Setup and Rule Diagrams

[Dave Killoran]

Setup and Rule Diagram Explanation

This is a Grouping: Undefined game.

The number of variables being selected—birds in this case—is left open, and so the game is classified as Undefined. Although a maximum of six birds can be in the forest (remember, there are only six birds total), prior to consideration of the rules there could be anywhere from zero to six birds in the forest. This uncertainty increases the difficulty of the game and is an element that must be tracked throughout the game. Of course, since it cannot be determined exactly how many birds are in the forest, there is no static “selection group” diagram as in a Defined game. Here is the setup:

D00_Game_#2_setup_diagram 1.png

Like many Undefined Grouping games, this one contains a large number of conditional rules. By using basic linkage, we can draw a slew of inferences. Let us examine each in greater detail:

• 1. J G. This inference results from linking the first two rules.
  
  2. M G. This inference results from linking the first two rules.
  
  3. W H. This inference results from linking the first and third rules.
  
  4. J W. This inference results from linking the first inference and the third rule. Note how the first inference has been recombined or “recycled” with the original rules.
  
  5. M W. This inference results from linking the second inference and the third rule. The third rule here refers to the rules as listed in the game.
  
  6. S J H G , W. The final rule is tricky and bears further analysis. When J is not in the forest, then S must be in the forest. Via the contrapositive, when S is not in the forest, then J must be in the forest. In each case, the absence of one of the birds forces the other bird to appear in the forest (hence J or S is always in the forest). This type of “omission” rule appears infrequently on LSAT games, but when it does, it tends to cause problems. It is easy to forget that the absence of a variable forces another variable to be present. In this case, when S is not in the forest, then J must be in the forest, and from the second rule, when J is in the forest, it follows that H must be in the forest. Of course, from the first rule and third inference, when H is in the forest, then G cannot be in the forest and W cannot be in the forest.
  
  7. W J S. From the fourth inference it is known that W and J cannot be in the forest together. Thus, when W is in the forest, then J cannot be in the forest, and from the last rule it follows that S must be in the forest (W S). This is another classic example of recycling an inference.
  
  8. G J S. Similar to the previous inference, when G is in the forest, then J cannot be in the forest, and from the last rule it follows that S must be in the forest
  (G S).

In light of all these inferences, the bigger question becomes, “When do you know you have made all of the inferences?” In this case the application of basic linkage creates a large number of inferences, and then the recycling of those inferences leads to even more inferences. At some point the time pressure of this section demands that you move on to the questions. Although in our diagram we could continue to make inferences (for example, if H is not in the forest, then J is not in the forest and S must be in the forest), there comes a point when you must ask yourself, “Do I have enough information to effectively attack the questions?” The answer here is undeniably “yes.” It may be that you do not discover every inference in the game, but when you feel you have exhausted all the obvious routes of inference-making, it is time to move on to the questions. The challenge in the questions then becomes keeping track of all the information at your disposal.

[Thejonesgroupe]

Need help understanding inference on pg.324 the third inference references linking rules 1 and 3 but I am not making the connection. Any assistance would be greatly appreciated....thank you

[Dave Killoran]

Hi Jones,

Thanks for the question! Let's start by looking at the two rules cited in the making of that inference.

The first rule involves H and G in a negative grouping rule:


H G


The third rule involves W and G in a straightforward conditional relationship:


W G


Because both rules involve G, they can be connected in the following fashion:


W G H


That chain connection elicits the inference that:


W H


And that is the inference referenced on page 324

If you wish to explore the reasoning behind that inference in more detail, take a look back at pages 244-245, where that same type of chain is discussed. You'll see this particular chain a lot in Logic Games, so make sure you are comfortable with it.

Please let me know if that explains that one for you. Thanks!

[florbonita]

When I first attempted to do this problem in the LG Bible awhile ago, I failed miserably and was completely stumped. After much practice from the course, I retried it. I am happy that I only missed 3 this time, but I missed one of the key inferences! J and s can be together! Consequently, I missed 3 problems that required this info. Question: How to avoid this in the future? What is the best way to diagram problems such as this? Should I spend more time trying to write ALL of the inferences (as the explanation is in the LG Bible and the HW book) or should I just "diagram" the rules with the appropriate arrows and read my inferences that way? I hope this question makes sense. Thank you for your help.

[Emily Haney-Caron]

Hi Flor,

Congrats on your improvement with this tricky game! With problems such as this one, the inferences you make are so critical. Because of that, I'd really recommend always writing out the contrapositives, in order to make sure you don't miss key conclusions. Then, be careful to stick to what you know - if you diagram the J-S rule, and then the contrapositive, all you know is what happens when either J or S is NOT in the group; so you want to avoid drawing conclusions about what happens when one of them IS in the group.

I hope that helps!
Emily

[mv2484]

Hi Powerscore - first, I want to say that I am in absolute awe of this series! It is truly amazing, and I can't wait to continue to dig into it.

I have been a little thrown by the Bird-watchers question on page 323. I was able to arrive at all of the inferences correctly, but the answer choices floored me a bit. I suspect that I am reading into each question a bit too much, and therefore am making over-assumptions - but here goes:

#6 - this was fine.
#7 - M & H are in the forest, which means that W is not (b/c M :dblline:W) and G is not (b/c
H G). If W is not, then J either is or isn't (b/c W J) and S is also a maybe (depending on J). In this scenario, it makes sense that J & S could also be in the forest, and so the answer is E. Correct? So, I'd have to account for the fact that in W J, both could not be present.
#8 - (must be false/cannot be true) If jays are not in the forest, then G are in the forest (b/c J G) and if G are in the forest then M are not (b/c M G). As I kept going - I came to arrive at: J, M, H are not in the forest, and S, G, M are. Why isn't the answer (A) - martins are in the forest. Answer (C) says that neither martins nor shrikes are in the forest. In actuality, martins are not in the forest and shrikes are in the forest - isn't one part of this question true and the other part false? I'm just really stuck on understanding this one.
#9 - this was fine.
#10 - this was fine.
#11 - if grosbeaks are in the forest, then I came to arrive at: G, W, and S are in the forest and J, M, and H are not in the forest. I realize that as according to the final inference, G S, the answer is A, but couldn't it be B as well - that the forest contains both wrens and shrikes? Because if there is G, then there is no H, M, or J, and so I jumped and thought there would then be W & S - but I suppose there could be no W as well (since J W, which means that both can be absent) which would then allow (A) to be the MUST be true, and (C) to be the could be true. Is that correct?
#12 - for this question, I got the correct answer, but arrived at it differently, from what is explained in the answer key, and I want to make sure my method isn't incorrect. If S H, then that can be combined with H:dblline:W - so that gives us S H W, which means S W, which is why/how I arrived at the answer (B). Is that also a fair method?

Thanks in advance for all of your help!

[Dave Killoran]

Hi MV,

Thanks for the nice words about the book! I'm glad it is helping you out so far

Let's take a look at this game and see if we can clarify a few of these questions:

• #7: Your analysis is spot on. Good job!
  
  #8: You ran into a problem here, so let's look at this more closely. First, you stated that, "If jays are not in the forest, then G are in the forest (b/c J G)" but that's not how it works. This rule is only activated if J are in the forest; but here they are out, so we don't know about G. Because of that, the rest of your inference chain breaks down.
  
  Second, all we know is that J S. So, when the jays aren't in the forest, the shrikes must be (which you correctly identified). Knowing that S is in the forest makes (D) a problem, because D says that M and S are not in the forest. But we know S has to be, so (D) cannot be true.
  
  #11: I think you made an error with the third rule here, which states that W G. Watch out for the Mistaken Reversal! Just because G is in the forest does not mean that W is there!
  
  #12: Not quite. The relationship of S H W doesn't mean that S W has to occur; it just means that S and W could be together in the forest. That may seem like a small distinction, but S H W doesn't allow for an actual inference (that is, something that must be true). All we know is that S and W both hate H, which means they might be ok being together, or maybe not.

Good work overall on your part, and please do not hesitate to let me know if you have any additional questions about all this. Thanks!

[mv2484]

It's all crystal clear now! You're amazing - thanks for your help

[Smdickso]

7) In question #7 both martins and harriers are in the forest. The correct answer is E) "There are at most two other kinds of birds in the forest." Is the key to this answer the "at most" portion of the answer E? Since having harriers in the forest would prevent grosbeaks & if wrens are in the forest so must be grosbeaks, and if jays are in the forest shrikes are not. Therefore it would seem that the possible options for birds in the forest would be (martins, harriers, jays) or (martins, harriers). It appears to be that three is the maximum number of birds in the forest. There seems to be a discrepancy because rule 4 prohibits Jays and shrikes from being in the forest together, yet the explanation on page 135 is saying that jays and shrikes can be in the forest together. Can you please explain this?

9)Answer is listed as C) which is that there can at most be 4 birds in the forest and in the explanation for the answer it lists out that Jays, harriers, martins, and shrikes can all be in the forest to come up with the possibility of 4 birds at a time being the maximum. This response again as in question 7 is confusing me since based on rule 4 jays and shrikes can not both be in the forest together. Can you please clarify this for me?

10) The correct answer for this question A) seems to be more dependent on the inference than on following rule #4 which would prohibit jays and shrikes from being in the forest together. Answer B) jays & Shrikes seem like the obvious choice based on rule #4. Can you please explain this?

Thank you,

Sheri

[Dave Killoran]

Hi Sheri,

Thanks for the questions! These three questions all revolve around the interpretation of the fourth rule, so if we can clear that up, I think the they will all make sense. I'll use #7 to discuss how that rule works.

Let's start with #7. First, it is possible for three birds (M, H, J/S) or four birds to be in the forest in this question (M, H, J, and S). So, in this sense, the "at most" is key since it covers the three and four options. Fro your question, I can see that the relationship between J and S is causing you some problems, so let's hone in on that rule.

The fourth rule in this game states that, "If jays are not in the forest, then shrikes are." The placement of the "not" here is critical, because it's not where it usually is. In this case, it's on the sufficient condition, not the necessary condition. Thus, the diagram is:


J S


As detailed on page 268 and 269, this presentation—where the sufficient condition is negative but not the necessary—is dangerous. Many students take this rule to mean that J and S can't be in the forest together, but the meaning is actually the polar opposite. The meaning of the rule as given comes down to: when one of J or S is absent, the other must be there. Thus, at least one of J or S is always selected, possibly both.

So, in question #7, when we get down to M and H are in, and that G and W are out, only J and S are left, and you could have either one or both. Hence, answer choice (E) is correct, and the "at most" includes the possibility of four birds.

Here's the good news: now that you've seen this "negative sufficient condition" rule presentation, you won't fall for it again. And that's one of the great values of preparing—you get a chance to see what they can do and how you should best handle it. This rule type has been appearing with greater frequency recently, and so it's important to know how tricky it can be.

Please let me know if that helps. Thanks!