# Reply To: Drill Ordering Games Set 2

September 29, 2016 at 4:09 pm #2749
Mike Kim
Keymaster

Hey Julie —

If you were able to get through the mastery drill well but ran into trouble with so many of the ordering games, I wonder if it’s specifically ordering rules/ inferences holding you back — see if that’s the case and, if so, see if brushing up on those helps —

Another thing you may want to look into is whether you could have framed those games and if framing would have been helpful.

Finally, one last suggestion is to, in your review, think of your work in terms of three components — visualizing / diagramming the game, making inferences, and answering q’s efficiently — obviously these areas are all related to one another, but the more you can categorize your challenges along those lines, the easier it becomes, I believe, to see the path toward improvement —

Just by coincidence, I happened to have played and written up 55-2 recently for another project — I just have a rough draft (so forgive me if I missed an inference or anything) but I figured you might find it useful — specifically, I thought it might help you to read my step-by-step thought process and compare it to yours —

Here’s what my paper looked like

And the written explanation —

PT 55, Game 2; 7 – 12 mins, frames? Recommended but not necessary.

Setup

This is an ordering game with a very limited number of elements and some numbers uncertainty. We have three elements—H, J, and L—each of which can be placed either once or twice. This, combined with the second rule, means that we must have a minimum of four and a maximum of six messages.

I initially started my diagram by combining the first two rules like this:

(image)

The third rule is easy enough to notate.

Finally, I chose to frame off of the last rule: J can go in 1, 2, or 3, and when it does, it cannot go in the other two of those positions.

These frames were not necessary, but I thought it might be helpful considered in conjunction with the fact that we know the first and last element must be H or J.

When J is in 1, it must also be in the last position, and when it is not, H must be first and last.

I chose to mark the spots I wasn’t sure would be occupied with …., but that’s completely optional — it’s just important that you remember there are potential slots to be filled.

(image)

Then I thought about how the HJ rule could be satisfied in the frames. In the first frame, it could only be satisfied if H is next to last. In the second frame, the rule is already satisfied and there isn’t another chance for it to come into play so we can forget it about it.

I then realized that the third rule cannot be satisfied in the third frame, which allows us to cross it out and actually leaves us with just two possible frames.

(image)

7. This is a Rules question and we can go down the list of rules and eliminate answers one at a time.

The first rule allows us to eliminate (A).
The second rule allows us to eliminate (C).
The third rule allows us to eliminate (B).
The fourth rule allows us to eliminate (E).

And so (D) is the correct answer.

8.

If, per this quest stem, J is before H, it must be the first frame. In this case, the most messages Sukanya can receive after the first J can and before getting one from H is two (both from L).

Thus the correct answer is (C).

9.

We can use our frames to evaluate the consequences of knowing there are four messages. Our key concerns are that we have to fit in an L, and we need to make sure we have an HJ ordering.

In order to make that happen in the first frame, the only possible order is J – L – H – J.

In the second frame, the only possible order is H – J – L – H.

In either frame, (A) must be true.

(B) doesn’t have to be true in the second frame.
(C) doesn’t have to be true in the second frame.
(D) doesn’t have to be true in the second frame.
(E) doesn’t have to be true in the second frame.

10.

We know that since L can’t leave the first message, L can’t leave the last. Thus, the correct answer is (E).

11.

If L is fifth, we know we are dealing with the second frame. In that frame, J can’t be third, and so L must be third and J fourth.

And based on those inferences, all other answers must be false and the only answer that must be true is (D).

12.

Since we already have three items placed in both frames, and the placed items happen to be at the two extremes in terms of order, we know that the two L’s must fit somewhere in between, and the maximum space between two L’s could only either be 1 or 0.

Let’s make sure 1 is a possibility.

Using the first frame, we can put L in 2, H in 3, and L in 4 without violating any of the rules.

Again, we know that we can’t have greater than 1 space between the L’s in the second frame, and so there is no need to consider it.