I need to duplicate list in prolog.
I have list:
L = [a(string1,value1),a(string2,value2),a(string3,value3),a(string4,value4)].
Output will be: L = [string1, string2, string3, string4].
How can I do this?
I can copy whole list by code:
copy([],[]).
copy([H|L1],[H|L2]) :- copy(L1,L2).
I have tried something like:
copy2([],[]).
copy2([H|L1],[K|L2]) :- member(f(K,_),H), copy2(L1,L2).
But it does not work properly.
But I need only strings from my original list. Can anyone help?
pattern matching is used to decompose arguments: you can do
copy([],[]).
copy([a(H,_)|L1],[H|L2]) :- copy(L1,L2).
It is uncommon to use a structure a/2 for this purpose. More frequently, (-)/2 is used for this. Key-Value is called a (key-value) pair.
Also the name itself is not very self-revealing. This is no copy at all. Instead, start with a name for the first argument, and then a name for the second. Lets try: list_list/2. The name is a bit too general, so maybe apairs_keys/2.
?- apairs_keys([a(string1,value1),a(string2,value2)], [string1, string2]).
Here are some definitions for that:
apairs_keys([], []).
apairs_keys([a(K,_)|As], [K|Ks]) :-
apairs_keys(As, Ks).
Or, rather using maplist:
apair_key(a(K,_),K).
?- maplist(apair_key, As, Ks).
Or, using lambdas:
?- maplist(\a(K,_)^K^true, As, Ks).
Declarative debugging techniques
Maybe you also want to understand how you can quite rapidly localize the error in your original program. For this purpose, start with the problematic program and query:
copy2([],[]).
copy2([H|L1],[K|L2]) :-
member(f(K,_),H),
copy2(L1,L2).
?- copy2([a(string1,value1),a(string2,value2),a(string3,value3),a(string4,value4)], [string1, string2, string3, string4]).
false.
Now, generalize the query. That is, replace terms by fresh new variables:
?- copy2([a(string1,value1),a(string2,value2),a(string3,value3),a(string4,value4)], [A, B, C, D]).
false.
?- copy2([a(string1,value1),a(string2,value2),a(string3,value3),a(string4,value4)], L).
false.
?- copy2([a(string1,value1),B,C,D], L).
false.
?- copy2([a(string1,value1)|J], L).
false.
?- copy2([a(S,V)|J], L).
false.
?- copy2([A|J], L).
A = [f(_A,_B)|_C], L = [_A|_D]
; ... .
So we hit bottom... It seems Prolog does not like a term a/2 as first argument.
Now, add
:- op(950,fx, *).
*_.
to your program. It is kind of a simplistic debugger. And generalize the program:
copy2([],[]).
copy2([H|L1],[K|L2]) :-
member(f(K,_),H),
* copy2(L1,L2).
Member only succeeds with H being of the form [_|_]. But we expect it to be a(_,_).
Related
My wanted output is this:
?- mirror ([1,2], [] , X ).
X= [1,2,2,1]
What I have so far:
mirror(L,R,X):- L is R , [R| revertList(L,X)] .
I cant think of how this works, please help me
It is not very different from reversing a list, but how you write it is not going to work. I googled "Prolog is" and after maybe 10 seconds I see that is/2 is for arithmetic expressions. I also don't know how you think that you can put predicate but maybe it is not possible? If you want to just append then you can use append to append the mirror reversed list to the end of the original list to get the final "mirror" result:
mirror(X, Y) :- reverse(X, R), append(X, R, Y).
but this is too easy? So I wonder maybe there is more to this question? I don't know why you have three arguments when you only need two arguments? Maybe you thought that you can use an accumulator to reverse the list because to reverse a list you use accumulator like this?
list_rev(L, R) :- list_rev(L, [], R).
list_rev([], R, R).
list_rev([X|Xs], Ys, R) :-
list_rev(Xs, [X|Ys], R).
But this is very easy to google, I just googled it and found it, so maybe you googled it too and you didn't like it? To get "mirrored" you just need to keep the original list too, like so:
list_mirrored(L, M) :- list_mirrored(L, [], M).
list_mirrored([], M, M).
list_mirrored([X|Xs], Ys, [X|Zs]) :-
list_mirrored(Xs, [X|Ys], Zs).
I wasn't sure if this is correct and I googled "Prolog append" and this is how it is done.
To describe lists in Prolog, always also consider DCG notation.
For example, in this concrete case:
mirror([]) --> [].
mirror([M|Ms]) --> [M], mirror(Ms), [M].
Your test case:
?- phrase(mirror([1,2]), Ls).
Ls = [1, 2, 2, 1].
It also works in the other direction. For example:
?- phrase(mirror(Ls), [a,b,c,c,b,a]).
Ls = [a, b, c] ;
false.
The most general query yields:
?- phrase(mirror(Ls), Ms).
Ls = Ms, Ms = [] ;
Ls = [_5988],
Ms = [_5988, _5988] ;
Ls = [_5988, _6000],
Ms = [_5988, _6000, _6000, _5988] ;
Ls = [_5988, _6000, _6012],
Ms = [_5988, _6000, _6012, _6012, _6000, _5988] ;
etc.
See dcg for more information.
Note that with the definition above, we have:
?- phrase(mirror(Ls), [a,b,a]).
false.
I leave generalizing this definition (if necessary) as an easy exercise.
This would get you the desired "output":
mirror(_,_,[1,2,2,1]).
That probably won't work for most inputs, but since you haven't explained the relationship between input and output for anything but this one case, that's a good as I can do.
I want to create a predicate in Prolog which will check if a list A is a sublist of a list B. Moreover I do not want my program to consider an empty list as a subset of another one.
E.g. included_list([1,4],[1,2,3,4,5]).
true.
included_list([2,3],[1,2,3,4,5]).
true.
included_list([1,6],[1,2,3,4,5]).
false.
included_list([],[1,2,3,4,5]).
false.
and so on...
So, I have written the following code so far:
member(X,[X|Tail]).
member(X,[Head|Tail]):- member(X,Tail).
included_list([X],_).
included_list([Head|Tail],List):- member(Head,List), included_list(Tail,List).
But the above code seems to be wrong, because in one specific case it throws true, instead of throwing wrong. I wish I'd made it clear having presented the following screenshot:
As you might have noticed the fifth(5th) sentence gives true, instead of wrong. That is, when I write a sentence of the form:
included_list([x,y],[w,x,v,z]).
whereas only x is included in the second list(and not y) the program gives me true(and this is wrong).
In general, if the first argument of the first list is included in the second list then, no matter if the rest of the former are included in the latter, the program gives me true.
In any other case the program gives me the right result(true or false).
What do I do wrong?
I will be waiting for your answers!
Thank you in advance!
Your problem is the first clause of included_list/2. This:
included_list([X], _).
What does it mean? It means, "If the first argument is a list with one element, succeed, ignoring the second argument."
A short aside: if you would not ignore compiler warnings, you would have caught this mistake already. You should get a loud and clear "Singleton variable" warning, hinting that the code you have written does not do what you think it does.
What you actually mean is more along the lines of:
subset_list([X|Xs], Ys) :-
subset_list_1(Xs, X, Ys).
subset_list_1([], X, Ys) :-
member(X, Ys).
subset_list_1([X|Xs], X0, Ys) :-
member(X0, Ys),
subset_list_1(Xs, X, Ys).
But I don't know why you don't simply use the available subset/2, and simply add a requirement that the subset is not an empty list:
subset_list(Subset, List) :-
Subset = [_|_], % a list with at least one element
subset(Subset, List).
Despite what the documentation claims, the second argument to subset/2 does not have to be a true "set", but it does expect that both lists are ground (do not contain any free variables). You can see the source code here.
In this answer we let meta-predicate maplist/2 handle recursion and define:
all_included(Sub, Es) :-
same_length(Es, Xs),
Sub = [_|_], % minimum length: 1
append(_, Sub, Xs), % maximum length: as long as `Es`
maplist(list_member(Es), Sub).
Let's run the queries the OP gave!
First up, use-cases we expect to succeed:
?- member(Xs, [[1,4],[2,3],[2,3,5],[3,4]]), all_included(Xs, [1,2,3,4,5]).
Xs = [1,4]
; Xs = [2,3]
; Xs = [2,3,5]
; Xs = [3,4]
; false.
Next up, some use-cases we expect to fail:
?- member(Xs, [[],[2,6],[1,6]]), all_included(Xs, [1,2,3,4,5]).
false.
?- all_included([3,5], [1,2,5]).
false.
I'm trying to write a predicate to remove the head from every list in list of lists and add the tails to a new list. The resulting list should be returned as the second parameter.
Here's the attempt:
construct_new(S,New) :-
New = [],
new_situation(S,New).
new_situation([],_).
new_situation([H|T], New) :-
chop(H, H1),
new_situation(T, [H1|New]).
chop([_|T], T).
You would call it like this:
construct_new([[x,x],[b,c],[d,e,f]],S).
This, however, only produces output true..
Step-by-step execution
Your query is construct_new(Input,Output), for some instanciated Input list.
The first statement in construct_new/2 unifies Output (a.k.a. New) with the empty list. Where is the returned list supposed to be available for the caller? Both arguments are now unified.
You call new_situation(Input,[])
You match the second clause new_situation([H|T],[]), which performs its task recursively (step 4, ...), until ...
You reach new_situation([],_), which successfully discards the intermediate list you built.
Solutions
Write a simple recursive predicate:
new_situation([],[]).
new_situation([[_|L]|T],[L|R]) :-
new_situation(T,R).
Use maplist:
construct_new(S,R) :-
maplist(chop,S,R).
Remark
As pointed out by other answers and comments, your predicates are badly named. construct_new is not a relation, but an action, and could be used to represent almost anything. I tend to like chop because it clearly conveys the act of beheading, but this is not an appropriate name for a relation. repeat's list_head_tail(L,H,T) is declarative and associates variables to their roles. When using maplist, the other predicate (new_situation) doesn't even need to exist...
...even though guillotine/3 is tempting.
This could be done with a DCG:
owth(Lists, Tails) :-
phrase(tails(Tails), Lists).
tails([]) --> [].
tails([T|Tails]) --> [[_|T]], tails(Tails).
Yielding these queries:
| ?- owth([[x,x],[b,c],[d,e,f]], T).
T = [[x],[c],[e,f]] ? ;
no
| ?- owth(L, [[x],[c],[e,f]]).
L = [[_,x],[_,c],[_,e,f]]
yes
(owth = Off with their heads! or, if used the other direction, On with their heads!)
If you also want to capture the heads, you can enhance it as follows:
owth(Lists, Heads, Tails) :-
phrase(tails(Heads, Tails), Lists).
tails([], []) --> [].
tails([H|Hs], [T|Tails]) --> [[H|T]], tails(Hs, Tails).
We use meta-predicate maplist/[3-4] with one of these following auxiliary predicates:
list_tail([_|Xs],Xs).
list_head_tail([X|Xs],X,Xs).
Let's run some queries!
?- maplist(list_head_tail,[[x,x],[b,c],[d,e,f]],Heads,Tails).
Heads = [x,b,d],
Tails = [[x],[c],[e,f]].
If you are only interested in the tails, use maplist/4 together with list_head_tail/3 ...
?- maplist(list_head_tail,[[x,x],[b,c],[d,e,f]],_,Tails).
Tails = [[x],[c],[e,f]].
... or, even simpler, maplist/3 in tandem with list_tail/2:
?- maplist(list_tail,[[x,x],[b,c],[d,e,f]],Tails).
Tails = [[x],[c],[e,f]].
You can also use the somewhat ugly one-liner with findall/3:
?- L = [[x,x],[b,c],[d,e,f]],
findall(T, ( member(M, L), append([_], T, M) ), R).
R = [[x], [c], [e, f]].
(OK, technically a two-liner. Either way, you don't even need to define a helper predicate.)
But definitely prefer the maplist solution that uses chop as shown above.
If you do the maplist expansion by hand, and name your chop/2 a bit better, you would get:
lists_tails([], []).
lists_tails([X|Xs], [T|Ts]) :-
list_tail(X, T),
lists_tails(Xs, Ts).
And since you can do unification in the head of the predicate, you can transform this to:
lists_tails([], []).
lists_tails([[_|T]|Xs], [T|Ts]) :-
lists_tails(Xs, Ts).
But this is identical to what you have in the other answer.
Exercise: why can't we say:
?- maplist(append([_]), R, [[x,x],[b,c],[d,e,f]]).
I need to get all list element in out side of the list using prolog.
['1','2','3',a,b,c] to become X='1','2','3',a,b,c
I need the result X as shown above.
is it possibile to get the result?
This will literally display what you're asking for (I'm assuming you're interested in the result strictly for display purposes).
print_list(L) :-
write('X='),
print_list_aux(L), !.
print_list_aux([H1,H2|T]) :-
print(H1),
write(','),
print_list_aux([H2|T]).
print_list_aux([X]) :-
print(X),
nl.
print_list_aux([]) :- nl.
Example:
?- print_list(['1', '2', a, b]).
X='1','2',a,b
true.
?-
If you use write/1 instead of print/1, you eliminate the quotes on the numbers in the output, so it would be: X=1,2,a,b.
How can I check if an element in the list is an empty list: [] ?
I've got the following:
display_degrees([A,B,C,D]):- write(B).
display_degrees([A,B,C,D]):- B==[], nl,write('has no degree'), nl, !.
When I enter in something like:
display_degrees([1,[],3,4]).
I just get: [] instead of 'has no degree'. Is my syntax wrong? Can I not add a clause to this predicate like this?
You're getting this behavior because proof search stops when a goal has succeeded. When you type
display_degrees([1,[],3,4]).
the first rule unifies, and it writes B. Since it was a success, it stops. You can ask Prolog to keep searching, and then it will find the second clause. In swipl, I get
?- [foo].
?- display_degrees([1,[],3,4]).
[]
true r % I type 'r' there
has no degree
true.
If you're just learning Prolog, I suggest you avoid the cut operator ! for some time. Also, doing IO is not the most intuitive thing. I would try some exercises with defining things like natural numbers and recursive functions. E.g., plus:
plus(z, X, X).
plus(s(X), Y, s(Z)) :- plus(X, Y, Z).
The problem with what you have is that the more general rule will fire first. You could switch the order:
display_degrees([A,[],C,D]) :- nl, write('has no degree'), nl, !.
display_degrees([A,B,C,D]) :- write(B).
I could just as well have written for the first predicate:
display_degrees([A,B,C,D]) :- B == [], nl, write('has no degree'), nl, !.
But the "shortcut" I show initially is more idiomatic for a Prolog predicate like this.
I kept the cut since you know you deterministically want one choice. The first rule will match if and only if the second list element is [].
| ?- display_degrees([1,[],3,4]).
has no degree
yes
| ?- display_degrees([1,2,3,4]).
2
yes
| ?-