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LSAT Prep Course - Games (Part 2)

teacher avatar Jeff Kolbly

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Taught by industry leaders & working professionals
Topics include illustration, design, photography, and more

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Taught by industry leaders & working professionals
Topics include illustration, design, photography, and more

Lessons in This Class

101 Lessons (2h 25m)
    • 1. 130 SEQUENTIAL GAMES - Chronological Games

      1:07
    • 2. SEQUENTIAL GAMES - INTRODUCTION

      0:23
    • 3. Chronological Game Example 1

      3:53
    • 4. Chronological Game Example 2

      0:51
    • 5. Chronological Game Example 3

      1:05
    • 6. Chronological Game Example 4

      0:28
    • 7. Chronological Game Example 5

      0:56
    • 8. Points to Remember

      0:41
    • 9. Solo Exercise 1

      2:31
    • 10. Solo Exercise 2

      0:44
    • 11. Solo Exercise 3

      1:01
    • 12. Solo Exercise 4

      0:53
    • 13. Solo Exercise 5

      0:18
    • 14. Solo Exercise 6

      1:37
    • 15. Solo Exercise 7

      1:22
    • 16. Solo Exercise 8

      0:48
    • 17. Solo Exercise 9

      1:08
    • 18. GENERATING FORMULAS - Introduction 1

      1:20
    • 19. Generating Formulas - Introduction 2

      1:04
    • 20. Generating Formula Game Example 1

      1:48
    • 21. Generating Formula Game Example 2

      0:53
    • 22. Generating Formula Game Example 3

      2:45
    • 23. Generating Formula Game Example 4

      1:44
    • 24. Generating Formula Game Example 5

      2:50
    • 25. Points to Remember

      0:37
    • 26. Solo Exercise 1

      2:02
    • 27. Solo Exercise 2

      1:29
    • 28. Solo Exercise 3

      1:16
    • 29. Solo Exercise 4

      1:34
    • 30. PATHS AND FLOW CHARTS Introduction 1

      0:54
    • 31. Paths and Flow Charts Introduction 2

      1:13
    • 32. Paths and Flow Charts Introduction 3

      1:27
    • 33. If then Drill

      0:36
    • 34. Solutions to If then Drill

      1:27
    • 35. Flow Chart Game Example 1

      1:41
    • 36. Flow Chart Game Example 2

      1:58
    • 37. Flow Chart Game Example 3

      1:49
    • 38. Circuit Game Example 1

      4:19
    • 39. Circuit Game Example 2

      0:32
    • 40. Points to Remember

      1:00
    • 41. Mentor Exercise 1

      3:10
    • 42. Mentor Exercise 2

      3:04
    • 43. Mentor Exercise 3

      3:35
    • 44. Mentor Exercise 4

      3:30
    • 45. Mentor Exercise 5

      3:14
    • 46. Mentor Exercise 6

      3:31
    • 47. Solo Exercise 1

      1:00
    • 48. Solo Exercise 2

      1:02
    • 49. Solo Exercise 3

      1:40
    • 50. Solo Exercise 4

      1:10
    • 51. Solo Exercise 5

      1:14
    • 52. Solo Exercise 6

      1:41
    • 53. Solo Exercise 7

      2:42
    • 54. Solo Exercise 8

      1:31
    • 55. Solo Exercise 9

      2:32
    • 56. Solo Exercise 10

      2:21
    • 57. Solo Exercise 11

      2:55
    • 58. Solo Exercise 12

      1:38
    • 59. Solo Exercise 13

      1:01
    • 60. GROUPING GAMES Introduction

      0:51
    • 61. Selection Game Example 1

      3:02
    • 62. Selection Game Example 2

      0:53
    • 63. Selection Game Example 3

      1:12
    • 64. Selection Game Example 4

      0:46
    • 65. Selection Game Example 5

      0:32
    • 66. Grouping by Threes Example 1

      1:33
    • 67. Grouping by Threes Example 2

      1:18
    • 68. Grouping by Threes Example 3

      1:27
    • 69. Grouping by Threes Example 4

      1:09
    • 70. Grouping by Threes Example 5

      1:03
    • 71. Points to Remember

      0:33
    • 72. Warm Up Drill 1

      2:56
    • 73. Warm Up Drill 2

      3:37
    • 74. Solo Exercise 1

      1:35
    • 75. Solo Exercise 2

      0:53
    • 76. Solo Exercise 3

      0:50
    • 77. Solo Exercise 4

      1:30
    • 78. Solo Exercise 5

      1:28
    • 79. Solo Exercise 6

      0:33
    • 80. Solo Exercise 7

      2:11
    • 81. Solo Exercise 8

      0:31
    • 82. ASSIGNMENT GAMES Introduction 1

      1:11
    • 83. Assignment Games Introduction 2

      0:15
    • 84. Assignment Games Introduction 3

      0:28
    • 85. Elimination Grid Example 1

      0:43
    • 86. Elimination Grid Example 2

      0:19
    • 87. Elimination Grid Example 3

      1:04
    • 88. Elimination Grid Example 4

      0:49
    • 89. Assignment Games Introduction 4

      0:58
    • 90. Multiple Choice Game Example 1

      2:32
    • 91. Multiple Choice Game Example 2

      0:40
    • 92. Multiple Choice Game Example 3

      0:28
    • 93. Multiple Choice Game Example 4

      0:38
    • 94. Assignment Games Introduction 5

      0:20
    • 95. Points to Remember

      0:46
    • 96. Solo Exercise 1

      1:02
    • 97. Solo Exercise 2

      0:28
    • 98. Solo Exercise 3

      0:51
    • 99. Solo Exercise 4

      0:23
    • 100. Solo Exercise 5

      1:48
    • 101. Solo Exercise 6

      0:47
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About This Class

Every year, students pay $1,000 and more to test prep companies to prepare for the LSAT. Now you can get the same preparation in an online course. LSAT Prep Course provides the equivalent of a 2-month, 50-hour course.

The LSAT is an aptitude test.¬† Like all aptitude tests, it must choose a medium in which to measure intellectual ability.¬† The LSAT has chosen logic.¬† Although this makes the LSAT hard, it also makes the test predictable ‚ÄĒ¬†it is based on fundamental principles of logic. ¬†LSAT Prep Course¬†analyzes and codifies these basic principles: the contrapositive, the if-then, pivotal words, etc.¬† Armed with this knowledge, you will have the ability to greatly increase your score.¬†

Features:

  • Videos!¬†Hundreds of videos explaining the text, examples, and exercises in step-by-step detail.
  • Analytical Reasoning:¬†Learn powerful diagramming techniques and step-by-step strategies to solve every type of game question that has appeared on the LSAT.¬†
  • Logical Reasoning:¬†Discover the underlying simplicity of these problems and learn the principles of logic these questions are based on.¬†¬†
  • Reading Comprehension:¬†Develop the ability to spot places from which questions are likely to be drawn as you read a passage (pivotal words, counter-premises, etc.).¬†
  • Mentor Exercises:¬†These exercises provide hints, insight, and partial solutions to ease your transition from seeing LSAT problems solved to solving them on your own.¬†

Meet Your Teacher

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Jeff Kolbly

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I have been teaching, writing, and publishing in the test prep field for 25 years.

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Transcripts

1. 130 SEQUENTIAL GAMES - Chronological Games: chronological games, chronological games, order elements at a time sequence. For example, James was born before George, who was born before Kim, who was born before Sarah. In the land of games that we studied earlier, the elements were ordered specially in chronological ordering games. The elements are ordered sequentially. This is true of many of the games that we will study in this chapter. Because these games are sequential in nature. There diagrams can be quite different from those used to salt, spatial and hybrid games. One of the most common efficient types of diagrams is the flow chart. In these diagrams, elements are connected by arrows. Now that we have a second way to diagram linear ordering games, we need, of course, some means of deciding which method to use in general game with no fixed elements should be solved using a flow chart. In constructing a full chart, follow these guidelines one. Look for a condition that starts the flow to build the chart around the element that occurs in the greatest number of conditions. Three. Keep the chart flexible. It will probably have to evolve with the changing conditions 2. SEQUENTIAL GAMES - INTRODUCTION: sequential games. Unlike spatial and hybrid games, sequential games do not order elements in space. Sequential games can be classified according to the criteria used to order the elements chronological before, after etcetera, quantifiable size, height, etcetera, ranking 1st 2nd etcetera. 3. Chronological Game Example 1: this game does not have any fixed elements such as t was born in 1972. So the flow chart method is indicated will use an Errol to indicate that one person is older than another. The condition w is older than the is nationally symbolized as w V. The condition s is younger than both. Why and he gives us why yes. And the S T is not younger than why gives is t one and Z is younger than why but older than you gives us y z and z you. Now we construct the diagram falling. The guidelines look for the element that occurs in the greatest number of conditions. It is why so we build a short around. Why start with condition for adding condition too? We get well, they're going from white to s and then adding condition five. We get an arrow going from y to Z, adding the third condition we get and they were going from Vito s and then adding the first condition. We could an arrow going from W to v, finally adding the six condition we get an arrow going from sea to you. There are no conditions on the element acts so it cannot be placed in the diagram. However, we will know below the diagram that it is wild. Two properties of the diagram should be noted before turning to the questions. First of two elements are in different rows and no sequence of arrows connects them. Then the diagram does not tell us which one is older. For example, since WN why are in different rows, W is in the bottom row and why is in the middle room and are not connected by a sequence of arrows. The diagram does not tell us who is larger. However, the diagram does tell us that he is older than you because the arrows flow from t toe Why to z to you. Second, the diagram tells us that only T W or acts can be the oldest because he starts a row and w starts a row and X is wild and likewise that only you best or X can be the youngest because you ends a row, S n zero and X is. While this first problem involves straightforward elimination choices, B and C are not possible sequences because the diagram shows that Z must be older than you , but be has you before, see? And so does he. So eliminate B and E c is not a possible sequence because the diagram shows that t must be older than you. The aero flows from ti toa Why? To z to you so eliminate c finally D is not a possible sequence because the diagram shows that he must be older than s The arrow flows from the s So we eliminate d. It's by process of elimination. The answer is a 4. Chronological Game Example 2: first, let's reintroduce the diagram derived earlier. This diagram shows that T y, w and V not necessarily in that order were all born before us because in the middle row we have t why pointing to us in the bottom row we have W v pointing toe s so they must been born in the year 71 through 74. This gives a falling possible diagram because one of the birth s is fixed, it is now more convenient to use a lineup diagram. Clearly, this diagram shows that Z must have been born after 1975 because only 76 77 78 are available choice either four makes unnecessarily false statement and the answer is E. 5. Chronological Game Example 3: first, let's reintroduce the previously derived diagram. Since two births deaths index air fixed, we again revert to a lineup diagram and we please x in 73 s at 76. The original diagram shows that W v t. And why we're all born before s so they must be placed to the left of s of the new diagram . However, we cannot uniquely determine their positions. WMD could have been born in 71 72 respectively, or tea and why could have been so One possible diagram is WV and T y. Clearly this diagram forces e and you to have been born in the year 77 78 respectively. Because those are only two years remaining. It's only two other births can be determined, and the answer is C. 6. Chronological Game Example 4: first, let's reintroduce the previously derived diagram and the new condition s was born before acts to the diagram and we get an arrow going from S two X In this diagram, You and s are in different rows. They're not connected by a sequence of arrows. So you could have been born before us. The answer therefore his see. 7. Chronological Game Example 5: first, let's reintroduce the previously derived diagram. The condition s is younger than both y and the anchored the diagram. Without it, we get the following diagram from the top we get T. Why then up to Z and then to you and the bottom gives us just w v and now both x and s or wild. Because the two parts of this diagram or independent they're not conducted by a sequence of arrows W and V could have been born before tea or after you. Instead, your birth cannot be determined for any of the people, and the answer is a 8. Points to Remember: points to remember one. The three types of sequential games are chronological before, after etcetera, quantifiable size, height, etcetera. Ranking 1st 2nd etcetera to most sequential games can be solved most efficiently with a flow chart. Three. In general, a game with no fixed elements should be solved using a flow chart for when constructing a flow chart. Follow these guidelines. One. Look for a condition that starts to flow to build the chart around the element that occurs in the greatest number of conditions. Three. Keep the chart flexible. It will probably have to evolve with the changing conditions. 9. Solo Exercise 1: since this game has no fixed elements, that is, none of the heights of the Children are given a flow chart is indicated. The condition Jane is taller than dawn can be symbolized as J. Polar than dawn, where the arrow stands for is taller than the condition. Pink and Irene are the same height, and both are shorter than dawn can be symbolized as the condition. Anna and Emily are taller than Gina can be symbolized. As Irene was taller than Bob can be symbolized has this and then, finally, both Charles and Frank are shorter than Gina can be symbolized as now, we construct a flow chart from these conditions. De occurs in two conditions and G occurs in two conditions. This indicates that there will probably be too charts for this problem. Start with the first condition next at the second condition, the rectangle around each and I indicates that they are the same height, then had the condition. I is taller than be the third and fifth. Conditions cannot be added to the diagram, so we build a second diagram for them Now for this question. If two elements were not conducted by a sequence of arrows or, if they are in different diagrams than their heights are independent of each other. Now Jean is in the first diagram, and Emily is in the second diagram, so either could be shorter than the other, and the answer is C. 10. Solo Exercise 2: first, let's reintroduce the diagram we derived earlier. The first diagram generates four different heights. Remember, H and I are the same height. You have one height here to three four. Since the second diagram is independent of the first A, E and J could all have the same height. D and G could have the same height and be C and F could all have the same height. So the second diagram does not necessarily add any more distinct heights, and the answer is B. 11. Solo Exercise 3: first, let's reintroduce the diagram derived earlier. The new condition Gina is the same height as I re ING combines the two original diagrams as follows Gene and now will be here. It will extend the box down around her, meaning that h i n g are all the same height now from the second diagram A and he are still taller than G. So put a here and pointing towards the box because I was taller than all three of these people. And E is also taller than all three people. And since G is taller than CNF, will have H i and G all taller than C B enough now. Since D and E are not connected by a sequence of arrows, either could be taller than the other. That is, Don is not necessarily taller than Emily, and the answer is a 12. Solo Exercise 4: first, let's introduce the previously derived diagram From the second diagram, we see that Frank must be shorter than Emily. The arrows flow from Emily to Gina to Frank, from Emily to Gina to Frank in statement. One is contained in the original information. This eliminates B, C and E. They don't contain statement one. No, that only A and D remain and neither contained statement to unless we need not check statement to as the statement. Three. It is false In the diagram. CNF are not conducted by a sequence of arrows, so either could be taller than the other. Hence, by process of elimination, the answer is a. 13. Solo Exercise 5: first, let's reintroduce the diagram we derived earlier from the first diagram, we see that J. D. H and I are all taller than be. This makes four Children taller than be. Therefore be cannot be the third tallest child, and the answer is E. 14. Solo Exercise 6: the condition. Both items Eman l must be placed in the carton before age can be symbolized as him before age and l before a the condition. I must be placed in the carton after H and K can be symbolized as h first, then I and K first, then I, Adam K cannot be placed in the cart. Next. Adam J can be naturally symbolized as not JK and the last condition can be paraphrased as red. I did before non red items, when possible, combining the first and third conditions we get adding the second condition we get adding the fourth condition we get now For this question, choices A and E violate the condition, not J. K. So it eliminated a and e from the diagram. I cannot come before him. The flow of the arrows is from M two h toe I, which eliminates choice. Be choice de violates the first condition him before age. So eliminate d. It's by process of elimination. The answer is C 15. Solo Exercise 7: first, let's reintroduce the previously Dr Diagram choice. A clearly violates the first condition because ages before him. It has to come after him. Choice B has K before age, which violates the condition Red items before non red items impossible. Eliminate. Be choice de violates the condition. K Before I eliminate choice, he has Jay before H, which violates the condition. Red items before non red times when possible, eliminate. It's by process of elimination. The answer is C. It is also instructed to solve this problem directly. Since H and L are red and l comes before age. L must be first so the original diagram becomes bell before him before age before I in K Still before I since H is red K must come after age and still before I so the diagram becomes finally. Since J is not red and cannot be next to K J. Must be last. This uniquely determined order is choice. See 16. Solo Exercise 8: first, let's reintroduced the diagram we originally derived since M is red and l comes before age . The original diagram becomes in before L before age before I in case still before I now, since I is Red J must be placed after I hence Jay's last, and the answer is a note. Since there is no sequence of arrows connecting K with either L or Age K can come before L directly after L or after age. Hence, choices B and C are not necessarily true. 17. Solo Exercise 9: first, let's reintroduce the diagram we derived earlier. Begin with choice A since K and M a red. They must come first and the original diagram becomes where the box around M and K indicates that they are first and second, not necessarily in that order. Now Jay could be placed last in this diagram without violating any condition. His choice A can be true. Eliminate choice. A turning the choice be I must be read since neither to non red items is I. Now the original diagram shows that I am age okay and ill. All proceeds I and there from us all be read hence J and on Lee Jae can be non red. Therefore, choice be cannot be true and therefore is the answer. 18. GENERATING FORMULAS - Introduction 1: generating formulas. The previous linear order in games we studied were static and finite. We were given a fixed number of elements and were asked questions about the possible order rings. Generating formulas, however, tend to be dynamic in the sense that a basic sequences given that is used to generate other sequences by repeated applications of the formulas. Because the formulas can be applied indefinitely, the sequences often have no end don't. Typically, we are interested in only the beginning of the sequence. Example. A particular computer code uses only the letters A, B, C and D. A word is forming the code. According to the following rules, ABC is the basic word from which all other words are constructed. Do you must appear in a word more than once, if at all, interchange in the first and last letters in a word creates a new word. Adding a pair of D's to the end of a word creates another word. Notice that the third and fourth conditions air permissive, that is, they could be applied but don't have to be note with permissive conditions. The contra positive rule of logic does not apply. The second condition on the other hand is mandatory. If d occurs in a word, it must occur at least once more. Note. With mandatory conditions, the Contra positive does apply. 19. Generating Formulas - Introduction 2: There are only two basic types of questions to these games one those that ask you to derive a new sequence from a basic sentence. In the previous game, for example, you may be given the word ABC and then asked to arrive a new word by applying the fourth and third rules in that order to those that ask you to discover from where a sequence was derived in the previous game, for example, you may be asked from which word was the word D B C D e a drive. The latter type of question tends to be more difficult, since there are many past you can retrace, only one of which will lead to the correct answer. Because working backwards is often difficult. Look for opportunities to reverse the direction by using the Contra positive, but apply the Contra positive only to mandatory conditions. Generally, formula games are one of the few types of games for which it is not advisable to draw a diagram. In fact, typically you cannot draw a diagram. Nevertheless, you may want to symbolize the rules for easy reference 20. Generating Formula Game Example 1: notice that rules two and three or permissive because they can be applied but need not be. No diagram could be drawn. Nor are there any readily derived conditions. So he turned to the questions. Let's use elimination on this question. A B C. D is the base word so we can eliminate pay now applying rule to which says that if e immediately follows, be then see can be moved to the front of the word we get. See moves in front of a and that gives us see a be de so we can eliminate. Answer choice. See? No. Rule three says that one letter of the same time can be added immediately after an A, a B or a C so adding and a after the base word and a B after the B and a C after the sea, we get answer choice D so we can eliminate that. Remember, we're looking for a sequence that is not a code word. No applying rule to to the base word we get. See moving, see in front of the A and the B. Now the supply rule three twice to the sea that gives a C C C A B D, which is Choice e. So we eliminate e. It's by elimination. The answer is B 21. Generating Formula Game Example 2: the letter. Achon start a word because it starts the base word from Rule one. This eliminates choices B and C since they do not contain statement one. The letter C can start the word since C A B D is formed from the base. Word by using rule to that is C can be moved immediately in front of the A or immediately in front of the word, which will give us C A B D. This eliminates answer choices A and D since they don't contain statement three. Hence, the answer is E and there is no need to check statement to. 22. Generating Formula Game Example 3: this question is hard because we don't know to which letters in the base word to apply the rules. Furthermore, there is more than one way to generate the word. But of course, only one of those ways is listed as an answer choice. We can, however, narrowed the number of answer choices by analyzing the given word notice that C occurs three times in a occurs twice, so Rule three must have been applied three times twice to see him once to a. This eliminates choices be and D, since neither has three threes. Next. Turning to choice A. We apply rule to to the base word, which will move the see in front of the A giving See a B D now rule to cannot be applied to this word again. Since C is now not immediately after B, this eliminates a next choice. See seems at first glance to be plausible. It begins the same way, as does Choice E, but notice and choice See that rule to is applied twice in a row. A little fiddling shows that if this is done than to seize in a row, would come at the beginning of the word so eliminate C in spite process of elimination. The answer is E as a matter of test taking strategy, this would be sufficient analysis of the question. However, it is instructive to verify that the answer is e. To that end, apply rule three to the sea in the base word, which gives we add another sea following the first occurrence of the sea. Now since see immediately follows B, we can use rule to to move the first occurrence of sea to the beginning of the word that gives a C a. Be this see here. Then D finally applied Rule three to A, which adds in a immediately after A and then apply Rule three to the sea, which adds a C immediately following. See this derives the word and we have applied the rules in the order Rule three rule to rule three and then rule three again. Hence the answers e 23. Generating Formula Game Example 4: start with the base word A B C D dropping the d we get ABC now playing Rule A We can move the sea to the front of the word and we get C A B. Now we can drop the b by using the new rule and we get C A hints Some words can end within a so I is true And this eliminates answer choice Be now for the second statement, start with the base word A B C D. Apply rule three to see which gives us a B C in the new C D. Now apply rule to with this See here to move it to the front of the word we have C A B this . See here then D Now apply the new rule to delete the D and we get C A B C. So at least one word could start with a C and in with C, so statement to is true and this eliminates answer choices A and D. Unfortunately, we have to check the third statement from the base word. All other words must be derived now in the base word. The letter D must be dropped before the b can. Thus a CD cannot be formed. This eliminates answer choice E and therefore the answer is C 24. Generating Formula Game Example 5: this problem can be solved by either deriving the four words offered or by finding the word that violates one or more of the rules. We shall solve it both ways. Choice A is a word because the last four letters are the reverse sequence of the 1st 4 letters, which is the base word. You have a D than a see that it be that in a and then it has been fixed to the end of the word. So we eliminate choice. A choice B is a word because you can apply rule three First to be that adds this newbie here and then to see to add a new see here so we can eliminate be noticed. This new rule is also permissive, so it need not be applied for choice. Be to be a word choice, See is a word because it could be derived by applying rule three to the base word a B c D. So a plane Rule three we get a B see, and then a new see here de now applying rule to to the first See, you can come to the front of the word and we get C A B this see here and then D now reversing the sequence of letters we get d first, then see, then be then a and then finally see. But this is the sequence of letters and choice. See, So C is a word and we eliminate C finally choice diesel word because it can be drive by a plane rule to to the base word which gives see a be de that implying the new rule gives de be a see So we eliminate d therefore, by process of elimination, the answer is e turning to the other method. We now show that e violates one of the rules. The only way that they can in the word is if the reverse sequence of the word is added to the word itself. But D. C. B A is not the reverse sequence of C b C D, which isn't even a word. Although the latter method was faster than driving the four words, it can be deceptively hard to spot the choice that violates one or more of the rules 25. Points to Remember: points to remember one with permissive conditions. The contra positive rule of logic does not apply to with mandatory conditions. The Contra positive does apply. Three. There are only two basic types of questions to these games one those that ask you to derive a new sequence from a basic sentence to those that ask you to discover from where a sequence was derived for generating formula games or one of the few types of games for which it is not advisable to draw a diagram never left. You may want to symbolize the rules for easy reference. 26. Solo Exercise 1: We will not use a diagram to solve this game. However, we will symbolize the conditions for easy reference and to better understand them. The first condition G cannot be the first or last letter in a word can be symbolized as she does not equal. First slash Last The condition I and J cannot be next. Each other is naturally symbolized as not I j the condition If J occurs in a word than it occurs an odd number times can be symbolized As if J then Jay is equal toe odd number of times. Finally, the condition h cannot begin A word in Les que ends award can be symbolized as if h is first then okay His last note that all the conditions in this game are mandatory. So we may safely apply the contra positive to any of them. A is not a word because it violates the condition not I. J b is not a word because it violates the condition g not first or last. If there is only one element in a word than it both begins and ends, the word see is a word. G does not begin or in the word I is not next to Jay J appears only once, which is an odd number of times in the condition. If HS first and K is last does not apply. Since H is not first, it's the answer is C. 27. Solo Exercise 2: first, let's introduce the symbols we derived earlier. Start with age. No h i k j But putting h in the box is not a word because ages first Que is not last. So it violates this condition. Here. This eliminates A and D next place I in the box, which gives us All right. All right. Okay, J This sequence of letters satisfies all the conditions. So it is a word. G is not first or laughs. Check. I is not next to Jerry Check. J does appear and it appears only once, which is an odd number. Times check in the final condition is satisfied by default. Since H is not first eyes, this eliminates choice. See, unfortunately, we have the check statement three place J in the box and we get J All right. Okay, J This, however, is no word. Since J appears twice violating the condition that J must occur an odd number of times. This eliminates E And the answer is B 28. Solo Exercise 3: first, let's reintroduce the symbols we derived earlier. G cannot be made into a word by adding letters to its right in because G could never be the first letter of a word. It violates the first condition. This eliminates a next I J cannot be made in toward by adding letters to its right in because I can never be next to Jay. In a word, this violates the second condition now for choice. D. H J J G is not a word because H is first, but K is not last. Additionally, J appears twice. But if we add a J and then a K in that order, then there will be an odd number of Jay's 123 and K will be the last letter, which makes this a word. Hence the answer is D. Don't make the mistake of choosing C or E. Both are already words. Jay is a word because it appears an odd number times once, and none of the other conditions air violated. Remember, we're looking for a string the letters that is not a word that can be made into one by adding one or more letters to its right in 29. Solo Exercise 4: first, let's reintroduce the symbols we derived earlier. The letters and statement one can be made into a word by first adding another J, which gives us an odd number, jays three and then moving I to the front of word to get it away from J and to prevent G from being the first letter in the word. This gives all right G age Today, J. J here G is not first or last check eyes not next to jade check J occurs three times an odd number check h is not first there for this final condition is satisfied by default. Next statement to I j cannot be made into a word by adding another J because that would result in to Jay's an even number. Finally, the letters and statement three could be made into a word first at a J to the right of G, then flip flop H and K. So we get K A g j. This word satisfies all the conditions. G is not first or last check. I is not next to J check. J occurs once, which is an odd number Times check and again the final condition is satisfied by default because H is not first, hence the answers D 30. PATHS AND FLOW CHARTS Introduction 1: paths and flow charts. Although flow charts and pass are not, strictly speaking, ordering games, they have many of the properties found in ordering games. In fact, as we saw with sequential ordering, many games can be solved more easily and efficiently using flow charts. Thus, it is natural to analyze them here. Flow charts and passed tend to be highly determinative. Once the chart has been constructed, the questions tip if he can be answered with little additional thought. Often the answers can be discerned by merely reading the chart. The catch is that the chart may not be easy to derive because this type of game typically has many conditions. The chart can easily get out of control. Charting is an art, however. There are some guidelines that will help one look for a condition that starts the flow or contains a lot of information to look for an element that occurs in many conditions. Three. Keep the chart flexible. It will probably have to evolve with the changing conditions 31. Paths and Flow Charts Introduction 2: before we start, we need to address some of the hazards and symbols common to these games because flow charts and pass evolve a flowing of information. If then. Symbol is the workhorse for these games because the information and often flow in both directions. The if and only if symbol also comes into play. A slash through a symbol indicates that information cannot flow in that direction. For example, a slash B means information cannot flow from A to B as you work through these games to be alert to any opportunity to apply the Contra positive rule of logic, often negative conditions can be expressed more clearly by reverting them in the Contra positive. For example, the statement, if it is not Sonny than Biff is not going to the beach can be rewarded more directly as if Biff is going to the beach, then it is sunny. It is not necessary that both parts of the if then statement be negative for this technique to be effective. For example, the statement, if Lindus hired then rolling is not can be recast as if ruling is hired, then Linda is not, although in this case the Contra positive statement is no simpler than original, it may and often does open up connections to other conditions 32. Paths and Flow Charts Introduction 3: we need to review to common fallacies associated with the Contra positive from the statement. If a then b, we can conclude using the Contra positive, if not be than not A. It would be fallacious, however, to conclude if not a then not be or if b than a. Also note that some Ming's at least one and perhaps all, up until this point, our discussion has been out of chronological order. We have discussed how to solve flow and path games without discussing how to identify them . Path games are easy to identify. Typically, they involve the actual movement of an element or of information. Some examples are four cities air connected by six roads. A memo can be passed from Sarah Toe Helen, but not from Sarah to John. If a litigant filed this case in federal court and loss that he may appeal to the fourth District Court, and from there to the Supreme Court, flow charts are harder to identify. Them pass. In fact, that could be quite cryptic over a game with many if then conditions, it's often a tip off to a flow chart game. Unfortunately, the if then thought is often embedded in other equivalent structures. For example, the sentence All A's are B's can be rewarded, as if X is an A and X is a B. For a more subtle example. Take the sentence. Linda and Sarah are not both hired. It can be recast as if Linda is hired than Sarah's not, or if Sarah's hired, then Linda is not. 33. If then Drill: If then drill. The following drill will help you identify embedded. If then statements directions translate each of the following conditions into an equivalent . If then statement one. No. A is a B to Alice. Will go to the party only if Bobby goes three. Anyone who is not in a cannot be four on liaise Arby's five of two light switches A and B A and B cannot both be on six of two light switches. A and B A is off when B is off, a is on when B is on. 34. Solutions to If then Drill: solutions to If then drill one. No, A is B is equivalent to. If a did not be too Alice will go to the party. Only if Bobby goes is equivalent to. If Alice goes, then Bobby goes explanation. This common structure causes students much confusion. It states only that if Alice is at the party, then Bobby must also be at the party. Note. This condition is not reciprocal. This statement if Bobby is at the party then Alice is also is not necessarily true. Three. Anyone who is not in a cannot be a B is equivalent to, if not a that not be explanation. The Contra positive further simplifies this too. If B, then a four on liaise Arby's is equivalent to, if be than a five of two light switches A and B and B cannot both be on is equivalent to If a is on then B is not explanation. If B is on that day is not will also suffice but is not necessary to state both. One is the contra positive. Of the other six of two light switches A and B A is often be is off is on when B is on is equivalent to a is off if and only if B is off. Explanation A is on if and only if b is on will also suffice. But again, it is not necessary to state both. 35. Flow Chart Game Example 1: as you analyze a flow chart, look for loops that connect groups of element. An example will illustrate in this game, Alison likes every girl, so we start the flow chart with her next. Every girl likes Bridget, but she does not like any of them. So we in the flow chart with Bridget, north of the diagram of balls. Next. Since Dominique likes three girls, two of whom are neither Courtney nor friends scene she must like both Alison and Emily in addition to Bridget, adding this result to the third condition. Courtney likes Dominique to the diagram gives. Finally, since Emily and Friends seen each like only one girl, and everyone likes Bridget, Emily and friends seen each like Bridget Onley, so there is nothing else to add to the diagram. Note. A CD forms a loop because from a the arrows can be fall around to C to D and then back to a . But a D E does not form a loop because from a you cannot get back to a whether you first go to D. So a to D. E. But then you can't go any further from there or whether you go from a to E. Now you stop immediately because you can't go up towards D in the chart. Only URLs from a and D point to E. Yes, we have a narrow here, going from a to D And we have a narrow here going from a t e so on Lee, Allison and Dominique like Emily hence the answer is D. 36. Flow Chart Game Example 2: first list introduced the diagram derived earlier. There is only one click in the chart. A two way aero connects a nd right here between A and D, so they form a clique. The Loop A C D. Does not form a clique because it's not two way a like C. But that feeling is not reciprocal because there's no arrow going back from sea to A it's the answer is B. In case you would like to review the derivation of the original diagram, it will now be repeated as you analyze a flow chart, look for loops that connect groups of element. An example will illustrate in this game, Alison likes every girl, so we start the flow chart with her next. Every girl likes Bridget, but she does not like any of them. So we in the flow chart with Bridget notice how the diagram of balls next. Since Dominique likes three girls, two of whom are neither Courtney nor Francine, she must like both Alison and Emily in addition to Bridget, adding this result to the third condition, Courtney likes Dominique to the diagram gives. Finally, since Emily and Friends seen each like only one girl and everyone likes Bridget, Emily and friends seen each like Bridget Onley. So there is nothing else to add to the diagram. Note. A CD forms a loop because from a the arrows can be fall around to C to D and then back to a But a D E does not form a loop because from a you cannot get back to a whether you first go to d so a to d e. But then you can't go any further from there or whether you go from a to E. Now you stop immediately because you can't go up towards D. 37. Flow Chart Game Example 3: first, let's reintroduce the flow chart. Originally derived in the chart, there are five arrows pointing to be so five girls like be. There are no arrows emanating from B. So none of those feelings a reciprocal. Hence, the answer is D. In case you would like to review the derivation of the original diagram, it will now be repeated as you analyze a flow chart, look for loops that connect groups of element. An example will illustrate in this game, Alison likes every girl, so we start the flow chart with her next. Every girl likes Bridget, but she does not like any of them. So we in the flow chart with Bridget, north of the diagram of balls. Next. Since Dominique likes three girls, two of whom are neither Courtney nor Francine, she must like both Alison and Emily in addition to Bridget, adding this result to the third condition, Courtney likes Dominique to the diagram gives. Finally, since Emily and Friends seen each like only one girl, and everyone likes Bridget, Emily and friends seen each like Bridget Onley, so there is nothing else to add to the diagram. Note. A CD forms a loop because from a The arrows can be fall around to C to D and then back to a but a D e does not form a loop because from a you cannot get back to a whether you first go to d so a to d e. But then you can't go any further from there or whether you go from a to E. Now you stop immediately because you can't go up towards D. 38. Circuit Game Example 1: flow chart games can bring to the foursome settle issues as the following Difficult game illustrates to keep the notation simple that the letter itself stand for the light is on and place a tilde before the letter to indicate the light is off the condition If K is on Ellis off is naturally symbolized as K implies not ill. Miller's K is on implies that l is off the condition J and end cannot both be on means that if one is on the others off so we can symbolise. That is if in is on then Jay is off. We could also say that if J is on then in his off. But we just need one of them, not both The condition m is off if and only if either jay or in is on is nationally symbolized as M is off if and only if J or in is on finally the condition. If oh is on Amazon and it goes off in his off means that oh is on if it only of Amazon. So oh, is on if and only if in his own now we come to the crucial decision with which conditions should we start a chart following the guidelines? Look for the element that occurs in the greatest number of conditions it is in of the three conditions that contained in the condition Oh, on if and only if Amazon is the most restrictive. So we start the flow chart with it, adding the second condition we get not J, adding the third condition we get No, because this condition involves in or weaken. Send an arrow from n down to not him. If it were an and instead of an or then we would not be able to do it because then both conditions would have to be satisfied, J and and would both have to be true before we could put a narrow. Their nurse of the first condition is independent of all the other conditions. So write it down as a second diagram. We'll use this to part diagram to answer the following questions. Since this is a counting problem, we anticipate it will be hard. It won't disappoint us from the bottom chart. We see that both K and L can be off. This condition states Onley that if K is on than l is off it says nothing about the case when K is off. So both K and L could be off. Furthermore, since the bottom chart is independent of the top chart, the status of K and L, whether on or off, does not affect the other lights next from the top chart. If both o and M are on, then immediately we have that Jay is off and him is off. Thus, it is possible for only two lights o an end to be on which eliminates a D and E. Next we check the alternative circumstance where both Oh, and end our off from the top chart. We see that if m is off, then either Jay or in must be on that were already assuming in is off therefore, J must be on and from this condition here. If m is on that, we negate this side and we negate the other side applying the Contra positive which will tell us that both J and in is off in particular. J cannot be on combining these cases em on or off. We see that one and only one of M J and end must be on. Hence it is possible for only one element to be on which eliminates. See, finally we check whether all lights could be off again. The top chart shows that this is not possible if M is off than from this. If and only if statement either jay or end must be on which eliminates a at least one light must be on. Hence, the answer is B. 39. Circuit Game Example 2: notice that answer Choices A, B, C and D all contained either in or oh, but not both. However, the fourth condition, if oh is on that in, is on. And if I was off that in is off is equivalent to saying that Oh is on if and only if in his own. In other words, they are always own at the same time, this eliminates a B c nd instant answers E. 40. Points to Remember: points to remember when constructing a flow chart used the falling guidelines. A look for a condition that starts the flow be looked for an element that occurs in many conditions. See keep the chart flexible. Be aware of the fallacies associated with the Contra positive from, if a then B, you can conclude. If not be that not a again. This is the contra positive, but you cannot conclude either, if not a that not be or if b, the name some means at least one. And perhaps all the following statements contained embedded. If then statements statement. All A's are B's is equivalent to if X isn't A than X, is a B A and B or not both Sees is equivalent to If a is a C than be is not. No a is a B is equivalent to if a then not be on liaise. Arby's is equivalent to, if be than a 41. Mentor Exercise 1: This is a rather hard game. Its underlying structure is actually simple, but there's lots of information to wade through. We start by symbolizing the conditions will use an equal sign to indicate that a person competes in a particular event. The first condition Ben competes on the horse. If it only of carry competes on the vault can be symbolized as B equals age for horse if and only if Kerry is on the vault. The second condition if Darlene does not compete on the parallel bars. So we have Darlene is not on the parallel bars. Then Fred computes in the vault, so F equals V. This in turn community recast using the Contra positive. So we negate both sides of the implication. So F is not in the on the vault implies that Darlene is on the parallel bars. Finally, the last condition. If Emily competes in the vault, then Fred does not can be symbolized. As Emily was in the vault, then Fred is not on the vote to start the flow chart, look for the element that occurs in the greatest number of conditions. It is out. So build a chart around it. Start with the final condition here. Now we know from the previous condition. If Frank Fred is not on the vault, then Darlene is the parallel bars. Adding that to this diagram we get D is equal to p. Finally, the first condition cannot be added to the diagram, so we will listed as a separate diagram. So we have been on the horse if and only if Carrie is on the vault. So these two conditions or these two diagrams, will form our basis for answering the questions. No, that is wild. Since it is not contained in diagram that will use this diagram to answer the questions for the 1st 1 we're told that bin competes on the horse. So from the bottom half of the diagram, we know immediately that carry must be on the vault. Furthermore, if Darlene does not compete on the parallel bars that applying the Contra positive to the top part of the diagram we see that Fred competes in the vault. So going from here, by applying the Contra positive we have, Darlene is done on the parallel bars. Then Fred will have to be on the vote. This feels both slots for the vault so no one else can compete in that event. These restrictions are sufficient to eliminate choices a be see and E. Hence, the answer is D. 42. Mentor Exercise 2: first, let's reintroduce the diagram we derived earlier hit now apply the Contra positive to the top half of the diagram were given that Darlene does not compete on the parallel bars, so that negates the statement here. And then by using the Contra positive, we change the direction of the implication and then negate this, which means that Fred will be on the vault now, continuing to use the Contra positive, you should be able to work down the steps and show that the answer is D. In case you would like to review the construction of the diagram, it will now be repeated. This is a rather hard game. Its underlying structure is actually simple, but there's lots of information to wade through. We start by symbolizing the conditions will use an equal sign to indicate that a person competes in a particular event. The first condition Ben competes on the horse. If it only of carry competes on the vault can be symbolized as B equals age for horse if and only if Kerry is on the vault. The second condition, if Darlene does not compete on the parallel bars, so we have Darlene is not on the parallel bars. Then Fred competes in the vault, so F equals V. This in turn community recast using the contra positive. So we negate both sides of the implication. So F is not in the on the vault implies that Darlene is on the parallel bars. Finally, the last condition. If Emily competes in the vault, then Fred does not can be symbolized as Emily was in the vault, then Fred is not on the vote to start the flow chart, look for the element that occurs in the greatest number of conditions. It is out. So build a chart around it. Start with the final condition here. Now we know from the previous condition. If Frank Fred is not on the vault, then Darlene is the parallel bars, adding that to this diagram we get D is equal to p. Finally, the first condition cannot be added to the diagram, so we will listed as a separate diagram. So we have been on the horse if and only if Carrie is on the vault. So these two conditions, or these two diagrams, will form our basis for answering the questions. Note that a is wild, since it is not contained in diagram that will use this diagram to answer the questions 43. Mentor Exercise 3: first, let's reintroduce the original diagram that we drive from the bottom. Condition been is on the horse if and only if Kerry is on the vault. We know that Ben and Carry must both compete on the parallel bars as his statement. One. Since Albert is an independent element, we intuitively expect he could compete on the horse. But you should verify this as to statement to If Emily competes in the vault, then from the diagram, Darlene must compete on the parallel bars. Emily on the vault implies that Fred is not on the vault, which in turn implies that Darlene is on the parallel bars. This, however, puts three people in the parallel bar event, contradicting the condition that there are two people in each event as to statement three. If Darlene does not compete in the parallel bars that applying the Contra positive to the top diagram shows that Fred must compete in the vault and Emily cannot complete in the vault, so we now use the contra posit to go in the opposite direction. From here to here. Now, it's easy to work out the schedule with ease restrictions. The answer is D In case you would like to review the construction of the diagram, it will now be repeated. This is a rather hard game. Its underlying structure is actually simple, but there's lots of information to wade through. We start by symbolizing the conditions will use an equal sign to indicate that a person competes in a particular event. The first condition Ben competes on the horse. If it only of carry competes on the vault can be symbolized as B equals age for horse if and only if Kerry is on the vault. The second condition If Darlene does not compete on the parallel bars, so we have Darlene is not on the parallel bars. Then Fred competes in the vault, so F equals V. This in turn community recast using the Contra positive. So we negate both sides of the implication. So F is not in the on the vault implies that Darlene is on the parallel bars. Finally, the last condition. If Emily competes in the vault, then Fred does not can be symbolized. As Emily was in the vault, then Fred is not on the vote to start the flow chart. Look for the element that occurs in the greatest number of conditions. It is out. So build a chart around it. Start with the final condition here. Now we know from the previous condition. If Frank Fred is not on the vault, then Darlene is the parallel bars, adding that to this diagram we get D is equal to p. Finally, the first condition cannot be added to the diagram, so we will listed as a separate diagram. So we have been on the horse if and only if Carrie is on the vault. So these two conditions or these two diagrams, will form our basis for answering the questions. No, that is wild, since it is not contained in diagram that will use this diagram to answer the questions. 44. Mentor Exercise 4: first, let's reintroduce the original diagram that we drive. This question is difficult because there are six different ways to assign different events to carry and Emily Additionally, the string of inferences needed to answer the question is quite long to begin at the new condition. If Kerry does not compete in the vault than Emily does to the diagram. So we have Carry does not Compete in the vault implies that Emily does no. One choice. A. It says that Ben does not compete on the horse. So, looking at the bottom half of our diagram, we can say immediately that if Ben does not compete in the horse, then Carry does not compete on the vault, plugging this into the top of the diagram. We can then work through the steps to show that this leads to a contradiction, namely, that Darlene also competes on the parallel bars. Hence, the answer is a. In case you would like to review the construction of the diagram, it will now be repeated. This is a rather hard game. Its underlying structure is actually simple, but there's lots of information to wade through. We start by symbolizing the conditions will use an equal sign to indicate that a person competes in a particular event. The first condition Ben competes on the horse. If it only of carry competes on the vault can be symbolized as B equals age for horse if and only if Kerry is on the vault. The second condition If Darlene does not compete on the parallel bars. So we have Darlene is not on the parallel bars. Then Fred competes in the vault, so F equals V. This in turn community recast using the contra positive. So we negate both sides of the implication. So F is not in the on the vault implies that Darlene is on the parallel bars. Finally, the last condition. If Emily competes in the vault, then Fred does not can be symbolized. As Emily was in the vault, then Fred is not on the vote to start the flow chart, look for the element that occurs in the greatest number of conditions. It is out. So build a chart around it. Start with the final condition here. Now we know from the previous condition. If Frank Fred is not on the vault, then Darlene is the parallel bars, adding that to this diagram we get D is equal to p. Finally, the first condition cannot be added to the diagram, so we will listed as a separate diagram. So we have been on the horse if and only if Carrie is on the vault. So these two conditions or these two diagrams, will form our basis for answering the questions. No, that is wild, since it is not contained in diagram that will use this diagram to answer the questions. 45. Mentor Exercise 5: first, let's introduce the diagram that we originally derived. Now we're told that Darlene competes in the vault, so that negates the conclusion here. So we complied. The Contra positive, and we get that Fred is in the vault event and Emily is not now been. Cannot compete on the horse because we would then have immediately carry is on the vault. And we've already derived that. Fred is on the vault, and we're given that Darlene is on the vault, which contradicts the statement that to compete in the vault from these conditions, you should be able to work out three valid schedules. Hence the answer is B. In case you would like to review the construction of the diagram, it will now be repeated. This is a rather hard game. Its underlying structure is actually simple, but there's lots of information to wade through. We start by symbolizing the conditions will use an equal sign to indicate that a person competes in a particular event. The first condition Ben competes on the horse if it only of carry competes on the vault can be symbolized as B equals age for horse. If and only if Kerry is on the vault the second condition. If Darlene does not compete on the parallel bars. So we have. Darlene is not on the parallel bars. Then Fred competes in the vault, so F equals V. This in turn, community recast using the contra positive. So in the gate, both sides of the implication so f is not in the on the vault implies that Darlene is on the parallel bars. Finally, the last condition. If Emily competes in the vault, then Fred does not can be symbolized as Emily was in the vault, then Fred is not on the vote to start the flow chart, look for the element that occurs in the greatest number of conditions. It is out. So build a chart around it. Start with the final condition here. Now we know from the previous condition. If Frank Fred is not on the vault, then Darlene is the parallel bars, adding that to this diagram we get D is equal to p. Finally, the first condition cannot be added to the diagram, so we will listed as a separate diagram. So we have been on the horse if and only if Carrie is on the vault. So these two conditions or these two diagrams will form our basis for answering the questions. Note that A is wild, since it is not contained in diagram that will use this diagram to answer the questions. 46. Mentor Exercise 6: first, let's reintroduce the diagram that we originally derived. This question is hard or at least long because it actually contains five questions, adding the new condition. If Fred does not compete in the vault than Emily does, just slightly changes the diagram. You just get a new era here. Going from Fred not in the vault implies that Emily is in the vault. Start with a If Ben competes on the horse, then from the bottom diagram, we know that Carry competes in the vault, turning to top diagram. Clearly, Emily cannot compete in the vault, since that would put three people in the vault. Emily Carrie in Albert. But if Emily does not compete in the vault that immediately by the Contra positive, we have that Fred does compete in the vault, which leads to the same contradiction. It's the answer is a. In case you would like to review the construction of the diagram, it will now be repeated. This is a rather hard game. Its underlying structure is actually simple, but there's lots of information to wade through. We start by symbolizing the conditions will use an equal sign to indicate that a person competes in a particular event. The first condition Ben competes on the horse. If it only of carry competes on the vault can be symbolized as B equals age for horse if and only if Kerry is on the Volt. The second condition If Darlene does not compete on the parallel bars. So we have. Darlene is not on the parallel bars. Then Fred competes in the vault, so F equals V. This in turn, community recast using the contra positive. So we negate both sides of the implication. So F is not in the on the vault implies that Darlene is on the parallel bars. Finally, the last condition. If Emily competes in the vault, then Fred does not can be symbolized as Emily was in the vault. Then Fred is not on the vote to start the flow chart, look for the element that occurs in the greatest number of conditions. It is out. So build a chart around it. Start with the final condition here. Now we know from the previous condition. If Frank Fred is not on the vault, then Darlene is the parallel bars, adding that to this diagram we get D is equal to p. Finally, the first condition cannot be added to the diagram, so we will listed as a separate diagram. So we have been on the horse if and only if Carrie is on the vault. So these two conditions, or these two diagrams, will form our basis for answering the questions. Note that A is wild, since it is not contained in diagram that will use this diagram to answer the questions. 47. Solo Exercise 1: diagramming makes this a fairly easy game start, replacing the seven towns in a circle as follows. This is not the most efficient way to arrange the towns, but it is the most natural, that is, it's the first scheme that were likely to think of next. Add Route one to the diagram, then add route to finally add Route three from the chart. There were three roads going to L one red and to blue, and no other town has more than two roads going toe it. For example. K has only one road Jay has to green roads. I also has two rows to blue roads, as does age in, has a green road in a red road, and M has to read roads. Hence the answer is E. 48. Solo Exercise 2: Let's reintroduce the chart that we derive The problem. One again from the charter is only one road going to K, and every other town has at least two roads going to it. So K has a fewest number of connections. So K has one. Jay has to. I has to. Each has two in Has to m has to, and l has one, 23 roads. Hence the answer is B. In case you would like to review the solution set up, it will now be repeated. Diagramming makes this a fairly easy game. Start replacing the seven towns in a circle as follows. This is not the most efficient way to arrange the towns, but it is the most natural. That is, It's the first scheme that were likely to think of next. Add Route one to the diagram, then add route to finally add Route three 49. Solo Exercise 3: adding the new road connecting I and J we get now there are three ways to get from belt. Okay, we can go from lto M that's one than to end to From end to J three and from J. Decay is four. We could also go from L. A to H one toe I to to j 32 K 4 and the third path is from El Toe I. That's one to J to decay three. So this is a shortest path because there's only three segments, and neither in nor AM is on that shortest path. So the answer is e no. Are drawing is a little misleading because it looks like the blue line and the Green Line intersect here. They don't. There's a bridge where one road goes over. The other to avoid that we could drawn are going from in two J, which would make it clear, but it was simply more direct and quicker to draw a line crossing the to. In case you would like to review the solution set up, it will now be repeated. Diagramming makes this a fairly easy game start, replacing the seven towns in a circle as follows. This is not the most efficient way to arrange the towns, but it is the most natural. That is, it's the first scheme that were likely to think of. Next. Add Route one to the diagram, then add route to finally add Route three. 50. Solo Exercise 4: no adding a new road directly connecting H and K to the original flow chart we get now. There are three ways to get from L to J without passing through any town more than once, as shown, we can go from L. A to M to end to Jay. That's one round. Well, we can go from bell toe age two K to Jay. That's the second round. Or we can go from l toe I to age two K and then to Jay. That's the third route. Inst answer is C. In case you would like to review the solution set up, it will now be repeated. Diagramming makes this a fairly easy game. Start by placing the seven towns in a circle as follows. This is not the most efficient way to arrange the towns, but it is the most natural, that is, It's the first scheme that were likely to think of next. Add Route one to the diagram, then add route to finally add Route three 51. Solo Exercise 5: adding the new roads to the original diagram gives now the route and choice e directly connects either m or in okay. If we look at the chart, there is no direct route. There is no path going directly from M decay, and there's no direct path queen from into K. In order to get Gok from M, you have to take an intermediary path, say, from M toe l two j and then down to K. And from in you could go from into J two k, but you can't go directly from in decay. Hence the answer is E. In case you would like to review the solution set up, it will now be repeated. Diagramming makes this a fairly easy game. Start replacing the seven towns in a circle as follows. This is not the most efficient way to arrange the towns, but it is the most natural. That is, it's the first scheme that were likely to think of next. Add Route one to the diagram, then add route to finally add Route three 52. Solo Exercise 6: let the letter itself stand for the light is on and place it Tilda before the letter to indicate that the light is off. We begin the chart with the third condition if oh is on than in is on and it goes off in his off. Which means that always on if and only if in his on next, the condition J and N cannot both be on can be symbolized, as if in is on. Then Jay is off. Or we could say, If J is on, then in is off. But we'll go with the 1st 1 And in that the diagram we get finally, the condition if K is on then in is on is naturally symbolized, as if K is on than in his own. And we'll put the K below the end and draw the arrow up to in No. In the diagram, there is a sequence of air old starting a K to end toe. Oh, so if K is on, then almost beyond. But that is not listed as one of her answer choices. But there is also a sequence of arrows going from K to end to not J. So if K is on, then J must be off, and that is offered as choice. See, hence, the answer is C. 53. Solo Exercise 7: let's reintroduce the diagram created in the previous problem. The fact that Jay is on promises to apply the Contra positive to the diagram. Since here we have not J we turn that into J and by the Contra positive we change the direction of the implication and we negate the end, which is the conclusion and applying the country positive. Here again, we now in the gate the oh, and since we have not in we apply the country positive again in the arrows lips direction and we negate the K. Since we're given that Jay is on, this diagram immediately shows that in is off and oh is off. And it also shows that que is off. So that gives us three lights off. Oh, in and K now, no conditions apply to either l or m. So their status whether honor office independent of the other lights hence both could be off, which gives us two more lights that could be off. Plus the three lights that we will read ride that must be off for a total of five lights. That could be off. Hence, the answer is E In case you would like to review the solution set up, it will now be repeated. Let the letter itself stand for the light is on and place a tilde before the letter to indicate that the light is off. We begin the chart with the third condition if oh is on than in is on and it goes off in his off. Which means that always on if and only if in his on next, the condition J and N cannot both be on can be symbolized as if in is on. Then Jay is off. Or we could say, If J is on, then in is off. But we'll go with the 1st 1 And in that the diagram we get finally, the condition if K is on then in is on is naturally symbolized, as if K is on than in his own. And we'll put the K below the end and draw the arrow up to in 54. Solo Exercise 8: the 1st 3 conditions say that rumors can go from G to H i N J So our diagram becomes G h g toe I, and from G to J Now the Fourth Conditions says that rumors can go between H and I, so we get a two way a row between eight and nine. The fifth condition says that rumors can pass from H two j, but not vice versa. So we have a narrow going from H J. The sixth condition says a rumor can go from Jato I but not vice versa. Finding the last condition says that rumors can pass either way between J and K. So we have a two way Aargh between J and K. No, I room was starting at. I can only go to H and from h. It can only go to J can't go this way cause the arrows pointing towards H not towards G and from J. You can go to K. It can also go back upto I which would just take it in a circle so everyone will get the rumor except for G. And the answer is B 55. Solo Exercise 9: first, let's introduce the original diagram that we dried. Begin with choice. A original diagram shows that a rumor can spread directly from age to J along this path here. So this is unlikely to be the longest path because it has only one segment turning the choice be. The diagram shows that a rumor can pass directly from J to K, since there cannot be two answers thistle inmates both A and B turning the choice. See, the diagram clearly shows that no rumor can spread from H two g can on legal from G H because that's the direction of the arrow for Choice D. The diagram shows that a rumor originating at I can spread to directly to age and then from H two j. This rumor consists of two segments. Finally, the path taken and Choice E from My decay includes the path in choice D plus an additional segment from J Decay. So we go from I to age and from H Down to J and then from J. Decay. And this is three segments. Hence the answer is E. In case you would like to review the solution set up, it will now be repeated The 1st 3 conditions say that rumors can go from G to H I N J. So our diagram becomes G H g toe I, and from G to J. Now the fourth condition says that rumors can go between H and I. So we get a two way a row between H and I. The fifth condition says that rumors can pass from H two j, but not vice versa. So we have a arrow going from H J. The sixth condition says a rumor can go from Jato I, but not vice versa. Finding the last condition says that rumors can pass either way between J and K, so we have a two way Aargh between J and K. 56. Solo Exercise 10: first, let's reintroduce the original diagram. The original diagram shows arrows leading directly from H two eye and from age to J, and only these two. Hence, the answer is C. As to the other. Choice is both A and B are incomplete, since we just showed that there are correct past, leading from H to both I and J Choice D is inaccurate, since a rumor cannot pass from H two K without passing through J. Now, a rumor could start here at H and go toe I. But then that's a dead end because there are no arrows going from I to any of the other letters other than age. But it can go from H down to J and then from J. K. Hence, a rumor needs at least two segments to reach K from age. Finally, choice E is inaccurate. No arrow points from H nor from any other person to G. All the arrows go from G to other letters G to H G, T I and G, T, J and Snow. Rumors can pass from age nor any other person to G. In case you would like to review the solution set up it will now be repeated. The 1st 3 conditions say that rumors can go from G to H i N J So our diagram becomes G H g toe I, and from G to J Now the Fourth Conditions says that rumors can go between H and I. So we get a two way a row between H and I. The fifth condition says that rumors can pass from H two j, but not vice versa. So you have ah, arrow going from H J. The sixth condition says a rumor can go from Jato I but not vice versa. Finding the last condition says that rumors can pass either way between J and K, so we have a two way Aargh between J and K. 57. Solo Exercise 11: First, let's introduce the originally derived diagram drawing a two way aero from G and K. We get the falling now look a choice. D If there were segment from I two g, no words, we'd fill in the arrow here. Then I could spread rumors to both h I would go to G and then G would go to age. And, of course, I is spreading the rumor to G and could receive route rumors from both h N g h goes toe I and G goes to die further. H could spread rumors to both I and J H tie again as original and also from H two J as a ritual diagram and could receive rumors from both G and I. He could spread rumors to both I because there goes from Jato I and Decay because there's an arrow going from J to K and could receive rumors from both G this arrow right here from G to J. And of course, from the original diagram, Jay could receive a rumor from K from this area here. Que could spread rumors to G from the new era we have here and from the original diagram it can spread rumors to J and could receive rumors from both G because it's a two way Errol here and from J. Because there's a two way arrow here. Finally, G could spread rumors to both I from the ritual diagram and K from the new Arab to weigh Errol here between G and K and could receive rumors from both I and K. Again, we have this new arrow here and then the new two way aero here. Hence the answer is Choice D. In case you would like to review the solution set up, it will now be repeated. The 1st 3 conditions say that rumors can go from G to h i N J. So our diagram becomes G h g toe I, and from G to J Now the Fourth Conditions says that rumors can go between H and I. So we get a two way a row between H and I. The fifth condition says that rumors can pass from H two j, but not vice versa. So you have ah, arrow going from H J. The sixth condition says a rumor can go from Jato I, but not vice versa. Finally, the last condition says that rumors can pass either way between J and K, so we have a two way Aargh between J and K. 58. Solo Exercise 12: the first condition. If Candace hired is an editor than both John and Mary are hired as gaffers. And if both John and Mary are hired as gaffers than Kentis, hired as an editor means that Kent has hired isn't editor if it only of both John and Mary are hired as gaffer's. This can be symbolized as Qin is an editor if and only if both John and Mary our gaffer's the second condition. If nor is not hired as an actor than Thomas hired as a gaffer can be symbolized as Nora, not an actor, then Tom is a gaffer. If Kent is hired as an editor, then we know from this diagram we can immediately say that John and Mary or gaffer's Look a Choice D it has Tom hired as a gaffer. This forces three people Tom, John and Mary to be gaffer's, violating the fact that only two people are hired as gaffer's. Hence, the answer is D 59. Solo Exercise 13: since Kimpton. Olivia fill the to gaffer positions. No one else is hired as a gaffer, only two. Now consider the second condition. If Nora is not hired as an actor, then Tom is hired as a gaffer, taking the Contra positive of this diagram. We negate both sides. Nora equals an actress, and we changed the direction of the implication. And Tom is not a gaffer now, since Tom cannot be hired as a gaffer, those two positions were already filled by Kent and Olivia. Nora must be hired as an actor because Tom is not a gaffer. Means that premise of this if then statement is true. Therefore the conclusion must be true. Instead, answers E. 60. GROUPING GAMES Introduction: grouping games, We have thoroughly studied the various ways to order elements. In this chapter, we will study various ways to group elements. We got a taste of the task involved in grouping elements when he studied hybrid games, which both order and group items. Because grouping games partition elements into sets, the number of elements is often an issue as mentioned before accounting can be challenging . This tends to make grouping games more difficult than ordering games. Pay close attention to the maximum or minimum number of elements in a group. This is often the heart of the game. Grouping games can be classified as those that partition elements into two groups and those that partition elements into three or more groups. The former sometimes called selection games because they select elements from a pool dividing the pool into two groups, those selected and those not selected, the phone example. A little street 61. Selection Game Example 1: this problems rather convoluted, because not only are their direct conditions on the players such as Drexel and Bryant do not both start, but there are also constraining the miracle conditions, such as exactly three fast break specialists must be chosen. It is best asshole this problem without a diagram, however, we will still symbolize the conditions for clarity and easy reference. The condition James starts on Lee if Bryant starts, implies that if James is starting than Bryant must be starting as well. So we symbolise it as it's James, then Bryant. The condition Drexler and Bryant do not both start means that if one starts than the other does not we symbolize this as if Drexler starts, then Bryant does not. We could also say, if Bryant starts, then Drexler does not, but we don't need both. Students often misinterpret this condition to mean that neither of them starts to state. Neither starts put both at the beginning of the sentence. As such, both Drexler and Bryant do not start the condition. If James starts them alone, does not, is naturally symbolize as it's James, they're not Malone. It tells us that if J starts that M does not, but tells us nothing when M does not start such a condition, where the two parts of an if then statement do not similarly affect each other is called a non reciprocal condition. On the other hand, a condition such as James if and only if not Monroe, affects J and M equally. In this case, we're told that if J starts that M does not as before, but we are told, additionally, that if M does not start, then Jay does. It is important to keep the distinction between reciprocal and non reciprocal relations clear. Become mistake is to interpret a non reciprocal relation as reciprocal. No, Ewing, Leitner and Robertson are independent because there are no conditions that refer directly to them. We now turn to the questions from the condition. If James, then Bryant, we know that if James starts, then Bryant must start as well. Now both James and Brian are fastbreak specialists, and three of the four fastbreak specialist must be chosen. So at least one of the remaining fastbreak specialist, Johnson or Pippen, must be chosen. Hence, the answer is D 62. Selection Game Example 2: we shall use the method of indirect proved to solve this problem that is, assuming particular answer choice is true. Then check whether it leads to a contradiction or an impossible situation. If so, it is the answer. If not, then select another answer choice and repeat the process until a contradiction is found. Begin with Choice A. Both Ewing and Rex Lor are from Group A, so the remaining three starters must be chosen from Group B. Additionally, they must all be fastbreak specialists, since neither Ewing nor Drexler is, and there are exactly three fast break specialists in Group B, so the third fastbreak specialists cannot be chosen. The answer, therefore, is a. This type of question can be time consuming because you may have to check all the answer choices. Save these questions for last. 63. Selection Game Example 3: this problem is both long and hard again. We will use an indirect proof start with choice A. Both Johnson and Bryant are from Group A, and both are fastbreak specialists. So the remaining three starters must be Children from Group B, one of which must be a fastbreak specialist. Now if James Robertson and Leitner are chosen, then there will be three fast break specialists, and none of the initial conditions will be violated. So is not necessarily false, and we eliminate it. Next, we check choice be. Both Leitner and Malone are from Group B, and neither is a fastbreak specialist, so the three remaining starters must all be fastbreak specialists, and two of them must be from Group A, Johnson and Bryant. This leaves only James and pitons to choose from. James cannot be chosen because Malone has already been chosen. Remember the fourth condition? If James then not Malone, and from the new condition hip and cannot be chosen because Bryant has already been chosen . Hence, the answer is B 64. Selection Game Example 4: James cannot start with Malone, according to the fourth condition. If James starts them alone, does not to play three fast break specialists. Therefore, Johnson, Bryant and Pippen are all required to start. There are choices for the fastbreak artists. We've already eliminated James, and since three have to be selected, the three remaining or all selected Johnson, Bryant and Pippen. Since Johnson and Brian are from Group A and exactly two players from that group start, these two players comprise the complete list of starters for Groupe when Malone also starts , since the answer is C. 65. Selection Game Example 5: Let's solve this problem by Contradiction supposed Bryant does not start. Then the three fast break specialist must be Johnson, James and Pippen. We have eliminated Bryant by assumption, but then, from the second condition, if James starts, then Bryant must start. But we've assumed that Bryant does not start. So this is a classic contradiction. In spring, it must always start, and the answer is he. 66. Grouping by Threes Example 1: we start by symbolizing the conditions. The condition G serves with M is naturally symbolized as G equals M. The condition f does not serve with them is civilized as F does not equal them. The condition L serves with only one other person means that l is on the committee of to We symbolize this as Belle equals two. The diagram will consist of three compartmentalized boxes. This gives the following schematic before turning to the questions too readily. Derive conditions should be noted first, since G serves with them and f does not serve with them. F cannot serve with G second since L sers on the two person committee, they'll cannot serve with G or M. Otherwise, l would be on a three person committee now for the first question. Answer. Choice A is not a committee because l must serve on a committee of two and we have three people here listed. Eliminate A B is not a committee because G cannot serve with F. That's one of her first rarely drive conditions. Neither see nor D is a committee, since G must serve with them. It's by process of elimination. The answers E 67. Grouping by Threes Example 2: first, let's reintroduce the diagram derived earlier. We shall use an indirect proof start with a if f serves with l, then G and M could serve on committee. One. Que could serve on committee to and the remaining people conserve at random without violating any conditions so F could serve with L. This eliminates a next test. Be if if serves with J on committee one, then G and M would have to serve on a committee to, and the remaining people could be placed as follows. This diagram does not violate any initial condition so F could serve with Jay. This eliminates be next test choice. See, there are two possible places for the pair. G and M Committee one and Committee to If G. N m. Serve on committee one than F would have to serve on a committee to as shown. Clearly, this diagram leaves no room for K. Since K cannot serve with either M or F. The case with the pair G and M serving on committee to leads to a similar result hence, L cannot serve with age, and the answer is C. 68. Grouping by Threes Example 3: first, let's reintroduce the diagram derived earlier if H serves with K on committee one than G and M must serve on committee to because G and M always served together and there's only one space left for committee. Three. This gives the following diagram. The diagram with H and K on committee, too, is not presented because it generates the same results again. We client indirect proof. Start with a if f serves with K, then from the diagram f must serve on committee one, and we can place I and J on committees two and three, respectively, as follows. This diagram does not violate any of initial conditions so F could serve with K G. Is with him in committee to shock. F is not with him because F is here in committee one with H and K Check and L is in committee three, which is a committee of just two people shot. This eliminates choice. A next test choice be J and F cannot serve on committee one, since from the above diagram, H and K are already there. Likewise, J and F cannot serve on committees two and three. Hence, the answer is B 69. Grouping by Threes Example 4: first, let's reintroduce the diagram derived earlier. We shall construct counter examples for four. The answer choice is the one for which we cannot construct The counter example will be the answer. Start with twice a Suppose K serves on committee one and G serves on committee to then from the condition G equals m. We know that M must also serve on committee to and the remaining people can be placed without violating any of the initial conditions as follows. This diagram is the counter example not only to a but to see an E as well. This eliminates a see and e next best choice be Suppose I serves on committee one with G and M Then the remaining people can be grouped as follows. This diagram does not violate any initial condition. So it is a counter example to choice B, which eliminates be hence, by process of elimination. The answer is D 70. Grouping by Threes Example 5: first, let's reintroduce the diagram we derived earlier. The first counter example in the previous problem shows that K can serve on a three person committee. This eliminates a next during the choice Be supposed AM serves with H on committee one. This forces G to also serve on committee one. Because G always serves with him the first condition. No place F on committee three and remaining people as follows. This diagram does not violate any initial condition. So M serves with age is consistent with the initial conditions. This eliminates be next turning to choice. See, suppose m. H and I served together on committee one. But since M must serve with G, there were before people on committee one. The same result occurs when M. H and G R in committee to and C is inconsistent with the initial conditions. And the answer is C. 71. Points to Remember: points to remember one. Pay close attention to the maximum or minimum number of elements in a group, for this is often the part of the game. Two grouping games were classified as those that divide elements into two sets selection games and those that divided elements into three or more sets. Three reciprocal condition effects. Both elements equally. Four. Don't interpret non reciprocal conditions as reciprocal for the method of indirect proof is often used with grouping games. 72. Warm Up Drill 1: we conform three single letter groups A be and see and we could also form three double letter groups A B BC and A C. And we conform one triple letter group. All three letters A, B and C. So we have three here, three here and one there for a total of seven rooms. Let's select the letter and then calculate the number of pairings for A We compare it off with B or we compare it off with C. What we compare it off with? D? No, for B. We compare it off with a but we've already done that. So be can be paired with C and B can be paired with D moving to see We've already paired it with a and we've already paired it with B. But we haven't yet paired it with D, and we have a total of six ways. Three plus two plus one It six ways of grouping for Question three. Let's start with the first set since two elements must be selected from each set, we must select a, B, A, C and B C. Now for a B. We have three choices from the other set. We can select d e de of and eat off. But we're told that if a is selected, then on Lee D could be selected so we can eliminate the third option here and for a C. We have three ways of selecting We have D e again d f again and e f. And once again we have the eliminate. PdF Finally, for BC, we get d e again d f and e A and in this case, E f is possible because A has not been selected. This gives is a total of one to skipping this 134 skipping this 1567 So the answer is seven . 73. Warm Up Drill 2: for Question one. We're told that some elements must be selected from the set A B so it may be a It may be be that is selected where it may be both A and B that are selected no, just a selected then from the second set, C, D and E. We cannot select both C and D, so it must be either see and E or D and e likewise that just be a selected. Then we could select C e or de. Finally, it's a MB are selected. Then we can select only one element from the other set, and it could be a C four a D or Annie. And to add these up, we get one to three four 567 for the second question, the condition the is not selected unless easiest selected is equivalent to if the then Z Now the condition why and w cannot both be selected is equivalent to. If y is selected, then W is not the condition you is selected on Lee. If V is not, is equivalent to if you is selected, then V is not the final condition says that either the or why is selected, but not both. So we have two scenarios either via selected or why is selected if via selected then from this condition we completely say that Z is selected. And from this condition here we can take the Contra positive. Not not the is V and then change the direction of the implication and we get not you. So we know that you is not selected. And from this condition here we know that why is not selected now Combining the pair these e with the two remaining elements W and X, we get V z w and these e x now if why is selected then of course, V cannot be selected and immediately w cannot be selected from this condition here. So why can be paired with the three remaining elements? Why, you x? Why u z and why xy so we have one to 345 pairings 74. Solo Exercise 1: the first condition either G or I must be selected can be symbolized as G or I. The condition hrk must be selected can be symbolized as H or K. The condition neither k nor I can be selected with age can be symbolized as if age is selected. I think a is not. And if h is selected and I is not the condition, neither L nor G can be selected unless the other is also selected. Simply means that if either L or G is selected, then both must be selected. In other words, L if and only if she choice A does not contain either h or K which violates the second condition. So eliminate a same thing for D. It does not contain H or okay, choice be it violates the condition if age not Kay. Because we have an h here and we have a K here choice See violates the condition l if, and only if g. We have a G in the choice, but we don't have an l. So eliminate. See, it's by process of elimination. The answer is e 75. Solo Exercise 2: first, let's reintroduce the diagram derived earlier. Begin with choice. A. Selecting both h and J will satisfy all the conditions. Insulin a choice a turning to choice be. Since H is selected, I cannot be selected. That's from this condition right here. If age than not I it's from the condition G or I G must be selected now from the condition l If it only of G, we must select l. This scenario has five items being selected the given H, J and M and then the L and G that we derived. And that violates the fact that only four items air selected. Hence the answer is B. 76. Solo Exercise 3: first, let's reintroduce the diagram we drive earlier. Since I have selected the condition, H implies not. I prevents h from being selected. We can take the Contra positive this expression so not not I would be I. And then we change the direction of the implication and we get not h hence the condition H or K forces K to be selected. Since we've already derived that we cannot select age no from the last condition. Neither L nor G can be selected since they must be selected together, which would deal a group of five. This leads only J to be selected. Hence the answer is D. 77. Solo Exercise 4: first, let's reintroduce the diagram dr. Earlier. Begin with choice A. If I has actually selected, then the four items selected would be fully determined. But choice A does not require that I be selected. Suppose gs selected. Remember either G or I must be selected from the first condition. Then l must be selected. Since G and L must be selected together, that's our last condition. Now we can satisfy all the conditions by selecting either H and M or H and J. It's the item selected are not fully determined. This eliminates choice a turning the choice be Since G is selected, we know that l must be selected from this condition Now since I has been selected, H cannot be selected again. That's the Contra positive of this condition here, negating not, I get I which implies change the direction of the implication and then not H hence from the condition H or K, we know that K must be selected. Thus the four items air uniquely determined the I in the G that we were given and the l and the K that we derived instead, answer is B 78. Solo Exercise 5: our strategy here is just the check each answer choice against the given conditions. Choice A is not possible. It violates the condition. The number of items in Group two is less than or equal to the number of items in Group three here. Group to has three items which is not less than Group three, which has only one item. Choice be is not an acceptable grouping because it violates the condition the and W cannot be in the same group. Choice C is not an acceptable grouping because it violates the condition. X Kim B in Group three Onley f wise in Group three and we have X here in group three and know why with it. And we have the same problem in answer choice e. It's by elimination. The answer is D, but let's verify that the four conditions are satisfied. Group one has the same number of items, one as group to which has one as well, so the first condition is satisfied. The number of items in Group two is less than the number of items in Group three four versus one, so the second condition is satisfied. V is in group one and W isn't Group three, so the third condition is satisfied and we have X in Group three along with why so the fourth condition is satisfied. 79. Solo Exercise 6: ex cannot be in Group one because we're given that group one contains only the item. Why now? Suppose excess in Group three then why must also be in Group three since X can be in Group three? Onley of why is in Group three? This is the last condition. But this violates the fact that why is given to be in Group One. Since we have shown that X cannot be in either Group One or Group Three, it must be in Group two hence, the answer is E. 80. Solo Exercise 7: choice A as possible. The following grouping satisfies all the conditions and has W and Y in Group two, Group one, Group two and Group three, Please Exit Group one W Why in Group two as it must be and then we could put u V and Z in Group three. Choice See is possible. The following grouping satisfies all the conditions. It has only X in Group one Choice. Dia's possible. The following grouping satisfies all the conditions and has you in Group three W Why at dizzy and group too. And then the U and Group three Choice is possible. The falling grouping satisfies all the conditions and has two items in Group three. We have W why in group one, xz and group too and the U in Group three. It's by process of elimination. The answer is B, but let's verify that Place V and Group three and place you in Group One. Since V and W cannot be in the same group, W must be in group to remember. There's only one item now in Group one. Further since W and wire in the same group. Why must also be in Group two now? Since the number of items in Group two is less than or equal to the number of items and Group three All the remaining items must be in Group three. However, this diagram violates the condition. X Kim being Group three Onley f Why is in group three? Well, we have an X here and there's no why with it. Hence again, the answer is B. 81. Solo Exercise 8: since Group two contains only one item, Group one must contain only one item. That's from the first condition the number of items in Group one. It's less than or equal to the number of items in Group two. Now. Group three cannot contain both the NW from the third condition because V and W cannot be in the same group. Hence, either the or W must be in Group One, as neither can be in Group two. Hence, the answer is a. 82. ASSIGNMENT GAMES Introduction 1: assignment games. We have discussed various ways to order elements, ordering games and various ways to group elements grouping games. Now we will discuss various ways to assign characteristics. Two elements. These air assignment games assignment games will wind up our discussion of the three major types of games. They tend to be the hardest games, so it's wise to save them for last, a seven games magic characteristic with an element of a game. For example, you may be asked to sign a schedule Bob works holding Monday, Tuesday or Friday. Or you may be told that person to see their Democrat or Republican. Because the characteristics are typically assigned in groups of elements, assignment games can look very similar to grouping games. Additionally, in grouping games, the groups are often identified by their characteristics. Over in assignment games, you pair each element with one or more characteristics, whereas in grouping games you partition elements into two or more groups. It is important to identify the type of game you're dealing with because different methods are needed to solve each time. The following examples illustrate the distinction between these two types of games. The first from the Group Games chapter is the Olympic Dream Team game. The second is an assignment game 83. Assignment Games Introduction 2: This is the Olympic Dream Team grouping game from the grouping game chapter. In this game, the goal is to select the starting lineup, thereby dividing the elements into two groups those selected and those not notice how the goal in the following game differs. 84. Assignment Games Introduction 3: in this game. The team has already been selected. Now the goal is to assign a position characteristic to each player and decide whether he is a free agent characteristic. No said the conditions, such as all the guards are free agents apply to groups of individuals. This makes the game, at first glance, appear to be a grouping game. Many assignment games can be sold very efficiently by using elimination grid. An example will illustrate this method of diagramming. 85. Elimination Grid Example 1: We indicate that a teacher does not work at a particular time by placing an Exxon elimination grid. Placing the two conditions warrant cannot teach on Monday or Thursday, and Warren can teach only evening classes on the grid gives placing the remaining conditions in like manner gives to answer the following questions. We will refer only to the grid, not the original problem. The grid clearly shows that all three can work on Tuesday night. Warn does not have index in the nighttime position. Nor does Doeren. Nor does Emerson. It's the answer is C. 86. Elimination Grid Example 2: first, let's reintroduce the elimination grid we derived earlier. Now Dornin and Novak are the only people can work Monday evening, and three classes are always in session, so extra help will be needed for Monday evenings. Therefore, the answer is a. 87. Elimination Grid Example 3: first, let's reintroduce the elimination rid derived earlier. The condition Dean Peterson cannot teach him Wednesday if Novak teaches on Thursday and Novak teaches on Thursday. If Dean Peterson cannot teach on Wednesday can be symbolized as P does not equal w if, and only if in equals Thursday. Now, if Novak works every day of the week except Wednesday, then in particular, he works Thursday. So from the condition that we just arrived, we know that Dean Peterson cannot work on Wednesday. But from the grid, this leads only Novak and Emerson to teach the three Wednesday morning classes because Dean Peterson cannot work. On Wednesday, we put it X here for Wednesday morning. Now we can clearly see that only Novak and Emerson are available on Wednesday morning. Inst answers d. 88. Elimination Grid Example 4: if you remember to think oven if and only if statement as an equality that this will be an easy problem. Remember the condition we drive before Peterson does not work on Wednesday? If and only if Novak works on Thursday. Taking the Contra positive of both sides of this if and only if statement. We get P works on Wednesday if and only if Novak does not work on Thursday. But the question states that Novak does not work on Thursday. Therefore, we can immediately conclude that Peterson works on Wednesday and the answer is D. 89. Assignment Games Introduction 4: caution. Not all scheduling games lend themselves to an elimination grid. It's sweet when this method can be applied, because the answer is typically can be read directly from the grid with little thought. Only one through the Simon Games, however, can be solved this way. Most often. The game will require a more functional diagram, and you'll need to spend more time tinkering with it. When you first reading assignment game, you need to quickly decide whether or not to use an elimination grip. You may decide to use a grid, then spend three minutes trying to set it up, only to realize that you have taken the wrong path and have wasted three minutes. Unfortunately, exact criteria cannot be given for when to use an elimination grid. But this much can be said if only two options characteristics are available to the elements . Yes, no on off etcetera, Then elimination grid is probably indicated in the next game, which is considerably harder. More than two options are available to each element. Is there for a game of multiple choice 90. Multiple Choice Game Example 1: in this game. The goal is to sign one of more characteristics. Feels a practice to each element partners. Hence it is a multiple choice game, and therefore an elimination grid is unwarranted. The diagram for this game will consist of four compartments, one for each of the partners D E E F N h as show. Let the letter C W and P stand for practices in criminal law practices and worker's comp and practices and patent law. In that order. Placing the condition deep practices and Worker's Comp and Patent law on the diagram gives next the condition D and E do not practice in the same field means that E practices only criminal law and D practices only worker's comp and patent law. Otherwise, they would practice in some of the same fields, adding this to the diagram yield. Next, the condition D and F Both practice and at least one of the same fields means that F must practice either workers comp or patent law or both. Adding this to the diagram along with the condition F practices and only two fields gives finally, the condition H and F do not practice in the same field means that if F practices Workman's comp than H does not and if if practices, patent law than H does not, in other words, not W or not p In the diagram, we use an arrow to indicate this conditional relationship between FN Age as follows. No, because of practices and two fields, h N f do not practice in the same fields h can practice and only one field. This diagram is a bit more restrictive than the situation warrants. F could practice criminal law. A more precise diagram would be, however, the previous diagram is sufficient for answering the questions that follow. For this question, let's reintroduce the previous growing. The diagram clearly shows that a, B, D and E are true and that c is false. Thus the answer is C for a F practices in exactly two fields. Check for B H practices in exactly one field ship for D E practices and only one field check for E D practice and exactly two fields check and for C E practices. And more than one field this false He practicing exactly one field 91. Multiple Choice Game Example 2: first, let's reintroduce the previously drawn diagram from the diagram. We can see that D and H cannot practice in exactly the same fields because deep practices in two fields and H practices in one field. This dismisses choice A. A similar analysis dismisses choices B, C and E as a matter of test taking strategies. This would be sufficient to mark the answer D, but it is instructed to work out a possible assignment. You should verify that the following diagram is consistent with all the initial conditions . 92. Multiple Choice Game Example 3: first, let's reintroduce the previously derived diagram from the diagram. We know that F must practice either workman's comp or patent law, but because of the new conditions, she cannot practice both. Otherwise, she would practice in the same fields as D, so F must practice criminal law, and the answer is D. 93. Multiple Choice Game Example 4: first, let's reintroduce the previously Dr Diagram again from the diagram we see the F and H practice and mutually exclusive feels That's what the Arrow is reminding us of. Furthermore, F practices in two fields and H and one field. So between them, they practice in all three fields. But we're told that the new partner practices and only two fields since he cannot practice in as many fields as do F N H combined. Hence the answer is E. 94. Assignment Games Introduction 5: you probably have noticed that once the diagram has been constructed assignment games or somewhat manageable, however, the diagram may not be easy to construct, and it may require considerable inspiration to figure out what kind of diagram to use as you work through. As you work the exercises and examples in this section, you will develop more intuition in this regard. 95. Points to Remember: points to remember one. Assignment games tend to be the hardest, so save them for last to the summit. Games pair each element with one or more characteristics, whereas grouping games partitioned elements into two or more groups. Three. It is important that you identify whether you're dealing with an assignment game or a group in game, because different methods are used to solve each type of Game four elimination grades are very effective in that can be applied, which is about 1/3 of the time. Five exact criteria cannot be given for when to use an elimination grid. But if only two options characteristics are available to the elements, yes, no on off etcetera, then elimination grid is probably indicated. 96. Solo Exercise 1: This is a scheduling game of medium difficulty. We shall use elimination grid to solve it to indicate that a person does not work on a particular day. Place an X on the grid to indicate that a person does work on a particular day. Police A w on the grid Police in the conditions on the grid yields Choice A is not a possible work schedule, since Drake and Edwards must work on consecutive days. And we have Drake here and we do not have Edwards either before or after. Likewise, for D here. We have Drake on Friday, but Edwards is not immediately before him, so we can eliminate both D and A since Adams works on Mondays. And Wednesday's only Choice B is not a possible schedule because it has Adams on Tuesday. Since Bates will not work on Fridays, choice See is not a possible schedule because it has baits. On Friday, it's by elimination. The answers E 97. Solo Exercise 2: adding the new condition to the grid. And we calling that only one person works each day. Heels clearly from the grid. Either great or Edwards must work Friday because the other three people are ext out on Fridays. Further, since Drake and Edwards work on consecutive days, they must work Thursday and Friday, so Drake cannot work Wednesday, and the answer is E. 98. Solo Exercise 3: begin with choice A at Cox Works on Thursday to the greatest falls. Now, the condition Drake and Edwards work on consecutive days generates two grids, one with Drake and Edwards working on Monday and Tuesday not necessarily in that order, and one with Drake and Edwards working on Tuesday and Wednesday not necessarily in that order as follows. Clearly, the first diagram leaves no room for Bates. The work, because every day is X doubt and diagram. Two forces Adams abates toe work together on Monday because that is the only day available for them, violating the condition that only one employee works at a time. Hiscox cannot work on Thursday, and the answer is a. 99. Solo Exercise 4: having the condition. Bates works Thursday to the grid yields. Since Drake and Edwards must work on consecutive days, neither can work on Friday. So we put X is there. This leaves only one person available for Friday, which is Cox. Hence the answer is C. 100. Solo Exercise 5: begin with choice A Place Cox on the grid as follows. Now Drake and Edwards could work on Thursday and Friday, respectively. So drink us here on Thursday will use a W and then X out the other days, because again, each person works only one day a week and Edwards is on Friday. We also x out Thursday for Bates because Drake is working on Thursday. Clearly, this diagram forces babes toe work on Monday because it's the only day available. And then that will force Adams to work on Wednesday. And we excel Monday for Adams, this diagram satisfies all the initial conditions, and Adams and Bates are not working on consecutive days. They're working on Monday for Bates and Wednesday for Adams, turning to choice be placed Edwards on the grid as follows since Drake and Edwards must work on consecutive days. We know that Drake must work on Tuesday because we already have Edwards assigned to Monday So w here for Drake on Tuesday and accept the rest of the days of the week because again, each person works only one day a week and we can x out Tuesday for Cox and Bates because we've already assigned Drake for Tuesday. Now the only day available for Absa work is Wednesday, and the only day available for Bates Toe work is Thursday. But this violates the new condition that Adams and Bates cannot work on consecutive days, and we haven't working on Wednesday and Thursday consecutive days. Hence the answer is B. 101. Solo Exercise 6: before starting, you should scan the answer choices for one that eliminates many positions for Bates. Now, if Edwards works Wednesday that Bates cannot work either Tuesday or Thursday. So we begin with Edwards working Wednesday Choice. See as follows again. This prevents Bates from working either Tuesday or Thursday, so we X out. Those days begin because the new conditions says that base cannot work either immediately, before or after Edwards and we have Edwards assigned toe Wednesday. Clearly, this diagram does not leave room for both Adams and Bates, since they cannot work on the same day. And the only day available for Adams's Monday And the only day available for Bates is Monday. Hence, the answer is C.