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Standard notation for plays

Posted By: Timothy Chow
Date: Friday, 30 May 2014, at 3:29 p.m.

In Response To: Standard notation for plays (Tom Keith)

Tom Keith wrote:

Suggested clarification: "There should be exactly one asterisk for each blot sent to the bar. If the asterisk could appear in more than one place, place it as early in the play as possible."

6/5, 6/5*/3 → 6/5*, 6/3

My suggestion would be: "When combining sequences, start with the move that has the highest point number." So in your example, start with 7/6. Combine that with 6/5, giving 7/5. 7/5 still has the highest point number, so combine that with 5/4.

7/6, 6/5, 6/5, 5/4 → 7/4, 6/5

There is a further potential ambiguity here. Suppose I have

7/6, 6/5*, 6/5, 5/4.

Following your first rule, I have created only one asterisk. Now when I start merging, the ambiguity is whether I should merge 7/6 with 6/5* or with 6/5. If we merge 7/6 with 6/5* then we get 7/5*/4 6/5. If we merge 7/6 with 6/5 then we get 7/4 6/5*. Now, 7/4 6/5* definitely looks weird, and I believe it contradicts what you intended to say about hits appearing as early as possible. So I'm assuming 7/5*/4 6/5 is what you want. However, how are you going to express this rule? You could say that in serial mergers, asterisked expressions take precedence. But I think that rule would give the wrong answer with

6/5* 6/5 5/4 4/3

since you would get 6/5*/3 6/5 which you have explicitly said should be eschewed in favor of 6/5* 6/3.

Come to think of it, I don't really understand your example

6/5, 6/5*/3 → 6/5*, 6/3

If you get as far as 6/5 6/5*/3 then what rule are you using to get to 6/5* 6/3? It doesn't look like a serial merger is possible. A parallel merger might be possible, but you surely don't want 6/5*(2)/3 so you'd have to break it up into 6/5*(2) 5/3, but your parallel merger rule doesn't give instructions for this, not to mention the fact that we're supposed to be getting 6/5* 6/3 in the end anyway.

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