I have a List with Integers and anonymous variables and I try to find the index of a special values. Problem is as soon I'm using nth1/3 to find the indices Prolog assigns values to the anonymous variables and therefore I find way too indices.
Example:
List = [1,\_,1], where I want as result X = 1, X = 3 from nth1(X,List,1), but as stated before I get X = 1, X = 2, X = 3.
There is a somewhat problematic issue hidden in your requirements: They violate an important declarative property called monotonicity. By this we mean that adding constraints can at most make the solution more specific, never more general.
For example, with the solution you posted, we get:
?- list_el_index([_], 1, N).
false.
Now I add a constraint by imposing an additional requirement on the hitherto free anonymous variable:
?- Var = 1, list_el_index([Var], 1, N).
Var = 1,
N = 0 .
I mean: Come on! We have added a constraint, and as a result get more solutions than before? Such a result is unfortunate and prevents us from reasoning in a logical way about this program.
The program also fails us in other respects. For example, let us ask: Which solutions are there at all?
?- list_el_index(Ls, El, I).
nontermination
Ideally, we would like the program to generate solutions in such cases! This generality is one of the foremost attractions of logic programming, and distinguishes it from more low-level paradigms.
One way to solve such issues is to symbolically distinguish the different kinds of elements that appear in your list.
For example, let us use:
u for an unknown value.
i(I) for an integer I.
With this new representation, your solution becomes:
list_el_index([i(I)|_], I, 0).
list_el_index([_|Tail], Element, Index) :-
list_el_index(Tail, Element, Index0),
Index #= Index0+1.
I have also taken the liberty to replace (is)/2 by (#=)/2, to advertise and stick to more general integer arithmetic that lets us more freely reorder the goals, if necessary. Depending on your Prolog implementation, you may have to import a library to benefit from (#=)/2.
With this representation, your initial case becomes:
?- list_el_index([i(1),u,i(1)], 1, Index).
Index = 0 ;
Index = 2 ;
false.
This works as desired!
Importantly, we can use the predicate also more generally, namely to generate possible answers:
?- list_el_index(Ls, El, I).
Ls = [i(El)|_2994],
I = 0 ;
Ls = [_2992, i(El)|_3000],
I = 1 ;
Ls = [_2992, _2998, i(El)|_3006],
I = 2 ;
Ls = [_2992, _2998, _3004, i(El)|_3012],
I = 3 .
Due to the program's monotonicity, we can fairly enumerate solutions by iterative deepening:
?- length(Ls, _), list_el_index(Ls, El, I).
Ls = [i(El)],
I = 0 ;
Ls = [i(El), _4812],
I = 0 ;
Ls = [_4806, i(El)],
I = 1 ;
Ls = [i(El), _4812, _4818],
I = 0 ;
etc.
This has become possible by using a representation that lets us distinguish the cases by pattern matching. Consider using this approach to make your programs usable in all directions, and to make logical reasoning applicable. It is quite easy to apply by using the appropriate wrapper or constant, and greatly increases the generality of your programs.
This works :
- L = [1,_,1], nth1(X, L, Y), ground(Y), Y= 1.
L = [1,_310914,1],
X = Y, Y = 1 ;
L = [1,_310914,1],
X = 3,
Y = 1.
Thanks to lurkers hint, I came up with this solution.
list_el_index([El1|_], El2, 0) :-
El1 == El2.
list_el_index([_|Tail], Element, Index) :-
list_el_index(Tail, Element, Index1),
Index is Index1+1.
Related
I've written some predicates which take the length of a list and attaches some constraints to it (is this the right vocabulary to be using?):
clp_length([], 0).
clp_length([_Head|Rest], Length) :-
Length #>= 0, Length #= Length1 + 1,
clp_length(Rest, Length1).
clp_length2([], 0).
clp_length2([_Head|Rest], Length) :-
Length #= Length1 + 1,
clp_length2(Rest, Length1).
The first terminates on this simple query, but the second doesn't:
?- Small in 1..2, clp_length(Little, Small).
Small = 1,
Little = [_1348] ;
Small = 2,
Little = [_1348, _2174] ;
false.
?- Small in 1..2, clp_length2(Little, Small).
Small = 1,
Little = [_1346] ;
Small = 2,
Little = [_1346, _2046] ;
% OOPS %
This is strange to me, because Length is pretty clearly greater than 0. To figure that out you could either search, find the zero, and deduce that adding from zero can only increase the number, or you could propagate the in 1..2 constraint down. It feels like the extra clause is redundant! That it isn't means my mental model of clpfd is pretty wrong.
So I think I have two questions (would appreciate answers to the second as comments)
Specifically, why does this additional constraint cause the query to work correctly?
Generally, is there a resource I can use to learn about how clpfd is implemented, instead of just seeing some examples of how it can be used? I'd prefer not to have to read Markus Triska's thesis but that's the only source I can find. Is that my only option if I want to be able to answer questions like this one?
1mo, there is the issue with naming. Please refer to previous answers by
mat
and me recommending relational names. You won't go far using inappropriate names. So list_length/2 or list_fdlength/2 would be an appropriate name. Thus we have list_fdlength/2 and list_fdlength2/2.
2do, consider the rule of list_fdlength2/2. Nothing suggests that 0 is of relevance to you. So that rule will be exactly the same if you are using 0 or 1 or -1 or whatever as base case. So how should this poor rule ever realize that 0 is the end to you? Better, consider a generalization:
list_fdlength2(fake(N), N) :- % Extension to permit fake lists
N #< 0.
list_fdlength2([], 0).
list_fdlength2([_Head|Rest], Length) :-
Length #= Length1 + 1,
list_fdlength2(Rest, Length1).
This generalization shows all real answers plus fake answers. Note that I have not changed the rule, I added this alternative fact only. Thus the fake solutions are actually caused by the rule:
?- list_fdlength2(L, 1).
L = [_A]
; L = [_A, _B|fake(-1)]
; L = [_A, _B, _C|fake(-2)]
; ... .
?- list_fdlength2(L, 0).
L = []
; L = [_A|fake(-1)]
; L = [_A, _B|fake(-2)]
; ... .
Each clause tries to contribute to the solutions just in the scope of the clause. But there is no way to derive (by the built-in Prolog execution mechanism) that some rules are no longer of relevance. You have to state that explicitly with redundant constraints as you did.
Now, back to your original solution containing the redundant constraint Length #>= 0. There should not be any such fake solution at all.
list_fdlength(fake(N), N) :-
N #< 0.
list_fdlength([], 0).
list_fdlength([_Head|Rest], Length) :-
Length #>= 0,
Length #= Length1 + 1,
list_fdlength(Rest, Length1).
?- list_fdlength(L, 1).
L = [_A]
; L = [_A, _B|fake(-1)] % totally unexpected
; false.
?- list_fdlength(L, 0).
L = []
; L = [_A|fake(-1)] % eek
; false.
There are fake answers, too! How ugly! At least, they are finite in number. But, you could have done it better by using
Length #>= 1 in place of Length #>=0. With this little change, there are no longer any fake solutions when N is non-negative and thus also your original program will be better.
If I have X #> 3. Prolog will give the residual goal X in 4..sup.. How can I assign X one of the possible values from 4 to sup? Any random value will suffice.
In SWI-Prolog,you can get random value assign with using secret option random_value(Seed).
see this:How does `random_variable `random_value` work in SWI-Prolog's labeling/2?
:- use_module(library(clpfd)).
random_labeling_test(A):-
A in 1..5,
labeling([random_value(332)],[A]).
?- random_labeling_test(A).
A = 5 ;
A = 1 ;
A = 2 ;
A = 4 ;
A = 3.
332 is meaningless.Set random seed into here.current milliseconds is often used .
I'm not sure this option is safe.
But in your case,this cannot work because upper bounds is infinite.It is reasonable.
In ECLiPSe, labeling variable selection and assigning value selection is completly controllable with using indomain and delete.
see this:http://eclipseclp.org/doc/tutorial/tutorial088.html
One naive way to solve this is to use a predicate like this:
enumeration(Z) :-
length(_, N),
enumeration_(N, Z).
enumeration_(N, N).
enumeration_(N0, N) :- N #= -N0.
You can now solve your task as follows:
?- X #> 3, enumeration(X).
X = 4 ;
X = 5 ;
X = 6 ;
X = 7 ;
X = 8 .
Pros:
a quite simple solution
portable to different Prolog systems.
Con:
potentially quite slow.
Example:
?- X #> 2^100, enumeration(X).
[waiting...]
To make this more efficient, you need to take into account the actual domains of variables.
There are at least two ways to do this:
(a) Use your solver's reflection predicates
Reflection predicates let you reason about the actual domains of variables by making them available as Prolog terms.
For example, where available, you can use fd_inf/2 to obtain the infimum of a constrained variable's domain:
?- X #> 3, fd_inf(X, Inf).
Inf = 4,
X in 4..sup.
You can use this as a starting point for the enumeration. I leave this as a challenge.
See your Prolog system's manual for more information about its reflection predicates.
(b) Inspect the residual goals
Alternatively, you can inspect the residual goals as Prolog terms. This is also a reflection mechanism, though one that only needs a single predicate for all kinds of different constraint solvers.
For example, where available, you can use copy_term/3 to obtain the residual goals as a list of Prolog goals:
?- X #> 3, copy_term(X, X, Gs).
Gs = [clpfd:(X in 4..sup)],
X in 4..sup.
From these goals, it is again straight-forward to deduce the infimum of X's domain.
Note that the top-level can use this same mechanism to actually produce the residual goals.
I'm starting learning Prolog and I want a program that given a integer P gives to integers A and B such that P = A² + B². If there aren't values of A and B that satisfy this equation, false should be returned
For example: if P = 5, it should give A = 1 and B = 2 (or A = 2 and B = 1) because 1² + 2² = 5.
I was thinking this should work:
giveSum(P, A, B) :- integer(A), integer(B), integer(P), P is A*A + B*B.
with the query:
giveSum(5, A, B).
However, it does not. What should I do? I'm very new to Prolog so I'm still making lot of mistakes.
Thanks in advance!
integer/1 is a non-monotonic predicate. It is not a relation that allows the reasoning you expect to apply in this case. To exemplify this:
?- integer(I).
false.
No integer exists, yes? Colour me surprised, to say the least!
Instead of such non-relational constructs, use your Prolog system's CLP(FD) constraints to reason about integers.
For example:
?- 5 #= A*A + B*B.
A in -2..-1\/1..2,
A^2#=_G1025,
_G1025 in 1..4,
_G1025+_G1052#=5,
_G1052 in 1..4,
B^2#=_G406,
B in -2..-1\/1..2
And for concrete solutions:
?- 5 #= A*A + B*B, label([A,B]).
A = -2,
B = -1 ;
A = -2,
B = 1 ;
A = -1,
B = -2 ;
etc.
CLP(FD) constraints are completely pure relations that can be used in the way you expect. See clpfd for more information.
Other things I noticed:
use_underscores_for_readability_as_is_the_convention_in_prolog instead ofMixingTheCasesToMakePredicatesHardToRead.
use declarative names, avoid imperatives. For example, why call it give_sum? This predicate also makes perfect sense if the sum is already given. So, what about sum_of_squares/3, for example?
For efficiency sake, Prolog implementers have choosen - many,many years ago - some compromise. Now, there are chances your Prolog implements advanced integer arithmetic, like CLP(FD) does. If this is the case, mat' answer is perfect. But some Prologs (maybe a naive ISO Prolog compliant processor), could complain about missing label/1, and (#=)/2. So, a traditional Prolog solution: the technique is called generate and test:
giveSum(P, A, B) :-
( integer(P) -> between(1,P,A), between(1,P,B) ; integer(A),integer(B) ),
P is A*A + B*B.
between/3 it's not an ISO builtin, but it's rather easier than (#=)/2 and label/1 to write :)
Anyway, please follow mat' advice and avoid 'imperative' naming. Often a description of the relation is better, because Prolog it's just that: a relational language.
I am attempting a past paper question for a Prolog exam. I drew a 'tree' for how I believed Prolog ought to behave given the program and a certain goal. However, Prolog does not behave as I expected, and given a query for which I believed it would return 'true', it actually returned 'false'.
Here is my program:
sum(Term,N) :- Term = 0, N = 0.
sum(Term,N) :- Term = f(M,Subterm), number(M), sum(Subterm,N-M).
My query and search tree are as follows (goals are bracketed and in bold):
[ sum(f(1,0),1) ]
Using Rule 1, let Term = 0, N = 0, tries to unify [ 1 = 0, 1 = 0 ] fail.
Redo: using Rule 2, let Term = f(1,0), N=1 [ f(1,0) = f(M,Subterm), number(M), sum(Subterm,1-1) ]
Unifying, let M=1 and Subterm=0 [ number(1), sum(0,0) ]
Using Rule 1, this should succeed. However (SWI) Prolog says 'false'.
If someone can point out to me why my reasoning is flawed (and how I can learn from this in future), I would be very grateful.
Since your program is almost a pure1 one, you can locate the error in a systematic manner without using a debugger. The idea is to generalize your program by removing goals, one-by-one. I came up with the following pure generalization which I obtained by "commenting" out some goals like so:
:- op(950, fy, *).
*(_).
sum(Term,N) :-
Term = 0,
N = 0.
sum(Term,N) :-
* Term = f(M,Subterm),
* number(M),
sum(Subterm,N-M).
?- sum(Term, N).
Term = 0, N = 0
; false.
Also the query above is more general than yours. This is a very useful technique in Prolog: Instead of thinking about concrete solutions, we
first let Prolog do all the work for us.
The answer was quite clear: There is exactly one solution to this relation, even if the relation is now generalized.
So the problem must be somewhere in the remaining visible part. Actually, it's the -. Why not write instead:
:- use_module(library(clpfd)).
sum(0, 0).
sum(Term, N0) :-
Term = f(M, Subterm),
N0 #= M+N1,
sum(Subterm, N1).
I find that program much easier to understand. If I read a name sum, I immediately look for a corresponding +. Of course, if you insist, you could write N0-M #= N1 instead. It would be exactly the same, except that this requires a bit more thinking.
Fine print you don't need to read
1) Your original program used number/1 which is not pure. But since the problem persisted by removing it, it did not harm our reasoning.
To be more accurate, the first rule tries to unify f(1,0) = 0 and 1 = 0, which of course fails.
Analysis of rule 2 is also incorrect. Partly, it's because Prolog does not evaluate arithmetic expressions inline. The term N-M is just a term (short-hand for '-'(N, M). It does not result in M being subtracted from M unless the evaluation is done explicitly via is/2 or an arithmetic comparison (e.g., =:=/2, =</2, etc).
The analysis of rule 2 would go as follows. Step 5 is where your logic breaks down due to the above.
Call sum(f(1,0), 1) results in Term = f(1,0) and N = 1.
In rule 2, Term = f(M, Subterm) becomes f(1,0) = f(M, Subterm) which results in M = 1 and Subterm = 0.
number(N) becomes number(1) and succeeds (since 1 is a number)
The call sum(Subterm, N-M) becomes sum(0, 1-1).
Prolog matches sum(0, 1-1) with the head of rule 1 sum(Term, N) :- Term = 0, N = 0., but it fails because 1-1 = 0 (which is the same as '-'(1, 1) = 0 unification fails.
Prolog matches sum(0, 1-1) with the head of rule 2, and unifies Term = 0 and N = 1-1 (or N = '-'(1, 1)).
Term = f(M, Subterm) becomes 0 = f(M, Subterm) which fails because 0 cannot match the term f(M, Subterm).
No more rules to attempt, so the predicate call fails.
The easy fix here is a common, basic Prolog pattern to use a new variable to evaluate the expression explicitly:
sum(Term,N) :-
Term = f(M,Subterm),
number(M),
R is N - M,
sum(Subterm, R).
You can also tidy up the code quite a bit by unifying in the heads of the clauses. So the clauses could be rewritten:
sum(0, 0).
sum(f(M, Subterm), N) :-
number(N),
R is N - M,
sum(Subterm, R).
EDIT: My answer is intended to guide you through a walk through of your existing logic. Other than correcting the misunderstanding regarding expression evaluation, I did not analyze your solution for overall correctness.
just started programming with prolog and I'm having a few issues. The function I have is supposed to take a value X and copy it N number of times into M. My function returns a list of N number of memory locations. Here's the code, any ideas?
duple(N,_,M):- length(M,Q), N is Q.
duple(N,X,M):- append(X,M,Q), duple(N,X,Q).
Those are not memory adresses. Those are free variables. What you see is their internal names in your prolog system of choice. Then, as #chac pointed out (+1 btw), the third clause is not really making sense! Maybe you can try to tell us what you meant so that we can bring light about how to do it correctly.
I'm going to give you two implementations of your predicate to try to show you correct Prolog syntax:
duple1(N, X, L) :-
length(L, N),
maplist(=(X), L).
Here, in your duple1/3 predicate, we tell prolog that the length of the resulting list L is N, and then we tell it that each element of L should be unified with X for the predicate to hold.
Another to do that would be to build the resulting list "manually" through recursion:
duple2(0, _X, []).
duple2(N, X, [X|L]) :-
N > 0,
NewN is N - 1,
duple1(NewN, X, L).
Though, note that because we use >/2, is and -/2, ie arithmetic, we prevent prolog from using this predicate in several ways, such as:
?- duple1(X, Y, [xyz, xyz]).
X = 2,
Y = xyz.
This worked before, in our first predicate!
Hope this was of some help.
I suppose you call your predicate, for instance, in this way:
?- duple(3,xyz,L).
and you get
L = [_G289, _G292, _G295] ;
ERROR: Out of global stack
If you try
?- length(X,Y).
X = [],
Y = 0 ;
X = [_G299],
Y = 1 ;
X = [_G299, _G302],
Y = 2 ;
X = [_G299, _G302, _G305],
Y = 3 ;
X = [_G299, _G302, _G305, _G308],
Y = 4 .
...
you can see what's happening:
your query will match the specified *M*, displaying a list of M uninstantiated variables (memory locations), then continue backtracking and generating evee longer lists 'til there is stack space. Your second rule will never fire (and I don't really understand its purpose).
A generator is easier to write in this way:
duple(N,X,M) :- findall(X,between(1,N,_),M).
test:
?- duple(3,xyz,L).
L = [xyz, xyz, xyz].