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by: Amy Brogan

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# Week 12: Apportionment Part 3 and Cryptography Part 1 MATH 1014

Marketplace > University of Cincinnati > Mathematics (M) > MATH 1014 > Week 12 Apportionment Part 3 and Cryptography Part 1
Amy Brogan
UC
GPA 3.7

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## About this Document

Introduction of the Hill-Huntington Method in apportionment and a review. Introduction to Cryptography with the Caesar Shift Cipher and Vignere Cipher.
COURSE
Mathematics of Social Choice
PROF.
Mary Koshar
TYPE
Class Notes
PAGES
9
WORDS
CONCEPTS
Apportionment; Hill-Huntington; Cryptography; Caesar Cipher; Vignere Cipher
KARMA
25 ?

## Popular in Mathematics (M)

This 9 page Class Notes was uploaded by Amy Brogan on Saturday April 2, 2016. The Class Notes belongs to MATH 1014 at University of Cincinnati taught by Mary Koshar in Spring 2016. Since its upload, it has received 11 views. For similar materials see Mathematics of Social Choice in Mathematics (M) at University of Cincinnati.

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## Reviews for Week 12: Apportionment Part 3 and Cryptography Part 1

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Date Created: 04/02/16
Koshar Amy Brogan March 28,30 & April 1, 2016 Apportionment Part 3 Review: Spread 10 calculators among 4 classes by population using the following methods. Class A: 87 Class B: 45 Class C: 96 Class D: 62 Hamilton Method Class A Class B Class C Class D Jefferson Method Class A Class B Class C Class D Webster Method Class A Class B Class C Class D Write out the results in full sentences. (Answers at the end of the Notes) Hill-Huntington Method  Arithmetic Mean o The product of two numbers that have been added together; the average o (a*b)/2  Geometric Mean o The square root of the product of two numbers o √ ???? ∗ ???? In the Hill-Huntington Method, the quotas are rounded by the Geometric Mean instead of the usual way. When the quota is a decimal, for example 3.14, then we would take the square root of 3*4 to see if the amount would round up or down. The Geometric Mean is usually close to the .5 value, but the smaller the numbers the further the distance from the midpoint, such as in the table below. Numbers Usual round limit Geometric Mean Equation Geometric Cut-Off (Approximate) 1-2 1.5 √1 ∗ 2 1.41 2-3 2.5 √2 ∗ 3 2.45 5-6 5.5 √5 ∗ 6 5.48 10-11 10.5 √10 ∗ 11 10.488 50-51 50.5 50.498 √50 ∗ 51 100-101 100.5 √100 ∗ 101 100.499 1000-1001 1000.5 √ 1000 ∗ 1001 1000.499875 5000-5001 5000.5 5000.499975 √5000 ∗ 5001 The Geometric Mean values will stay the same, but the quota they are being compared to will either round up or down depending on the values involved. Ex: Quota values 1.521, 2.461, and 10.433 Quota Usual Round Limit Geometric Limit Round up/down? 1.521 1.5 1.41 Up 2.461 2.5 2.45 Up 10.433 10.5 10.488 Down The Standard devisor and Quota are reached the same as in the earlier methods, but the way we round up or down is different. Once we down round, the standard devisor is altered like in the Jefferson and Webster methods to meet the number of items being divided out. Use the front page’s example with this method and write your answer in a full sentence: Hill-Huntington Quota Geometric Mean Rounded Value End Result Method Class A: 87 3 Class B: 45 1.55 Class C: 96 3.31 Class D: 62 2.14 Total Pop: 290 ______/10 ______/10 SD: 290/10= 29 Check your answers: Quota Geometric Mean Rounded Value End Result Hill-Huntington Method (Approximate) Class A: 87 3 3 3 Class B: 45 1.55 2 2 √ 1 ∗ 2 = 1.41 Class C: 96 3.31 √ 3 ∗ 4 = 3.46 3 3 Class D: 62 2.14 √ 2 ∗ 3 = 2.45 2 2 Total Pop: 290 10 / 10 SD: 290/10= 29 Class A will receive 3 calculators, Class B will receive 2 calculators, Class C will receive 3 calculators, and Class D will receive 2 calculators. So which method is the best? It has been agreed that the best should:  Be unbiased by size o The Jefferson Method favors larger populations (probably due to his state being the largest one in the continental congress at the time of creation).  Minimize difference in representative shares o (apportionment/population) o Using this ^ equation, the Hill-Huntington Method shows to much of a difference between groups, making end results uneven  There should not be a loss of representatives for a group when the number of available seats increases o Hamilton Paradox For general use, the Webster Method is the best. Practice 1: With the following example, use each of the methods we have learned thus far: Spread 12 managers among 3 new plants by number of employees. Plant 1: 93 Plant 2: 76 Plant 3: 145 Review Answers Spread 10 calculators among 4 classes by population using the following methods. Class A: 87 Class B: 45 Class C: 96 Class D: 62 Total Population: 290 Standard Devisor: 290/10 = 29 Hamilton Quota Round Down Add/Upgrade End Result Method Class A 3 3 3 Class B 1.55 1 +1 2 Class C 3.31 3 3 Class D 2.138 2 2 Total 9/10 10 / 10 By the Hamilton Method, Class A will receive 3 calculators, Class B will receive 2 calculators, Class C will receive 3 calculators, and Class D will receive 2 calculators. Jefferson Quota Round Down New Standard Round Down End Result Method Devisor- Ex: 23 Class A 3 3 3.78 3 3 Class B 1.55 1 1.957 1 1 Class C 3.31 3 4.17 4 4 Class D 2.138 2 2.696 2 2 9/10 10 / 10 By the Jefferson Method, Class A will receive 3 calculators, Class B will receive 1 calculator, Class C will receive 4 calculators, and Class D will receive 2 calculators. Webster Quota Round Up Final Result Method Class A 3 3 3 Class B 1.55 2 2 Class C 3.31 3 3 Class D 2.138 2 2 10 / 10 By the Webster Method, Class A will receive 3 calculators, Class B will receive 2 calculators, Class C will receive 3 calculators, and Class D will receive 2 calculators. Because the Hamilton and Webster Method results agree, Class A will receive 3 calculators, Class B will receive 2 calculators, Class C will receive 3 calculators, and Class D will receive 2 calculators. Practice 1 Answers Hamilton Quota Round Down Add/Upgrade End Result Method Plant 1: 93 3.555 3 +1 4 Plant 2: 76 2.905 2 +1 3 Plant 3: 145 5.543 5 5 Total: 314 12/12 314/12 = 26.166 By the Hamilton Method, Plant 1 will receive 4 managers, Plant 2 will receive 3, and Plant 3 will receive 5 managers. Jefferson Quota Adjusted Standard Round End Result Devisor: 24 Method Plant 1: 93 3.555 3.875 3 3 Plant 2: 76 2.905 3.166 3 3 Plant 3: 145 5.543 6.041 6 6 Total: 314 12/12 314/12 = 26.166 By the Jefferson Method, Plant 1 will receive 3 managers, Plant 2 will receive 3, and Plant 3 will receive 6 managers. Quota Round up Adjust: Round up End Result Webster Method 26.5 Plant 1: 93 3.555 4 3.509 4 4 Plant 2: 76 2.905 3 2.867 3 3 Plant 3: 145 5.543 6 5.471 5 5 Total: 314 13/12 12/12 314/12 = 26.166 By the Webster Method, Plant 1 will receive 4 managers, Plant 2 will receive 3, and Plant 3 will receive 5 managers. (Same as Hamilton) Hill-Huntington Quota Geometric Mean Rounded Adjusted Value: Rounded End Result Method (Approximate) Value 28 Values Plant 1: 93 3.555 3 ∗ 4 = 3.46 4 3.32 4 4 √ Plant 2: 76 2.905 √2 ∗ 3 = 2.45 3 2.71 3 3 Plant 3: 145 5.543 √5 ∗ 6 = 5.48 6 5.17 5 5 Total: 314 13/12 12/12 314/12 = 26.166 By the Hill-Huntington Method, Plant 1 will receive 4 managers, Plant 2 will receive 3, and Plant 3 will receive 5 managers. (Same as Hamilton and Webster methods.) Since three methods agree on answers, Plant 1 will receive 4 managers, Plant 2 will receive 3, and Plant 3 will receive 5 managers. Koshar Amy Brogan April 1, 2016 Week 12: Cryptography – Part 1 Why would a person ever need to send a secret message? Some possible answers are because the information is crucial to the military, a business deal needs to be known to only a few, academic findings need to be protected from being stolen, and electronic information (like bank statements and social security) also needs to be protected. How would we keep this information protected? By coding it, or Cryptography (Kripto = hidden/secret, -graphy = writing). Coding could be anything from the binary coding in our electronics, to barcode/ UPC (universal product code) labels on the items we buy, to zip codes. Coding advanced the most probably in World War II when the Germans invented the Enigma Machine, which was unbreakable until a group of Polish mathematicians banded together. It was finally broken by the Polish Cipher Bureau, beginning in December 1932. This success was a result of efforts by three Polish cryptologists, Marian Rejewski, Jerzy Różycki and Henryk Zygalski, working for Polish military intelligence. Rejewski reverse-engineered the device, using theoretical mathematics and material supplied by French military intelligence. After that, the Germans stepped up the complexities of their enigma machines and it became harder and harder to decode their messages. Anywho, this class will not be covering the complexities of modern coding and de-coding. We will be covering the simple methods of cryptography, some of which started as far back as Julius Caesar of Rome. Look at these two coded messages: Message 1: D W W D F N I U R P V R X W K Message 2: S L L S U C X J G E K G M L Z How would you start decoding?  Looking for frequent letters o The most commonly used letters in the alphabet (like A E R S and O) will show up with more frequency in coded messages than less used letters (like Z X J and Q)  Looking for patterns o Looking for double letters and the letters surrounding them will help with identifying un-coded words with the same patterns Caesar Shift Cipher Julius Caesar (beware the Ides of March) used this cipher to code his messages so they couldn’t be intercepted by the enemy, or read by the messenger. By moving a certain amount of spaces to the left or right down the alphabet, it was simple code. Example 1: Shift Right 5 Take the following message and move five spaces to the right of each letter. L E A V E T O M O R R O W becomes Q J F A J Y T R T W W T B If your letter is closer to the end of the alphabet and runs out of places, continue through the beginning of the alphabet. Example: X does not have five places to move, so we go back to the beginning of the alphabet at continue counting spaces. X Y Z A B C 1 2 3 4 5 Advantages  Simple, easy to use/teach  No equipment required to code or de-code  Effectively hides the message Disadvantages  Simple to decode  Only one level of coding  The receiver needs to know the key (number of spaces to shift) o Caesar was brilliant to create this method, but a bit unimaginative to change the key. He would always use R3 (shift right 3) to code his messages. So unless otherwise specified, for a Caesar Cipher, use R3 to code or de-code the message/text. Vignere Cipher The Vignere Cipher is similar to the Caesar Cipher, but there is another layer of coding added. Adding number values to the letters (Where A is equal to zero since 26 will just start the alphabet over) will provide this layer. This cipher also takes care of patterns of double letters in codes as you will see. A B C D E F G H I J K L M 0 1 2 3 4 5 6 7 8 9 10 11 12 N O P Q R S T U V W X Y Z 13 14 15 16 17 18 19 20 21 22 23 24 25 (I suggest keeping a chart similar to this close at hand for quick decoding. It will be better than writing it out each time or trying to do it in your head) After this, a key composed of a word with three to five letters is needed. Message: L E A V E T O M O R R O W Key: D E W 3 4 22 Place each value of the code under the message, and add the number values of each letter to encode. L E A V E T O M O R R O W 11 4 0 21 4 19 14 12 14 17 17 14 22 D E W D E W D E W D E W D 3 4 22 3 4 22 3 4 22 3 4 22 3 14 8 22 24 8 41 17 16 36 2 21 36 25 Then fill in the alphabet by the number, like if the added value is 2, put B. L E A V E T O M O R R O W 11 4 0 21 4 19 14 12 14 17 17 14 22 D E W D E W D E W D E W D 3 4 22 3 4 22 3 4 22 3 4 22 3 14 8 22 24 8 41 17 16 36 2 21 36 25 *15 *10 *10 O I W Y I P R Q K U V K Z (For the stared ‘*’ values, subtract that number by 25 until you get a number within the alphabet range of 0-25) Practice 1: Try coding the message, but with the Key: SUM B U R N C A P I T A L 11 4 0 21 4 19 14 12 14 17 17 Answer: B U R N C A P I T A L 11 4 0 21 4 19 14 12 14 17 17 S U M S U M S U M S U 18 20 12 18 20 12 18 20 12 18 20 19 40 29 21 22 12 33 28 31 18 31 * * * * * * T O D V W M H C F S F Review Looking back at the messages from the beginning, de-code them using the Caesar Cipher with the knowledge that the first one is a standard coding, and the second is L8. Message 1: D W W D F N I U R P V R X W K Message 2: S L L S U C X J G E K G M L Z Keep in mind that when you are decoding, you will have to move the opposite direction of the code. So like with the first message, as it is R3 you will have to move 3 letters to the left to de-code. Answers: Message 1 D W W D F N I U R P V R X W K A T T A C K F R O M S O U T H Message 2 S L L S U C X J G E K G M L Z A T T A C K F R O M S O U T H

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