I have been trying to solve a pathfinding problem in Prolog.where the predicates are
edge(a,b).
edge(a,c).
edge(b,d).
edge(c,d).
edge(d,e).
edge(d,f).
edge(f,g).
the rules is
edge(X,Y) :- edge(X,Z), edge(Z,Y).
then when I compiled and run the query
| ?- edge(a,X).
it is showing
Fatal Error: local stack overflow (size: 8192 Kb, environment variable used: LOCALSZ)
then I searched for the solution and found that including atom(x).,atom(y). in our rule can solve the stack overflow problem . i.e the new rule is
edge(X,Y) :- atom(X), atom(Y), edge(X,,Z), edge(Z,Y). and yes it did solved the stack overflow problem .but,I would like to know how exactly this (atom/1)
predicate is solving my problem here?and what does it do to our variables X,Y to solve the StackOverflow problem?
I am a newbie to Prolog any help would be appreciated
thank you. :)
First on naming, the edge/2 name doesn't describe your predicate very well. You probably really want path/2 which consists of one or more edges.
Does atom/1 really solve your problem? In other words, does edge(X, Y) really now provide all of the correct solutions to a query? All that atom/1 does is ensure that its argument is an atom, so it cannot be an unbound variable. So edge(X, Y) does not provide all of the correct solutions. It only yields those solutions that you have direct facts for, since the predicate edge(X, Y) as currently defined always fails with either X or Y unbound.
| ?- edge(a, Y).
Y = b ? ;
Y = c ? ;
no
Where is the solution Y = d for example? edge(X, Y) is only picking up the solutions that are given in your edge/2 facts, but no solutions that include multiple connected edges.
Your original problem is due to infinite recursion, which is a result of edge/2 calling itself unnecessarily. Naming can actually be important here since it makes the logic more precise and correct. We can say that edge(X, Y) means that X and Y form an edge (X and Y are directly connected). We can say that path(X, Y) means there's a path from X to Y via one or more edges. In other words, a path from x to y can either be an edge from x to y, or it can be an edge from x to z and a path from z to y.
path(X, Y) :- edge(X, Y).
path(X, Y) :- edge(X, Z), path(Z, Y).
Now we get:
| ?- path(a, X).
X = b ? a
X = c
X = d
X = e
X = f
X = g
X = d
X = e
X = f
X = g
(1 ms) no
| ?-
There are duplicates since there may be multiple ways of getting from a to e for example. If you included an argument that showed the traversed path, this would become evident.
This solution isn't the end of the story, however. Your current facts are such that there are no "circuitous" paths (paths that eventually revisit the same node if followed). To handle that, you need a predicate argument to keep track of what nodes/edges you've traversed already and avoid traversing them again.
Related
I am a beginner in Prolog and I have a task to do.
I need to check if the graph is connected.
For now I have that...
graph(
[arc(a,b)],
[arc(a,f)],
[arc(b,c)],
[arc(c,d)],
[arc(c,e)],
[arc(e,d)],
[arc(f,c)],
[arc(f,e)],
[arc(f,g)],
[arc(g,c)],
[arc(c,a)]).
edge(X,Y):-arc(X,Y);arc(Y,X).
path(X,Y):-edge(X,Y).
path(X,Y):-edge(X,Z),path(Z,Y).
triangle(X,Y,Z):-arc(X,Y),arc(Y,Z),arc(Z,X).
cycle(X):-arc(X,Y),path(Y,X).
connectivity([]):-forall(member(edge(X,Y)),path(X,Y)).
Check:
connectivity(graph).
upper I have arc(x,y) and I need check if every pair is connected.
Could u help me ?
Since you changed the question after I was almost done I will post what would solve the question before the change and you can figure out how to change it to meet your update.
arc(a,b).
arc(a,f).
arc(b,c).
arc(c,d).
arc(c,e).
arc(e,d).
arc(f,c).
arc(f,e).
arc(f,g).
arc(g,c).
arc(c,a).
edge(X,Y) :-
arc(X,Y), !.
edge(X,Y) :-
arc(Y,X).
path_prime(Visited,X,Y) :-
\+ member(X,Visited),
edge(X,Y), !.
path_prime(Visited,X,Y) :-
\+ member(X,Visited),
edge(X,Z),
path_prime([X|Visited],Z,Y).
path(X,X) :-
ground(X), !.
path(X,Y) :-
path_prime([],X,Y).
nodes(Nodes) :-
setof(A,B^arc(A,B),Starts),
setof(B,A^arc(A,B),Ends),
union(Starts,Ends,Nodes).
connected(X,Y) :-
nodes(Nodes),
member(X,Nodes),
member(Y,Nodes),
path(X,Y).
The first thing that has to be done is to get a list of the unique nodes which will be a set.
This can be done using
nodes(Nodes) :-
setof(A,B^arc(A,B),Starts),
setof(B,A^arc(A,B),Ends),
union(Starts,Ends,Nodes).
Notice that both the start and the end node of an arc are done separately. In particular notice that the node d is only in the destination of an arc.
Since you included edge(X,Y):-arc(X,Y);arc(Y,X). in your question, this indicated that the arcs should not be directional and so it is possible to get cycles. To avoid the cycles the list of visited nodes is added to the argument list and checked before proceeding.
As no test cases or examples of a correct solution were given, some times a node connected to itself is valid and so the clause
path(X,X) :-
ground(X), !.
was added.
This is by no means an optimal or best way to do this, just to give you something that works.
Partial run
?- connected(X,Y).
X = Y, Y = a ;
X = a,
Y = b ;
X = a,
Y = c ;
X = a,
Y = d ;
X = a,
Y = e ;
X = a,
Y = f ;
X = a,
Y = g ;
X = b,
Y = a ;
X = Y, Y = b ;
X = b,
Y = c ;
...
As I often comment, you should do problems with pen an paper first before writing code. If you don't know exactly what the code will be before you start typing the first line of code then why are you typing in code?
Questions from comments:
And setof ,union ,whats mean? Im rly beigneer and I don't understand that language and predicates.
setof/3 collects all of the values from arc/2. Since only one of the two values is needed, ^ tells setup/3 not to bind the variable in the Goal, or in beginner terms to just ignore the values from the variable.
union/3 just combines the to sets into one set; remember that a set will only have unique values.
I am developing a path finding algorithm in Prolog, giving all nodes accessible by a path from a starting node. To avoid duplicate paths, visited nodes are kept in a list.
Nodes and neighbors are defined as below:
node(a).
node(b).
node(c).
node(d).
node(e).
edge(a,b).
edge(b,c).
edge(c,d).
edge(b,d).
neighbor(X,Y) :- edge(X,Y).
neighbor(X,Y) :- edge(Y,X).
The original algorithm below works fine:
path2(X,Y) :-
pathHelper(X,Y,[X]).
pathHelper(X,Y,L) :-
neighbor(X,Y),
\+ member(Y,L).
pathHelper(X,Y,H) :-
neighbor(X,Z),
\+ member(Z,H),
pathHelper(Z,Y,[Z|H]).
This works fine
[debug] ?- path2(a,X).
X = b ;
X = c ;
X = d ;
X = d ;
X = c ;
false.
however, when changing the order of the two clauses in the second definition, such as below
pathHelper(X,Y,L) :-
\+ member(Y,L),
neighbor(X,Y).
When trying the same here, swipl returns the following:
[debug] ?- path2(a,X).
false.
The query doesn't work anymore, and only returns false. I have tried to understand this through the tracecommand, but still can't make sense of what exactly is wrong.
In other words, I am failing to understand why the order of neighbor(X,Y)and \+ member(Y,L)is crucial here. It makes sense to have neighbor(X,Y) first in terms of efficiency, but not in terms of correctness to me.
You are now encountering the not so clean-cut borders of pure Prolog and its illogical surroundings. Welcome to the real world.
Or rather, not welcome! Instead, let's try to improve your definition. The key problem is
\+ member(Y, [a]), Y = b.
which fails while
Y = b, \+ member(Y,[a]).
succeeds. There is no logic to justify this. It's just the operational mechanism of Prolog's built-in (\+)/1.
Happily, we can improve upon this. Enter non_member/2.
non_member(_X, []).
non_member(X, [E|Es]) :-
dif(X, E),
non_member(X, Es).
Now,
?- non_member(Y, [a]).
dif(Y,a).
Mark this answer, it says: Yes, Y is not an element of [a], provided Y is different from a. Think of the many solutions this answer includes, like Y = 42, or Y = b and infinitely many more such solutions that are not a. Infinitely many solutions captured in nine characters!
Now, both non_member(Y, [a]), Y = b and Y = b, non_member(Y, [a]) succeed. So exchanging them has only influence on runtime and space consumption. If we are at it, note that you check for non-memberness in two clauses. You can factor this out. For a generic solution to this, see closure/3. With it, you simply say: closure(neighbor, A,B).
Also consider the case where you have only edge(a,a). Your definition fails here for path2(a,X). But shouldn't this rather succeed?
And the name path2/2 is not that fitting, rather reserve this word for an actual path.
The doubt you have is related to how prolog handle negation. Prolog uses negation as failure. This means that, if prolog has to negate a goal g (indicate it with not(g)), it tries to prove g by executing it and then, if the g fails, not(g) (or \+ g, i.e. the negation of g) succeeds and viceversa.
Keep in mind also that, after the execution of not(g), if the goal has variables, they will not be instantiated. This because prolog should instantiate the variables with all the terms that makes g fail, and this is likely an infinite set (for example for a list, not(member(A,[a]) should instantiate the variable A with all the elements that are not in the list).
Let's see an example. Consider this simple program:
test:-
L = [a,b,c],
\+member(A,L),
writeln(A).
and run it with ?- trace, test. First of all you get a Singleton variable in \+: A warning for the reason i explained before, but let's ignore it and see what happens.
Call:test
Call:_5204=[a, b]
Exit:[a, b]=[a, b]
Call:lists:member(_5204, [a, b])
Exit:lists:member(a, [a, b]) % <-----
Fail:test
false
You see at the highlighted line that the variable A is instantiated to a and so member/2 succeeds and so \+ member(A,L) is false.
So, in your code, if you write pathHelper(X,Y,L) :- \+ member(Y,L), neighbor(X,Y)., this clause will always fail because Y is not sufficiently instantiated. If you swap the two terms, Y will be ground and so member/2 can fail (and \+member succeeds).
I'm currently learning SWI-Prolog. I want to implement a function factorable(X) which is true if X can be written as X = n*b.
This is what I've gotten so far:
isTeiler(X,Y) :- Y mod X =:= 0.
hatTeiler(X,X) :- fail,!.
hatTeiler(X,Y) :- isTeiler(Y,X), !; Z is Y+1, hatTeiler(X,Z),!.
factorable(X) :- hatTeiler(X,2).
My problem is now that I don't understand how to end the recursion with a fail without backtracking. I thought the cut would do the job but after hatTeilerfails when both arguments are equal it jumps right to isTeiler which is of course true if both arguments are equal. I also tried using \+ but without success.
It looks like you add cuts to end a recursion but this is usually done by making rule heads more specific or adding guards to a clause.
E.g. a rule:
x_y_sum(X,succ(Y,1),succ(Z,1)) :-
x_y_sum(X,Y,Z).
will never be matched by x_y_sum(X,0,Y). A recursion just ends in this case.
Alternatively, a guard will prevent the application of a rule for invalid cases.
hatTeiler(X,X) :- fail,!.
I assume this rule should prevent matching of the rule below with equal arguments. It is much easier just to add the inequality of X and Y as a conditon:
hatTeiler(X,Y) :-
Y>X,
isTeiler(Y,X),
!;
Z is Y+1,
hatTeiler(X,Z),
!.
Then hatTeiler(5,5) fails automatically. (*)
You also have a disjunction operator ; that is much better written as two clauses (i drop the cuts or not all possibilities will be explored):
hatTeiler(X,Y) :- % (1)
Y > X,
isTeiler(Y,X).
hatTeiler(X,Y) :- % (2)
Y > X,
Z is Y+1,
hatTeiler(X,Z).
Now we can read the rules declaratively:
(1) if Y is larger than X and X divides Y without remainder, hatTeiler(X,Y) is true.
(2) if Y is larger than X and (roughly speaking) hatTeiler(X,Y+1) is true, then hatTeiler(X, Y) is also true.
Rule (1) sounds good, but (2) sounds fishy: for specific X and Y we get e.g.: hatTeiler(4,15) is true when hatTeiler(4,16) is true. If I understand correctly, this problem is about divisors so I would not expect this property to hold. Moreover, the backwards reasoning of prolog will then try to deduce hatTeiler(4,17), hatTeiler(4,18), etc. which leads to non-termination. I guess you want the cut to stop the recursion but it looks like you need a different property.
Coming from the original property, you want to check if X = N * B for some N and B. We know that 2 <= N <= X and X mod N = 0. For the first one there is even a built-in called between/2 that makes the whole thing a two-liner:
hT(X,B) :-
between(2, X, B),
0 is (X mod B).
?- hT(12,X).
X = 2 ;
X = 3 ;
X = 4 ;
X = 6 ;
X = 12.
Now you only need to write your own between and you're done - all without cuts.
(*) The more general hasTeiler(X,X) fails because is (and <) only works when the right hand side (both sides) is variable-free and contains only arithmetic terms (i.e. numbers, +, -, etc).
If you put cut before the fail, it will be freeze the backtracking.
The cut operation freeze the backtracking , if prolog cross it.
Actually when prolog have failed, it backtracks to last cut.
for example :
a:- b,
c,!,
d,
e,!,
f.
Here, if b or c have failed, backtrack do not freeze.
if d or f have failed, backtrack Immediately freeze, because before it is a cut
if e have failed , it can backtrack just on d
I hope it be useful
This Prolog program de fines the third argument to be the maximum value of the fi rst two numeric arguments:
max(X, Y, X) :- X >= Y, !.
max(X, Y, Y).
I think that this program works just fine. But I am told that it can give incorrect result. Can you tell when and why?
This is a textbook example.
?- max(5,1,1).
true.
Homework: Why is the program wrong? How do we make the program correct?
EDIT
max(X, Y, X) :- X >= Y, !.
max(X, Y, Y).
Our intention is to say:
If X is greater than Y, then Max is X. Otherwise, Max must be Y.
Instead, what is say is:
When the first and third arguments (X and Max) can be unified, and X is greater than Y, succeed. Otherwise, if the second and third arguments (Y and Max) can be unified, succeed.
The obvious problem arises then the first and third arguments cannot be unified, but the second and the third can.
Instead:
max(X, Y, X) :- X >= Y.
max(X, Y, Y) :- X < Y.
or
max(X, Y, Max) :- X >= Y, !, Max = X.
max(_, Max, Max).
It does work fine, provided the third argument is uninstantiated. The danger here would be if there were a way to backtrack into the second rule, or if the third argument is instantiated to the same value as the second. It's not particularly safe looking because max(X, Y, Y). is equal to max(_, Y, Y) which just sets the result to the second value without any thought. The cut at the end of the first rule effectively ensures that backtracking will not commence if X >= Y, so the second rule should only be entered when X < Y and Z is not already equal to Y.
Though it mostly works, it's not a good habit to get into. People new to Prolog tend to think procedurally and making use of the cut like this to ensure a particular result through procedural trickery ultimately holds you back and leads to convoluted Prolog that cannot be driven in different and interesting ways. There are several other ways of writing this predicate that work just as well but do not rely on the cut to ensure their behavior, for instance:
max(X, Y, X) :- X >= Y.
max(X, Y, Y) :- X < Y.
or
max(X, Y, Z) :- X >= Y -> Z = X ; Z = Y.
Neither of these is vulnerable to the problem of the third being instantiated. Interestingly, this is a great illustration of the difference between a red cut and a green cut. Your code has a red cut, where the behavior is dependent on the cut, but if I simply change my first solution to this:
max(X, Y, X) :- X >= Y, !.
max(X, Y, Y) :- X < Y.
That's a green cut, because the behavior is not dependent on the cut, but Prolog's performance may improve slightly since it won't backtrack into the second clause to try it. Here we're explicitly telling Prolog, don't both making the next check because we know it will fail. With a red cut, there's no other check which will fail.
It's unfortunate that stating the condition twice feels redundant but relying on a single rule feels clunky. In practice, my experience is that scenarios like these are not ultimately all that common; usually you have atoms or structures you can match in the head of the clause that create behavior like we have in my first substitute, but without needing a body. For example:
perform(scan(target, X, Y)) :- ...
perform(scan(calibration, X)) :- ...
This has the same effect: Prolog will backtrack until it unifies successfully, then it will back track again, but the exclusive nature of the matching will prevent another body from being executed. If we find out it's spending too much time backtracking we can add cuts to improve the performance, but in practice it's unlikely to be a problem.
I have got some values H, and I would like to find the maximum one using \+, how can i do it?
maxValue(X) :-
Get(Id, X),
\+( Get(Id, Y), X < Y ).
don't have a clue....please help, thanks!
Using negation is one way to find the maximum. And it really works.
Here is an example:
p(2).
p(1).
p(3).
?- p(X), \+ (p(Y), Y > X).
X = 3
But the complexity will be O(n*n) where n is
the number of facts. But the maximum can be
determined in O(n). So maybe the following is
more efficient for large fact bases:
:- dynamic(the_max/1).
update_max(X) :-
the_max(Y), X>Y, !, retract(the_max(Y)), assertz(the_max(X)).
update_max(_).
find_max(X) :-
assertz(the_max(0)),
(p(Y), update_max(Y), fail; true),
retract(the_max(X)).
?- find_max(X).
X = 3
But watch out, when you use it from multiple threads,
you need to adapt it a little, i.e. make the_max
thread local.
Best Regards
See also these questions/answers:
Prolog query to find largest element in database?
Max out of values defined by prolog clauses