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- Tue Oct 20, 2015 11:00 pm
#40583
Setup and Rule Diagram Explanation
This is an Advanced Linear: Balanced game.
The game scenario establishes that a naturalist will give lectures on five birds, one after the other. Each lecture is also given in exactly one of two locations: G or H. The order of the lectures (1-5) should be the base, and there should be rows for birds and the locations stacked above the base set:
With the basic structure in place, let’s examine each rule.
The first two rules are similar, and both are very straightforward. The first rule establishes that the first lecture is in G, and the second rule establishes that the fourth lecture is in H:
The third rule indicates that exactly three of the lectures are in G, which then means that the other two lectures are in H. Thus, the composition of the locations variable set is established at G G G H H5. We’ll show that fact in the final diagram.
The fourth and fifth rules are both similar, and involve sequences. Because this is an Advanced Linear game, references to birds and locations should be shown as vertical blocks:
Fourth rule:
Fifth rule:
These last two rules create several Not Laws, and further inferences. Let’s consider the fourth rule first:
Further, from the fourth rule, we know that the lecture on S is given at H, so if S is not the fourth lecture, then wherever S appears will be the “other” lecture given at H. So, if the second lecture was on S, and thus given at H), then the first, third, and fifth lectures would be given at G.
This relationship is particularly interesting if the other H appears fifth. If the fifth lecture is given at H, then by applying the fourth rule, S must be fourth (because it cannot be fifth) and O must be fifth (because it must be given later than S). This is a fairly deep inference, and having it prior to starting the questions is not necessary. But, what is necessary—and what should be drawn from this discussion—is that the limited number of Hs and the second and fourth rules interact in a way that you should track during the game. Thus, where S and O appear will play an important role in the game.
Let’s show the final setup for the game, and now include the fact that there are only two Hs and three Gs (also, because one of each has been placed, those will be marked out). R is a random, and could be noted as such, but it is also limited by the dual-option produced by the Not Laws on the first lecture, and thus we will leave it unmarked as a random.
This is an Advanced Linear: Balanced game.
The game scenario establishes that a naturalist will give lectures on five birds, one after the other. Each lecture is also given in exactly one of two locations: G or H. The order of the lectures (1-5) should be the base, and there should be rows for birds and the locations stacked above the base set:
With the basic structure in place, let’s examine each rule.
The first two rules are similar, and both are very straightforward. The first rule establishes that the first lecture is in G, and the second rule establishes that the fourth lecture is in H:
The third rule indicates that exactly three of the lectures are in G, which then means that the other two lectures are in H. Thus, the composition of the locations variable set is established at G G G H H5. We’ll show that fact in the final diagram.
The fourth and fifth rules are both similar, and involve sequences. Because this is an Advanced Linear game, references to birds and locations should be shown as vertical blocks:
Fourth rule:
Fifth rule:
These last two rules create several Not Laws, and further inferences. Let’s consider the fourth rule first:
- Because the first lecture is in G (from the first rule), and the fourth rule establishes that S is given in H, we can infer that S cannot be first. Then, from the sequence in the fourth rule, we can infer that O cannot be first or second (O cannot be second, because S can never be first). Looking at the other end of the diagram, from the fourth rule we can infer that S can never be fifth (because O must always be given later). Thus, from the fourth rule (and from considering the first rule), four Not Laws are produced:
- Because T must always be given earlier than P, P can never be first. Conversely, T can never be given fifth, because P must always be given later than T. And, because P is given in G, and the fourth lecture is given in H (from the second rule), P can never be fourth. This adds three more Not Laws to the diagram. Additionally, because the first lecture cannot be S, O, or P, it must be on R or T:
Further, from the fourth rule, we know that the lecture on S is given at H, so if S is not the fourth lecture, then wherever S appears will be the “other” lecture given at H. So, if the second lecture was on S, and thus given at H), then the first, third, and fifth lectures would be given at G.
This relationship is particularly interesting if the other H appears fifth. If the fifth lecture is given at H, then by applying the fourth rule, S must be fourth (because it cannot be fifth) and O must be fifth (because it must be given later than S). This is a fairly deep inference, and having it prior to starting the questions is not necessary. But, what is necessary—and what should be drawn from this discussion—is that the limited number of Hs and the second and fourth rules interact in a way that you should track during the game. Thus, where S and O appear will play an important role in the game.
Let’s show the final setup for the game, and now include the fact that there are only two Hs and three Gs (also, because one of each has been placed, those will be marked out). R is a random, and could be noted as such, but it is also limited by the dual-option produced by the Not Laws on the first lecture, and thus we will leave it unmarked as a random.
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