Example of my CLP problem (this is a small part of a larger problem which uses the clpfd library):
For a list of length 5, a fact el_sum(Pos,N,Sum) specifies that the N consecutive elements starting from position Pos (index from 1) have sum equal to Sum. So if we have
el_sum(1,3,4).
el_sum(2,2,3).
el_sum(4,2,5).
Then [1,2,1,4,1] would work for this example since 1+2+1=4, 2+1=3, 4+1=5.
I'm struggling with how to even start using the el_sum's to find solutions with an input list [X1,X2,X3,X4,X5]. I'm thinking I should use findall but I'm not really getting anywhere.
(My actual problem is much bigger than this so I'm looking for a solution that doesn't just work for three facts and a small list).
Thanks!
You are mixing here the monotonic world of constraints with some non-monotonic quantification. Don't try to mix them too closely. Instead, first transform those facts into, say, a list of terms.
el_sums(Gs) :-
G = el_sum(_,_,_),
findall(G, G, Gs).
And then, only then, start with the constraint part that will now remain monotonic. So:
?- el_sums(Gs), length(L5,5), maplist(l5_(L5), Gs).
l5_(L5, el_sum(P, N, S)) :-
length([_|Pre], P),
length(Cs, N),
phrase((seq(Pre),seq(Cs),seq(_)), L5),
list_sum(Cs,S).
seq([]) --> [].
seq([E|Es]) --> [E], seq(Es).
Not sure this will help, I don't understand your workflow... from where the list do come ? Anyway
:- [library(clpfd)].
el_sum(Pos,N,Sum) :-
length(L, 5),
L ins 0..100,
el_sum(Pos,N,Sum,L),
label(L), writeln(L).
el_sum(P,N,Sum,L) :-
N #> 0,
M #= N-1,
Q #= P+1,
el_sum(Q,M,Sum1,L),
element(N,L,T),
Sum #= Sum1 + T.
el_sum(_P,0,0,_L).
yields
?- el_sum(1,2,3).
[0,3,0,0,0]
true ;
[0,3,0,0,1]
true ;
...
Related
I want to write predicate which can count all encountered number:
count(1, [1,0,0,1,0], X).
X = 2.
I tried to write it like:
count(_, [], 0).
count(Num, [H|T], X) :- count(Num, T, X1), Num = H, X is X1 + 1.
Why doesn't work it?
Why doesn't work it?
Prolog is a programming language that often can answer such question directly. Look how I tried out your definition starting with your failing query:
?- count(1, [1,0,0,1,0], X).
false.
?- count(1, Xs, X).
Xs = [], X = 0
; Xs = [1], X = 1
; Xs = [1,1], X = 2
; Xs = [1,1,1], X = 3
; ... .
?- Xs = [_,_,_], count(1, Xs, X).
Xs = [1,1,1], X = 3.
So first I realized that the query does not work at all, then I generalized the query. I replaced the big list by a variable Xs and said: Prolog, fill in the blanks for me! And Prolog did this and reveals us precisely the cases when it will succeed.
In fact, it only succeeds with lists of 1s only. That is odd. Your definition is too restricted - it correctly counts the 1s in lists where there are only ones, but all other lists are rejected. #coder showed you how to extend your definition.
Here is another one using library(reif) for
SICStus|SWI. Alternatively, see tfilter/3.
count(X, Xs, N) :-
tfilter(=(X), Xs, Ys),
length(Ys, N).
A definition more in the style of the other definitions:
count(_, [], 0).
count(E, [X|Xs], N0) :-
if_(E = X, C = 1, C = 0),
count(E, Xs, N1),
N0 is N1+C.
And now for some more general uses:
How does a four element list look like that has 3 times a 1 in it?
?- length(L, 4), count(1, L, 3).
L = [1,1,1,_A], dif(1,_A)
; L = [1,1,_A,1], dif(1,_A)
; L = [1,_A,1,1], dif(1,_A)
; L = [_A,1,1,1], dif(1,_A)
; false.
So the remaining element must be something different from 1.
That's the fine generality Prolog offers us.
The problem is that as stated by #lurker if condition (or better unification) fails then the predicate will fail. You could make another clause for this purpose, using dif/2 which is pure and defined in the iso:
count(_, [], 0).
count(Num, [H|T], X) :- dif(Num,H), count(Num, T, X).
count(Num, [H|T], X) :- Num = H, count(Num, T, X1), X is X1 + 1.
The above is not the most efficient solution since it leaves many choice points but it is a quick and correct solution.
You simply let the predicate fail at the unification Num = X. Basically, it's like you don't accept terms which are different from the only one you are counting.
I propose to you this simple solution which uses tail recursion and scans the list in linear time. Despite the length, it's very efficient and elegant, it exploits declarative programming techniques and the backtracking of the Prolog engine.
count(C, L, R) :-
count(C, L, 0, R).
count(_, [], Acc, Acc).
count(C, [C|Xr], Acc, R) :-
IncAcc is Acc + 1,
count(C, Xr, IncAcc, R).
count(C, [X|Xr], Acc, R) :-
dif(X, C),
count(C, Xr, Acc, R).
count/3 is the launcher predicate. It takes the term to count, the list and gives to you the result value.
The first count/4 is the basic case of the recursion.
The second count/4 is executed when the head of the list is unified with the term you are looking for.
The third count/4 is reached upon backtracking: If the term doesn’t match, the unification fails, you won't need to increment the accumulator.
Acc allows you to scan the entire list propagating the partial result of the recursive processing. At the end you simply have to return it.
I solved it myself:
count(_, [], 0).
count(Num, [H|T], X) :- Num \= H, count(Num, T, X).
count(Num, [H|T], X) :- Num = H, count(Num, T, X1), X is X1 + 1.
I have decided to add my solution to the list here.
Other solutions here use either explicit unification/failure to unify, or libraries/other functions, but mine uses cuts and implicit unification instead. Note my solution is similar to Ilario's solution but simplifies this using cuts.
count(_, [], 0) :- !.
count(Value, [Value|Tail],Occurrences) :- !,
count(Value,Tail,TailOcc),
Occurrences is TailOcc+1.
count(Value, [_|Tail], Occurrences) :- count(Value,Tail,Occurrences).
How does this work? And how did you code it?
It is often useful to equate solving a problem like this to solving a proof by induction, with a base case, and then a inductive step which shows how to reduce the problem down.
Line 1 - base case
Line 1 (count(_, [], 0) :- !.) handles the "base case".
As we are working on a list, and have to look at each element, the simplest case is zero elements ([]). Therefore, we want a list with zero elements to have no instances of the Value we are looking for.
Note I have replaced Value in the final code with _ - this is because we do not care what value we are looking for if there are no values in the list anyway! Therefore, to avoid a singleton variable we negate it here.
I also added a ! (a cut) after this - as there is only one correct value for the number of occurrences we do not want Prolog to backtrack and fail - therefore we tell Prolog we found the correct value by adding this cut.
Lines 2/3 - inductive step
Lines 2 and 3 handle the "inductive step". This should handle if we have one or more elements in the list we are given. In Prolog we can only directly look at the head of the list, therefore let us look at one element at a time. Therefore, we have two cases - either the value at the head of the list is the Value we are looking for, or it is not.
Line 2
Line 2 (count(Value, [Value|Tail],Occurrences) :- !, count(Value,Tail,TailOcc), Occurrences is TailOcc+1.) handles if the head of our list and the value we are looking for match. Therefore, we simply use the same variable name so Prolog will unify them.
A cut is used as the first step in our solution (which makes each case mutually exclusive, and makes our solution last-call-optimised, by telling Prolog not to try any other rules).
Then, we find out how many instances of our term there are in the rest of the list (call it TailOcc). We don't know how many terms there are in the list we have at the moment, but we know it is one more than there are in the rest of the list (as we have a match).
Once we know how many instances there are in the rest of the list (call this Tail), we can take this value and add 1 to it, then return this as the last value in our count function (call this Occurences).
Line 3
Line 3 (count(Value, [_|Tail], Occurrences) :- count(Value,Tail,Occurrences).) handles if the head of our list and the value we are looking for do not match.
As we used a cut in line 2, this line will only be tried if line 2 fails (i.e. there is no match).
We simply take the number of instances in the rest of the list (the tail) and return this same value without editing it.
I want to solve this problem in Prolog. i want give a list of natural numbers to find all the elements in the list that satisfy this condition:
All elements on the left of it are smaller than it and all the elements on the right of it are larger than it.
For example give a list [3,2,4,1,5,7,8,9,10,8] the answer would be 5,7
So far I've manage to make this function that given an element of the list it returns true or false if the element satisfises the condition described above.
check(Elem, List) :-
seperate(Elem, List, List1, List2),
lesser(Elem, List1, X1),
bigger(Elem, List2, X2),
size(X1, L1),
size(X2, L2),
size(List, L3),
match(L1, L2, L3),
Now I want to make another predicate that given a list, it does the above computations for each element of the list. Due to the fact that more than one element may satisfy it I want to create another list with all the elements that satisfy the problem.
The question would be something like ?-predicate_name([[3,2,4,1,5,7,8,9,10,8],N). and the result would be a list of elements.
Sry If I am not using the right terms of Prolog. I will describe what I want to do in sequential logic language to be more specific although it's not a good idea to think like that. If we consider a the predicate check as a function that given a list and element of the list it would return true or false whether or not the element satisfied the conditions of the problem. Now I want to parse each element of the list and for each one of it call the function check. If that would return true then I would add the element in another list a.k.a result. I want to do this in Prolog but I don't know how to iterate a list.
Here is a version using DCGs and assuming we want to compare arithmetically.
list_mid(L, M) :-
phrase(mid(M), L).
mid(M) -->
seq(Sm),
[M],
{maplist(>(M),Sm)},
seq(Gr),
{maplist(<(M),Gr)}.
seq([]) -->
[].
seq([E|Es]) -->
[E],
seq(Es).
Often it is not worth optimizing this any further. The first seq(Sm) together with the subsequent maplist/2 might be merged together. This is a bit tricky, since one has to handle separately the cases where Sm = [] and Sm = [_|_].
mid(M) -->
( [M]
| max(Mx),
[M],
{Mx < M}
),
min(M).
max(M) -->
[E],
maxi(E, M).
maxi(E, E) -->
[].
maxi(E, M) -->
[F],
{G is max(F,E)},
maxi(G, M).
min(_) -->
[].
min(M) -->
[E],
{M < E},
min(M).
I'm going to take a different approach on the problem.
We want to find all of the values that meet the criteria of being a "mid" value, which is one defined as being greater than all those before it in the list, and less than all those after.
Define a predicate mid(L, M) as meaning M is a "mid" value of L:
mid([X|T], X) :- % The first element of a list is a "mid" if...
less(X, T). % it is less than the rest of the list
mid([X|T], M) :- % M is a "mid" of [X|T] if...
mid(T, X, M). % M is a "mid" > X
% (NOTE: first element is not a "mid" by definition)
mid([X|T], LastM, X) :- % X is a "mid" > Y if...
X > LastM, % X > the last "mid"
less(X, T). % X < the rest of the list, T
mid([X|T], LastM, M) :- % Also, M is a "mid" if...
Z is max(X, LastM), % Z is the larger of X and the last "mid"
mid(T, Z, M). % M is the "mid" of T which is > Z
less(X, [Y|T]) :- % X is less than the list [Y|T] if...
X < Y, % X < Y, and
less(X, T). % X < the tail, T
less(_, []). % An element is always less than the empty list
Each query will find the next "mid":
| ?- mid([3,2,4,1,5,7,8,9,10,8], M).
M = 5 ? ;
M = 7 ? ;
no
Then they can be captured with a findall:
mids(L, Ms) :-
findall(M, mid(L, M), Ns).
| ?- mids([3,2,4,1,5,7,8,9,10,8], Ms).
Ms = [5,7]
yes
| ?- mids([2], L).
L = [2]
(1 ms) yes
This is probably not the most computationally efficient solution since it doesn't take advantage of a couple of properties of "mids". For example, "mids" will all be clustered together contiguously, so once a "mid" is found, it doesn't make sense to continue searching if an element is subsequently encountered which is not itself a "mid". If efficiency is a goal, these sorts of ideas can be worked into the logical process.
ADDENDUM
With credit to #false for reminding me about maplist, the above predicate call less(X, T) could be replaced by maplist(<(X), T) eliminating the definition of less in the above implementation.
Basically, I need to create a predicate of the form sublist(S,M,N,L), where S is a new list formed from the elements of L between index M and index N, inclusive.
Here's where I've gotten:
sublist([],_,_,[]).
sublist([],M,N,_) :- (M > N).
sublist(S,M,N,L) :- sublist2(S,M,N,L,-1).
sublist2([H|T],St,En,[H2|T2],Idx) :-
(Idx2 is Idx + 1,
St =< Idx2,
En >= Idx2,
H = H2,
sublist2(T,St,En,T2,Idx2);
Idx2 is Idx + 1,
sublist2(T,St,En,T2,Idx2)).
As with all my prolog problems, I feel I'm making it way more complicated than it should be. I've got the base cases right, but anything else evaluates to false. Any advice for this problem, and just general approach to prolog? I understand the language for the most part, but I can't seem to see the simple solutions.
Simple solutions follow simple outlook. For lists it's recursion. Recursive programming is simple - just imagine you already have your function, following the given interface/requirements, and so you get to use it whenever you feel like it (but better, in the reduced cases).
sublist(S,M,N,[_A|B]):- M>0, M<N, sublist(S,M-1,N-1,B).
think of it as stating a law of sublists: sublist in a shorter list starts at decreased index.
sublist(S,M,N,[A|B]):- 0 is M, M<N, N2 is N-1, S=[A|D], sublist(D,0,N2,B).
and,
sublist([],0,0,_).
it is exclusive in the second index. tweak it. :)
There is the possibility to handle indexing in a way similar to more traditional languages:
sublist(L, M, N, S) :-
findall(E, (nth1(I, L, E), I >= M, I =< N), S).
or equivalently
sublist(L, M, N, S) :-
findall(E, (between(M, N, I), nth1(I, L, E)), S).
nth1/3 is for indexing from 1, otherwise nth0/3 allows C style - start from 0. I've placed the sublist as last argument. It's a common convention in Prolog to place output parameters after input.
Here a (cumbersome) recursive definition
sublist(L,M,N,S) :- sublist2(1,L,M,N,S).
sublist2(_,[],_,_,[]).
sublist2(I,[X|Xs],M,N,[X|Ys]) :-
between(M,N,I),
J is I + 1,
!, sublist2(J,Xs,M,N,Ys).
sublist2(I,[_|Xs],M,N,Ys) :-
J is I + 1,
sublist2(J,Xs,M,N,Ys).
I need to modify the vanilla meta-interpreter in order to make a search with limited depth. I'm using the following code for testing my sollution:
value(wire1,1).
connected(wire2, wire1).
connected(wire3, wire2).
connected(wire4, wire3).
connected(wire5, wire4).
connected(wire6, wire5).
connected(wire7, wire6).
connected(wire8, wire7).
connected(wire9, wire8).
value(W,X):-connected(W,V), value(V,X).
And the target is that something like:
solve(value(w9,X), 3). /*depth =3, it should return false*/
solve(value(w9,X), 20). /*depth=20 is enought for returning X=1*/
By the way my code is
solve(true,_):-!.
solve((A,B),D) :-!, solve(A,D), solve(B,D).
solve(A,D) :- clause(A, B),solve(B,D2),D=D2+1,D>0).
But it don't work property. Can you help me? Thanks a lot in advance
An interesting page on metaprogramming came from a good developer: Markus Triska.
Here (A Couple of Meta-interpreters in Prolog) you find both theory and practice. For instance:
... Another group of extensions aims to improve the incomplete default computation strategy. We start from an MI that limits the depth of the search tree:
mi_limit(Goal, Max) :-
mi_limit(Goal, Max, _).
mi_limit(true, N, N).
mi_limit((A,B), N0, N) :-
mi_limit(A, N0, N1),
mi_limit(B, N1, N).
mi_limit(g(G), N0, N) :-
N0 > 0,
mi_clause(G, Body),
N1 is N0 - 1,
mi_limit(Body, N1, N).
You were almost there. Only the last clause needs a slight rearranging:
solve(A, D) :- clause(A, B), D1 is D - 1, D1 > 0, solve(B, D1).
?- solve(value(wire9, X), 9). ===> false.
?- solve(value(wire9, X), 10). ===> X = 1.
dls(X,X,[X],L):-
L >0 goal(X).
dls(X,Y,[A|p],L):-
L > 0 ,goal(Y) ,
move(X,Y),
L1 is L - 1 ,
dls(Z,Y ,P,L1).
Me and a friend are writing a program which is supposed to solve a CLP problem. We want to use minimize to optimize the solution but it won't work, because it keeps saying that the number we get from sum(P,#=,S) is between two numbers (for example 5..7). We haven't been able to find a good way to extract any number from this or manipulate it in any way and are therefore looking for your help.
The problem seems to arise from our gen_var method which says that each element of a list must be between 0 and 1, so some numbers come out as "0..1" instead of being set properly.
Is there any way to use minimize even though we get a number like "5..7" or any way to manipulate that number so that we only get 5? S (the sum of the elements in a list) is what we're trying to minimize.
gen_var(0, []).
gen_var(N, [X|Xs]) :-
N > 0,
M is N-1,
gen_var(M, Xs),
domain([X],0,1).
find([],_).
find([H|T],P):- match(H,P),find(T,P).
match(pri(_,L),P):-member(X,L), nth1(X,P,1).
main(N,L,P,S) :- gen_var(N,P), minimize(findsum(L,P,S),S).
findsum(L,P,S):- find(L,P), sum(P,#=,S).
I've slightly modified your code, to adapt to SWI-Prolog CLP(FD), and it seems to work (kind of). But I think the minimum it's always 0!
:- use_module(library(clpfd)).
gen_var(0, []).
gen_var(N, [X|Xs]) :-
N > 0,
M is N-1,
gen_var(M, Xs),
X in 0..1 .
find([], _).
find([H|T], P):-
match(H, P),
find(T, P).
match(pri(_,L),P):-
member(X, L),
nth1(X, P, 1).
findsum(L,P,S) :-
find(L, P),
sum(P, #=, S).
main(N, L, P, S) :-
gen_var(N, P),
findsum(L, P, S),
labeling([min(S)], P).
Is this output sample a correct subset of the expected outcome?
?- main(3,A,B,C).
A = [],
B = [0, 0, 0],
C = 0 ;
A = [],
B = [0, 0, 1],
C = 1 ;