Setup and Rule Diagrams

[Dave Killoran]

Setup and Rule Diagram Explanation

This is an Advanced Linear: Unbalanced: Underfunded game.

This game is reminiscent of the third game on the September 1995 LSAT. That game featured houses in one of three styles—ranch, split-level, and Tudor—on opposite sides of a street. In this game, stores on opposite sides of the street are decorated with lights of one of three colors, leading to the following initial setup:

D00_Game_#4_setup_diagram 1.png

Because only three colors of lights are available for each store, we can infer that when one color is unavailable there is only a dual-option remaining for that store. This realization is one of the keys to the game.

The first two rules establish that no adjacent stores have lights of the same color, and no stores facing each other have lights of the same color. This creates horizontal and vertical not-blocks for each color, which will be represented with “C” for color (this is a more efficient representation than drawing out all six not-blocks):

D00_Game_#4_setup_diagram 2.png

These two rules prove to have a significant effect on the game.

The third rule limits the use of yellow on each side of the street:


Side Exactly 1 Yellow

The fourth and fifth rules both place specific colors on specific stores:

D00_Game_#4_setup_diagram 3.png

Applying the third rule, no other store on the north side of the street can be yellow. Applying the second rule, store 6 cannot be yellow and store 3 cannot be red. Applying the first rule, stores 2 and 6 cannot be red. These inferences can all be shown as Not Laws:

D00_Game_#4_setup_diagram 4.png

Of course, when one color is eliminated from consideration for a store, a dual-option remains. When two colors are eliminated, the store must be decorated in lights of the remaining color. Thus, store 3 must be green, and store 6 must be green. Consequently, from the first rule, store 1 cannot be green, and must therefore be red, and store 8 cannot be green:

D00_Game_#4_setup_diagram 5.png

Filling in the remaining dual-options brings us to the final setup for the game:

D00_Game_#4_setup_diagram 6.png

[srcline@noctrl.edu]

Hello

I initially did not find this game to be all that difficult, but I dont really know where I am going wrong with this game.

I set this game up with two vertical "stacks" with N and S being the base, N was numbered 1-9 and S was numbered 2-10 with both "stack" being adjacent to one another facing one another. Would this be the right set up?

The last two rules create not laws and I concluded that since red lights decorate store 4 store 3 cant be red and since Yellow lights decorate store 5, store 6 cant be yellow. I am lost for that point on

Maybe I am misunderstanding the first two rules? Also I don't think that this game was in the logic games, was this part of a drill in the course books? Please help

Thankyou
Sarah

[Lucas Moreau]

Hello, srcline,

It may be problematic that you're making the game in two vertical stacks. Horizontal stacks might be more appropriate, like this:

1 3 5 7 9
2 4 6 8 0

You are correct in how the rules interact as far as stores 3 and 6. But there's one last trick to be played - Rule 2 is very powerful for this game. It limits the number of yellow lights we have, which lets us fill in more parts of the game. Right now it looks like this:

_ _ Y _ _
_ R _ _ _

But since we know that there are no more yellow lights on the north side of the street, we can now safely say that store 3 is green - it can't be yellow and it can't be red due to store 4. That gives us this:

_ G Y _ _
_ R _ _ _

That's as far as we can get with just the basic rules - sometimes you just can't get much further before you get to the questions. The information found in the questions should be sufficient to spell out the rest of what you need. If you have any specific questions about the specific questions, feel free to reply with them.

Hope that helps,
Lucas Moreau

[srcline@noctrl.edu]

Hello Lucas

Thankyou for clearing that up. However, when I went back and did the set up and inferences this is what I got

N: _ G Y _(7 is not Y) _
S;_(2 is not R) R G _ _

Is this correct ?

Thankyou
Sarah

[Robert Carroll]

Sarah,

You are on your way, but there is a bit more you can tell.

As Lucas said, the rules about 4 being R and 5 being Y give us:

_ G Y _ _

_ R _ _ _

Because there is exactly one Y on each side of the street, you can put not-laws for Y not only in 7, as you have, but on 1 and 9 also. This is because the Y on the north side is definitely 5, per the rule, so there are no more Y in odd-numbered spaces.

You are right that 2 is not R, due to the first rule. 1 is not G for the same reason, and we know it's not Y because of what I talked about in the previous paragraph, so 1 is R. Thus:

R G Y _ _

_ R _ _ _

7 and 9 must be different colors, so:

R G Y R/G G/R

_ R _ _ _

And 6 is G because it cannot be Y or R by the first two rules. So:

R G Y R/G G/R

_ R G _ _

This is not much more than you had, and only includes a few more inferences. Does it all make sense?

Robert Carroll