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Given a list of possible summands I want to determine which, if any, can form a given sum. For example, with [1,2,3,4,5] I can make the sum of 9 with [4,5], [5,3,1], and [4,3,2].
I am using GNU Prolog and have something like the following which does not work
numbers([1,2,3,4,5]).
all_unique(_, []).
all_unique(L, [V|T]) :-
fd_exactly(1, L, V),
all_unique(L, T).
fd_sum([], Sum).
fd_sum([H|T], Sum):-
S = Sum + H,
fd_sum(T, S).
sum_clp(N, Summands):-
numbers(Numbers),
length(Numbers, F),
between(1, F, X),
length(S, X),
fd_domain(S, Numbers),
fd_domain(Y, [N]),
all_unique(S, Numbers),
fd_sum(S, Sum),
Sum #= Y,
fd_labeling(S).
I think the main problem is that I am not representing the constraint on the sum properly? Or maybe it is something else?
Just in case you're really interested in CLP(FD), here is your corrected program.
numbers([1,2,3,4,5]).
% note: use builtins where available, both for efficiency and correctness
%all_unique(_, []).
%all_unique(L, [V|T]) :-
% fd_exactly(1, L, V),
% all_unique(L, T).
fd_sum([], 0). % sum_fd_SO.pl:8: warning: singleton variables [Sum] for fd_sum/2
fd_sum([H|T], Sum):-
% note: use CLP(FD) operators and the correct operands
Sum #= S + H,
fd_sum(T, S).
sum_clp(N, S):- % sum_fd_SO.pl:13-23: warning: singleton variables [Summands] for sum_clp/2
numbers(Numbers),
length(Numbers, F),
between(1, F, X),
length(S, X),
fd_domain(S, Numbers),
%fd_domain(Y, [N]),
%all_unique(S, Numbers),
fd_all_different(S),
fd_sum(S, N),
%Sum #= Y,
fd_labeling(S).
test
?- sum_clp(3,L).
L = [3] ? ;
L = [1,2] ? ;
L = [2,1] ? ;
no
I think mixing the code for sublist into clp code is causing some confusion. GNU-Prolog has a sublist/2 predicate, you can use that.
You seem to be building the arithmetic expression with fd_sum but it is incorrectly implemented.
sum_exp([], 0).
sum_exp([X|Xs], X+Xse) :-
sum_exp(Xs, Xse).
sum_c(X, N, Xsub) :-
sublist(Xsub, X),
sum_exp(Xsub, Xe),
N #= Xe.
| ?- sum_exp([A, B, C, D], X).
X = A+(B+(C+(D+0)))
yes
| ?- sum_c([1, 2, 3, 4, 5], 9, X).
X = [4,5] ? ;
X = [2,3,4] ? ;
X = [1,3,5] ? ;
(1 ms) no
| ?- length(X, 4), sum_c(X, 4, [A, B]), member(A, [1, 2, 3]).
A = 1
B = 3
X = [_,_,1,3] ? ;
A = 2
B = 2
X = [_,_,2,2] ? ;
A = 3
B = 1
X = [_,_,3,1] ?
yes
I'm trying to rewrite code from Haskell to Prolog.
count :: Eq a => a -> [a] -> Int
count x = length . filter (x==)
f :: [Integer] -> [Integer]
f [] = []
f list = filter (\x -> count x list == 1) list
This code return list that contains elements that appears only once in the list.
So if I have list [1,1,2,2,3,4,4,5] this function returns [3,5]
I tried to find filter construction in Prolog but seems there no such thing. How can I make similar function in Prolog ?
To the existing answers, I would like to add an answer that is quite general in the sense that you can use it in multiple directions.
Building block: list_element_number/3
I start with the following predicate, defining a relation between:
a list Ls0
an element E
the number N of occurrences of E in Ls0
Here it is:
list_element_number(Ls0, E, N) :-
tfilter(=(E), Ls0, Ls),
length(Ls, N).
This solution uses tfilter/3 from library(reif). The predicate subsumes the function count you have posted. The main benefit of this predicate over the function is that the predicate can be used not only in those cases that even Haskell can do easily, such as:
?- list_element_number([a,b,c], a, N).
N = 1.
No, we can use it also in other directions, such as:
?- list_element_number([a,b,c], X, 1).
X = a ;
X = b ;
X = c ;
false.
Or even:
?- list_element_number([a,b,E], X, 2).
E = X, X = a ;
E = X, X = b ;
false.
Or even:
?- list_element_number([A,B,C], X, 3).
A = B, B = C, C = X ;
false.
And even in the most general case, in which all arguments are fresh variables:
?- list_element_number(Ls, E, N).
Ls = [],
N = 0 ;
Ls = [E],
N = 1 ;
Ls = [E, E],
N = 2 ;
Ls = [E, E, E],
N = 3 .
We can fairly enumerate all answers like this:
?- length(Ls, _), list_element_number(Ls, E, N).
Ls = [],
N = 0 ;
Ls = [E],
N = 1 ;
Ls = [_160],
N = 0,
dif(E, _160) ;
Ls = [E, E],
N = 2 .
Main predicate: list_singletons/2
Using this building block, we can define list_singletons/2 as follows:
list_singletons(Ls, Singles) :-
tfilter(count_one(Ls), Ls, Singles).
count_one(Ls, E, T) :-
list_element_number(Ls, E, Num),
cond_t(Num=1, true, T).
This uses cond_t/3 and (again) tfilter/3 from library(reif).
Sample queries
Here are a few sample queries. First, the test case you have posted:
?- list_singletons([1,1,2,2,3,4,4,5], Singles).
Singles = [3, 5].
It works as desired.
Now a case involving variables:
?- list_singletons([A,B], Singles).
A = B,
Singles = [] ;
Singles = [A, B],
dif(A, B).
On backtracking, all possibilities are generated: Either A = B holds, and in that case, there is no element that occurs only once. Or A is different from B, and in that case both A and B occur exactly once.
As a special case of the above query, we can post:
?- list_singletons([A,A], Singles).
Singles = [].
And as a generalization, we can post:
?- length(Ls, _), list_singletons(Ls, Singles).
Ls = Singles, Singles = [] ;
Ls = Singles, Singles = [_7216] ;
Ls = [_7216, _7216],
Singles = [] ;
Ls = Singles, Singles = [_7828, _7834],
dif(_7828, _7834) ;
Ls = [_7216, _7216, _7216],
Singles = [] ;
Ls = [_7910, _7910, _7922],
Singles = [_7922],
dif(_7910, _7922) .
Enjoy the generality of this relation, obtained via logical-purity.
A more simple version :
filter_list(L,OutList):-findall(X, (select(X,L, L1),\+member(X, L1)) , OutList).
?- filter_list([1,1,2,2,3,4,4,5],L).
L = [3, 5].
Without findall, you can try
filter_list(In, Out) :- filter_list(In, _, Out).
filter_list([], [], []).
filter_list([H|T], L1, L2) :-
filter_list(T, LL1, LL2),
( member(H, LL1)
-> L1 = LL1, L2 = LL2
; (select(H, LL2, L2)
-> L1 = [H|LL1]
; L1 = LL1, L2 = [H|LL2])).
without counting...
filter_uniques([],[]).
filter_uniques([H|T],F) :-
delete(T,H,D),
( D=T -> F=[H|R],S=T ; F=R,S=D ),
filter_uniques(S,R).
a more direct rewrite of your code, with library(yall) support for inlining of the filter predicate (the first argument to include/3)
filt_uniq(L,F) :-
include({L}/[E]>>aggregate(count,member(E,L),1),L,F).
A simple version:
(see the comment by #false below, the version with findall/3 has some inconsistency problems in more complex queries but second version looks ok however it is definitely not so efficient ).
filter_list(L,OutList):-findall(X, (member(X,L),count(X,L,N),N=:=1) , OutList).
count(_,[],0).
count(X,[X|T],N):-count(X,T,N1),N is N1+1.
count(X,[X1|T],N):-dif(X,X1),count(X,T,N).
The predicate filter_list/2 uses findall/3 and simply states find all X that belong to the list L and count returns 1 and store them in OutList.
Example:
?- filter_list([1,1,2,2,3,4,4,5],L).
L = [3, 5].
You could write filter_list/2 without using findall/3 like:
filter_list(L,OutList):- filter_list(L,OutList,L).
filter_list([],[],_).
filter_list([H|T],[H|T1],L):-count(H,L,N), N=:=1, filter_list(T,T1,L).
filter_list([H|T],T1,L):-count(H,L,N), N > 1, filter_list(T,T1,L).
I got a problem with lists. What I need to do is to split one list [1,-2,3,-4], into two lists [1,3] and [-2,-4]. My code looks like the following:
lists([],_,_).
lists([X|Xs],Y,Z):- lists(Xs,Y,Z), X>0 -> append([X],Y,Y) ; append([X],Z,Z).
and I'm getting
Y = [1|Y],
Z = [-2|Z].
What am I doing wrong?
If your Prolog system offers clpfd you could preserve logical-purity. Want to know how? Read on!
We take the second definition of lists/3 that #CapelliC wrote in
his answer as a starting point, and replace partition/4 by tpartition/4 and (<)/2 by (#<)/3:
lists(A,B,C) :- tpartition(#<(0),A,B,C).
Let's run a sample query!
?- As = [0,1,2,-2,3,4,-4,5,6,7,0], lists(As,Bs,Cs).
As = [0,1,2,-2,3,4,-4,5,6,7,0],
Bs = [ 1,2, 3,4, 5,6,7 ],
Cs = [0, -2, -4, 0].
As we use monotone code, we get logically sound answers for more general queries:
?- As = [X,Y], lists(As,Bs,Cs).
As = [X,Y], Bs = [X,Y], Cs = [ ], X in 1..sup, Y in 1..sup ;
As = [X,Y], Bs = [X ], Cs = [ Y], X in 1..sup, Y in inf..0 ;
As = [X,Y], Bs = [ Y], Cs = [X ], X in inf..0 , Y in 1..sup ;
As = [X,Y], Bs = [ ], Cs = [X,Y], X in inf..0 , Y in inf..0 .
Here you have. It splits a list, and does not matter if have odd or even items number.
div(L, A, B) :-
append(A, B, L),
length(A, N),
length(B, N).
div(L, A, B) :-
append(A, B, L),
length(A, N),
N1 is N + 1,
length(B, N1).
div(L, A, B) :-
append(A, B, L),
length(A, N),
N1 is N - 1,
length(B, N1).
Refer this:
domains
list=integer*
predicates
split(list,list,list)
clauses
split([],[],[]).
split([X|L],[X|L1],L2):-
X>= 0,
!,
split(L,L1,L2).
split([X|L],L1,[X|L2]):-
split(L,L1,L2).
Output :
Goal: split([1,2,-3,4,-5,2],X,Y)
Solution: X=[1,2,4,2], Y=[-3,-5]
See, if that helps.
Just for variety, this can also be done with a DCG, which is easy to read for a problem like this:
split([], []) --> [].
split([X|T], N) --> [X], { X >= 0 }, split(T, N).
split(P, [X|T]) --> [X], { X < 0 }, split(P, T).
split(L, A, B) :-
phrase(split(A, B), L).
As in:
| ?- split([1,2,-4,3,-5], A, B).
A = [1,2,3]
B = [-4,-5] ? ;
no
It also provides all the possible solutions in reverse:
| ?- split(L, [1,2,3], [-4,-5]).
L = [1,2,3,-4,-5] ? ;
L = [1,2,-4,3,-5] ? ;
L = [1,2,-4,-5,3] ? ;
L = [1,-4,2,3,-5] ? ;
L = [1,-4,2,-5,3] ? ;
L = [1,-4,-5,2,3] ? ;
L = [-4,1,2,3,-5] ? ;
L = [-4,1,2,-5,3] ? ;
L = [-4,1,-5,2,3] ? ;
L = [-4,-5,1,2,3] ? ;
(2 ms) no
Gaurav's solution will also do this if the cut is removed and an explicit X < 0 check placed in the third clause of the split/3 predicate.
There are several corrections to be done in your code.
If you enjoy compact (as readable) code, a possibility is
lists([],[],[]).
lists([X|Xs],Y,Z) :-
( X>0 -> (Y,Z)=([X|Ys],Zs) ; (Y,Z)=(Ys,[X|Zs]) ), lists(Xs,Ys,Zs).
But since (SWI)Prolog offers libraries to handle common list processing tasks, could be as easy as
lists(A,B,C) :- partition(<(0),A,B,C).
I'm kinda new to Prolog so I have a few problems with a certain task. The task is to write a tail recursive predicate count_elems(List,N,Count) condition List_Element > N, Count1 is Count+1.
My approach:
count_elems( L, N, Count ) :-
count_elems(L,N,0).
count_elems( [H|T], N, Count ) :-
H > N ,
Count1 is Count+1 ,
count_elems(T,N,Count1).
count_elems( [H|T], N, Count ) :-
count_elems(T,N,Count).
Error-Msg:
ERROR: toplevel: Undefined procedure: count_elems/3 (DWIM could not correct goal)
I'm not quite sure where the problem is. thx for any help :)
If you want to make a tail-recursive version of your code, you need (as CapelliC points out) an extra parameter to act as an accumulator. You can see the issue in your first clause:
count_elems(L, N, Count) :- count_elems(L,N,0).
Here, Count is a singleton variable, not instantiated anywhere. Your recursive call to count_elems starts count at 0, but there's no longer a variable to be instantiated with the total. So, you need:
count_elems(L, N, Count) :-
count_elems(L, N, 0, Count).
Then declare the count_elem/4 clauses:
count_elems([H|T], N, Acc, Count) :-
H > N, % count this element if it's > N
Acc1 is Acc + 1, % increment the accumulator
count_elems(T, N, Acc1, Count). % check the rest of the list
count_elems([H|T], N, Acc, Count) :-
H =< N, % don't count this element if it's <= N
count_elems(T, N, Acc, Count). % check rest of list (w/out incrementing acc)
count_elems([], _, Count, Count). % At the end, instantiate total with accumulator
You can also use an "if-else" structure for count_elems/4:
count_elems([H|T], N, Acc, Count) :-
(H > N
-> Acc1 is Acc + 1
; Acc1 = Acc
),
count_elems(T, N, Acc1, Count).
count_elems([], _, Count, Count).
Also as CapelliC pointed out, your stated error message is probably due to not reading in your prolog source file.
Preserve logical-purity with clpfd!
Here's how:
:- use_module(library(clpfd)).
count_elems([],_,0).
count_elems([X|Xs],Z,Count) :-
X #=< Z,
count_elems(Xs,Z,Count).
count_elems([X|Xs],Z,Count) :-
X #> Z,
Count #= Count0 + 1,
count_elems(Xs,Z,Count0).
Let's have a look at how versatile count_elems/3 is:
?- count_elems([1,2,3,4,5,4,3,2],2,Count).
Count = 5 ; % leaves useless choicepoint behind
false.
?- count_elems([1,2,3,4,5,4,3,2],X,3).
X = 3 ;
false.
?- count_elems([1,2,3,4,5,4,3,2],X,Count).
Count = 0, X in 5..sup ;
Count = 1, X = 4 ;
Count = 3, X = Count ;
Count = 5, X = 2 ;
Count = 7, X = 1 ;
Count = 8, X in inf..0 .
Edit 2015-05-05
We could also use meta-predicate
tcount/3, in combination with a reified version of (#<)/2:
#<(X,Y,Truth) :- integer(X), integer(Y), !, ( X<Y -> Truth=true ; Truth=false ).
#<(X,Y,true) :- X #< Y.
#<(X,Y,false) :- X #>= Y.
Let's run above queries again!
?- tcount(#<(2),[1,2,3,4,5,4,3,2],Count).
Count = 5. % succeeds deterministically
?- tcount(#<(X),[1,2,3,4,5,4,3,2],3).
X = 3 ;
false.
?- tcount(#<(X),[1,2,3,4,5,4,3,2],Count).
Count = 8, X in inf..0 ;
Count = 7, X = 1 ;
Count = 5, X = 2 ;
Count = 3, X = Count ;
Count = 1, X = 4 ;
Count = 0, X in 5..sup .
A note regarding efficiency:
count_elems([1,2,3,4,5,4,3,2],2,Count) left a useless choicepoint behind.
tcount(#<(2),[1,2,3,4,5,4,3,2],Count) succeeded deterministically.
Seems you didn't consult your source file.
When you will fix this (you could save these rules in a file count_elems.pl, then issue a ?- consult(count_elems).), you'll face the actual problem that Count it's a singleton in first rule, indicating that you must pass the counter down to actual tail recursive clauses, and unify it with the accumulator (the Count that gets updated to Count1) when the list' visit is done.
You'll end with 3 count_elems/4 clauses. Don't forget the base case:
count_elems([],_,C,C).
I want to make a Prolog program.
Predicate will be like this:
name(name, failedCourse, age)
Database of the program is:
name(george, math, 20).
name(steve, phys, 21).
name(jane, chem, 22).
I want to implement the predicate nameList(A, B). A means list of names, B means number of names on the list. For example:
nameList([george, steve],2). returns true
nameList([george, steve],X). returns X=2
nameList(X,2). returns X=[george, steve]; X=[george, jane]; X=[steve, jane]
nameList([martin],1). returns false (because martin is not included database.)
I wanted to make a list that includes all names on the database. For that reason I made a findall.
descend(X,Y,A) :- name(X,Y,A).
descend(X,Y,A) :- name(X,Z,A),descend(Z,Y,A).
findall(director(X),descend(Y,X),Z).
?- findall(B,descend(B,X,Y),A). returns A = [george, steve, jane].
But I could not figure it out to use list A in predicates :( I cannot search the list for A in the nameList.
If you help me, I will be very grateful.
The main thing you need is a predicate that calculates combinations of a given length and of a given list:
comb(0, _, []).
comb(N, [X | T], [X | Comb]) :-
N > 0,
N1 is N - 1,
comb(N1, T, Comb).
comb(N, [_ | T], Comb) :-
N > 0,
comb(N, T, Comb).
Usage:
?- comb(2, [a, b, a], Comb).
Comb = [a, b] ;
Comb = [a, a] ;
Comb = [b, a] ;
false.
(See more here.)
Now you can just apply this predicate on your data:
name(george, math, 20).
name(steve, phys, 21).
name(jane, chem, 22).
name_list(L, N) :-
findall(X, name(X, _, _), Xs),
length(Xs, Len),
between(0, Len, N),
comb(N, Xs, L).
Usage examples:
?- name_list(L, N).
L = [],
N = 0 ;
L = [george],
N = 1 ;
L = [steve],
N = 1 ;
L = [jane],
N = 1 ;
L = [george, steve],
N = 2 ;
L = [george, jane],
N = 2 ;
L = [steve, jane],
N = 2 ;
L = [george, steve, jane],
N = 3 ;
false.
?- name_list([george, steve], N).
N = 2 ;
false.
?- name_list(L, 2).
L = [george, steve] ;
L = [george, jane] ;
L = [steve, jane] ;
false.
name(george, math, 20).
name(steve, phys, 21).
name(jane, chem, 22).
name_list(Name_List,N) :-
integer(N),
findall(Name,name(Name,_,_),L),
combination(L,N,Name_List).
name_list(Name_List,N) :-
var(N),
findall(Name,name(Name,_,_),L),
length(L,Len),
for(1,N,Len),
combination(L,N,Name_List).
combination(X,1,[A]) :-
member(A,X).
combination([A|Y],N,[A|X]) :-
N > 1,
M is N - 1,
combination(Y,M,X).
combination([_|Y],N,A) :-
N > 1,
combination(Y,N,A).