Setup and Rule Diagrams

[Dave Killoran]

Setup and Rule Diagram Explanation

This is a Basic Linear: Unbalanced: Overloaded game.

J93_Game_#1_setup_diagram 1.png

This setup has been created vertically to make it easier to show that Monday and Wednesday have only one student. The bar at the end of the slots on each of those days signifies that only one student can be assigned to each day.

This game is Unbalanced: Overloaded because there are six students but only five available days. The game scenario establishes that all five days are used and that all students are tutored, and that one of the days receives two students. Thus, there is a 2-1-1-1-1 numerical distribution of students to the days, where the instructor coaches one student every day, except for one of the days where the instructor coaches two students. Although this distribution is in general quite manageable, this game turns out to be difficult.

The fourth rule establishes that neither Monday nor Wednesday can be the day with two students, and thus either Tuesday, Thursday, or Friday has two students. Within this group of three, Thursday is the most restricted because the addition of one more variable satisfies the two students-on-one-day rule. Thus, Monday, Wednesday, and Thursday should be scrutinized closely in this game because the assignment of any available student to one of those days “closes off” that day to further students.

Note that one approach to this game would be to quickly sketch out all three scenarios—one where Tuesday is assigned two students, one where Thursday is assigned two students, and one where Friday is assigned two students.

The first rule creates an H Z sequence. Because Monday can be assigned only one student, a Z Not Law is created on Monday. However, because Friday can be assigned two students, no Not for H can be created on Friday, although Friday is the only day on which both H and Z could possibly be coached by the instructor.

The third rule creates a rotating block between K and O. While this block initially appears unimportant, note that the block action in combination with the limitation of only one day having two students creates certain inferences involving K and O. For example, if two students are assigned to Tuesday, that pair must include either K or O (if Tuesday did not have K or O, then K or O would have to go on Thursday, but this would violate the rule about exactly one day being assigned two students).

[rameday]

For game 10 on page 3-47 I am not sure how they got the answer to question 1 & 4. I understand that k/O have to be together but for example number 4 I thought the answer was E and for number 1 I had no idea how to go about answering that question.


A

[BethRibet]

Hi Rameday,

Thanks for posting.

K and O can not be together, they have to be next to each other, in either order. (KO/OK). So for question #1, that immediately rules out answer choice C. H and Z are also not together, because according to the first rule: H > Z. That rules out B. Since K and O need to be right next to each other (rule 3), and I has to be on Thursday (rule 2), we also can think about the overall sequence.


That is: __ __ __ _I_ __
M T W R F

Notice that with I on Thursday, the only way that K or O could be on Friday would be if Thursday has two people. But in the local role for #1, 2 people are on Tuesday. With 6 people fit into 5 days (and one per day except for one day with two students), only one day will have two people. That means only I can be on Thursday, so K and O must come before I in order to be placed together. K & O could go Monday or Tuesday, or Tuesday and Wednesday, to comply with the rule. Notice that either way, one of them will go on Tuesday (with the other on either Monday or Wednesday). Therefore we can rule out any answer choice that doesn't include either K or O on Tuesday. That is, the right answer choice can't have both K & O, but must have one of them. That makes it pretty clear, the only remaining answer is D.

For question #4:

We need to find a day which, if it includes Z, will fully determine the schedule. The only rule we have about Z is that Z comes some time after H. That is Z is not on Monday. So that rules out answer choice A.
You can check each remaining answer choice to see which one places U for sure. My instinct however is to start by checking what happens if Z is on Thursday, because putting Z with I on Thursday again will give us information about where K & O have to be.

__ __ __ Z+I __
M T W R F

Again, because K & O need to be next to one another (rule 2), we know neither can go on Friday, since Thursday is already full. We also know H precedes Z, based on rule 1, so it can't go on Friday. To make sure that one person is seen every day then, the only person left who can go on Friday is U. Therefore if Z is on Thursday, that determines that U is on Friday. So the correct answer choice is D.

Hope that helps!
Beth

[kim4956]

Can somebody explain to me how to arrive at the answer for question 3? Do we just have to plug in the answer choices? It seems pretty time-consuming, so I'm wondering if there's a quicker way to solve problems like this.

Also, I must be missing something, but when I tried plugging in some of the answer choices for this question, (D) wasn't the only possible answer. So I think I'm either missing an inference or did not carry out one of the rules...

Clarification appreciated!

[Adam Tyson]

Ah, the dreaded "5 if" question! Yes indeed, those can be a time consuming pain in the neck, and for that reason they may be good candidates for skipping and saving until the end of the game so that you can apply whatever you have learned from all the other questions.

That said, the approach is typically to do exactly what you thought, just test them all one at a time to see which one is right (although you may be able to eliminate some wrong answers quickly without testing just by looking at the rest of your work to see if you've already disproven them). However, in a question like this, where they are asking which one MUST be true, you don't want to test them by seeing if they COULD be true. Instead, see if you can violate them. For example, answer A - if U is on Monday, does H have to be Tuesday? Test it by putting H somewhere other than Tuesday (I tried it on Thursday, with I, forcing Z to Friday and the OK block fits in either order at Tuesday/Wednesday). If you can find a solution that violates the answer, then that answer does not have to be true, and is therefore a loser.

D is the right answer because once you place U on Thursday, there is no way for Z to go anywhere other than Friday. Try putting it somewhere else and you will find that either the OK block won't fit or else H is not before Z.

Again, test those Must Be True questions (whether or not they are of the "5 if" variety, with every answer starting with the word "if") by seeing if you can find an alternative. If you can, then it doesn't have to be true, but if you cannot find an alternative then it must be true and is the right answer.

I hope that helps!

[lsatjourneygirl]

Correct me if I am wrong, but according to the first rule with H-Z, would it be possible for a scenario as follows
H
U K O I Z

since even though there is one day in which 2 students are taught, they are separate times?


thank you!

[Adam Tyson]

The H-Z sequencing rule, taken in isolation, would allow for that possibility - they could be on the same day as long as H was the earlier session - but the last rule of the game prohibits that particular solution because neither Monday nor Wednesday can be the day for which two students are scheduled. Be sure that any scenarios you try out comply with ALL of the rules, and you will stay on the right track.

[lsatjourneygirl]

Oh sorry! I meant


U K O I HZ

HZ being in the 5th slot

would this be possible?

[Clay Cooper]

Hi lsatjourneygirl,

Yes, that permutation is possible; it obeys all the rules.

Thanks for your question!

[Kristina Moen]

The random U and H-Z seem to be giving me the most problems.

Does the H-Z mean that they cannot be scheduled on the same day? I'm inclined to say that it's possible, as long as H goes before Z.

The only real laws/ not-laws that I have in my diagram are that Z cannot go on Monday, either K or O will always go on Tuesday, and I must go on Thursday.

Is this an optimal set-up or did I miss something because the questions seem too challenging for me to have and optimal set up.

Thank you for your help!

To start with, this is an overloaded game with six students (HIKOUZ) and five days (MTWTHF). Since all five days are used and all students are tutored, there is "one day when the instructor will coach two students in separate but consecutive sessions." This means that your interpretation of the H-Z rule is correct. They could go on the same day, as long as it's a day where two students are tutored, and H is tutored before Z. So since you know that there's only one student that can be tutored on Monday, you know that Z can't be on that day which gives you a not-block.

The third rule creates a rotating block between K and O. They do not necessarily have to go on Tuesday! Note that Thursday could be the day with two students. However, if any OTHER days have two students (meaning Thursday does not have two students), then you are correct. But you can't make that global inference.

This was a very challenging game. Try it again without the inference that K/O have to go on Tuesday. Making an incorrect inference like that would make the game very hard! Let us know how it goes.