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Setup and Rule Diagram Explanation

This is a Grouping: Partially Defined, Numerical Distribution, Identify the Templates game.

This game is extremely challenging, and it requires you to fully understand the implications of some relatively standard rules.
In the game scenario, an artisan is creating three stained glass windows from a selection of five colors. Each color is used at least once, and each window contains at least two colors. This leads to an initial setup as follows:
D10_game #2_setup_diagram 1.png
D10_game #2_setup_diagram 1.png (2.52 KiB) Viewed 3598 times
However, a major point of concern within this scenario is the use of “at least” in relation to both the use of the colors and the number of colors within each window. The use of “at least” opens the door to many more combinations than would exist if the word “exactly” had been used in each instance. For example, without considering the rules, initially each of the three windows could be made from all five colors. This use of “at least” is the first sign that this game may be harder than average.

The first rule establishes that exactly one window will contain both green and purple glass:
D10_game #2_setup_diagram 2.png
D10_game #2_setup_diagram 2.png (1.61 KiB) Viewed 3598 times
Carefully note the wording in this rule however; it does not state that one of the windows will be only green and purple glass, just that exactly one of the windows contains both green and purple glass, which means that other colors could be present. Again, this allows for more options than would otherwise be the case.

The second rule is fairly straightforward, and it establishes that exactly two of the windows contain rose glass:

..... ..... ..... ..... ..... ..... ..... ..... ..... ..... 2R

This rule means that our minimum color set is now: G O P R R Y.

The third rule is actually two rules in one, and each rule is a negative grouping rule. Because we have a vertical component to our Grouping diagram, we will show these two rules as vertical not-blocks:
D10_game #2_setup_diagram 3.png
D10_game #2_setup_diagram 3.png (2.58 KiB) Viewed 3598 times
One of the implications of this rule is that none of Y, G, and O can be used three times. Because each color must be used in at least one window, if, for example, O was used three times (once in each window), then Y could not be used in any window. Thus, Y, G, and O are each used in one or two windows, and this numerical fact plays an important role in the game.

The final rule has a powerful effect on the game. This rule contains a negative sufficient condition, and most students initially diagram it as follows:
D10_game #2_setup_diagram 4.png
D10_game #2_setup_diagram 4.png (1.19 KiB) Viewed 3598 times
Of course, the contrapositive of this rule is:
D10_game #2_setup_diagram 5.png
D10_game #2_setup_diagram 5.png (1.17 KiB) Viewed 3598 times
When both representations are considered together, the rule means that when one of O or P is not used in a window, the other color must be used. Because every window must then contain O or P (or possibly both), this can be represented directly on the diagram as an O/P option for each window:
D10_game #2_setup_diagram 6.png
D10_game #2_setup_diagram 6.png (4.1 KiB) Viewed 3598 times
Note that while P could be used in all three windows, O can only be used in at most two windows (see the discussion of the third rule above), so one window will always contain P and not O.

This is an incredibly powerful rule that must be accounted for in the making of each window, and by itself it answers question #8. Consider also what this rule means for Y. From the third rule Y cannot be used in a window that contains O, and thus whenever Y is used in a window, P must also be used:
D10_game #2_setup_diagram 7.png
D10_game #2_setup_diagram 7.png (975 Bytes) Viewed 3598 times
While this inference is not directly tested in the questions (how unfortunate!), it does play a role in answering questions such as #12.

At this juncture, our setup appears as follows:
D10_game #2_setup_diagram 8.png
D10_game #2_setup_diagram 8.png (10.79 KiB) Viewed 3598 times
At this point, most students use this as the final setup and move on to the questions. And, with the information already derived from our cursory examination of the rules, they will answer questions #7 and #8 with relative ease. Plus, the inferences derived during our discussion will carry them to overall success on the game.

But, there is a further step that can be taken, and it is one that rests on recognizing that the specific window numbers are not that critical in this game. Take a moment to glance at the rules and the seven questions. You may notice that the composition of the windows is not attached to specific window numbers: none of the rules references a specific window number, and only question #7 references specific window numbers. Thus, we can also attack this came by considering the composition of each window irrespective of the window numbers. While this approach is not necessary for game success, it is worthwhile to consider from a strategic standpoint, and as an instructive lesson for future games that might be similar.

When the game is considered without referencing window numbers, the initial scenario would appear as follows:
D10_game #2_setup_diagram 9.png
D10_game #2_setup_diagram 9.png (3.88 KiB) Viewed 3598 times
But, some of the rules—such as the first rule—establish further restrictions on the color combinations, such as the following:
D10_game #2_setup_diagram 10.png
D10_game #2_setup_diagram 10.png (3.68 KiB) Viewed 3598 times
In addition, there is also the fact that Y is a very powerful variable in this game. Y must be used at least once, but it cannot be used in all three windows due to the third rule. Thus, Y must be used once or twice (and it correspondingly eliminates certain colors from being used and also requires P to be used). From this standpoint, the use of Y suggests a templating approach, one when Y is used once, and one when Y is used twice. Let’s examine both:

Template #1: Y used in exactly one window

When Y is used in exactly one window, then the GP block from the first rule must be used in a different window (remember, from the third rule G and Y cannot be used in the same window). The “third” window then retains the choice of O or P. The use of each color must still be tracked, and the second rule involving R must also be tracked. This template thus provides a base, but it still allows for multiple options.
D10_game #2_setup_diagram 11.png
D10_game #2_setup_diagram 11.png (4.98 KiB) Viewed 3598 times
Template #2: Y used in exactly two windows

This template provides a bit more detailed information. Two of the windows contain Y, and thus they also contain P. The other window contains the GP block from the first rule, as well as O. The use of each color must still be tracked, and the second rule involving R must also be tracked.
D10_game #2_setup_diagram 12.png
D10_game #2_setup_diagram 12.png (6.1 KiB) Viewed 3598 times
Is having the templates necessary to solving this game or even to solving it quickly? No, the rules and the resulting inferences will allow for success (thus, in the questions, we will generally use the rules to solve each problem and only occasionally reference the templates). But, again, this approach is slightly more efficient and gives you a better overall picture of the possible directions the game can take. Therefore, templates are useful to consider, especially because a higher percentage of recent games have hinged more on group composition rather than on particular group names or group numbers, and this feature at times benefits from a template approach.

Overall, this is a very challenging game, and it qualifies as one of the hardest games to ever appear on the modern LSAT. Almost every student struggles to complete this game within the time constraint, so do not be discouraged if you had trouble with this game. Instead, study each rule and its implications so that the next time the test makers attempt to use these ideas to create difficulty you will be prepared.
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I found this game to be pretty difficult. During the test - I wasn't able to create a good set up or make many inferences, and ended up skipping it after getting through just questions 7 and 8 (I could tell I was stuck and losing time so wanted to move on to the next games). But even going back after the test to solve it, it took me 12 minutes or so to complete - and I still can't figure out #9!

Is this considered a difficult game or am I maybe missing something in the set up/inferences? Is there a key inference to understanding the colors? Would it be possible to see what the correct set up is? And how do you answer question 9?

Thanks so much for any help!
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 Dave Killoran
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This game is definitely considered very hard, and after the D10 test a lot of people were talking about how much this game negatively affected their exam.

Jon Denning, one of the senior instructors here, actually wrote a blog post about this game a while back. Check out ... #more-1380 and let me know if that gives you some help here.

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Thanks, Dave. Good to know I was not alone. Thanks for directing me to the blog post!
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I'm having a lot of trouble with this game. Can you explain how you would go about the general setup and what inferences you can make?


 Jon Denning
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Hey Andrew - thanks for the question! This game caused problems for many test takers in December 2010, but upon review it turns out that if you caught a couple of powerful (and fairly common) inferences the game itself really wasn’t all that difficult. Let’s take a look at those inferences and hopefully you’ll see how to better attack this one.

To me, the rule that immediately stands out as significant is the last rule. What this rule essentially tells you is that the absence of either O or P forces the other one to be used. So if O is out then P must be in, and vice versa. This exact conditional rule has appeared in a number of games previously, and the ramifications are consistent and HUGE: either O or P must ALWAYS be used, in every window, every time! You simply can’t get rid of both of them at once. Note that you can use both of them at once, but the key is that at least one of them has to be there. That answers question #8 (can’t be just G and R because you have to have either O or P), nearly answers #9 (knowing that the two windows must have R gives you O+R or P+R), and rules out some answers for other questions, too.

And there's even more. Consider what that rule means for Y: since Y and O cannot be in the same window (rule 3), when you see Y you know you’ve also got P. This inference never gets directly tested, but it does really help with a question like #12, where you can quickly eliminate every option but P (the other four cannot go in all three windows because they either get specifically limited to only two windows like R, or they conflict with another variable like O, G, and Y).

Let’s look at one more, question #13. Again, knowing you must have two Rs (rule 2), and that for this question they can’t be with O, then you must have two RP windows. One RP window has to have G to make your single GP group (rule 1), and the other will be RPY since Y must be used. And that’s that: RPY must be a window and you’ve got your correct answer.

Slowly but surely you see how the O/P rule controls other variables and plays a vital role in your ability to attack this game. And the great news is that this type of rule, even this EXACT rule, has come up many times on past LSATs, so recognizing its significance could very likely pay off in the future, as well.

I hope that helps!

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Hello Jon,

You explanation helped a great deal, I cant believe I missed the 4rth rule inference, it makes sense because in both instances O and P are on the necessary side, could represent this with a double arrow? Also I wanted to make sure my inference for rule 3 was correct I had :

Y :arrow: not G nor O
(+) G and O :arrow: not Y

So: G :dblline: Y and G :dblline: O

Also, O and P are the power variables and since this show up twice, would this mean that they are the most restricted?

 Nikki Siclunov
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Hi Sarah,

Jon is on the road at the moment, so I thought I'd jump in an answer your question.

Your interpretation of the third rule is partially correct, but it contains an error. It's actually a negative grouping rule, suggesting that Y cannot be in the same window as either G or O. We can diagram this rule in several ways:
  • ..... ..... ..... G

    Y ..... :arrow: ..... +

    ..... ..... ..... O
  • G

    or ..... :arrow: ..... Y

Personally, I'd just use the Double-Not arrow:
  • Y :dblline: G
    Y :dblline: O
Note that G and O could be in the same window as each other, as long as Y is not there. One of the implications of this rule is that none of Y, G, and O can be used three times. Because each color must be used in at least one window, if, for example, O was used three times (once in each window), then Y could not be used in any window. Thus, Y, G, and O are each used in one or two windows - a numerical inference that plays an important role in the game.
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This was an extremely difficult game for me. I missed a number of the inferences that you made but did get others so I was 1/2 and 1/2 on the inferences.

My question is how did you consider that -P :arrow: O and the contrapositive being -O :arrow: P was the most powerful inference? I diagrammed it correctly but did not get the inference that O or P would be in all three windows. I'm not exactly sure how you inferred that and I missed it so I wanted to know your strategy that next time I'm faced with a scenario like such, I would know how to attack it. What was the thing that lead you to that thought when you saw the rule? When I saw the rule, I did not think to apply it to all three windows in some way.

Additionally, for the -Y/G and -Y/O inference, I wrongly inferred that Y :arrow: Y + P or Y + R. Your inference was Y :arrow: P. How did you come up with that and why was Y :arrow: Y + R incorrect?
 Adam Tyson
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That rule is one of the more confusing conditional rules you'll find in logic games, and rules like it pop up frequently enough that we wrong a whole blog post about it, which you can find here: ... rule-lsat/

The short answer is this: when a conditional rule says "if this is out, this other thing must be in," you can immediately infer that one or the other of those two variables must always be included. Having both is possible, but having neither is forbidden, because as soon as one goes out the other must come in!

As for the inference about Y and P, it comes from combining two rules. If Y is in, O is out:

Y :arrow: O

And if O is out, P is in (the contrapositive of that last, tricky rule):

O :arrow: P

Combine the two into a chain and you get:

Y :arrow: O :arrow: P

which allows us to make the shortcut inference of Y :arrow: P!

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