Setup and Rule Diagrams

[Dave Killoran]

This game is also discussed in our Podcast: LSAT Podcast Episode 70: The May 2020 LSAT-Flex Logic Games Section

Setup and Rule Diagram Explanation

This is a Basic Linear: Unbalanced: Underfunded game.

As far as final games in a section go, this is not an unreasonable ending. The second half of the questions in this game are time-consuming because of the need to draw out Local scenarios, but the difficulty is not off the charts.

The setup for this game is fairly straightforward: 6 variables available for 8 spaces. This is easy enough to diagram:

• G H J L M O 6
  
  
  
  ___ ___ ___ ___ ___ ___ ___ ___
   1     2    3    4    5    6   7   8

Ultimately it will turn out that G and L are randoms, which you should note on your diagram.

You should immediately notice that there are more spaces than variables; this means that the numerical aspect of the game will not be a simple 1-to-1 relationship. You might initially be inclined to begin working out all the numerical distributions of spaces to variables, but you should wait to see if the rules reveal further information that might be useful. And, as luck would have it, the first two rules do exactly that. So, we'll talk about the numerical distribution after we address those rules.

The first two rules establish a connection between pairs of spaces:

• 1 5
  
  2 7

While these diagrams get the point across, the better representation is shown below, in Kelsey's post where there is a visual link made between the pairs with lines.

By pairing these spaces, the game effectively doubles the variables that appear in these spaces. Thus, we know that numerically, two of the variables are doubled, which starts off our distribution of spaces (8) to variables (6) as:

• 2-2

We now know that there are 4 spaces left (8 - 4 = 4), and there are still 4 variables left to address (6 - 2 = 4). Therefore, the remaining distribution must be 1-1-1-1 (4 into 4, one time each), bringing us to a final distribution of spaces to variables in this game of:

• 2-2-1-1-1-1

So, to recap, two of the variables will appear twice (in the spaces designated in the first two rules), and every other variable will appear once. This means that any variable that appears in spaces 3, 4, 6, or 8 (the spaces not in the first two rules) cannot be doubled. If that doesn't sound like it makes sense, try putting a variable in 3, 4, 6, or 8 and doubling it elsewhere: you will run out of spaces to fit all of the variables.

The third rule establishes a rotating not-block with M and O:

• OM
  
  MO

This rule does not initially allows us to make any inferences, although you can immediately see that if M or O is used in one of the first two rules, several Not Laws would automatically follow.

The fourth rule is less standard than the first three. In this rule, at some point in the game there must be an HM block. This does not mean that every time H appears M must be behind it, and so the rule should not be diagrammed as simply HM. Instead, make some type of note that HM has to appear at least once, as in:

• HM at least 1

In this case, as it turns out, the HM block can only occur once because, as Kelsey explains below, the two plays that are performed twice cannot be next to each other each time (one is 5th and one is 7th). However, thinking about the diagramming here (with the "at least once") is important because you'll see this language in other games and you want to handle it correctly each time.

The fifth and final rule is another less-than-standard rule. Here a performance of J must precede any performance of H. This can be diagrammed in various ways, so choose the method that is clearest to you. Here are two examples:

• J H first
  
  at least one J all H

In reviewing this rule, note the language carefully: it does not say that every J is ahead of every H, just that there is always at least one J ahead of the first appearance of H. Thus, a configuration such as J H J is acceptable.

This last rule also causes some additional restrictions for J and H. Because a J must always precede any H, there is no way for H to be performed 1st. Applying the first rule, that also means that H cannot be performed 5th.

There's also no way for J to be performed 8th, although this is much more difficult to see at a glance. Here's why: if J is performed 8th, then a second J will need to be performed earlier in order to be ahead of H. But, the only variables that can be doubled must fit in the 1-5 and 2-7 spaces in order to conform with the first two rules. So, there's no way for J to be performed last and conform with the rules of the game.

Combining all of this information leads to three Not Laws on the diagram:

• G H J L M O 6
  
  
  
  ___ ___ ___ ___ ___ ___ ___ ___
   1     2    3    4    5    6   7   8
  H H      J

As you can see, this setup is pretty wide open! Thus, you'd have to expect going in that you'd see a reasonable number of Local questions, and that a good amount of individual diagramming and solution-making would be required. So don't be concerned when that starts happening.

[kassierimel]

Could someone please post the setup and inferences to this game?

[KelseyWoods]

Hi kassierimel!

Here's how I would set this game up:

Screen Shot 2021-01-04 at 2.52.17 PM.png

Inferences:

• Since each play must be performed and we know that the 1st play is also 5th and the 2nd play is also 7th, we know that the numerical distribution is 2 - 2 - 1 - 1 - 1 - 1. This means that the plays that are performed 2, 4, 6, and 8 are each performed only once.

• The HM block can only occur once because the two plays that are performed twice cannot be next to each other each time (one is 5th and one is 7th).

• From the last rule, H cannot be 1st, which means it can also not be 5th.

• Since at least one J must be before all of the Hs and the 8th play can only be performed once, J cannot be 8th.

That's pretty much it for the inferences that we can make upfront. This game has several local questions for which you will need to construct mini-diagrams. The big thing to remember when making these diagrams is those links between slots 1 and 5 and slots 2 and 7.

Hope this helps!

Best,
Kelsey

[kassierimel]

Kelsey,

This was very helpful. Thank you!

[rlouis1993]

Hi there, I have a few questions about the stimulus:

(1) Where in the stimulus does it suggest all the pieces have to go at least once? I didn't make this assumption because it was not explicitly stated in the stimulus.

(2) For the first two rules (1,5 and 2,7) I read it as a conditional statement in which 1 ---> 5 and 2 ---> 7. This is obviously the wrong interpretation and that they should be bi-conditionals instead (i.e. 1 <--> 5 and 2 <---> 7). Again, where int he language does it suggest a bi-conditional relationship?

For me, not understanding these two concepts was detrimental to my success.

Thank you for any clarification you can provide.

[boehmejayne@gmail.com]

Hi rlouis,
The prompt says "exactly six plays... will be performed". That's how we can know that all six of the plays will be performed at the eight day festival. All the names of the plays are given in the middle of the sentence which probably made it more difficult to see.

[Jeff Wren]

Hi rlouis,

For your first question, boehmejayne's answer is exactly right.

In the scenario (the first paragraph that describes the game) , we are told that "exactly six plays ... will be performed." This means that six different plays, the six that are listed, will all be performed. For example, if the same one play was performed all 8 days, you wouldn't' describe that as six plays, you'd describe that as 1 play that ran for 8 days.

This wording can be easy to miss, but it is critical that you read everything very, very carefully. When reading the scenario and rules, you'll want to slow down and take your time, don't rush the setup. We recommend that students read the entire game scenario and rules twice, once to get a good understanding of the game before diagramming, and a second time when you actually diagram the rules/setup. Misunderstanding a key element of the game or mis-diagramming a rule is one of the worst things that can go wrong on test day because it can throw off your entire LG section.

For your second question, the first rule says that whatever goes first must also go 5th. In this case, we know that something will go first (in other words, it's not possible to leave the first space empty in this game). That variable will also go 5th. Because it is the same variable which will go in spaces 1 and 5, the reverse is also true, that whatever variable goes 5th will go 1st.

This is different from a more typical condition rule, such as "If A goes 1st, then B goes 5th." In that situation, B can go 5th without A having to go 1st.

The applies for the second rule.

[TootyFrooty]

This game is also discussed in our Podcast: LSAT Podcast Episode 70: The May 2020 LSAT-Flex Logic Games Section

Setup and Rule Diagram Explanation

This is a Basic Linear: Unbalanced: Underfunded game.

As far as final games in a section go, this is not an unreasonable ending. The second half of the questions in this game are time-consuming because of the need to draw out Local scenarios, but the difficulty is not off the charts.

The setup for this game is fairly straightforward: 6 variables available for 8 spaces. This is easy enough to diagram:

• G H J L M O 6
  
  
  
  ___ ___ ___ ___ ___ ___ ___ ___
   1     2    3    4    5    6   7   8

Ultimately it will turn out that G and L are randoms, which you should note on your diagram.

You should immediately notice that there are more spaces than variables; this means that the numerical aspect of the game will not be a simple 1-to-1 relationship. You might initially be inclined to begin working out all the numerical distributions of spaces to variables, but you should wait to see if the rules reveal further information that might be useful. And, as luck would have it, the first two rules do exactly that. So, we'll talk about the numerical distribution after we address those rules.

The first two rules establish a connection between pairs of spaces:

• 1 5
  
  2 7

While these diagrams get the point across, the better representation is shown below, in Kelsey's post where there is a visual link made between the pairs with lines.

By pairing these spaces, the game effectively doubles the variables that appear in these spaces. Thus, we know that numerically, two of the variables are doubled, which starts off our distribution of spaces (8) to variables (6) as:

• 2-2

We now know that there are 4 spaces left (8 - 4 = 4), and there are still 4 variables left to address (6 - 2 = 4). Therefore, the remaining distribution must be 1-1-1-1 (4 into 4, one time each), bringing us to a final distribution of spaces to variables in this game of:

• 2-2-1-1-1-1

So, to recap, two of the variables will appear twice (in the spaces designated in the first two rules), and every other variable will appear once. This means that any variable that appears in spaces 3, 4, 6, or 8 (the spaces not in the first two rules) cannot be doubled. If that doesn't sound like it makes sense, try putting a variable in 3, 4, 6, or 8 and doubling it elsewhere: you will run out of spaces to fit all of the variables.

The third rule establishes a rotating not-block with M and O:

• OM
  
  MO

This rule does not initially allows us to make any inferences, although you can immediately see that if M or O is used in one of the first two rules, several Not Laws would automatically follow.

The fourth rule is less standard than the first three. In this rule, at some point in the game there must be an HM block. This does not mean that every time H appears M must be behind it, and so the rule should not be diagrammed as simply HM. Instead, make some type of note that HM has to appear at least once, as in:

• HM at least 1

In this case, as it turns out, the HM block can only occur once because, as Kelsey explains below, the two plays that are performed twice cannot be next to each other each time (one is 5th and one is 7th). However, thinking about the diagramming here (with the "at least once") is important because you'll see this language in other games and you want to handle it correctly each time.

The fifth and final rule is another less-than-standard rule. Here a performance of J must precede any performance of H. This can be diagrammed in various ways, so choose the method that is clearest to you. Here are two examples:

• J H first
  
  at least one J all H

In reviewing this rule, note the language carefully: it does not say that every J is ahead of every H, just that there is always at least one J ahead of the first appearance of H. Thus, a configuration such as J H J is acceptable.

This last rule also causes some additional restrictions for J and H. Because a J must always precede any H, there is no way for H to be performed 1st. Applying the first rule, that also means that H cannot be performed 5th.

There's also no way for J to be performed 8th, although this is much more difficult to see at a glance. Here's why: if J is performed 8th, then a second J will need to be performed earlier in order to be ahead of H. But, the only variables that can be doubled must fit in the 1-5 and 2-7 spaces in order to conform with the first two rules. So, there's no way for J to be performed last and conform with the rules of the game.

Combining all of this information leads to three Not Laws on the diagram:

• G H J L M O 6
  
  
  
  ___ ___ ___ ___ ___ ___ ___ ___
   1     2    3    4    5    6   7   8
  H H      J

As you can see, this setup is pretty wide open! Thus, you'd have to expect going in that you'd see a reasonable number of Local questions, and that a good amount of individual diagramming and solution-making would be required. So don't be concerned when that starts happening.

Hi,

I read the last rule as a single occurrence, as in "at least once" but not always. What makes it become always?

I read it this way because it said "A" performance, as in, "a performance" before...

[Dana D]

Hey Tooty,

You are correct in reading the rule as 'at least once' and not always. As Dave's explanation points out, however, for this specific game there is only one time that this HM block can be used due to other constraints, so it is also a rule that applies for 'every' time this HM block is used (which is only once).

[willwants170]

I made the inference that M cannot be in position 6 because H cannot be in position 1 and 5, and if M went 6, it cannot go twice (because the two variables that go twice are in 1,5 and 2,7) and consequently would not fulfill the HM block. Would that be correct?