Setup and Rule Diagrams

[Administrator]

Setup and Rule Diagram Explanation

This is a Grouping: Balanced, Defined-Moving, Numerical Distribution, Identify the Templates game.

The game scenario establishes that each of seven trees must be planted on exactly one of three lots:

June15_game_2_diagram_1.png

Since exactly seven trees are being planted on the three lots, but it is yet unknown how many trees are being planted on each lot, the game is Defined-Moving. In addition, since each tree must be planted on exactly one lot, the game Balanced. Notice the absence of any overt indication as to the minimum number of trees per lot.

The first rule states that the trees planted on one of the lots are H, O, and exactly one other tree:

June15_game_2_diagram_2.png

The second rule states that M and W cannot be planted on the same lot.

June15_game_2_diagram_3.png

The third rule indicates that either L or W, but not both, is planted on lot 1:

June15_game_2_diagram_4.png

The fourth rule states that either M or O, but not both, is planted on lot 2:

June15_game_2_diagram_5.png

Finally, the last rule requires that the number of trees on lot 3 is greater than the number of trees on lot 1:

• lot 3 lot 1

This is an excellent reminder to examine the Numerical Distributions that govern the assignment of the seven trees to each lot. Here’s what we know so far:

• 1. Although the scenario does not specify a minimum number of trees per lot, the rules clearly demand at least one tree per lot: there must be at least one tree on lot 1 (L/W), and at least one on lot 2 (M/O). And, since there must be more trees on lot 3 than on lot 1, the minimum number of trees on lot 3 must be two:
  
  June15_game_2_diagram_7.png
  2. Once the minimum requirement for each group is satisfied, we are left with 3 trees to work with. Our distribution needs to conform to the following additional conditions:
  
  There must be exactly three trees on one of the lots (first rule).
  There must be more trees on lot 3 than on lot 1 (last rule)

The easiest way to start creating our distributions would be to test which lots can accommodate the group of three trees (H, O, and one other tree). If we plated them on lot 1, then we would need to plant at least four trees on lot 3 in order to comply with the last rule. This would end up using all seven trees, leaving nothing for lot 2 (in violation of the fourth rule). Therefore, the group of three trees cannot be planted on lot 1:

June15_game_2_diagram_8.png

If the group of three trees is planted on lot 2, we would be left with exactly four trees to distribute between lots 1 and 3. And, since there must be more trees on lot 3 than on lot 1 (3 > 1), the four trees cannot be split evenly between the two lots. Therefore, lot 3 will end up with three of the trees, while lot 1 will end up with only one tree. Due to the fixed nature of this distribution, it would be best to represent it with a Template:

June15_game_2_diagram_9.png

This Template can be taken a step further. Recall that either M or O, but not both, must be planted on lot 2 (rule 4). So, if the group of three trees—H, O, and one other tree—is planted on lot 2, then M cannot be planted there. However, M cannot be planted on lot 1 either, because that lot contains only one tree in this distribution, and that tree must be either L or W (rule 3). Therefore, in the 1-3-3 distribution, M must be planted on lot 3. And, since W cannot be planted on the same lot as M (second rule), W cannot be planted on lot 3:

June15_game_2_diagram_10.png

Next, we need to examine the distribution where the group of three trees (H, O and one other tree) is planted on lot 3. We are left with four trees to distribute between lots 1 and 2, but need to ensure that lot 3 > lot 1. With three trees planted on lot 3, lot 1 can have at most 2 trees. This creates two separate distributions:

June15_game_2_diagram_11.png

In each of these two distributions, M must be planted on lot 2 in compliance with the fourth rule. And, since M and W cannot be planted on the same lot, W cannot be planted on lot 2:

June15_game_2_diagram_12.png

The game epitomizes the central proposition that is inherent in a Templates approach: you will spend a bit more time in the setup, but this time will be regained in the lightning-fast execution of the questions. Even though our templates are not fully developed (each of them contains a fair number of empty spaces), they do show how the rules operate together, and reveal important inferences that you may not have been able to make otherwise (for instance, you may notice that M is not planted on lot 1 in any of the templates).

This approach also highlights the close link between Numerical Distribution and Templates. Granted, not all distribution-driven games can be solved with Templates, and not all Template-driven games contain Numerical Distributions. Nevertheless, there is a significant association between the two.

Notice, also, that the decision to examine the Numerical Distributions (and Templates) could not have been made without thoroughly analyzing the rule set first. Students often make the mistake of creating distributions immediately upon noticing a numerical imbalance in the scenario, which could potentially lead to the creation of many more distributions than the rules ultimately allow. So, examine your rules first, and then decide if a Numerical Distribution (that may or may not lead to the creation of Templates) is worth analyzing.

The final diagram for the game should look like this:

June15_game_2_diagram_13.png

June15_game_2_diagram_14.png

[ncolicci11]

Hi Powerscore,

I might be completely overlooking something in the rules, but is it possible to have a 4-2-1? Say HOPS in lot 3, LM in lot 2, and W in lot one?

Thanks for your help!!

[Jeremy Press]

Hi n!

No, that's not possible, because of the language of the rule including H and O, which says that "the trees planted on one lot are the hickory, the oak, and exactly one other tree." Your proposed distribution would have two other trees with H and O.

Thanks!

Jeremy

[ncolicci11]

Hi Jeremy!

Thank you. I knew I was missing something- it turns out I was completely overlooking that. Thanks again!!

[sbose]

Hi!

I just had a general question for grouping games...should you always analyze what the Numerical Distribution is of the game when it's a grouping game? If no, what in the game indicates that you should determine what the Numerical Distribution of the game is?

Thank you in advance for your help!

[Jeremy Press]

Hi sbose,

Distribution is a fundamental element of every game, not just grouping games. But sometimes, the distributional possibilities are more obvious, so I don't have to think hard about them. For example, if the scenario tells me that 9 students are assigned to 3 classes, each class getting 3 students and each student assigned to exactly 1 class, the scenario itself has done the distributional work for me (3-3-3). On the other hand, some scenarios are so open-ended that it's probably not worth trying to determine every single distributional option in the game. So, for example, in an "Undefined Grouping Game," they might tell me that "Some of 6 fruits are selected to be included in a fruit stand." That's pretty open-ended (no numerical requirement is attached to the fruits; and very little numerical requirement is given as to how many are included in the stand), so I'm probably not going to go through the process of saying: well, it could be 6-0, or 5-1, or 4-2, or 3-3, etc.

In grouping games specifically, when one of the variable sets is fully or nearly-fully specified, and the other set is unspecified (or only partially determined), then it makes sense to think about distributional possibilities. So, in this game, the trees are fully specified: each tree goes to exactly one lot. However, the numbers of trees assigned to the lots are, from the scenario language, unspecified (and, from the rules, they are only partially determined). In that situation there's enough of a constraint on the numbers (even though they're not completely determined for me) that it makes sense to write out all the possibilities. That makes this and similar games the ideal kind of game to take a moment to determine the available distributions.

I hope this helps!

Jeremy

[sbose]

Hi Jeremy,

Yes, that was super helpful! Thanks so much!

[NR 2020]

How did you draw the P/S in 1 inference in Template C?

[Luke Haqq]

Hi NR2020!

Happy to address your question. Since your question is about template C, we can start by winnowing down from the fact that this is an identify the templates game. Of the possible distribution of 7 variables over 3 base variables (1-3-3, or 2-2-3), template C in the above walk-through is a 2-2-3 distribution.

These distributions are largely dictated by the rule that the number of variables in lot 3 are greater than the number in lot 1. If lot 1 has two variables, for example, then we know the 3rd lot has three. That is the base for template C. Within that framework, we know that one of the spots in lot 1 is occupied by L/W because of the third rule in the game.

We can then ask what other possible variables could fit in lot 1. Since we know from the first rule that there is an H-O block, we can eliminate the possibility that either H or O could occur in lot 1. We also know from the fourth rule that either M or O, but not both, occurs in lot 2. We can infer that it must be M of these two that goes in lot 2, because if O were placed there, the first rule would come into conflict with the fifth rule--that would involve a 2-3-2 distribution (not possible, according to the fifth rule). We must be dealing with a 2-2-3 distribution.

With that foundation, we know that L/W occupies one of two spots in lot 1. In lot 2, we know that M is one of the two spots, and in lot 3, we know that H and O occupy two of the three spots.

Finally, that leaves us with both of the variables in P/S, as well as one of the variables in L/W (since we're already given that one of L/W is in lot 1). The second rule ends up being helpful here. Since M can't be with W according to that rule, then that narrows down the possibilities. Instead of trying both L/W and P/S in lot 2, we can eliminate W as an option because M and W can't be together. If one places L in lot two, then we know that W and P/S are in lot 1, and a P/S is in lot 3. Alternatively, if L is not in lot 2, the only alternative is P/S. With that P/S in lot 2, we can infer that another P/S must go in lot 1, and an L/W in lot 3 (since L and W can't be together on lot 1). In both of these two possibilities, there is a P/S spot reserved in lot 1. This is how one can infer the P/S inference in lot 1 in template C.

Hope that explanation helps!