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This is a Basic Linear: Balanced, Identify the Templates game.
This game is remarkably similar to the second game on this test. In this game, six vehicles are serviced over the course of six days, one vehicle per day. Thus, this is a Balanced Basic Linear game with the following initial setup:
Similar to the prior game, there are again four rules. Let us examine each in detail.
The first rule can cause some diagramming confusion. The rule suggests a sequence is in play, as at least one vehicle is always serviced later in the week than the hatchback. However, instead of diagramming the rule as H ___, the best representation is to show the functional effect of the rule, namely that H can never be serviced last:
The second rule creates a standard chain sequence:
V R H
As with all three-variable chain sequences, six Not Laws are created:
However, because we know that H cannot be serviced last from the first rule, the latest that H can be serviced is Friday. Operationally, the effect of this rule combination is that the last three Not Laws from the second rule “slide” over one day, to Thursday and Friday:
Saturday, which already cannot be a day when H is serviced, is also off-limits to V and R. Adding these three Not Laws back into the diagram creates the full Not Law set from the first two rules:
Note that already the last position is becoming restricted, and is down to the triple-option of L/P/S.
The third rule creates an unusual block, one where P is always serviced consecutively with S or V, but not both. This creates four separate, exclusive block possibilities:
Essentially, then, there is either a rotating PS block, or a rotating PV block, but SPV and VPS are precluded. This can be diagrammed in various ways, including:
No Not Laws can be drawn from this rule.
The fourth rule creates two mutually exclusive possibilities:
Two inferences can immediately be drawn from this rule, namely that S can never be serviced first (because S is always serviced later than P or later than L) and S can never be serviced last (because S is always serviced before L or before P):
When added to the prior two rules, this results in the inference that only L or P can be serviced last:
As in the last game, let’s examine the scenarios that result under each possibility.
Template #1: P S L
Under this option, L must be serviced last because only L and P could be serviced last, and P must be serviced earlier than L in this scenario. With L being serviced last, the first five days must meet the following conditions:
- 1. V R H
2. P S
3. PV or VP or PS (SP is impossible since P S)
Template #2: L S P
Under this option, P must be serviced last because only L or P could be serviced last, and L must be serviced earlier than P in this scenario. When P is serviced last, then either V or S must be serviced on Friday, to satisfy the third rule. However, because V must be serviced earlier than R and H from the second rule, V cannot be serviced on Friday, and thus S must be serviced on Friday:
The remaining four variables are H, L, R and V. From the second rule, H, R, and V are in this sequence: V R H. L acts as a random as it already satisfies the L S P sequence because S and P are serviced on Friday and Saturday, respectively. This results in the complete diagram for this template:
Combining all of the prior information creates the final setup for this game:
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