I'm new to Prolog and as an exercise I want to make an list invertion predicate. It uses the add_tail predicate that I made earlier—some parts might be redundant, but I don't care:
add_tail(A, [], A) :-
!.
add_tail([A|[]], H, [A,H]) :-
!.
add_tail([A|B], H, [A|C]) :-
add_tail(B,H,C).
It works same as builtin predicate append/3:
?- add_tail([a,b,c], d, A).
A = [a, b, c, d].
?- append([a,b,c], [d], A).
A = [a, b, c, d].
When I use append in my invert predicate, it works fine, but if I use add_tail, it fails:
invert([], []).
invert([A|B], C) :-
invert(B, D),
append(D, [A], C).
invert2([], []).
invert2([A|B], C) :-
invert2(B, D),
add_tail(D, A, C).
?- invert([a,b,c,d], A).
A = [d, c, b, a].
?- invert2([a,b,c,d], A).
false. % expected answer A = [d,c,b,a], like above
What exactly is my mistake? Thank you!
The implementation of add_tail/3 does not quite behave the way you expect it to.
Consider:
?- append([], [d], Xs).
Xs = [d].
?- add_tail([], d, Xs).
false.
That's bad... But it gets worse! There are even more issues with the code you presented:
By using (!)/0 you needlessly limit the versatility of your predicate.
Even though [A|[]] maybe correct, it obfuscates your code. Use [A] instead!
add_tail is a bad name for a predicate that works in more than one direction.
The variable names could be better, too! Why not use more descriptive names like As?
Look again at the variables you used in the last clause of add_tail/3!
add_tail([A|B], H, [A|C]) :-
add_tail(B, H, C).
Consider the improved variable names:
add_tail([A|As], E, [A|Xs]) :-
add_tail(As, E, Xs).
I suggest starting over like so:
list_item_appended([], X, [X]).
list_item_appended([E|Es], X, [E|Xs]) :-
list_item_appended(Es, X, Xs).
Let's put list_item_appended/3 to use in list_reverted/2!
list_reverted([], []).
list_reverted([E|Es], Xs) :-
list_reverted(Es, Fs),
list_item_appended(Fs, E, Xs).
Sample query:
?- list_reverted([a,b,c,d], Xs).
Xs = [d, c, b, a].
It is difficult to pinpoint your exact mistake, but the first two clauses of add_tail/3, the ones with the cuts, are wrong (unless I am misunderstanding what the predicate is supposed to do). Already the name is a bit misleading, and you should should care that you have redundant code.
list_back([], B, [B]).
list_back([X|Xs], B, [X|Ys]) :-
list_back(Xs, B, Ys).
This is a drop-in replacement for your add_tail/3 in your definition of invert/2. But as you are probably aware, this is not a very clever way of reversing a list. The textbook example of how to do it:
list_rev(L, R) :-
list_rev_1(L, [], R).
list_rev_1([], R, R).
list_rev_1([X|Xs], R0, R) :-
list_rev_1(Xs, [X|R0], R).
First try the most general query, to see which solutions exist in the most general case:
?- add_tail(X, Y, Z).
yielding the single answer:
X = Z,
Y = []
That's probably not the relation you intend to define here.
Hint: !/0 typically destroys all logical properties of your code, including the ability to use your predicates in all directions.
The first clause of add_tail/3 has a list as second argument, so it will never apply to your test case. Then we are left with 2 clauses (simplified)
add_tail([A],H,[A,H]):-!.
add_tail([A|B],H,[A|C]) :- add_tail(B,H,C).
You can see that we miss a matching clause for the empty list as first argument. Of course, append/3 instead has such match.
based on previous answer of "#mat" the problem is residue in the first two lines
your predicate add_tail is not like append because
with append i get this
| ?- append(X,Y,Z).
Z = Y,
X = [] ? ;
X = [_A],
Z = [_A|Y] ? ;
X = [_A,_B],
Z = [_A,_B|Y] ? ;
X = [_A,_B,_C],
Z = [_A,_B,_C|Y] ? ;
X = [_A,_B,_C,_D],
Z = [_A,_B,_C,_D|Y] ? ;
X = [_A,_B,_C,_D,_E],
Z = [_A,_B,_C,_D,_E|Y] ? ;y
and unfortunately with ur add_tail i get this result
| ?- add_tail(X,Y,Z).
Z = X,
Y = [] ? ;
X = [_A],
Z = [_A|Y] ? ;
X = [_A|_B],
Y = [],
Z = [_A|_B] ? ;
X = [_A,_B],
Z = [_A,_B|Y] ? ;
X = [_A,_B|_C],
Y = [],
Z = [_A,_B|_C] ?
X = [_A,_B,_C],
Z = [_A,_B,_C|Y] ? y
yes
after a simple modification in your add_tail code i obtained your expected result
code
% add_tail(A,[],A):-! . comment
add_tail([],H,H) :-!.
add_tail([A|B],H,[A|C]) :- add_tail(B,H,C).
test add_tail
| ?- add_tail(X,Y,Z).
Z = Y,
X = [] ? ;
X = [_A],
Z = [_A|Y] ? ;
X = [_A,_B],
Z = [_A,_B|Y] ? ;
X = [_A,_B,_C],
Z = [_A,_B,_C|Y] ? ;
X = [_A,_B,_C,_D],
Z = [_A,_B,_C,_D|Y] ? ;
X = [_A,_B,_C,_D,_E],
Z = [_A,_B,_C,_D,_E|Y] ? y
yes
finaly
i test ur invert predicate without modification
| ?- invert([_A,_B,_C],L).
L = [_C,_B,_A] ? ;
no
I hope this post help you to explain how the predicate done inside
enjoy
I can add a single object to a list with this code and query:
makelist(A, B, C, X) :- append([1, 2, 3],A, X).
?- makelist(a, b, c, X).
X = [1, 2, 3|a].
However rather than the usual separator comma (,) there's a vertical line separator (|) and I cannot add another object to the same list:
makelist(A, B, C, X) :- append([1, 2, 3],A, X), append(X,B, X).
?- makelist(a, b, c, X).
false.
There are several misunderstandings. First, lists and their elements are confused which leads to the "dotted pair" that is |a]. Then, it seems that the role of variables is not clear to you.
In the goal append([1,2,3],A,X) both A and X are supposed to be lists. However, you set A to a which is not a list. The problem behind is that append/3 accepts any term for the second argument. To see this, just look at its answers:
?- append([1,2,3],A,X).
X = [1,2,3|A].
So the A can be anything, but it should be rather only a list.
?- A = a, append([1,2,3],A,X).
A = a, X = [1,2,3|a].
Note that append/3 insists that the first argument is a list (or a partial list). Then once you have a bad list, you can no longer append it further. That is:
?- A = a, append([1,2,3],A,X), append(X, _, _).
false.
Note that I did not use your definition literally. Instead, I replaced certain arguments by _ since they cannot help.
The second problem stems from the goal append(X, B, X). This goal, can only be true if B = []. Let's try this out:
?- append(X, B, X).
X = [], B = []
; X = [_A], B = []
; X = [_A,_B], B = []
; X = [_A,_B,_C], B = []
; ... .
And:
?- B = [_|_], append(X, B, X).
loops.
While there are many answers for X, B is always [].
append appends lists. So you want to make a list with a single element to append it to another list:
makelist(A, B, C, X) :- append([1, 2, 3], [A], X). % What's the purpose of B and C, btw?
Also you can add items to the beginning of a list more effectively:
makelist(A, B, C, [A | [1, 2, 3]]).
I have a list [a, b, a, a, a, c, c]
and I need to add two more occurrences of each element.
The end result should look like this:
[a, a, a, b, b, b, a, a, a, a, a, c, c, c, c]
If I have an item on the list that is the same as the next item, then it keeps going until there is a new item, when it finds the new item, it adds two occurrences of the previous item then moves on.
This is my code so far, but I can't figure out how to add two...
dbl([], []).
dbl([X], [X,X]).
dbl([H|T], [H,H|T], [H,H|R]) :- dbl(T, R).
Your code looks a bit strange because the last rule takes three parameters. You only call the binary version, so no recursion will ever try to derive it.
You already had a good idea to look at the parts of the list, where elements change. So there are 4 cases:
1) Your list is empty.
2) You have exactly one element.
3) Your list starts with two equal elements.
4) Your list starts with two different elements.
Case 1 is not specified, so you might need to find a sensible choice for that. Case 2 is somehow similar to case 4, since the end of the list can be seen as a change in elements, where you need to append two copies, but then you are done. Case 3 is quite simple, we can just keep the element and recurse on the rest. Case 4 is where you need to insert the two copies again.
This means your code will look something like this:
% Case 1
dbl([],[]).
% Case 2
dbl([X],[X,X,X]).
% Case 3
dbl([X,X|Xs], [X|Ys]) :-
% [...] recursion skipping the leading X
% Case 4
dbl([X,Y|Xs], [X,X,X|Ys]) :-
dif(X,Y),
% [...] we inserted the copies, so recursion on [Y|Xs] and Ys
Case 3 should be easy to finish, we just drop the first X from both lists and recurse on dbl([X|Xs],Ys). Note that we implicitly made the first two elements equal (i.e. we unified them) by writing the same variable twice.
If you look at the head of case 4, you can directly imitate the pattern you described: supposed the list starts with X, then Y and they are different (dif(X,Y)), the X is repeated 3 times instead of just copied and we then continue with the recursion on the rest starting with Y: dbl([Y|Xs],Ys).
So let's try out the predicate:
?- dbl([a,b,a,a,a,c,c],[a,a,a,b,b,b,a,a,a,a,a,c,c,c,c]).
true ;
false.
Our test case is accepted (true) and we don't find more than one solution (false).
Let's see if we find a wrong solution:
?- dif(Xs,[a,a,a,b,b,b,a,a,a,a,a,c,c,c,c]), dbl([a,b,a,a,a,c,c],Xs).
false.
No, that's also good. What happens, if we have variables in our list?
?- dbl([a,X,a],Ys).
X = a,
Ys = [a, a, a, a, a] ;
Ys = [a, a, a, X, X, X, a, a, a],
dif(X, a),
dif(X, a) ;
false.
Either X = a, then Ys is single run of 5 as; or X is not equal to a, then we need to append the copies in all three runs. Looks also fine. (*)
Now lets see, what happens if we only specify the solution:
?- dbl(X,[a,a,a,b,b]).
false.
Right, a list with a run of only two bs can not be a result of our specification. So lets try to add one:
?- dbl(X,[a,a,a,b,b,b]).
X = [a, b] ;
false.
Hooray, it worked! So lets as a last test look what happens, if we just call our predicate with two variables:
?- dbl(Xs,Ys).
Xs = Ys, Ys = [] ;
Xs = [_G15],
Ys = [_G15, _G15, _G15] ;
Xs = [_G15, _G15],
Ys = [_G15, _G15, _G15, _G15] ;
Xs = [_G15, _G15, _G15],
Ys = [_G15, _G15, _G15, _G15, _G15] ;
Xs = [_G15, _G15, _G15, _G15],
Ys = [_G15, _G15, _G15, _G15, _G15, _G15] ;
[...]
It seems we get the correct answers, but we see only cases for a single run. This is a result of prolog's search strategy(which i will not explain in here). But if we look at shorter lists before we generate longer ones, we can see all the solutions:
?- length(Xs,_), dbl(Xs,Ys).
Xs = Ys, Ys = [] ;
Xs = [_G16],
Ys = [_G16, _G16, _G16] ;
Xs = [_G16, _G16],
Ys = [_G16, _G16, _G16, _G16] ;
Xs = [_G86, _G89],
Ys = [_G86, _G86, _G86, _G89, _G89, _G89],
dif(_G86, _G89) ;
Xs = [_G16, _G16, _G16],
Ys = [_G16, _G16, _G16, _G16, _G16] ;
Xs = [_G188, _G188, _G194],
Ys = [_G188, _G188, _G188, _G188, _G194, _G194, _G194],
dif(_G188, _G194) ;
[...]
So it seems we have a working predicate (**), supposed you filled in the missing goals from the text :)
(*) A remark here: this case only works because we are using dif. The first predicates with equality, one usually encounters are =, == and their respective negations \= and \==. The = stands for unifyability (substituting variables in the arguments s.t. they become equal) and the == stands for syntactic equality (terms being exactly equal). E.g.:
?- f(X) = f(a).
X = a.
?- f(X) \= f(a).
false.
?- f(X) == f(a).
false.
?- f(X) \== f(a).
true.
This means, we can make f(X) equal to f(a), if we substitute X by a. This means if we ask if they can not be made equal (\=), we get the answer false. On the other hand, the two terms are not equal, so == returns false, and its negation \== answers true.
What this also means is that X \== Y is always true, so we can not use \== in our code. In contrast to that, dif waits until it can decide wether its arguments are equal or not. If this is still undecided after finding an answer, the "dif(X,a)" statements are printed.
(**) One last remark here: There is also a solution with the if-then-else construct (test -> goals_if_true; goals_if_false, which merges cases 3 and 4. Since i prefer this solution, you might need to look into the other version yourself.
TL;DR:
From a declarative point of view, the code sketched by #lambda.xy.x is perfect.
Its determinacy can be improved without sacrificing logical-purity.
Code variant #0: #lambda.xy.x's code
Here's the code we want to improve:
dbl0([], []).
dbl0([X], [X,X,X]).
dbl0([X,X|Xs], [X|Ys]) :-
dbl0([X|Xs], Ys).
dbl0([X,Y|Xs], [X,X,X|Ys]) :-
dif(X, Y),
dbl0([Y|Xs], Ys).
Consider the following query and the answer SWI-Prolog gives us:
?- dbl0([a],Xs).
Xs = [a,a,a] ;
false.
With ; false the SWI prolog-toplevel
indicates a choicepoint was left when proving the goal.
For the first answer, Prolog did not search the entire proof tree.
Instead, it replied "here's an answer, there may be more".
Then, when asked for more solutions, Prolog traversed the remaining branches of the proof tree but finds no more answers.
In other words: Prolog needs to think twice to prove something we knew all along!
So, how can we give determinacy hints to Prolog?
By utilizing:
control constructs !/0 and / or (->)/2 (potentially impure)
first argument indexing on the principal functor (never impure)
The code presented in the earlier answer by #CapelliC—which is based on !/0, (->)/2, and the meta-logical predicate (\=)/2—runs well if all arguments are sufficiently instantiated. If not, erratic answers may result—as #lambda.xy.x's comment shows.
Code variant #1: indexing
Indexing can improve determinacy without ever rendering the code non-monotonic. While different Prolog processors have distinct advanced indexing capabilities, the "first-argument principal-functor" indexing variant is widely available.
Principal? This is why executing the goal dbl0([a],Xs) leaves a choicepoint behind: Yes, the goal only matches one clause—dbl0([X],[X,X,X]).—but looking no deeper than the principal functor Prolog assumes that any of the last three clauses could eventually get used. Of course, we know better...
To tell Prolog we utilize principal-functor first-argument indexing:
dbl1([], []).
dbl1([E|Es], Xs) :-
dbl1_(Es, Xs, E).
dbl1_([], [E,E,E], E).
dbl1_([E|Es], [E|Xs], E) :-
dbl1_(Es, Xs, E).
dbl1_([E|Es], [E0,E0,E0|Xs], E0) :-
dif(E0, E),
dbl1_(Es, Xs, E).
Better? Somewhat, but determinacy could be better still...
Code variant #2: indexing on reified term equality
To make Prolog see that the two recursive clauses of dbl1_/3 are mutually exclusive (in certain cases), we reify the truth value of
term equality and then index on that value:
This is where reified term equality (=)/3 comes into play:
dbl2([], []).
dbl2([E|Es], Xs) :-
dbl2_(Es, Xs, E).
dbl2_([], [E,E,E], E).
dbl2_([E|Es], Xs, E0) :-
=(E0, E, T),
t_dbl2_(T, Xs, E0, E, Es).
t_dbl2_(true, [E|Xs], _, E, Es) :-
dbl2_(Es, Xs, E).
t_dbl2_(false, [E0,E0,E0|Xs], E0, E, Es) :-
dbl2_(Es, Xs, E).
Sample queries using SWI-Prolog:
?- dbl0([a],Xs).
Xs = [a, a, a] ;
false.
?- dbl1([a],Xs).
Xs = [a, a, a].
?- dbl2([a],Xs).
Xs = [a, a, a].
?- dbl0([a,b,b],Xs).
Xs = [a, a, a, b, b, b, b] ;
false.
?- dbl1([a,b,b],Xs).
Xs = [a, a, a, b, b, b, b] ;
false.
?- dbl2([a,b,b],Xs).
Xs = [a, a, a, b, b, b, b].
To make above code more compact, use control construct if_/3 .
I was just about to throw this version with if_/3 and (=)/3 in the hat when I saw #repeat already suggested it. So this is essentially the more compact version as outlined by #repeat:
list_dbl([],[]).
list_dbl([X],[X,X,X]).
list_dbl([A,B|Xs],DBL) :-
if_(A=B,DBL=[A,B|Ys],DBL=[A,A,A,B|Ys]),
list_dbl([B|Xs],[B|Ys]).
It yields the same results as dbl2/2 by #repeat:
?- list_dbl([a],DBL).
DBL = [a,a,a]
?- list_dbl([a,b,b],DBL).
DBL = [a,a,a,b,b,b,b]
The example query by the OP works as expected:
?- list_dbl([a,b,a,a,a,c,c],DBL).
DBL = [a,a,a,b,b,b,a,a,a,a,a,c,c,c,c]
Plus here are some of the example queries provided by #lambda.xy.x. They yield the same results as #repeat's dbl/2 and #lambda.xy.x's dbl/2:
?- dif(Xs,[a,a,a,b,b,b,a,a,a,a,a,c,c,c,c]), list_dbl([a,b,a,a,a,c,c],Xs).
no
?- list_dbl(X,[a,a,a,b,b]).
no
?- list_dbl(L,[a,a,a,b,b,b]).
L = [a,b] ? ;
no
?- list_dbl(L,DBL).
DBL = L = [] ? ;
DBL = [_A,_A,_A],
L = [_A] ? ;
DBL = [_A,_A,_A,_A],
L = [_A,_A] ? ;
DBL = [_A,_A,_A,_A,_A],
L = [_A,_A,_A] ? ;
...
?- list_dbl([a,X,a],DBL).
DBL = [a,a,a,a,a],
X = a ? ;
DBL = [a,a,a,X,X,X,a,a,a],
dif(X,a),
dif(a,X)
?- length(L,_), list_dbl(L,DBL).
DBL = L = [] ? ;
DBL = [_A,_A,_A],
L = [_A] ? ;
DBL = [_A,_A,_A,_A],
L = [_A,_A] ? ;
DBL = [_A,_A,_A,_B,_B,_B],
L = [_A,_B],
dif(_A,_B) ? ;
DBL = [_A,_A,_A,_A,_A],
L = [_A,_A,_A] ?
dbl([X,Y|T], [X,X,X|R]) :- X \= Y, !, dbl([Y|T], R).
dbl([H|T], R) :-
T = []
-> R = [H,H,H]
; R = [H|Q], dbl(T, Q).
The first clause handles the basic requirement, adding two elements on sequence change.
The second one handles list termination as a sequence change, otherwise, does a plain copy.