### Game Setup

**Time on first attempt: **14 minutes

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Boy, was I slow on this one. This happens to me fairly often on newer LG sections; I go *extremely *fast on 2-3 games, then quite slowly on another game. I usually get the whole section right, but I need the extra time to get through the harder one. This is why it’s very important to finish easy games *fast. *You’ll appreciate the extra time on hard ones.

(I was slow because I missed a key deduction I should have made.)

This is a grouping game. And here’s the secret I missed:

- Each group needs two people. And there are only three options: G, I and one of F/H
- Deduction: Either G or I must be in a group. If one is out, the other is in.

They test you on this deduction in the final question. It’s phrased as a rule substitution question, but the answer is also a “must be true” deduction that *really *makes the game easier if you get it earlier.

On to the rules. Here’s how you should set up the game:

I’ve added rule 4: G can’t visit Sydney. I’ve also added the constraint from the setup that each city must have exactly two visitors. This becomes *very *important. (And I missed drawing it in my own setup, which was part of the reason I was slow)

Next, we can draw the other rules. Then we’ll think about deductions:

When looking for deductions, it helps to think of the most restricted point. Here, that’s one of the three cities: S. Normally, there are four people who can go in a city: G, I, F and H

But, S only has three: I, F and H. And, thanks to rule 2, it can only be *one *of F and H. So actually, the people who can go in S are: I, F/H

This is important. I *has *to go in S:

This is the major deduction of the game. I actually didn’t get it until partway through, so don’t feel bad if you didn’t spot it.

The other major deduction is that either G or I has to be in each group. That’s because, again, only G, I, or F/H can go. Since we need two, that means you need at least one of G and I (since only one of F/H can be in)

You can draw this as an additional rule as a reminder:

Note that we *could *have both G and I in a city. The rule above only means we need at least one. It does *not *mean we can’t have both. (“G —> not I” is the “not both” format of a rule. This is “not I —> G”)

Now let’s consider the other rules. We know that Ibanez must go in exactly two places. Since they *have *to go in Sydney, that means we can split the game into two scenarios: one with Ibanez in Manila, and one with Ibanez in Tokyo:

I’ve added two other things:

- G must go in the other city, since that’s the only way to have two people in each group without I (G + F/H).
- In the second scenario, rule 3 applies. Since G is in Manila, then H is in Tokyo.

That’s pretty much it for the scenarios. Remember that you *have *to place both F and H at least once in each scenario, as well. (But not together)

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