The question is : The salt has been stolen! Well, it was found that the culprit was either the Caterpillar,
Bill the Lizard or the Cheshire Cat. The three were tried and made the following
statements in court:
CATERPILLAR: Bill the Lizard ate the salt.
BILL THE LIZARD: That is true!
CHESHIRE CAT: I never ate the salt.
As it happened, at least one of them lied and at least one told the truth. Who ate the
salt?
I know for sure if bill is true, than all statements are true, and if cheshire is true, then all are false, so it must be the caterpillar.
Looking at in predicate calculus and programming it, it would be something like this right:
suspect(caterpillar).
suspect(lizard).
suspect(cat).
:- suspect(cat), suspect(lizard).
:- suspect(cat), suspect(caterpillar).
:- suspect(lizard), suspect(caterpillar).
%where these imply not more than one of these can be true or returned in our set
But then further describing this in predicate logic I don't how I would describe the descriptions or plea's they have made. And how that if one statement is true can imply that others may be falses.
One nice thing about this puzzle is that you do not even need first-order predicate logic to model it: It suffices to use propositional logic, because whether a suspect lies or tells the truth can be indicated with a Boolean variable, and the statements themselves are also only statements over Boolean variables.
Thus, consider using a constraint solver over Boolean variables when solving this task with Prolog. See clpb for more information about this.
Here is a sample solution, using SICStus Prolog or SWI:
:- use_module(library(clpb)).
solution(Pairs) :-
Suspects = [_Caterpillar,Lizard,Cat],
pairs_keys_values(Pairs, [caterpillar,lizard,cat], Suspects),
Truths = [CaterpillarTrue,LizardTrue,CatTrue],
% exactly one of them ate the salt
sat(card([1], Suspects)),
% the statements
sat(CaterpillarTrue =:= Lizard),
sat(LizardTrue =:= Lizard),
sat(CatTrue =:= ~Cat),
% at least one of them tells the truth:
sat(card([1,2,3], Truths)),
% at least one of them lies:
sat(card([1,2,3], [~CaterpillarTrue,~LizardTrue,~CatTrue])).
And from this the unique solution is readily determined without search:
?- solution(Pairs).
Pairs = [caterpillar-1, lizard-0, cat-0].
Related
In many Prolog guides the following code is used to illustrate "negation by failure" in Prolog.
not(Goal) :- call(Goal), !, fail.
not(Goal).
However, those same tutorials and texts warn that this is not "logical negation".
Question: What is the difference?
I have tried to read those texts further, but they don't elaborate on the difference.
I like #TesselatingHeckler's answer because it puts the finger on the heart of the matter. You might still be wondering, what that means for Prolog in more concrete terms. Consider a simple predicate definition:
p(something).
On ground terms, we get the expected answers to our queries:
?- p(something).
true.
?- \+ p(something).
false.
?- p(nothing).
false.
?- \+ p(nothing).
true.
The problems start, when variables and substitution come into play:
?- \+ p(X).
false.
p(X) is not always false because p(something) is true. So far so good. Let's use equality to express substitution and check if we can derive \+ p(nothing) that way:
?- X = nothing, \+ p(X).
X = nothing.
In logic, the order of goals does not matter. But when we want to derive a reordered version, it fails:
?- \+ p(X), X = nothing.
false.
The difference to X = nothing, \+ p(X) is that when we reach the negation there, we have already unified X such that Prolog tries to derive \+p(nothing) which we know is true. But in the other order the first goal is the more general \+ p(X) which we saw was false, letting the whole query fail.
This should certainly not happen - in the worst case we would expect non-termination but never failure instead of success.
As a consequence, we cannot rely on our logical interpretation of a clause anymore but have to take Prolog's execution strategy into account as soon as negation is involved.
Logical claim: "There is a black swan".
Prolog claim: "I found a black swan".
That's a strong claim.
Logical negation: "There isn't a black swan".
Prolog negation: "I haven't found a black swan".
Not such a strong a claim; the logical version has no room for black swans, the Prolog version does have room: bugs in the code, poor quality code not searching everywhere, finite resource limits to searching the entire universe down to swan size areas.
The logical negation doesn't need anyone to look anywhere, the claim stands alone separate from any proof or disproof. The Prolog logic is tangled up in what Prolog can and cannot prove using the code you write.
There are a couple of reasons why,
Insufficient instantiation
A goal not(Goal_0) will fail, iff Goal0 succeeds at the point in time when this not/1 is executed. Thus, its meaning depends on the very instantiations that happen to be present when this goal is executed. Changing the order of goals may thus change the outcome of not/1. So conjunction is not commutative.
Sometimes this problem can be solved by reformulating the actual query.
Another way to prevent incorrect answers is to check if the goal is sufficiently instantiated, by checking that say ground(Goal_0) is true producing an instantiation error otherwise. The downside of this approach is that quite too often instantiation errors are produced and people do not like them.
And even another way is to delay the execution of Goal_0 appropriately. The techniques to improve the granularity of this approach are called constructive negation. You find quite some publications about it but they have not found their way into general Prolog libraries. One reason is that such programs are particularly hard to debug when many delayed goals are present.
Things get even worse when combining Prolog's negation with constraints. Think of X#>Y,Y#>X which does not have a solution but not/1 just sees its success (even if that success is conditional).
Semantic ambiguity
With general negation, Prolog's view that there exists exactly one minimal model no longer holds. This is not a problem as long as only stratified programs are considered. But there are many programs that are not stratified yet still correct, like a meta-interpreter that implements negation. In the general case there are several minimal models. Solving this goes far beyond Prolog.
When learning Prolog stick to the pure, monotonic part first. This part is much richer than many expect. And you need to master that part in any case.
I'm working through Clocksin and Mellish to try and finally go beyond just dabbling in Prolog. FWIW, I'm running SWI-Prolog:
SWI-Prolog version 7.2.3 for x86_64-linux
Anyway, I implemented a diff/2 predicate as part of exercise 1.4. The predicate is very simple:
diff(X,Y) :- X \== Y.
And it works when used in the sister_of predicate, like this:
sister_of(X,Y) :-
female(X),
diff(X,Y),
parents(X, Mum, Dad ),
parents(Y, Mum, Dad ).
in that, assuming the necessary additional facts, doing this:
?- sister_of(alice,alice).
returns false as expected. But here's the rub. If I do this instead:
?- sister_of(alice, Who).
(again, given the additional facts necessary)
I get
Who = edward ;
Who = alice;
false
Even though, as already shown, the sister_of predicate does not treat alice as her own sister.
On the other hand, if I use the SWI provided dif/2 predicate, then everything works the way I would naively expect.
Can anyone explain why this is happening this way, and why my diff implementation doesn't work the way I'm expecting, in the case where I ask for additional unifications from that query?
The entire source file I'm working with can be found here
Any help is much appreciated.
As you note, the problem stems from the interplay between equality (or rather, inequality) and unification. Observe that in your definition of sister_of, you first find a candidate value for X, then try to constrain Y to be different, but Y is still an uninstantiated logic variable and the check is always going to succeed, like diff(alice, Y) will. The following constraints, including the last one that gives a concrete value to Y, come too late.
In general, what you need to do is ensure that by the time you get to the inequality check all variables are instantiated. Negation is a non-logical feature of Prolog and therefore potentially dangerous, but checking whether two ground terms are not equal is safe.
I got a database that looks like
hasChild(person1, person2).
hasChild(person1, person3).
hasChild(person4, person5).
Which means that (for example) person1 has child named person2.
I then create a predicate that identifies if the person is a parent
parent(A):- hasChild(A,_).
Which identifies if the person is a parent, i.e. has any children
Then I try to create a predicate childless(A) that should return true if the user doesn't have any children which is basically an inverse of parent(A).
So I have 2 questions here:
a) is it possible to somehow take an "inverse" of a predicate, like childless(A):-not(parent(A)). or in any other way bypass this using the hasChild or any other method?
b) parent(A) will return true multiple times if the person has multiple children. Is it possible to make it return true only once?
For problem 1, yes. Prolog is not entirely magically delicious to some because it conflates negation and failure, but you can definitely write:
childless(X) :- \+ hasChild(X, _).
and you will see "true" for people that do not have children. You will also see "true" for vegetables, minerals, ideologies, procedures, shoeboxes, beer recipes and unfeathered bipeds. If this is a problem for you, a simple solution is to improve your data model, but complaining about Prolog is a very popular alternative. :)
For problem 2, the simplest solution is to use once:
parent(A) :- once(hasChild(A, _)).
This is a safer alternative to using the cut operator, which would look like this:
parent(A) :- hasChild(A, _), !.
This has a fairly significant cost: parent/1 will only generate a single valid solution, though it will verify other correct solutions. To wit:
?- parent(X).
X = person1.
Notice it did not suggest person4. However,
?- childless(person4).
true.
This asymmetry is certainly a "code smell" to most any intermediate Prolog programmer such as myself. It's as though Prolog has some sort of amnesia or selective hearing depending on the query. This is no way to get invited to high society events!
I would suggest that the best solution here (which handles the mineral/vegetable problem above as well) is to add some more facts about people. After all, a person exists before they have kids (or do they?) so they are not "defined" by that relationship. But continuing to play the game, you may be able to circumvent the problem using setof/3 to construct a list of all the people:
parent(Person) :-
setof(X, C^hasChild(X, C), People),
member(Person, People).
The odd expression C^hasChild(X, C) tells Prolog that C is a free variable; this ensures that we get the set of all things in the first argument of hasChild/2 bound to the list People. This is not first-order logic anymore folks! And the advantage here is that member/2 will generate for us as well as check:
?- parent(person4).
true.
?- parent(X).
X = person1 ;
X = person4.
Is this efficient? No. Is it smart? Probably not. Is it a solution to your question that also generates? Yes, it seems to be. Well, one out of three ain't bad. :)
As a final remark, some Prolog implementations treat not/1 as an alias for \+/1; if you happen to be using one of them, I recommend you not mistake compatibility with pre-ISO conventions for a jovial tolerance for variety: correct the spelling of not(X) to \+ X. :)
Here's another way you could do it!
Define everything you know for a fact as a Prolog fact, no matter if it is positive or negative.
In your sample, we define "positives" like person/1 and "negatives" like childless/1. Of course, we also define the predicates child_of/2, male/1, female/1, spouse_husband/2, and so on.
Note that we have introduced quite a bit of redundancy into the database.
In return, we got a clearer line of knowns/unknowns without resorting to higher-order constructs.
We need to define the right data consistency constraints:
% There is no person which is neither male nor female.
:- \+ (person(X), \+ (male(X) ; female(X))).
% Nobody is male and female (at once).
:- \+ (male(X), female(X)).
% Nobody is childless and parental (at once).
:- \+ (childless(X), child_of(_,X)).
% There is no person which is neither childless nor parental.
:- \+ (person(X), \+ (childless(X) ; child_of(_,X))).
% There is no child which is not a person.
:- \+ (child_of(X,_), \+ person(X)).
% There is no parent which is not a person.
:- \+ (child_of(_,X), \+ person(X)).
% (...plus, quite likely, a lot more integrity constraints...)
This is just a rough sketch... Depending on your use-cases you could do the modeling differently, e.g. using relations like parental/1 together with suitable integrity constraints. YMMY! HTH
The matter of deterministic success of some Prolog goal has turned up time and again in—at least—the following questions:
Reification of term equality/inequality
Intersection and union of 2 lists
Remove duplicates in list (Prolog)
Prolog: How can I implement the sum of squares of two largest numbers out of three?
Ordering lists with constraint logic programming)
Different methods were used (e.g., provoking certain resource errors, or looking closely at the exact answers given by the Prolog toplevel), but they all appear somewhat ad-hack to me.
I'm looking for a generic, portable, and ISO-conformant way to find out if the execution of some Prolog goal (which succeeded) left some choice-point(s) behind. Some meta predicate, maybe?
Could you please hint me in the right direction? Thank you in advance!
Good news everyone: setup_call_cleanup/3 (currently a draft proposal for ISO) lets you do that in a quite portable and beautiful way.
See the example:
setup_call_cleanup(true, (X=1;X=2), Det=yes)
succeeds with Det == yes when there are no more choice points left.
EDIT: Let me illustrate the awesomeness of this construct, or rather of the very closely related predicate call_cleanup/2, with a simple example:
In the excellent CLP(B) documentation of SICStus Prolog, we find in the description of labeling/1 a very strong guarantee:
Enumerates all solutions by backtracking, but creates choicepoints only if necessary.
This is really a strong guarantee, and at first it may be hard to believe that it always holds. Luckily for us, it is extremely easy to formulate and generate systematic test cases in Prolog to verify such properties, in essence using the Prolog system to test itself.
We start with systematically describing what a Boolean expression looks like in CLP(B):
:- use_module(library(clpb)).
:- use_module(library(lists)).
sat(_) --> [].
sat(a) --> [].
sat(~_) --> [].
sat(X+Y) --> [_], sat(X), sat(Y).
sat(X#Y) --> [_], sat(X), sat(Y).
There are in fact many more cases, but let us restrict ourselves to the above subset of CLP(B) expressions for now.
Why am I using a DCG for this? Because it lets me conveniently describe (a subset of) all Boolean expressions of specific depth, and thus fairly enumerate them all. For example:
?- length(Ls, _), phrase(sat(Sat), Ls).
Ls = [] ;
Ls = [],
Sat = a ;
Ls = [],
Sat = ~_G475 ;
Ls = [_G475],
Sat = _G478+_G479 .
Thus, I am using the DCG only to denote how many available "tokens" have already been consumed when generating expressions, limiting the total depth of the resulting expressions.
Next, we need a small auxiliary predicate labeling_nondet/1, which acts exactly as labeling/1, but is only true if a choice-point still remains. This is where call_cleanup/2 comes in:
labeling_nondet(Vs) :-
dif(Det, true),
call_cleanup(labeling(Vs), Det=true).
Our test case (and by this, we actually mean an infinite sequence of small test cases, which we can very conveniently describe with Prolog) now aims to verify the above property, i.e.:
If there is a choice-point, then there is a further solution.
In other words:
The set of solutions of labeling_nondet/1 is a proper subset of that of labeling/1.
Let us thus describe what a counterexample of the above property looks like:
counterexample(Sat) :-
length(Ls, _),
phrase(sat(Sat), Ls),
term_variables(Sat, Vs),
sat(Sat),
setof(Vs, labeling_nondet(Vs), Sols),
setof(Vs, labeling(Vs), Sols).
And now we use this executable specification in order to find such a counterexample. If the solver works as documented, then we will never find a counterexample. But in this case, we immediately get:
| ?- counterexample(Sat).
Sat = a+ ~_A,
sat(_A=:=_B*a) ? ;
So in fact the property does not hold. Broken down to the essence, although no more solutions remain in the following query, Det is not unified with true:
| ?- sat(a + ~X), call_cleanup(labeling([X]), Det=true).
X = 0 ? ;
no
In SWI-Prolog, the superfluous choice-point is obvious:
?- sat(a + ~X), labeling([X]).
X = 0 ;
false.
I am not giving this example to criticize the behaviour of either SICStus Prolog or SWI: Nobody really cares whether or not a superfluous choice-point is left in labeling/1, least of all in an artificial example that involves universally quantified variables (which is atypical for tasks in which one uses labeling/1).
I am giving this example to show how nicely and conveniently guarantees that are documented and intended can be tested with such powerful inspection predicates...
... assuming that implementors are interested to standardize their efforts, so that these predicates actually work the same way across different implementations! The attentive reader will have noticed that the search for counterexamples produces quite different results when used in SWI-Prolog.
In an unexpected turn of events, the above test case has found a discrepancy in the call_cleanup/2 implementations of SWI-Prolog and SICStus. In SWI-Prolog (7.3.11):
?- dif(Det, true), call_cleanup(true, Det=true).
dif(Det, true).
?- call_cleanup(true, Det=true), dif(Det, true).
false.
whereas both queries fail in SICStus Prolog (4.3.2).
This is the quite typical case: Once you are interested in testing a specific property, you find many obstacles that are in the way of testing the actual property.
In the ISO draft proposal, we see:
Failure of [the cleanup goal] is ignored.
In the SICStus documentation of call_cleanup/2, we see:
Cleanup succeeds determinately after performing some side-effect; otherwise, unexpected behavior may result.
And in the SWI variant, we see:
Success or failure of Cleanup is ignored
Thus, for portability, we should actually write labeling_nondet/1 as:
labeling_nondet(Vs) :-
call_cleanup(labeling(Vs), Det=true),
dif(Det, true).
There is no guarantee in setup_call_cleanup/3 that it detects determinism, i.e. missing choice points in the success of a goal. The 7.8.11.1 Description draft proposal only says:
c) The cleanup handler is called exactly once; no later than
upon failure of G. Earlier moments are:
If G is true or false, C is called at an implementation
dependent moment after the last solution and after the last
observable effect of G.
So there is currently no requirement that:
setup_call_cleanup(true, true, Det=true)
Returns Det=true in the first place. This is also reflected in the test cases 7.8.11.4 Examples that the draf proposal gives, we find one test case which says:
setup_call_cleanup(true, true, X = 2).
Either: Succeeds, unifying X = 2.
Or: Succeeds.
So its both a valid implementation, to detect determinism and not to detect determinism.
Clue
Four guests (Colonel Mustard, Professor Plum, Miss Scarlett, Ms. Green) attend a dinner party at the home of Mr. Boddy. Suddenly, the lights go out! When they come back, Mr Boddy lies dead in the middle of the table. Everyone is a suspect. Upon further examination, the following facts come to light:
Mr Boddy was having an affair with Ms. Green.
Professor Plum is married to Ms. Green.
Mr. Boddy was very rich.
Colonel Mustard is very greedy.
Miss Scarlett was also having an affair with Mr. Boddy.
There are two possible motives for the murder:
Hatred: Someone hates someone else if that other person is having an affair with his/her spouse.
Greed: Someone is willing to commit murder if they are greedy and not rich, and the victim is rich.
Part A: Write the above facts and rules in your Prolog program. Use the following names for the people: colMustard, profPlum, missScarlet, msGreen, mrBoddy. Be careful about how you encode (or don’t encode) symmetric relationships like marriage - you don’t want infinite loops! married(X,Y) :- married(Y,X) % INFINITE LOOP
?-suspect(Killer,mrBoddy)
Killer = suspect_name_1
Killer = suspect_name_2
etc.
Part B: Write a predicate, suspect/2, that determines who the suspects may be, i.e. who had a motive.
?-suspect(Killer,mrBoddy)
Killer = unique_suspect.
Part C: Add a single factto your database that will result in there being a unique suspect.
Clearly indicate this line in your source comments so that it can be removed/added for
grading.
?-suspect(Killer,mrBoddy)
Killer = unique_suspect.
Whenever I type in
suspect(Killer,mrBoddy).
I get
suspect(Killer,mrBoddy).
Killer = profPlum
I'm missing
Killer = colMustard.
Here's my source.
%8) Clue
%facts
affair(mrBoddy,msGreen).
affair(missScarlett, mrBoddy).
affair(X,Y) :- affair(X,Y), affair(Y,X).
married(profPlum, msGreen).
married(X,Y) :- married(X,Y), married(Y,X).
rich(mrBoddy).
greedy(colMustard).
%rules
hate(X,Y) :- married(X,Spouse), affair(Y,Spouse).
greed(X,Y) :- greedy(X), not(rich(X)), rich(Y).
%suspect
suspect(X,Y):- hate(X,Y).
suspect(X,Y):- greed(X,Y).
There are two kinds of problems with your program. One is on the procedural level: you observed that Prolog loops; the other is on the logical level — Prolog people call this rather the declarative level. Since the first annoying thing is this endless loop, let's first narrow that down. Actually we get:
?- suspect(Killer,mrBoddy).
Killer = profPlum ;
ERROR: Out of local stack
You have now several options to narrow down this problem. Either, go with the other answer and call up a tracer. While the tracer might show you the actual culprit it might very well intersperse it with many irrelevant steps. So many that your mind will overflow.
The other option is to manually modify your program by adding goals false into your program. I will add as many false goals as I can while still getting a loop. The big advantage is that this way you will see in your source the actual culprit (or to be more precise one of potentially many such culprits).1 After trying a bit, this is what I got as failure-slice:
?- suspect(Killer,mrBoddy), false.
married(profPlum, msGreen) :- false.
married(X,Y) :- married(X,Y), false, married(Y,X).
hate(X,Y) :- married(X,Spouse), false, affair(Y,Spouse).
suspect(X,Y):- hate(X,Y), false.
suspect(X,Y):- false, greed(X,Y).
All remaining parts of your program were irrelevant, that is, they are no longer used. So essentially the rule
married(X,Y) :- married(X,Y), married(Y,X).
is the culprit.
Now, for the declarative part of it. What does this rule mean anyway? To understand it, I will interpret :- as an implication. So provided what is written on the right-hand side is true, we conclude what is written on the left-hand side. In this case:
Provided X is married to Y and Y is married to X
we can conclude that
X is married to Y.
This conclusion concluded what we have assumed to be true anyway. So it does not define anything new, logically. You can just remove the rule to get same results — declaratively. So married(profPlum, msGreen) holds but married(msGreen, profPlum) does not. In other words, your rules are not correct, as you claim.
To resolve this problem, remove the rule, rename all facts to husband_wife/2 and add the definition
married(M,F) :- husband_wife(M,F).
married(F,M) :- husband_wife(M,F).
So the actual deeper problem here was a logical error. In addition to that Prolog's proof mechanism is very simplistic, turning this into a loop. But that is not much more than a welcome excuse to the original logical problem.2
Footnotes:1 This method only works for pure, monotonic fragments. Non-monotonic constructs like not/1 or (\+)/1 must not appear in the fragment.
2 This example is of interest to #larsmans.
The problem is the recursive rules of the predicates affair/2 and married/2. Attempting to use them easily leads to an endless loop (i.e. until the stack memory is exhausted). You must use a different predicate in each case to represent that if X is having an affair with Y, then Y is having an affair with X. You also need to change your definition of the suspect/2 predicate to call those new predicates.
To better understand why you get an endless loop, use the trace facilities of your Prolog system. Try:
?- trace, suspect(Killer, mrBoddy).
and go step by step.