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 Dave Killoran
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#71225
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 Grouping Game: Defined-Fixed, Balanced, Numerical Distribution

This was an infamous game and many students walked away from the September 2019 LSAT complaining about the difficulty of the section overall, and this game in particular. The reasons are obvious: there are a number of moving parts, rules which don't appear to be well-connected, and a lot of spaces to fill. To master this game, you have to have a reasoned approach, follow the indicators the test makers give you, and focus on the remaining open spaces. If you can do that, then the game isn't as difficult as advertised.

To begin with, there are three variable sets:

  • The five flower arrangers: S, T, U, W, and Z
    The four individual flowers in each arrangement
    The four kinds of flowers: G, H, L, and R
None of these variable sets has an inherent sense of order, so we are looking at a Grouping game. In this case, we will choose the individual flower arrangers as the base, since it's easy to visualize each group, and the rules clearly use that variable set to distribute the other variables. the four individual flower spaces will then be stacked above each person, and those spaces will be filled in by the actual flowers:


  • ___ ..... ___ ..... ___ ..... ___ ..... ___

    ___ ..... ___ ..... ___ ..... ___ ..... ___

    ___ ..... ___ ..... ___ ..... ___ ..... ___

    ___ ..... ___ ..... ___ ..... ___ ..... ___
     S .....      T .....  U .....   W .....   Z
From an initial setup, this is intimidating! There are twenty open spaces at this point. The one advantage is that each group is defined as four spaces, and that is fixed throughout the game. So, there are no moving parts to further complicate things.

The rules will now allocate flowers to each arranger. We'll show the allocation and then talk about each rule:

  • ___ ..... ___ ..... ___ ..... ___ ..... ___

    ___ ..... ___ ..... ___ ..... ___ ..... ___

    _R_ ..... _G_ ..... _H_ ..... _G_ ..... _R_

    _R_ ..... _H_ ..... _R_ ..... _G_ ..... _H_
     S .....      T .....       U .....   W .....    Z
Note that immediately this game is far more manageable: there aren't twenty open spaces, but just ten, and we have a wealth of information about the relationships governing those opens spaces.

Let's look at those relationships, while keeping in mind that each arrangers can only use three kinds of flowers. This use of three kinds of flowers for four spaces results in a 2-1-1 numerical distribution of flowers-to-types. For example, perhaps an arranger uses two hyacinths, one gardenia, and one rose in their arrangement. That's three types and four total flowers in a 2-1-1 alignment.

Rule #1: Solomon is the only arranger to use exactly two Rs. This means that Solomon needs to use two other kinds of flowers (one of each) to fill out the arrangement. In this case, that means Solomon's arrangement must be filled by two kinds of G, H, and L.

The exclusivity aspect to Solomon being the only one to use two Rs will also impact other arranger's options, you can be sure.

Rule #2: Tabitha uses exactly one H, and while the forum doesn't allow us to show this well due to the limitations of the graphical tools, on a diagram we'd use a block around the H to show that it's the only one Tabitha can choose. You can add H Not Laws under the upper tiers as well if that helps cement the idea visually.

Rule #3: Ursula uses exactly one R, and while the forum doesn't allow us to show this well due to the limitations of the graphical tools, on a diagram we'd use a block around the R to show that it's the only one Ursula can choose. You can add R Not Laws under the upper tiers as well if that helps cement the idea visually.

Rule #4: Will uses two Gs, and thus like Solomon will need to have two other flower types (of one flower each) to fill out the arrangement. In this case, that means Will's arrangement must be filled by two kinds of H, L, and R.

Rule #5: Zepi has double-exclusivity: not only is Zepi using just a single H and a single R (which we'd individually block), that combination is also stated to be unique across the arrangements.

Note that this rule is critically important because with Zepi established as using two separate types just once each, the remaining two spaces must be filled by two flowers of a third kind. Since H and R have already been used, that means the remaining two spaces in Zepi's arrangement must be filled by two Gs or two Ls:

  • G or L

    G or L

         _R_

         _H_
         Zepi

This inference is critical, and you can see how the test makers buried it in the fifth of six rules.

Rule #6: This rule establishes that one person is using exactly two Ls (and from the prior rule inference we can see that Z is one of the candidates).

At this point, many students were feeling overwhelmed. There are a lot of rules and variables floating around, and all the "exactly" and "no one else" references are tough to track. Remember, when things seem like there are too many options, it's means there are still limitations in the game that you haven't seen! Stop for a moment and examine what the test makers have given you, starting with restricted points or variables that seem to connect to others.

We made some minor inferences about remainder options with rules #1, #4, and #5, but the process does not end there. Let's now examine the restrictions that result from the "exactly" and "no one else" conditions:

Rule #5 is a good place to start, since the exclusive one R and one H restriction plays so well with the remaining options for Will. If you recall, Will must have two flowers from the group of H, L, and R (on top of the two Gs already used). However, if Will were to select H and R, then Will and Zepi would both have HR, a violation of the fifth rule. Thus, we can infer that Will selects one L, and then the remainder is H/L:

  •      H/R

         _L_

         _G_

         _G_
         Will
As a side note, the use of two Gs in Will's arrangement immediately means Will could not use two Ls, and cannot satisfy the last rule. More on this as we work through the remainder of the inferences.

Let's continue on and look at Ursula. Ursula uses exactly one R and at least one H. What are the options for the two remaining flowers? On the surface it looks wide open, but its not. Could Ursula use another R? No, because that would violate Rule #1 which states that Solomon is the only arranger two use two Rs. So, the choices are H, G, and L in some configuration. But wait, there's more :-D Ursula's base configuration is H and R, as is Zepis, but Zepi is the only one who can have one H and one R. Thus, Ursula can't have just those two, and must have one more R or one more H. But we just ruled out an extra R, so we can infer that Ursula must have another H. That then leaves the choice of G or L as the fourth flower:

  •      G/L

         _H_

         _H_

         _R_
    Ursula
As a side note, the use of two Hs in Ursula's arrangement immediately means Ursula could not use two Ls, and cannot satisfy the last rule.

Continuing the trend of moving backwards through the arrangers, Tabitha is next up. We know that Tabitha use exactly one H, so no H can be used in the remainder of the arrangement. That leaves choices among G, L, and R for the two open spaces. Let's consider what happens with each choice, focusing on the flower that could be doubled:

  • Can Tabitha choose two Gs? Yes, Tabitha could choose one H, two Gs, and—this is a tough inference—one L. Why can't Tabitha choose an R instead of an L? Because if that occurred, the arrangement would contain one H and one R, a violation of Rule #5. Whoa, that's tricky and very involved! Hard to make that deduction.

    Can Tabitha choose two Hs? No, as established above rule #2 states that Tabitha uses exactly one H.

    Can Tabitha choose two Ls? Yes, Tabitha could choose one H, one G, and two Ls. This would satisfy the last rule, meaning Tabitha and Zepi are so far the only two arrangers who could meet that final rule.

    Can Tabitha choose two Rs? No, that would cause a violation of Rule #1.
Note that a larger inference can be drawn from the above, which is that Tabitha must use at least one L, leaving the following options:

  •      G/L

         _L_

         _G_

         _H_
    Tabitha

The remaining arranger is Solomon, who uses exactly two Rs. Thus, Solomon must use two separate kinds of flowers from the group of G, H, and L.


  • G/H/L

    G/H/L

         _R_

         _R_
    Solomon

Solomon also cannot be a destination for the two Ls required by the last rule, however, as that would force Solomon to use two Rs and two Ls, which is only two kinds of flowers (and the scenario stipulates that three are required). Thus only Tabitha or Zepi can be the destination for the two Ls, and one of them must have two Ls in any solution to the game.


Overall, an intimidating game for the outset, but by using Zepi to crack open the restrictions on the remaining spaces, the game becomes far more manageable. Is it easy? No, not at all. But it's more doable than it seems at first.
 mallie
  • Posts: 7
  • Joined: Nov 28, 2019
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#72337
Hi there! I had a lot of trouble with this game from the get go. I classified it as a grouping game, although it seemed like a more complex grouping game than usual. I used the first letter of each name (Solomon, Tabitha, Ursula, Will, and Zepi) as the main variable set, and from there added each flower type according to the rules.

Based on the rules, my set up looked something like this:


R G H G H
R H R G R
S T U W Z

And from the rules I got the following deductions:
1. S cannot have any additional Rs
2. T cannot have any Hs or Rs, so the remaining two flowers must be G or L.
3. U cannot have any Rs and must have another H. The remaining spot can be H/G/L
4. Z cannot have another R or H, so the remaining spots must be G/L

Considering that I was able to make deductions, I am rattled that I can’t click in to this game to figure it out, assuming that my deductions aren’t way off base. Any help is appreciated!
 Paul Marsh
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#72346
Hi Mallie! This setup for this game is a bit unusual. Your inferences and setup all look correct, but there are some additional inferences that we can make here as well.

So we have 5 people (S, T, U, W, Z) each making an arrangement of four flowers (g, h, l, r). We also know that each person uses 3 different flowers - a bit of quick math leads us to the deduction that each person will use one flower twice, and two other flowers once. (Remember to always do this kind of math in grouping games!)

So my setup looked like this:

S: _ _ _ _
T: _ _ _ _
U: _ _ _ _
W: _ _ _ _
Z: _ _ _ _

Note: some students might consider making three spaces for each person (one big space for the flower that gets used twice, and two little spaces for the two flowers that get used once), but this would likely make things more difficult, since for several of the people we don't know which flower is being used twice and which flowers are being used once.

Ok, onto the rules.

Rule 1: S and nobody else uses 2 r's

Rule 2: T uses exactly one h and at least one g

Rule 3: U uses exactly one r and at least one h

Rule 4: W uses 2 g's

Rule 5: Z and nobody else uses exactly one h and exactly one r

Rule 6: Exactly one person uses 2 l's

First, we want to plop these rules right into our master diagram, like so:

S: r r _ _
T: h g _ _
U: r h _ _
W: g g _ _
Z: h r _ _


Now, I want to make some inferences and consider my Not Laws (what can't go in each person's arrangement). Let's go person by person. For S, we already know which flower is being used twice: r. For those last two spots, there really isn't much else to say. We know that it will be two different flowers, each being used once. Any of h, l, or g could take up those last two spots. So my final diagram for S will be:

S: r r (h/l/g) (h/l/g)

How about T - what can go in those last two spots? Not h (Rule 2). Not r either (nice job making that inference!), because Rule 1 says T can't have 2 r's and Rule 5 tells us that T can't have 1 r (since it already has 1 h). So our last two spots have to be filled up with g and l. They can't both be g (since T already has a g), so one of them must be an l. Our final diagram for T will be:

T: h g l (l/g)

Moving on to U. From Rule 5, we know that U can't use just one r and just one h. But Rule 3 tells us that U uses only one r. So we know that U uses a second h. Either l or g will work in the last spot. Final diagram for U:

U: r h h (l/g)

Now W. W uses g twice already, so we're looking for 2 flowers to be used once each for these last two spots. We can't use r and h for those last two spots because that would violate Rule 5. That means l must be in one of those spots, and the other is either h or r. Final diagram for W:

W: g g l (h/r)

Last one - Z! We know from Rule 5 that h and r are used exactly once. So we need a different flower to be used twice. From Rule 1, we know that flower can't be r. So the last two spots of Z will be filled by g (twice) or l (twice). Final diagram for Z:

Z: h r (gg / ll)

Those diagrams do a good job of accounting for all the rules, with the exception of Rule 6. Applying Rule 6, we know that either T or Z (but not both) will need to have 2 l's. Putting an asterisk next to the l's for T and Z is a helpful tool to remember Rule 6. Putting everything together, the master diagram for this game will look like this:

S: r r (h/l/g) (h/l/g)
T: h g l (l*/g)
U: r h h (l/g)
W: g g l (h/r)
Z: h r (gg / ll*)

Once we have that diagram, the questions for this game become a lot simpler. Hope that helps!
 lathlee
  • Posts: 655
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#72490
Hi. After knowing certain tricks and rules operate (took me 3 re-readinging), This game ain't that hard; I admit it and I see it. but i think considering every LSAT takers write this game in time sensitive environment, I think its' fair to include this game as one of the hardest LG all time Powerscore-TM list?

As in I think this game is more difficult than ( https://blog.powerscore.com/lsat/bid-15 ... -all-time/ )
PT22, June 1997, Game #4: Juggling Teams; PT21, December 1996, Game #4: Product Advertising Advanced Linear: Unbalanced: Underfunded; PT77, December 2015, Game #3: Employee and Offices
 Claire Horan
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#72501
Hi lathlee,

Sure, people can differ if which games they find most difficult. Powerscore generally chooses the statistically most difficult games, but also looks for a sample of games that may be difficult for a variety of reasons. In any case, you should pat yourself on the back for conquering a very difficult game!

Happy studying!
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 Dave Killoran
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#73362
lathlee wrote:As in I think this game is more difficult than ( https://blog.powerscore.com/lsat/bid-15 ... -all-time/ )
PT22, June 1997, Game #4: Juggling Teams; PT21, December 1996, Game #4: Product Advertising Advanced Linear: Unbalanced: Underfunded; PT77, December 2015, Game #3: Employee and Offices
Hey Lathlee,

I bet this will be added this at some point to the all-time list but I'm waiting for the stats to build up on it. And, as a point of interest, I find all of the games above harder than this one :) This game is hard in my opinion, but also it has clear avenues of attack that make it easier. So, it's a great learning game!

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