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- Date: Tue, 30 Oct 2007 15:08:43 +0000
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http://www.w3.org/Bugs/Public/show_bug.cgi?id=5183 ------- Comment #6 from mike@saxonica.com 2007-10-30 15:08 ------- >So we have these possible partitions into groups: 1. {Fl}, {De}, {Do} 2. {Fl,De}, {Do} 3. {Fl}, {De, Do} I don't think solution (1) satisfies rule 2: De and Do don't contain a pair of values that are unequal. So I think (2) and (3) are the only possible partitionings. I tried to devise the rules on the basis that it would always be possible to select one value from each group in the second step; I can't easily prove that I have succeeded. Mike
Received on Tuesday, 30 October 2007 15:08:50 UTC