- Mon Jul 11, 2016 11:27 am
#26954
Setup and Rule Diagram Explanation
This is an Advanced Linear Game: Unbalanced: Underfunded.
The five-year period forms a natural base for this game, and the car purchases and graduations should be stacked on top of the base. The remaining variable set—the three friends—should be used to fill in the spaces in the car and graduation rows, as such:
The friends variable provides uncertainty in this game, namely because the number of variables is fewer than the available slots. At first, it appears that the three friends will fill three spaces in each row (the graduation row and the car row), leaving two empty space variables (E’s) to fill the two remaining slots in each row. If this was the case, the Underfunded aspect of the game would be easily deflected. But, the second-to-last rule makes it clear that R and S will graduate in the same year, meaning that there are three E’s in the graduation row (along with an RS block and variable S). In the car row there are 2 or 3 E’s, depending on when R buys his car (R could conceivably buy a car in the same year as S or T). Let’s look at each rule, with the exception of the last rule, which can be represented right on the diagram (subscripts are used to represent the car/graduation elements):
These four rules can be combined into one super-sequence:
This sequence is immensely powerful in this game because it limits the placement of a number of variables. In fact, the sequence is so powerful that there are only two possibilities for the graduation row. However, other than mentioning this fact, we tend not to diagram out those options because the sequence is so strong that we do not need to expend the time diagramming out the options. Let us instead look at the complete diagram to the game, adding in Not Laws and the dual options created by the super-sequence:
From the sequence, the graduations of the three friends must be after Sue bought her car but before Taylor bought his car, and thus 1991 and 1995 in the graduation row must be empty (E). In addition, because three separate yearly events occur after S bought her car, Sue can only have bought her car in 1991 or 1992 (and this is represented by a dual-option). Similarly, because three separate yearly events occur before T bought his car, Taylor can only have bought his car in 1994 or 1995 (represented by another dual-option).
Note that the last rule, which states that “at least one of the friends graduated in 1993,” is represented by an E Not Law under 1993 in the graduation row.
UPDATE per some discussion below: this is a game that can definitely be attacked/solved with two Templates (based on the gradation options—RS or T—in 1993), so if you approached it with that strategy you're totally in the clear!
Here's what they look like:
Template 1 (RS graduate in 1993):
Car: _S/_ _/S_ ____ ____ _T__ Rc?
Grad: ____ ____ _RS_ _T__ ____
1 2 3 4 5
Template 2 (T graduates in 1993):
Car: _S__ ____ ____ _T/_ _/T_ Rc?
Grad: ____ _RS_ _T__ ____ ____
1 2 3 4 5
Hope this helps!
This is an Advanced Linear Game: Unbalanced: Underfunded.
The five-year period forms a natural base for this game, and the car purchases and graduations should be stacked on top of the base. The remaining variable set—the three friends—should be used to fill in the spaces in the car and graduation rows, as such:
The friends variable provides uncertainty in this game, namely because the number of variables is fewer than the available slots. At first, it appears that the three friends will fill three spaces in each row (the graduation row and the car row), leaving two empty space variables (E’s) to fill the two remaining slots in each row. If this was the case, the Underfunded aspect of the game would be easily deflected. But, the second-to-last rule makes it clear that R and S will graduate in the same year, meaning that there are three E’s in the graduation row (along with an RS block and variable S). In the car row there are 2 or 3 E’s, depending on when R buys his car (R could conceivably buy a car in the same year as S or T). Let’s look at each rule, with the exception of the last rule, which can be represented right on the diagram (subscripts are used to represent the car/graduation elements):
These four rules can be combined into one super-sequence:
This sequence is immensely powerful in this game because it limits the placement of a number of variables. In fact, the sequence is so powerful that there are only two possibilities for the graduation row. However, other than mentioning this fact, we tend not to diagram out those options because the sequence is so strong that we do not need to expend the time diagramming out the options. Let us instead look at the complete diagram to the game, adding in Not Laws and the dual options created by the super-sequence:
From the sequence, the graduations of the three friends must be after Sue bought her car but before Taylor bought his car, and thus 1991 and 1995 in the graduation row must be empty (E). In addition, because three separate yearly events occur after S bought her car, Sue can only have bought her car in 1991 or 1992 (and this is represented by a dual-option). Similarly, because three separate yearly events occur before T bought his car, Taylor can only have bought his car in 1994 or 1995 (represented by another dual-option).
Note that the last rule, which states that “at least one of the friends graduated in 1993,” is represented by an E Not Law under 1993 in the graduation row.
UPDATE per some discussion below: this is a game that can definitely be attacked/solved with two Templates (based on the gradation options—RS or T—in 1993), so if you approached it with that strategy you're totally in the clear!
Here's what they look like:
Template 1 (RS graduate in 1993):
Car: _S/_ _/S_ ____ ____ _T__ Rc?
Grad: ____ ____ _RS_ _T__ ____
1 2 3 4 5
Template 2 (T graduates in 1993):
Car: _S__ ____ ____ _T/_ _/T_ Rc?
Grad: ____ _RS_ _T__ ____ ____
1 2 3 4 5
Hope this helps!
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