Time on first attempt: 14 minutes
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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