I have a list L and I need to split each element into a separate list and again append them together. This is the code I made for the same.
split([],[]).
split([H|T],Ls):-split(T,Ls),splist(H,[]).
make(Val,[H1|List],[H1|Res]):- make(Val,List,Res). make(Val, List,[Val|List]).
splist(H,L2):- make(Sum,[],L1),append(L1,L2,NewL).
When I use this code, each element of L is passed recursively from split() to splist() and made into a list L1 with single element by make(). I need append to keep concatenating L1 and L2. But it does not so so
For example, I have L=[1,2,3]. Now I need the following process to be done.
H=1, L1=[1] and L2=[1]. Next H=2, L1=[2] and L2=[1,2]. Next H=3, L1=[3] and L2=[1,2,3].
I need the output as mentioned above, but this is what my code does.
H=1, L1=[1], and L2= [1]. Next H=2, L1=[2] and L2=[2]. Next H=3, L1=[3] and L2=[3].
I can't make any sense out of your code. make definition is incomplete. As is, it does nothing and then fails.
Your split is equivalent to split(X,[]):- reverse(X,R), maplist(spl([]),R). with spl(B,A):-splist(A,B)., i.e. it tries splist(H,[]) for each element H of the input list X, backwards, to see whether it fails or not - that's its only outcome, as the arguments are fixed - H and [].
naming your predicates split and splist is a very bad idea - we humans are wired to distinguish words from their start, and the only different letter in these names is hidden way far near the end. IOW the two names are very similar, and it is very easy to misread and mistype them.
lastly, for splist(H,L2):- make(Sum,[],L1),append(L1,L2,NewL)., since make cn only fail, so will splist. But even if make were to produce something in L1 out of thin air - Sum starts out uninstantiated mind you - what does it say about L2? That it can be appended to the list L1? Any list can be appended to any other, saying that is saying nothing.
?? :)
Related
I'm trying to write Prolog logic for the first time, but I'm having trouble. I am to write logic that takes two lists and checks for like elements between the two. For example, consider the predicate similarity/2 :
?- similarity([2,4,5,6,8], [1,3,5,6,9]).
true.
?- similarity([1,2,3], [5,6,8]).
false.
The first query will return true as those two lists have 5 and 6 in common. The second returns false as there are no common elements between the two lists in that query.
I CANNOT use built in logic, such as member, disjoint, intersection, etc. I am thinking of iterating through the first list provided, and checking to see if it matches each element in the second list. Is this an efficient approach to this problem? I will appreciate any advice and help. Thank you so much.
Writing Prolog for the first time can be really daunting, since it is unlike many traditional programming languages that you will most likely encounter; however it is a very rewarding experience once you've got a grasp on this new style of programming! Since you mention that you are writing Prolog for the first time I'll give some general tips and tricks about writing Prolog, and then move onto some hints to your problem, and then provide what I believe to be a solution.
Think Recursively
You can think of every Prolog program that you write to be intrinsically recursive in nature. i.e. you can provide it with a series of "base-cases" which take the following form:
human(John). or wildling(Ygritte) In my opinion, these rules should always be the first ones that you write. Try to break down the problem into its simplest case and then work from there.
On the other hand, you can also provide it with more complex rules which will look something like this: contains(X, [H|T]):- contains(X, T) The key bit is that writing a rule like this is very much equivalent to writing a recursive function in say, Python. This rule does a lot of the heavy lifting in looking to see whether a value is contained in a list, but it isn't complete without a "base-case". A complete contains rule would actually be two rules put together: contains(X, [X|_]).
contains(X, [H|T]):-contains(X, T).
The big takeaway from this is to try and identify the simple cases of your problem, which can act like base cases in a recursive function, and then try to identify how you want to "recurse" and actually do work on the problem at hand.
Pattern Matching
Part of the great thing about Prolog is the pattern matching system that it has in place. You should 100% use this to your advantage whenever you can -- it is especially helpful when trying to do anything with lists. For example:
head(X, [X|T]).
Will evaluate to true when called thusly: head(1, [1, 2, 3]) because intrinsic in the rule is the matching of X. This sort of pattern matching on the first element of a list is incredibly important and really the key way that you will do any work on lists in Prolog. In my experience, pattern matching on the head of a list will often be one of the "base-cases" that I mentioned beforehand.
Understand The Flow of the Program
Another key component of how Prolog works is that it takes a "top-down" approach to reading code. What I mean by that is that every time a rule is called (except for definitions of the form king(James).), Prolog starts at line 1 and continues until it reaches a rule that is true or the end of the file. Therefore, the ordering of your rules is incredibly important. I'm assuming that you know that you can combine rules together via a comma to indicate logical AND, but what is maybe more subtle is that if you order one rule above another, it can act as a logical OR, simply because it will be evaluated before another rule, and can potentially cause the program to recurse.
Specific Example
Now that I've gotten all of my general advice out of the way, I'll actually reference the given problem. First, I'd write my "base-case". What would happen if you are given two lists whose first elements are the same? If the first element in each list is not the same, then they have to be different. So, you have to look through the second list to see if this element is contained anywhere in the rest of the list. What kind of rule would this produce? OR it could be the case that the first element of the first list is not contained within the second at all, in which case you have to advance once in the first list, and start again with the second list. What kind of rule would this produce?
In the end, I would say that your approach is the correct one to take, and I have provided my own solution below:
similarity([H|_], [H|_]).
similarity(H1|T1], [_|T2]):- similarity([H1|T1], T2).
similarity([_|T1], [H2|T2]):- similarity(T1, [H2|T2]).
Hope all of this helps in some way!
I just have a little question about the following code I've just written.
permu([],[]).
permu([H|T],R) :- same_length([H|T],R), select(H,R,Rz), permu(T,Rz).
Result: gives out all permutation (doesn't delete redundancies, but that's not the matter; by using findall and defining another predicate I can fix that easily)
My initial code:
permu([],[]).
permu([H|T],R) :- select(H,R,Rz), permu(T,Rz).
Result: gives out the first result and ends up in endless loop, when clicking "Next". But when I decided to precisely define their length (with same_length as shown above) it suddenly worked.
I have already traced both codes, but I still don't understand why the length of both lists HAS to be defined under any circumstances.
Looking at my initial code, both lists lose one element in each step of the recursion. The list on the left gets its head cut off each run and my R variable is being shortened because of select. The recursion process ends, when both lists run out of elements in the same recursion run => due to select, both lists need to share the same number and type of elements.
What does prevent Prolog from realizing it? What makes him fall into the endless loop?
Taking this code as an example I don't have to define the length but it still works:
permu([], []).
permu([H|T], R) :- permu(T, Rz), select(H, R, Rz).
In contrast to my initial code here the select is just put after the recursion call (permu).
I learn Prolog at university and keep stumbling on someting rather odd during the home excercises. I wrote following Prolog clauses, which are part of a much bigger program:
edges(X,Edges):-
findall(Edge,(highway(X,Y,Edge);highway(Y,X,Edge)),Edges).
edgesList([],_).
edgesList([node(X)|InL],OutL):-
member((node(X),Edges),OutL),
edges(X,Edges),
edgesList(InL,OutL).
Which use following facts:
highway(1,2,yellow).
highway(2,3,blue).
highway(1,3,yellow).
You can see highway as a fact that describes two nodes in the first two arguments and an edge in the third. All facts together form a connected graph.
With the clause edgesList, I want to list the edges per node e.g.
Result = [(node(1),[yellow,yellow]),(node(2),[blue,yellow]),(node(3),[blue,yellow])]
But when I write my query:
edgesList([node(1),node(2),node(3)],List).
I get following result:
List = [(node(1),[yellow, yellow]), (node(2),[blue, yellow]), (node(3),[blue, yellow])|_G610]
For some reason, Prolog won't unify the tail of the result-list with the empty list, despite the fact that the member-predicate is used correct, I assume. It's something that happend a few times now in different excercises and it would be good to know what I did wrong...
The problem is in the clause:
edgesList([],_).
because in the end it will fill the list with an uninstantiated tail (|_G610).
One solution is :
edges(X,Edges):-
findall(Edge,(highway(X,Y,Edge);highway(Y,X,Edge)),Edges).
edgesList([],[]).
edgesList([node(X)|InL],[(node(X),Edges)|T]):-
edges(X,Edges),
edgesList(InL,T).
I have this very simple problem: write a Prolog program that implement the append Prolog function, that concatenate two strings and that work in the following way:
append([a,b],[c,d],X). ---> X = [a,b,c,d]
append([a,b],X,[a,b,c,d]). ---> X = [c,d]
append([a,b],[X,d],[a,b,c,d]). ---> X=c
append(X,Y,[a,b,c,d]). ---> X=[] and Y=[a,b,c,d)
So I have the following two solutions and I am not so sure if my declarative interpretation is correct:
1) SOLUTION 1:
myappend1([],L,L).
myappend1([X|L1],L2,[X|L3]) :- myappend1(L1,L2,L3).
I think that I can read it in a declarative way as following:
The fact say that: if the first list (L1) is empty and the second list (L2) is not empty then it is TRUE that the concatenation of L1*L2 is L2
If the fact it is not true it means that the first list is not empty and so the concatenation of the first list and the second list it is not true that is the second list
So, let me call the first list L1, the second list L2 and the third list L3 then the rule responds TRUE if L3 is the concatenation of L1 and L2, false otherwise
I think that the declarative meaning of this rule is that: the head of the rule is true if the body of the rule is true.
In the head extract the first X element from the L1 list and from L3 list (and try to unify, if it matching go ahead, otherwise it means that the third list it is not the concatenation for the first and the second list)
In the body call the function on the first list without X element, the second list and the L3 list (that represent the concatenation)
When it reach the base case in which I have the demonstrated fact myappend1([],L,L). that is true, the program do backtracking at the previous past and because the X element of the first list unified with the X element of the third list it can do that also this computational pass it is TRUE and go back until reach the first assertion
Is this a correct declarative interpretation?
2) SECOND SOLUTION:
myappend2([],L,L).
myappend2(L1,L2,L3) :- L1=[X|T], % Dimostra questo predicato AND
L3=[X|L4], % Dimostra questo predicato AND
myappend2(T,L2,L4). % Dimostra questa funzione
As in the previous solution the fact simply say that: if the first list (L1) is empty and the second list (L2) is not empty then it is TRUE that the concatenation of L1*L2 is L2
If the fact it is not true it means that the first list is not empty and so the concatenation of the first list and the second list it is not true that is the second list
If the fact it is not true Prolog call the rule and this rule means that: the head of the rule is true if the body of the rule is true.
In this case I can read it in this way:
The concatenation of L1 and L2 is L3 is TRUE if it is true that:
The current first X element of L1 unifies with the current first element of concatenation list and myappend2 called on the first sublist, L2 and the third sublist it is true
Is it correct?
for me it is so difficult reasoning in declarative way :-(
Like last time, you're adding restrictions that aren't present in the code. Don't feel bad about it, Prolog is very different and it will take time to get used to it.
Let's start.
append([], L, L).
You said:
If the first list (L1) is empty and the second list (L2) is not empty then it is TRUE that the concatenation of L1*L2 is L2
In fact this rule says nothing about whether L2 is empty--or even a list!--or not. It simply says that the empty list appended to something else is that something else. Observe:
?- append([], foo, X).
X = foo.
The declarative reading here is "the empty list appended to L is L."
If the fact it is not true it means that the first list is not empty and so the concatenation of the first list and the second list it is not true that is the second list
Yes, this is correct, but Prolog isn't probing that deeply into the body. It just says "the first list is not empty, so this rule does not match; moving on."
The next rule:
myappend1([X|L1], L2, [X|L3]) :- myappend1(L1,L2,L3).
Your commentary seems excessively complex to me. I would say that this rule says: "myappend1 of the list [X followed by L1] to L2 is the list [X followed by L3], if myappend1 of the list L1 to L2 is L3." The consequences of this reading, however, are exactly as you describe.
Your understanding of what is happening in the first version is, therefore, correct.
The second solution is, mechanically, exactly the same as the first solution. The only difference is that we have moved the unification from the head of the clause into the body. This version is, to my eyes, clearly inferior, because all it has done is create extra work for the reader.
I think the problem you're having, so far, is that your declarative reasoning is intimately tied up with Prolog's engine of computation. A purer declarative reading like the ones I have supplied are simpler and look more like what the Prolog is saying (and have less to do with how it is evaluated).
It will take practice for you to separate these notions, but I think it will help you get better (clearly it's something you're concerned about). In the meantime there's nothing wrong with coming here and asking for help like you've been doing when you get confused. :)
Let me know if I can help more!
When you try to figure out the declarative meaning of a predicate, you are asking: For which solutions does this predicate hold?
Prolog's1 clauses contribute to the set of solutions independently. So making any connections between the clauses needs some extra scrutiny. It is very easy to make some assumptions that are not the case:
myappend1([],L,L).
If the fact it is not true it means that the first list is not empty and so ...
Consider a goal, myappend1([],[],[a]). The fact does not apply, still the first list is empty. Here, you are attempting to operationalize the meaning of the clause. It is very tempting to do this since the largest part of programming languages can only be understood by imagining how something happens step-by-step. The difficulty in Prolog lies in trying to ignore such details, without entirely ignoring procedural aspects.
myappend1([X|L1],L2,[X|L3]) :- myappend1(L1,L2,L3).
To read rules, in particular recursive rules, it is helpful to look at the :- which is a 1970s rendering of ← . So it is an arrow, but it goes from right-to-left. Therefore, you can read this rules as follows, starting with the right-hand-side:
Provided that myappend(L1,L2,L3) holds, now cross the :- over to the left side also myappend([X|L1],L2,[X|L3]) holds.
Sometimes, an even better way to read such a rule is to cover the head completely and ask
??? :- myappend1(L1,L2,L3).
Assume, I know some L1, L2, L3 that hold for myappend1(L1,L2,L3). what can I conclude out of this? Is there anything interesting? Is there anything related I can construct easily out of those 3 values?
This is something which is in the beginning a bit revolting, because you might say: But how do I know that such exists? Well, you don't. You are only assuming it exists. If it will never exist, then you will never be able to make that conclusion.
Many try to read the rules left-to-right, but while Prolog is actually executing them left-to-right, the meaning they cover is easier to understand going in the direction of the conclusion. When Prolog executes a rule left-to-right it does not know if this will work out or not. So the execution might be of entirely speculative nature. Think of append(L1,[z],[a,b,c,d,e]). Here, Prolog will apply this rule for each element of the list. But all such application is in vain. That is, ultimately it will fail.
Fine print
1 Actually, the pure, monotonic subset of Prolog.
Hi i'm trying to insert an element in a list but it is very important from my program that the result is stored in the original list and not in a new one.
Any code that i have written or found on the internet only succeeds if you create a new list in which the end result is kept.
So my question is can anyone tell me how to define a function: insert(X,L) where X is an element and L is a list?
No, Prolog just doesn't work that way. There is no such thing as "modifying" a value. A variable can be unified with a specific value, but if it was already [1,3], it won't ever be [1,2,3] later.
As aschepler says, you cannot add or make any change to a proper list, i.e. a list in which every element is already bound. The only "modifying" we can do is unifying one expression with another.
However there is a concept of a partial list to which additional elements can be "added" at the end. This is typically known as a difference list, although that nomenclature may not be immediately understandable.
Suppose we start, not with an empty list, but with a free variable X. One might however think of subtracting X from X and getting "nothing". That is, an empty difference list is represented by X - X. The minus "-" here is a purely formal operator; no evaluation of the difference is intended. It's just a convenient syntax as you see from how difference lists can be used to accomplish what you (probably) want to do.
We can add an element to a difference list as follows:
insertDL(M,X-Y,X-Z) :- Y = [M|Z].
Here M is the new element we want to add, X-Y is the "old" difference list, and X-Z is the "new" difference (to which M has been added, by unifying the previously free variable Y with the partial list [M|Z], so that Z becomes the "open" tail of partial list X).
When we are finally done inserting things into our difference list, we can turn X into a proper list by setting the "free tail" at that point to the empty list [ ]. In this sense X is the "same" variable as when we first began, just unified by incremental steps from free variable to proper list.
This is a very powerful technique in Prolog programming, and it takes some practice to feel comfortable using it. Some links to further discussion on the Web:
[From Prolog lists to difference lists]
http://www.irisa.fr/prive/ridoux/ICLP91/node8.html
[Implementing difference lists in Prolog]
http://www.cl.cam.ac.uk/~jpw48/difflists.pdf
[Lecture Notes: Difference Lists]
http://www.cs.cmu.edu/~fp/courses/lp/lectures/11-diff.pdf
Some prologs provide the setarg/3 predicate in order to modify terms in place.
In order to use it over lists, you only need to consider that they are just a nice representation of chains of compound terms with functor '.'/2
In any case, when you need to use setarg/3 in Prolog, it probably means you are doing something wrong.