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CS 621 Artificial Intelligence Lecture 8 - 19/08/05 Prof. Pushpak Bhattacharyya

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.

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CS 621 Artificial Intelligence Lecture 8 - 19/08/05 Prof. Pushpak Bhattacharyya

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  1. CS 621 Artificial IntelligenceLecture 8 - 19/08/05Prof. Pushpak Bhattacharyya Fuzzy Logic Application Prof. Pushpak Bhattacharyya, IIT Bombay

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

  3. 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. Prof. Pushpak Bhattacharyya, IIT Bombay

  4. 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 Prof. Pushpak Bhattacharyya, IIT Bombay

  5. Example (Contd 2) d) X is very heavy or old or not highly educated is fairly truee) 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) Prof. Pushpak Bhattacharyya, IIT Bombay

  6. Basic Profile Prof. Pushpak Bhattacharyya, IIT Bombay

  7. To Adjust For Different Linguistic Variables Prof. Pushpak Bhattacharyya, IIT Bombay

  8. To Find k1 and k2 we need < x1, y1 > , <x2, y2> x1 = 0 y1 = 0.01 x2 = ? y2 = 0.99 Prof. Pushpak Bhattacharyya, IIT Bombay

  9. 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 Prof. Pushpak Bhattacharyya, IIT Bombay

  10. Solving for k1 & k2 k1(x-k2) = ln α So, k1 x – k1 k2 = ln αat < x1, y1 >k1 x1 – k1 k2 = ln α1 --- (1) where α1 = y1/1-y1 Prof. Pushpak Bhattacharyya, IIT Bombay

  11. 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)) Prof. Pushpak Bhattacharyya, IIT Bombay

  12. 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 Prof. Pushpak Bhattacharyya, IIT Bombay

  13. Solving for k1 & k2 (Contd 3) k2 = x2 /2 k1 = 9.2 / x2 k2 = x2 / 2 Prof. Pushpak Bhattacharyya, IIT Bombay

  14. 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)) Prof. Pushpak Bhattacharyya, IIT Bombay

  15. 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)) Prof. Pushpak Bhattacharyya, IIT Bombay

  16. 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)) Prof. Pushpak Bhattacharyya, IIT Bombay

  17. 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 )) Prof. Pushpak Bhattacharyya, IIT Bombay

  18. Old 1/(1+e-k1(x-k2)) Tall Heavy Educated Prof. Pushpak Bhattacharyya, IIT Bombay

  19. True y = 1/(1+e-k1(x-k2)) k1 and k2 values are chosen arbitrarily Prof. Pushpak Bhattacharyya, IIT Bombay

  20. μ for Different Configurations - 1 a) X is highly educated and not very young is very true l = Level of education μ2true (μ) μ = min [μ2educated (l), 1- ( 1- μold(age)) ) 2 Prof. Pushpak Bhattacharyya, IIT Bombay

  21. 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- μheavy(wt), (μedu(L))1/2) Prof. Pushpak Bhattacharyya, IIT Bombay

  22. 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)) Prof. Pushpak Bhattacharyya, IIT Bombay

  23. μ for Different Configurations - 4 d) X is very heavy or old or not highly educated is fairly true (μtrue(μ) )1.5 μ = max (μold(age), μ2heavy(wt), 1 - μ2edu (l)) Prof. Pushpak Bhattacharyya, IIT Bombay

  24. μ for Different Configurations - 5 e) X is short, not very young and highly educated is very true μ2true(μ) μ = min [1 – (1 - μold(age))2, μ2edu (l), 1 - μtall(ht) ] Prof. Pushpak Bhattacharyya, IIT Bombay

  25. Question : How to actually read off values John: age: 35 ht : 5.8’ wt : 75 Kg education l : College Prof. Pushpak Bhattacharyya, IIT Bombay

  26. Prof. Pushpak Bhattacharyya, IIT Bombay

  27. Fuzzy Inferencing Closely related to Fuzzy Expert Systems Expert Systems: Rules : Antecedent Consequent p q p1 p2 p3 …. Pn qi Prof. Pushpak Bhattacharyya, IIT Bombay

  28. Inferencing Inferencing Forward Chaining Backward Chaining Supposed to prove the fact F Prof. Pushpak Bhattacharyya, IIT Bombay

  29. 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. Prof. Pushpak Bhattacharyya, IIT Bombay

  30. 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. Prof. Pushpak Bhattacharyya, IIT Bombay

  31. Fuzzy Expert System Rules are in forms of linguistic variables. Example of “Inverted pendulum control” Prof. Pushpak Bhattacharyya, IIT Bombay

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