Related
Eg:
List[1,2,3,4,5,6] with N equal to 6 should print true because there are exactly 3 values that add upto 6. 1+2+3.
List[2,5,7,9] with N equal to 12 should print false as there are no 3 elements that add upto 12.
Let's maybe start with a more general predicate that describes the relation between a list, a sublist of said list and the sum of numbers in the sublist. To make it obvious which argument is what, it is opportune to chose a descriptive name for the predicate, say sum_ofsub_fromlist/3. Now let's observe that if the first argument is the sum of the numbers in the sublist, then successively subtracting those numbers from the sum yields zero, e.g.: X=A+B → X-A-B=0. So there will be a base case that contains 0 as the sum and [] as the sublist (rule 1) and a recursive rule that subtracts the elements of the sublist from the sum (rule 2). And since a sublist does not contain all elements of the list it's taken from in general, there will be a recursive rule for skipping elements of the list that do not occur in the sublist (rule 3). This rule is only needed as long as there are still elements in the sublist, so a constraint would be beneficial, that prevents this rule from succeeding once the sublist is empty. These ideas can be realized in Prolog like so:
sum_ofsub_fromlist(0,[],_L). % rule 1
sum_ofsub_fromlist(X,[A|Bs],[A|As]) :- % rule 2
X0 is X-A,
sum_ofsub_fromlist(X0,Bs,As).
sum_ofsub_fromlist(X,Bs,[_A|As]) :- % rule 3
dif(Bs,[]), % constraint: sublist not empty
sum_ofsub_fromlist(X,Bs,As).
You can query this predicate to assure yourself that it delivers all sublists for the given sum in your examples:
?- sum_ofsub_fromlist(6,S,[1,2,3,4,5,6]).
S = [1, 2, 3] ;
S = [1, 5] ;
S = [2, 4] ;
S = [6] ;
false.
?- sum_ofsub_fromlist(12,S,[2,5,7,9]).
S = [5, 7] ;
false.
Building on this you can then write a calling predicate that only succeeds for sublists of length three:
sum_oftriple_fromlist(S,T,L) :-
T=[_,_,_], % T has to be a triple
sum_ofsub_fromlist(S,T,L).
This predicate yields the answers you desire:
?- sum_oftriple_fromlist(6,T,[1,2,3,4,5,6]).
T = [1, 2, 3] ;
false.
?- sum_oftriple_fromlist(12,T,[2,5,7,9]).
false.
Note that the predicate is also working with negative numbers:
?- sum_oftriple_fromlist(6,T,[-5,-3,-1,2,4,7,8,9]).
T = [-5, 2, 9] ;
T = [-5, 4, 7] ;
T = [-3, 2, 7] ;
false.
?- sum_oftriple_fromlist(-6,T,[-6,-5,-4,-3,-2,-1,2,4]).
T = [-6, -4, 4] ;
T = [-6, -2, 2] ;
T = [-5, -3, 2] ;
T = [-3, -2, -1] ;
false.
However, due to the use of is/2, the predicate only works if the first and the third arguments are ground:
?- sum_oftriple_fromlist(S,T,[1,2,3,4,5,6]).
ERROR: is/2: Arguments are not sufficiently instantiated
Exception: (7) sum_ofsub_fromlist(_G918, [_G1016, _G1019, _G1022], [1, 2, 3, 4, 5, 6]) ?
?- sum_oftriple_fromlist(6,T,[A,B,C,D,E,F]).
ERROR: is/2: Arguments are not sufficiently instantiated
Exception: (7) sum_ofsub_fromlist(6, [_G2121, _G2124, _G2127], [_G1945, _G1948, _G1951, _G1954, _G1957, _G1960]) ?
If that's fine with you, you can stop here. Alternatively, you could opt to make the predicate more versatile by using CLP(FD). Just apply these minor changes to your code:
:- use_module(library(clpfd)). % <- new
sum_oftriple_fromlist(S,T,L) :-
T=[_,_,_],
sum_ofsub_fromlist(S,T,L).
sum_ofsub_fromlist(0,[],_L).
sum_ofsub_fromlist(X,[A|Bs],[A|As]) :-
X0 #= X-A, % <- change
sum_ofsub_fromlist(X0,Bs,As).
sum_ofsub_fromlist(X,Bs,[_A|As]) :-
dif(Bs,[]),
sum_ofsub_fromlist(X,Bs,As).
Now the above queries deliver answers:
?- sum_oftriple_fromlist(S,T,[1,2,3,4,5,6]).
S = 6,
T = [1, 2, 3] ;
S = 7,
T = [1, 2, 4] ;
S = 8,
T = [1, 2, 5] ;
. % another
. % seventeen
. % results here
The second query, however, yields residual goals (see documentation for details) as results:
?- sum_oftriple_fromlist(6,T,[A,B,C,D,E,F]).
T = [A, B, C],
_G2424+A#=6,
C+B#=_G2424 ;
T = [A, B, D],
_G2424+A#=6,
D+B#=_G2424 ;
.
.
.
To get actual numbers, you have to restrict the range of the numbers and subsequently label the variables in the list:
?- L=[A,B,C,D,E,F], sum_oftriple_fromlist(6,T,L), L ins 1..6, label(L).
L = [1, 1, 4, 1, 1, 1],
A = B, B = D, D = E, E = F, F = 1,
C = 4,
T = [1, 1, 4] ;
L = [1, 1, 4, 1, 1, 2],
A = B, B = D, D = E, E = 1,
C = 4,
F = 2,
T = [1, 1, 4] ;
.
.
.
Possibly you are only interested in lists where every number only appears once:
?- L=[A,B,C,D,E,F], all_distinct(L), sum_oftriple_fromlist(6,T,L), L ins 1..6, label(L).
L = [1, 2, 3, 4, 5, 6],
A = 1,
B = 2,
C = 3,
D = 4,
E = 5,
F = 6,
T = [1, 2, 3] ;
L = [1, 2, 3, 4, 6, 5],
A = 1,
B = 2,
C = 3,
D = 4,
E = 6,
F = 5,
T = [1, 2, 3] ;
.
.
.
Or maybe you don't even want to restrict the sum:
?- L=[A,B,C,D,E,F], all_distinct(L), sum_oftriple_fromlist(S,T,L), L ins 1..6, label(L).
L = [1, 2, 3, 4, 5, 6],
A = 1,
B = 2,
C = 3,
D = 4,
E = 5,
F = S, S = 6, % sum = 6
T = [1, 2, 3] ;
.
.
.
L = [1, 2, 4, 3, 5, 6],
A = 1,
B = 2,
C = 4,
D = 3,
E = 5,
F = 6,
S = 7, % sum = 7
T = [1, 2, 4] ;
.
.
.
As you can see the CLP(FD) version of the predicate resembles a true relation as opposed to the non-CLP(FD) version. And of course your example queries yield the same answers with both versions.
Your code only considers the first 3 items in the list, and not any other combinations.
The most natural way to structure a solution involving a list is to base your recursion on the structure of the list. So:
If the first element of the list (say, X) is to be included in the 3 values that sum to N, we need to find a way to find 2 values in the rest of the list that sum to N-X.
If it isn't, just try to solve the problem using the rest of the list.
Note that you may need a "helper" version of your predicate that allows you to add other parameters. In this case, knowing how many values you need to add up would be helpful.
Given a list L of natural numbers I need to create a list containing the sums of the elements of all the subsets of L. For example if L=[1,3,6] I want to obtain the list [0,1,3,4,6,7,9,10].
I tried to use this code
subsetSums(List,Sums) :- findall(Sum,(subset(Sub,List),sum_list(Sub,Sum)),Sums).
but with the following query I get [0] as the only result instead of [0,1,2,3]
?- subsetSums([1,2],Sums).
Where am I wrong?
EDIT: I'm working on SWI Prolog and subset/2 should be a native predicate.
As suggested in the comment, you have to write your own subset/2 predicate and then use findall/3 on this predicate, like this:
subset([], []).
subset([E|T], [E|T1]):-
subset(T, T1).
subset([_|T], L):-
subset(T, L).
subsetSums(List,Sums) :-
findall(S,(subset(List,Sub),sumlist(Sub,S)),Sums).
?- subsetSums([1,2],L).
L = [3, 1, 2, 0]
?- subsetSums([1,2,3],L).
L = [6, 3, 4, 1, 5, 2, 3, 0]
where the output of subset/2 is:
subset([1,2,3],L).
L = [1, 2, 3]
L = [1, 2]
L = [1, 3]
L = [1]
L = [2, 3]
L = [2]
L = [3]
L = []
Basically what I want to achieve is :
Given a list of domain Variables set these Variables with a domain relative to a List of Numbers. Example:
......
List=[A1,A2,A3],
domain(List,1,5],
setDomain(List,[1,2]),
labeling([],List).
Result:
A1=1, A2=1, A3=1 or
A1=1, A2=1, A3=2 or
A1=1, A2=2, A3=1
and so on...
What I have tried:
setDomain(List,ListIntegers):-
element(X, List, Element),
member(Element,ListIntegers),
main(List):-
List=[A1,A2,A3],
domain(List,1,5],
setDomain(List,[1,2]),
labeling([],List).
but not success...
Can anyone help understand how can I accomplish this?
In your solution, you're using labeling/2 but haven't defined its arguments using CLP(FD), so it doesn't do anything for you. It's not very clear from your question or simple example, but it sounds like you want a list of a given length whose elements are each taken from a domain consisting of an arbitrary list of elements?
You could do so with something like this:
member_(List, Element) :- member(Element, List).
domain_list(Length, Domain, List) :-
length(List, Length),
maplist(member_(Domain), List).
This would give:
6 ?- domain_list(3, [1,3], L).
L = [1, 1, 1] ;
L = [1, 1, 3] ;
L = [1, 3, 1] ;
L = [1, 3, 3] ;
L = [3, 1, 1] ;
L = [3, 1, 3] ;
L = [3, 3, 1] ;
L = [3, 3, 3].
7 ?-
This also works for any kind of elements:
7 ?- domain_list(3, [tom, a(b)], L).
L = [tom, tom, tom] ;
L = [tom, tom, a(b)] ;
L = [tom, a(b), tom] ;
L = [tom, a(b), a(b)] ;
L = [a(b), tom, tom] ;
L = [a(b), tom, a(b)] ;
L = [a(b), a(b), tom] ;
L = [a(b), a(b), a(b)].
8 ?-
If you wanted to use CLP(FD), you'd need to keep a couple of things in mind. CLP(FD) is for integer domains, and CLP(FD) has its own way of specifying domains, which is not in list form.
For instance, if you wanted a list of length N whose elements were in the domain described by [1,2,3,5,6,8], you would write it as:
length(List, N),
List ins 1..3 \/ 5..6 \/ 8,
label(List).
Which would result in, for example:
2 ?- length(List, 3), List ins 1..3 \/ 5..6 \/ 8, label(List).
List = [1, 1, 1] ;
List = [1, 1, 2] ;
List = [1, 1, 3] ;
List = [1, 1, 5] ;
List = [1, 1, 6] ;
List = [1, 1, 8] ;
List = [1, 2, 1] ;
List = [1, 2, 2]
...
Using ECLiPSe Prolog you can write:
:-lib(fd).
applyDomain([],_).
applyDomain([H|T],D):-
var_fd(H,D),
applyDomain(T,D).
domainList(ListDomain,LengthList,ListOutput):-
length(ListOutput,LengthList),
list_to_dom(ListDomain,Domain),
applyDomain(ListOutput,Domain).
Query:
?- domainList([2,3,5,7,8],5,L).
L = [_530{[2, 3, 5, 7, 8]}, _547{[2, 3, 5, 7, 8]}, _564{[2, 3, 5, 7, 8]}, _581{[2, 3, 5, 7, 8]}, _598{[2, 3, 5, 7, 8]}]
Yes (0.00s cpu)
The output means that each variable (in this case _530, _547 and so on) in the list L has the specified domain {[2, 3, 5, 7, 8]}. If you want to label the list you can simply add
labeling(ListOutput).
as last line of domainList/3 and you get:
?- domainList([2, 3, 5, 7, 8], 5, L).
L = [2, 2, 2, 2, 2]
Yes (0.00s cpu, solution 1, maybe more)
L = [2, 2, 2, 2, 3]
Yes (0.00s cpu, solution 2, maybe more)
L = [2, 2, 2, 2, 5]
Yes (0.00s cpu, solution 3, maybe more)
and so on... If you want that all the list will be different, just add
alldifferent(ListOutput),
before labeling/1, and you'll get
?- domainList([2, 3, 5, 7, 8], 5, L).
L = [2, 3, 5, 7, 8]
Yes (0.00s cpu, solution 1, maybe more)
L = [2, 3, 5, 8, 7]
Yes (0.00s cpu, solution 2, maybe more)
L = [2, 3, 7, 5, 8]
Yes (0.00s cpu, solution 3, maybe more)
I usually don't use SWI prolog for clpfd problems, so i don't know if there is a similar solution in SWI...
I am trying to generate all possible combinations of lists of three elements where all are distinct. I am using the CLPFD library to define the domain of the variable.
I defined the following
listDif(F,X):-F ins 1..3,findall(F,all_distinct(F),X).
And the answers to the queries are
?- listDif([1,_,2],X).
X = [[1, 3, 2]].
?- listDif([1,_,_],X).
X = [[1, _7374, _7380]],
_7374 in 2..3,
all_distinct([1, _7374, _7380]),
_7380 in 2..3.
?-
How do I display the lists with the possible integer values?
If you want to generate lists with CLPFD then you need to use lists. :) Your code is just using individual integers.
list3(F) :-
length(F, 3), % F is a list of length 3
F ins 1..3, % Elements of F are in the range 1..3
all_distinct(F). % F has distinct elements
Now you have a predicate which succeeds for unique lists consisting of 1, 2, 3:
?- list3(F), label(F).
F = [1, 2, 3] ;
F = [1, 3, 2] ;
F = [2, 1, 3] ;
F = [2, 3, 1] ;
F = [3, 1, 2] ;
F = [3, 2, 1].
Then you can use findall/3 if you want to have a list of all of these lists:
?- findall(F, (list3(F), label(F)), AllList3).
AllList3 = [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]].
I'm trying to figure out how to generate a list of sets, where each set has a length of N and the sum of each set is X.
I found this code:
num_split(0,[]).
num_split(N, [X | List]):-
between(1,N,X),
plus(X,Y,N),
num_split(Y,List).
And I can use that to get a list of sets with sum X:
num_split(6,List),length(List,5).
List = [1, 1, 1, 1, 2] ;
List = [1, 1, 1, 2, 1] ;
List = [1, 1, 2, 1, 1] ;
List = [1, 2, 1, 1, 1] ;
List = [2, 1, 1, 1, 1] ;
false.
The problem is that those are all permutations, and I'm looking for combinations. The output I'm looking for should be something like get_combos(Sum,Length,List):
get_combos(6,2,List).
List = [5,1];
List = [4,2];
List = [3,3];
false.
Any pointers?
If you have access to a CLP(FD) library, you can use this code:
:- [library(clpfd)].
get_combos(Sum, Length, List) :-
length(List, Length),
List ins 1 .. Sum,
% all_distinct(List), not really useful here
sum(List, #=, Sum),
chain(List, #<),
label(List).
test:
?- get_combos(10,3,L).
L = [1, 2, 7] ;
L = [1, 3, 6] ;
L = [1, 4, 5] ;
L = [2, 3, 5] ;
Maybe I misunderstood your question. Use this chain
...
chain(List, #=<),
....
to get possible duplicates values:
?- get_combos(10,3,L).
L = [1, 1, 8] ;
L = [1, 2, 7] ;
L = [1, 3, 6] ;
L = [1, 4, 5] ;
L = [2, 2, 6] ;
L = [2, 3, 5] ;
L = [2, 4, 4] ;
L = [3, 3, 4] ;
false.
Enforce an "equal or greater" restriction between successive values in the array.
You can add it on as another predicate:
is_combination([]).
is_combination([_]).
is_combination([A,B|List]) :- A =< B, is_combination([B|List]).
get_combos(Sum, Length, List) :-
num_split(Sum, Length, List),
is_combination(List).
Unfortunately, tacking it on the end of the num_split/3 does not necessarily increase its performance, so adding it directly into the algorithm would be marginally better:
get_combos(_, 0, []).
get_combos(Sum, 1, [Sum]).
get_combos(Sum, Length, [A, B|List]) :-
between(1, Sum, A),
plus(A, NextSum, Sum),
plus(1, NextLength, Length),
get_combos(NextSum, NextLength, [B|List]),
A =< B.
I'm not sure just how much more performance this gets, as the comparison has to be after the recursion, due to the less-than-or-equals operator (=<) requiring both operands to be fully instantiated for it to work.