PR7.ppt

UNIVERSITY OF SOUTH CAROLINA
Department of Computer Science and Engineering
More Built-in Predicates
Notes for Ch.7 of Bratko
For CSCE 580 Sp03
Marco Valtorta

Testing the Type of 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
data objects (terms)
simple objects Compound
objects
(structures)
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,Ageâge(Child,Age),L).
• ?- bagof( Age,Childâge(Child,Age),L).
• ?- setof( Age, Childâge(Child,Age),L).
• ?- setof( Age:Child,age(Child,Age),L).
• ?-findall( Child,age( Child,Age),L).

PR7.ppt

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PR7.ppt