because all of its vertices and edges were part of Graph G . Diagram L is not a graph at all, because there is a line extending from vertex a that does not connect a to another vertex. So, Diagram L cannot be a subgraph. Diagram M is not a subgraph of Graph G because it contains a vertex f , which is not part of G . Identifying and Naming Cycles The water cycle begins and ends with water. (credit: "Diagram of the water cycle" by NASA, Public Domain) When you think of a cycle in everyday life, you probably think of something that begins and ends the same way. For example, the water cycle ( ) begins with water in a lake or ocean, which evaporates into water vapor, condenses into clouds, and then returns to earth as rain or some other form of precipitation that settles into lakes or oceans. A cycle in graph theory is similar in that it begins and ends in the same way: It is a series of connected edges that begin and end at the same vertex but otherwise never repeat any vertices. In a cycle, there are always the same number of vertices as edges, and all vertices must be of degree 2. Cycles are often referred to by the number of vertices. For example, a cycle with 5 vertices can be called a 5-cycle. Cycles can also be named after polygons based on the number of edges. For example a 5-cycle is also called a pentagon. lists these names for cycles of size 3 through size 10. Cycle Category Number of Edges Example triangle 3 quadrilateral 4 pentagon 5 hexagon 6 heptagon 7 octagon 8 nonagon 9 decagon 10 Categories of Cycles There are many more polygon names, including a megagon that has a million edges and a googolgon that has 10 100 edges, but usually we just say n -gon when the number n is past 10. For example, a cycle with 11 edges could be called an 11-gon. Notice that the 10-cycle, or decagon, appears to cross over itself in . Remember, graphs can be drawn differently but represent the same connections. In , the same decagon is transformed into a graph that does not appear to overlap itself. We have done this without changing any of the connections so both diagrams represent the same relationships, and both diagrams are considered decagons. Transformation of a Decagon Googol to Google Did the name "googolgon" ring a bell for you? If so, there is a good reason for it. The creators of Google.com renamed their search engine Google, a misspelled reference to the number "googol" alluding to the enormous number of calculations their algorithm can tackle. A googol is a 1 followed by 100 zeroes, or 10 100 . (William L. Hosch, "Google, American Company," https://www.britannica.com/topic/Google-Inc, Encyclopedia Britannica online) Cyclic Subgraphs and Cliques When cycles appear as subgraphs within a larger graph, they are called cyclic subgraphs . Cyclic subgraphs are named by listing their vertices sequentially. The vertex where you begin is not important. Graph K in has two triangle cycles ( g , h , j ) and ( h , i , j ), and one quadrilateral cycle ( g , h , i , j ). Cycles in Graph K The same cycle can be named in many ways, but we must keep in mind that listing two vertices consecutively indicates an edge exists between them. For example, (g, h, i, j) can also be named (h, i, j, g) or (j, i, h, g). However, you cannot name it (g, i, h, j,) which doesn't reflect the correct order of the vertices. We can see that the order (g, i, h, j) is incorrect because the cycle has no edges gi or hj. Identifying Cyclic Subgraphs Identify the types of cyclic subgraphs in Graph H in and name them. Graph H In order to avoid missing a cycle, begin by analyzing the vertex a , then proceed in alphabetical order. Vertex a is part of two cycles, the quadrilateral cycle ( a , b , f , g ,) and the hexagonal cycle ( a , b , c , d , e , f ). Next look at vertex b . It is part of the hexagonal cycle ( b , c , d , e , f , g ). After checking each of the remaining vertices, we see there are no other cycles. In order to avoid naming the same cycle twice, consider naming cycles beginning with the letter closest to the beginning of the alphabet and then progressing clockwise. We have seen that sociologists use graphs to study the structures of social networks. In sociology, there is a principle known as Triadic Closure. It says that if two individuals in a social network have a friend in common, then it is more likely those two individuals will become friends too. Sociologists refer to this as an open triad becoming a closed triad. This concept can be visualized as graphs in . (Chakraborty, Dutta, Mondal, and Nath, "Application of Graph Theory in Social Media, International Journal of Computer Sciences and Engineering , 6(10):722-729) In the open triad in , person a and person b each has a friendship with person c . In the closed triad, person a and person b have also developed a friendship. Notice that the graph representing the closed triad is a three-cycle, or a triangle, in graph theory terms. Graphs of Open Triad and Closed Triad Another topic of interest to sociologists, as well as computer scientists and scientists in many fields, is the concept of a clique . In a social group, a clique is a subgroup who are all friends. A triad is an example of a clique with three people, but there can be cliques of any size. In graph theory, a clique is a complete subgraph. Identifying Triads and Cliques The graphs in YOUR TURN 12.6 represent social communities. The vertices are individuals and the edges are social connections. Use the graphs to answer the questions. Which graph has the most triads? Name the triangles. Which graph has the fewest triads? Name the triangles. How might an increase in the number of triads in a graph affect the resilience of a community? Explain your reasoning. Identify a clique with more than three vertices in Graph E by naming its vertices. Graph E has the most triads. The triangles are ( a, b, c ), ( a, c, e ), ( a, c, f ), ( a, c, g ), ( a, e, g ), ( c, e, g ), ( e, g, h ), ( e, h, i ), and ( f, g, i ). Graph D has the fewest triads. There are no triangles in the graph. More triads means vertices of greater degree, which indicates greater resilience. Specifically, if an edge is removed from a closed triad, there the two individuals who are no longer adjacent are still connected by way of the third member of the triad. The vertices a, e, c, and g form a clique with four vertices. Three-Cycles in Complete Graphs Just as complete graphs have a predictable number of edges, complete graphs have a predictable number of cyclic subgraphs. Let’s look at the three-cycles within complete graphs with up to six vertices, which are shown in . Complete Graphs with Up to Six Vertices Let's list the names of all the triangles in each graph. Since every pair of vertices is adjacent, any three vertices on a complete graph form a triangle . There is only one triangle in the complete graph with three vertices, ( a , b , c ). For the rest of the graphs, it is important that we take an organized approach. Start with the vertex that is first alphabetically, listing any triangles that include that vertex also in alphabetical order. Then, proceed to the next vertex in the alphabet, and list any triangles that include that vertex, except those that are already listed. Keep going in this way as shown in . Complete Graph With: 3 Vertices 4 Vertices 5 Vertices 6 Vertices All a triangles ( a , b , c ) ( a, b, c ), ( a, b, d ), ( a, c, d ) ( a, b, c ), ( a, b, d ), ( a, b, e ) ( a, c, d ), ( a, c, e ) ( a, d, e ) ( a, b, c ), ( a, b, d ), ( a, b, e ), ( a, b, f ) ( a, c, d ), ( a, c, e ), ( a, c, f ) ( a, d, e ), ( a, d, f ) ( a, e, f ) Other b triangles None ( b, c, d ) ( b, c, d ), ( b, c, e ), ( b, d, e ) ( b, c, d ), ( b, c, e ), ( b, c, f ) ( b, d, e ), ( b, d, f ) ( b, e, f ) Other c triangles None None ( c, d, e ) ( c, d, e ), ( c, d, f ) ( c, e, f ) Other d triangles None None None ( d, e, f ) Total 1 ( 2 + 1 ) + 1 = 3 + 1 = 4 ( 3 + 2 + 1 ) + ( 2 + 1 ) + 1 = 6 + 3 + 1 = 10 ( 4 + 3 + 2 + 1 ) + ( 3 + 2 + 1 ) + ( 2 + 1 ) + 1 = 10 + 6 + 3 + 1 = 20 Listing Triangles in Complete Graphs Look at the last row in . Do you see a pattern emerge for counting triangles in a complete graph? Without drawing a complete graph with 7 vertices, we can predict that it will have ( 5 + 4 + 3 + 2 + 1 ) + ( 4 + 3 + 2 + 1 ) + ( 3 + 2 + 1 ) + ( 2 + 1 ) + 1 = 35 triangles inside it. This pattern also appears in a famous diagram known to Western mathematicians as "Pascal’s Triangle." displays the first 11 rows of Pascal’s Triangle. Row 0 of Pascal’s Triangle only has the number 1 in it. The first and last entries in each of the other rows are also 1s. Otherwise, all the entries are formed by adding two entries from the previous row. For example, in row 6, entry 1 is 6, which was found by adding 1 and the 5 from the previous row, and entry 2 is 15, which was found by adding the 5 and the 10 from the previous row, as shown in . Pascal’s Triangle IMPORTANT! When you count the rows and entries of Pascal’s Triangle begin with row 0 and entry 0, not row 1 and entry 1. The Mathematical Secrets of Pascal's Triangle by Wajdi Mohamed Ratemi If we begin to list the third entries in each row of Pascal's triangle from the top down, we see 1, 4, 10, 20, 35, and so on. Notice that these values are exactly the number of triangles in a complete graph of sizes 3, 4, 5, 6, and 7, respectively. Specifically, entry 3 in row 7 is 35, the number of triangles in a complete graph of size 7. Let's practice using Pascal’s triangle to find the number of triangles in a complete graph of a particular size. STEPS TO FIND THE NUMBER OF TRIANGLES IN A COMPLETE GRAPH OF SIZE n Step 1 Calculate the entries in row n of Pascal's Triangle using sums of pairs of entries in previous rows as needed. Step 2 The number of triangles is entry 3 in row n . Remember to count the entries 0, 1, 2, and 3, from left to right. Using Pascal’s Triangle to Find the Number of Triangles in a Complete Graph Use Pascal’s Triangle in to find the number of triangles in a complete graph with 11 vertices. Step 1: Identify the row number and calculate its entries. Since the complete graph has 11 vertices, we will need to look in row 11 of Pascal’s Triangle. Use row 10 in to find the entries in row 11 as shown in . Rows 10 and 11 of Pascal’s Triangle Step 2: Find entry number 3. The number of triangles is in entry 3. Remember to count the entries beginning with entry 0 as shown in . Identifying an Entry in Pascal’s Triangle Entry 3 in row 11 is 165. So, there are 165 triangles in a complete graph with 11 vertices. Since the entries in Pascal’s Triangle are useful in many applications, many resources such as online and traditional calculator functions have been developed to assist in calculating the values. The website dCode.xyz has a tool that automatically calculates entries in Pascal’s Triangle using these steps: Step 1: Make sure to select Index 0. Step 2: Select your preference, display the triangle until a particular row (line), display only one row (line), or calculate the value at coordinates (which means give the value of a single entry). Step 3: Enter row number (line number) you are interested, or enter the row number and entry number in parentheses: (row number, entry number). Step 4: Select Calculate. Step 5: Retrieve the answer from the results box. Pascal's Triangle Tool on dCode.xyz Check Your Understanding Key Terms complete subgraph cycle cyclic subgraph clique Key Concepts The sum of the degrees of the vertices in a graph is twice the number of edges. In a complete graph every pair of vertices is adjacent. A subgraph is part of a larger graph. Cycles are a sequence of connected vertices that begin and end at the same vertex but never visit any vertex twice. Formulas For the Sum of Degrees Theorem, sum of the degrees = 2 × number of edges or number of edges = sum of degrees 2 The number of edges in a complete graph with n vertices is the sum of the whole numbers from 1 to n − 1 , 1 + 2 + 3 + ⋯ + ( n − 1 ) . The number of edges in a complete graph with n vertices is 1 + 2 + 3 + ⋯ + ( n − 1 ) = n ( n − 1 ) 2 . Videos The Mathematical Secrets of Pascal's Triangle by Wajdi Mohamed Ratemi

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