can ban ve prolog
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Chương 3
Tri thức và lập luận
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Nội dung chính chương 3
I. Logic – ngôn ngữ của tư duy
II. Logic mệnh đề (cú pháp, ngữ nghĩa, sức mạnh biểu diễn, các thuật toán suy diễn)
III. Prolog (cú pháp, ngữ nghĩa, lập trình prolog, bài tập và thực hành)
IV. Logic cấp một (cú pháp, ngữ nghĩa, sức mạnh biểu diễn, các thuật toán suy diễn)
Lecture 1
Lecture 2
Lecture 3,4
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Lecture 2:Prolog, Lập trình prolog
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Lecture2-outline
I. Cơ bản về Prolog, lập trình prolog
II. Suy diễn trong Prolog (kiểm tra,Suy diễn lùi, đệ qui)
III. CUT và Phủ định
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I. Cơ bản về ngôn ngữ Prolog, lập trình prolog
27/09/04 AIPP Lecture 2: Prolog Fundamentals 5
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16:10 23/09/04 Lecture 1: An Introduction 6
Chương trình và thực hiện chương trình trong prolog
• Program is a database of facts and rules.– Some are always true (facts):
father( john, jim).– Some are dependent on others being true (rules):
parent( Person1, Person2 ) :- father( Person1, Person2 ).
• To run a program, we ask questions about the database.
parent(john,jim).parent(john,X).
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16:10 23/09/04 Lecture 1: An Introduction 7
Prolog in EnglishExample Database:
John is the father of Jim.Jane is the mother of Jim.
Jack is the father of John.
Person 1 is a parent of Person 2 if Person 1 is the father of Person 2 or Person 1 is the mother of Person 2.
Person 1 is a grandparent of Person 2 if some Person 3 is a parent of Person 2 and
Person 1 is a parent of Person 3.
Example questions:
Who is Jim's father?Is Jane the mother of Fred?
Is Jane the mother of Jim? Does Jack have a grandchild?
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16:10 23/09/04 Lecture 1: An Introduction 8
Prolog in PrologExample Database:
father( john, jim ).mother( jane, jim ).father( jack, john ).
parent( Person1, Person2 ) :- father( Person1, Person2 ).
parent( Person1, Person2 ) :- mother( Person1,
Person2 ).
grandparent( Person1, Person2 ) :- parent( Person3,
Person2 ), parent( Person1,
Person3 ).
Example questions:
?- father( Who, jim ).?- mother( jane, fred ).?- mother( jane, jim ).?- grandparent( jack, _ ).
Example Database:
John is the father of Jim.Jane is the mother of Jim.
Jack is the father of John.
Person 1 is a parent of Person 2 if Person 1 is the father of Person 2 or Person 1 is the mother of Person 2.
Person 1 is a grandparent of Person 2 if some Person 3 is a parent of Person 2 and
Person 1 is a parent of Person 3.
Example questions:
Who is Jim's father?Is Jane the mother of Fred?
Is Jane the mother of Jim? Does Jack have a grandchild?
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 9
Cú pháp• Prolog program consist of clauses.
• A clause has a head and a body (Rule) or just a head (Fact).
• A head consists of a predicate name and arguments.
• A clause body consists of a conjunction of terms.
• Terms can be constants, variables, or compound terms.
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 10
Clauses - Câu
• Prolog program consist of clauses.A clause = An individual definition (whether it be a
fact or rule).
e.g. mother(jane,alan). = Fact
parent(P1,P2):- mother(P1,P2). = Rule
• A clause consists of a head• and sometimes a body.
– Facts don’t have a body because they are always true.
head body
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 11
• A predicate denotes a property or relationship between objects
• ‘Predicate’ is the name given to the word occurring before the bracket in a fact or rule:
parent(jane,alan).
• By defining a predicate you are specifying which information needs to be known for the property denoted by the predicate to be true.
Predicate – Vị từ
Predicate name
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 12
Arguments – Các tham số của vị từ
• A predicate head consists of a predicate name and sometimes some arguments contained within brackets and separated by commas.
mother(jane,alan).
• A body can be made up of any number of subgoals (calls to other predicates) and terms.
• Arguments also consist of terms, which can be:– Constants e.g. jane,– Variables e.g. Person1, or– Compound terms (explained in later lectures).
Predicate name Arguments
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 13
Terms: ConstantsHạng thức hằng
Constants can either be:• Numbers:
– integers are the usual form (e.g. 1, 0, -1, etc), but
– floating-point numbers can also be used (e.g. 3.0E7)
• Symbolic (non-numeric) constants:– always start with a lower case alphabetic character and
contain any mixture of letters, digits, and underscores (but no spaces, punctuation, or an initial capital).
• e.g. abc, big_long_constant, x4_3t).
• String constants:– are anything between single quotes e.g. ‘Like this’.
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 14
Terms: VariablesHạng thức biến
• Variables always start with an upper case alphabetic character or an underscore.
• Other than the first character they can be made up of any mixture of letters, digits, and underscores.
e.g. X, ABC, _89two5, _very_long_variable
• There are no “types” for variables (or constants) – a variable can take any value.
• All Prolog variables have a “local” scope:– they only keep the same value within a clause; the same
variable used outside of a clause does not inherit the value (this would be a “global” scope).
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 15
Naming tips – Một số qui ước khi đặt các tên
• Use real English when naming predicates, constants, and variables.e.g. “John wants to help Somebody.”
Could be: wants(john,to_help,Somebody).
Not: x87g(j,_789).
• Use a Verb Subject Object structure: wants(john,to_help).
• BUT do not assume Prolog Understands the meaning of your chosen names!– You create meaning by specifying the body (i.e.
preconditions) of a clause.
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 16
Using predicate definitions Xây dựng vị từ
• Command line programming is tediouse.g. | ?- write(‘What is your name?’), nl, read(X),
write(‘Hello ‘), write(X).
• We can define predicates to automate commands:
greetings:- write(‘What is your name?’),
nl,
read(X),
write(‘Hello ‘),
write(X).
| ?- greetings.What is your name?|: tim.Hello timX = tim ?yes
Prolog Code Terminal
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 17
Using multiple clauses (Định nghĩa vị từ bởi nhiều câu)
• Different clauses can be used to deal with different arguments.
greet(hamish):-
write(‘How are you doin, pal?’).
greet(amelia):-
write(‘Awfully nice to see you!’).
= “Greet Hamish or Amelia” = a disjunction of goals.
| ?- greet(hamish). | ?- greet(amelia).
How are you doin, pal? Awfully nice to see you!
yes yes
• Clauses are tried in order from the top of the file.• The first clause to match succeeds (= yes).
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 18
Re-trying Goals – Câu truy vấn có nhiều đáp số
• When a question is asked with a variable as an argument (e.g. greet(Anybody).) we can ask the Prolog interpreter for multiple answers using: ;
| ?- greet(Anybody).
How are you doin, pal?
Anybody = hamish? ; “Redo that match.”
Anybody = amelia? ; “Redo that match.”
no “Fail as no more matches.”
• This fails the last clause used and searches down the program for another that matches.
• RETURN = accept the match• ; = reject that match
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 19
Ordering of clauses (Thứ tự các câu là quan trọng)
• The order of multiple clauses is important.
greet(hamish):-
write('How are you doin, pal?').
greet(amelia):-
write('Awfully nice to see you!').
• The most specific clause should always be at the top.• General clauses (containing variables) at the bottom.
| ?- greet(hamish).Hullo hamish?yes
greet(Anybody):-write('Hullo '), write(Anybody).
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 20
Ordering of clauses (thứ tự các câu)
• The order of multiple clauses is important.
greet(hamish):-
write('How are you doin, pal?').
greet(amelia):-
write('Awfully nice to see you!').
• The most specific clause should always be at the top.• General clauses (containing variables) at the bottom.
| ?- greet(hamish).How are you doin, pal?.yes
greet(Anybody):-write('Hullo '), write(Anybody).
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 21
Unification (hợp nhất hai hạng thức)• When two terms match we say that they unify.
– Their structures and arguments are compatible.• This can be checked using =/2
|?- loves(john,X) = loves(Y,mary).
X = mary, unification leads to instantiationY = john? yes Terms that unify Outcome
fred = fred. yes.‘Hey you’ = ‘Hey you’. yesfred=X. X=fred.X=Y. Y = X.foo(X) = foo(bar). X=bar.foo(N,N) = foo(bar,X). N=bar, X=bar.foo(foo(bar)) = foo(X) X = foo(bar)
Terms that don’t unifyfred = jim.‘Hey you’ = ‘Hey me’.frou(frou) = f(frou).foo(bar) = foo(bar,bar).foo(N,N) = foo(bar,rab).
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 22
Asking questions of the databaseTruy vấn cơ sở dữ liệu
We can ask about facts directly:
|?- mother(X,alan).
X = jane?
Yes
Or we can define rules that prove if a property or relationship holds given the facts currently in the database.
|?- parent(jane,X).
X = alan?
yes
mother(jane,alan).
father(john,alan).
parent(Mum,Child):-
mother(Mum,Child).
parent(Dad,Child):-
father(Dad,Child).
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27/09/04 AIPP Lecture 2: Prolog Fundamentals 23
Summary• Prolog program consist of clauses.• A clause has a head and a body (Rule) or just a head (Fact).• A head consists of a predicate name and arguments.• A clause body consists of a conjunction of terms.• Terms can be constants, variables, or compound terms.• We can set our program goals by typing a command that unifies
with a clause head.• A goal unifies with clause heads in order (top down).• Unification leads to the instantiation of variables to values.• If any variables in the initial goal become instantiated this is
reported back to the user.
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II. Suy diễn trong prolog (kiểm tra, suy diễn lùi, đệ qui)
27/09/04 AIPP Lecture 2: Prolog Fundamentals 24
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 25
Tests – Kiểm tra• When we ask Prolog a question we are asking for
the interpreter to prove that the statement is true. ?- 5 < 7, integer(bob).yes = the statement can be proven.no = the proof failed because either
– the statement does not hold, or– the program is broken.
Error = there is a problem with the question or program. *nothing* = the program is in an infinite loop.
• We can ask about:– Properties of the database: mother(jane,alan).– Built-in properties of individual objects: integer(bob).– Absolute relationships between objects:
• Unification: =/2• Arithmetic relationships: <, >, =<, >=, =:=, +, -, *, /
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 26
Arithmetic Operators – Các phép toán số• Operators for arithmetic and value comparisons are
built-in to Prolog= always accessible / don’t need to be written
• Comparisons: <, >, =<, >=, =:= (equals), =\= (not equals)
= Infix operators: go between two terms.=</2 is used
• 5 =< 7. (infix) • =<(5,7). (prefix) all infix operators can also be prefixed
• Equality is different from unification=/2 checks if two terms unify
=:=/2 compares the arithmetic value of two expressions ?- X=Y. ?- X=:=Y. ?-X=4,Y=3, X+2 =:= Y+3.
yes Instantiation error X=4, Y=3? yes
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 27
| ?- X is 5+4*2.X = 13 ? yes
Arithmetic Operators (2)• Arithmetic Operators: +, -, *, /
= Infix operators but can also be used as prefix.– Need to use is/2 to access result of the arithmetic
expression otherwise it is treated as a term:
|?- X = 5+4. |?- X is 5+4.
X = 5+4 ? X = 9 ?
yes yes
(Can X unify with 5+4?) (What is the result of 5+4?)
• Mathematical precedence is preserved: /, *, before +,- • Can make compound sums using round brackets
– Impose new precedence– Inner-most brackets first
| ?- X is (5+4)*2.X = 18 ? yes
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 28
Tests within clauses – Các phép toán bool trong các câu
• These operators can be used within the body of a clause:– To manipulate values,sum(X,Y,Sum):-
Sum is X+Y.
– To distinguish between clauses of a predicate definition
bigger(N,M):-
N < M, write(‘The bigger number is ‘), write(M).
bigger(N,M):-
N > M, write(‘The bigger number is ‘), write(N).
bigger(N,M):-
N =:= M, write(‘Numbers are the same‘).
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 29
Backtracking – Suy diễn lùi|?- bigger(5,4).
bigger(N,M):-
N < M,
write(‘The bigger number is ‘), write(M).
bigger(N,M):-
N > M,
write(‘The bigger number is ‘), write(N).
bigger(N,M):-
N =:= M,
write(‘Numbers are the same‘).
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 30
Backtracking – Suy diễn lùi|?- bigger(5,4).
bigger(5,4):-
5 < 4, fails
write(‘The bigger number is ‘), write(M).
bigger(N,M):-
N > M,
write(‘The bigger number is ‘), write(N).
bigger(N,M):-
N =:= M,
write(‘Numbers are the same‘).
Backtrack
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 31
Backtracking – Suy diễn lùi|?- bigger(5,4).
bigger(N,M):-
N < M,
write(‘The bigger number is ‘), write(M).
bigger(5,4):-
5 > 4,
write(‘The bigger number is ‘), write(N).
bigger(N,M):-
N =:= M,
write(‘Numbers are the same‘).
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 32
|?- bigger(5,4).
bigger(N,M):-
N < M,
write(‘The bigger number is ‘), write(M).
bigger(5,4):-
5 > 4, succeeds, go on with body.write(‘The bigger number is ‘), write(5).
The bigger number is 5
yes
|?-
Backtracking – Suy diễn lùi
Reaches full-stop = clause succeeds
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 33
Backtracking – Suy diễn lùi|?- bigger(5,5). If our query only matches the final clause
bigger(N,M):-
N < M,
write(‘The bigger number is ‘), write(M).
bigger(N,M):-
N > M,
write(‘The bigger number is ‘), write(N).
bigger(5,5):-
5 =:= 5,
write(‘Numbers are the same‘).
Is already known as the first two clauses failed.
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 34
Backtracking – Suy diễn lùi|?- bigger(5,5). If our query only matches the final clause
bigger(N,M):-
N < M,
write(‘The bigger number is ‘), write(M).
bigger(N,M):-
N > M,
write(‘The bigger number is ‘), write(N).
bigger(5,5):-
write(‘Numbers are the same‘).
Numbers are the same
yes
Satisfies the same conditions.
Clauses should be ordered according to specificityMost specific at top Universally applicable at bottom
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 35
Satisfying Subgoals – Đích trung gian
• Most rules contain calls to other predicates in their body. These are known as Subgoals.
• These subgoals can match:– facts,– other rules, or– the same rule = a recursive call
1) drinks(alan,beer).2) likes(alan,coffee).3) likes(heather,coffee).
4) likes(Person,Drink):-drinks(Person,Drink). a different subgoal
5) likes(Person,Somebody):-likes(Person,Drink), recursive subgoalslikes(Somebody,Drink).
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 36
Representing Proof using TreesCây biểu diễn chứng minh
To help us understand Prolog’s proof strategy we can represent its behaviour using AND/OR trees.
1. Query is the top-most point (node) of the tree.
2. Tree grows downwards (looks more like roots!).
3. Each branch denotes a subgoal.1. The branch is labelled with the number of the matching clause and
2. any variables instantiated when matching the clause head.
4. Each branch ends with either:1. A successful match ,
2. A failed match , or
3. Another subgoal.
|?- likes(alan,X).
2 X/coffee
X = coffee
1st solution= “Alan likes coffee.”
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 37
Representing Proof using Trees (2)
|?- likes(alan,X).
X/coffee
X = coffee
• Using the tree we can see what happens when we ask for another match ( ; )
2
4
drinks(alan,X).
1 X/beer
X = beer
2nd solution = “Alan likes beer because Alan drinks beer.”
1st match is failed and forgotten
Backtracking
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 38
Recursion using TreesCây đệ qui
|?- likes(alan,X).
X/coffee
X = coffee
• When a predicate calls itself within its body we say the clause is recursing
2
4
1 X/beer
X = beer
5
likes(alan,Drink)
Conjoined subgoals
likes(Somebody,Drink)
drinks(alan,X).
X/coffee 2
X = coffee
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 39
|?- likes(alan,X).
X/coffee
X = coffee
2
4
1 X/beer
X = beer
3rd solution = “Alan likes Alan because Alan likes coffee.”
5
likes(alan,coffee)
likes(Somebody,coffee)
drinks(alan,X).
X/coffee 2
X = coffee
Somebody/alan
2
Somebody = alan
Recursion using Trees (2)• When a predicate calls itself within its body we say the clause is recursing
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 40
|?- likes(alan,X).
X/coffee
X = coffee
2
4
1 X/beer
X = beer
4th solution = “Alan likes Heather
because Heather likes coffee.”
5
likes(alan,coffee) likes(Somebody,coffee)
drinks(alan,X).
X/coffee 2
X = coffee
Somebody/alan
2
Somebody = alan
Somebody / heather3
Somebody = heather
• When a predicate calls itself within its body we say the clause is recursing
Recursion using Trees (3)
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30/09/04 AIPP Lecture 3: Rules, Results, and Backtracking 41
Infitite Recursive Loop• If a recursive clause is called with an incorrect goal it will loop as it can neither prove it nor disprove it.
likes(Someb,coffee)
2Somebody = alan
3
Somebody = heather
5likes(Someb,coffee)
Someb = alan
2 likes(coffee,coffee)
likes(coffee,X) likes(coffee,X)
likes(coffee,X2)
likes(coffee,X3)
likes(X,X2)
likes(X2,X3)
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II. CUT và phủ định và findall
27/09/04 AIPP Lecture 2: Prolog Fundamentals 42
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CUTa(X, Y) :- b(X), !, c(Y).
b(1). b(2). b(3).
c(1). c(2). c(3).
------------------
Kết quả truy vấn thế nào?
?- a(Q, R).
Q = 1,
R = 1 ;
Q = 1,
R = 2 ;
Q = 1,
R = 3.
Tai sao?
27/09/04 AIPP Lecture 2: Prolog Fundamentals 43
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CUT
a(X) :- b(X), !, c(X).
b(1). b(2). b(3).
c(2).
----------------------
?- a(Q).
false.
?- a(2).
true.
Tại sao?
27/09/04 AIPP Lecture 2: Prolog Fundamentals 44
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14/10/04 AIPP Lecture 7: The Cut 45
Green Cuts ! (Sử dụng CUT khi nào?)
f(X,0):- X < 3, !.
f(X,1):- 3 =< X, X < 6, !.
f(X,2):- 6 =< X.
|?- trace, f(2,N).
1 1 Call: f(2,_487) ?
2 2 Call: 2<3 ?
2 2 Exit: 2<3 ? ?
1 1 Exit: f(2,0) ?
N = 0 ? ;
noIf you reach this point don’t bother trying any other clause.
• Notice that the answer is still the same, with or without the cut.– This is because the cut does not alter the logical behaviour of the
program.
– It only alters the procedural behaviour: specifying which goals get checked when.
• This is called a green cut. It is the correct usage of a cut.• Be careful to ensure that your clauses are actually mutually
exclusive when using green cuts!
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14/10/04 AIPP Lecture 7: The Cut 46
• Because the clauses are mutually exclusive and ordered we know that once the clause above fails certain conditions must hold.
• We might want to make our code more efficient by removing superfluous tests.
Red Cuts ! (Không sử dụng CUT khi nào?)
| ?- f(7,N).
1 1 Call: f(7,_475) ? 2 2 Call: 7<3 ? 2 2 Fail: 7<3 ? 3 2 Call: 3=<7 ? 3 2 Exit: 3=<7 ? 4 2 Call: 7<6 ? 4 2 Fail: 7<6 ? 5 2 Call: 6=<7 ? 5 2 Exit: 6=<7 ? 1 1 Exit: f(7,2) ?
N = 2 ?
yes
f(X,0):- X < 3, !.
f(X,1):- 3 =< X, X < 6, !.
f(X,2):- 6 =< X.
Redundant?
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Phủ địnhparents(a,b).
parents(c,d).
parents(b,c).
parents(a,e).
nochild(X):- \+ parents(X,_).
------------------------
?- parents(f,X).
false.
?- nochild(a).
false.
?- nochild(e).
true.
27/09/04 AIPP Lecture 2: Prolog Fundamentals 47
\+ X means not(X) that is the way to implement negation in Prolog; however not(X) does not mean that X is false, it means that X can't be proved true from the database.
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Phủ định (tiếp)• Kiểm tra một số có là nguyên tố?.
is_prime(2).
is_prime(3).
is_prime(P):-integer(P), P>3, P mod 2 =\= 0, \+ has_factor(P,3).
has_factor(P,N):- P mod N =:=0.
has_factor(P,N):- L is N+2, L*L<P, has_factor(P,L).
------------------------
?- is_prime(97).
true.
?- is_prime(18).
false.
27/09/04 AIPP Lecture 2: Prolog Fundamentals 48
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findallparents(a,b).
parents(c,d).
parents(b,c).
parents(a,e).
---------------
?- parents(a,X).
X = b ;
X = e.
?- findall(X,parents(a,X),F).
F = [b, e]
?- findall([X,Y],parents(X,Y),F).
F = [[a, b], [c, d], [b, c], [a, e]].
27/09/04 AIPP Lecture 2: Prolog Fundamentals 49
Sử dụng dấu “;” nếu muốn Prolog in các kết quả khác
Vị từ xây dựng sẵn trong Prolog cho phép tìm tập F gồm tất cả các X thỏa mãn parents(a,X).
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Bài tập chữa trên lớp
1. Tim phan tu cuoi cung cua mot danh sach.
Chu y: [X|R]: X la 1 phan tu, R la 1 list.
2. Hoanvi
3. Sap xep danh sach
4. Giai phuong trinh bac nhat
5. Hop va giao 2 tap hop.
6. Thay the 1 ky tu boi 1 ky tu khac trong danh sach
7. Trich xau con tu vi tri K den L cua mot xau.
8. Tim thua so chung lon nhat 2 so
9. Do thi va tim duong di trong do thi.
27/09/04 AIPP Lecture 2: Prolog Fundamentals 50