multi_available(X):-
member(X,
[grilled_chicken,
jambo_beef,
grilled_cheese_sandwich,roast_beef,potato_salad,chicken_rice,
jambo_beef,
service_dish,service_dish,
beef_stew,potato_corn,grilled_chicken,roast_beef,mushroom_rice,
service_dish,
jambo_beef]).
member(X,[X|_]).
member(X,[_|T]):-member(X,T).
How can I check if an element is repeated in a list in Prolog?
Using Prologue of Prolog:
member(X, L) is true if X is an element of the list L.
select(X, Xs, Ys) is true if X is an element of the
list Xs and Ys is the list Xs with one occurrence of X removed.
I get the following simple solution:
:- use_module(library(basic/lists)).
multi_available(L, X) :- select(X, L, R), member(X, R).
It might not be the most efficient. But it works:
Jekejeke Prolog 2, Runtime Library 1.2.5
(c) 1985-2017, XLOG Technologies GmbH, Switzerland
?- multi_available([a,b,c,b,a,d],X).
X = a ;
X = b ;
X = b ;
X = a ;
No
I'm not sure if you just want to know whether an element occurs more than once or whether you need how many times it occurs, so here's both!
See if the List contains Element, +1 to Count for every occurrence
occurrences(Element,[Head|Tail], Count, OutputCount) :-
Element = Head,
NewCount is Count + 1,
occurrences(Element, Tail, NewCount, OutputCount).
Element does not match the first element of the List? Don't add 1, go to the next element.
occurrences(Element, [Head|Tail], Count, OutputCount) :-
Element \= Head,
occurrences(Element, Tail, Count, OutputCount).
If the List is empty, return the final count
occurrences(Element,[],Count,Count).
Since the Count can't be negative, just pass it on as Zero and only return the result.
occurrences(Element, List, Count) :- occurrences(Element, List, 0, Count).
Check whether the result is higher than 1
moreThanOnce(Element, List) :-
occurrences(Element, List, Count),
Count > 1.
Does Element occur more than once in List?
?- moreThanOnce(1, [1,2,3,1,1,1]).
How many times does it occur?
?- occurrences(1, [1,2,3,1,1,1], Count).
Note that if a list is empty, [Head|Tail] will automatically fail and List = [] automatically succeeds. This means that OutputCount remains empty throughout the program, UNTIL the end condition is met, which is only reached after [Head|Tail] fails, meaning the list is empty. I hope this is clear enough, be sure to ask for clarifications if it isn't.
Related
I'm having trouble with the following predicate:
treeToList(void, []).
treeToList(arbol(X, HI1, HD1), L) :-
treeToList(HI1, L1),
treeToList(HD1, L2),
append(L1, [X|L2], L).
maximumInList([X], X).
maximumInList([A|L], X) :-
maximumInList(L,X1),
(A > X1 -> X = A; X = X1).
maxNodeInTree(arbol, N) :-
treeToList(arbol, L),
maximumInList(L, N).
TreeToList gets a tree and returns a list with all of its nodes. Meanwhile maximumInList gets a list and returns the maximum element in the list.
Both of these predicates work fine individually, however the last one, maxNodeInTree, is supposed to first get the list L using treeToList which then will be passed to maximumInList and it'll return the maximum element in the whole tree. And yet Prolog returns false.
Any tips are appreciated!
You have a typo in the last predicate (arbol instead of Arbol). Try:
maxNodeInTree(Arbol, N) :-
treeToList(Arbol, L),
maximumInList(L, N).
I am trying to make a list from a tuple of variable size. However, I am having trouble figuring out how to represent an empty tuple (or a tuple with a single value in it), which I need as my end case.
This is what I have right now which, judging by the trace, does create a list (reversed however but it's not really a problem) but it fails at the very end.
tuple_to_list((), []).
tuple_to_list((X, ()), [X]).
tuple_to_list((X, XS), List) :-
tuple_to_list(XS, [X|List]).
Just :
tuple_to_list((X, XS),[X | List]) :-
tuple_to_list(XS, List).
tuple_to_list((X), [X]):-
X \= (_,_).
Last clause X \= (,). because of
?- (X) = (a,b).
X = (a, b).
I want to write predicate which can count all encountered number:
count(1, [1,0,0,1,0], X).
X = 2.
I tried to write it like:
count(_, [], 0).
count(Num, [H|T], X) :- count(Num, T, X1), Num = H, X is X1 + 1.
Why doesn't work it?
Why doesn't work it?
Prolog is a programming language that often can answer such question directly. Look how I tried out your definition starting with your failing query:
?- count(1, [1,0,0,1,0], X).
false.
?- count(1, Xs, X).
Xs = [], X = 0
; Xs = [1], X = 1
; Xs = [1,1], X = 2
; Xs = [1,1,1], X = 3
; ... .
?- Xs = [_,_,_], count(1, Xs, X).
Xs = [1,1,1], X = 3.
So first I realized that the query does not work at all, then I generalized the query. I replaced the big list by a variable Xs and said: Prolog, fill in the blanks for me! And Prolog did this and reveals us precisely the cases when it will succeed.
In fact, it only succeeds with lists of 1s only. That is odd. Your definition is too restricted - it correctly counts the 1s in lists where there are only ones, but all other lists are rejected. #coder showed you how to extend your definition.
Here is another one using library(reif) for
SICStus|SWI. Alternatively, see tfilter/3.
count(X, Xs, N) :-
tfilter(=(X), Xs, Ys),
length(Ys, N).
A definition more in the style of the other definitions:
count(_, [], 0).
count(E, [X|Xs], N0) :-
if_(E = X, C = 1, C = 0),
count(E, Xs, N1),
N0 is N1+C.
And now for some more general uses:
How does a four element list look like that has 3 times a 1 in it?
?- length(L, 4), count(1, L, 3).
L = [1,1,1,_A], dif(1,_A)
; L = [1,1,_A,1], dif(1,_A)
; L = [1,_A,1,1], dif(1,_A)
; L = [_A,1,1,1], dif(1,_A)
; false.
So the remaining element must be something different from 1.
That's the fine generality Prolog offers us.
The problem is that as stated by #lurker if condition (or better unification) fails then the predicate will fail. You could make another clause for this purpose, using dif/2 which is pure and defined in the iso:
count(_, [], 0).
count(Num, [H|T], X) :- dif(Num,H), count(Num, T, X).
count(Num, [H|T], X) :- Num = H, count(Num, T, X1), X is X1 + 1.
The above is not the most efficient solution since it leaves many choice points but it is a quick and correct solution.
You simply let the predicate fail at the unification Num = X. Basically, it's like you don't accept terms which are different from the only one you are counting.
I propose to you this simple solution which uses tail recursion and scans the list in linear time. Despite the length, it's very efficient and elegant, it exploits declarative programming techniques and the backtracking of the Prolog engine.
count(C, L, R) :-
count(C, L, 0, R).
count(_, [], Acc, Acc).
count(C, [C|Xr], Acc, R) :-
IncAcc is Acc + 1,
count(C, Xr, IncAcc, R).
count(C, [X|Xr], Acc, R) :-
dif(X, C),
count(C, Xr, Acc, R).
count/3 is the launcher predicate. It takes the term to count, the list and gives to you the result value.
The first count/4 is the basic case of the recursion.
The second count/4 is executed when the head of the list is unified with the term you are looking for.
The third count/4 is reached upon backtracking: If the term doesn’t match, the unification fails, you won't need to increment the accumulator.
Acc allows you to scan the entire list propagating the partial result of the recursive processing. At the end you simply have to return it.
I solved it myself:
count(_, [], 0).
count(Num, [H|T], X) :- Num \= H, count(Num, T, X).
count(Num, [H|T], X) :- Num = H, count(Num, T, X1), X is X1 + 1.
I have decided to add my solution to the list here.
Other solutions here use either explicit unification/failure to unify, or libraries/other functions, but mine uses cuts and implicit unification instead. Note my solution is similar to Ilario's solution but simplifies this using cuts.
count(_, [], 0) :- !.
count(Value, [Value|Tail],Occurrences) :- !,
count(Value,Tail,TailOcc),
Occurrences is TailOcc+1.
count(Value, [_|Tail], Occurrences) :- count(Value,Tail,Occurrences).
How does this work? And how did you code it?
It is often useful to equate solving a problem like this to solving a proof by induction, with a base case, and then a inductive step which shows how to reduce the problem down.
Line 1 - base case
Line 1 (count(_, [], 0) :- !.) handles the "base case".
As we are working on a list, and have to look at each element, the simplest case is zero elements ([]). Therefore, we want a list with zero elements to have no instances of the Value we are looking for.
Note I have replaced Value in the final code with _ - this is because we do not care what value we are looking for if there are no values in the list anyway! Therefore, to avoid a singleton variable we negate it here.
I also added a ! (a cut) after this - as there is only one correct value for the number of occurrences we do not want Prolog to backtrack and fail - therefore we tell Prolog we found the correct value by adding this cut.
Lines 2/3 - inductive step
Lines 2 and 3 handle the "inductive step". This should handle if we have one or more elements in the list we are given. In Prolog we can only directly look at the head of the list, therefore let us look at one element at a time. Therefore, we have two cases - either the value at the head of the list is the Value we are looking for, or it is not.
Line 2
Line 2 (count(Value, [Value|Tail],Occurrences) :- !, count(Value,Tail,TailOcc), Occurrences is TailOcc+1.) handles if the head of our list and the value we are looking for match. Therefore, we simply use the same variable name so Prolog will unify them.
A cut is used as the first step in our solution (which makes each case mutually exclusive, and makes our solution last-call-optimised, by telling Prolog not to try any other rules).
Then, we find out how many instances of our term there are in the rest of the list (call it TailOcc). We don't know how many terms there are in the list we have at the moment, but we know it is one more than there are in the rest of the list (as we have a match).
Once we know how many instances there are in the rest of the list (call this Tail), we can take this value and add 1 to it, then return this as the last value in our count function (call this Occurences).
Line 3
Line 3 (count(Value, [_|Tail], Occurrences) :- count(Value,Tail,Occurrences).) handles if the head of our list and the value we are looking for do not match.
As we used a cut in line 2, this line will only be tried if line 2 fails (i.e. there is no match).
We simply take the number of instances in the rest of the list (the tail) and return this same value without editing it.
I want to solve this problem in Prolog. i want give a list of natural numbers to find all the elements in the list that satisfy this condition:
All elements on the left of it are smaller than it and all the elements on the right of it are larger than it.
For example give a list [3,2,4,1,5,7,8,9,10,8] the answer would be 5,7
So far I've manage to make this function that given an element of the list it returns true or false if the element satisfises the condition described above.
check(Elem, List) :-
seperate(Elem, List, List1, List2),
lesser(Elem, List1, X1),
bigger(Elem, List2, X2),
size(X1, L1),
size(X2, L2),
size(List, L3),
match(L1, L2, L3),
Now I want to make another predicate that given a list, it does the above computations for each element of the list. Due to the fact that more than one element may satisfy it I want to create another list with all the elements that satisfy the problem.
The question would be something like ?-predicate_name([[3,2,4,1,5,7,8,9,10,8],N). and the result would be a list of elements.
Sry If I am not using the right terms of Prolog. I will describe what I want to do in sequential logic language to be more specific although it's not a good idea to think like that. If we consider a the predicate check as a function that given a list and element of the list it would return true or false whether or not the element satisfied the conditions of the problem. Now I want to parse each element of the list and for each one of it call the function check. If that would return true then I would add the element in another list a.k.a result. I want to do this in Prolog but I don't know how to iterate a list.
Here is a version using DCGs and assuming we want to compare arithmetically.
list_mid(L, M) :-
phrase(mid(M), L).
mid(M) -->
seq(Sm),
[M],
{maplist(>(M),Sm)},
seq(Gr),
{maplist(<(M),Gr)}.
seq([]) -->
[].
seq([E|Es]) -->
[E],
seq(Es).
Often it is not worth optimizing this any further. The first seq(Sm) together with the subsequent maplist/2 might be merged together. This is a bit tricky, since one has to handle separately the cases where Sm = [] and Sm = [_|_].
mid(M) -->
( [M]
| max(Mx),
[M],
{Mx < M}
),
min(M).
max(M) -->
[E],
maxi(E, M).
maxi(E, E) -->
[].
maxi(E, M) -->
[F],
{G is max(F,E)},
maxi(G, M).
min(_) -->
[].
min(M) -->
[E],
{M < E},
min(M).
I'm going to take a different approach on the problem.
We want to find all of the values that meet the criteria of being a "mid" value, which is one defined as being greater than all those before it in the list, and less than all those after.
Define a predicate mid(L, M) as meaning M is a "mid" value of L:
mid([X|T], X) :- % The first element of a list is a "mid" if...
less(X, T). % it is less than the rest of the list
mid([X|T], M) :- % M is a "mid" of [X|T] if...
mid(T, X, M). % M is a "mid" > X
% (NOTE: first element is not a "mid" by definition)
mid([X|T], LastM, X) :- % X is a "mid" > Y if...
X > LastM, % X > the last "mid"
less(X, T). % X < the rest of the list, T
mid([X|T], LastM, M) :- % Also, M is a "mid" if...
Z is max(X, LastM), % Z is the larger of X and the last "mid"
mid(T, Z, M). % M is the "mid" of T which is > Z
less(X, [Y|T]) :- % X is less than the list [Y|T] if...
X < Y, % X < Y, and
less(X, T). % X < the tail, T
less(_, []). % An element is always less than the empty list
Each query will find the next "mid":
| ?- mid([3,2,4,1,5,7,8,9,10,8], M).
M = 5 ? ;
M = 7 ? ;
no
Then they can be captured with a findall:
mids(L, Ms) :-
findall(M, mid(L, M), Ns).
| ?- mids([3,2,4,1,5,7,8,9,10,8], Ms).
Ms = [5,7]
yes
| ?- mids([2], L).
L = [2]
(1 ms) yes
This is probably not the most computationally efficient solution since it doesn't take advantage of a couple of properties of "mids". For example, "mids" will all be clustered together contiguously, so once a "mid" is found, it doesn't make sense to continue searching if an element is subsequently encountered which is not itself a "mid". If efficiency is a goal, these sorts of ideas can be worked into the logical process.
ADDENDUM
With credit to #false for reminding me about maplist, the above predicate call less(X, T) could be replaced by maplist(<(X), T) eliminating the definition of less in the above implementation.
I have a problem where I have a list like this:
[[el1, el2, el3],
[el4, el5, el6],
[[el7, el8, el9], [el10, el11, el12], ..... , [elxx, elyy, elzz]],
[el, el, el]...]]
I want to pattern match the inner list of lists, the
[el7, el8, el9], [el10, el11, el12], ..... , [elxx, elyy, elzz]
How can this be done?
As of now I patternmatch the other elements with
my_method([[El1, El2, El3] | Rest]).
UPDATE
I want to pattern match if the next item of the list is a list of lists - I will be iterating over this list, removing item after item. There can be any number of lists of lists, and they can contain any number of items. They can also contain lists of lists. In fact, I will recursively call the same processing method whenever I come upon a list of lists.
All bottom level lists will have three elements, however these elements might be different:
[1, p, neg(5,6)]
[5, neg(7,6), assumption]
You said "I will be iterating over this list, removing item after item", so here's code that does just that, assuming an "item" is a three-element list of non-lists.
nested_member(X,X) :-
X = [A,_,_],
\+ is_list(A).
nested_member(X,[L|_]) :-
nested_member(X,L).
nested_member(X,[_|L]) :-
nested_member(X,L).
This can be used to backtrack over the "items":
?- nested_member(X,[[el1, el2, el3], [el4, el5, el6],
[[el7, el8, el9], [el10, el11, el12],[elxx, elyy, elzz]]]).
X = [el1, el2, el3] ;
X = [el4, el5, el6] ;
X = [el7, el8, el9] ;
X = [el10, el11, el12] ;
X = [elxx, elyy, elzz] ;
false.
I you want, you can even find out how deep in the list the items were found:
nested_member(X,L,D) :-
nested_member(X,L,0,D).
nested_member(X,X,D,D) :-
X = [A,_,_],
\+ is_list(A).
nested_member(X,[L|_],D0,D) :-
D1 is D0+1,
nested_member(X,L,D1,D).
nested_member(X,[_|L],D0,D) :-
nested_member(X,L,D0,D).
You can use predicates similar to the following.
qualify([], []).
qualify([H|T], [HN|TN]) :- qualify_one(H, HN), qualify(T, TN).
qualify_one([H|_], N) :- qualify_one(H, N1), N is N1 + 1, !.
qualify_one(_, 0).
What qualify does is for each member of the list to find out on what level of the scale “not a list”, “simple list”, “list of lists”, … it is, based on the first item.
Example:
?- qualify([1,[2,3,3],[[4,5,6], [7,8,9]]], NS).
NS = [0, 1, 2].