[a,b,c,d] 15/9/2013. [a,b,c,d] List. List. List. List is one of the most important Prolog data structures.

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1 is one of the most important Prolog data structures. A list is an ordered sequence of zero or more terms written between square brackets and separated by commas. [a,b,c,d] September 1 st Session 2013/2014 (A131) Wan Hussain Wan Ishak School of Computing UUMCollege of Arts and Sciences Universiti Utara Malaysia (P) (F) (E) hussain@uum.edu.my (U) Lecture notes, Operators and Arithmetic Arithmetic Constructing Expressions Processing Character Strings and Structures [a,b,c,d] Elements The elements of a list can be any kind of Prolog terms, including other lists. [1,2,3,4] [a1,a2,a3] [sawi, kangkung] [kucing(comel), kucing(hitam)] [[satu,dua],[tiga,empat]] 1

2 The empty list is written: [] Please note one element list [a] is not equivalent to the atom a.? [a]=a. no The list can be divided into head and tail by the symbol. [H T] Thefirstelementistheheadandtherestarethetail. [a [b,c,d,e]] can be constructed or decomposed through unification. Unify with [a,b,c] = X [X,b,Z] = [a,y,c] [[a,b],c] = [X,Y] Result X = [a,b,c] X = a, Y = b, Z = c X = [a,b], Y = c [a(b),c(x)] = [Z,c(a)] X = a, Z = a(b) [a [b,c,d,e]] The tail of a list is always a list, the head of a list is an element. Everynonemptylisthasaheadandatail. [a,b,c,d] =[a [b,c,d]] [a]=[a []] More examples Built-in predicates Unify with Result [X Y] = [a,b,c,d] X = a, Y = [b,c,d] [X Y] = [a] X = a, Y = [] [X,Y Z] = [a,b,c] X = a, Y = b, Z = [c] [X,Y,Z A] = [a,b,c] X = a, Y = b, Z = c, A = [] [X,Y Z] = [a W] X = a, W = [Y Z] Predicate append/3 length/2 member/2 member/3 remove/3 removeall/3 reverse/2 Function join or split lists get the length of a Prolog list get or check a member of a list get or check a member of a list and its position remove an element from a list remove all occurrences of an item from a list check or get the reverse of a list 2

3 append/3 append(first, Second, Whole) member/3 member(element,, Position) Join list Splitting list?-append([a,b],[c,d],whole). Whole =[a,b,c,d]?-append([a,b],second,[a,b,c,d]). Second =[c,d]?-append(first,[c,d],[a,b,c,d]). First =[a,b] Element position Get Element at given position?- member(c, [a,b,c,d], Position). Position=3;?- member(element, [a,b,c,d], 2). Element=b length/2 member/3(more example): length(term, Length)?- length([a,b,c,d], Length). Length=4 Get the element and its position?- member(element, [a,b,c,d], Position). Element =a, Position = 1; Element =b, Position = 2; Element =c, Position = 3; Element =d, Position = 4; member/2 remove/3?-member(a, [a,b,c,d]). remove(element,, Remainder) member(element, )?- member(element, [a,b,c,d]). Element = a ; Element = b ; Element = c ; Element = d ; Takeout an element Element missing?????- remove(b,[a,b,c,d], Remainder). Remainder =[a,c,d];?- remove(element,[a,b,c,d],[a,b,d]). Element=c; 3

4 remove/3(more example): Reverse/2?- remove(element, [a,b,c,d], Remainder). Element=a, Remainder =[b,c,d]; Element=b, Remainder =[a,c,d]; Element=c, Remainder =[a,b,d]; Element=d, Remainder =[a,b,c]; What element can be removed Reverse the list Causes error reverse(, Revlist)?- reverse([a,b,c,d], Revlist). Revlist =[d,c,b,a]?- reverse(, [d,c,b,a]). =[a,b,c,d] ; Error 4, Heap Space Full, Trying ewrite/1 removeall/3 - Exercise What is the output? removeall(item,, Remainder)?-append([ab],[b,c,d], X).?-reverse([b,c,d],R),append([ab],R,X).?- reverse([b,c,d], R), member(f, R, 1), remove(f, R, B), append([ab], B, X). removeall/3 (Example): Remove all repeated elements Remove all elements that match with the first element?- removeall(a, [a,b,a,b,a], Remainder). Remainder =[b,b]?- removeall(item, [a,b,a,b,a], Remainder). Item =a, Remainder =[b,b]?- removeall(item, [b,a,b,a], Remainder). Item =b, Remainder =[a,a] append/3 length/2 member/2 member/3 remove/3 removeall/3 reverse/2 How to define or construct the predicates? How they are working/how they manipulate the list? 4

5 append/3 addon/3 addon(first, Second, Whole) Definition: addon([],x,x). addon([h T1],X,[H T2]):- addon(t1,x,t2). stoptherecursion addelement into thelist 3 append/3 addon/3 From 2, (2.1) addon([],[c],t2). Call (1) addon([],x,x). true []=] Stop at 3 rule (1). Rule (2) will not execute. (1) addon([],x,x). addon/3 (2) addon([h T1],X,[H T2]):- (2.1) addon(t1, X, T2). X=[c] append/3 addon/3?-addon([a,b],[c],w). (1) addon([],x,x). addon/3 (2) addon([h T1],X,[H T2]):- (2.1) addon(t1, X, T2).?- addon([a,b],[c],w). W =[a,b,c] 1 (2) addon([a [b]],[c],[a T2]):- (2.1) addon([b],[c],t2). [a [b,c]] T2 = [b,c] (1) addon([],x,x). addon/3 (2) addon ([H T1],X,[H T2]):- (2.1) addon (T1, X, T2). 1 Call (1) addon([],x,x). Call (2) addon([h T1],X,[H T2]). Call (2.1) addon(t1,x,t2). false [a,b]\=[] addon([b],[c],t2). go to 2 true [a,b]=[h T1] [c]=x W=[H T2] H=a T1=[b] X=[c] W=[a T2] T2=_ 2 (2) (2.1) 3 (1) addon([b []],[c],[b T2]):- addon([],[c],t2). addon([],[c],[c]). [b [c]] T2 = [c] append/3 addon/3 From 1, (2.1) addon([b],[c],t2). (1) addon([],x,x). addon/3 (2) addon([h T1],X,[H T2]):- (2.1) addon(t1, X, T2). length/2 noofterms/2 noofterms(term, Length) Call (1) addon([],x,x). false [b]\=[] 2 Call (2) addon([h T1],X,[H T2]). true [b]=[h T1] H=b [c]=x T1=[] W=[H T2] X=[c] Call (2.1) addon(t1,x,t2). W=[b T2] T2=_ addon([],[c],t2). go to 3 Definition: noofterms([],0). noofterms([h Tail],K):- noofterms(tail,j), KisJ+1. stoptherecursion calculatethelength 5

6 length/2 noofterms/2 noofterms/2 (1) noofterms([],0). length/2 noofterms/2 noofterms/2 (1) noofterms([],0). 1?-noOfTerms([a,b,c],T). Call (1) noofterms([],0). Call (2) noofterms([h Tail],K). Call (2.1) noofterms(tail,j). (2) noofterms([h Tail],K):- (2.1) noofterms(tail,j), false [a,b,c]\=[] true [a,b,c]=[h Tail] T=K H=a Tail=[b,c] T=K=_ 4 From 3, (2.1) noofterms([],j). Call (1) noofterms([],0). true []=[] J=0 (2) noofterms([h Tail],K):- (2.1) noofterms(tail,j), Prologwill now return to 3 to execute 2.2. Followed by 2 and 1. noofterms([b,c],j). go to 2 Call (2.2) pending. (At the moment Sub goal 2.2 will not be called.) 2 length/2 noofterms/2 From 1, (2.1) noofterms([b,c],j). Call (1) noofterms([],0). Call (2) noofterms([h Tail],K). Call (2.1) noofterms(tail,j). (1) noofterms([],0). noofterms/2 (2) noofterms([h Tail],K):- (2.1) noofterms(tail,j), false [b,c]\=[] true [b,c]=[h Tail] J=K noofterms([c],j). go to 3 Call (2.2) pending. (At the moment Sub goal 2.2 will not be called.) H=b Tail=[c] J=K=_? - noofterms([a,b,c],t). 1 (2) noofterms([a [b,c]],k):- K=3 (2.1) noofterms([b,c],j), J=2 (2.2) 2 K is J + 1. (2) noofterms([b [c]],k):- K=2 (2.1) noofterms([c],j), J=1 (2.2) 3 K is J + 1. (2) noofterms([c []],K):- (2.1) noofterms([],j), (2.2) 4 K (1) is J noofterms([],0) (2.2) K is J + 1. K is K is (1) noofterms([],0). noofterms/2 (2) noofterms([h Tail],K):- (2.1) noofterms(tail,j), K=1 J=0 In console T = 3 K is length/2 noofterms/2 From 2, (2.1) noofterms([c],j). Call (1) noofterms([],0). Call (2) noofterms([h Tail],K). Call (2.1) noofterms(tail,j). (1) noofterms([],0). noofterms/2 (2) noofterms([h Tail],K):- (2.1) noofterms(tail,j), false [c]\=[] true [c]=[h Tail] J=K noofterms([],j). go to 4 Call (2.2) pending. (At the moment Sub goal 2.2 will not be called.) H=c Tail=[] J=K=_ noofterms([a [b,c]],k). K is noofterms([b [c]],k) J = K = 2 K is noofterms([c []],K). J = K = 1 K is noofterms([],0). J = K = 0 6

7 member/2 is_in/2 remove/3 take_out/3 is_in(element, ) Definition: is_in(x,[x T]). is_in(x,[h T]):- is_in(x, T). checkifthe ElementisHeadofthe list ifnot,traverse therest ofthelist(tail) take_out (Element,, Remainder) Definition: take_out (X,[X R],R). element found take_out (X,[F R],[F S]):- take_out (X,R,S). traverse the list,atthe same time create new list is_in(x,[a,b,c]). reverse/2 overturn/2 overturn(, Revlist) X = a Return X = a is_in(x,[b,c]). Definition: overturn(,rev):- overturn2(,[],rev). X = b is_in(x,[c]). Return X = b X = c is_in(x,[]). Return X = c overturn2([h T],Z,W):- overturn2(t,[h Z],W). theelementposition overturn2([],x,x). is_in(c,[a,b,c]). Character Strings Three ways to represent a string of characters in Prolog: c \= a is_in(c,[b,c]). As an atom atoms are compact but hard to takeapart or manipulate. C \= b is_in(c,[c]). As a list of ASCII codes can use standard list processing techniques on them. c = c Return is_in(c,[]). As a list of one-character atoms - can use standard list processing techniques on them. 7

8 Character Strings Use write/1 with double quotes.?-write( abc ). [97,98,99] Character Strings Example(Prolog codes): splitnprint(ayat):- name(ayat,ayat), write_str(ayat). write_str([h T]):- put(h), write_str(t). write_str([]). Output:?- splitnprint(ahmad). ahmad Character Strings Use put/1 or putb/1. write_str([h T]):- put(h), write_str(t). write_str([]). Character Strings (Exercise) Write acode/rule toprintoutasfollows:?- print_splits("prolog"). p rolog pr olog pro log prol og prolo g prolog print_split([],word). write_str([]). Character Strings Convert between an atom or number and a byte list - name/2. name(atomic, )?- name(makan,). = [109,97,107,97,110]?- name(atomic, [109,97,107,97,110]). Atomic = makan Character Strings (Exercise) Write a code/rule to check whether the word end with s or not:?- end_with_s("flowers").?- end_with_s("car"). no?- end_with_s("cars"). print_splits(word):- print_split(word,word). print_split([_ T],Word):- append(h,t, Word), write_str(h), write(` `), write_str(t),nl, print_split(t, Word). write_str([h T]):- put(h), write_str(t). end_with_s():- length(, L), member(115,, L). 8

9 Operators Notation Data structure in Prolog: functor(arg1, arg2,, argn). Functor can become operator example: eat(ali,rice). Functor Canbewritten as: Linguistic Operator ali eat rice. Linguistic Operator Precedence value for operator Thevaluebetween The precedence value can influence the structure meaning: low precedence meaning high priority. operator with 33 is given higher priority compare to35. operator with 33 will be first executed. Operators Notation Types of operators Operator Example Infix Prefix +(3,4) Postfix 8 factorial Linguistic Operator Precedence Value Thevaluebetween The precedence value can influence the structure meaning: low precedence meaning high priority. operator with 33 is given higher priority compare to 35. operator with 33 will be first executed. Linguistic Operator Operator can be defined using op/3. op/3 - Declares an operator with a given type and precedence. op(precedence, Type, Name) Integer Name of the operator Atom or list of atoms Linguistic Operator Precedence Value Example use display/1 To view the standard syntax Change the precedence value Error Wrongly use operator?- op(35,xfx,is_in).?- op(33,fx,room).?- display(apple is_in room kitchen). is_in(apple,room(kitchen))?- op(36,fx,room).?- display(apple is_in room kitchen).! ! Error 42 : Syntax Error! Goal : eread(_11848,_11850)?- display(room kitchen is_in house). room(is_in(kitchen,house)) Defining operator is_in and room 9

10 Linguistic Operator Precedence Value Example Using Linguistic Operator Examine?- op(35,xfx,is_in).?- op(33,fx,room).?- display(apple is_in room kitchen). is_in(apple,room(kitchen)) Standard Syntax % Rules if binatang(x) and makan(x,ikan) then kucing(x). if(then(and(binatang(_13214),makan(_13214,ikan)), kucing(_13214)))?- op(36,fx,room).?- display(room kitchen is_in house). room(is_in(kitchen,house)) Linguistic Operator Defining Type Ifthe precedencevalue isthesame Prologwillreadeitherfrom lefttoright or righttoleftbased onthetype. Type for operator Types fx fy xf yf xfx xfy yfx Meaning non-associative prefix operator right associativeprefix operator non-associative postfix operator left associativepostfix operator non-associative infix operator right associativeinfix operator left associativeinfix operator prefix postfix infix Arithmetic 2 Operation + 1 =? Arithmetic how we do calculation in Prolog. Use predicate is/2 not =/2. Using Linguistic Operator Example cth.pl % Linguistic operators :- op(800, fx, if), op(700, xfx, then), op(300, xfy, or), op(200, xfy, and). % Facts binatang(comel). binatang(tompok). makan(comel,ikan). Makan(tompok,ikan). mengiau(comel). mengiau(tompok). % Rules if binatang(x) and makan(x,ikan) then kucing(x). Arithmetic Operation is/2 takes an arithmetic expression on its right, evaluates it, and unifies the result with its argument on the left. Format: X is X is 9. A = 1, X is A is <argument > is <arithmetic expression>. 10

11 Arithmetic Operation Wrong expression: 2+1is3. Typeerror-notintheright format. Arithmetic Operation Other arithmetic functions supported in LPA Prolog 2is1+X Instantiationerror valueofxis unknown. Arithmetic Operation Different with is/2 and =/2 : is/2 evaluate its right argument and unifies it with itsargumentontheleft docalculation. =/2 is to test if both arguments on the left and right are unify do unification. Arithmetic Operators The precedence of operators: Xis1 2*5^2/8+6 Equivalent to X is (1 (((2 * (5 ^ 2))/ 8)+ 6)) Where ^ is performed first, then*and/, and finally+ and-. Arithmetic Operators Common arithmetic operators(infix operators) Operator & Symbol Example & Output + Addition X is Subtraction X is * Multiplication X is 3 * 2. 6 / Floating-point division X is 3 / // Integer division X is 3 // 2. 1 Using Arithmetic Operators?- papar(5) printout(n):- 2. printout(1, N) printout(n, N) printout(x, N):- 7. printout_num(x), 8. nl, 9. NewX is X + 1, 10. printout(newx, N) printout_num(0) printout_num(x):- 15. write(x), 16. NewX is X - 1, 17. printout_num(newx). 11

12 Constructing Expression Different between Prolog an other programming languages is: Other programming language evaluate arithmetic expression s whenever they occur Prolog evaluates them only in specific places. Practical Calculation When using expressions in Prolog, we are mixing the styles of encoding knowledge. Normal syntax: Canbewrittenas:? X is ?-sum(2, 3, X). Constructing Expression of built-in Predicates that evaluate expressions. R is Ex Evaluates Ex and unifies result with R Ex1 =:= Ex2 test if the results of two expressions are equal Ex1 =\= Ex2 test if the results of two expressions are not equal Ex1 > Ex2 Ex1 < Ex2 test if the result of one expression is greater than another test if the result of one expression is less than another Ex1 >= Ex2 test if one expression is greater than or equal to another Ex1 =< Ex2 test if one expression is less than or equal to another Practical Calculation doublevalue(x, Z):- ZisX*X. areaofsquare(p, L, Area):- Area is P*L. Exercise Explain which of the following succeed, fail, or raise error conditions, and why: 1.?- 5 is ?- 5 =:= ?- 5 = ?- 5 is 2 + What. 5.?- 4+1 = ?- 4+1 is ?- 4+1 =:= ?- What is ?- What =:= ?- What is ?- What = ?- What =\=