6.0 prolog lab using visual prolog version 5.2 example 1: assume the following “likes” facts...
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6.0 PROLOG LAB Using visual PROLOG version 5.2
Example 1: Assume the following “likes” facts knowledge baselikes (ali, football).likes (ali, tennis).likes (ahmad, tennis).likes (ahmad, handball).likes (samir, handball).likes (samir, swimming).likes (khaled, horseriding).
defines non-deterministic predicates that can backtrack and generate multiple solutions. Predicates declared with the keyword nondeterm can fail and, in this case, do not produce any solution.
A sequence of characters, implemented as a pointer to an entry in a hashed symbol-table, containing strings. The syntax is the same as for strings.
To represent the likes facts in VPROLOG
• FileNewnoname.pro• Then in the predicate section write the
declaration of the used predicates:PREDICATESnondeterm likes (symbol,symbol)• Then in the clauses section write the facts:CLAUSESlikes (ali,football).likes (ali,tenis).likes (ahmad,tenis).likes (ahmad,handball).likes (samir,handball).likes (samir,swimming).likes (khaled,horseriding).
1 .Queries as goals in PROLOG
• To supply a query in PROLOG put it in the goal section as follows:
GOAL
likes (ali, football).• Then press Cntrl+G
The output will be Yes or No for concrete questions
1. Concrete questions: (queries without variables)
Example:
GOAL
likes(samir, handball). Yes
Likes (samir,football). No
2 .Queries with variables• To know all sports that ali likes:
likes(ali,What).
• To know which person likes teniss
likes (Who,tenis).
• To know who likes what (i.e all likes facts)
likes (Who,What)
3.Compound queries with one variable
1. To list persons who likes tennis and football, the goal will be
likes( Person, tennis ),likes (Person, football).
2. To list games liked by ali and ahmad
likes (ali,Game),likes (ahmad,Game).
Game=tennis
Person=ali
4-Compound queries with multiple variables
• To find persons who like more than one game:
likes(Person, G1),likes (Person,G2),G1<>G2.
• To find games liked by more than one person:
likes (P1,Game),likes(P2,Game),P1<>P2.
5. Facts containing variables• Assume we add the drinks facts to the likes
database as follows:PREDICATESdrinks(symbol, symbol)CLAUSES drinks(ali, pepsi).drinks (samir, lemonada).drinks (ahmad, milk).• To add the fact that all persons drink water:drinks (Everyone, water).• If we put a goal like:drinks (samy,water). drinks (ahmad,water).
The answer will be: yes
6.Rules• Rules are used to infer new facts from
existing ones.
• A rule consists of two parts: head and body separated by the symbol (:-) .
• To represent the rule that express the facts that two persons are friends if they both like the same game:
friends( P1,P2):-
likes (P1,G),likes (P2,G),P1<>P2. The head of the rule
The body of the rule
Rule’s symbol
7.Backtracking in PROLOG• For PROLOG to answer the query :friends (ali, P2).PROLOG will do the following matches and backtracking to
the friends rule:
P1GP2P1<>P2friends (ali,P2)
alifootballalifalsefail
alitennisalifalsefail
alitennisahmadtruesucceed
8.The domains section in VPROLOG
• It is used to define domains other than the built-in ones, also to give suitable meanings to predicate arguments.
• For example to express the likes relation:likes(person, game)
We should first declare person in the domains section as follows:
domainsperson,game= symbol predicateslikes (person, game)
Person and game are unknown domains(error)
This declaration will allow prolog to detect type errors.Example: the query likes(X,Y), likes(Y,Z) will result in a type error, because Y will be matched to a game constant and can not replace a person variable in the second sub goal.
9 .Built-in domainsdomainNumber of bitsrange
short16 bits-32768..32767
ushort16 bits0..65535
long32 bits-2147483648..
2147483648
ulong32 bits0..4294967295
integer16 bit platform-32768..32767
integer32 bit platform-2147483648..
2147483648
byte8 bits0..255
word16 bit0..65535
dword32 bit0.. ..4294967295
10. Basic standard domainsDomaindescription
char8-bits surrounded by single quotation: ‘a’
realFloating point number:
42705, 9999, 86.72, 911.98e237
Range 1e-307..1e+308
stringPointer to 0-terminated byte arraey
telephone, “Telephone”,t2,”T2”
symbolAs strings but implemented as a pointer to hashed symbol tablefaster than strings
11. How to do arithmetic operationsDOMAINSnumber=integerPREDICATESaddnum(number,number,number)multnum(number,number,number)CLAUSESaddnum(X,Y,S):-S=X+Y.multnum(X,Y,P):-P=X*Y.If the goal is:Addnum(5,7,X).OrMultnum(5,7,X).
X=12
X=35
Note:The compound goal:
multnum(7,5,X),addnum(X,X,Answer). Will result in
X=35, Answer=70
12. More programs to test domains
• The isletter(X) predicate gives yes if X is a letter:PREDICATESisletter(char)CLAUSESisletter(Ch):-‘a’<=Ch,Ch<=‘z’.isletter(Ch):-‘A’<=Ch,Ch<=‘Z’.
To test this program, give it the following goals:a- isletter(‘x’). yesb- isletter(‘2’). noc- isletter(“hallo”). type errord- isletter(a). Type errore- isletter(X) free variable message
13. Multiple arity predicates overloading
• The arity of a predicate is the number of arguments that it takes.
• You can have two predicates with the same name but with different arities.
• You must group different arity versions of a given predicate name in both the predicate and clauses sections.
• The different arities predicates are treated as different predicates.
14. Example: for multiple arity predicates
DOMAINSperson=symbolPREDICATES
nondeterm father (person)
nondeterm father (person, person)CLAUSESfather (Man):- father (Man,_).father (ali, ahmad).father (samy, khaled).
Anonymous variable can match anything.
A person is said to be a father if he is a father of any one.
15. A complete expert system that decides how a person can buy a car.• Assuming
– A person is described by the two relations:
salary (person, money)
savings (person, money)– A car for sale is described by the two relations:
cash (cartype, money)
takseet(cartype,money)
• A person can buy a car
-with cash money if one third of his savings is greater or equal to the cash price of the car.
-with takseet if one third of his salary is greater or equal to the price of the car in case of cash divided by 30.
- If the person can buy a car using cash or takseet the system will advice him to buy with cash
The VPROLOG program DOMAINSperson, car =symbolmoney=realway=stringPREDICATESsalary (person, money)savings (person, money)cash (car, money)takseet (car, money)canbuy (person, car,way)howcanbuy (person, car,way)
CLAUSEScash (cressida,15000).cash (camri,55000).cash (caprise82,5000).cash (caprise90,8000).cash (landcruizer2003,100000).takseet (Car, TPrice):-cash (Car, Price), TPrice=Price*1.2.canbuy ( Person, Car, “cash”):-savings (Person, M), cash( Car, N), M/3>=N.canbuy ( Person,Car, “takseet”):-salary (Person, M), takseet(Car,T),M/3>=T/30.
howcanbuy (Person,Car,Way):-
canbuy (Person,Car,Way),Way="cash";
canbuy (Person, Car, Way), Way="takseet", NOT(canbuy (Person,Car,"cash")).
Should be placed around expressions which include only constants and /or bounded variables.
OR
These variables will be bounded before reaching the “not” expression containing them
16 .Controlling Backtracking• The fail Predicate:• V Prolog begins
backtracking when a call fails.
• The fail predicate is used to force backtracking.
• The following program uses fail to get all solutions.
• Without fail we will get only one solution.
DOMAINS
name=symbol
PREDICATES
nondeterm father(name,name)
everybody
CLAUSES
father (ali, salem).
father (ahmad, ibrahim).
father (ahmad, zaynab).
everybody:-
father (X,Y), write (X,” is ” ,Y,”’s father\n” ), fail.
everybody.
GOAL everybody.
To make the goal succeed at the end
• Preventing backtracking by the cut (!)• It is impossible to backtrack across a cut.• Green cut: it is a cut used to prevent
solutions which does not give meaningful solutions.
• Red cut : it is a cut used to prevent alternate subgoals.
• Example: to prevent backtracking to previous subgoals:
r1:- a, b, !, c.r1:-d.
Only first solution to a and b is considered with many solutions for c
this clause for r1 will not be considered if a solution is found in the above rule.
17. Highway Map modelling and recursive rules
• To represent the shown map we use the predicate
link ( node, node, distance)
• To find a path from s to d– get it from mentioned facts :
path (S,D, TDist):-link (S, D, TDist).– Or find a node x which has a link
from S to X and a path from X to D as follows:
path (S,D, TDist):-
link (S,X,DS1),path (X,D,DS2), TDist=DS1+DS2.
a b
c
d
4
2
56
g
link(a,b,4).
link (a,c,2).
link (b,g,5).
link (c ,g,6).
link (c,d,5).
link (d,g,3).
5
3
Total distance
DOMAINSnode=symboldistance= integerPREDICATESnondeterm link (node, node, distance)nondeterm path ( node, node, distance)CLAUSESlink(a,b,4).link (a,c,2).link (b,g,5).link (c ,g,6).link (c,d,5).link (d,g,3).path (S,D, TDist):-link (S, D, TDist).path (S,D, TDist):-
link (S, X, TD1 ),path (X,D,TD2), TDist=TD1+TD2.GOAL path (a, g, TotalDistance).
Recursive rule
The complete path distance finder program
Facts that model the road map
TotalDistance=9TotalDistance=8TotalDistance=103 Solutions
output
18. Rules which behave like procedures
• A rule that when its head is found in a goal will print something.
• Rules used like case statements:
Only rules with matching arguments will be executed, others will be tested but will fail.
greet-:write(“ASalamo Alykom”),n1.
PREDICATESnondeterm action (integer)CLAUSESaction(1):- nl, write (“N=1”),nl.action(2):- nl, write (“N=2”),nl.action(3):- nl, write (“N=3”),nl.action (N):- nl,N<>1,N<>2,N<>3,write (“N=?”),nl.GOALwrite (“Type a number 1->3”), readint (N) , action(N).
• Since rules that has unmatched arguments will be tested and will fail. This will slow the program. To speed up the system use the cut as shown.
• If you want to test a range x>5 and <9 place the cut after the test sub goals.
PREDICATESnondeterm action (integer)CLAUSESaction(1):-!, nl, write (“N=1”),nl.action(2):- !,nl, write (“N=2”),nl.action(3):- !,nl, write (“N=3”),nl.action (_):- nl,,write (“unknown number ?”),nl.GOALwrite (“Type a number 1->3”), readint (N) , action(N).
Action(X):- X>5,X<9,!,write(“5<N<9”),n1.
To make a rule return a value
• To define a rule which classifies a number either positive, negative or zero:
PREDICATESnondeterm classify (integer,symbol).CLAUSESclassify( X,pos):- X>0.classify( X,neg):- X<0.classify( X,zero):- X=0.GoalClassify ( 5,What)Classify (-4,What)Classify (X,What)
What=pos
What=nig
error
What carries the returned
result
What carries the returned
result
19. Compound data objects• Compound data objects allow you to treat
several pieces of information as a single item (like structures in c, c+, or c#).
• Example: to represent a date object:DOMAINSdate_cmp= date(string, unsigned, unsigned)Hence in a rule or goal you can write:..,D=date(“March”,1,1960). Here D will be treated as a single item.
called : functor
• The arguments of a compound object can themselves be compound:
Example : to represent the information for the birthday of a person:-
birthday(person(“ahmad”,”ali”),date(1,”march”,1960)(
BIRTHDAY
PERSON DATE
Ahamd Ali March1 1960
DOMAINSbirthday= birthday (person, date)person = person (name, name)date= date (day, month, year)name, month =stringday, year= unsigned
19.1 A family birthday programDOMAINSperson = person (name, name)birthdate= bdate (day, month, year)name, month =stringday,year= unsignedPREDICATESnondeterm birthday (person, bdate)CLAUSESbirthday(person("amin","mohamad"),bdate(1,"March",1960)).birthday(person("mohamd","amin"),bdate(11,"Jan",1988)).birthday(person("abdo","mohamad"),bdate(11,"Oct",1964)).birthday(person("ali","mohamad"),bdate(1,"Feb",1950)).birthday(person("suzan","antar"),bdate(1,"March",1950)).GOAL%birthday(X,Y).%birthday(X,date(_,"March",_)).%birthday(person(X,"mohamad"),bdate(_,_,Y).(
Compound objects
Lists persons whose second name is “mohamd” with year of birth
Lists all birthday’s person, and date objects
Lists persons born on March
19.2 Using system dateDOMAINS
person = person (name, name)
bdate= bdate (day, month, year)
name=string
day ,month, year= unsigned
PREDICATES
nondeterm birthday (person, bdate)
get_birth_thismonth
testnow (unsigned, bdate)
write_person(person)
CLAUSES
birthday(person("amin","mohamad"),bdate(1,3,1960)).
birthday(person("mohamd","amin"),bdate(11,5,1988)).
birthday(person("abdo","mohamad"),bdate(11,5,1964)).
birthday(person("ali","mohamad"),bdate(1,6,1950)).
birthday(person("suzan","antar"),bdate(1,3,1982)).
get_birth_thismonth:- date (_,Thism ,_), write ( “now month is", Thism), nl, birthday (P, D), testnow(Thism, D), write_person(P), fail.
get_birth_thismonth:-write (“Press any key to continue"), nl, readchar(_).
testnow(TM,bdate (_,BM,_)):- TM=BM.
write_person(person(Fn, Ln)):-write(" ", Fn, "\t\t ", Ln), nl.
GOAL get_birth_thismonth.
19.3 Using alternatives in domain declaration
• Assume we want to declare the following statements:– Ali owns a 3-floar house– Ali owns a 4-door car– Ali owns a 3.2 Ghz computer– To define a thing to be either house, car, or computer:
DOMAINSthing= house( nofloars); car (nodoors); computer(ghertz)nofloars,nodoors=integerghertz=realperson= person(fname, lname)fname, lname=symbolPREDICATESnondeterm owns(person, thing)CLAUSESowns( person(mohamad, ali), house(3)).owns(person(mohamd,ali),computer(3.2)).
Use or(;) to separate alternatives.
goalowns (P,X).2 solutions
19.4 Lists• To declare the subjects
a teacher might teach:PREDICATESteacher (symbol, symbol, symbol)CLAUSESteacher (ahmad, ezz,cs101).teacher (ahmad, ezz, cs332).teacher (amin, mohamad, cs435).teacher (amin,mohamad, cs204).teacher (amin,mohamad, cs212).teacher (reda, salama, cs416).teacher (reda, salama,cs221).
Here, the teacher name is repeated many times.
We need a variable length data structure that holds all subjects that a teacher can teach
DOMAINSsubject=symbol *PREDICATESteacher (symbol,symbol,subject)CLAUSESteacher(ahmad, ezz,[cs101,cs332] )
Solution use a list data structure for the subject
Solution: use a list data structure for the subject
teacher (amin,mohamad, [cs204, cs435, cs212] ).teacher (reda,salama, [cs416, cs221] ).
20 .Repetition and Recursion• Repetition can be expressed in PROLOG
in procedures and data structures.
• Two kinds of repetition exist in PROLOG :
Backtracking: search for multiple solutions Recursion: a procedure calls itself
Normal recursion:
Takes a lot of memory and time
Tail recursion: fast and less memory
Compiled into iterative loops in M/C language
When a sub goal fails, PROLOG returns to the most recent subgoal that has an untried alternative.Using the fail predicate as the end of repeated subgoals we enforce PROLOG to repeat executing different alternatives of some subgoals.
20.1 Repetition using backtracking
PREDICATES
nondeterm country (symbol)
print_countries
CLAUSES
country (“Egypt”).
country (“SaudiArabia”).
country ( “Seria”).
country (“Sudan”).
print_countries:- country(X), write(X), nl, fail.
print_countries.
GOAL
print_countries.
•Example: Using fail to print all countries:
country(X)write (X)
nlfailprint_countries
X=“Egypt”okokfailfail
X=“SaudiArabia”
okokfailfail
X=“Seria”okokfailfail
X=“Sudan”okokfailfail
20.1.2 Pre and post actions
• PREDICATES• nondeterm country (symbol)• nondeterm print_countries• CLAUSES• country ("Egypt").• country ("SaudiArabia").• country ( "Seria").• country ("Sudan").• print_countries:-• write("Some Arabic
countries are"),nl,fail.• print_countries:-
country(X), write(X," and "), fail.
• print_countries:-• nl, write("there are others"),
nl.• GOAL• Print_countries.
•To do pre-actions (before the loop )or post-actions (after the loop), write different versions of the same predicate.
country(X)write(X, ” and “)
failPrint_countries
X=“Egypt”Okfailfail
X=“SaudiArabia”Okfailfail
X=“Seria”Okfailfail
X=“Sudan”Okfailfail
Some Arabic countries areEgypt and SaudiArabia and Seria and SudanThere are others
Pre-action
Post-actionMain-action
20.2 Implementing backtracking with loops• The following repeat
predicate tricks PROLOG and makes it think it has infinite solutions:
repeat.
repeat:- repeat.
The shown program uses repeat to keep accepting a character and printing it until it is CR
Multiple solutions attract backtracking when fail occurs
PREDICATES
repeat
typewriter
CLAUSES
repeat.
repeat:- repeat.
trypewriter:-
repeat,
readchar(C),
write(C) ,
C=‘\r’,! . /* if CR then cut
1st solution
2nd solution
Nth solution
(N+1)th solution
20.3 Recursive procedures• To find the factorial of a
number N : FACT(N)
IF N=1 then FCT(N)=1
Else
FACT(N)=N* FACT(N-1)
• Using PROLOG, the factorial is implemented with two versions of the factorial rule as shown here.
PREDICATES
factorial (unsigned, real)
CLAUSES
factorial (1,1):-!.
factorial (N, FactN):-
M=N-1,
factorial (M, FactM),
FactN=N*FactM.
Goal
factorial (5,F).
F=1201 Solution
20.4 Tail Recursion• Occurs when a procedure calls itself as the last step. • Occurs in PROLOG ,when
1. The call is the very last subgoal in a clause.
2. No backtracking points exist before this call.
3. The recursive predicate does not return a value.
4. Only one version of this predicate exists.
• Example:
count (N):-
write(N), nl,
NewN=N+1,
count (NewN).
GOAL
count(0).
Write numbers from 0 to infinity ( will get unexpected values due to overflow)
(1) Last call no need to save state of the previous call stack is free no extra memory is needed small memory and fast due to no need to push state
No alternative solutions hereNo alternative solutions here
)2 (No alternative solutions exist here no backtracking points
(3)N is bound to a constant value and no other free variable no return value
You can use a cut(!) before the last call to be sure of (2).
20.5 Using arguments as loop variables• To implement
factorial with iterative procedure using C-language:
long fact;long p=1;unsigned i=1;while(i<=n){p=p*i;i=i+1;}fact=p;
PREDICATES
factorial (unsigned, long)
factorial_r(unsigned, long, unsigned, long)
CLAUSES
factorial (N, FactN):-
factorial_r(N, FactN,1,1).
factorial_r( N, FactN, I, P):-
I<=N,
! ,
NewP=P*I,
NewI=I+1,
factorial_r(N,FactN,NewI,NewP).
factorial_r(N, FactN ,I ,P):- I>N, FactN=P.
Initialize arguments I,P
Test end condition to go to other version if not satisfied.
To prevent backtracking and to allow tail recursion.
21 .Lists and Recursion• List processing allows handling objects that
contain an arbitrary number of elements.• A list is an object that contains an arbitrary
number of other objects.• A list that contains the numbers 1,2 and 3 is
written as [1,2,3] .• Each item contained in a list is know as an
element.• To declare the domain of a list of integers:
– Domains– intlist= integer * Means list of integers
Could be any other name ( ilist, il,…)
• The elements of a list must be of a single domain.
• To define a list of mixed types use compound objects as follows:
• Domains– elementlist= element *– element=ie (integer); re ( real); se (string)
• A list consists of two parts:– The head : the first element of the list– The tail : a list of the remaining elements.
• Example: if l=[a,b,c] the head is a and the tail is [b, c].
The HeadtThe Tail
• If you remove the first element from the tail of the list enough times, you get down to the empty list ([ ]).
• The empty list can not be broken into head and tail.
• A list has a tree structure like compound objects.• Example: the tree structure of
[a, b,c, d] is drawn as shown
in the figure.
list
a list
b list
c list
d [ ]
Note : the one element list [a] is not as the element a since [a] is really the compound data structure shown here
[a]
a [ ]
21.1 List processing• Prolog allows you to treat the head and tail
explicitly. You can separate the list head and tail using the vertical bar (|).
• Example :
• [a, b, c ]
≡[a | [b, c ] ]
≡[a | [b| [c ] ] ]
≡[a | [b| [c| [] ] ] ]
≡[a , b| [c ] ]
• 21.2 List unification: the following table gives examples of list unification:
list1list2Variable binding
[X, Y, Z][ book, ball, pen ]X=book, Y=ball, Z=pen
[7][X|Y]X=7,Y=[ ]
[1, 2, 3, 4][X, Y|Z ]X=1,Y=2,Z=[3,4]
[1,2][3|X]fail
• 21.3 Using lists– Since a list is really a
recursive compound data structure, you need recursive algorithms to process it.
– Such an algorithm is usually composed of two clauses:-
• One to deal with empty list.• The other deals with
ordinary list (having head and tail).
• Example: the shown program writes the elements of an integer list.
DOMAINS
list= integer *
PREDICATES
write_a_list( list )
CLAUSES
write_a_list([ ]).
write_a_list ( [ H | T ] ):-
write(H),
nl,
write_a_list( T).
GOAL
write_a_list([ 1, 2, 3]).
Do nothing but report success.
Write the head
Write the Tail list
T=[ ]
• Two logical rules are used to determine the length of a list
1. The length of [ ] is 0.
2. The length of the list [X|T] is 1+the length of T.
• The shown Prolog program counts the number of elements of a list of integers (can be used for any type).
DOMAINS
list= integer *
PREDICATES
length_of ( list, integer)
CLAUSES
length_of ( [], 0).
Length_of ( [ H | T ],L ):-
length_of ( T ,M ),
L=M+1.
GOAL
length_of([ 1, 2, 3],L).
T=[ ]
21.4 Counting list elements
L=3
Homework: modify the length program to calculate the sum of all elements of a list
21.5 Modifying the list• The following
program adds 1 to each element of a list L1 by making another list L2:– If L1 is [ ] then L2=[ ]– If L1=[H1|T1]
assuming L2=[H2|T2] then
• H2=H1+1.• Add 1 to T1 by the
same way
DOMAINSlist= integer *PREDICATESadd1 ( list, list)CLAUSESadd1 ( [], []).
add1 ( [ H1 | T1 ],[H2|T2] ):-H2=H1+1,add1( T1,T2).GOALadd1( [ 1, 2, 3], L ).
L=[ 2, 3, 4]
21.6 Removing Elements from a list• The following
program removes negative elements from a list L1 by making another list L2 that contains non negative numbers of L1 :– If L1 is [ ] then L2=[ ]– If L1=[H1|T1] and
H1<0 then neglect H1 and process T1
– Else make head of L2 H2=H1 and Repeat for T1.
DOMAINSlist= integer *PREDICATESdiscard_NG ( list, list)CLAUSESdiscard_NG ( [], []).
discard_NG ( [ H1 | T1 ],L2 ):-H1<0,!,discard_NG(T1,L2).
discard_NG ( [ H1 | T1 ],[H1|T2] ):-discard_NG(T1,T2).GOALdiscard_NG( [ 1, -2, 3], L ).
L=[ 1, 3]
21.7 List Membership
• To detect that an element E is in a list L:– If L=[H|T] and E=H
then report success
Else– search for E in T
DOMAINS
namelist= name*
name= symbol
PREDICATES
nondeterm member (name, namelist)
CLAUSES
member (E ,[E | _ ] ).
member (E, [ _ | T]):-
member (E,T).
GOAL
member (ali, [samy, salem,ali]).
Yes
Try the goalmember (X, [samy, salem,ali]).What happens if we put a cut at the first clause as follows:member (E ,[E | _ ] ):- ! .
21.8 Appending one list to another
• To append L1 to L2 we get the result in L3 as follows
• Append (L1,L2,L3):– If L1=[ ] then L3=L2.– Else– Make H3=H1
and– make T3 =T1 +L2
DOMAINS
intlist= integer*
PREDICATES
append (intlist, intlist, intlist)
CLAUSES
append ([], L2,L2 ).
append ([H | T1], L2,[H|T3] ):-
append (T1, L2, T3).
GOAL
append([1,2,3],[5,6],L).
L= [1,2,3,5,6].Try the goals:1-append ([1,2],[3],L), append( L, L,LF).2-append ( L,[5,6], [1,2,3,5,6]).3-append (L1,L2, [1,2,3]).
21.9 Tracing the append goal
DOMAINS
intlist= integer*
PREDICATES
append (intlist, intlist, intlist)
CLAUSES
append ([], L2,L2 ).
append ([H | T1], L2,[H|T3] ):-
append (T1, L2, T3).
GOAL
append([1,2,3],[5,6],L).
L= [1,2,3,5,6].
append([1,2,3],[5,6],L).
append ([1 | [2,3] ],[5,6], [1|T3] ):-
append ([2,3 ],[5,6], [T3] ):-
append ([2 | [3] ],[5,6], [2|T3’] ):-
append ([3 ],[5,6], T3’ ):-
append ([3 | [ ] ],[5,6], [3|T3’’] ):-
append ([ ],[5,6], T3’’ ). T3’’ =[5,6]
T3’ =[3,5,6]
T3 =[2,3,5,6]
L =[1,2,3,5,6]
22 VPROLOG facts Section• The facts section is declared to allow a
programmer to add or remove facts at run time.• Facts declared in fact section are kept in tables
to allow modification, while normal facts are compiled into binary code for speed optimization.
1. To declare a fact in a fact section:• DOMAINS• person=string• FACTS -up• father (person, person)• CLAUSES• father (“samy”, “ali”).
Name of this fact section
22.1Updating the facts section• To add a fact to the facts section use
• asserta or assert to insert the new fact at the beginning of the fact section
• Example:– asserta( father(“ali”,”ahmad”)).
• assertz to add the fact at the end of its fact section.– assertz( father(“ahmad”,”khalil”)).
If we execute the goal father ( X,Y).,
The output will be as shown here:
Will be added before the fact:father (“samy”, “ali”).
Will be added after the fact:father (“samy”, “ali”).
X=ali, Y=ahmadX=samy, Y=aliX=ahmad, Y=khalil
• There is no automatic check in VPROLOG if you insert a fact twice, to prevent this you can declare the following predicate which tests a fact before adding it as follows:
• PREDICATES
• uassert(up)
• CLAUSES
uassert (father(F,S)):- father(F,S), ! ;
assert (father(F,S)).
Search for the fact if found breakElse assert it
22.2Saving and loading facts.
• To save the facts in a file you can use the save predicate as follows:
save (filename).Or
Save (filename, factsectionname).
• To load a fact database use consult as follows:
consult (filename, factsectionname).
DOMAINSperson=stringFACTS - up father (person, person)predicatesuassert (up)addfathersrepeat CLAUSESfather ("samy", "ali").uassert (father(F,S)):- father(F,S), ! ; assert (father(F,S)). repeat. repeat:- repeat. addfathers:- repeat,readln (F),readln (S), uassert(father(F,S)),F="",!,save("c:\\facts",up).
goalconsult(“c:\\facts”), addfathers, father(X,Y).
Removing facts• To remove a fact use the predicate retract as
follows: retract(<the fact>[,factsectionname]).
• Example to remove the fact father(“samy”,“ali”):
retract(father(“samy”,”ali”).
• To remove all facts about “samy” as a father:retract (father(“samy”,_)).Note: the previous program adds a father whose name
is null, to remove all facts that has a null string name modify the add father predicate as follows:
addfathers:- repeat,readln (F),readln (S), uassert(father(F,S)),F="",!, retract(father(F,S)), save("c:\\facts",up).
To remove the last entered fact with null strings
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