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can Mathematica be useful for this?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg56545] can Mathematica be useful for this?
*From*: Pedrito <pedrito6 at softhome.net>
*Date*: Thu, 28 Apr 2005 02:40:47 -0400 (EDT)
*References*: <718e40e89ee7126bf169d5585b776d5c@akikoz.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Thanks a lot Andrzej. Your idea of using Mathematica for the logic
algebra simplifications is great.
I have been trying different systems of rules for evaluation and finally
I found that Mathematica is able to obtain numeric results just with
these eight rules:
p[True] = 1; p[False] = 0;
p[A_] := p[LogicalExpand[A]] /; ! A === LogicalExpand[A];
p[! A_] := 1 - p[A];
p[! A_ && B_] := p[B] - p[A && B];
p[A_ && ! B_] := p[A] - p[A && B];
p[A_ || B_] := p[A] + p[B] - p[A && B];
p[A_, B_] := p[A && B]/p[B];
Now, it's possible to solve the previous probability problem this way:
initialData = {p[ S] == 1/3, p[F] == 26/100, p[G] == 2/10, p[F && S] ==
15/100, p[F && G] == 5/100, p[G && S] == 1/10,
p[F && G && S] == 2/100};
FullSimplify[p[(F && !G && !S) || ( !F && G && !S) || ( !F && !G && S)], initialData]
-->19/75
FullSimplify[p[F || G || S], initialData]
-->77/150
I have found a case where FullSimplify doesn't evaluate numerically:
FullSimplify[p[F], p[F] == 1/10]
-->p[F]
So I have to write:
FullSimplify[p[F]+0., p[F] == 1/10]
0.1
Is there another way for obtaining a numeric result with FullSimplify?
Thanks in advance!
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