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# Math for Social Choice Final Review MATH 1014

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This 22 page Study Guide was uploaded by Amy Brogan on Tuesday April 12, 2016. The Study Guide belongs to MATH 1014 at University of Cincinnati taught by Mary Koshar in Spring 2016. Since its upload, it has received 198 views. For similar materials see Mathematics of Social Choice in Mathematics (M) at University of Cincinnati.

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Koshar Amy Brogan Math for Social Choice Final Review Voting Systems Terms: Borda Count: point system for ranking vote o Last place gets zero points, increasing by one up from there Hare “Elimination”Method o Eliminate the choice with the least amount of 1 place votes and choose from there o Number of rounds = number of candidates Plurality o Most votes in first place o Can have a tie because it’s not for majority Majority =more than 50% When the vote is between two people, Majority and Plurality are the same Plurality Run-Off o Only two rounds Top two 1 place candidates (eliminateall the rest) Choose from the new most first place Condercet’s “Head-to-head” Method: o Compare two choices at a time, looking for one who is the best overall o When comparing two, eliminate allothers (A vs.B only, then B vs. C only, etc) RankMethod o Spread out possible points ApprovalVoting: each voter can give 1 vote to asmany/few candidates as they choose, not a rank o Approval voting Drawbacks: No distinction between top choice and other choices Approving everybody advances no one Equal to not voting Compare with plurality:approval is not the majority, but possibly acceptable DisapprovalVoting: o each voter can disapprove (anti-vote) as many/few candidates as they want Coombs: o elimination method, but deleting the candidate with the most last place votes Permutation: o an arrangement of objects (such as candidates) where order is important o n! Combination: o an arrangement where order is not important o 2^n Weighted Voting Systems: o voters do not necessarily have the sameweight (voter situations where voters vote yes or no to a motion.) Dictator: o has enough to pass the vote by themselves, and can’t passthe vote without them Dummy: o an unnecessary voter who has no power in a weighted system, the vote can pass without them in any situation Veto Power: o Does not have the power to pass a vote on its own, but motion need their vote to pass every time Shapely-Shubik Power Index (1954) o Power: a voter has power when he/she is pivotal in some voting permutation o Pivotal: the first person who’s input puts the vote up to or over the margin Preference list ballot: o A ballot that ranks the candidates from most preferred to least preferred, with noties May’s Theorem: o For two alternatives and an odd number of voters, majority rule is the only voting system that satisfies the three natural properties All votes are treated equally If B beats A, and more people switch to B in a recast, then B should still win A tie cannot occur unless there is an even number of voters Desirable Traits of Voting - Should be a winner - Every Ballot is Equal- Equal candidates (no secret states) - Advantage for actual want vs. manipulation (voting only for political party, or just to beat one candidate) Condorcet winner: candidate beats all others in head to headmatches Monotonicity: if B wins the election, and a recastmakesB’sranking higher, B should still win Independent Irrelevant Alternatives (IIA): change in rankings of irrelevant alternates should not change the outcome o If B wins election 1, and inthe recast C beats A, B should still win Pareto Condition: everyone prefers A to B, then B should not win Drawbacks: Condorcet: Possible to have NO winner because no one beats everyone head-to-head Hare: winner doesn’t beat every other candidate Plurality Run-off: not the same outcome as Condorcet Plurality: If B&C switchranking, the original winner (A)may not stay the winner Sequential Pairwise: does not satisfy Pareto condition Kenneth Arrow: Arrow’s Impossibility Theorem: 1951:Proved that finding anabsolute fair and decisive voting system is impossible With 3+ candidates and any number of voters, there does not exist there never will exist a voting system that always producesa winner, satisfies Pareto condition, is independent of irrelevant alternatives, and is not a dictatorship Notation for quota and voting weights:[q: w1, w2, …, wn] Permutations; n factorial:where order matters for n voters = n! outcomes (7! = 7x6x5x4x3x2x1) Practice Part 1: Do Borda Count, Hare’sMethod, Condorcet, Plurality, and Plurality Run-Off for chart 1 and 2. Chart1 (5) (4) (3) (3) (2) st 1 E A C D B 2nd B B B B C 3rd C C D C D 4th D D A A A th 5 A E E E E Chart2 (17) (12) (9) (4) (4) (2) (1) First Q T E W Q E Q Second W R Y R T W E Third E Q R Y R Q T Fourth R W W Q Y R Y Fifth T Y T T E T W Sixth Y E Q E W Y R Example 3: How would you chart this? 26% want A astheir first choice, but would also approve of B 25% want A astheir first choice and approve neither B or C 15% want B as their first choice and approve neither A or C 18% want C as their first choice, but would alsoapprove of B 16% want C as their first choice and approve neither A or B Part 2: Chart3: Do both Coombs and Hare’sMethod with this chart and compare. Nst Voters 1 1 1 1 1 1 1 1 C D C B E D C 2nd A A E D D E A rd 3th E E D A A A E 4 B C A E C B B 5th D B B C B C D Examples 2: Permutation or Combination? 1. Rob andMaryare planning a trip around the world. They are deciding on which 9 cities to visit. 2. Batting order for 7 players on a twelve-person team? 3. A committee of 10 members wants to choose a chair andassistant chair. 4. All possible ways to build a Subway sandwich. 5. All possible outcomes when ranking 3 candidates 6. Choosing 3 applicants out of 445 people who applied for a programming position Sets and Power: Identify any dictators, dummies, or veto power holders in these sets. [15: 11, 5, 5, 2] [15: 16, 7, 5] [14: 9, 6, 5, 5, 3, 1] [6:5, 3, 1] For these sets calculate the power of each voter using the Shapely-Shubik Index: [6:3, 1, 1, 1, 1, 1, 1, 1, 1] [7:5, 4, 3, 1] Calculate the Banzhaf Power Index for the following set: [6:4, 3, 2, 1] Terms List Chapter 11: Criticalvoter: a voter who is needed in a combination voting system for themotion to pass o Order does not matter Coalition: a set of votes in agreement for the motion to pass, “yes” votesgiven bythe weighted voters o winning coalition: a voting set/coalition that is equal or to or greater than the quota, the motion passes o minimalwinning coalition: a winning set/coalition where every voter is critical Combinations: an unordered arrangement of item (votes) Banzhaf Power Index: o a count of all voting combinations in which a voter can cast a decisive vote o the measure of the voting power of the voters Chapter 13: Adjusted Winner Procedure: o Works for only two parties o Each is allocated100 points to distribute between the items in question. The higher points for the item, the more the party wants that item. o An item may have to be split or shared by each side to end up gaining items thatadd up to the same number of points Equitable: each player believes s/he received the same fractional part of the total value Envy-free: each party would not feel like they would not be happier with what the other received Pareto-optimal: no other allocation, arrived at byanymeans, can make either party better off without making the other partyworse off Knaster Inheritance Procedure: each party bids for one item and the highest bid gain it, and has to “buy-out” the others by paying them a portion of their share Taking Turns: each party takes turns choosing items until all are given out o Strategies Bottom-Up: if one party knows the preference of the other, they could work from the least desired items to the most to find out who selects first so that each side is better off than if a coin had been flipped for first choice. Divide and Choose: o First person divides, the second chooses o Works best with fewer parties. The more parties there are, the more complex the cutting and selecting process becomes. VickreyAuctions: a sealed-bid auction in which the highest bidder wins, but pays only the amount of the second-highest bid Practice Chapter 11 1. Consider the set [7:5, 4, 2, 1, 1]. A) Determine all winning coalitions (that include weight 5 voter) in which weight-5 voter is not critical. B) If possible, list three coalitions in which the weight-2 voter is critical. 2. Calculate: C513 and 7P 3 3. Determinea weighted voting system that satisfies all of the following, if possible*: a. is a 4-voter system b. has a simple majority c. has a voter with veto power d. does not have a dictator * It is possible. It might take you a while, or none at all, but you can fulfill all requirements. Chapter 13 1. Mike and Philare collegeroommates and encounter serious conflicts during thefirst week of school. They use the Adjust Winner Procedure to resolve the dispute. The chart shows their points distributed. Who controls what? Issue Mike Phil Stereo Levels 4 22 Smoking Rights 10 20 Room Party Policy 50 25 Cleanliness 6 3 Alcohol Use 15 15 Phone Time 1 8 Lights-Out Time 10 2 Visitor Policy 4 5 2. John andMaryare set to inherit their parent’s house. If they bid as follows, who gets the house and how much does the other get paid using the Knaster inheritance Proceedure John Mary $55,900 $59,100 3. Boband Carol rank a series of objects from most preferred to least. Using the Bottom-Up strategy, If Bob knows each of their preferences, who should he let choose first for the most gain? Bob Carol car boat investments investments MP3 player car boat Washer-dryer television television washer-dryer MP3 player Apportionment and Cryptography Terms to know, understand, andapply: Chapter 14: Apportionment problem o Any problem wherean object or group of people has to be given out in whole values Standard divisor o Total population / number of items being given out o Used to find the quota for each group (School A, B, C) Standard and adjusted quota o Standard: group population / standard devisor o Adjusted: group populations/ a devisor larger or smaller tomake the rounded outcome add up to the number of items being given out Upper and lower quota o Upper: when the quota is rounded up by largest fractional value (3.67 rounded up over 3.66) o Lower: when the quota is rounded down (as in the Hamilton Method) Geometricmean o The square root of the rounded down quota and the next highest number o √???? ⋅ ???? 3.65 =√ 3 ∗4 Absolute and relative differences o Absolute: the result of subtracting a smaller number from a larger one o Relative: the result of subtracting a smaller number from a larger one, then expressing it as a percentage 120-100 = 20% Alabama paradox o A state lost a representative seat solely because the House increased. o Only possible in the Hamiltonmethod. Population paradox o An instance when one group’sapportionment decreases even though the population has increased; while another group loses population and gains more apportionment A school has 30 teaching assistants to spreadamong 4 classes by population. College Algebra has188 students, Geometry has138, Pre-calculus has142, andCalculus I has 64. Hamilton method o Method that assigns each state either its lower quota or its upper quota Hamilton Population Quota Round Down Adjust End Algebra 188 Geometry 138 Pre-Calculus 142 Calculus 64 Total 532 SD: ___ students per TA 4/4 * Known for the Alabama Paradox Hill-Huntington method o A devisor method that minimizes the relative differences between groups o Uses Geometricmean to round quotas, and often adjusted quotas H-H Population Quota Geometricmean Round Adjust: SD= End Result Algebra 188 1.413 √ 1 ∗2 = 1.414 Geometry 138 1.037 Pre-Calculus 142 1.067 Calculus 64 .48 √ 0 ∗1 = 0 Total 532 SD: 133 students per TA Jefferson method o Method based on rounding all quotas down; often uses adjusted quota Jefferson Population Quota Round Down Adjust: End Result Algebra 188 1.413 Geometry 138 1.037 Pre-Calculus 142 1.067 Calculus 64 .48 Total 532 SD: 133 students per TA *Favors larger populations. Webster method o Method rounding quota the usual way; reduced absolute differences; often uses adj. q Webster Population Quota Round Adjust: End Result Algebra 188 1.413 Geometry 138 1.037 Pre-Calculus 142 1.067 Calculus 64 .48 Total 532 SD: 133 students per TA *Used today in Congress Chapter 17, section 4: Cryptography o The study of how to make and break secret codes o Cryptosystem:a method used to encrypt a code and decrypt byworking backwards Caesar cipher o A cryptosystem used by JuliusCaesar where each letter is shift the same amount of spaces o Standard Caesar Shift: Right 3 (Decode: Left 3) Encode the message: GO WEST using R5 Decrypt: V HQ G K H O S knowing it was encoded with R3 Decimation cipher o A cryptosystem that usesmultiplication by a fixed value to shift each letter Encode “EXPLODE” with D.3 (Multiply everything by3) Decrypt: I UM M AE U using D.5 Vignere cypher o A cryptosystem that uses a key word to determine howmuch a letter is shifted Encode “DANCEOFF” with the key “Dust” Decrypt: L E L Q FY X K using KEYas the key Modular arithmetic o Addition and multiplication involving mondulo n 4 mod77 ≡ ____ 5 mod58 ≡ ____ 6 mod36 ≡ ____ 12 mod ____ ≡ 7 24 mod ____ ≡ 15Make these work both to solve and create What problems do you run into when trying to make“3 mod ___ ≡ 13”?What could the outcome be by switching 3 and 13? Public Key Cryptography (not the algorithm itself) o RSA Public Key Encryption Scheme o A method of encoding that permits each person to announce publically the meansby which secret messages are to be sent to him or her. National Geographicvideo Enigma machine o https://youtu.be/JOL6J13hHyU?t=30m46s Answers Voting Systems Part 1: Chart 1: Hare Method: eliminate those with least amount of 1 place votes A B C D E st 4 2 (least 1 place) 3 3 st 5 4 / 5 3 (least new 1 ) 5 4 (least 1 ) / 8 / 5 / / 12 *winner / 5 Condorcet: Chart 3: 17 votes = 9 votes for majority E (5) vs. B (4+3+3) B B (5+4) vs. C B hasmajorityalready, C eliminated B (5+4) vs D D Eliminated B (5+3+3+2) vs. A(4) B iswinner overall Plurality Run-off: A B C D E 4* top 2 2 3 3 5* top 2 4+3+3+2 = 9* win 5 Plurality Problems: 6 4 3 1st A C B nd 2rd B A C 3 C B A A wins with 6 votes, but 6 4 3 st 1 A C C 2nd B A B 3rd C B A If in the recast, 3 people vote C over B, thenC wins with 7 votes total instead of A(6) Chart 2: Condorcet: Head to Head Comparisons Q W Q(24) does not beat R(25) T Q 17 9 So let’s try it with R: 12 17 12 4 9 4 4 2 R E 4 12 17 1 2 Q wins againstW’s15 1st 4 9 1 votes with 34 4 3 T (21) was beaten by Q(28), R(20) does not beat E(29), Q E who was already beaten by and since Q wasbeaten 17 9 earlier, there is not a 12 2 W, soR and E can’t be the Condorcet winner. Condorcet winner. 4 1 T Y Q (38) beats E (11) 17 9 Q R 12 4 17 12 4 4 9 2 2 4 1 1 T (36) beats Y(13) Hare’s EliminationMethod: Top votes Q W E R T Y 22 4 11 0 (eliminate) 12 0 (eliminate) 22 4 (eliminate) 11 12 26 11 (eliminate) 12 28 21 Q wins with Hare’s method. Remember:We are only looking at who is the top vote per round, per column of votes. As candidates are eliminated their votes have to go toa new candidate because there will be a new candidate in the top ranking. Borda Count: st nd rd th th th Borda Count Points: 1 place = 5;2 = 2; 3 =3; 4 = 2; 5 = 1;6 = 0 (Votes in Column x Borda count for rank) + (votes in next column x Borda count for rank) + etc. (17x_) + (12x_) + (9x_) + (4x_) + (4x_) + (2x_) + (1x_) = ___ Q: (17x5) + (12x3) + (9x0) + (4x2) + (4x5) + (2x3) + (1x5) =160 W: (17x4) + (12x2) + (9x2) + (4x5) + (4x0) + (2x4) + (1x1) = 139 E: (17x3) + (12x0) + (9x5) + (4x0) + (4x1) + (2x5) + (1x4) = 114 R: (17x2) + (12x4) + (9x3) + (4x4) + (4x3) + (2x2) + (1x0) =141 T: (17x1) + (12x5) + (9x1) + (4x1) + (4x4) + (2x1) + (1x3) = 111 Y: (17x0) + (12x1) + (9x4) + (4x3) + (4x2) + (2x0) + (1x2) = 70 To check your math, you canadd up all the totals you got here^ and compare it with the total Borda points. st nd Borda points for 1 place + Borda Points for 2 + … all multiplied by total number of voters 5+4+3+2+1 = 15 TotalVoters: 17+12+9+4+4+2+1= 49 15 x49 =735 Borda points 160+139+114+141+111+70 = 735 Plurality: You only need to look at the first place ranking: (17) (12) (9) (4) (4) (2) (1) First Q T E W Q E Q Q has22 first placevotes, and the most, soQ wins by Plurality. Plurality Run-Off As there wasn’t a tie with Plurality, there is no need for a run-off. We would onlyneed to do a run-off if there was a tie of two or more candidates. Example 3: since a percentage is out of 100, we canmake each1% the same as1 vote: 26 25 15 18 16 A X X B X X X C X X A: 26 +25 =51 B: 26 +15 +18 =59 *wins by approval voting C: 18 + 16 = 34 But when we remember that ApprovalVoting doesn’t take into account which is the top vote, problems arise. Check with plurality: 26 25 15 18 16 A X (1 ) X (1 ) B X X X st st C X (1 ) X (1 ) A: 51 *most 1 place votes B: 15 – fewest first place votes C: 39 This is why Approval Voting is not morewidely used. Answers Part 2: Chart 3: Hare’sMethod: Coomb’sMethod: Starting in 5 and moving up A B C D E A B C D E 0 1 3 2 1 0 3 2 2 0 / 1 3 2 1 1 / 4 2 0 / / 3 4* / 3 / / 2 2 D is the most liked / / / 4 3* E is the least disliked? Hare over Coombs: it’s widelyaccepted that it’s better to eliminate from the top than to eliminate candidates placed on the bottom possibly for emotional dislike. Examples 2: 1. Since they are just choosing their countries to visit, it’s combination. Mary can write down China, and then Rob can add Brazil. The only time when this example could be permutation is when they are planning their route. (England to France to Italy to Israel to India toChina etc. If you mixed these up, the flying back and forth would be time consuming and expensive.) 2. This one is permutation because it asking for the order of the batters in a team. 3. Choosing a chair and assistant chair is permutation because the order matters. If they were simply making a sub-committee, then it would be a combination. 4. In class we came to the decision that this is combination because a BLT is still a BLT no matter if the bacon goes above or below the tomato. 5. Because we areranking candidates, it is permutation. 6. Choosing 3 candidates would still be the same three, no matter what order their names were called, so this is a combination. Sets: [15: 11, 5, 5, 2] – dummy(2) – The last voter will never be needed to pass a motion. [15: 16, 7, 5] – dictator (16) – The first voter can pass the motion by him/herself and the others can’t vote together to pass it [14: 9, 6, 5, 5, 3, 1] - none [6:5, 3, 1] – veto (5) – The first voter is needed in every possibility to pass the motion, the other two can’t do it without them, but they can’t pass the motion by themselves. Power: {1, 1, 1, 3, 1, 1, 1, 1, 1} and {1, 1, 1, 1, 3, 1, 1, 1, 1} and{1, 1, 1, 1, 1, 3, 1, 1, 1} W3 =3/9 = 1/3, and all the rest have a total of 2/3 power combined. (2/3) x (1/8)= 2/24 = 1/12. This makes[6:3, 1, 1, 1, 1, 1, 1, 1, 1] powerful by (1/3, 1/12, 1/12, …, 1/12) Ex: [7: 5, 4, 3, 1] [Q:A, B, C, D] Permutations Pivotal Voter A(B)CD – 5+4=9 > 7 B ABDC B ACBD C ACDB C AD(B)C – 5+1 = 6+3=9 > 7 C ADCB B BCDA A BCAD A BDCA C BDAC C BACD C BADC A CDAB A CDBA A CADB B CABD B CBAD B CBDA A DABC B DACB C DCAB C DCBA A DBCA B DBAC A I have filled in a couple of the Permutation blanks so you can see how to get the pivotal voter. Once the total reaches 7 or more, the last voter added in is the pivotal one. A: 8; B: 8;C:8; D: 0 out of 24: (8/24, 8/24, 8/24, 0). This can also be written with simplified fractions or in decimal form. Banzhaf: [6: 4, 3, 2, 1] Winning sets (combinations, order does not matter): {4, 3} {4, 2} {3, 2, 1} {4, 3, 2, 1} 4: 2 3: 2 2: 2 1: 1 2+2+2+1=8 For the set [6: 4,3, 2, 1] the Banzhaf power index is (1/4, ¼, ¼, 1/8). Answers: Chapter 11 1. [7: 5, 4, 2, 1a, 1b]. A) {5, 4, 2, 1a, 1b} {5, 4, 2, 1a} {5, 4, 2, 1b} B) If possible, list three coalitions in which the weight-1a voter is critical. {5, 4, 2, 1a, 1b} & {5, 4, 2, 1a}(only two are possible) 2. 15C13 =105 nC k = n!/[k!(n-k)!] 15! 15! = 15x14x13x12x11x10x9x8x7x6x5x4x3x2x1 = 15 x 14 = 210 =105 13!(15-13)! 13!(2!) 13x12x11x10x9x8x7x6x5x4x3x2x1 x2x1 2 2 There are 105 combinations of 13 voters out of a system with15 voters. 7P 3 =210 nP k = n!/(n-k)! 7! = 7x6x5x4x3x2x1 = 7x6x5 = 210 (7-3)! 4x3x2x1 3. a. is a 4-voter system b. hasa simplemajority c. hasa voter with veto power d. does not have a dictator Use [q: A, B, C, D] where A = q-1 and B+C+D = A to double check your systems Possible sets: [21: 20, 9, 9, 2] [62: 61, 22, 20, 19] [13: 12, 4, 4, 4] [7:6, 3, 2, 1] Chapter 13 1. Issue Mike Phil Stereo Levels 4 22 * Smoking Rights 10 20 * Room Party Policy 50 * 25 Cleanliness 6 * 3 Alcohol Use 15 15 Phone Time 1 8 * Lights-Out Time 10 * 2 Visitor Policy 4 5 * Step 1: decide who gets what initially 1.2: They bothgave the same amount of points for who decides the issue of Alcohol Use. Count up the points as are andgive it to the person with the least. Mike Phil 50 +6 + 10 = 66 22 +20 +8 + 5 =55 +15 =70 Step 2: Find an object, or a portion of an object, that can be switched to the other side to make their awarded points even. Item Mike Phil Ratio Stereo Levels 4 22 * 22/4 = 5.5 Smoking Rights 10 20 * 20/10 =2 Room Party Policy 50 * 25 50/25 =2 Cleanliness 6 * 3 6/3 =2 Alcohol Use 15 15* 15/15 =1 * Phone Time 1 8 * 8/1 =8 Lights-Out Time 10 * 2 10/2 = 5 Visitor Policy 4 5 * 5/4 =1.25 Step 3: with the item whose ratio is closest to 1, divide it between Mike and Phil 66 +15x = 70 – 15x 30x = 4 x = 4/30 = 0.133 =13.3% Step 4: Distribute the items, including the portion of any split/shared item. Mike controls: room party policy, cleanliness, lights-out time, and 13% of alcoholuse Phil controls: Stereo levels, smoking rights, phone time, visitor policy, and87% ofalcohol use 2. Bid for: House John Mary Who bid the most? $55,900 $59,100 * Divide the bids by the Divide by 2 number of parties 27,950 29,550 Kitty: number of shares Kitty: 1 x 29,550 as parties outside of the “winner” Kitty paid out to other 29,550 parties - 27,950 1,600 Divide out the 1,600/2 = remainder among all + 800 + 800 parties End results $28, 750 House, pays $28,750 3. Bob Carol car boat investments investments MP3 player car boat Washer-dryer television television washer-dryer MP3 player Bob Chooses first: Bob: boat(4 )h car (1st) MP3 player (3 )rd Carol: investments(2 )nd television (5th) washer-dryer (4 )th CarolChoses first: Bob: car(1 ) television(5 )h MP3 player (3 )rd Carol: investments(2 )nd boat(1 )t washer-dryer (4 )th Answers Apportionment Part2: Hamilton Population Quota Round Down Adjust End Algebra 188 1.4135 1 (.41) 1 Geometry 138 1.03759 1 (.03) 1 Pre-Calculus 142 1.06767 1 (.06) 1 Calculus 64 .48 0 +1 (.48) 1 Total 532 SD: 133 students per TA 3/4 4/4 H-H Population Quota Geometricmean Round Adjust: SD= End Result Algebra 188 1.413 √1 ∗2 = 1.414 1 Unrequired 1 Geometry 138 1.037 1 1 Pre-Calculus 142 1.067 1 1 Calculus 64 .48 √0 ∗1 = 0 1 1 Total 532 SD: 133 students 4/4 per TA Jefferson Population Quota Round Down Adjust: SD = 90End Result Algebra 188 1.413 1 2.08 – 2 2 Geometry 138 1.037 1 1.53 – 1 1 Pre-Calculus 142 1.067 1 1.57 – 1 1 Calculus 64 .48 0 0.70 – 0 0 Total 532 SD: 133 students per TA3/4 4/4 4/4 Webster Population Quota Round Adjust: 128 End Result Algebra 188 1.413 1 1.468 –1 1 Geometry 138 1.037 1 1.078 –1 1 Pre-Calculus 142 1.067 1 1.109 –1 1 Calculus 64 .48 0 0.500 –1 1 Total 532 SD: 133 students per TA 3/4 4/4 Cryptography: Caesar Cipher G O W E S T +5 L T B J X Y V H Q G K H O S -3 S E N D H E L P Decimation Coding E X P L O D E 4 23 5 11 14 3 4 *3 12 69 45 33 42 9 12 ≡17 ≡19 ≡7 ≡16 M R T H Q J M I U M M A E U 8 20 12 12 0 4 20 +26+26=60 +26+26+26=90 +26=30 /5 12 4 18 18 0 6 4 M E S S A G E VignereCipher D A N C E O F F 3 0 13 2 4 14 5 5 D U S T D U S T +3 +20 +18 +19 +3 +20 +18 +19 6 20 31≡5 21 7 34≡8 23 24 G U F V H I X Y L E L Q F Y X K 11 4 11 16 5 24 23 10 +26=37 K E Y K E Y K E -10 -4 -24 -10 -4 -24 -10 -4 1 0 13 6 1 0 13 6 B A N G B A N G Modular Arithmetic 77 4 mod77 ≡ __1__ 4 = 19.25 4*19=76 77-76=1 5 mod58 ≡ __3__ 58= 11.6 5*11 = 55 58-55=3 5 6 mod36 ≡ __0__ 36/6 =6 12 mod __295__ ≡ 7 (possible answer) 7 + 12 * X = _____ 24 mod __87__≡ 15 (possible answer) ?. What problems do you run into when trying to make “3 mod ___ ≡ 13”? What would the outcome be by switching 3 and 13? When coming up with a value for the blank, it won’t work when reversing to solvefor 13’sspot. That is because the outcome is large than the number the mod is being solved by. By switching 13 and3, it is possible to get an equation that works both ways. --> but 166/3=55.33, not a value starting with51, so the equation is wrong.

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