ARTIFICIAL INTELLIGENCE

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					           CS 621 Artificial Intelligence

               Lecture 8 - 19/08/05

           Prof. Pushpak Bhattacharyya
             Fuzzy Logic Application

19-08-05           Prof. Pushpak Bhattacharyya, IIT   1
                              Bombay
           NLP -> Fuzzy Logic
Language statements imprecise
Linguistic variable -> Adjective
Hedges -> Adverb




19-08-05           Prof. Pushpak Bhattacharyya, IIT   2
                              Bombay
              Example Problem
   Assume 4 types of Fuzzy Predicates applicable to
   persons (age, height, weight and level of education).
   The membership functions are of the basic form
   1/(1+e-x), but of appropriate shape and orientation.




19-08-05           Prof. Pushpak Bhattacharyya, IIT        3
                              Bombay
           Example (Contd 1)
Determine the truth values of :
  a) A person X is highly educated and not very young
  is very true.
  b) X is very young, tall, not heavy and somewhat
  educated is true
  c) X is more of less old or highly educated is fairly
  true




19-08-05           Prof. Pushpak Bhattacharyya, IIT       4
                              Bombay
            Example (Contd 2)
   d) X is very heavy or old or not highly educated is
   fairly true

   e) X is short, not very young and highly educated is
   very true

   ( assume that the level of education has 4 values:
   Elementary school, High school, College, PHD)


19-08-05            Prof. Pushpak Bhattacharyya, IIT      5
                               Bombay
           Basic Profile




19-08-05   Prof. Pushpak Bhattacharyya, IIT   6
                      Bombay
To Adjust For Different Linguistic
           Variables




19-08-05   Prof. Pushpak Bhattacharyya, IIT   7
                      Bombay
           To Find k1 and k2 we need
              < x1, y1 > , <x2, y2>




             x1 = 0 y1 = 0.01 x2 = ? y2 = 0.99
19-08-05           Prof. Pushpak Bhattacharyya, IIT   8
                              Bombay
                        Finding α
y = 1/(1+e-k1(x-k2))

   => (1-y)/y = e-k1(x-k2)

   => k1(x - k2) = ln(y/1-y)

   call y/1-y = α




19-08-05               Prof. Pushpak Bhattacharyya, IIT   9
                                  Bombay
              Solving for k1 & k2
k1(x-k2) = ln α
So, k1 x – k1 k2 = ln α

   at < x1, y1 >

   k1 x1 – k1 k2 = ln α1      --- (1)
   where α1 = y1/1-y1




19-08-05           Prof. Pushpak Bhattacharyya, IIT   10
                              Bombay
      Solving for k1 & k2 (Contd 1)
k1 x2 - k1 k2 = ln α2               ---(2)
α2 = y2 / 1- y2
Solving from k1 and k2
k1 = (1/(x2- x1) ) ln ( ((1-y1)/y1)/((1-y2)/y2))




19-08-05             Prof. Pushpak Bhattacharyya, IIT   11
                                Bombay
      Solving for k1 & k2 (Contd 2)
k2 = x2 – (ln α2) / k1
Lets use x1= 0, y1= 0.01, x2=?, y2= 0.99

k1         = 2/x2 ln 99
           = 2* 4.6 / x2
           = 9.2 /x2




19-08-05                   Prof. Pushpak Bhattacharyya, IIT   12
                                      Bombay
      Solving for k1 & k2 (Contd 3)
k2 = x2 /2

k1 = 9.2 / x2
k2 = x 2 / 2




19-08-05        Prof. Pushpak Bhattacharyya, IIT   13
                           Bombay
                    Profile of Old
x1 = 0, y1 = 0.01
x2 = 80 yrs
y2 = 0.99
k1 = 9.2 / 80 = 0.1
k2 = 40

Profile of old y = 1/(1+e - 0.1(x - 40))



19-08-05              Prof. Pushpak Bhattacharyya, IIT   14
                                 Bombay
                   Profile of Tall
x1 = 0, y1 = 0.01
x2 =6ft, y2 = 0.99

k1 = 9.2 / 6 = 1.5
k2 = 6/2 = 3

Profile of Tall, y = 1/(1+e -1.5(x - 3))


19-08-05              Prof. Pushpak Bhattacharyya, IIT   15
                                 Bombay
                Profile of Heavy
x1 = 0, y1 = 0.01
x2 = 100kg yrs, y2 = 0.99

k1 = 9.2 / 100 = 0.1
k2 = 100/2 = 50

Profile of Heavy, y = 1/(1+e - 0.1(x - 50))


19-08-05             Prof. Pushpak Bhattacharyya, IIT   16
                                Bombay
                Profile of Educated
school = 0.25,
high School = 0.5,
college 0.75,
PhD = 1.00

x1 = 0, y1 = 0.01, x2 =1 yrs, y2 = 0.99

Profile:
y = 1/(1+e - 9.2 (x – 0.5 ))

19-08-05                Prof. Pushpak Bhattacharyya, IIT   17
                                   Bombay
      Old                                        1/(1+e-k1(x-k2))




      Tall



      Heavy



Educated


19-08-05      Prof. Pushpak Bhattacharyya, IIT                  18
                         Bombay
                                True
y = 1/(1+e-k1(x-k2))

k1 and k2 values are chosen arbitrarily




19-08-05               Prof. Pushpak Bhattacharyya, IIT   19
                                  Bombay
μ for Different Configurations - 1
   a) X is highly educated and not very young is very
      true
   l = Level of education
   μ 2true (μ)
                                                  2
   μ = min [μ educated (l), 1- ( 1- μold(age) ) )
               2




19-08-05           Prof. Pushpak Bhattacharyya, IIT     20
                              Bombay
    μ for Different Configurations - 2
  b) X is very young, tall, not heavy and somewhat
     educated is true
  μ true (μ)

  μ = min ( ( 1 - μold(age))2, μtall(ht), 1- μ heavy(wt),
    (μedu(L))1/2)




19-08-05              Prof. Pushpak Bhattacharyya, IIT      21
                                 Bombay
    μ for Different Configurations - 3
  c) X is more or less old or highly educated is fairly
     true

  (μtrue(μ ))1.5

  μ = max ( (μold(age))1/2, μ2edu (l))




19-08-05            Prof. Pushpak Bhattacharyya, IIT      22
                               Bombay
     μ for Different Configurations - 4
 d) X is very heavy or old or not highly educated is fairly
    true

 (μ true(μ) )1.5

 μ = max (μold(age), μ2heavy(wt ), 1 - μ2edu (l))



19-08-05            Prof. Pushpak Bhattacharyya, IIT    23
                               Bombay
     μ for Different Configurations - 5

  e) X is short, not very young and highly educated is
     very true
  μ2true(μ)

  μ = min [1 – (1 - μ old(age))2, μ2edu (l), 1 - μtall(ht) ]




19-08-05              Prof. Pushpak Bhattacharyya, IIT         24
                                 Bombay
    Question : How to actually read off
                 values
   John:
            age: 35

            ht : 5.8’

            wt : 75 Kg

            education l : College

19-08-05        Prof. Pushpak Bhattacharyya, IIT   25
                           Bombay
19-08-05   Prof. Pushpak Bhattacharyya, IIT   26
                      Bombay
                 Fuzzy Inferencing
Closely related to Fuzzy Expert Systems

Expert Systems:
                         Rules :

           Antecedent               Consequent
                p                        q

p1 p2 p3 …. Pn                              qi
19-08-05                Prof. Pushpak Bhattacharyya, IIT   27
                                   Bombay
                Inferencing
                Inferencing

  Forward Chaining                   Backward Chaining

  Supposed to prove the fact F




19-08-05        Prof. Pushpak Bhattacharyya, IIT         28
                           Bombay
            Forward Chaining
              ( Data Driven)

Given Facts are matched with LHS of rules

RHS of satisfied rules are added to the fact base

Stop when the required F comes to the fact base.


19-08-05         Prof. Pushpak Bhattacharyya, IIT   29
                            Bombay
           Backward Chaining
             ( Goal Driven)

Start from F to see if it matches the RHS of any
  rule.

LHS of matched Rule becomes the new goal.

Stop when a fact is hit.

19-08-05         Prof. Pushpak Bhattacharyya, IIT   30
                            Bombay
           Fuzzy Expert System


Rules are in forms of linguistic variables.

Example of “Inverted pendulum control”




19-08-05        Prof. Pushpak Bhattacharyya, IIT   31
                           Bombay

				
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