from Mata 2005...
If both descriptors, R and S, are used as first descriptors (see Fig. 5),
> they should be independently and sequen- tially used to form pairs of
> descriptors. Then all pairs sit- uated at the same rank level are compared.
> The first difference encountered is used to rank ligands. This means that
> in the examples in this figure, pairs 1,2 and 2,1 (the order is not
> important) should be simultaneously compared in both ligands and the number
> of l and u pairs evaluated
[image: Inline images 1]
On 19 April 2017 at 20:37, John Mayfield <john.wilkinson...@gmail.com>
wrote:
>
> On 19 April 2017 at 18:00, Robert Hanson <hans...@stolaf.edu> wrote:
>
>> I don't know what ChemSketch is doing here, but it is missing two trivial
>> centers in Sphere I. Although you have not indicated them, they have to be
>> something simple, at least for Jmol! But let's say that is still possible
>> to not have them defined.
>
>
> That's CDK not ChemSketch (although ChemSketch labels the same as I said),
> I deliberately contrived the example and missed those off on purpose to
> simplify it - replace with a N if you like. In practice such situations a
> very rare.
>
> It's been a long time but IIRC the power set/perminations is when
> everything is tired for something like this. Maybe they've replaced this
> now:
>
> S - R OH R - S
> \ | /
> R - S - N - CH - N - S - R
> / \
> S - R S - R
>
> the branches equal so the middle is unspecified. To prove it you need
> generate all possible sequences of like/unlike descriptors. Even if they're
> not equal... I can't remember exactly if the right here the reference is
> now always S but even so we still need to generate all the way they can be
> visited on the left and the right now has two ways:
>
> S - R OH R - S
> \ | /
> R - S - N - CH - N - S - S
> / \
> S - R S - R
>
> right (which has priority) has two combinations:
> SRSSSR -> lulllu
> SSRRSS -> lluull (best)
>
> Rule 5 is the place - the only place -- that assigns "r" and "s". Right?
>
>
> Yes
>
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