#18 - Global, Justify

[Jon Denning]

Question 18 is a Justify question, which is interesting because it's the second appearance of a Justify in this section (the first being question #11 in Game 2). If you'll recall from that earlier question, the correct answer will result in every spot being filled definitively, with no movement and no uncertainty.

Here's (a close copy of) what I wrote for question 11 about this particular task:

"The right answer in a Justify question will be so powerful, so impactful, as to completely fill in every single position with absolute certainty.

In a way then this is a lot like Must Be True, as the correct answer will be the only one that's fixed—in this case completely fixed in terms of its results, where all of the positions will be known for sure.

So you can do one of two things here (or both, potentially).

First, if you've noticed something in the game that creates a completed setup you can quickly scan the answers for it. This is hit or miss though, since there may be multiple triggers that fill in all the spots, or if you're setup is somewhat thin (as ours is here) you may have zero insight into what that trigger could be.

Second, and a bit more reliably, you can start to churn through the answers, aggressively seeking one that's so profound in its implications that the chain reaction it starts will run to completion, with all the pieces slotting into place."

Let's try that approach, aggressively (!) moving through the options trying to avoid following a single, set path. (Remember if anything more than a single outcome is possible—even we can have even the slightest amount of uncertainty—the answer is wrong)

Answer choice (A): F goes twice, meaning G and H go once each. But there are a number of ways to place just the Fs, much less the rest: we could have an F on M and on S, with a G on M and H and I on T. We could also have the two Fs on M and T, with an I on M, H on S, and G on T. That's too much variability.

Answer choice (B): If G goes twice then H and F are singles, and G must be on M and T, since G cannot be on S (final rule). G on M puts H on T, so T is full (G and H). No we have an F and an I left. F must go to S since otherwise we would have two Is on S, with the final I on M with G. That's every spot, filled with certainty. It looks like this:

I F H
G I G
M S T

So (B) is our answer.

We can quickly cycle through the rest though.

Answer choice (C): like answer choice (A), H going twice allows for uncertainty. That is, we could Hs on M and S: M has HI, S has HI, T has GF. Or we could Hs on M and T: M has GH, S has FI, T has HI.

Answer choice (D): Putting F and G on T only tells you that G cannot got to M (T would need an H if so). But M could still be HI or FI: if HI then S could also be HI or FI; if FI then S would be HI. Lots of possibilities.

Answer choice (E): like (D), filling T with G and H does nothing definitively. M could be GI with S as FI, of M could be FI with S as GI or HI...we can't know.