Diagram

[carmenmarrs]

I have a question about the setup for condition #3 & 4..i couldnt figure out how to put them in the diagram & if you change the & to or AND negate the letters for the contropositive?

I thought you did the both but the explanations in the superprep book are confuzing me!

Thanks so much!!

Carmen

[stackoutawinner]

Diagramming the exclusive or can be a little tricky. I've never had the benefit of seeing how someone else does it, but generally I'll just fill in the diagram normally and then denote the exclusivity next to that part of the diagram, you could do the circle with the plus in it, or write out XOR... whatever works.

~ Y --> L XOR O
~ L v ~ O --> F & S

As you can see, the final constraint is fairly straight forward. If either L or O is omitted, it triggers the chunk F & S. Typically in my diagrams, I write the chunk by having ONE arrow leaving the element and then splitting into two telling me it's both... whereas two separate statements I have two separate arrows.

But I agree... can we please see how one of the forum mods would diagram the XOR on a binary grouping??? Thanks in advance

[dan]

Hey guys. Mike has written a full diagram/solution to this game. I'll attach it here.

Best of luck!

dan

[cdc3d]

What is the fastest time that anyone has completed this problem? I find that this problem takes me a minimum of 7 minutes to set up, and I know that is way too long! I could tell that it was automatically a binary problem, but diagramming the constraints took me forever, and then I ended up with a diagram that was difficult to read. I tried doing the modified set up, which made it easier to read, but it took even longer. Any tips?

[dan]

I just went through and diagrammed this from scratch. It took me 2:30 to get through the diagram. Granted, I've seen this game before. That said, that shouldn't matter much. If you're very comfortable with conditional logic, your speed should only be limited by how long it takes you to read the rules and draw the lines. In other words, there shouldn't be much thinking involved (except for on those really tough rules, such as the fourth one for this game).

Time saving tips:

1. If you can see immediately that the third constraint needs to be noted off to the side instead of in the Logic Chain picture (b/c it contains an "or" in the outcome), that should save some time. Maybe it took you a while to think through this one and decide what to do with it? You can save some time by getting a sense for those rare constraints that don't fit into the logic chain picture. Just note them to the side and move on. Generally, any constraint that has an "or" in the conditional outcome (If John goes, then Sally or Peter goes) or any constraint that has an "and" in the conditional trigger (If John and Sally go, then Peter goes) is best noted to the side. These are very rare, but if you can spot them quickly you won't waste a whole bunch of time thinking about what to do with them.

2. Become a conditional master. Can you go from reading "If firs are in the park, pines are not" straight to two quick arrows in 10 seconds or less? If you're writing the logic out like this first: F ---> -P, P ---> -F, and then transferring to the diagram, you could probably pick up some time by eliminating the middle step. At first, it's good to translate each constraint before diagramming it, but as you become more and more comfortable with conditional logic, diagramming a constraint like that should feel as comfortable as writing your name.

Hope that helps!

dan

[carmenmarrs]

Thank you so very much! You guys are great! This game was just terrible. I've been doing lots of tests, but I'm going through the games now & this forum has really helped!

Thanks again,

Carmen

[cdc3d]

Dan,

Thanks for the info. I knew immediately that it was a binary game, and I was immediately able to diagram the first two constraints. Once I recognized the third constraint needed to be diagrammed on the side, that significantly reduced some time. However, I kept stumbling over the fourth constraint. That darn "not both" construction took me much longer than it should have to figure out the diagram and its contrapositive. I looked at the set up that you posted, and it helped me realize that I was including an additional, superfluous step. Essentially, I was diagramming "not both laurels and oaks" as: (1) L and ~O --> F + S; (2) ~L and O --> F + S; and (3) ~L and ~O --> F + S. Once I visually reproduced those diagrams on my binary chart, my chart was unreadable! Bah!

So... I had to step back and think, does "not both" necessarily imply "neither"? And is it necessary to include "L and ~O" and "~L and O," or can I simplify and synthesize the two inferences to create one "~L OR ~O" inference? By the time I had finished the metacognitive debate, I had used the allotted time and had a headache, too!

That said, I think I have it. I spent an entire afternoon on this one problem, and finally by nightfall had managed to become sufficiently comfortable with the "not both" construction that I could complete the entire game in 7 minutes, 30 seconds. Nonetheless, I still maintain the right to despise the LSAT for its use of the "not both" construction!

Thanks!
-Chelsea

[haeaznboiyoung]

I'm still having a little problem understanding how to diagram rules 3 and 4.

For rule 3, the diagram said:
~Y -> ~L AND O

OR

~Y -> L AND ~O

From there, I'm not seeing how the contrapositive was derived:

L AND O -> Y

OR

~L AND ~O -> Y

For rule 4, I see in the diagram that the rule was essentially split up:

~L -> F AND S
~O -> F AND S

I'm reading the rule as "If ~L AND ~O -> F AND S"

I would understand splitting up the rule if it read ~L OR ~O, but it explicitly states the condition can only work if both L AND O are not included.


My understanding of contras and negating OR's and AND's are completely being ignored here. Can someone please clarify?

[ManhattanPrepLSAT1]

You're having the same issue with this game as so many others. In fact one day I spent all morning with people coming to me with the exact same question. It must have been an assigned practice test or something, but everyone had the exact same question all morning.

Here's how I see the 3rd and 4th constraints...

3rd:

~Y ---> L or O (not both)

you could write it another way

~Y ---> (L or O) + (~L or ~O)

If you take the contrapositive

(L + O) or (~L + ~O) ---> Y

back into English

If both L's and O's are in, or L's and O's are out, then Y's are in.


4th:

Not both can be expressed

~(L + O) ---> F + S

using the distributive property

~L or ~O ---> F + S


So, I can see why taking the contrapositive of the 3rd constraint can be tricky, but if you think about it on an intuitive level, you can see that the stated contrapositive above is accurate. The 4th constraint though is a bit tough. Your interpretation would have been correct had the constraint read, "if neither laurels nor oaks are in the park, then it contains firs and spruces."

Does that clear this one up?

[farhadshekib]

You're having the same issue with this game as so many others. In fact one day I spent all morning with people coming to me with the exact same question. It must have been an assigned practice test or something, but everyone had the exact same question all morning.

Here's how I see the 3rd and 4th constraints...

3rd:

~Y ---> L or O (not both)

you could write it another way

~Y ---> (L or O) + (~L or ~O)

If you take the contrapositive

(L + O) or (~L + ~O) ---> Y

back into English

If both L's and O's are in, or L's and O's are out, then Y's are in.


4th:

Not both can be expressed

~(L + O) ---> F + S

using the distributive property

~L or ~O ---> F + S


So, I can see why taking the contrapositive of the 3rd constraint can be tricky, but if you think about it on an intuitive level, you can see that the stated contrapositive above is accurate. The 4th constraint though is a bit tough. Your interpretation would have been correct had the constraint read, "if neither laurels nor oaks are in the park, then it contains firs and spruces."

Does that clear this one up?

Hey Mshermn... Thanks for the great explanation.

I had a really tough time w/ this game because I did not understand the last 2 rules correctly; this actually led me to bomb the games section, and consequently my confidence went down the drain for the entire test.

Anyhow, I have 2 quick questions:

1) Are there are other games that have similar constraints? I've done all the games from 1-39 at least once, and I do not remember coming across rules that gave me this much trouble...

2) What is the correct contrapositive of the last rule? The diagram you provided suggests:

~ L --> F + S
or
~ O --> F +S

Contrapositive:

~F or ~ S --> L + O

But the super-prep, on page 240, states that "if the park doesn't have firs AND spruces, then it must have both laurels AND Oaks".

I am assuming that they mean: ~ F + ~S --> L + O.

Thanks for your help.

[timmydoeslsat]

Rule 1:

M ---> ~Y

Rule 2:

F ---> ~P

Rule 3:

Can be linked up with the first rule, so I will tag it on there.

M ---> ~Y ---> One of L and O

Rule 4:

~ [L and O] ---> [F and S]



I want to tell you something that will make these rules seem so second nature to you.

Think about what this hypothetical rule means:

~A ---> B

Look at its contrapositive:

~B ---> A

So, you tell me, could there ever be a situation where A and B are BOTH OUT?

No!

If A is out, then B is in.

If B is out, then A is in.

We could, however, have both in. Nothing stops us from having both in.

What happens if A is in? We do not know! We know what happens when A is out, but we know nothing about A being in. Same with B.

Look at the placements of the negation in our conditional statement.

If you have this...

~A ---> B

Look at the necessary condition, one must always be in.

If you have this...

A ---> ~B

Look at the necessary condition, one must always be out.

Because the contrapositive of that statement is:

B ---> ~A

Could A and B BOTH BE IN at the same time? NO!

If A is in, B is out.

If B is in, A is out.

Both could be out at the same time however. We know nothing of ~A and ~B. We know what happens if A or if B.


So let us go back to that fourth rule for a moment, I will bring it back down for us to talk about:

~ [L and O] ---> [F and S]

This mirrors our ~A ---> B concept.

We know that at least one has to be in! Both could be in, but we know for a fact that both CANNOT be out.

That rule is telling us:

If L and O are not in together, then F and S are in together.

If F and S are not in together, then L and O are in together.

We could have a situation where we have both LO and FS groups.

We know that it is impossible to not have at least one of those two groups.

The third rule, in my opinion is much easier to state as this:

M ---> ~Y ---> One of L and O

We know that the contrapositive is when we negate the necessary condition and sufficient condition and reverse them in the arrow.

Think about how we can negate the concept of "One of L and O"

Would both L and O negate the concept of "One of L and O" YES!

Would having no L AND no O negate the concept of "One of L and O" YES!

We can indicate this in the following way:

Contrapositive:

[No L and No O] OR [L and O]---> Y ---> ~M

[ManhattanPrepLSAT1]

1) Are there are other games that have similar constraints? I've done all the games from 1-39 at least once, and I do not remember coming across rules that gave me this much trouble...

I think the game that might give you the best practice with similar conditionals would be

PT31, S1, G2

though I'll admit that it doesn't help practice with that last constraint.

2) What is the correct contrapositive of the last rule? The diagram you provided suggests:

~ L --> F + S
or
~ O --> F +S

Contrapositive:

~F or ~ S --> L + O

But the super-prep, on page 240, states that "if the park doesn't have firs AND spruces, then it must have both laurels AND Oaks".

I am assuming that they mean: ~ F + ~S --> L + O.

When the SuperPrep says that "if the park doesn't have both firs and spruces, then it must have both laurels and oaks it means that

~(F + S) --> L + O

"not having both" is equivalent to missing at least one of the two. When you use the distributive property and bring the ~ inside the sufficient condition you get...

~F or ~S --> L + O

Tricky use of the word "and," huh?

[farhadshekib]

1) Are there are other games that have similar constraints? I've done all the games from 1-39 at least once, and I do not remember coming across rules that gave me this much trouble...

I think the game that might give you the best practice with similar conditionals would be

PT31, S1, G2

though I'll admit that it doesn't help practice with that last constraint.

2) What is the correct contrapositive of the last rule? The diagram you provided suggests:

~ L --> F + S
or
~ O --> F +S

Contrapositive:

~F or ~ S --> L + O

But the super-prep, on page 240, states that "if the park doesn't have firs AND spruces, then it must have both laurels AND Oaks".

I am assuming that they mean: ~ F + ~S --> L + O.

When the SuperPrep says that "if the park doesn't have both firs and spruces, then it must have both laurels and oaks it means that

~(F + S) --> L + O

"not having both" is equivalent to missing at least one of the two. When you use the distributive property and bring the ~ inside the sufficient condition you get...

~F or ~S --> L + O

Tricky use of the word "and," huh?

Hah, that's what I figured after reading more of their explanation. That is interesting. Thank you - and to timmydoeslsat - for the clear and vivid explanations. Keep up the great work.

[yying_wong]

I hate to keep beating this dead horse, but I do have a question on SuperPrep explanation on Page 241, paragraph #2: "...if there are no spruces in the park, then the pairing of spruces with firs is also not in the park."

Rule 4 says: ~L or ~ O --> F+S
The contrapositive is ~F or ~ S --> L + O

My understanding is that either F or S is absent, then L and O are in. Then let's move on the question # 8, the SuperPrep explanation says if S is out, F must be out as well. I don't see any indication that is true. The contrapositive says if one of F and S is out, then L+O are in. How do they link F and S as a pair that must be both in or both out? What am I missing here? Please help!!!

[timmydoeslsat]

I hate to keep beating this dead horse, but I do have a question on SuperPrep explanation on Page 241, paragraph #2: "...if there are no spruces in the park, then the pairing of spruces with firs is also not in the park."

Rule 4 says: ~L or ~ O --> F+S
The contrapositive is ~F or ~ S --> L + O

My understanding is that either F or S is absent, then L and O are in. Then let's move on the question # 8, the SuperPrep explanation says if S is out, F must be out as well. I don't see any indication that is true. The contrapositive says if one of F and S is out, then L+O are in. How do they link F and S as a pair that must be both in or both out? What am I missing here? Please help!!!

I am reading that explanation and it does not state that if S is out then F is out.

The rules in question are these: Rules 3 and 4

Rule 3:

~Y ---> 1 L/O in, 1 L/O out

Rule 4:

~ [L and O] ---> [F and S]


So the last rule we can see as the "at least one rule."

If we do not have the [L and O] block, we have the [F and S] block. We could have both blocks, but we could never have neither block.

Question 8 gives us a local question of M and S being out of the park and it asks us what could be true.

In:

Out: M S

I know that I can invoke the contrapositive of the last rule. This park will not have [F and S], so I know it will have [L and O]

In: L O

Out: M S

I know that since my L and O is together, that this invokes the contrapositive of the third rule. I will not be able to have L and O split up. This means that Y is in the park.

In: L O Y

Out: M S

I have F and P left.

The only relevant rules governing its placement now is: F --->~P

This can be thought of as the "not both" rule.

This means that at least one of those two must be out.

In: L O Y

Out: M S F/P


P/F is free to float.

So we know that it could be true that four types of trees are in the park. Five is impossible and L, O, Y all must be in the park. A is the answer.