1. 6.0 PROLOG LAB Using visual PROLOG version 5.2 Example 1: Assume the following “likes” facts knowledge base likes (ali, football). likes (ali, tennis). likes (ahmad, tennis). likes (ahmad, handball). likes (samir, handball). likes (samir, swimming). likes (khaled, horseriding).

2. 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 File  New  noname.pro Then in the predicate section write the declaration of the used predicates: PREDICATES nondeterm likes (symbol,symbol) Then in the clauses section write the facts: CLAUSES likes (ali,football). likes (ali,tenis). likes (ahmad,tenis). likes (ahmad,handball). likes (samir,handball). likes (samir,swimming). likes (khaled,horseriding).

3. 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

4. 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)

5. 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

6. 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.

7. 5. Facts containing variables Assume we add the drinks facts to the likes database as follows: PREDICATES drinks(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

8. 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

9. 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: friends (ali,P2) P1<>P2P2GP1 fail falsealifootballali fail falsealitennisali succeed trueahmadtennisali

10. 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: domains person,game= symbol predicates likes (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.

11. 9. Built-in domains rangeNumber of bitsdomain -32768..3276716 bitsshort 0..6553516 bitsushort -2147483648.. 2147483648 32 bitslong 0..429496729532 bitsulong -32768..3276716 bit platforminteger -2147483648.. 2147483648 32 bit platforminteger 0..2558 bitsbyte 0..6553516 bitword 0....429496729532 bitdword

12. 10. Basic standard domains descriptionDomain 8-bits surrounded by single quotation: ‘a’ char Floating point number: 42705, 9999, 86.72, 911.98e237 Range 1e-307..1e+308 real Pointer to 0-terminated byte arraey telephone, “Telephone”,t2,”T2” string As strings but implemented as a pointer to hashed symbol table  faster than strings symbol

13. 11. How to do arithmetic operations DOMAINS number=integer PREDICATES addnum(number,number,number) multnum(number,number,number) CLAUSES addnum(X,Y,S):- S=X+Y. multnum(X,Y,P):- P=X*Y. If the goal is: Addnum(5,7,X). Or Multnum(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

14. 12. More programs to test domains The isletter(X) predicate gives yes if X is a letter: PREDICATES isletter(char) CLAUSES isletter(Ch):- ‘a’<=Ch, Ch<=‘z’. isletter(Ch):- ‘A’<=Ch, Ch<=‘Z’. To test this program, give it the following goals: a- isletter(‘x’).  yes b-isletter(‘2’).  no c-isletter(“hallo”).  type error d-isletter(a).  Type error e-isletter(X)  free variable message

15. 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.

16. 14. Example: for multiple arity predicates DOMAINS person=symbol PREDICATES nondeterm father (person) nondeterm father (person, person) CLAUSES father (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.

17. 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)

18. 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

19. The VPROLOG program DOMAINS person, car =symbol money=real way=string PREDICATES salary (person, money) savings (person, money) cash (car, money) takseet (car, money) canbuy (person, car,way) howcanbuy (person, car,way)

20. CLAUSES cash (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.

21. 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

22. 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

23. 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.

24. 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 5 6 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

25. DOMAINS node=symbol distance= integer PREDICATES nondeterm link (node, node, distance) nondeterm path ( node, node, distance) CLAUSES link(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=9 TotalDistance=8 TotalDistance=10 3 Solutions output

26. 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. PREDICATES nondeterm action (integer) CLAUSES action(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. GOAL write (“Type a number 1->3”), readint (N), action(N).

27. 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. PREDICATES nondeterm action (integer) CLAUSES action(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. GOAL write (“Type a number 1->3”), readint (N), action(N). Action(X):- X>5,X<9,!,write(“5<N<9”),n1.

28. To make a rule return a value To define a rule which classifies a number either positive, negative or zero: PREDICATES nondeterm classify (integer,symbol). CLAUSES classify( X,pos):- X>0. classify( X,neg):- X<0. classify( X,zero):- X=0. Goal Classify ( 5,What) Classify (-4,What) Classify (X,What) What=pos What=nig error What carries the returned result

29. 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: DOMAINS date_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

30. 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 PERSONDATE AhamdAliMarch11960 DOMAINS birthday= birthday (person, date) person = person (name, name) date= date (day, month, year) name, month =string day, year= unsigned

31. 19.1 A family birthday program DOMAINS person = person (name, name) birthdate= bdate (day, month, year) name, month =string day,year= unsigned PREDICATES nondeterm birthday (person, bdate) CLAUSES birthday(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

32. 19.2 Using system date DOMAINS 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.

33. 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: DOMAINS thing= house( nofloars); car (nodoors); computer(ghertz) nofloars,nodoors=integer ghertz=real person= person(fname, lname) fname, lname=symbol PREDICATES nondeterm owns(person, thing) CLAUSES owns( person(mohamad, ali), house(3)). owns(person(mohamd,ali),computer(3.2)). Use or(;) to separate alternatives. goal owns (P,X).  2 solutions

34. 19.4 Lists To declare the subjects a teacher might teach: PREDICATES teacher (symbol, symbol, symbol) CLAUSES teacher (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 DOMAINS subject=symbol * PREDICATES teacher (symbol,symbol,subject) CLAUSES teacher(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] ).

35. 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 solutionsRecursion: 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.

36. 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: print_co untries failnlwrite (X) country(X) fail ok X=“Egypt” fail ok X=“SaudiA rabia” fail ok X=“Seria” fail ok X=“Sudan”

37. 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. Print_cou ntries failwrite(X, ” and “) country(X) fail OkX=“Egypt” fail OkX=“SaudiArabia” fail OkX=“Seria” fail OkX=“Sudan” Some Arabic countries are Egypt and SaudiArabia and Seria and Sudan There are others Pre-action Post-action Main-action

38. 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 1 st solution 2 nd solution N th solution (N+1) th solution

39. 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=120 1 Solution

40. 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 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).

41. 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.

42. 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,…)

43. 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 Headt The Tail

44. 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 b c 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 [ ]

45. 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 ] ]

46. 21.2 List unification: the following table gives examples of list unification: Variable bindinglist2list1 X=book, Y=ball, Z=pen[ book, ball, pen ][X, Y, Z] X=7,Y=[ ][X|Y][7] X=1,Y=2,Z=[3,4][X, Y|Z ][1, 2, 3, 4] fail[3|X][1,2]

47. 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=[ ]

48. 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

49. 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 DOMAINS list= integer * PREDICATES add1 ( list, list) CLAUSES add1 ( [], []). add1 ( [ H1 | T1 ],[H2|T2] ):- H2=H1+1, add1( T1,T2). GOAL add1( [ 1, 2, 3], L ). L=[ 2, 3, 4]

50. 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. DOMAINS list= integer * PREDICATES discard_NG ( list, list) CLAUSES discard_NG ( [], []). discard_NG ( [ H1 | T1 ],L2 ):- H1<0, !, discard_NG(T1,L2). discard_NG ( [ H1 | T1 ],[H1|T2] ):- discard_NG(T1,T2). GOAL discard_NG( [ 1, -2, 3], L ). L=[ 1, 3]

51. 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 goal member (X, [samy, salem,ali]). What happens if we put a cut at the first clause as follows: member (E,[E | _ ] ):- !.

52. 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]).

53. 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]

54. 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

55. 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=ahmad X=samy, Y=ali X=ahmad, Y=khalil

56. 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 break Else assert it

57. 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). DOMAINS person=string FACTS - up father (person, person) predicates uassert (up) addfathers repeat CLAUSES father ("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). goal consult(“c:\\facts”), addfathers, father(X,Y).

58. Removing facts To remove a fact use the predicate retract as follows: retract( [,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