I'm trying (failing) to understand an exercise where I'm given the following clauses;
pterm(null).
pterm(f0(X)) :- pterm(X).
pterm(f1(X)) :- pterm(X).
They represent a number in binary, eg. f0(null) is equivalent to 0, f1(null) is equivalent to 1, etc.
The objective is to define a predicate over pterm such that one is the successor of the other when true. It seems like a relatively simple exercise but I'm struggling to get my head around it.
Here is the code I've written so far;
incr(X,Y) :- pterm(f0(X)), pterm(f1(Y)).
incr(X,Y) :- pterm(f0(f1(X))), pterm(f1(f1(Y))).
Having tested this I know it's very much incorrect. How might I go about inspecting the top level arguments of each pterm?
I've made minimal progress in the last 4 hours so any hints/help would be appreciated.
1)
I'll start with the "how to inspect" question, as I think it will be the most useful. If you're using swi-prolog with xpce, run the guitracer:
?- consult('pterm'). % my input file
% pterm compiled 0.00 sec, 5 clauses
true.
?- guitracer.
% The graphical front-end will be used for subsequent tracing
true.
?- trace. % debugs step by step
true.
[trace] ?- pterm(f0(f1(null))). % an example query to trace
true.
A graphical interface will come up. Press the down arrow to unify things step by step. What's going on should make sense fairly quickly.
(use notrace. and nodebug. appropriately to exit trace and debug modes afterwards).
2) You seem to misunderstand how predicates work. A predicate is a logical statement, i.e. it will always return either true or false. You can think of them as classical boolean functions of the type "iseven(X)" (testing if X is even) or "ismemberof(A,B)" (testing if A is a member of B) etc. When you have a rule like "pred1 :- pred2, pred3." this is similar to saying "pred1 will return true if pred2 returns true, and pred3 returns true (otherwise pred1 returns false)".
When your predicates are called using constants, checking its truth value is a matter of checking your facts database to see if that predicate with those constants can be satisfied. But when you call using variables, prolog goes through a wild goose chase, trying to unify that variable with all the allowable stuff it can link it to, to see if it can try to make that predicate true. If it can't, it gives up and says it's false.
A predicate like incr(X,Y) is still something that needs to return true or false, but, if by design, this only becomes true when Y is the incremented version of X, where X is expected to be given at query time as input, then we have tricked prolog into making a "function" that is given X as input, and "returns" Y as output, because prolog will try to find an appropriate Y that makes the predicate true.
Therefore, with your example, incr(X,Y) :- pterm(f0(X)), pterm(f1(Y)). makes no sense, because you're telling it that incr(X,Y) will return true for any X,Y, as long as prolog can use X to find in the fact database any pterm(f0(X)) that will lead to a known fact, and also use Y to find a pterm(f1(Y)) term. You haven't made Y dependent on X in any way. This query will succeed for X = null, and Y = null, for instance.
Your first clause should be something like this.
incr(X,Y) :- X = pterm(f0(Z)), Y = pterm(f1(Z)).
where = performs unification. I.e. "find a value for Z such that X is pterm(f0(Z)), and for the same value of Z it also applies that Y = pterm(f1(Z))."
In fact, this could be more concisely rewritten as a fact:
incr( pterm(f0(Z)), pterm(f1(Z)) ).
3)
Your second clause can be adapted similarly. However, I'm not sure if this is correct in terms of the logic of what you're trying to achieve (i.e. binary arithmetic). But I may have misunderstood the problem you're trying to solve.
My assumption is that if you have (0)111, then the successor should be 1000, not 1111. For this, I would guess you need to create a predicate that recursively checks if the incrementation of the digits below the currently processed one results in a 'carried' digit.
(since the actual logic is what your assignment is about, I won't offer a solution here. but hope this helps get you into grips with what's going on. feel free to have a go at the recursive version and ask another question based on that code!)
Related
Say we were to execute the following, and SWI Prolog responds:
?- write(X).
_13074
true.
What is _13074? Is this an address? Is this an ID of some sort? I notice that we'll get a different value each time. Furthermore, why does Prolog report true.? Is this Prolog saying that anything can be unified with X? Why do these appear in the order they do?
If we were to unify X with, say, 1, then the result is different.
?- X = 1, write(X).
1
X = 1.
What is _13074?
The simplest answer is that it represents an uninstantiated variable.
To be more precise from the Prolog Standard
anonymous variable: A variable (represented in a term or Prolog text by _) which differs from every other variable (and anonymous
variable) (see 6.1.2, 6.4.3)
instantiated: A variable is instantiated with respect to substitution if application of the substitution yields an atomic term or a compound term.
A term is instantiated if any of its variables are instantiated.
uninstantiated: A variable is uninstantiated when it is not instantiated.
variable: An object which may be instantiated to a term during execution.
named variable: A variable which is not an anonymous variable (see 6.1.2, 6.4.3)
So obviously all of that is self referential but in short by the standard there are anonymous variables _ and named variables, E.g. X, Y, Ls.
Note that the standard does not say what is the difference between _ and variables with numbers in the suffix, E.g. _13074. Those are implementation specific.
The standard does note for writing a term,
7.10.5 Writing a term
When a term Term is output using write-term/3 (8.14.2) the action which is taken is defined by the rules below:
a) If Term is a variable, a character sequence representing that variable is output. The sequence begins with _ (underscore) and the remaining characters are implementation dependent. The same character sequence is used for each occurrence of a particular variable in Term. A different character sequence is used for each distinct variable in Term.
Since you specifically mention SWI-Prolog there are other variable caveats to be aware of:
named singleton variables AKA auxiliary variables
Named singletons start with a double underscore (__) or a single underscore followed by an uppercase letter, E.g., __var or _Var.
Attribute variables - provide a technique for extending the Prolog unification algorithm Holzbaur, 1992 by hooking the binding of attributed variables. There is no consensus in the Prolog community on the exact definition and interface to attributed variables. The SWI-Prolog interface is identical to the one realised by Bart Demoen for hProlog Demoen, 2002. This interface is simple and available on all Prolog systems that can run the Leuven CHR system (see chapter 9 and the Leuven CHR page).
Global variables - are associations between names (atoms) and terms.
I don't plan to dive deeper into variables as one has to start looking at SWI-Prolog C level source code to really get a more accurate understanding, (ref). I also don't plan to add more from the standard as one would eventually have to reproduce the entire standard here just to cover all of the references.
For more definitions from the Prolog standard see: Is this Prolog terminology correct? (fact, rule, procedure, predicate, ...) The answer is a community wiki so most users can add to it and the OP does not get the points, so upvote all you want.
Is this an address?
No
Sometimes you will also see logic variable used but I don't plan to expand on that here, however for the record SWI-Prolog is NOT based on WAM it is based on A Portable Prolog Compiler.
See above 7.10.5 Writing a term
Is this an ID of some sort?
I would not argue with that in a causal conversation about SWI-Prolog but there is enough problems with that simple analogy to split hairs and start a discussion/debate, E.g. can a blob be assigned to a variable? What is numbervars?
See above 7.10.5 Writing a term
I notice that we'll get a different value each time.
The Prolog standard uses the word occurrence.
See above 7.10.5 Writing a term
why does Prolog report true.?
Prolog is a logic language which executes queries (goal) that result in either true or false or the instantiated values of variables, however there can be side effects such as writing to a file, throwing exceptions, etc.
The Prolog standard states
A.2.1.1 The General Resolution Algorithm
The general resolution of a goal G of a database P is defined by the following non-deterministic algorithm:
a) Start with the initial goal G which is an ordered conjunction of
predications.
b) If G is the singleton true then stop (success).
c) Choose a predication A in G (predication-choice)
d) If A is true, delete it, and proceed to step (b).
e) If no renamed clause in P has a head which unifies with A then stop (failure).
f) Choose a freshly renamed clause in P whose head H unifies with A (clause-choice) where σ = MGU(H, A) and B is the body of the clause,
g) Replace in G the predication A by the body B, flatten and apply the substitution σ.
h) Proceed to step (b).
Also see:
Resolution
MGU
Is this Prolog saying that anything can be unified with X?
For very simple Prolog implementations (ref) then the question would make sense. In the real world and even more so with SWI-Prolog were the rubber meets the road I would have to say not in all cases.
For most Prolog code syntactic unification is what is driving what is happening. See: A.2.1.1 The General Resolution Algorithm above. However if you start to think about things like blobs, attributes, threads, exceptions, and so on then you really have to look at what is a variable, even the kind of variable and what that variable can do , E.g.
?- X is true.
ERROR: Arithmetic: `true/0' is not a function
ERROR: In:
ERROR: [10] _4608 is true
ERROR: [9] toplevel_call(user:user: ...) at c:/program files/swipl/boot/toplevel.pl:1117
?- trie_new(A_trie).
A_trie = <trie>(0000000006E71DB0).
?- write(X).
_13074
true.
Why do these appear in the order they do?
write(X). is the goal entered by the user.
The goal is executed which in this case has the side effect of writing to the current output stream stream, E.g. current_output/1, the variable X which for SWI-Prolog for this occurrence of X is uninstantiated and is displayed as _13074.
The logic query ends and the result of the query being logical is either true or false. Since the query executed successfully the result is true.
If we were to unify X with, say, 1, then the result is different.
?- X = 1, write(X).
1
X = 1.
I will presume you are asking why there is no true at the end.
IIRC with SWI-Prolog, if the query starts with a variable and then the query succeeds with the variable being instantiated that will be reported and no true or false will then appear, E.g.
?- X = 1.
X = 1.
?- current_prolog_flag(double_quotes,V).
V = string.
?- X = 1, Y = 1, X = Y.
X = Y, Y = 1.
If however the query succeeds and no variable was instantiated then the query will report true E.g.
?- 1 = 1.
true.
?- current_prolog_flag(double_quotes,string).
true.
If the the query fails the query will report false E.g.
?- X = 1, Y = 2, X = Y.
false.
?- current_prolog_flag(quotes,String).
false.
I suspect this much information will now have you asking for more details but I won't go much deeper than this as SO is not a place to give a lecture condensed into an answer. I will try to clarify what is written but if it needs a lot more detail expect to be requested to post a new separate question.
I know the info from the standard presented here leaves lots of lose ends. If you really want the details from the standard then purchase the standard as those of us who have have done. I know it is not cheap but for questions like this it is the source of the answers.
Looking at the code below:
multiple(X,0).
multiple(X,Y) :- lt(0,X), lt(0,Y), diff(Y,X,D), multiple(X,D).
There happens to be something wrong. For your reference:
lt/2 is whether the first argument is less than the second.
diff/3 is whether the third argument is equal to the first argument minus the second.
lt/2 and diff/3 are defined correctly.
Is there a logical mistake in the definition? Is assuming that 0 is the multiple of every number problematic or is the logical mistake somewhere else? I get correct answers but the query goes to infinite loop I think.
EDIT:
here are the other definitions.
natNum(0).
natNum(s(X)) :- natNum(X).
lt(0,s(X)) :- natNum(X).
lt(s(X),s(Y)) :- lt(X,Y).
sum(0,X,X).
sum(s(X),Y,s(Z)) :- sum(X,Y,Z).
diff(X,Y,Z) :- sum(Z,Y,X).
?- multiple(X, s(s(s(s(s(s(0))))))).
where s(0) is 1, s(s(0)) is 2 etc. It gives all the desired answers for X but after the last answer, it gets stuck. I assume in an infinite recursive loop?
What is happening in your program? Does it loop forever, or does it only take some time since you haven't updated your hardware in recent decades? We cannot tell. (Actually, we could tell by looking at your program, but that is much too complex for the moment).
What we can do with ease is narrow down the source of this costly effort. And this, without a deep understanding of your program. Let's start with the query:
?- multiple(X, s(s(s(s(s(s(0))))))).
X = s(0)
; X = s(s(0))
; X = s(s(s(0)))
; X = s(s(s(s(s(s(0))))))
; loops. % or takes too long
Isn't there an easier way to do this? All this semicolon typing. Instead, simply add false to your query. In this manner the solutions found are no longer shown and we can concentrate on this annoying looping. And, if we're at it, you can also add false goals into your program! By such goals the number of inferences might be reduced (or stays the same). And if the resulting fragment (called a failure-slice) is looping, then this is a reason why your original program loops:
multiple(_X,0) :- false.
multiple(X,Y) :- lt(0,X), false, lt(0,Y), diff(Y,X,D), multiple(X,D).
natNum(0) :- false.
natNum(s(X)) :- natNum(X), false.
lt(0,s(X)) :- natNum(X), false.
lt(s(X),s(Y)) :- false, lt(X,Y).
?- multiple(X, s(s(s(s(s(s(0))))))), false.
loops.
Do your recognize your program? Only those parts remained that are needed for a loop. And, actually in this case, we have an infinite loop.
To fix this, we need to modify something in the remaining, visible part. I'd go for lt/2 whose first clause can be generalized to lt(0, s(_)).
But wait! Why is it OK to generalize away the requirement that we have a natural number? Look at the fact multiple(X,0). which you have written. You have not demanded that X is a natural number either. This kind of over-generalizations often appears in Prolog programs. They improve termination properties at a relatively low price: Sometimes they are too general but all terms that additionally fit into the generalization are not natural numbers. They are terms like any or [a,b,c], so if they appear somewhere you know that they do not belong to the solutions.
So the idea was to put false goals into your program such that the resulting program (failure-slice) still loops. In the worst case you put false at a wrong place and the program terminates. By trial-and-error you get a minimal failure-slice. All those things that are now stroked through are irrelevant! In particular diff/3. So no need to understand it (for the moment). It suffices to look at the remaining program.
I've created a simple Prolog program (using GNU Prolog v1.4.4) with a single fact:
sunny.
When I run the following query:
sunny.
I get:
yes
As I'd expect. When I run this query:
X.
I get:
uncaught exception: error(instantiation_error,top_level/0)
when I expected to get:
X = sunny
Anyone know why?
Prolog is based on first-order logic but X is a second order logic query (the variable stands for a rule head / fact, not only a term): you ask "which predicates can be derived?" or in other words "which formulas are true?". Second order logic is so expressive that we lose many nice properties of first-order logic (*). That's why a second order variable must be sufficiently instantiated to know which rule to try at the time it is called (that's what the error message means). For instance the queries
?- X=member(A,[1,2,3]), X.
and
?- member(A,[1,2,3]).
still allow Prolog to try the definition of the member predicate (in fact the two definitions are equivalent) but
?- X, X=member(A,[1,2,3]).
will throw an exception because at the time X should be derived, we don't know that it's supposed to become the predicate member(A,[1,2,3]).
Your case is much simpler though: you can wrap sunny as a term into a predicate such that Prolog knows which rules to try. The facts
weather(sunny).
weather(rainy).
define the predicate weather such that now we only have a first-order variable as argument in our query:
?- weather(X).
X = sunny ;
X = rainy.
Now that we are talking about the term level, everything works as you expected.
(*) Although the problem of finding out if a formula is valid is undecidable in both cases, in first order logic at least all true formulas can be eventually derived but if a formula is false, the search might not terminate (i.e. first-order logic is semi-decidable). For second order logic there are formulas that can neither be proved not disproved. What is worse is that we cannot even tell if a second-order formula belongs to this category.
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'm taking a crack at Prolog (using SWI-Prolog) and everything works like I want it to, i.e., the logic is calculated correctly and it finds the right solutions but the whole backtracking thing is screwing with me.
Here's the code:
tall(X) :- skinny(X) ; eatless(X).
eatless(X) :- playsmore(X).
playsmore(X) :- hasxbox(X) ; hasplaystation(X).
skinny(a).
vegetarian(a).
hasxbox(b).
eatsburger(c).
hasplaystation(d).
list_all_tall :- forall(tall(Tall), writeln(Tall)).
Very basic stuff. Here's what I get as a result of my queries:
?- tall(a).
true ; % Note 1
false.
?- tall(b).
true ; % Note 2
false.
?- tall(c).
false.
?- tall(d).
true.
As you can see from Notes 1 and 2, it waits for me to hit ; to move on and then considers the first solution as null and eventually outputs false.
I can use cuts to control this behavior better but I also want the following commands to work properly:
?- tall(X).
X = a ;
X = b ;
X = d.
And:
?- list_all_tall.
a
b
d
true.
These two commands give the solution exactly the way I want. Its just the ones for Notes 1 and 2 that are driving me up the wall. Is there a way that I can keep the functionality as it is right now for tall(X). and list_all_tall., while fixing the functionality of tall(a). and tall(b). to my liking, i.e., the program should exit with a true. after I ask tall(a). or tall(b).
I'd appreciated it if instead of giving straight answers someone could actually explain how I could go about fixing it myself because maybe my way of thinking in Prolog is all bassackwards.
PS: No offense intended to tall, skinny, fat, burger eating, video game playing, vegetarian folks.
Just to supplement Daniel's well-explained answer (+1) for your specific case, consider:
tall(a).
Prolog will look at the first match, which is through:
tall(X) :- skinny(X) ; eatless(X).
This will succeed because skinny(a) will succeed. However, there's a disjunction ; leaving a choice point for Prolog that it hasn't explored yet. Because skinny(a) succeeds and the choice point is pending, you get true but prompted to seek more. Prolog then backtracks to the choice point and tries to satisfy eatless(a) but fails. Thus, you get:
?- tall(a).
true ; % because `skinny(a)` succeeded
false. % because `eatless(a)` failed
Taking another example:
tall(d).
Again, this matches the tall/1 predicate, but this time, skinny(d) fails and prolog moves right on (due to the disjunction) to eatless(d) which succeeds. However, there are no more choice points after that success, so you get:
?- tall(d).
true. % There were no choice points available after success
The best thing to do is not worry about it, because you're not always going to be able to prevent it.
Prolog doesn't ever know that there will be another answer. It just knows that there may be another answer. This is called a choice point. Whenever Prolog reaches an alternative, it creates a choice point and then follows the first option. If that option doesn't work out, it backs up to the most recent choice point and tries the next alternative. If it runs out of alternatives without finding an answer, you get no or false.
You can try to write your code so that you don't get a choice point if you know there are no more items. member/2, for instance, in some Prologs you get false after the last item and in others you do not. But it isn't a composition problem to have a dud choice point after all your solutions. Your user interface probably won't show users Prolog's prompts directly. You can use setof/3 and the other extralogical predicates to get all the solutions. The false won't "leak" out into the world. It's a little unnerving at first, but just trust it and don't worry too much about it.
It is possible to run the same predicate, tall/1 in this case, in different modes based on different instantiation patterns.
When you run ?- tall(a). you instantiate the argument (i.e., X=a) and you want to receive either true or false (and no choicepoints, indicated by ;).
In Prolog this mode is called semi-deterministic.
You can force your predicate to be semi-deterministic for this specific instantiation pattern in the following way:
tall(X):- (ground(X) -> once(tall0(X)) ; tall0(X)).
Here ground(X) succeeds just in case X is fully instantiated.
Fully instantiated means that it is not a variable nor is it a compound term containing a variable.
tall0(X) is your original predicate.
The second mode you want to use is ?- tall(X).
Here you expect all results to be given subsequently, using ;.
This mode is called non-deterministic in Prolog.
The complete code for your example is:
tall(X):- (ground(X) -> once(tall0(X)) ; tall0(X)).
tall0(X):- skinny(X) ; eatless(X).
eatless(X):- playsmore(X).
playsmore(X):- hasxbox(X) ; hasplaystation(X).
skinny(a).
hasxbox(b).
hasplaystation(d).
Now the single predicate tall/1 can be called in the two modes, producing the behavior you want. Semi-deterministic usage:
?- tall(a).
true.
Non-deterministic usage:
?- tall(X).
X = a ;
X = b ;
X = d.
Hope this helps!