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#90735
Setup and Rule Diagram Explanation

This is a Basic Linear game.

This setup is still in progress. Please post any questions below!
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 srr021
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#91102
Does the first rule imply that O cannot be last?
 Rachael Wilkenfeld
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#91586
Hi srr,

Absolutely correct. The rule says that whenever it's an O performance, the next day must be a J performance. That means there has to be a next day to perform. So O cannot be last.

Great work!
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 Sammi00
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#92005
Hi,

Could you please do a setup for this problem? I was really stuck on some of the questions and forced to do trial and error, which took way too long. Could we possibly do a SS or a JJ in this situation? The stimulus did not disclose whether this was right or wrong.

My set up consisted of (horizontally)

1__(same).
2_
3_ (same).
4_
5_
6_ ( Cannot be O).

1:O
2: J
3:O
4:J
5:S
6:J
Unfortunately this is all I had, with two other scenarios that only yielded results for 2 questions.

Thank you!
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 ashpine17
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#92022
I had three boards, one with J in spots one and three and I did the same for O and S because we only had three variables. Does it not matter how many times each variable repeats so long as we get all three of them appearing once? Could an instructor confirm this?

If J is in spots one and three, then J/O/S could go in spots 2 and I left the other spots empty. If O is in spots 1 and 3, then S cannot go in spot 2 because it would violate the rule /SO and so only J can be in that spot because if O were there then spot three would have to be J. I put S in a bubble floating above spots 3-5 and left the others empty. If S were in spots 1 and 3, O cannot be in spot 2 and it can only be J or S, then I put an OJ block floating above spots 3-5.
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 ashpine17
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#92023
I also forgot to mention that when O are in spots 1 and 3, doesn't it mean both spots 2 and 4 are J's? That means there has to be either two S's or one S and one J in either spots 4 or 5.
 Adam Tyson
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#92108
As confirmed template junkie, I like ashpine's suggestion here of setting up three diagrams based around which of the three variables is in the 1/3 spots. And ashpone, your second comment is also correct: when O is in those two positions, we know for sure that J is in both 2 and 4, and we could not have another O in the game. The last two spaces would either both be S or else one S and one J (and Sammi, you're correct that there is no problem with either a JJ block or an SS block!)

We should also think about numeric distributions for this game, since there are three variables to fill six positions. The most extreme of those would be a 4-1-1 distribution, meaning one of the variables goes 4 times with the others each going just once. 3-2-1 is also possible, as is 2-2-2. In a game like this I expect at least one question to test those numeric ideas (and in fact, question 8 does just that).

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