SWI-Prolog dynamic/1 vs dynamic/2 - prolog

I am teaching myself Prolog and have been given a handful of examples.
One of which uses the dynamic/1 built-in directive:
:- dynamic(items/1).
I get the idea of dynamic. That one can modify predicates via the assert, and retract predicates.
However, the program also uses the following in places:
:- dynamic(location/2).
What is the difference between the two /1 and /2, is their also a /3 .... /n?

In Prolog, predicates are identified by their name (or functor) and their number of arguments (or arity). Thus, items/1 denotes a predicate with functor items and arity 1 while location/2 denotes a predicate with functor location and arity 2. Two predicates with the same functor but different arities are different predicates.

Related

What does '/1' stands for in Prolog?

I recently started studying and working with Prolog for an application where logic programming is very suited for. In particular, I am woroking with SWI-Prolog (v. 7.6.4, amd64) and Etalis (v. 1.1), an extension for prolog.
What I am not understading is the meaning of: /0, /1, and even /2, /3, that I found in examples during the definition of custom predicates. Most of the prolog directives are defined by directive_name/1.
So, what do these numbers stand for?
It's simply the number of arguments.
For example, append/3 means append(A1, A2, A3).
This number is called the arity, it is the number of arguments a predicate, or functor takes. A constant is in fact a fuctor with zero arity, so /0.
The name of a predicate together with the arity identify a specific predicate. For example the member/2 predicate is diffrent from the append/2 predicate, and the append/3 predicate is different from the append/2 predicate.

why in-built prolog predicates are written with a slash(/)

for example: aggregate_all/3 , findall/3, aggregate/4 and so on.
What is the difference between aggregate/3 and aggregate/4.
The number is the so-called arity of the predicate and indicates the number of arguments of the predicate.
This is useful because there are often several variants of predicates that share the same name, but differ in their number of arguments.
Examples: findall/3 and findall/4, append/3 and append/2 etc.
You also often see the notation (Pred)/2. Example: (#=)/2.This is because #= (for example) is also an infix operator, and the parentheses turn (#=)/2 into a valid Prolog term.
The slash (/) symbol is not used only in built in predicates but in all predicates ant it states the number of parameters (arity) of the predicate for example aggregate/3 is a predicate with 3 parameters while aggregate/4 is a predicate with 4 parameters.

How to call a predicate from another predicate in Prolog?

So I just started Prolog and I was wondering two things:
1) Is there built in functions (or are they all called predicates?) for simple things like max of 2 numbers, or sine of a number, etc... If so, how do I access them?
2) How can I call a predicate from another one? I wrote two predicates called car and cdr. car returns the head of a list and cdr returns the list without the head. But now I want to call car on the cdr. Here are some examples for clarification:
car([3,4,5,5], H). would return H = 3
cdr([3,4,5,5],L). would return L = [4,5,5]
and what I am asking is how can I do this:
car(cdr[3,4,5,5]))
??
As others have pointed out, the predicates in Prolog are called that for a reason: they really aren't functions. Many newcomers to Prolog start out by trying to map the functionality they know in other languages over to Prolog and it generally fails. Prolog is a very different programming tool than most other languages. So it's a bit like using a variety of hammers for a long time, then having someone hand you a wrench, and you wonder why it doesn't make a good hammer.
In Prolog, predicates are a means of declaring relations between entities. If you say foo(a, b) it means there's a relationship between a and b called foo. You've probably seen the examples: knows(joe, jim). and knows(jim, sally). And you can define a relation, like:
remotely_acquainted(X, Y) :- knows(X, Z), knows(Z, Y), \+ knows(X, Y).
Or something like that.
A predicate does not return a value. It either succeeds or it fails. If you have a sequence of predicates separated by commas (an "and" relationship) and Prolog encounters a predicate that fails, it backs up (backtracks) to the nearest prior predicate which it can make succeed again with different instantiation of its arguments and moves forward again.
Just to add a little to the confusion, there are some predicates in Prolog designed specifically for the evaluation of arithmetic expressions. These act like functions, but they are special case. For example:
X is Y / gcd(Z, 4).
Here, gcd of Z and 4 is computed an its value returned, and then Y is divided by that value and the result is instantiated into X. There are a variety of other functions as well, such as max/2, sin/1, etc. You can look them up in the documentation.
Arithmetic comparative operators function this way as well (using =:=/2, >/2, </2, etc with numeric expressions). So if you say:
X < Y + Z
The Prolog will consider numerical evaluation of these arguments and then compare them.
So having said all that, Prolog does allow embedding of term structures. You could have something like:
car(cdr([1,2,3]))
as a term. Prolog will not interpret it. Interpretation is left up to the programmer. I could then create a predicate which defines an evaluation of such terms:
car([H|_], H).
cdr([_|T], T).
proc_list(car(X), Result) :-
proc_list(X, R1),
car(R1, Result), !.
proc_list(cdr(X), Result) :-
proc_list(X, R1),
cdr(R1, Result), !.
proc_list(X, X).
The cut in the above clauses prevents backtracking to proc_list(X, X) when I don't want it.
Then:
| ?- proc_list(car(cdr([1,2,3])), R).
R = 2
yes
| ?- proc_list(car(cdr(cdr([1,2,3]))), R).
R = 3
yes
| ?-
Note this is a simple case and I may not have captured all of the subtleties of doing a proper sequence of car and cdr. It can also be made more general using =.. and call, etc, instead of discrete terms car and cdr in the parameters. For example, a slightly more general proc_list might be:
proc_list(Term, Result) :-
Term =.. [Proc, X], % Assumes terms have just one argument
member(Proc, [car, cdr]), % True only on recognized terms
proc_list(X, R1), % Recursively process embedded term
ProcCall =.. [Proc, R1, Result], % Construct a calling term with Result
call(ProcCall), !.
proc_list(X, X).
This technique of processing a term does step away from relational behavior which Prolog is best at, and leans into functional behavior, but with an understand of how Prolog works.
Prolog has a really different attitude to computing...
You don't define functions, but relations among arguments. The most similar and well known language I'm aware of is SQL. Think of predicates as tables (or stored procedures, when some computation not predefined by database engine is required).
car([H|_],H).
cdr([_|T],T).
car_of_cdr(L, Car) :- cdr(L, Cdr), car(Cdr, Car).
but since lists' syntax is a core part of the language, a better definition could be
car_of_cdr([_,X|_], X).
Anyway, I think you should spend some time on some Prolog tutorial. SO info page has much more information...
:- use_module(support).
This means the module will use predicates written in other modules.
<module_name>:<predicate_name>(<atoms / Variables>).
This way you can call a predicate in another module.

In Prolog, is a fact the same as a functor?

If you have for instance next line of Prolog declaration:
move(state(middle, onbox, middle, hasnot),
grasp,
state(middle, onbox, middle, has)).
Are both move and state functors?
I'm kind off confused by facts, functors, terms, ...
In Prolog functors are syntactic elements we use to build structures (compound terms) from simpler ones.
Think of a hierarchy of bound Prolog terms, with the base containing the simplest "atomic" cases, i.e. atoms and numbers. Add to these Prolog variables, which may be bound or not depending on context. The rules for Prolog functor names (identifiers) are the same as for Prolog atoms
Functors are syntactic units that have a finite number of arguments ("arity"), and if a functor is supplied with terms for those arguments, then we get a compound term. In your example there is a principal functor move with three arguments, so its arity is 3. The functor name and arity are often combined, since technically Prolog treats the same functor name with two distinct arities as different functors, and so we might refer to move/3 as the outer functor of your compound term.
Note that the first and third arguments in your example of a term are themselves compound terms, built using functor state/4:
move(state(middle, onbox, middle, hasnot),
grasp,
state(middle, onbox, middle, has))
Here I removed the period from the end of your example. Likely in its original context this was indeed a Prolog "fact", although it might also have been a query. Periods in Prolog can serve terminate input of a term.
The key point here is that Prolog predicates are special cases of functors. If move/3 is a predicate (taking three arguments), then your example could be a fact (if it is asserted someplace in your application) or a query (e.g. if posed as a top-level goal). In accordance with what we said above about a functor name with two distinct arities, Prolog treats a predicate name with different arities as different predicates!
Prolog's use of functors to express facts (and rules) makes it a fairly slick and powerful metaprogramming environment. Prolog has a special built-in predicate (operator =..) called univ that unpacks a compound term into a list whose head is the principal functor's name and whose remaining items are the arguments given to that functor in the specific compound term. For example:
?- X = state(middle, onbox, middle, hasnot), X =.. List.
X = state(middle, onbox, middle, hasnot)
List = [state, middle, onbox, middle, hasnot]
yes
This allows us to convert between Prolog lists and compound terms in both directions, and to build Prolog goals "dynamically" if need be (using a predicate name for the principal functor of a term).
Yes, move and state are functors. A functor is F in F(Term1, ...). But they aren't facts: in your case, there is only one fact, which is the complete line.
Functors describe a term, but are not the term itself (like method signatures in imperative languages) and consist of the structure's or predicate's name and arity.
In you example, the functors are move/3 and state/4.
Notice that foo(a) and foo(a,b) have different functors, foo/1 and foo/2.

binary predicate to square list and sublists in Prolog

I am new to prolog and was trying to create a binary predicate which will give
a list in which all numbers are squared, including those in sublists.
e.g.
?-dcountSublists([a,[[3]],b,4,c(5),4],C).
C=[a,[[9]],b,c(5),16]
Can anyone guide me how i can do this.
Thank You. Answer with a snippet is appreciated
This is easily achieved using recursion in Prolog. Remember that everything in Prolog is either a variable, or a term (atoms are just 0-arity terms), so a term like the following:
[a,[[3]],b,4,c(5),4]
...is easily deconstructed (also note that the list syntax [..] is sugar for the binary predicate ./2). Prolog offers a range of predicates to test for particular types of terms as well, such as numbers, strings, or compound terms (such as compound/1).
To build the predicate you're after, I recommend writing it using several predicates like this:
dcountSublists(In, Out) :-
% analyze type of In
% based on type, either:
% 1. split term into subterms for recursive processing
% 2. term cannot be split; either replace it, or pass it through
Here's an example to get you started which does the hard bit. The following recognizes compound terms and breaks them apart with the term de/constructor =../2:
dcountSublists(In, Out) :-
% test if In has type compound term
compound(In),
% cut to exclude backtracking to other cases below this predicate
!,
% deconstruct In into functor and an argument list
In =.. [Func|Args],
% apply dcountSublists/2 to every argument, building new args
maplist(dcountSublists, Args, NewArgs),
% re-construct In using the new arguments
Out =.. [Func|NewArgs].
dcountSublists(In, Out) :-
% test if In has type atom
atom(In), !,
% pass it through
Out = In.
Testing:
?- dcountSublists([a,[[e]],b,a,c(s),a], L).
L = [a, [[e]], b, a, c(s), a].
Note that this fails if the input term has numbers, because it doesn't have a predicate to recognize and deal with them. I'll leave this up to you.
Good luck!
SWI-Prolog has the predicate maplist/[2-5] which allows you to map a predicate over some lists.
Using that, you only have to make a predicate that will square a number or the numbers in a list and leave everything else the same. The predicates number/1, is_list/1 are true if their argument is a number or a list.
Therefore:
square(N,NN):-
integer(N),
NN is N*N.
square(L,LL):-
is_list(L),
dcountSublists(square,L,LL).
square(Other,Other):-
\+ number(Other),
\+ is_list(Other).
dcountSublists(L,LSquared):-
maplist(square,L,LSquared).
with the negation in the final predicate we avoid multiple (wrong) solutions:
for example dcountSublists([2],X) would return X=[4] and X=[2] otherwise.
This could be avoided if we used an if-then-else structure for square or once/1 to call square/2.
If this is homework maybe you should not use maplist since (probably) the aim of the exercise is to learn how to build a recursive function; in any case, I would suggest to try and write an equivalent predicate without maplist.

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