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This is a Grouping Game: Partially Defined.
The scenario in this game lists six technicians, each one of which is repairing between one and three types of machines. Thus, each technician is a separate “group,” and one of the tasks of the game is to attempt to determine how many machines each technician repairs.
The rules contain a great deal of numerical information, which is not surprising given the uncertainty in the scenario about group sizes. We know from the scenario that each technician repairs at least one type of machine, so let’s look at each of the six rules and see what other information we can gather about group sizes:
- This rule establishes that exactly four technicians (including X) must repair R.
- This rule establishes that Y repairs at least two types of machines, T and V.
- This rule starts off a run of four rules that all address the size of the groups in the game. In this case, S and Y cannot repair the same type of machine. Because each technician must repair at least one type of machine, this rule effectively eliminates S and Y from each repairing all three types of machines. When combined with the second rule, then, we can deduce that Y repairs exactly two types of machine, and those machines are T and V. Of course, if Y repairs T and V, then S cannot repair T and V, and thus S must repair exactly one machine, R.
Note that the circled numbers above S and Y indicate that those technicians are fixed at those group sizes. Of course, this is only a partial setup based on the first three rules. Let’s continue on by examining the remaining rules:
- This rule establishes that Z repairs more types of machines than Y. Because we have already established that Y repairs exactly two types of machines, we can now infer that Z repairs three types of machines, and since there are only three types of machines in total, Z must repair R, T, and V:
- This rule is similar to the third rule, but in this instance W and S do not repair any of the same types of machines. We can thus infer that W does not repair R, and that W repairs either one or two machines (either T or V or both):
- The final rule establishes that U repairs exactly two types of machines. And, due to the constraints created by the first rule, we can determine that U must repair R, and that the second machine U repairs is then T or V. The only remaining point of indeterminacy in the game is X. We know X must repair R, but it is also possible (but not necessary) that X repairs one or two additional machines. This uncertainty is identified with a 1/2/3 notation above X: