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 = _G209Y = [_G209|_G270] ;

X = _G209Y = [_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 withrecursive 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 = 4X = 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 +(4, 2)

can force evaluation using 'is'

?- X = 4 + 2. ?- X is 4 + 2.

X = 4+2 X = 6

Yes Yes

?- foo = foo.

Yes

?- X = foo.

X = foo

Yes

?- [H|T] = [a,b,c].

H = aT = [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 =\= YX > Y ?- 12 =:= 6+6.X >= Y Yes X < YX =< 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) 2 + 3

user defined? likes(dave, cubs) dave likes cubs

operators have the following characteristics position of appearance

prefix e.g., -3

infix e.g., 2 + 3

postfix e.g., 5!

precedence2 + 3 * 4 2 + (3 * 4)

associativity8 – 5 - 2 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 formxf, yf (postfix)fx, fy (prefix)xfx, xfy, yfx, yfy (infix)

the location of f denotes the operator positionx means only operators of lower precedence may appear

herey 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)

1200 xfx -->, :-

1200 fx :-, ?-

1150 fx dynamic, multifile, module_transparent, discontiguous, volatile, initialization

1100 xfy ;, |

1050 xfy ->

1000 xfy ,

954 xfy \

900 fy \+

900 fx ~

700 xfx <, =, =.., =@=, =:=, =<, ==, =\=, >, >=, @<, @=<, @>, @>=, \=, \==, is

600 xfy :

500 yfx +, -, /\, \/, xor

500 fx +, -, ?, \

400 yfx *, /, //, <<, >>, mod, rem

200 xfx **

200 xfy ^

IQ Test choose the most likely

answer by analogyis to as is to

A) B) C)

Question 1:

is to as is to

A) B) C)

Question 2:

is to as is to

A) B) C)

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) ;

Answer = inside(small sized square, large sized circle) ;

No

?- analogy(q2, Answer).

Answer = above(large sized triangle, small sized triangle) ;

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 either1. refine the transformations (difficult)2. use heuristics to pick most likely

answer (approach taken by Green)

is to as is to

A) B) C)