June 2007 LSAT, Game 4 Setup, Rivertown Recycling Centers

Game Setup

Time on second attempt: 6:30

See “repeating games” at bottom of section

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This is a grouping game. It’s also comparatively difficult, about as difficult as game 3. There are three groups, which you should draw vertically:

I added a few rules to this diagram. The setup says the centers have at least two, but no more than three materials. So I drew two slots. Then, on the right there is a blank space a vertical line: this shows that each group has space for one more item – but only one more.

These vertical lines are a thing you can use on this type of game to show when a group is closed off. For example, if a group had max two items, you’d draw the vertical line directly to the right of the spaces.

The arrow represents rule 2. Anything in center 2 is also recycled by center 1. Note that there are two possibilities here:

• Center 2 has three materials: center 1 has exactly the same materials.
• Center 2 has two materials: center 1 has those two, but might have a third as well.

Next, you should draw the other rules. W leads to N, and P can only appear once and not with glass:

The second item means that plastic can only appear once, and it can’t go with glass. There are a couple of other ways you could draw this rule. For example:

(You’d also need to list “P = 1”)

Choose whichever version is most intuitive for you. I like the one I chose even though it’s normally a diagram used for linear games (indicating that PG can’t go beside each other, in either order). In my experience I’ve found there’s never been an issue with using this diagram style for both purposes; it feels easiest to me.

I circled both T and N. This is normally done for variables with no rules. So you might wonder why I circled N, since N is related to W.

The reason is that while W places N, N doesn’t actually have any restrictions. So N is very easy to place. Even easier than T, in fact! If you place N, you can place either W or N. Whereas if you place T, you can’t place W.

This “N is easy to place” deduction comes up in many places, and makes the game much simpler.

The other thing to notice is that P and G are hard to place. There are actually only two options for P: it can go in center 1, or center 3. It can’t go in center 2, because anything in 2 also goes in 1. (Meaning P would go twice: in both 1 and 2. But P can only go once.)

Since P can only go in center 1 or 3, we can create two diagrams: One with P in center 1, and one with P in center 3. Let’s draw P in center 1 first:

Now, we know P and G can’t go together. That makes G hard to place – where can we put them?

It turns out that if P is in center 1, then G must be in center 3. Why not 2? Because anything in 2 is in 1, and then we would have P and G together.

Now, why have I drawn P in the third position in center 1? Because there is only one P. Which means P isn’t in center 2, but center 1 shares both of the two things from center 2. Which means there must be two open slots in center 1, and only two materials in center 2.

As for the second scenario, it’s simpler: P is in center 3, and G is somewhere in center 1. Centers 1 and 2 could either have 2 or 3 materials.

G can’t be in 3, because G and P can’t go together. So, G must be in either 1 or 2. But, there’s a twist: anything in 2 must be in 1. So, either way, G is in 1!

Another notable deduction: if we are in a scenario with three materials in center 2, it must be in the second scenario.

There’s actually a bit more we can deduce about the first scenario. Who else can go in the two spots shared between centers 1 and 2? Not P, and not G: those are already covered. So the only ones left are W, N and T.

If we place W, we have N. If we place T, who can we place? Not W (that would require N as well), so we have to place N. So whether you place T or W, N is always there:

The second slot in center 3 is relatively open. Anyone except P can go there. Just remember all materials need to be recycled somewhere. So, if T is recycled in 1 and 2, then W (and N) will have to be recycled in center 3.

You don’t need to get these scenarios in advance, of course. But they help. I should note that in my case I actually hadn’t figured out the scenarios explicitly. But, I knew the following two things:

• N seemed easy to place and often had to be placed
• P and G were restricted, and seemed to split between 1 and 3.

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Repeating Games

I’ve written elsewhere about the benefits of repeating games, to solidify your intuition for deductions. Note that the purpose of repeating games is to prove the answers right, so it doesn’t matter if you remember the right answer.

I repeated this game about three days after I first saw it, by which time I had forgotten the answers. I’ve written how long it took me on the second attempt. That time, or a couple minutes above it, is roughly the standard you should be aspiring to – a lot of people take 8-9 minutes on a repeat attempt, get everything right, and pat themselves on the back. But that’s too slow. The faster you go when repeating, the faster you’ll learn to go the first time you see a game.

(I say “a couple minutes above” my time because, after years of teaching the LSAT, I’m really, really fast. You should be almost as fast as me, but you don’t exactly need to match my pace to score -0.)

Time on second attempt: 6:30