I'm exploring dependent structures of constraints like this one:
assign(X,Y) :-
X in 1..5,
((X mod 2 #= 1) #=> Y in 2..3),
((X mod 2 #= 0) #=> Y #= 5).
What I'm looking for is a representation of X's and Y's domains that is as sparse as possible - in this case it would be something along the lines of X in {1,3,5} and Y in {2,3} or X in {2,4} and Y = 5.
One way of doing that would be to detect all variables on the left side of the #=>, enumerate all their values and collect and merge them together, something like ?- assign(X, Y), findall(X-D, (indomain(X),fd_dom(Y,D)), C), do stuff with C, but maybe there is a more efficient way?
I've also encountered an error trying to label([X,Y]): Arguments are not sufficiently instantiated that goes away when I add another constraint on Y's domain.
When should I expect this error to occur? I feel I have a poor understanding of clpfd's mechanisms and limitations, are there any resources I could learn from? I already know the basics of constraint programming, arc consistency etc.
To keep clpfd enumeration predicates (like indomain/1, label/1, labeling/2, etc.) from ever throwing instantiation errors, simply ensure that all variables have been assigned some finite domain before any enumeration predicates is executed.
So how about directly translating what you wrote to code?
assign(X,Y) :- X in 1\/3\/5, Y in 2..3. % X in {1,3,5} and Y in {2,3}
assign(X,Y) :- X in 2..4, Y in 5. % or X in {2,4} and Y = 5
A simple query (with SWI-Prolog):
?- assign(X,Y), labeling([],[X,Y]).
X = 1, Y = 2
; X = 1, Y = 3
; X = 3, Y = 2
; X = 3, Y = 3
; X = 5, Y = 2
; X = 5, Y = 3
; X = 2, Y = 5
; X = 3, Y = 5
; X = 4, Y = 5.
Related
Trying to calculate the triangular number sequence in Prolog.
This is my solution:
where X is the nth position of the sequence and Y is the result.
triang(1, 1).
triang(X, Y) :-
X>0,
A is X - 1,
triang(A, B),
Y is B + X.
?- triang(5,X).
X = 15
But when i try to do for example triang(X,10) I receive an error
Arguments are not sufficiently instantiated.
I guess this is because X is not defined in the consult.
is there any recommendation how to solve this problem,thank you.
First of all, the result you got is not that bad. It says: sorry, I am unable to come to a conclusion and before producing an incorrect result, I prefer to produce an error.
The actual reason is the following goal
?- X > 0.
error(instantiation_error,(is)/2).
So here we ask for X that are greater than zero. And there are many, in fact infinitely many. There is no direct way to enumerate that set for this built-in and thus it prefers the error.
However, with library(clpz) or clpfd, there is a better way:
:- use_module(library(clpz)). % use clpfd for SWI instead
:- op(150, fx, #).
triang(0, 0).
triang(X, Y) :-
#X #>0,
#Y #>0,
#A #= #X - 1,
#Y #= #B + #X,
triang(A, B).
?- triang(X,15).
X = 5
; false.
?- triang(X,14).
false.
?- triang(X,X).
X = 0
; X = 1
; false.
?- triang(X,Y).
X = 0, Y = 0
; X = 1, Y = 1
; X = 2, Y = 3
; X = 3, Y = 6
; X = 4, Y = 10
; X = 5, Y = 15
; X = 6, Y = 21
; ... .
?- #X #> 0.
clpz:(X in 1..sup).
So now there is an answer to #X #> 0. The answer is often called a constraint. In this case it tells us that X must be in the interval 1 up to (kind of) infinity.
Doing exercise 10.4 on learnprolognow, could someone explain to me or help me visualize why for ?- p(X),p(Y) we get:
X=1,Y=1; X=1,Y=2; X=2, Y=1; X=2, Y=1.
And not just
X=1, Y=1; X=1, Y=2.
I think I'm misunderstanding how the cut happens, when it's in the ruleset instead of the query - because I think I can visualize it for ?- p(X),!,p(Y)., where it actually behaves as I thought the last one would...
Edit: From the website
% database
p(1).
p(2):-!.
p(3).
% Queries
p(X). % returns: X=1; X=2.
p(X),p(Y). % returns: X=1,Y=1; X=1, Y=1; X=2, Y=2. (?)
p(X),!,p(Y). % returns X=1, Y=1; X=1, Y=2.
To understand this problem you can imagine a tree with X in as first level and Y as second level (prolog uses sld resolution that can be well described with a tree). Consider this problem:
p(1).
p(2):-!.
p(3).
sol(X,Y):-
p(X),
p(Y).
I've added the predicate solve/2 to make it more clear. Run the query:
?- solve(X,Y).
First of all you have to choose the value for X. Prolog uses depth first search from the top to the bottom, from left to right. So it evaluates p(x): p(1) succeed (because is the first clause, if you write p(2) above p(1), p(2) will succeed) and so X = 1. Then evaluates p(Y): p(1) succeed and so you have the first solution:
X = Y, Y = 1.
If you click on more, then prolog does a backtrack (you can imagine this as a step above on the tree) and tries another value for p(Y). In this case p(2) succeed, the predicate is true and you get:
X = 1, Y = 2.
Now, if you click on more, due to the fact there is a cut (!) in the body of p(2) (a general rule in prolog has the form head :- body), prolog will not go more in depth and p(3) is ignored. So there's no more solution to p(Y). So there is another backtracking and this time, for p(X), p(2) succeed and X = 2 and for p(Y), p(1) succeed and you get:
X = 2, Y = 1.
If you click on more, you get:
X = Y, Y = 2.
Now, due to the fact there is a cut after p(2), there are no more solutions available for both X and Y (! cuts everything below p(2)).
If you remove the cut you get all the possible solutions:
X = Y, Y = 1
X = 1,
Y = 2
X = 1,
Y = 3
X = 2,
Y = 1
X = Y, Y = 2
X = 2,
Y = 3
X = 3,
Y = 1
X = 3,
Y = 2
X = Y, Y = 3
Keep in mind that the order of the clauses is important. If you write
p(2).
p(1):-!.
p(3).
You get
X = Y, Y = 2
X = 2,
Y = 1
X = 1,
Y = 2
X = Y, Y = 1
You can check this behaviour using the tracer. In SWI or SWISH you can write ?-
trace, solve(X,Y).
If you have a situation like this:
p(1).
p(2).
p(3).
sol(X,Y):-
p(X),
!,
p(Y).
prolog will tests all the possible values for Y and only one value for X because the cut stops the exploration of the tree (ideally you have 3 branches for X (1,2,3) and 3 for Y (1,2,3), ! cuts 2 and 3 from X) and you get:
X = Y, Y = 1
X = 1,
Y = 2
X = 1,
Y = 3
Sorry for the long post, hope to be clear.
test(X, Y) :-
X ins 1..3,
Y ins 1..3,
X #\= Y.
Here is my attempt at doing it. The goal would be to type this into SWI-Prolog so that this output comes out.
?- test(X, Y).
X = 1
Y = 2 ;
X = 2,
Y = 1;
X = 3,
Y = 1 ;
... etc.
I'm actually trying to solve the 8-queens problem using prolog and have this so far.
eight_queens(Qs, L) :-
Qs = [ [X1,Y1], [X2, Y2], [X3, Y3], [X4, Y4], [X5, Y5], [X6, Y6], [X7, Y7], [X8, Y8], [X9, Y9] ],
Qs ins 1..9,
X1 #\= X2,
X1 #\= X3,
...
etc.
But I keep getting this error: "Arguments are not sufficiently instantiated" for both the test function and the eight_queens problem.
Besides the observation about in/2 and ins/2 posted by #coder, that solve your imminent problem, I would add the following points that are good to keep in mind when using CLP(FD):
1. Always make labeling the last goal
First let's observe the answers for the variant marked as 2nd way using ins in #coder's post but without the goal label/1:
test(X, Y) :-
[X,Y] ins 1..3,
X #\= Y.
?- test(X,Y).
X in 1..3, % residual goal
X#\=Y, % residual goal
Y in 1..3. % residual goal
Since there is no unique answer to the query, Prolog answers with residual goals (see section A.8.8 of the CLP(FD) manual) for more information). These residual goals are constraints that are being propagated and with every additional (non-redundant) constraint the domain is narrowed. If this does not lead to a unique solution like in the example above you can get concrete values by labeling the constrained variables (e.g. with label/1). This observation suggests to use labeling as the last goal:
?- test(X,Y), label([X,Y]).
X = 1,
Y = 2 ;
X = 1,
Y = 3 ;
X = 2,
Y = 1 ;
X = 2,
Y = 3 ;
X = 3,
Y = 1 ;
X = 3,
Y = 2.
This is obviously the same result as with #coders version but the three pairs (X,Y) = (1,1) ∨ (2,2) ∨ (3,3) are not considered when labeling due to the constraint X#\=Y being posted before the goal label([X,Y]). In #coder's version it is the other way around: label([X,Y]) is delivering all three pairs as possible solutions and the last goal X#\=Y is eliminating them subsequently. To see this just leave the last goal as a comment and query the predicate:
test(X,Y):- [X,Y] ins 1..3, label([X,Y]). %, X#\=Y.
?- test(X,Y).
X = Y, Y = 1 ; % <- (1,1)
X = 1,
Y = 2 ;
X = 1,
Y = 3 ;
X = 2,
Y = 1 ;
X = Y, Y = 2 ; % <- (2,2)
X = 2,
Y = 3 ;
X = 3,
Y = 1 ;
X = 3,
Y = 2 ;
X = Y, Y = 3. % <- (3,3)
The difference is minuscule in this example, so there's nothing wrong with #coder's version. But in general this might lead to a big difference if the constraints posted after labeling exclude a lot of candidates. So it's good practice to always put labeling as the last goal.
2. Separate labeling from the actual relation
Coming from the previous observations it is opportune to divide the predicate into a core relation that is posting all the constraints and labeling. Consider the restructured predicate test/2 as a template:
test(X,Y) :-
test_(X,Y,L), % the core relation
label(L). % labeling
test_(X,Y,L) :-
L=[X,Y], % variables to be labeled in a flat list
L ins 1..3,
X#\=Y.
The predicate test_/3 is describing the actual relation by posting all the necessary constraints and has a list as an additional argument that contains all the variables to be labeled. Obtaining the latter might not be trivial, depending on the data structures your arguments come with (consider for example a list of lists as an argument that you want to turn into a flat list for labeling). So the predicate test/2 is only calling test_/3 and subsequently the labeling goal. This way you have a clean and easily readable separation.
3. Try different labeling strategies
The goal label(L) is the simplest way to do labeling. It is equivalent to labeling([],L). The first argument of labeling/2 is a list of options that gives you some control over the search process, e.g. labeling([ff],L) labels the leftmost variable with the smallest domain next, in order to detect infeasibility early. Depending on the problem you are trying to solve different labeling strategies can lead to results faster or slower. See the documentation of labeling/2 for available labeling strategies and further examples.
ins is used for lists, in is used for single variable so in your example:
test(X, Y) :-
X ins 1..3,
Y ins 1..3,
X #\= Y.
X,Y are assumed to be lists. This does not produces a syntax error, but produces error when trying to run it with X,Y not being lists.
Also when using in Low..High doesn't mean that the variable is int just X=<High and X>=Low. In order to put the constraint to be integers use label/1:
:- use_module(library(clpfd)).
%using in/
test(X,Y):- X in 1..3,Y in 1..3,label([X,Y]), X#\=Y.
%2nd way using ins
test(X,Y):- [X,Y] ins 1..3, label([X,Y]), X#\=Y.
Example:
?- test(X,Y).
X = 1,
Y = 2 ;
X = 1,
Y = 3 ;
X = 2,
Y = 1 ;
X = 2,
Y = 3 ;
X = 3,
Y = 1 ;
X = 3,
Y = 2 ;
false.
I was wondering whether in prolog it is possible to get it to brute force all the possible calculations for something like this:
6 is Z + Q
Z = 1 Q = 5
Z = 2 Q = 4
Z = 3 Q = 3
I suggest to use, if your Prolog support it, a Finite Domain solver.
I usually use GProlog and I can obtain what you ask with something like
fd_domain([A, B], 1, 100),
6 #= A + B,
fd_labeling([A, B]),
where fd_domain/3 set the domain for variables A and B (from 1 to 100), 6 #= A + B set the constraint (A + B is 6) and fd_labelling/1 get all possibles calculations.
In Swi-Prolog is a little different.
First of all, you have to load the CLP(FD) library with
:- use_module(library(clpfd)).
To set the variables and the domain, you can write
Vars = [A, B],
Vars ins 1..100,
Setting the constraint is equal
6 #= A + B,
and to get all possible combinations, you can write
label(Vars),
The generate-and-test approach also works. Of course, you still need some constraints, for example:
?- between(1, 6, X), % X is an integer between 1 and 6
between(1, 6, Y), % Y is an integer between 1 and 6
X =< Y, % X is not larger than Y
X + Y =:= 6. % the sum is 6
X = 1, Y = 5 ;
X = 2, Y = 4 ;
X = Y, Y = 3 ;
false.
The order of the subqueries is significant, so you could as well call it generate-then-test. If you are not afraid to hard-code some of the constraints, there might be ways to avoid generating some of the values, and make some of the tests unnecessary, for example:
?- between(1, 6, X), % X is an integer between 1 and 6
between(X, 6, Y), % Y is an integer between X and 6
X + Y =:= 6. % the sum is 6
X = 1, Y = 5 ;
X = 2, Y = 4 ;
X = Y, Y = 3 ;
false.
You should realize that going down that road far enough is about the same as implementing a constraint solver like CLP(FD) for example.
I am to write a program that does this:
?- pLeap(2,5,X,Y).
X = 2,
Y = 3 ;
X = 3,
Y = 4 ;
X = 4,
Y = 5 ;
X = 5,
Y = 5 ;
false.
(gives all pairs X,X+1 between 2 and 5, plus the special case at the end).
This is supposedly the solution. I don't really understand how it works, could anyone guide me through it?
pLeap(X,X,X,X).
pLeap(L,H,X,Y) :-
L<H,
X is L,
Y is X+1.
pLeap(L,H,X,Y) :-
L=<H,
L1 is L+1,
pLeap(L1,H,X,Y).
I'd do it simply like this:
pLeap(L,H,X,Y) :-
X >= L,
X =< H,
Y is X+1.
Why doesn't it work (ignoring the special case at the end)?
You could use library clpfd for you problem.
:- use_module(library(clpfd)).
pLeap(L,H,X,Y) :-
X in L..H,
Y #= min(H, X+1),
label([X]).
Here is the output:
?- pLeap(2,5,X,Y).
X = 2,
Y = 3 ;
X = 3,
Y = 4 ;
X = 4,
Y = 5 ;
X = 5,
Y = 5.
The >= and =< operators don't instantiate their arguments, and you can only use them once the arguments have already been instantiated.
Put another way, in the given solution, X and Y are given values with is, and the < and =< operators are only used on L and H, whose values are given by the user. (On the given solution, try pLeap(L,H,2,3) and you'll get the same problem as you're having.)
In your case, though, you try to use >= and =< on X, which has no value yet, and so the interpreter complains.