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Hi Jan,
On 30 January 2001 at 15:55:01 -0500 (which was 20:55 where I live)
Jan Rifkinson wrote and made these points:
Marck>> If you have 3 filters on the first page (A, B, C) and 2 sets on the
Marck>> alternatives pages (D, E) and (F) then the logic is:
Marck>> (A AND B AND C) OR (D AND E) OR (F).
Marck>> Is that any clearer?
JR> Now I'm confused. I thought it would have been:
JR> (A & B & C) or (D or E or F)
JR> What happens if you 5 alternatives or more as I do on one filter?
Let me try and show you the same set first: On the first page is this
_________________________________
|A_________________|??_____|Yes___|
|B_________________|??_____|Yes___|
|C_________________|??_____|Yes___|
and then, on the Alternatives, we have two sets
_________________________________
|D_________________|??_____|Yes___|
|E_________________|??_____|Yes___|
_________________________________
|F_________________|??_____|Yes___|
That makes (A AND B AND C) OR (D AND E) OR (F).
To get (A & B & C) or (D or E or F) on the first page is this
_________________________________
|A_________________|??_____|Yes___|
|B_________________|??_____|Yes___|
|C_________________|??_____|Yes___|
and then, on the Alternatives, we have *three* sets
_________________________________
|D_________________|??_____|Yes___|
_________________________________
|E_________________|??_____|Yes___|
_________________________________
|F_________________|??_____|Yes___|
JR> What happens if you 5 alternatives or more as I do on one filter?
If they're in a block they're all ANDed. If they're all separate,
they're all ORed.
- --
Cheers -- .\\arck D. Pearlstone -- Moderator TBUDL / TBBETA / TBTECH
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[ Any opinions are my own and not those of RIT labs ]
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