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 Dave Killoran
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#72144
This game is discussed in our Podcast: LSAT Podcast Episode 31: The September 2019 LSAT Logic Games Section

Setup and Rule Diagram Explanation

This is a Advanced Linear: Unbalanced: Overloaded, Numerical Distribution game.


The scenario establishes that there are five animals in numbered pens, 1-5. This creates a Linear base for the game. Then, each pen contains an animal (puppy or kitten) and a name. This creates two rows above the five pens, for our Advanced Linear base:

  • K: F G H J 4

    P: R S T W 4




    Name:       ___ ..... ..... ___ ..... ..... ___ ..... ..... ___ ..... ..... ___



    P/K:           ___ ..... ..... ___ ..... ..... ___ ..... ..... ___ ..... ..... ___
    .....         1 ..... .....      2 ..... .....      3 ..... .....  4 ..... .....  5
While there are various numbers of kittens and puppies that could be selected, and initially the possible numerical relationships are:

  • 4 kittens - 1 puppy
    3 kittens - 2 puppies
    2 kittens - 3 puppies
    1 kitten - 4 puppies

We'll reserve further comment on this aspect until the rules are examined.


Rule #1: This is an extremely helpful rule, with two strong effects:

1. It limits the numerical combinations of kittens and puppies. Now the 1 kitten - 4 puppies option is eliminated:

  • 4 kittens - 1 puppy
    3 kittens - 2 puppies
    2 kittens - 3 puppies
    1 kitten - 4 puppies


2. It establishes the type of animal in the first and fifth pens:


  • Name:       ___ ..... ..... ___ ..... ..... ___ ..... ..... ___ ..... ..... ___



    P/K:           _K_ ..... ..... ___ ..... ..... ___ ..... ..... ___ ..... ..... _K_
    .....         1 ..... .....      2 ..... .....      3 ..... .....  4 ..... .....      5

Rule #2: Taffy—which is a ridiculous name for a dog—cannot be in a pen next to a kitten. When combined with the first rule, this rule turns out to be extremely limiting:
  • Taffy is a puppy, and so cannot be in the first or fifth pens.
    Since Taffy cannot be next to a pen with a kitten, Taffy cannot be in the second or fourth pens.
    Thus, if Taffy is to be used, Taffy would have to be in the third pen, with puppies in the second and fourth pens.

    This leads to two basic inferences:

    • 1. If Taffy is displayed, it's in the third pen and the numerical distribution is 2 kittens - 3 puppies:

      ..... T :arrow: T3 :arrow: 2K/3P


      2. Taffy, if displayed, would have to appear in the middle of a giant puppy block:

      • Name:       ___ ..... ..... ___ ..... ..... _T_ ..... ..... ___ ..... ..... ___



        P/K:           _K_ ..... ..... _P_ ..... ..... _P_ ..... ..... _P_ ..... ..... _K_
        .....         1 ..... .....      2 ..... .....      3 ..... .....      4 ..... .....      5


Rule #3: This rule states that Garnet or Honey, but not both, must be used. G and H are both kittens, and so now the maximum number of kittens that can be used is three:

  • Kittens: F, G/H, J


Thus, another numerical option can be eliminated:

  • 4 kittens - 1 puppy
    3 kittens - 2 puppies
    2 kittens - 3 puppies
    1 kitten - 4 puppies


Note that if only two kittens were displayed, one of those kittens would be G or H, and the other would have to be F or J.

Rule #4: The final rule connects the kitten and puppy groups, and states that if Wags is displayed, then Garnet is displayed in the second pen:

  • W :arrow: G2
Since Garnet appears in both of the final two rules, connect them!

  • If W is displayed, then G must be displayed, and consequently H cannot be displayed, resulting in:

    W :dblline: H

Since Garnet is a kitten, you can also connect this rule to the second rule:
  • If W is displayed, then G (a kitten) must be displayed second. This means T could not be displayed, resulting in the inference that W and T cannot be displayed together:

    W :dblline: T

    Since W and T are both puppies, this means the puppy selection options are: R, S, T/W
Last, note that the implication of W being selected also tell you that you are in the 3K/2P distribution:
  • When W is displayed, G is displayed second. This means kittens are in the first, second, and fifth pens (and those three kittens would be G in 2, and F and J in the other two pens). Since you can only have three kittens maximum, the fourth and fifth pens would have to be puppies (one of which is W and then the choice of R or S).

Note that, the game has only two remaining numerical options for the distribution of kitten and puppies into the five pens:

  • 3 kittens - 2 puppies — the kitten selection would be F, G/H, J

    2 kittens - 3 puppies — the puppy selection would be R, S, T/W; the two kittens would be one each from the G/H and F/J pairs
 Rachael Wilkenfeld
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#72149
This game is an overloaded advanced linear game. We have 8 animals that fit into 5 pens, and we also have to track an additional variable set of puppies/kittens. Whenever you see an aspect of a game that's overloaded or underfunded, you should start keeping track of numerical concepts.

We have two types of animals that will go in these pens, and we want to write the variables sorted into the two animal types.
Kittens: FGHJ4
Puppies: RSTW4

Because numbers are so critical here, I mark the number of puppies and kittens possible in my variable list. On first glance, without really looking to the rules, it looks like I could have 4 possible distributions.
4 puppies-1 kitten
3 puppies-2 kittens
2 puppies-3 kittens
1 puppy-4 kittens

We'll see that not all of those are possible, but at the beginning of the set-up, I want to have all the potential distributions written so that I can cross them out as they become impossible. Our basic diagram will start like this:

..... :longline: ..... :longline: ..... :longline: ..... :longline: ..... :longline: (kittens/puppies)

..... :longline: ..... :longline: ..... :longline: ..... :longline: ..... :longline: (FGHJRSTW)
1 ..... ..... 2 ..... ..... 3 ..... ..... 4 ..... ..... 5

By setting up two stacks, it makes it easier to track both the name of an animal and the species.

Let's turn to the rules. The first rule tells us that the first and fifth pens have to hold kittens. On first glance, we would just add that to our diagram.
..... k ..... ..... ..... ..... ..... ..... ..... ..... k .....

..... :longline: ..... :longline: ..... :longline: ..... :longline: ..... :longline: (kittens/puppies)

..... :longline: ..... :longline: ..... :longline: ..... :longline: ..... :longline: (FGHJRSTW)
1 ..... ..... 2 ..... ..... 3 ..... ..... 4 ..... ..... 5

But--remember our distributions? We can also eliminate the 4 puppies-1 kitten, because we now know we have at least 2 kittens. So go ahead and cross that out on your diagram.

Next rule---Taffy can't be displayed next to any kitten. You could do a not block for tk, kt. We know more than that, though. Since kittens have to be in pens 1 and 5, we know Taffy (not a kitten) can't be in 1 or 5, or 2 or 4 because they are next to kittens. That means if Taffy is in, Taffy is in pen 3. I'd write the inference below:

If T :arrow: T3

We can keep going with this inference though. If T can't be next to any kittens, then if T is in, 2 and 4 have to be puppies.

If T :arrow: P2T3P4

That also means if T is in, we are in a 3 puppy-2 kitten distribution.

Next rule. Either G or H, but not both, must be in a pen. We can do this as two contrapositives.
G :arrow: H
G :arrow: H

We can't just write G/H in a slot because we don't know which pen G/H belongs in. This rule does give us another inference though--and it's about the distribution. If only one of G and H can be in, so at most 3 of our 4 kittens can be there. We can eliminate ANOTHER distribution (the 1 puppy-4 kittens). That leaves us only with 3 puppies 2 kittens OR 2 puppies 3 kittens.

Finally we have if W is in, G has to be in 2. We write that as a straightforward conditional.
W :arrow: G2

What inferences can we pull from there? Well, if G is in H is out---we know that from the rule before.
W :arrow: H

What else do we know? What other rules involved slot 2? Our T rules! T can only be in if there are puppies in 2 and 4. G is not a puppy. So G in 2, eliminates T from the pens.
W :arrow: T

We could go a step further. Since we know that the distribution for W in means that we have 3 kittens (in 1, 2, and 5), 3 and 4 would both have to be puppies when W is in.

Whew. That was a lot of inferences, but not a ton on the diagram itself. That's ok! Going into this game, I'd focus on looking for scenarios involving T or W, because they are both so limiting.

Hope all that helps!
Rachael
 gwlsathelp
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#83539
Would this setup have benefited from diagramming the templates because it has some pretty restricted variables?
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 KelseyWoods
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#83598
Hi gwlsathelp!

Are there specific templates you have in mind? To me, it seems like there's still a lot of variability in terms of the exact order of the variables so I probably wouldn't use full templates here. But templates are somewhat down to personal preference so if you have templates you think would be helpful, you can let us know.

The main thing restrictions here really come down to the numerical distribution of puppies and kitties as well as the groups of specific pets that can be in/out based on these numerical distributions. Look at the last point that Dave made above:
Note that, the game has only two remaining numerical options for the distribution of kitten and puppies into the five pens:

3 kittens - 2 puppies — the kitten selection would be F, G/H, J

2 kittens - 3 puppies — the puppy selection would be R, S, T/W; the two kittens would be one each from the G/H and F/J pairs
I would add a couple of things here:

In the 3 kittens - 2 puppies scenario, T would have to be out (because there would be no way for it to not be next to a kitten) so you'd have two of R/S/W in and one out.

In the 2 kittens - 3 puppies scenario, W would have to be out (because when W is in, G is 2nd meaning we have 3 kittens). So the 3 puppies would be R S T, with T being 3rd and R/S in 2nd and 4th in some order.

Those numerical distribution scenarios give us a lot of info that is helpful for diagramming the local questions more quickly.

Hope this helps!

Best,
Kelsey
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 JocelynL
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#85752
Hello, when I came across this game I thought it was a grouping not linear based on 5 in and 4 out, which affected my diagraming. What would clue me into seeing this as advanced linear vs grouping?
 Adam Tyson
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#86264
I agree with you that this game has a strong Grouping component, JocelynL, and could be called a Grouping/Linear combination game. The clues to the linear element start with the fact that the cages are numbered and there are rules that reference cage numbers, but that alone isn't enough to really say it's linear. The real clue that you have to think about the exact order is in the rule about T, which cannot be next to any kitten. "Next to" is a linear concept, and means that the left to right placement could matter a great deal!

All this leads to setting up the base as 5 cages, lined up left to right and numbered 1 through 5.
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 JocelynL
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#86285
Adam Tyson wrote: Sat Apr 10, 2021 10:52 am I agree with you that this game has a strong Grouping component, JocelynL, and could be called a Grouping/Linear combination game. The clues to the linear element start with the fact that the cages are numbered and there are rules that reference cage numbers, but that alone isn't enough to really say it's linear. The real clue that you have to think about the exact order is in the rule about T, which cannot be next to any kitten. "Next to" is a linear concept, and means that the left to right placement could matter a great deal!

All this leads to setting up the base as 5 cages, lined up left to right and numbered 1 through 5.
Adam, thank you so much. Your replies are always so helpful!
 kupwarriors9
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#89495
I just wanted to confirm that either G or H always has to be selected since the rule says: " Next rule. Either G or H, but not both, must be in a pen." Or can both G and H be not selected, theoretically? Because, in the LG bible it says for :dblline: , the options are for G :dblline: H (for the example in the LGB) are:
1. only H is selected
2. both G and H are not picked
3. only G is picked
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 Dave Killoran
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#89496
Yes, as I say above: "Rule #3: This rule states that Garnet or Honey, but not both, must be used." Italics added for emphasis.

In many cases, this would be shown in two ways: one by showing G/H as a required selection, and then by using G :dblline: H to eliminate the possibility of both. The mere appearance of G :dblline: H doesn't invalidate the "at least one" part of the rule.

In a vacuum, it would be as you note in your post, but this isn't a vacuum since they told you one was being used :)

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