I trying to write a function getNumbers(List,Result) such that List is a list which his elements can be integer or list of lists , for example -
List = [1,[1,2,[3],[4]],2,[4,5]]
List = [1,[1,1,[1],[1]],1,[1,1,[[[[[1]]]]]]]
List = [[4,[[]],2],[[1],[],[1]]]
etc..
And the output should be all the numbers stored in that List , for example -
?- getNumbers([1,[1,2,[3],[4]],2,[4,5]],R).
R = [1,2,3,4,5].
?- getNumbers([1,[1,1,[1],[1]],1,[1,1,[[[[[1]]]]]]],R).
R = [1].
?- getNumbers([],R).
R = [].
?- getNumbers([[4,[[]],2],[[1],[],[1]]],R).
R = [1,2,4].
So far I tried the follow code -
getNumbers([],Result) :- Result=[],!.
getNumbers([H|Rest],Result) :- getNumbers(Rest,NewResultRest),
( atomic(H) ->
Result = [H|NewResultRest]
; getNumbers(H,NewResultHead),Result = [NewResultHead|NewResultRest] ).
But it gives wrong result , like -
getNumbers([[2],5,7,[3,6,5]],Result).
Result = [[2], 5, 7, [3, 6, 5]].
It seems that the function doesn't exludes 2 from [2] or any other numbers stored in nested list.
How could I fix my implementation ?
you need to append nested lists:
getNumbers([],Result) :- Result=[],!.
getNumbers([H|Rest],Result) :-
getNumbers(Rest,NewResultRest),
( atomic(H)
-> Result = [H|NewResultRest]
; getNumbers(H,NewResultHead),
append(NewResultHead, NewResultRest, Result) % only this change
).
note that [] is atomic: thus
?- getNumbers([[4,[[]],2],[[1],[],[1]]],R).
R = [4, [], 2, 1, [], 1].
from your description, you should use number/1 to test element' type. After the change
?- getNumbers([[4,[[]],2],[[1],[],[1]]],R).
R = [4, 2, 1, 1].
Related
I have a written a functional function that tells the user if a list is ordered or not, given the list inputted. However, if a user inputs a variable as the input instead of a list, I would like to output an infinite list. How can I go about this? Here is the current code
ordered([]).
ordered([_]).
ordered([X,Y|Ys]) :- X =< Y , ordered( [Y|Ys] ).
Here is some input
? ordered([1,2,3]).
true
? ordered([1,5,2]).
false
I also want for variables to creat infinite list like so
? ordered(L).
L = [];
L = [_1322] ;
L = [_1322, _1323] ;
L = [_1322, _1323, _1324] ;
L = [_1322, _1323, _1324, _1325].
The list should increase until the user exits as shown.
The list should increase until the user exits as shown.
Solution:
ordered([]).
ordered([_]).
ordered([X,Y|Ys]) :- X #=< Y , ordered( [Y|Ys] ).
EDIT:
SWI Prolog doc
The arithmetic expression X is less than or equal to Y. When reasoning over integers, replace (=<)/2 by #=</2 to obtain more general relations. See declarative integer arithmetic (section A.9.3).
What properties should the list of variables have? The currently accepted answer by Anton Danilov says that [3, 2, 1] is not an ordered list:
?- List = [A, B, C], List = [3, 2, 1], ordered(List).
false.
but it also says that [3, 2, 1] is an instance of an ordered list:
?- List = [A, B, C], ordered(List), List = [3, 2, 1].
List = [3, 2, 1],
A = 3,
B = 2,
C = 1 ;
false.
Viewed logically, this is a contradiction. Viewed procedurally, it is fine, but also the #=< relationship between the variables in the list is meaningless. The comparison of the unbound variables does not say anything about the relationship of the list elements if they are bound to values at some point.
You can use constraints to exclude future unordered bindings:
:- use_module(library(clpfd)).
ordered([]).
ordered([_]).
ordered([X, Y | Xs]) :-
X #=< Y,
ordered([Y | Xs]).
This way you cannot bind the variables in the list to incorrect numbers later on:
?- List = [A, B, C], List = [3, 2, 1], ordered(List).
false.
?- List = [A, B, C], ordered(List), List = [3, 2, 1].
false.
But later correct ordered bindings are still allowed:
?- List = [A, B, C], ordered(List), List = [1, 2, 3].
List = [1, 2, 3],
A = 1,
B = 2,
C = 3 ;
false.
This may not be the best solution, but I believe it can give you some idea of how to do what you need. In SWI-Prolog, the predicate freeze(+Var,:Goal) delays the execution of Goal until Var is bound.
ordered([]).
ordered([_]).
ordered([X,Y|R]) :-
freeze( X,
freeze( Y,
( X #=< Y,
ordered([Y|R]) ) ) ).
Here are some examples with finite lists:
?- ordered([1,2,3]).
true.
?- ordered([1,2,3,0]).
false.
?- ordered(L), L=[1,2,3].
L = [1, 2, 3] ;
false.
?- ordered(L), L=[1,2,3,0].
false.
For an infinite list, you will need to "take" its prefix:
take([]).
take([_|R]) :- take(R).
Here is an example with infinite list:
?- ordered(L), take(L).
L = [] ;
L = [_375396] ;
L = [_376366, _376372],
freeze(_376366, freeze(_376372, (_376366#=<_376372, ordered([])))) ;
L = [_377472, _377478, _377484],
freeze(_377472, freeze(_377478, (_377472#=<_377478, ordered([_377484])))) ;
L = [_378590, _378596, _378602, _378608],
freeze(_378590, freeze(_378596, (_378590#=<_378596, ordered([_378602, _378608])))) ;
L = [_379720, _379726, _379732, _379738, _379744],
freeze(_379720, freeze(_379726, (_379720#=<_379726, ordered([_379732, _379738, _379744]))))
Examples: ([1,2,3,7,6,9], 6). should print True, as 1+2+3=6.
([1,2,3,7,6,9], 5). should print False as there are no three numbers whose sum is 5.
([],N) where N is equal to anything should be false.
Need to use only these constructs:
A single clause must be defined (no more than one clause is allowed).
Only the following is permitted:
+, ,, ;, ., !, :-, is, Lists -- Head and Tail syntax for list types, Variables.
I have done a basic coding as per my understanding.
findVal([Q|X],A) :-
[W|X1]=X,
[Y|X2]=X,
% Trying to append the values.
append([Q],X1,X2),
% finding sum.
RES is Q+W+Y,
% verify here.
(not(RES=A)->
% finding the values.
(findVal(X2,A=)->
true
;
(findVal(X,A)->
% return result.
true
;
% return value.
false))
;
% return result.
true
).
It does not seem to run throwing the following error.
ERROR:
Undefined procedure: findVal/2 (DWIM could not correct goal)
Can someone help with this?
You can make use of append/3 [swi-doc] here to pick an element from a list, and get access to the rest of the elements (the elements after that element). By applying this technique three times, we thus obtain three items from the list. We can then match the sum of these elements:
sublist(L1, S) :-
append(_, [S1|L2], L1),
append(_, [S2|L3], L2),
append(_, [S3|_], L3),
S is S1 + S2 + S3.
Well, you can iterate (via backtracking) over all the sublists of 3 elements from the input list and see which ones sum 3:
sublist([], []).
sublist([H|T], [H|S]) :- sublist(T, S).
sublist([_|T], S) :- sublist(T, S).
:- length(L, 3), sublist([1,2,3,7,6,9], L), sum_list(L, 6).
I'm giving a partial solution here because it is an interesting problem even though the constraints are ridiculous.
First, I want something like select/3, except that will give me the tail of the list rather than the list without the item:
select_from(X, [X|R], R).
select_from(X, [_|T], R) :- select_from(X, T, R).
I want the tail, rather than just member/2, so I can recursively ask for items from the list without getting duplicates.
?- select_from(X, [1,2,3,4,5], R).
X = 1,
R = [2, 3, 4, 5] ;
X = 2,
R = [3, 4, 5] ;
X = 3,
R = [4, 5] ;
X = 4,
R = [5] ;
X = 5,
R = [] ;
false.
Yeah, this is good. Now I want to build a thing to give me N elements from a list. Again, I want combinations, because I don't want unnecessary duplicates if I can avoid it:
select_n_from(1, L, [X]) :- select_from(X, L, _).
select_n_from(N, L, [X|R]) :-
N > 1,
succ(N0, N),
select_from(X, L, Next),
select_n_from(N0, Next, R).
So the idea here is simple. If N = 1, then just do select_from/3 and give me a singleton list. If N > 1, then get one item using select_from/3 and then recur with N-1. This should give me all the possible combinations of items from this list, without giving me a bunch of repetitions I don't care about because addition is commutative and associative:
?- select_n_from(3, [1,2,3,4,5], R).
R = [1, 2, 3] ;
R = [1, 2, 4] ;
R = [1, 2, 5] ;
R = [1, 3, 4] ;
R = [1, 3, 5] ;
R = [1, 4, 5] ;
R = [2, 3, 4] ;
R = [2, 3, 5] ;
R = [2, 4, 5] ;
R = [3, 4, 5] ;
false.
We're basically one step away now from the result, which is this:
sublist(List, N) :-
select_n_from(3, List, R),
sumlist(R, N).
I'm hardcoding 3 here because of your problem, but I wanted a general solution. Using it:
?- sublist([1,2,3,4,5], N).
N = 6 ;
N = 7 ;
N = 8 ;
N = 8 ;
N = 9 ;
N = 10 ;
N = 9 ;
N = 10 ;
N = 11 ;
N = 12 ;
false.
You can also check:
?- sublist([1,2,3,4,5], 6).
true ;
false.
?- sublist([1,2,3,4,5], 5).
false.
?- sublist([1,2,3,4,5], 8).
true ;
true ;
false.
New users of Prolog will be annoyed that you get multiple answers here, but knowing that there are multiple ways to get 8 is probably interesting.
I want to write a predicate split/2 that generates all consecutive lists found inside another list.
Example: split([1,2,3,4],X) should return
X = [4], X = [2,3],X = [1,2], X = [1,2,3] etc.
So far I only have a predicate that returns all possible sublists of a list:
sublist([],[]).
sublist([H|T], [H|R]) :-
sublist(T,R).
sublist([_|T], R) :-
sublist(T,R).
However, with the query from the example this predicate includes unwanted answers like X = [2,4] and X = [1,3] that aren't found consecutively in [1,2,3,4].
Usually a problem is easier if you split it in subproblems. We can first construct a predicate that will construct all suffixes for a given list.
We can construct such predicate as follows:
suffix(_, []).
suffix([H|T], [H|T2]) :-
suffix(T, T2).
So for each point in the list, we can decide to stop (with the empty list), or emit the next item. For the given sample list, we thus get:
?- suffix([1,2,3,4],X).
X = [] ;
X = [1] ;
X = [1, 2] ;
X = [1, 2, 3] ;
X = [1, 2, 3, 4].
Now we only need to decide when we start the suffix. For each item in the list, we can decide to start at that point, and enumerate over all suffixes that we then append to that item:
split([H|T], [H|S]) :-
suffix(T, S).
split([_|T], S) :-
split(T, S).
For example:
?- split([1,2,3,4],X).
X = [1] ;
X = [1, 2] ;
X = [1, 2, 3] ;
X = [1, 2, 3, 4] ;
X = [2] ;
X = [2, 3] ;
X = [2, 3, 4] ;
X = [3] ;
X = [3, 4] ;
X = [4] ;
false.
The nice thing is that we got a second predicate "for free": we can also obtain all suffixes for a list.
We might want to include the empty list as well. I leave this as an exercise.
I need a solution that deletes elements that have pairs of occurrences from list.
I did it in haskell, but i don't have any ideas how to interpretate it in Prolog.
For example [1,2,2,2,4,4,5,6,6,6,6] -> [1,2,2,2,5]
Code in Haskell :
import Data.List
count e list = length $ filter (==e) list
isnotEven = (== 1) . (`mod` 2)
removeUnique :: [Int] -> [Int]
removeUnique list = filter (\x -> isnotEven (count x list) ) list
The following follows your Haskell code.
You need library(reif) for SICStus|SWI.
:- use_module(reif).
oddcount_t(List, E, T) :- % reified: last argument is truth value
tfilter(=(E), List, Eqs),
length(Eqs, Nr),
M is Nr mod 2,
=(M, 1, T).
removeevenocc(List, RList) :-
tfilter(oddcount_t(List), List, RList).
?- removeevenocc([1,2,2,2,4,4,5,6,6,6,6], R).
R = [1,2,2,2,5].
?- removeevenocc([1,X], R).
X = 1, R = []
; R = [1, X],
dif(X, 1).
Note the last question. Here, the list was not entirely given: The second element is left unknown. Therefore, Prolog produces answers for all possible values of X! Either X is 1, then the resulting list is empty, or X is not 1, then the list remains the same.
this snippet uses some of the libraries (aggregate,lists,yall) available, as well as some builtins, like setof/3, and (=:=)/2:
?- L=[1,2,2,2,4,4,5,6,6,6,6],
| setof(K,C^(aggregate(count,member(K,L),C),0=:=C mod 2),Ds),
| foldl([E,X,Y]>>delete(X,E,Y),Ds,L,R).
L = [1, 2, 2, 2, 4, 4, 5, 6, 6|...],
Ds = [4, 6],
R = [1, 2, 2, 2, 5].
edit
to account for setof/3 behaviour (my bug: setof/3 fails if there are no solutions), a possible correction:
?- L=[1],
(setof(K,C^(aggregate(count,member(K,L),C),0=:=C mod 2),Ds);Ds=[]),
foldl([E,X,Y]>>delete(X,E,Y),Ds,L,R).
L = R, R = [1],
Ds = [].
Now there is a choice point left, the correct syntax could be
?- L=[1],
(setof(K,C^(aggregate(count,member(K,L),C),0=:=C mod 2),Ds)->true;Ds=[]),
foldl([E,X,Y]>>delete(X,E,Y),Ds,L,R).
L = R, R = [1],
Ds = [].
I want to convert incomplete lists into difference lists and vice versa.
This is code to convert regular list to difference:
reg2diff(L,X-Y):-append(L,Y,X).
How do I go the other direction?
Incomplete to difference list:
inc2diff(L,Z):-
( nonvar(L)
-> ( L=[_|T] -> inc2diff(T,Z) ; L=[] -> Z=[] )
; L=Z
).
Use it as
23 ?- L=[1,2,3|_],inc2diff(L,X).
L = [1, 2, 3|X].
24 ?- L=[1,2,3|Z],inc2diff(L,X).
L = [1, 2, 3|X],
Z = X.
25 ?- L=[1,2,3],inc2diff(L,X).
L = [1, 2, 3],
X = [].
| ?- listing(dl_list),dl_list([a,b,c,d|X]-X,L).
% file: ...util.pg
dl_list(A - [], A).
L = [a,b,c,d]
X = []
yes