Setup and Rule Diagrams

[Dave Killoran]

Setup and Rule Diagram Explanation

This is an Advanced Linear Game: Defined-Balanced.

This game is challenging because with seven rules the setup can be tricky, and many students fail to draw the key inference about S.

The game scenario establishes that six dogs are being judged. The six dogs are a variable set, and the ribbons are another variable set. The ribbons make the most sense to choose as a base because there is an inherent sense of order, but there is a catch: only four ribbons are awarded. Thus, the first four spaces will be numbered 1 through 4, but the last two spaces will be listed as “NR” for No Ribbon. These last two spaces are interchangeable in that there is no fifth or sixth ribbon. We will also place a vertical divided bar after the fourth ribbon in order to emphasize this difference. On top of the ribbon base there will be a row for the six named dogs.

Let us examine each rule:

• First and Second Rules
  
  The first rule establishes a variable set wherein each dog is either G or L, but not both. To account for this variable set another row must be added to the diagram.
  
  The second rule establishes a variable set wherein each dog is either M or F. To account for this variable set another row must be added to the diagram.
  
  Neither of the first two rules addresses any specific variable, but they both create a greater number of elements to track. With these new variable sets, the basic game structure appears as follows:
  
  Oct 91__M12_game#4_L11_explanations_game#1_setup_diagram_1.png
  Third Rule
  
  This rule establishes that both female dogs win ribbons, meaning that the two dogs who do not win a ribbon are both male. In addition, exactly one of the female dogs is a labrador, meaning that the other female dog is a greyhound, which creates the following situation:
  
  Oct 91__M12_game#4_L11_explanations_game#1_setup_diagram_2.png
  Fourth Rule
  
  This is a very helpful rule. If only one labrador wins a ribbon, then the other three dogs that win a ribbon are greyhounds. Plus, because from the third rule we know that one of the females who wins a ribbon is a labrador, we can be certain that there are no male labradors that win a ribbon.
  
  
  Fifth Rule
  
  This rule establishes a powerful sequence:
  
  Oct 91__M12_game#4_L11_explanations_game#1_setup_diagram_3.png
  Even though U is not addressed in this rule, the sequence still produces several important inferences:
  
  
  1. Both P and R win ribbons. Because P and R each finish ahead of at least three dogs, P and R must both win ribbons.
  
  2. S wins a ribbon. Because S finishes ahead of at least two dogs, S must win a ribbon.
  
  3. At least one of Q and T does not win a ribbon.
  
  The biggest uncertainty, then, is where U places as that will directly affect Q or T.
  
  Sixth and Seventh Rules
  
  The final two rules connect certain dogs to specific dog types. Using subscripts for greyhound and labrador, let’s review the sequence again:
  
  Oct 91__M12_game#4_L11_explanations_game#1_setup_diagram_4.png
  U is also a labrador.
  
  With this new information, and the fact that exactly one labrador wins a ribbon, we can infer that S is the labrador that wins the ribbon, and that U can therefore not win a ribbon (as U is also a labrador). Thus, U must finish in one of the last two places (and thus behind S), and therefore S must place third. P and R must then place first and second in some order.
  
  Further, because S is the labrador that wins a ribbon, from the third rule S must be female. In addition, from the fourth rule, the other three ribbon-winning dogs are greyhounds, meaning the dogs in places 1, 2, and 4 are greyhounds. The information above leads to the following setup:
  
  Oct 91__M12_game#4_L11_explanations_game#1_setup_diagram_5.png
  To complete the setup, we need to examine the last two places. Remember, there is no true order to the last two places; that is, neither is necessarily “fifth” or “sixth.” One of the two dogs is U, which is a male labrador. The other is the remainder of Q or T, and will be either a greyhound or a labrador. Applying this information leads to the final setup:
  
  Oct 91__M12_game#4_L11_explanations_game#1_setup_diagram_6.png

With the information above, the questions should now be considerably easier.

[Mousey10]

Hi I needed some clarification on the dog show game classified as a killer game in the PowerScore course, particularly the rule, "Dogs P and R place ahead of dog S, and dog S places ahead of dogs Q and T". What kept me stumped before looking to the explanations was that P, R, and S MUST be awarded ribbons. I didn't interpret it as a sequencing game whereby it is place 1, 2, 3, 4 and two unordered no award dogs. Instead, I initially interpreted as more of a grouping game, meaning, place 1-4 are awarded ribbons and the other two are entirely excluded. This led to the inference that the above cited rule applies ONLY IF P, R, S, Q and T are awarded ribbons. For example, I created a hypothetical with the order of P, S, Q, and T, excluding U and R, and applying the ordering sequence ONLY to those awarded ribbons. Can you help me clarify my reasoning in interpreting the rule so that I avoid making a similar error in the future? Thank you .

[Dave Killoran]

Hi Mousey,

Thanks for the question (and for taking a course with us!). Let's see if we can't break this down a bit and make it clearer.

First, there's a reason this game is in the Killer section, and that's because it is tricky! The game starts out with what appears to be a Grouping arrangement: ribbon winners and non-ribbon winners (I'd say losers, but I like dogs and don't want to term them as losers ). However, the key to this game comes in the scenario where they note that the ribbons are awarded in a Linear order: "first, second, third, and fourth places." This creates a somewhat confusing situation because you have 4 ribbon winning dogs in spaces 1-4, and then two non-ribbon winning dogs (who effectively are equal, but serve as the 5th and 6th dogs). That arrangement creates the setup you see in the Lesson Answer Key.

Second, it appears that when you interpreted the fifth rule, you made the rule conform to your view of the game, instead of changing your view of the game to conform to the rule. The rule isn't stated in conditional terms, and thus you have to take them at face value when you read that P and R are ahead of S, and S is ahead of Q and T. the way that is stated makes it a global truth, not something that applies only to the ribbon-winning dogs. Right here was where you ran into trouble, so in reviewing this, consider the language they used in enunciating this rule: no conditionality, and stated as a flat fact. That is the indicator that this rule applies to all of the dogs, not just the ribbon-winning dogs.

I'll make a side note here about this game--which I've always found to be one of the most interesting games out there--and it relates to the setup and this rule. Because of the way the game is set up (4 ribbon winning dogs in linear order, two non-ribbon winning dogs who are effectively equal in position), from the wording of this rule alone you know that both S cannot be a non-ribbon winning dog. why? Because if S was a non-ribbon winning dog, there would be no way to "order" it, and have it place head of any other dog. A subtle point, for sure.

Looking at your hypothetical--P, S, Q, and T, excluding U and R--I'm sure that now you see why it violates the fifth rule, but for anyone else reading, the problem comes form the fact that when R is a non-ribbon winning dog, it "places" behind dog S, a rule violation. This is the second part of the issue you had, in that it is hard to argue that a non-ribbon winning dog placed ahead of a ribbon-winning dog.

So, I'd say there are two takeaways for you in reviewing this game:

• 1. Look at the rule language closely. Make sure your setup conforms to how the rule is worded.
  
  2. When you see elements of linearity, keep in mind that linear rule will then have to conform to that linearity. The same would go for any game element--Grouping, Circularity, Pattern, etc.

Please let me know if that helps out, and thanks for asking about this--it's a really interesting question!

[Mousey10]

Thanks for driving the point home Dave, I needed that .

[Jon Denning]

A question was posed earlier this afternoon by a student working with PowerScore's Logic Games: Game Type Training Volume I book (http://shop.powerscore.com/?action=prod ... 00O2WQWIA3) about a notoriously tricky game from the October 1991 LSAT. The final game on that exam deals with six dogs in a dog show, where first through fourth places are awarded ribbons, while fifth and sixth are not.

This probably sounds like a Basic Linear game (ordering 1-4, leaving space for 5 and 6), however this game came with a twist: rules also specified dog breeds (greyhound or labrador) and gender. That complicated things for a lot of people, hence my comment above about notoriety, and made this an Advanced Linear game.

In this post I'll give an overview of how to set this game up, along with the critical inferences and some insights into a question or two. Before I do, however, I should note that anyone interested in our Game Type Training series would be well-served by purchasing the accompanying Setups Encyclopedias (http://shop.powerscore.com/?action=prod ... 00O2WNWIA3), which contain full explanations—setups and inferences, as well as questions and answer choices—for every game in the corresponding Type Training. Those two sets of books pair beautifully and allow for intensive application and reinforcement, principles at the heart of games success!

Now, to the game.

So we've got our four variable sets—places (1-4, 5, 6), dogs, breed, gender—with places as the base and three stacks above it for the other three variable sets. For clarity, let’s have the six dogs as the bottom level, breed (G/L) as the middle, and gender (M/F) as the top.

The third rule tells us that both female dogs (F F) get ribbons, meaning 5 and 6 are male (M), and that one of those females is a labrador (L) and the other is a greyhound (G). Note that this doesn’t tell us about the breeds of the four male dogs! We also don’t know where any of the dogs goes yet.

The fourth rule combines with the third by saying that exactly one L wins a ribbon, and since we know a female lab wins a ribbon, the other winners are all G. So our winners are: 1 FL, 1 FG, 2 MG. Non-winners are both male, but breed unknown.

Rule 5 is a big one. It forms a chain involving five of the six dogs, with S after P and R, and preceding Q and T. That means P, R, and S definitely get ribbons, and one or both of Q and T is a non-ribbon dog.

Rule 6 isn’t terribly helpful at this point, as it just gives up breeds for P and R, two high-placing dogs.

Rule 7, on the other hand, puts of lot of pieces in place. If S is a labrador we can connect that with two prior rules: from rule 5 we know that S gets a ribbon, and from rule 3 we know that only one L, a female, gets a ribbon...meaning S is the only L in the first four ribbon spots, and must be a female! So what of U, another lab? It must be in either 5 or 6, since we can’t have a second L get a ribbon in 1-4! That also means U is a male, since we know from rule 3 the two females get ribbons (hence 5 and 6 are males).

With that information we can start to place our dogs definitively: S has P and R ahead of it, and we now know S has Q, T, and U after it. So S, a female lab, is in 3! P and R, two greyhounds, are in 1 and 2 although we don’t know their exact order or gender. The fourth and final ribbon spot must be either Q or T, and must be a greyhound. Again gender can’t be determined. 5 and 6, which are essentially the same thing since there’s no ordering of non-ribbon dogs, are U and either Q or T (whichever isn’t 4th), and are both males. U as we’ve been told in rule 7 is a lab.

As you can hopefully tell, we’ve just filled in a LOT of the spaces in this game! There’s still some uncertainly of course, but considering where we started, with 18 open spaces, things are now looking pretty good. Let’s do a couple of questions to drive the power of this setup home.

18. Easy enough. We know P and R are greyhounds, so (A) is gone. We know S and U are labs, although that doesn’t help here. So what of Q and T? There’s nothing that forces them to be labs, so they too can be greyhounds and (E) is the answer.

19. Move quickly through these choices looking for a violation/impossibility! (A) can work since we can have a female greyhound in 2. (B) is impossible since the only lab that wins a ribbon is S, and S is in 3! So (B) is the correct answer.

20. What dogs must be male? The two non-ribbon dogs, 5 and 6. And what dog must be in either 5 or 6? U! So (E) is your answer.


You know, I started this with the intention of only doing a few questions...but I’m enjoying myself so I’m going to do them all! If you’d rather go back and rework the game on your own before you read the rest of this by all means do. The final four question explanations will be waiting for you when you get back

21. A can be false question is simply asking for an answer that is not necessarily true. That means the four wrong answers must be true, while the correct answer doesn’t have to be. Let’s just look for a choice we can violate. (A) says P is ahead of R. Does that have to happen? Nope! P and R can swap spots in 1 and 2, so (A) isn’t necessarily true (can be false) and is correct! Aren’t entirely certain? Try the other four and you’ll see that each one must occur as presented.

22. If Q is female she can’t be in 5 or 6, since those dogs are both males. That means T and U are in 5 and 6, and Q must be in 4: she’s after S from the rules, and she gets a ribbon since both females do...she must be in the last ribbon spot, 4. We’re asked what can be false, meaning, again, what’s not necessarily true. As we saw in #21 the four incorrect answers will all be true, and we can find these quickly. (A) and (B) are true because the females are S and Q, and the other four are males. (C) is true as was just explained. (D) is also true since the only lab to win a ribbon is S, making P, R, and Q greyhounds. (E) doesn’t have to be true and is thus correct, as we don’t know the breed of T, a non-ribbon dog (we know T is male in this instance, but not his gender).

23. If T is 4, then 5 and 6 are Q and U (order unknown). We also know that P, R, and T must be greyhounds since no labs but S can win ribbons. What must be true? (B): Q is a male, since Q is in either 5 or 6 and both of those spots are males.

24. A broad could be true question. Move aggressively through the answers looking to either identify one that is clearly uncertain, or eliminating the four that break rules! (A), (B), and (C) are all clear impossibilities, as P, R, and S are the first three dogs and always win ribbons. (E) too is impossible since U is either 5 or 6 and wins nothing. So (D) is the correct answer: T can win a ribbon, but doesn’t have to, as it can swap in and out with Q.


So there you have it. A very tricky game (as my colleague Dave has explained in a separate thread here http://forum.powerscore.com/lsat/viewto ... game#p7555), but with the right strategies and insights it all comes together nicely. I hope this helps, and please do consider my advice about the Setups Encyclopedias—they’re extremely useful as you work through the Training Type guides. Thanks so much!

Jon

[MichaelJAG]

How do we know that we can expand the base from 1-4 to 1-6, so we can place all of the variables? I reread the Balanced versus Unbalanced Games section in the Logic Games Bible, and I did not see a mention of this tactic. The section says that we can use "E placeholder variables" for Underfunded Games, so I certainly understand the logic behind expanding the base. I thought that the game was initially an Overloaded Game wherein there would be ties: 6 into 4. However, the game scenario explicitly says that only one ribbon is awarded per place. I understood the game much better after I realized that we could expand the base. Are there many games like this one that call for an expansion of the base? Thank-you!

[Jon Denning]

Hey Michael,

Thanks for the questions! I don't see this as a case of "expanding" the base so much as just having a spot to account for all six dogs, whether they're ranked (1-4) or not (5 and 6). Essentially that balances the game out and makes it a 1-to-1 distribution, something that is nearly always desirable. And that's why I knew to do it the way I did: anytime I can create a base with a spot for every variable to occupy I'm going to do so!

A similar technique could be used for Overloaded Grouping games, where you have more variable options than "in" (selected) spots. In that case showing the "in" and "out" groups could effectively balance the distribution, and knowing the composition of either of those two groups would tell you the composition of the other (knowing who's not selected would tell you who is selected).

For instance, imagine a 5-person team being selected for from a group of 8 candidates. Here you'd have the "in" group of five people, but also an "out" group of the remaining three people. Showing all 8 spots with some sort of division representing the 5/3 split (like a vertical line to separate the two groups: _ _ _ _ _ | _ _ _ ) balances, and thus simplifies, the distribution.

And yes, that sort of thing does happen a fair amount on the LSAT, so always be on the lookout for it! As you note, providing a spot for all the variables to occupy, when possible, makes things a lot easier

I hope that helps!

[alexis.la]

Hello PowerScore. I think that with practice, my instinct on how to approach a game like this will get better, but this game really made me feel like I hadn't studied the whole LG bible already I just want to highlight the aspects that threw me off and maybe you could offer some insight.

1. The set up. When the rules added gender and dog type I had no idea how to diagram it. I first thought I should just add subscripts next to a dog's initials once I place them (ie. S *subscript F for female and L for lab*). But I thought dog type and gender may be important and should be more obvious, so I considered adding extra rows, but for some reason it didn't click to have a row for 'dog type' and 'gender'. Instead I thought I'd need separate rows for greyhounds, labs, females and males (4 rows total). I would then find the 3rd place slot, and put an S on the female row, and an S on the lab row, as a way of checking off their characteristics. INSTEAD of 'S' on the 3rd spot 'F' on the corresponding gender row, and 'L' on the corresponding dog type row. Dave's diagram is organized though, and it did help me to understand the rules more (especially seeing that 2 males will not be awarded ribbons). This is very confusing, but my brain was this scrambled during the practice test. Extra variables never fail to throw me off. In my first real LSAT this was a problem actually, I just don't seem to grasp how to account for variables that aren't inherently ordered, then I get flustered.

2. Originally, the P, R, S, Q, T rule threw me off, because I had a grouping mindset, thinking I only had 4 slots. And here is this rule saying the order for 5 dogs. So I was thinking "well then how do I decide which dog gets excluded from the chain"? Instead of seeing 5th and 6th as part of the linear placing positions. It wasn't clicking that the rule is upheld even if T doesn't get a ribbon. I was equating 'losing' as a standalone group, but even the losers are placed. I also had a bit of a conditionality mindset at first, and thought that if S didn't get a ribbon then the whole rule is void (because I was equating 'losing' with being a separate group, which is wrong) and then all the positions are up in the air. I do fully understand the rule now though.

Sorry for the rant. I think as I practice I'll get more confident, but do you have any advice?

[Adam Tyson]

In general, alexis, when you have multiple different sets of variables going into some order, you should create a different row for each set of variables. So "gender" is the set, with M and F as the variables in that set, while "breed" is another set, with G and L being the variables in that set.

Another way to think about it is like this: How many things do I need to know about each position in the order? In this case, the answer is I need to know which dog it is (that's one row), what gender it is (that's a second row), and what breed it is (that's the third row): three things I need to know for each position in the order, so three rows are required. There won't be a G and an L in the same position, so I don't need separate rows for G and L!

As to the out group here, yes, the dog that places 4th is placing ahead of the two dogs that don't get ribbons, so the "out" group is, in this game, part of the order. But that won't always be true in these kinds of games! In other games the situation might be something like "out of 6 dogs, 4 will be selected to be in the show, and those four will be ranked 1st through 4th." In that case you could not say that the dog ranked 4th is ranked ahead of the dogs that weren't selected, because they weren't in the order at all. That would change this game drastically! You do need to pay careful attention to whether the "out" group is truly out, or if they are still a part of the sequence. This won't happen very often, but you need to account for it when it does happen!

[ange.li6778]

Hi Powerscore, this game tripped me up because I, like many others, didn't think that dogs 5 and 6 would still be part of the set up. I see now that Rule 5, the PR-S-QT sequence, is a major flag that the 4 ribbon winners aren't the only dogs placed linearly. Once I understood that, setting up the diagram and answering the questions was a breeze. Which other games from the PTs fall into this category (seems like a grouping game but the "out" group is still placed linearly)? And in the scenario that this question showed up on my real test but I hadn't practiced a similar example before taking the test, should I be using something like Rule 5 to tip me off that I should be setting it up linearly rather than as a grouping game? Thank you!