- Thu Apr 21, 2016 7:50 pm
Thanks for your question. Let me see if I understand what you're saying: You believe that T can be assigned to table A, contrary to what we infer. This, according to you, is because T can be assigned to wait on more than one of the tables given the first rule in the game, so even if S cannot be assigned to table A, T can be - as long as S and T are both assigned together on another table in accordance with the last rule.
Your conclusion is based on a mistaken reading of the last rule. The rule states, "S and T are both assigned to the same table." That does not mean that they are assigned on at least one table as each other. The rule amounts to an absolute mandate to assign S and T to the same table as each other, removing any possibility that S can be assigned to a table without T (or vice versa). So, if S and T are always assigned to the same table as each other, and S cannot be assigned to table A, it follows that T cannot be assigned to table A either.
Let me know if this clears things up!
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