Matakuliah
Tahun
Versi
: H0383/Sistem Berbasis Pengetahuan
: 2005
: 1/0
Pertemuan 4
Bahasa Pemrograman Logika
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mendemonstrasikan bahasa
pemrograman untuk implementasi
predicate logic
2
Outline Materi
• LISP
• PROLOG
3
Bahasa LISP
• LISP menggunakan notasi prefix
– (+ 5 6 9)
–5+6+9
– Contoh: konversi 50 Celcius Fahrenheit:
– (+(*(/ 9 5) 50) 32)
• List dalam LISP
– (a b (c d) e f) adalah list
– (c d) adalah sublist
– c, d adalah top element dari sublist
4
Bahasa LISP
• List kosong = NIL
• Variable assignment:
– (setq x 10) artinya x = 10
– (setq x (+ 3 5)) == x=3+5
•
Basic list manipulation function
– (car’(a b c)) = a
– (cdr’(a b c)) = (b c)
– (cons’a’(b c))
= (a b c)
– (list ’a ’(bc)) = (a (b c))
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Bahasa LISP
• Function call:
– (function-name arg1 arg2 …)
• Additional manipulation list
– (append ’(a) ’(b c)) = (a b c)
– (last ’(a b c d)) = (d)
– (member ’b ’(a b d) = (b d)
– (reserve ’(a (b c) d)) = (d (b c) a)
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Bahasa LISP
• Defining function
– (defun name(parm1, parm2 …) body)
– Defun averagethree (n1 n2 n3) (/ (+ n1 n2 n3)
3))
– (averagethree 10 20 30) = 20
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Bahasa LISP
• Predicate call:
– (atom ‘aabb) = true; aabb is a valid atom
– (equal ‘a (car ‘(a b)) = true; a=a
– (greaterp 2 4 27)= true; (lessp 5 3 1 2)= nil
– (zerop .000001) = nil;
– (evenp 3) = nil; (oddp 3) = true;
– (number 10) = true;
– (listp ‘(a)) = true; a is a valid list
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Bahasa Lisp
• Conditional:
– (cond (<test1> <action1>)
(<test2> <action2>)
:
(<testk> <actionk>)
(defun maximum2 (a b)
(cod ((>a b) a)
(t b)))
maximum2(234 320) = 320
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Bahasa LISP
• Logical function: or, and, not, t(true), null
• Input: (+5 (read))
• If 6 is inserted by keyboard then result is
11
• Output: print’(a b c) ; print a list
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Bahasa LISP
• Iteration
– (do (<var1 val1><var-update1)
:
(<test> <return-value>)
(s-expression)
• Contoh : factorial:
(defun factorial (n)
(do ((count n (- count 1))
(product n (* product (-count 1)
((equal 0 count) product)))
11
Bahasa LISP
• Recursive:
• (defun factorial (n)
(cond (( zerop n) 1)
(t (* n (factorial (- n 1)))))
12
Bahasa LISP
• Property lists
– (putprop object value attribute)
– (putprop ’car ’ford ’make)
– (putprop ’car ’1988 ’year)
– (putprop ’car ’red ’color)
– (putprop ’car ’four-door ’style)
– (get ‘car ‘make) = Ford
– (get ‘car ‘color) = red
13
Bahasa LISP
• Arrays
– (setf myarray(make-array ’(10)))
– (setf (aref myarray 0) 25
– (setf (aref myarray 1) ’red
– (aref myarray 0) = 25
– (aref myarray 1) = red
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Bahasa LISP
• Mapping function:
– (mapcar ’1 +’(5 10 15 20 25)) = (6 11 16 21
26)
– (mapcar ’+ ’(1 2 3 4 5 6) ’(1 2 3 4)) = (2 4 6 8)
• Lambda function:
– (defun cubic(lists)
(mapcar #’(lambda (x) (*x x x)) lists))
– (cubic(1 2 3 4)) = (1 8 27 64)
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Bahasa PROLOG
• PROLOG: PROgramming in Logic
– sister(sue,bill)
– parent(ann,sam), parent(joe,ann)
– male(joe), female(ann)
If X is the parent of Y, and Y is the parent of
Z, and X is a male, then X is grandfather
of Z
grandfather(X,Z) :- parent(X,Y),
parent(Y,Z),male(X)
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Bahasa PROLOG
•
•
•
•
•
•
•
Query:
?- parent(X,sam)
X=ann
?- female(joe)
No
?-male(joe)
yes
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Bahasa PROLOG
•
•
•
•
•
•
List in PROLOG:
[tom,sue,joe,marry,bill]
?- [Head|Tail] = [tom,sue,joe,marry]
Head = tom, Tail=[sue,joe,marry]
List manipulation:
append, member, conc, add, delete
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Bahasa PROLOG
• member(X,[X|Tail])
• member(X,[Head|Tail]):-member(X,Tail)
• X is a member of the list L if X is the head
of L.
• X is a member of L if X is a member of the
tail of L
• ?- member(c,[a,b,c,d])yes
• ?- member(b,[a,[b,c],d])No
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Penutup
• Bahasa-bahasa pemrograman logika
dapat digunakan untuk implementasi
problem berbasis pengetahuan.
• Selain LISP dan PROLOG, terdapat
berbagai jenis lainnya misal CLIPS.
• Meski beragam, namun semantiknya
serupa.
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