How to validate commutativity involving the dif/2 constraint? - prolog

There is a lot of hype around the dif/2 constraint, especially as a rescue for some non-declarativity of (\=)/2 and (\==)/2. This non-declarativity is often characterized as non-monotonicity and examples of non-communtativity are given.
But what would be the means to test whether test cases involving dif/2 are commutative. Here is a meta explanation what I want to do:
I make a commutativity test, and I want to probe that both variants
give the same result:
?- A, B.
-- versus --
?- B, A.
So usually you can check monotonicity, if it boils down to checking commutativity, with the (==)/2 built-in predicate. Since this predicate follows instantiated variables.
But if you are testing cases that produce constraints, call_with_residue/2
is not enough, you need also to have equality of constraints. Which can
be tricky, as the following example shows:
Welcome to SWI-Prolog (Multi-threaded, 64 bits, Version 7.3.23)
Copyright (c) 1990-2015 University of Amsterdam, VU Amsterdam
?- dif(f(X,X),f(a(U,g(T)),a(g(Z),U))), X=a(g(Z),U).
X = a(g(Z), U),
dif(f(a(g(Z), U), U, Z, U, T), f(a(U, g(T)), g(Z), T, g(Z), Z)).
?- X=a(g(Z),U), dif(f(X,X),f(a(U,g(T)),a(g(Z),U))).
X = a(g(Z), U),
dif(f(U, T), f(g(Z), Z)).
Any ideas how to proceed?
Disclaimer, its a trap:
I don't endorse commutativity testing as a good testing method, where you can separate good and bad predicates versus a specification. Since usually both the good and bad predicates might have no problems with commutativity.
I am using commutativity testing as a vehicle to find out about methods for equality of dif/2 constraints. This equality can then be used in more traditional test cases as a validation point.

There are several ways. Maybe the easiest in this case is to simply re-post the collected residual constraints.
In this example, we get:
?- X = a(g(Z), U),
dif(f(a(g(Z), U), U, Z, U, T), f(a(U, g(T)), g(Z), T, g(Z), Z)).
X = a(g(Z), U),
dif(f(U, T), f(g(Z), Z)).
The goal is now much simpler!
You can collect residual goals with copy_term/3 and call_residue_vars/2.

Related

How do I determine if a prolog `compund term` has a particular `atom` inside?

I want a predicate to tell whether a particular atom (say x) appears inside a compound term, however deeply nested.
I tried to read about predicates given at https://www.swi-prolog.org/pldoc/man?section=manipterm. I think it will involve walking down the compound term using functor/3 and ../2 recusively. Is there a simpler approach, or some library that does this?
Carlo's elegant answer works on SWI-Prolog but is not portable as the ISO Prolog standard specification for the arg/3 predicate (which most Prolog systems implement) requires its first argument to be bound to an integer, preventing using it as a backtracable generator of compound term arguments. A more portable alternative would be:
haystack_needle(H, N) :-
H == N.
haystack_needle(H, N) :-
functor(H, _, A),
between(1, A, I),
arg(I, H, A),
haystack_needle(A, N).
The between/3 predicate is not specified in the ISO Prolog standard but it's a de facto standard predicate usually provided as either a built-in predicate or a library predicate.
Ignoring cyclic terms, I would use ==/2,compound/1 and arg/3
haystack_needle(H,N) :- H==N.
haystack_needle(H,N) :- compound(H),arg(_,H,A),haystack_needle(A,N).
haystack_needle/2 will succeed multiple times, then will allow for counting occurrences:
?- aggregate(count, haystack_needle(t(a,x,b,[x,x,x]),x), C).
C = 4.
Note, the needle does not need to be an atom...
You can use once/1 to tell if needle appears in haystack:
?- once(haystack_needle(t(a,x,b,[x,x,x]),x)).
true.
?- once(haystack_needle(t(a,x,b,[x,x,x]),z)).
false.
As noted by Paulo, arg/3 could not act appropriately in ISO compliant Prologs.
A portable alternative (with arguments swapped without other purpose than avoiding confusion) could be
needle_haystack(N,H) :- N==H.
needle_haystack(N,H) :- H=..[_|As],member(A,As),needle_haystack(N,A).
Of course, if member/2 is available :)

Duplicate constraints in CLP(FD) and with dif/2

In SWI-Prolog, the following query gives this result:
?- X mod 2 #= 0, X mod 2 #= 0.
X mod 2#=0,
X mod 2#=0.
While correct, there is obviously no need for the second constraint
Similarly:
?- dif(X,0), dif(X,0).
dif(X, 0),
dif(X, 0).
Is there no way to avoid such duplicate constraints? (Obviously the most correct way would be to not write code that leads to that situation, but it is not always that easy).
You can either avoid posting redundant constraints, or remove them with a setof/3-like construct. Both are very implementation specific. The best interface for such purpose is offered by SICStus. Other implementations like YAP or SWI more or less copied that interface, by leaving out some essential parts. A recent attempt to overcome SWI's deficiencies was rejected.
Avoid posting constraints
In SICStus, you can use frozen/2 for this purpose:
| ?- dif(X,0), frozen(X,Goal).
Goal = prolog:dif(X,0),
prolog:dif(X,0) ? ;
no
| ?- X mod 2#=0, frozen(X, Goal).
Goal = clpfd:(X in inf..sup,X mod 2#=0),
X mod 2#=0,
X in inf..sup ? ;
no
Otherwise, copy_term/3 might be good enough, provided the constraints are not too much interconnected with each other.
Eliminate redundant constraints
Here, a setof-like construct together with call_residue_vars/1 and copy_term/3 is probably the best approach. Again, the original implementation is in SICStus....
For dif/2 alone, entailment can be tested without resorting to any internals:
difp(X,Y) :-
( X \= Y -> true
; dif(X, Y)
).
Few constraint programming systems implement contraction also related to factoring in resolution theorem proving, since CLP labeling is not SMT. Contraction is a structural rule, and it reads as follows, assuming constraints are stored before the (|-)/2 in negated form.
G, A, A |- B
------------ (Left-Contraction)
G, A |- B
We might also extend it to the case where the two A's are derivably equivalent. Mostlikely this is not implemented, since it is costly. For example Jekejeke Minlog already implements contraction for CLP(FD), i.e. finite domains. We find for queries similar to the first query from the OP:
?- use_module(library(finite/clpfd)).
% 19 consults and 0 unloads in 829 ms.
Yes
?- Y+X*3 #= 2, 2-Y #= 3*X.
3*X #= -Y+2
?- X #< Y, Y-X #> 0.
X #=< Y-1
Basically we normalize to A1*X1+..+An*Xn #= B respectively A1*X1+..+An*Xn #=< B where gcd(A1,..,An)=1 and X1,..,Xn are lexically ordered, and then we check whether there is already the same constraint in the constraint store. But for CLP(H), i.e. Herbrand domain terms, we have not yet implemented contraction. We are still deliberating an efficient algorithm:
?- use_module(library(term/herbrand)).
% 2 consults and 0 unloads in 35 ms.
Yes
?- neq(X,0), neq(X,0).
neq(X, 0),
neq(X, 0)
Contraction for dif/2 would mean to implement a kind of (==)/2 via the instantiation defined in the dif/2 constraint. i.e. we would need to apply a recursive test following the pairing of variables and terms defined in the dif/2 constraint against all other dif/2 constraints already in the constraint store. Testing subsumption instead of contraction would also make more sense.
It probably is only feasible to implement contraction or subsumption for dif/2 with the help of some appropriate indexing technique. In Jekejeke Minlog for example for CLP(FD) we index on X1, but we did not yet realize some indexing for CLP(H). What we first might need to figure out is a normal form for the dif/2 constraints, see also this problem here.

How to properly express inequality in prolog?

TL;DR: sibling(a,X) succeeds with the answer X = a, but sibling(a,a) fails.
I have the following Prolog file:
children(a, c).
children(a, d).
children(b, c).
children(b, d).
sibling(X, Y) :-
X \== Y, A \== B,
children(X, A), children(X, B),
children(Y, A), children(Y, B).
It seems clear enough to me, two person are siblings if their parents are the same. Also, a person is not their own sibling.
But when I tried to run some queries on GNU Prolog, I get some strange results:
| ?- sibling(a, b).
true ? a
true
true
yes
This is the intended behavior. a and b are siblings. There are three results, which is a bit weird, but I assume Prolog is binding A = c, B = d and A = d, B = c.
| ?- sibling(a, a).
no
I think this means a and a are not siblings.
| ?- sibling(a, X).
X = a ? a
X = b
X = a
X = b
X = a
X = b
X = a
X = b
(15 ms) yes
This is where I got stuck: It says X = a, which means sibling(a,a) is true, but sibling(a,a) failed in the previous query!
I feel that I'm not understanding what \== actually does in Prolog.
What is happening, and how do I fix this?
TL;DR: Use prolog-dif—or iso_dif/2 (on iso-prolog conforming systems like gnu-prolog)!
Good question, +1!
In fact, it's a question I have asked myself, and the answer has to do with logical-purity: logical purity is a central aspect of what
makes Prolog as a language so special, as it enables you to:
describe—not prescribe
write code that is relational—not just functional
think in terms of problems / solutions—not the individual steps of the search process itself
operate on a higher level—not get lost in nitty-gritty details
Unlike many other programming languages, Prolog programs have both procedural semantics (defining the execution steps and their order) and declarative semantics (allowing you to state relations that should hold and let the Prolog processor find a proper way of execution by itself).
But, beware: Prolog has some features that, when used, ruin declarative semantics. To prevent this, try to structure your application into two parts: an impure shell for dealing with side-effects (input/output) and a logically pure base which comprises pure monotonic Prolog code.
Try moving the inequality to the end of the predicate. Maybe it gives you true because it's not instantiated already.
sibling(X,Y):- children(X, A), children(X, B),
children(Y, A), children(Y, B),
X \== Y, A \== B.

How to prevent Prolog from backtracking where it shouldn't

I'm trying to solve a CSP where I need to distribute cocktails over bartenders so that each bartender has at most one cocktail and all cocktails are given a bartender. I solved it by creating a list of clpfd variables,first giving them the full domain of all bartenders and then removing all bartenders that don't know how to make that cocktail.
My code works, but there is one problem: it's too slow. If I look in the profiler, remove_domain gets called 2000 times(for the input I'm giving my program), while it's Redo statistic is >100 000.
What do I need to change in one of these functions(or both) so that prolog doesn't need to backtrack?
produce_domains(_,_,[],[]) :- !.
produce_domains(Bartenders,NBartenders,[Cocktail|Cocktails],[Var|Vars]) :-
Var in 1..NBartenders,
remove_domain(Bartenders,NBartenders,Cocktail,Var),!,
produce_domains(Bartenders,NBartenders,Cocktails,Vars),!.
remove_domain([],0,_,_) :- !.
remove_domain([Bartender|Bartenders],NBartenders,Cocktail,Var) :-
(\+ member(Cocktail,Bartender) -> Var #\= NBartenders;!),!,
NNBartenders is NBartenders - 1,
remove_domain(Bartenders,NNBartenders,Cocktail,Var),!.
I have already read this related question, but I am using the latest Windows build of SWI-Prolog(5.10.5), so that shouldn't be the problem here.
You do not need so many !/0: Prolog can often tell that your predicates are deterministic.
Let me first offer the following version of your code. It uses names that are more relational, contains no !/0 and uses higher-order predicates to make the code shorter.
:- use_module(library(clpfd)).
bartenders_cocktails_variables(Bs, Cs, Vs) :-
length(Bs, LBs),
maplist(bartenders_cocktail_variable(Bs, LBs), Cs, Vs).
bartenders_cocktail_variable(Bs, N, C, V) :-
V in 1..N,
foldl(compatible_bartender(C,V), Bs, 1, _).
compatible_bartender(C, V, Cs, N0, N1) :-
( member(C, Cs) -> true
; V #\= N0
),
N1 #= N0 + 1.
Notice that I am counting upwards instead of downwards to enumerate the bartenders (which are just lists of cocktails they are able to mix), since this seems more natural. I was also able to omit a (\+)/1 by simply switching the branches of the if-then-else.
Example query, showing that the predicate is deterministic in this use case:
?- bartenders_cocktails_variables([[a,b],[a,b],[x,y]], [x,a,b], Vars).
Vars = [3, _G1098, _G1101],
_G1098 in 1..2,
_G1101 in 1..2.
We see: Cocktail x must be mixed by the third bartender etc.
I think this part of your program may not be responsible for the slow performance you are describing. Maybe other parts of your program are (unintentionally) not deterministic? Maybe try different labeling strategies or other constraints? We may be able to help you more if you post more context.

Are HiLog terms still useful in modern Prolog?

Are Hilog terms (i.e. compounds having as functors arbitrary terms) still regarded as a powerful feature in XSB Prolog (or any other Prolog) ?
Are there many XSB projects currently using this feature ? which of them for example ?
I ask since as far as I understand higher order programming is equally possible using the ISO built-in call/N.
Specifically, I would like to understand if XSB is using Hilog terms just for historical reasons or if Hilog terms have considerable advantages in comparison to the current ISO standard.
Within XSB, Hilog terms are very strongly connected to the module system which is unique to XSB. XSB has a functor based module system. That is, within the same scope length(X) might belong to one module, whereas length(L, N) might belong to another. As a consequence, call(length(L), N) might refer to one module and call(length(L, N)) to another:
[Patch date: 2013/02/20 06:17:59]
| ?- use_module(basics,length/2).
yes
| ?- length(Xs,2).
Xs = [_h201,_h203]
yes
| ?- call(length(Xs),2).
Xs = [_h217,_h219]
yes
| ?- use_module(inex,length/1).
yes
| ?- length(Xs,2).
Xs = [_h201,_h203]
yes
| ?- call(length(Xs),2).
++Error[XSB/Runtime/P]: [Existence (No module inex exists)] in arg 1 of predicate load
| ?- call(call(length,Xs),2).
Xs = [_h228,_h230];
It might be that in such a context there are differences between call/N and Hilog terms. I have, however, so far not found one.
Historically, Hilog terms have been introduced 1987-1989. At that point in time, call/N already existed as built-ins in NU and as library(call) in Quintus Prolog with only cursory documentation. It has been proposed 1984 by Richard O'Keefe. On the other hand, call/N was clearly unknown to the authors of Hilog, as is exemplified on p.1101 of Weidong Chen, Michael Kifer, David Scott Warren: HiLog: A First-Order
Semantics for Higher-Order Logic Programming Constructs. NACLP
1989. 1090-1114. MIT-Press.
... Generic transitive closure can also be defined in Prolog:
closure(R, X, Y) :- C =.. [R, X, Y], call(C).
closure(R, X, Y) :- C =.. [R, X, Z], call(C), closure(R, Z, Y).
However, this is obviously inelegant compared to HiLog (see Section 2.1), since this involves both constructing a term out of a list and reflecting this term into an atomic formula using "call". The point of this example is that the lack of theoretical foundations for higher-order constructs in Prolog resulted in an obscure syntax, which partially explains why Prolog programs involving such constructs are notoriously hard to understand.
Now, this can be done with call/N like so:
closure(R, X, Y) :- call(R, X, Y).
closure(R, X, Y) :- call(R, X, Z), closure(R, Z, Y).
Which is even more general than the (=..)/2-version because R is no longer restricted to being an atom. As an aside, I'd rather prefer to write:
closure(R_2, X0,X) :- call(R_2, X0,X1), closure0(R_2, X1,X).
closure0(_R_2, X,X).
closure0(R_2, X0,X) :- call(R_2, X0,X1), closure0(R_2, X1,X).
HiLog allows goals such as
foo(X(a, Y(b))).
and ISO Prolog does not. In ISO Prolog, you'd have to write
foo(T), T=..[X, a, R], R=..[Y, b].
which is less convenient and might be slower.

Resources