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CS 621 Artificial Intelligence Lecture 8 - 19/08/05 Prof. Pushpak Bhattacharyya. Fuzzy Logic Application. NLP -> Fuzzy Logic. Language statements imprecise Linguistic variable -> Adjective Hedges -> Adverb. Example Problem. - PowerPoint PPT Presentation
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19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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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 Bombay
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NLP -> Fuzzy Logic
Language statements impreciseLinguistic variable -> AdjectiveHedges -> Adverb
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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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 Bombay
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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 truec) X is more of less old or highly educated is fairly true
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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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 Bombay
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Basic Profile
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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To Adjust For Different Linguistic Variables
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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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 Bombay
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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
Finding α
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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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 Bombay
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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 Bombay
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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 Bombay
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Solving for k1 & k2 (Contd 3)
k2 = x2 /2
k1 = 9.2 / x2
k2 = x2 / 2
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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Profile of Old
x1 = 0, y1 = 0.01x2 = 80 yrsy2 = 0.99k1 = 9.2 / 80 = 0.1k2 = 40
Profile of old y = 1/(1+e - 0.1(x - 40))
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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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 Bombay
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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 Bombay
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Profile of Educatedschool = 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 Bombay
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Old
Tall
Heavy
Educated
1/(1+e-k1(x-k2))
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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Truey = 1/(1+e-k
1(x-k
2))
k1 and k2 values are chosen arbitrarily
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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a) X is highly educated and not very young is very true
l = Level of educationμ 2
true (μ)
μ = min [μ2educated (l), 1- ( 1- μold(age) ) ) 2
μ for Different Configurations - 1
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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μ for Different Configurations - 2b) X is very young, tall, not heavy and somewhat
educated is trueμ 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 Bombay
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μ for Different Configurations - 3c) 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 Bombay
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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 - μ2
edu (l))
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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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 Bombay
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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 Bombay
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19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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Fuzzy InferencingClosely related to Fuzzy Expert Systems
Expert Systems:Rules :
Antecedent Consequentp q
p1 p2 p3 …. Pn qi
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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Inferencing
Forward Chaining Backward Chaining
Supposed to prove the fact F
Inferencing
19-08-05 Prof. Pushpak Bhattacharyya, IIT Bombay
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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 Bombay
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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 Bombay
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Fuzzy Expert System
Rules are in forms of linguistic variables.
Example of “Inverted pendulum control”