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So I am currently learning prolog and I can't get my head around how this language works.
"It tries all the possible solutions until it finds one, if it doesn't it returns false" is what I've read that this language does. You just Describe the solution and it finds it for you
With that in mind, I am trying to solve the 8 queens problem ( how to place 8 queens on a chess board without anyone threatening the others).
I have this predicate, 'safe' that gets a list of pairs, the positions of all the queens and succeeds when they are not threatening each other.
When I enter in the terminal
?- safe([(1,2),(3,5)]).
true ?
| ?- safe([(1,3),(1,7)]).
no
| ?- safe([(2,2),(3,3)]).
no
| ?- safe([(2,2),(3,4),(8,7)]).
true ?
it recognizes the correct from the wrong answers, so it knows if something is a possible solution
BUT
when I enter
| ?- safe(L).
L = [] ? ;
L = [_] ? ;
it gives me the default answers, even though it recognizes a solution for 2 queens when I enter them.
here is my code
threatens((_,Row),(_,Row)).
threatens((Column,_),(Column,_)).
threatens((Column1,Row1),(Column2,Row2)) :-
Diff1 is Column1 - Row1,
Diff2 is Column2 - Row2,
abs(Diff1) =:= abs(Diff2).
safe([]).
safe([_]).
safe([A,B|T]) :-
\+ threatens(A,B),
safe([A|T]),
safe(T).
One solution I found to the problem is to create predicates 'position' and modify the 'safe' one
possition((0,0)).
possition((1,0)).
...
...
possition((6,7)).
possition((7,7)).
safe([A,B|T]) :-
possition(A),
possition(B),
\+ threatens(A,B),
safe([A|T]),
safe(T).
safe(L,X):-
length(L,X),
safe(L).
but this is just stupid, as you have to type everything explicitly and really really slow,
even for 6 queens.
My real problem here, is not with the code itself but with prolog, I am trying to think in prolog, But all I read is
Describe how the solution would look like and let it work out what is would be
Well that's what I have been doing but it does not seem to work,
Could somebody point me to some resources that don't teach you the semantics but how to think in prolog
Thank you
but this is just stupid, as you have to type everything explicitly and really really slow, even for 6 queens.
Regarding listing the positions, the two coordinates are independent, so you could write something like:
position((X, Y)) :-
coordinate(X),
coordinate(Y).
coordinate(1).
coordinate(2).
...
coordinate(8).
This is already much less typing. It's even simpler if your Prolog has a between/3 predicate:
coordinate(X) :-
between(1, 8, X).
Regarding the predicate being very slow, this is because you are asking it to do too much duplicate work:
safe([A,B|T]) :-
...
safe([A|T]),
safe(T).
Once you know that [A|T] is safe, T must be safe as well. You can remove the last goal and will get an exponential speedup.
Describe how the solution would look like and let it work out what is
would be
demands that the AI be very strong in general. We are not there yet.
You are on the right track though. Prolog essentially works by enumerating possible solutions and testing them, rejecting those that don't fit the conditions encoded in the program. The skill resides in performing a "good enumeration" (traversing the domain in certain ways, exploiting domain symmetries and overlaps etc) and subsequent "fast rejection" (quickly throwing away whole sectors of the search space as not promising). The basic pattern:
findstuff(X) :- generate(X),test(X).
And evidently the program must first generate X before it can test X, which may not be always evident to beginners.
Logic-wise,
findstuff(X) :- x_fulfills_test_conditions(X),x_fullfills_domain_conditions(X).
which is really another way of writing
findstuff(X) :- test(X),generate(X).
would be the same, but for Prolog, as a concrete implementation, there would be nothing to work with.
That X in the program always stands for a particular value (which may be uninstantiated at a given moment, but becomes more and more instantiated going "to the right"). Unlike in logic, where the X really stands for an unknown object onto which we pile constraints until -ideally- we can resolve X to a set of concrete values by applying a lot of thinking to reformulate constraints.
Which brings us the the approach of "Constraint Logic Programming (over finite domains)", aka CLP(FD) which is far more elegant and nearer what's going on when thinking mathematically or actually doing theorem proving, see here:
https://en.wikipedia.org/wiki/Constraint_logic_programming
and the ECLiPSe logic programming system
http://eclipseclp.org/
and
https://www.metalevel.at/prolog/clpz
https://github.com/triska/clpfd/blob/master/n_queens.pl
N-Queens in Prolog on YouTube. as a must-watch
This is still technically Prolog (in fact, implemented on top of Prolog) but allows you to work on a more abstract level than raw generate-and-test.
Prolog is radically different in its approach to computing.
Arithmetic often is not required at all. But the complexity inherent in a solution to a problem show up in some place, where we control how relevant information are related.
place_queen(I,[I|_],[I|_],[I|_]).
place_queen(I,[_|Cs],[_|Us],[_|Ds]):-place_queen(I,Cs,Us,Ds).
place_queens([],_,_,_).
place_queens([I|Is],Cs,Us,[_|Ds]):-
place_queens(Is,Cs,[_|Us],Ds),
place_queen(I,Cs,Us,Ds).
gen_places([],[]).
gen_places([_|Qs],[_|Ps]):-gen_places(Qs,Ps).
qs(Qs,Ps):-gen_places(Qs,Ps),place_queens(Qs,Ps,_,_).
goal(Ps):-qs([0,1,2,3,4,5,6,7,8,9,10,11],Ps).
No arithmetic at all, columns/rows are encoded in a clever choice of symbols (the numbers indeed are just that, identifiers), diagonals in two additional arguments.
The whole program just requires a (very) small subset of Prolog, namely a pure 2-clauses interpreter.
If you take the time to understand what place_queens/4 does (operationally, maybe, if you have above average attention capabilities), you'll gain a deeper understanding of what (pure) Prolog actually computes.
There is a very detailed Draft proposal for setup_call_cleanup/3.
Let me quote the relevant part for my question:
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.
And this example:
setup_call_cleanup(S=1,G=2,write(S+G)).
Succeeds, unifying S = 1, G = 2.
Either: outputs '1+2'
Or: outputs on backtracking '1+_' prior to failure.
Or (?): outputs on backtracking '1+2' prior to failure.
In my understanding, this is basically because a unification is a
backtrackable goal; it simply fails on reexecution. Therefore, it is
up to the implementation to decide whether to call the cleanup just
after the fist execution (because there will be no more observable
effects of the goal), or postpone until the second execution of the
goal which now fails.
So it seems to me that this cannot be used to detect determinism
portably. Only a few built-in constructs like true, fail, !
etc. are genuinely non-backtrackable.
Is there any other way to check determinism without executing a goal twice?
I'm currently using SWI-prolog's deterministic/1, but I'd certainly
appreciate portability.
No. setup_call_cleanup/3 cannot detect determinism in a portable manner. For this would restrict an implementation in its freedom. Systems have different ways how they implement indexing. They have different trade-offs. Some have only first argument indexing, others have more than that. But systems which offer "better" indexing behave often quite randomly. Some systems do indexing only for nonvar terms, others also permit clauses that simply have a variable in the head - provided it is the last clause. Some may do "manual" choice point avoidance with safe tests prior to cuts and others just omit this. Brief, this is really a non-functional issue, and insisting on portability in this area is tantamount to slowing systems down.
However, what still holds is this: If setup_call_cleanup/3 detects determinism, then there is no further need to use a second goal to determine determinism! So it can be used to implement determinism detection more efficiently. In the general case, however, you will have to execute a goal twice.
The current definition of setup_call_cleanup/3 was also crafted such as to permit an implementation to also remove unnecessary choicepoints dynamically.
It is thinkable (not that I have seen such an implementation) that upon success of Call and the internal presence of a choicepoint, an implementation may examine the current choicepoints and remove them if determinism could be detected. Another possibility might be to perform some asynchronous garbage collection in between. All these options are not excluded by the current specification. It is unclear if they will ever be implemented, but it might happen, once some applications depend on such a feature. This has happened in Prolog already a couple of times, so a repetition is not completely fantasy. In fact I am thinking of a particular case to help DCGs to become more determinate. Who knows, maybe you will go that path!
Here is an example how indexing in SWI depends on the history of previous queries:
?- [user].
p(_,a). p(_,b). end_of_file.
true.
?- p(1,a).
true ;
false.
?- p(_,a).
true.
?- p(1,a).
true. % now, it's determinate!
Here is an example, how indexing on the second argument is strictly weaker than first argument indexing:
?- [user].
q(_,_). q(0,0). end_of_file.
true.
?- q(X,1).
true ; % weak
false.
?- q(1,X).
true.
As you're interested in portability, Logtalk's lgtunit tool defines a portable deterministic/1 predicate for 10 of its supported backend Prolog compilers:
http://logtalk.org/library/lgtunit_0.html
https://github.com/LogtalkDotOrg/logtalk3/blob/master/tools/lgtunit/lgtunit.lgt (starting around line 1051)
Note that different systems use different built-in predicates that approximate the intended functionality.
?- X in 1..100.
X in 1..100.
?- X #> 0, X #< 101.
X in 1..100.
So far so good!
Now let's imagine that the sup bound of X depends on Y in 100..200.
?- Y in 100..200, X #> 0, X #=< Y.
Y in 100..200,
Y#>=X,
X in 1..200.
The domain of X propagates correctly to 1..200, with the constraint between X and Y.
But…
?- Y in 100..200, X in 1..Y.
ERROR: Arguments are not sufficiently instantiated
I thought that in was equivalent to the inequalities declaration, but it apparently is not.
Is there any hidden reason as to why in requires that both bounds are ground?
That's a very good question. It has a straight-forward answer that may at first appear deep, yet is really easy to understand once you are familiar with declarative debugging approaches that are beginning to become more widely available.
Declarative debugging relies on properties of logic programs that we know always hold. A key property of pure logic programs (see logical-purity) is monotonicity:
Generalizing a goal can at meast increase the set of solutions.
Here is an example:
?- append([a], [b], [c]).
false.
Why does this fail? We want to find an actual cause of failure, an explanation why this does not succeed. It is tempting to think in procedural terms and trace the execution, but this quickly gets extremely complex and does not really show us the essence of the reason.
Instead, we can get to the essence by generalizing the goals systematically. For example, the following succeeds:
?- append([a], [b], Cs).
I have generalized the query by replacing the term [c] with Cs, which subsumes it.
You can do this automatically, see Ulrich Neumerkel's library(diadem) and especially the GUPU system where these ideas are fully developed and help tremendously to locate the precise location of mistakes in Prolog programs.
To apply such techniques to their utmost extent, you must aim to preserve as many interesting properties as possible in your programs, essentially by staying in the pure monotonic core of Prolog. I say "essentially" because these techniques also work somewhat beyond this subset on a class of programs that is very hard to classify formally. Note also that much of the ongoing work on logic programming is aimed towards increasing the pure monotonic subset of Prolog as far as possible, so the restriction becomes less and less severe over time.
Consider now the specifics of (in)/2: The domain argument of (in)/2 carries within it a declarative flaw in that it uses a defaulty representation of domains. There are several ways to see this, and notably from the following situation: Suppose I tell you that a domain has the form inf..High, then what is High? It can be either of:
sup
an integer.
The fact that we cannot clearly distinguish these cases tells you that this representation is defaulty.
A clean representation of boundaries would look like this:
inf, sup to denote the infinities
n(N) to denote the integer N.
With such a representation, I could tell you that the domain looks like this: inf..n(N), and you know that N must be an integer.
Now another example:
?- X in 1..0.
false.
Why does this fail? Let us again generalize the arguments to find a reason:
?- X in Low..0.
Suppose now that (in)/2 accepted this as an admissible query. Then what is Low? Again, it can be either an integer, or the atom inf. Unfortunately, there is no sensible way to constrain Low in this way: Constraining it to integers would make CLP(FD) constraints most appropriate, making it tempting to answer with:
X in Low..0,
Low in inf..0
but that is of course too much, because it precludes Low = inf:
?- Low in inf..0, Low = inf.
Type error: `integer' expected, found `inf' (an atom)
Knowing that we cannot decide the type of the term Low in such situations, and lacking a sensible way to state this, an instantiation error is raised to indicate that such a generalization is inadmissible.
From these considerations, the overall answer is really simple:
(in)/2 cannot be easily generalized due to the defaulty representation of domains.
Invariably, there are only two reasons for using such representations:
they are easier to type
they are easier to read.
The hefty drawback of course is that they stand massively in the way of declarative reasoning over your programs.
Note that internally, the CLP(FD) system of SWI-Prolog does use a clean representation of domains. This can also be carried over to the user side, so that we could write:
?- X in n(Low)..0.
with the declaratively perfectly valid answer:
X in n(Low)..0,
Low in inf..0.
Over time, such cleaner notations will definitely find their way to users. For now, it would be a bit too much to ask, at least in my experience.
In the specific examples you cite, the domain boundary is already constrained to integers, so of course we can distinguish the cases and translate this automatically to inequalitites. However, this would not remove the core problem of defaultyness, and exchanging the goals would still raise an instantiation error.
So from what I understand about deterministic predicates:
Deterministic predicate = 1 solution
Non-deterministic predicate = multiple solutions
Are there any type of rules as to how you can detect if the predicate is one or the other? Like looking at the search tree, etc.
There is no clear, generally accepted consensus about these notions. However, they are usually based rather on the observed answers and not based on the number of solutions. In certain contexts the notions are very implementation related. Non-determinate may mean: leaves a choice point open. And sometimes determinate means: never even creates a choice point.
Answers vs. solutions
To see the difference, consider the goal length(L, 1). How many solutions does it have? L = [a] is one, L = [23] another... but all of these solutions are compactly represented with a single answer substitution: L = [_] which thus contains infinitely many solutions.
In any case, in all implementations I know of, length(L, 1) is a determinate goal.
Now consider the goal repeat which has exactly one solution, but infinitely many answers. This goal is considered non-determinate.
In case you are interested in constraints, things become even more evolved. In library(clpfd), the goal X #> Y, Y #> X has no solution, but still one answer. Combine this with repeat: infinitely many answers and no solution.
Further, the goal append(Xs, Ys, []) has exactly one solution and also exactly one answer, nevertheless it is considered non-determinate in many implementations, since in those implementations it leaves a choice point open.
In an ideal implementation, there would be no answers without solutions (except false), and there would be non-determinism only when there is more than one answer. But then, all of this is mostly undecidable in the general case.
So, whenever you are using these notions make sure on what level things are meant. Rather explicitly say: multiple answers, multiple solutions, leaves no (unnecessary) choice point open.
You need understand the difference between det, semidet and undet, it is more than just number of solutions.
Because there is no loop control operator in Prolog, looping (not recursion) is constructed as a 'sequence generating' predicate (undet) followed by the loop body. Also you can store solutions with some of findall-group predicates as a list and loop later with the member/2 predicate.
So, any piece of your program is either part of loop construction or part of usual flow. So, there is a difference in designing det and undet predicates almost in the intended usage. If you can work with a sequence you always do undet and comment it as so. There is a nice unit-test extension in swi-prolog which can check wheter your predicate always the same in mean of det/semidet/undet (semidet is for usage the same way as undet but as a head of 'if' construction).
So, the difference is pre-design, and this question should not be arised with already existing predicates. It is a good practice always comment the intended usage in a comment like.
% member(?El, ?List) is undet.
Deterministic: Always succeeds with a single answer that is always the same for the same input. Think a of a static list of three items, and you tell your function to return value one. You will get the same answer every time. Additionally, arithmetic functions. 1 + 1 = 2. X + Y = Z.
Semi-deterministic: Succeeds with a single answer that is always the same for the same input, but it can fail. Think of a function that takes a list of numbers, and you ask your function if some number exists in the list. It either does, or it doesn't, based on the contents of the list given and the number asked.
Non-deterministic: Succeeds with a single answer, but can exhibit different behaviors on different runs, even for the same input. Think any kind of math.random(min,max) function like random/3
In essence, this is entirely separate from the concept of choice points, as choice points are a function of Prolog. Where I think the Prolog confusion of these terms comes from is that Prolog can find a single answer, then go back and try for another solution, and you have to use the cut operator ! to tell it that you want to discard your choice points explicitly.
This is very useful to know when working with Prolog Unit Testing
I often end up writing code in Prolog which involves some arithmetic calculation (or state information important throughout the program), by means of first obtaining the value stored in a predicate, then recalculating the value and finally storing the value using retractall and assert because in Prolog we cannot assign values to variable twice using is (thus making almost every variable that needs modification, global). I have come to know that this is not a good practice in Prolog. In this regard I would like to ask:
Why is it a bad practice in Prolog (though i myself don't like to go through the above mentioned steps just to have have a kind of flexible (modifiable) variable)?
What are some general ways to avoid this practice? Small examples will be greatly appreciated.
P.S. I just started learning Prolog. I do have programming experience in languages like C.
Edited for further clarification
A bad example (in win-prolog) of what I want to say is given below:
:- dynamic(value/1).
:- assert(value(0)).
adds :-
value(X),
NewX is X + 4,
retractall(value(_)),
assert(value(NewX)).
mults :-
value(Y),
NewY is Y * 2,
retractall(value(_)),
assert(value(NewY)).
start :-
retractall(value(_)),
assert(value(3)),
adds,
mults,
value(Q),
write(Q).
Then we can query like:
?- start.
Here, it is very trivial, but in real program and application, the above shown method of global variable becomes unavoidable. Sometimes the list given above like assert(value(0))... grows very long with many more assert predicates for defining more variables. This is done to make communication of the values between different functions possible and to store states of variables during the runtime of program.
Finally, I'd like to know one more thing:
When does the practice mentioned above become unavoidable in spite of various solutions suggested by you to avoid it?
The general way to avoid this is to think in terms of relations between states of your computations: You use one argument to hold the state that is relevant to your program before a calculation, and a second argument that describes the state after some calculation. For example, to describe a sequence of arithmetic operations on a value V0, you can use:
state0_state(V0, V) :-
operation1_result(V0, V1),
operation2_result(V1, V2),
operation3_result(V2, V).
Notice how the state (in your case: the arithmetic value) is threaded through the predicates. The naming convention V0 -> V1 -> ... -> V scales easily to any number of operations and helps to keep in mind that V0 is the initial value, and V is the value after the various operations have been applied. Each predicate that needs to access or modify the state will have an argument that allows you to pass it the state.
A huge advantage of threading the state through like this is that you can easily reason about each operation in isolation: You can test it, debug it, analyze it with other tools etc., without having to set up any implicit global state. As another huge benefit, you can then use your programs in more directions provided you are using sufficiently general predicates. For example, you can ask: Which initial values lead to a given outcome?
?- state0_state(V0, given_outcome).
This is of course not readily possible when using the imperative style. You should therefore use constraints instead of is/2, because is/2 only works in one direction. Constraints are much easier to use and a more general modern alternative to low-level arithmetic.
The dynamic database is also slower than threading states through in variables, because it performs indexing etc. on each assertz/1.
1 - it's bad practice because destroys the declarative model that (pure) Prolog programs exhibit.
Then the programmer must think in procedural terms, and the procedural model of Prolog is rather complicate and difficult to follow.
Specifically, we must be able to decide about the validity of asserted knowledge while the programs backtracks, i.e. follow alternative paths to those already tried, that (maybe) caused the assertions.
2 - We need additional variables to keep the state. A practical, maybe not very intuitive way, is using grammar rules (a DCG) instead of plain predicates. Grammar rules are translated adding two list arguments, normally hidden, and we can use those arguments to pass around the state implicitly, and reference/change it only where needed.
A really interesting introduction is here: DCGs in Prolog by Markus Triska. Look for Implicitly passing states around: you'll find this enlighting small example:
num_leaves(nil), [N1] --> [N0], { N1 is N0 + 1 }.
num_leaves(node(_,Left,Right)) -->
num_leaves(Left),
num_leaves(Right).
More generally, and for further practical examples, see Thinking in States, from the same author.
edit: generally, assert/retract are required only if you need to change the database, or keep track of computation result along backtracking. A simple example from my (very) old Prolog interpreter:
findall_p(X,G,_):-
asserta(found('$mark')),
call(G),
asserta(found(X)),
fail.
findall_p(_,_,N) :-
collect_found([],N),
!.
collect_found(S,L) :-
getnext(X),
!,
collect_found([X|S],L).
collect_found(L,L).
getnext(X) :-
retract(found(X)),
!,
X \= '$mark'.
findall/3 can be seen as the basic all solutions predicate. That code should be the very same from Clockins-Mellish textbook - Programming in Prolog. I used it while testing the 'real' findall/3 I implemented. You can see that it's not 'reentrant', because of the '$mark' aliased.