Transcript PIB Pertemuan 12

Resolution & Refutation (review)
Short Quiz
Prolog Introduction & Exercise
   ,   
 
   ,   
  
(KB P  false)  KB  P
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Translate facts and rules into predicate logic
Tranform sentence into CNF (conjunctive
normal form) / horn clause / well formed
formula
p1  ... p j  .... pn1  r1  ...rn2
s1  .... s n3  q1  ... qk  .... qn 4
SUBST( , p1  ... p j 1  p j 1  .... pm  s1  ...  sn3 
r1  ...rn 2  q1  ... qk 1  qk 1  .... qn )
1.
2.
3.
4.
5.
6.
Eliminate implications & bi-conditionals
Move  inwards
Standardize variables: each quantifier has
different ones
Skolemize: choose a “fact” to eliminate 
Eliminate 
Distribute ^ over v
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Buktikan bahwa “Marcus membenci Caesar”
Facts
1.
2.
3.
4.
Marcus adalah seorang manusia
Marcus orang Pompei
Marcus mencoba membunuh Caesar
Caesar adalah seorang penguasa
1.
2.
Semua orang Pompei adalah orang Romawi
Semua orang Romawi setia pada Caesar atau membenci
Caesar
Setiap orang setia pada minimal 1 orang
Orang hanya mencoba membunuh penguasa yang
kepadanya mereka tidak setia
Rules
3.
4.
Facts:
1. man(marcus)
2. pompeian(marcus)
3. tryassassinate(marcus,caesar)
4. ruler(caesar)
Rules:
1. x pompeian(x)  roman(x)
2. x roman(x)  loyalto(x, caesar) v hate(x,caesar)
3. x y loyalto(x, y)
4. x y man(x) ^ ruler(y) ^ tryassassinate(x,y) 
loyalto(x,y)
Prove that: hate(marcus, caesar), use refutation.
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Make CNF from rules
pompeian(x) v roman(x)
roman(x) v loyalto(x,caesar) v hate(x,caesar)
loyalto(x,y)
(man(x) ^ ruler(y) ^ tryassassinate(x,y)) v
loyalto(x,y)
man(x) v ruler(y) v tryassassinate(x,y) v
loyalto(x,y)
Apply substitutions into variable, and try to change
rules into facts.
1.
2.
3.
4.
PROLOG = “Programmation en logique” (Marseille, 1972)
Declarative programming language with procedural elements
Used for problems of AI / knowledge-based (expert) systems
Motivation:
reconcile use of logic as declarative knowledge representation with
procedural representation of knowledge
Strengths:
Logical descriptions of problems, instead of HOW to solve them  let
computer work out the solution
Well-suited for problems involving search
Simple programs can be understood by reading the code
Limitations
Flow of control / procedural semantics
Prolog-program = collection of clauses = meta programming
Facts
Rules
Goals (queries), shaped liked facts
Facts describe properties of objects and relations between objects; comparable to
tables in a relational database
student
Name
interest
Name
Hans
Tina
Lars
Frida
Prolog notation:
Prolog notation:
student(hans).
interest(hans,math).
student(tina).
interest(tina,datalogi).
student(lars).
interest(lars,physics).
student(frida).
interest(frida,math).
Hans
Tina
Lars
Frida
Subject
Math
Datalogi
Physics
Math
Prolog notation (facts):
<predicate_name>(arg1, arg2…).
Simple IF-THEN statements
“Every reasonable student is interested in math.”
interest(X,math) :- student(X).
head
body
All specified conditions in the body (all clauses) must be true to
make the predicate in the head true.
Conjunctions (AND connected):
mother(X,Person) :- parent(X,Person),sex(X,female).
Disjunctions (OR connected):
interest(X,prolog) :- interest(X,artificial_intelligence).
interest(X,prolog) :- interest(X,logic).
Goals are queries
One ore more subgoals
?- student(thomas).
=> no
Pattern matching: a fact matches a goal if
Same predicate
Same corresponding arguments.
Goals can contain variables:
?- student(X).
=> X = hans ;
=> X = tina ;
=> X = lars ;
=> X = frida ;
=> no.
Variables are instantiated( included in rules),but cannot be declared!
Leftmost-depth-first search for solutions
Matching: either two terms are identical, or they become identical by variable
substitution (resolution based on pred.logic)
Processing of subgoals from left to right
Backtracking (from goal try to substitute variables into facts or rules)
1:
2:
3:
4:
5:
?- p(Z).
1
Z=a
s(Z)
s(a).
s(b).
q(a).
p(X) :- s(X).
p(Y) :- q(Y).
p(Z) 5
4
2
Z=b
q(Z)
3
Z=a
The Prolog interpreter uses backward chaining:
starting from a goal (theorem)
prove the goal by searching for rules whose ”head” (action part)
matches the goal
Given are the following rules:
1 E  AC
X
2 H  F C
3 H BE
4 CB
5 X H
F&C
B&E
facts
AND
AND
prove
A, B
H
OR
F
C
B
E
X
AND
B
A
C
B
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http://www.swi-prolog.org (ver. 5.6.62)
Graphical User Interface (XPCE)
Interface with Java, C++
Steps:
 Consult / Compile
 Goal search / Execute
 Debug > Graphical Debugger
 Trace: show you how the unification &
substitution occurs
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Use prolog to prove the “Marcus’s Case”
man(marcus).
pompeian(marcus).
ruler(caesar).
tryassassinate(marcus, caesar).
roman(X) :- pompeian(X).
hate(X, Y) :- man(X), ruler(Y),
tryassassinate(X, Y).
loyalto(X, Y) :- man(X), ruler(Y).
George
x Mum
Spencer
x Kydd
Elizabeth
Diana
x Charles
Anne
William
Harry
Peter
x Philip
x Mark
Zara
Margaret
Andrew
Beatrice
x Sarah
Eugenie
Edward
 Buat fakta berdasarkan pohon keluarga di atas
 Buat aturan untuk aturan:
 saudaraLaki(X,Y)  X berjenis kelamin laki-laki, X dan Y adalah anak dari
seseorang, X dan Y bukan orang yang sama
 saudaraPerempuan(X,Y)  X berjenis kelamin perempuan, X dan Y adalah
anak dari seseorang, X dan Y bukan orang yang sama
 bersaudaraKandung(X,Y)  X dan Y memiliki ayah dan ibu yang sama, X dan
Y bukan orang yang sama
 kakek(X,Y,Z)  X adalah ayah dari Y, Y adalah ayah dari Z
 nenek(X,Y,Z)  X adalah ibu dari Y, Y adalah ibu dari Z
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Simple data structure to process non-numeric
relation
A list is a set of ordered sequential elements
[2,2,1,1,4,4,5,6]
[b,u,d,i]
[budi, iwan, wati]
How to use a list?
 Simple case: empty []
 Ordered element: [head|tail]
▪ [2,2,1,1,4,4,5,6]  [ 2 | [2,1,1,4,4,5,6] ]
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[a,b,c] unifies with [Head|Tail] resulting in
Head=a and Tail=[b,c]
[a] unifies with [H|T] resulting in H=a and T=[]
[a,b,c] unifies with [a|T] resulting in T=[b,c]
[a,b,c] doesn't unify with [b|T]
[] doesn't unify with [H|T]
[] unifies with []. Two empty lists always can
be unified.
Predicate: member(X,List).
X is a member of List, in case of:
1. X is a head, or
2. X is a member of the tail
member(X,[X|Tail]).
member(X,[Head|Tail]):member(X,Tail).
reverse([],X,X).
reverse([X|Y],Z,W) :- reverse(Y,[X|Z],W).
reverse([1,2,3],[],A)
reverse([2,3],[1],A)
reverse([3],[2,1],A)
reverse([],[3,2,1],A)
true A = [3,2,1]
1. [a,b,c,d]=[a,[b,c,d]].
2. [a,b,c,d]=[a|[b,c,d]].
3. [a,b,c,d]=[a,b,[c,d]].
4. [a,b,c,d]=[a,b|[c,d]].
5. [a,b,c,d]=[a,b,c,[d]].
6. [a,b,c,d]=[a,b,c|[d]].
7. [a,b,c,d]=[a,b,c,d,[]].
8. [a,b,c,d]=[a,b,c,d|[]].
9. []=_.
10. []=[_].
11. []=[_|[]].
What should Prolog give as result?
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Develop a dictionary to translate number
1..10 from Indonesian to English and vice
versa, example:
 Translate([satu, sembilan],Y). Y = [one,nine]
 Translate(Z,[one]).  Z = [satu]
 Translate([],[]).