NetNews Usenet Archive 1993 #3/NN_1993_3.iso/spool/comp/ai/fuzzy/139

Text File | 1993-01-22 | 3.1 KB | 71 lines

Newsgroups: comp.ai.fuzzy
Path: sparky!uunet!math.fu-berlin.de!informatik.tu-muenchen.de!lan!hafner
From: [e-mail] (Walter 'madhouse' Hafner)
Subject: Re: What does "AND" mean?
In-Reply-To: [e-mail]'s message of Thu, 21 Jan 1993 22:54:23 GMT
References: <[e-mail]>
Sender: [e-mail] (USENET Newssystem)
Organization: Inst. fuer Informatik, TU Muenchen, Germany
Date: Fri, 22 Jan 1993 14:57:14 GMT
Message-ID: <[e-mail]>
Lines: 58

Disclaimer: Sorry for my english - But i think not all of you speak german :-)

In article <[e-mail]> [e-mail] (Glen C. Perkins) writes:

   Ultimately, it appears that the only way to model this deceptively
   simple and I think typical psychological rule: "the prettier and the
   nicer, the more interesting" would be for me to sit down and build a
   truth table. I'd probably need to make one axis PRETTY and one axis
   NICE in, say, increments of 0.1, and then ask myself how I'd feel
   about each combination. Then I'd try to figure out what the function,
   INTEREST(PRETTY, NICE) really was.

   This sort of seems like I'm missing the benefits of fuzzy logic's
   "ease of development." If I have to design a fancy algorithm for the
   AND in my one rule, why not just throw away the rule and just use
   the algorithm? I know fuzzy logic isn't the answer to every question,
   but it does seem as though it ought to be ideal for encoding
   something like my little rule.

   Am I missing something? (My marbles, perhaps?) I'd be interested in
   hearing a bit of a discussion if anyone is interested, so don't e-mail
   me personally (unless you want to introduce me to a (1.0, 1.0). I
   suppose that would be the '90s equivalent of a "10"!)

Well. There are the so called 'compensatoric operators' for these kind of
problems. The best known is the 'Lambda-Operator':

mC(x) := (mA(x) * mB(x)) ^ (1 - l) * (1 - (1 - mA(x)) * (1 - mB(x))) ^ l

By choosing a value for 'l' (Lambda), you can get any operator from 'and'
(l = 0) to 'or' (l = 1). Likewise you can do:

mC(x) := min(mA(x), mB(x)) ^ (1 - l) * max(mA(x), mB(x)) ^ l

A short table (min, max formula):

    PRETTY    NICE    'and' (min)    'or' (max)    l = 0.2        l = 0.7

1)    0.0    0.5    0.0        0.5        0.0        0.0
2)    0.3    0.5    0.3        0.5        0.33        0.43
3)    0.3    0.8    0.3        0.8        0.37        0.60
4)    0.3    1.0    0.3        1.0        0.38        0.70

There a problem, Glen did not mention can be seen: In case 2 and 3 the 'and'
value both is 0.3, but in real life ( :-) ) certainly girl 3 would be chosen
... compensatoric operators are capable to handle this.

The main disadvantage is the complexity of these operators.

Sorry, if i'm a bit short, but currently I don't have much time. For better
understanding look in the 'Zimmermann'-book.

-Walter
--
Walter Hafner,  Fichtenweg 10,  8901 Dinkelscherben,  Germany,  Tel: [phone]
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