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More Built-in Predicates. Notes for Ch.7 of Bratko For CSCE 580 Sp03 Marco Valtorta. Testing the Type of Terms. data objects (terms). var( X) succeeds if X is an uninstantiated variable nonvar( X) atom( X) succeeds if X currently stands for an atom integer( X) float( X)
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More Built-in Predicates Notes for Ch.7 of Bratko For CSCE 580 Sp03 Marco Valtorta
Testing the Type of Terms data objects (terms) • var( X) succeeds if X is an uninstantiated variable • nonvar( X) • atom( X) succeeds if X currently stands for an atom • integer( X) • float( X) • number( X) number (integer or float) • atomic( X) number or atom • compound( X) structure Compound objects (structures) simple objects constants variables atoms numbers
Type of Terms in SWI-Prolog • See section 4.5 of the manual • There a few more built-in predicates, such as • string(+Term) • ground(+Term) succeeds if Term contains no uninstantiated variables
Example: Counting Atoms • ch7_1.pl • We want to count actual occurrences of an atom, not terms that match an atom • count1(Atom,List,Number) counts terms that match Atom 3 ?- count1(a,[a,b,X,Y],Na). X = a Y = a Na = 3 • count2(Atom,List,Number) counts actual occurrences of Atom 5 ?- count2(a,[a,b,X,Y],Na). X = _G304 Y = _G307 Na = 1
Cryptarithmetic Puzzle • SEND + MORE = MONEY • Assign distinct decimal digits to distinct letters so that the sum is valid • Albert Newell and Herbert Simon studied in depth puzzle like these in their study of human problem solving (Human Problem Solving, Prentice Hall, 1972) • People use a mixture of trial and error and constraint processing: e.g., M must be 1, S must be 8 or 9, O must be 0, etc.
Cryptarithmetic Puzzle II • sum( N1,N2,N) if N is N1+N2 • Numbers are represented by lists of digits: the query is: • ?-sum( [S,E,N,D],[M,O,R,E],[M,O,N,E,Y]). • We need to define the sum relation; we generalize to a relation sum1, with • Carry digit from the right (before summing, C1) • Carry digit to the left (after summing, C) • Set of digits available before summing (Digits1) • Set of digits left after summing (Digits) • sum1( N1,N2,N,C1,C,Digits1,Digits)
Cryptarithmetic Puzzle III • Example: 1 ?- sum1([H,E],[6,E],[U,S],1,1,[1,3,4,7,8,9],Digits). H = 8 E = 1 U = 4 S = 3 Digits = [7, 9] • There are several (four, in fact) other answers. 1<- <-1 8 1 6 1 4 3
Cryptarithmetic Puzzle IV • We start off with all digits available, we do not want any carries at the end, and we do not care about which digits are left unused: • sum(N1,N2,N) :- sum1(N1,N2,N,0,0,[0,1,2,3,4,5,6,7,8,9],_). • We assume all three lists are of the same lengths, padding with zeros if necessary: • ?- sum([0,S,E,N,D],[0,M,O,R,E],[M,O,N,E,Y]).
Cryptarithmetic Puzzle V sum1( [], [], [], C, C, Digits, Digits). sum1( [D1|N1], [D2|N2], [D|N], C1, C, Digs1, Digs) :- sum1( N1, N2, N, C1, C2, Digs1, Digs2), digitsum( D1, D2, C2, D, C, Digs2, Digs). digitsum( D1, D2, C1, D, C, Digs1, Digs) :- del_var( D1, Digs1, Digs2), % Select an available digit for D1 del_var( D2, Digs2, Digs3), % Select an available digit for D2 del_var( D, Digs3, Digs), % Select an available digit for D S is D1 + D2 + C1, D is S mod 10, % Reminder C is S // 10. % Integer division
Cryptarithmetic Puzzle VI • Nonderministic deletion of variables del_var( A, L, L) :- nonvar(A), !. % A already instantiated del_var( A, [A|L], L). del_var( A, [B|L], [B|L1]) :- del_var(A, L, L1).
Constructing and Decomposing Terms • Clauses are terms too! • As in LISP, data and programs have a common representation: S-expressions (atoms or lists) in LISP, terms in Prolog 2 ?- [user]. |: a. |: a :- b. 3 ?- clause(a,X). X = true ; X = b ; No
Clauses are Terms • The principal functor of a clause is its neck (:- ) 4 ?- [user]. |: :-(b,c). |: % user compiled 0.00 sec, 32 bytes Yes 5 ?- listing(b). b :- c. Yes
univ • Term =.. L, if • L is a list containing the principal functor of Term, followed by Term’s argument • Example: • fig7_3.pl • substitute( Subterm,Term,Subterm1,Term1) • ?-substitute( sin(x),2*sin(x)*f(sin(x)),t,F).
univ • Example: • fig7_3.pl • substitute( Subterm,Term,Subterm1,Term1) • ?-substitute( sin(x),2*sin(x)*f(sin(x)),t,F). • An occurrence of Subterm in Term is something in Term that matches Subterm • If Subterm = Term, then Term1 = Subterm1 Else if Term is atomic, then Term1 = Term Else carry out the substitution on the arguments of Term
Dynamic Goals • Goals may be created at run time, as in: Obtain( Functor), Compute( Arglist), Goal =.. [ Functor|Arglist], Goal. % works in SWI; with some Prologs, call( Goal) • See ch7_2.pl for an example: try_call :- Goal =.. [ member| [a, [a,b,c]]], % Goal is a term Goal. % Goal is a predicate • This is also an example of Prolog’s ambiguous syntax: the first Goal is a term (a variable), while the second Goal is a predicate
Equality and Comparison • X = Y matching (unification) • X is E matches arithmetic value • (discouraged in SWI) • E1 =:= E2 arithmetic • E1 =\= E2 arithmetic inequality • T1 == T2 literal equality of terms • T1 \== T2 not identical • T1 @< T2 term comparison • See section 4.6.1 of SWI manual for standard ordering
Prolog Database Manipulation • asserta/1 • assertz/1 • Same as assert/1 • retract/1 • More, described in section 4.13 of SWI manual • Only dynamic predicates may be asserted, retracted, etc. • dynamic +Functor/+Arity, . . . • Section 4.14 of SWI manual
Control Facilities • once( P) • fail • true • not P (alternative syntax: \+ P) • call( P) • repeat, defined as: repeat. repeat :- repeat. dosquares example: ch7_3.pl
bagof, setof, and findall • bagof( X,P,L) if L is the list of Xs such that P(X) holds • setof( X,P,L) is like bagof, but without duplication • Read ^ as “there exists” in bagof and setof • findall( X,P,L) is like bagof but collects all objects X regardless of (possibly) different solutions for variables in P that are not shared with X • Code for findall is in fig7_4.pl • renamed findall1, because findall is built-in
bagof, etc. Examples • fig7_4.pl: • age( peter,7). age( ann,5). age( pat,8). age( tom,5). • ?- bagof( Child,age(Child,5),L). • ?- bagof( Child,age(Child,Age),L). • ?- bagof( Child,Age^age(Child,Age),L). • ?- bagof( Age,Child^age(Child,Age),L). • ?- setof( Age, Child^age(Child,Age),L). • ?- setof( Age:Child,age(Child,Age),L). • ?-findall( Child,age( Child,Age),L).