more Prolog test vs. find built-in predicates list operations: member, append, nth0, reverse, … not, default vs. logical negation comparison operators, arithmetic user-defined operators position, precedence, associativity Dave Reed

Test vs. find ?- member(X, [a, b, c]). X = a ; X = b ; X = c ; No ?- member(a, X). X = [a|_G260] ; X = [_G259,a|_G262] Yes ?- member(X, Y). X = _G209 Y = [_G209|_G270] ; X = _G209 Y = [_G275,_G209|_G278] Yes Prolog programs define relations: query w/ constants: "test" whether relation holds between the constants query w/ variables: "find" values for which the relation holds with a variable as first argument, the query is used to "find" members of the given list with a variable as the second argument, the query is used to "find" lists that have the given item as member with both variables, the query is used to "find" general schemas for list membership

Append predicate ?- append([a,b], [c,d], L). L = [a,b,c,d] Yes ?- append(L1, L2, [a,b,c,d]). L1 = [] L2 = [a,b,c,d] ; L1 = [a] L2 = [b,c,d] ; L1 = [a,b] L2 = [c,d] ; L1 = [a,b,c] L2 = [d] ; L1 = [a,b,c,d] L2 = [] ; No another useful predefined predicate for list manipulation append(L1,L2,L3) : L3 is the result of placing the items in L1 at the front of L2 can be used in reverse to partition a list again, we could define append ourselves: append([], L, L). append([H|T], L, [H|A]) :- append(T, L, A). as we saw with ancestor example, clause ordering is important with recursive definitions if put rule first, infinite loop possible

Other list predicates ?- L = [a,b,c,d], length(L, Len), nth1(Len, L, X). L = [a,b,c,d] Len = 4 X = d Yes ?- reverse([a,b,a,c], Rev), delete(Rev, a, Del). Rev = [c,a,b,a] Del = [c,b] Yes ?- select(a, [a,b,a,c], R). R = [b,a,c] ; R = [a,b,c] ; No ?- select(a, L, [b,c]). L = [a,b,c] ; L = [b,a,c] ; L = [b,c,a] ; No is_list(Term) : succeeds if Term is a list length(List,Len) : Len is the number of items in List nth0(Index,List,Item) : Item is the item at index Index of List (starting at 0) nth1(Index,List,Item) : Item is the item at index Index of List (starting at 1) reverse(List,RevList) : RevList is List with the items in reverse order delete(List,Item,NewList) : NewList is the result of deleting every occurrence of Item from List select(Item,List,Remain) : Remain is the List with an occurrence of Item removed

not predicate not defines the (default) negation of conditional statements when applied to relations with no variables, it is equivalent to logical negation ( ) in reality, it returns the opposite of whatever its argument would return if X would succeed as a query, then not(X) fails if X would fail as a query, then not(X) succeeds anything Prolog can't prove true it assumes to be false! ?- is_list([]). Yes ?- not(is_list([])). No ?- member(d, [a,b,c]). No ?- not(member(d, [a,b,c])). Yes ?- member(X, [a,b,c]). X = a Yes ?- not(member(X, [a,b,c])). No ?- X = d, not(member(X, [a,b,c])). X = d Yes ?- not(member(X, [a,b,c])), X = d. No

Programming exercises suppose we want to define a relation to test if a list is palindromic palindrome(List) : succeeds if List is a list whose elements are the same backwards & forwards palindrome([]). palindrome(List) :- reverse(List, Rev), List = Rev. suppose we want to define relation to see if a list has duplicates has_dupes(List) : succeeds if List has at least one duplicate element has_dupes([H|T]) :- member(H, T). has_dupes([_|T]) :- has_dupes(T). suppose we want the opposite relation, that a list has no dupes no_dupes(List) : succeeds if List has no duplicate elements no_dupes(List) :- not( has_dupes(List) ).

Built-in comparison operators X = Y does more than test equality, it matches X with Y, instantiating variables if necessary X \= Y determines whether X & Y are not unifiable –for ground terms, this means inequality –can think of as: not(X = Y) –again, doesn't make a lot of sense for variables Note: arithmetic operators (+, -, *, /) are not evaluated (4, 2) can force evaluation using 'is' ?- X = ?- X is X = 4+2X = 6Yes ?- foo = foo. Yes ?- X = foo. X = foo Yes ?- [H|T] = [a,b,c]. H = a T = [b,c] Yes ?- foo \= bar. Yes ?- X \= foo. No ?- 4+2 = 6. No

Arithmetic operations arithmetic comparisons automatically evaluate expressions X =:= Y X and Y must both be arithmetic expressions (no variables) X =\= Y X > Y ?- 12 =:= 6+6. X >= Y Yes X < Y X =< Y ?- X =:= 6+6. ERROR: Arguments are not sufficiently instantiated Example: sum_of_list(ListOfNums, Sum) : Sum is the sum of numbers in ListOfNums sum_of_list([], 0). sum_of_list([H|T], Sum) :- sum_of_list(T,TailSum), Sum is H + TailSum.

Programming exercise suppose we want to define relation to see how many times an item occurs in a list num_occur(Item,List,N) : Item occurs N times in List num_occur(_,[],0). num_occur(H, [H|T], N) :- num_occur(H, T, TailN), N is TailN+1. num_occur(H, [_|T], N) :- num_occur(H, T, TailN), N is TailN. is the first answer supplied by this relation correct? are subsequent answers obtained via backtracking correct?

User-defined operators it is sometimes convenient to write functors/predicates as operators predefined: +(2, 3) user defined? likes(dave, cubs) dave likes cubs operators have the following characteristics position of appearance prefixe.g., -3 infixe.g., postfixe.g., 5! precedence * (3 * 4) associativity 8 – – (5 – 2)

op new operators may be defined as follows :- op(Prec, PosAssoc, Name). Name is a constant Prec is an integer in range 0 – 1200 (lower number binds tighter) PosAssoc is a constant of the form xf, yf (postfix) fx, fy (prefix) xfx, xfy, yfx, yfy (infix) the location of f denotes the operator position x means only operators of lower precedence may appear here y allows operators of lower or equal precedence Example: :- op(300, xfx, likes).

Operator example %% likes.pro likes(dave, cubs). likes(kelly, and(java, and(scheme,prolog))). ?- likes(dave, X). X = cubs Yes ?- likes(Who, What). Who = dave What = cubs ; Who = kelly What = and(java, and(scheme,prolog)) ; No %% likes.pro :- op(300, xfx, likes). :- op(250, xfy, and). dave likes cubs. kelly likes java and scheme and prolog. ?- dave likes X. X = cubs Yes ?- Who likes What. Who = dave What = cubs ; Who = kelly What = java and scheme and prolog ; No by defining functors/predicates as operators, can make code more English-like

SWI-Prolog operators the following standard operators are predefined in SWI-Prolog to define new operators Name & Type are easy Precedence is tricky, must determine place in hierarchy Note: always define ops at top of the program (before use) 1200xfx -->, :- 1200fx :-, ?- 1150fx dynamic, multifile, module_transparent, discontiguous, volatile, initialization 1100xfy ;, | 1050xfy -> 1000xfy, 954xfy \ 900fy \+ 900fx ~ \=, \==, is 600xfy : 500yfx +, -, /\, \/, xor 500fx +, -, ?, \ 400yfx *, /, //, >, mod, rem 200xfx ** 200xfy ^

IQ Test choose the most likely answer by analogy Question 1: Question 2: Question 3:

Analogy reasoner want to write a Prolog program for reasoning by analogy (Evans, 1968) need to decide on a representation of pictures constants: small, large, triangle, square, circle functor/operator: sized sized(small, triangle) OR small sized triangle sized(large, square) OR large sized square functors: inside, above inside(small sized circle, large sized square). above(large sized triangle, small sized square). need to represent questions as well operators: is_to, as (bind looser than sized, so higher prec #) predicate: question question(Name, F1 is_to F2 as F3 is_to [A1,A2,A3]).

IQ test questions :- op(200, xfy, is_to). :- op(200, xfy, as). :- op(180, xfy, sized). question(q1, inside(small sized square, large sized triangle) is_to inside(small sized triangle, large sized square) as inside(small sized circle, large sized square) is_to [inside(small sized circle, large sized triangle), inside(small sized square, large sized circle), inside(small sized triangle, large sized square)]). question(q2, inside(small sized circle, large sized square) is_to inside(small sized square, large sized circle) as above(small sized triangle, large sized triangle) is_to [above(small sized circle, large sized circle), inside(small sized triangle, large sized triangle), above(large sized triangle, small sized triangle)]). question(q3, above(small sized square, large sized circle) is_to above(large sized square, small sized circle) as above(small sized circle, large sized triangle) is_to [above(large sized circle, small sized triangle), inside(small sized circle, large sized triangle), above(large sized triangle, small sized square)]).

Analogy transformations transform(invertPosition, inside(small sized Figure1, large sized Figure2) is_to inside(small sized Figure2, large sized Figure1)). transform(invertPosition, above(Size1 sized Figure1, Size2 sized Figure2) is_to above(Size2 sized Figure2, Size1 sized Figure1)). transform(invertSizes, inside(small sized Figure1, large sized Figure2) is_to inside(small sized Figure2, large sized Figure1)). transform(invertSizes, above(Size1 sized Figure1, Size2 sized Figure2) is_to above(Size2 sized Figure1, Size1 sized Figure2)). also need to represent transformations predicate: transform transform(Name, F1 is_to F2). Note: different but related transformations can have the same name

Analogy reasoner %%%%%%%%%%%%%%%%%%%%%%%%%%% %% analogy.pro Dave Reed 1/23/02 %% %% A program based on Evans' analogy reasoner. %%%%%%%%%%%%%%%%%%%%%%%%%%% :- op(200, xfy, is_to). :- op(200, xfy, as). :- op(180, xfy, sized). analogy(Question, Solution) :- question(Question, F1 is_to F2 as F3 is_to Answers), transform(Rule, F1 is_to F2), transform(Rule, F3 is_to Solution), member(Solution, Answers). %% questions (as before) %% transformations (as before) to find an answer: 1.look up the question based on its name 2.find a transformation that takes F1 to F2 3.apply that rule to F3 to obtain a potential solution 4.test to see if that solution is among the answers to choose from

Analogy reasoner ?- analogy(q1, Answer). Answer = inside(small sized square, large sized circle) ; No ?- analogy(q2, Answer). Answer = above(large sized triangle, small sized triangle) ; No ?- analogy(q3, Answer). Answer = above(large sized circle, small sized triangle) ; No Note: Questions 1 & 2 yield the same answer twice. WHY? Is it possible for a questions to have different answers?

Handling ambiguities Question 4: ?- analogy(q4, Answer). Answer = above(small sized triangle, large sized circle) ; Answer = above(small sized circle, large sized triangle) ; No it is possible for 2 different transformations to produce different answers must either 1.refine the transformations (difficult) 2.use heuristics to pick most likely answer (approach taken by Green)