Related
I want to multiply elements in a List with findall/3. Specifically I have two functions double(X,Y) which doubles X and square(X,Y) that returns the squared value of X. My problem is that it the operation works only for the first element of the list.
double(X,Y) :- Y is X*2.
square(X,Y) :- Y is X*X.
map_f(Operation,[H|List],[R|Results]) :-
Predicate=..[Operation,H,R],
call(Predicate),
findall(X,( member(X,List) ), Results).
For example, if I type map_f(double,[3,1,2,6,3,1,6],L). ,
I expect the output: L = [6,2,4,12,6,2,12],
but instead it shows:
?- map_f(double, [3, 1, 2, 6, 3, 1, 6], List).
List = [6, 1, 2, 6, 3, 1, 6]
Yes (0.00s cpu)
Any help will be very appreciated.
If you want to use findall/3, you'd have to write it like this:
?- Xs = [3,1,2,6,3,1,6], findall(Y, ( member(X, Xs), double(X, Y) ), Ys).
Xs = [3, 1, 2, 6, 3, 1, 6],
Ys = [6, 2, 4, 12, 6, 2, 12].
If you really want to pass the predicate as an argument and use =.., the logic is still the same, you'd just have to re-write your definition so that it does the right thing:
map_f(Pred_name, L1, L2) :-
Goal =.. [Pred_name, X, Y],
findall(Y, ( member(X, L1), Goal ), L2).
Then:
?- map_f(double, [3,1,2,6,3,1,6], R).
R = [6, 2, 4, 12, 6, 2, 12].
?- map_f(square, [3,1,2,6,3,1,6], R).
R = [9, 1, 4, 36, 9, 1, 36].
But, instead of:
Goal =.. [Pred_name, Arg1, Arg2], Goal
it is easier to use call/N+1:
call(Pred_name, Arg1, Arg2)
So your definition will become:
map_f(Pred_name, L1, L2) :-
findall(Y, ( member(X, L1), call(Pred_name, X, Y) ), L2).
But really, all of this is completely unnecessary if you only have lists. You can just use maplist/N+1, like that:
?- maplist(double, [3,1,2,6,3,1,6], R).
R = [6, 2, 4, 12, 6, 2, 12].
... which iterates over the lists instead of backtracking over them. You can see a maplist implementation for example here:
https://github.com/SWI-Prolog/swipl-devel/blob/2d20d4e8ac28adfcede7a9bd231ea0d9d12d0bbb/library/apply.pl#L195-L205
If your predicate is a real relation (so if it works both ways), you can also use maplist both ways. findall cannot do that! Here is one silly example:
?- maplist(succ, [1,2,3], R).
R = [2, 3, 4].
?- maplist(succ, R, [1,2,3]).
R = [0, 1, 2].
?- map_f(succ, [1,2,3], R).
R = [2, 3, 4].
?- map_f(succ, R, [1,2,3]).
ERROR: Arguments are not sufficiently instantiated
% appends an element to the beginning of a list.
append_element(X, T, [X|T]).
% append a list to another list to create a combined list,
% by breaking the first list apart, and using append_element.
append_list([], L, L).
append_list([H|T], L, NewList) :-
append_element(H, L, NL),
append_list(T, NL, NL).
When I try to run append_list,
?- append_list([1,2], [3, 4, 5], NL).
I get back false. Instead of
NL = [2, 1, 3, 4, 5].
Why?
my problem is:
While learning Prolog i wanted to make a NxN Sudoku solver.
This solver will get the input like
[[1,2,3,4],[3,4,1,2],[2,3,4,1],[4,1,2,3]]
Where some of them might be variables. The solver has to solve that Sudoku.
The problem is way smaller:
firstElementsOf([],_).
firstElementsOf([[X|_]|Rest2],Y) :-
firstElementsOf(Rest2,Y2),
append([X],[Y2],NotFlat),
flatten(NotFlat,Y).
This should be the beginning of checking, if every column has distinct numbers. The Y from firstElementsOf should contain only the first elements of the given rows. In the Example:
[1,3,2,4]
Sadly, thanks to append, it always adds another empty space to the Y list.
It gives:
[1,3,2,4,_1320]
Question1: Is there a way to get rid of that _1320?
Question2: Is this even right? Will there be a way to get the 2nd and 3rd elements of the Input with that?
For question 1: I suppose the error is in
firstElementsOf([],_).
I think should be
firstElementsOf([],[]).
Off topic: are you sure that you can't simply write the other clause as follows?
firstElementsOf([[X|_]|Rest2],[X|Y]) :-
firstElementsOf(Rest2,Y).
For question 2: I propose a more general predicate: the following getPosList/3 with support of getPosElem/3
getPosElem([H | _], 1, H).
getPosElem([_ | T], Pos, H) :-
Pos > 1,
Pm1 is Pos - 1,
getPosElem(T, Pm1, H).
getPosList([], _, []).
getPosList([H | T], Pos, [E | L]) :-
getPosElem(H, Pos, E),
getPosList(T, Pos, L).
It extract a list of all elements in position Pos, so
getPosList([[1,2,3,4],[3,4,1,2],[2,3,4,1],[4,1,2,3]], 1, L),
it's equivalent to firstElementOf([[1,2,3,4],[3,4,1,2],[2,3,4,1],[4,1,2,3]], L) and extract [1, 3, 2, 4],
getPosList([[1,2,3,4],[3,4,1,2],[2,3,4,1],[4,1,2,3]], 2, L),
extract [2, 4, 3, 1],
getPosList([[1,2,3,4],[3,4,1,2],[2,3,4,1],[4,1,2,3]], 3, L),
extract [3, 1, 4, 2],
getPosList([[1,2,3,4],[3,4,1,2],[2,3,4,1],[4,1,2,3]], 4, L),
extract [4, 2, 1, 3] and
getPosList([[1,2,3,4],[3,4,1,2],[2,3,4,1],[4,1,2,3]], 5, L),
or a number greather than 5, return false
I am a beginner in prolog. I tried this in swipl interpreter:
?- length(Lists, 3), ord_subset(Lists, [1, 2, 3, 4]).
false.
expecting to get all length-3 lists that are subsets of [1, 2, 3, 4] like [1, 2, 3] or [1, 2, 4]. Why do i get false?
Notice: both length and ord_subset are builtin functions (or whatever they are called) in SWI-Prolog.
You don't get a solution because the ord_subset/2 predicate only checks if a list is a subset of another list; it does not generate subsets.
Here is one simplistic way to define a predicate that does what you seem to be after:
subset_set([], _).
subset_set([X|Xs], S) :-
append(_, [X|S1], S),
subset_set(Xs, S1).
This assumes that these are "ordsets", that is, sorted lists without duplicates.
You will notice that the subset happens to be also a subsequence. We could have written instead:
subset_set(Sub, Set) :-
% precondition( ground(Set) ),
% precondition( is_list(Set) ),
% precondition( sort(Set, Set) ),
subseq_list(Sub, Set).
subseq_list([], []).
subseq_list([H|T], L) :-
append(_, [H|L1], L),
subseq_list(T, L1).
With either definition, you get:
?- length(Sub, 3), subset_set(Sub, [1,2,3,4]).
Sub = [1, 2, 3] ;
Sub = [1, 2, 4] ;
Sub = [1, 3, 4] ;
Sub = [2, 3, 4] ;
false.
You can even switch the order of the two subgoals in the example query, but this is probably the better way to write it.
However, the second argument must be ground; if it is not:
?- subset_set([A,B], [a,B]), B = a.
A = B, B = a ; Not a real set, is it?
false.
I have this list :
C = [[1,0],[2,3],[1,2],[1,3]]
I'll like find if the number 1 included in a sublist inside my list in position [1,_ ] and i like to save to a list Newlist the number of X ..... [1,X].
I will give an example... i have the list C and i am searching for sublist which first element it's 1 and give me the Newlist.
The Newlist must be : Newlist=[0,2,3]
It had the second element of the sublists who has the number 1 at the first element.
If you use SWI-Prolog with module lambda.pl, (you can find it at http://www.complang.tuwien.ac.at/ulrich/Prolog-inedit/lambda.pl) you can write
:- use_module(library(lambda)).
my_filter(V, L, R) :-
foldl(V+\X^Y^Z^(X = [V,W]
-> append(Y, [W], Z)
; Z = Y),
L, [], R).
nth0/3 allows to access list' elements by index:
?- C = [[1,0],[2,3],[1,2],[1,3]], findall(P, nth0(P, C, [1,_]), NewList).
C = [[1, 0], [2, 3], [1, 2], [1, 3]],
NewList = [0, 2, 3].
edit I'm sorry I didn't read the question right. nth0 is misleading. Could be instead
findall(E, member([1,E], C), NewList)
You need a "filter". This is what it could look like:
filter_1_at_pos_1([], []). % The new list is empty when the input list is empty
filter_1_at_pos_1([[1,X]|Sublist], [X|Xs]) :- % The first element is 1 so the
% second element belongs to the
% new list
!, filter_1_at_pos_1(Sublist, Xs). % filter the remainder of the list
filter_1_at_pos_1([[N,_]|Sublist], Xs) :-
N \== 1, % The first element is not 1, ignore the second element
!, filter_1_at_pos_1(Sublist, Xs).
As #mbratch suggested, just define the solution for one element of the input list for each possible condition, in this case 1) empty list 2) first element is 1 and 3) first element is not 1.
?- C = [[1,0],[2,3],[1,2],[1,3]], filter_1_at_pos_1(C, NewList).
C = [[1, 0], [2, 3], [1, 2], [1, 3]],
NewList = [0, 2, 3].
The cuts make the predicate deterministic. The cut in the last clause is not necessary.