Prolog programming - path way to a solution - prolog

Question

I am studying prolog at university and facing some problems. What I already found out is just solution to a problem. However, I'm more interested in the way to think, i.e. how to get such solution.
Can somebody give me an advise on this field. I would really appreciate your help.
I give an example I am coping with and also, found a solution on stackoverflow here, but what I looking for is how does he do that, how does he find the answer :)
Write a predicate flatten(List,Flat) that flatten a list, e.g. flatten([a,b,[c,d],[[1,2]],foo],X) will give X=[a,b,c,d,1,2,foo].
This is the answer I found on stackoverflow:
flatten(List, Flattened):-
flatten(List, [], Flattened).
flatten([], Flattened, Flattened).
flatten([Item|Tail], L, Flattened):-
flatten(Item, L1, Flattened),
flatten(Tail, L, L1).
flatten(Item, Flattened, [Item|Flattened]):-
\+ is_list(Item).
this answer belongs to user gusbro and asked by user Parhs, I have try to find a way to contact user gusbro to ask him how he can derive such answer but I cannot.
Thank you very much.

Answer 1

Well, all I can say is that the way to solve a problem depends largely on the problem itself. There is a set of problems which are amenable to solve using recursion, where Prolog is well suited to solve them.
In this kind of problems, one can derive a solution to a larger problem by dividing it in two or more case classes.
In one class we have the "base cases", where we provide a solution to the problem when the input cannot be further divided into smaller cases.
The other class is the "recursive cases", where we split the input into parts, solve them separately, and then "join" the results to give a solution to this larger input.
In the example for flatten/2 we want to take as input a list of items where each item may also be a list, and the result shall be a list containing all the items from the input. Therefore we split the problem in its cases.
We will use an auxiliary argument to hold the intermediate flattened list, and thats the reason why we implement flatten/3.
Our flatten/2 predicate will therefore just call flatten/3 using an empty list as a starting intermediate flattened list:
flatten(List, Flattened):-
flatten(List, [], Flattened).
Now for the flatten/3 predicate, we have two base cases. The first one deals with an empty list. Note that we cannot further divide the problem when the input is an empty list. In this case we just take the intermediate flattened list as our result.
flatten([], Flattened, Flattened).
We now take the recursive step. This involves taking the input list and dividing the problem in two steps. The first step is to flatten the first item of this input list. The second step will be to recursively flatten the rest of it:
flatten([Item|Tail], L, Flattened):-
flatten(Item, L1, Flattened),
flatten(Tail, L, L1).
Ok, so the call to flatten(Item, L1, Flattened) flattens the first item but passes as intermediate list an unbound variable L1. This is just a trickery so that at the return of the predicate, the variable L1 still remain unbounded and Flattened will be of the form [...|L1] where ... are the flattened items of Item.
The next step, which calls flatten(Tail, L, L1) flattens the rest of the input list and the result is bounded with L1.
Our last clause is really another base case, the one that deals with single items (which are not lists). Therefore we have:
flatten(Item, Flattened, [Item|Flattened]):-
\+ is_list(Item).
which checks whether item is a list and when it is not a list it binds the result as a list with head=Item and as tail the intermediate flattened list.

Answer 2

First, I'll show you my approach to the problem, then I've got some resources for learning to think recursively.
Here's my solution to the problem "flatten a list of lists (of lists ...)". I've annotated it to show how I got there:
First, let's define the public interface to our solution. We define flatten/2. It's body consists of a call to the internal implementation flatten/3, which takes an accumulator, seeded as an empty list.
flatten ( X , R ) :-
flatten ( X , [] , R ) ,
.
That was easy.
The internal predicate flatten/3 is a little more complex, but not very.
First, we have the boundary condition: the empty list. That marks the end of what we need to do, so we unify the accumulator with the result:
flatten( [] , X , X ).
The next (and only) other case is a non-empty list. For this, we examine the head of the list. Our rule here is that it needs to flattened and appended to the result. A good rule of programming is to write descriptive code, and Prolog is itself a descriptive, rather than procedural, language: one describes the solution to the problem and lets the inference engine sort things out.
So...let's describe what needs to happen now, and punt on the mechanics of flattening the head of the list:
flatten( [X|Xs] , T , Y ) :-
flatten_head(X,X1) ,
append( T,X1,T1) ,
flatten( Xs , T1 , Y )
.
That, too, was easy.
That's the essence of the entire solution, right there. We've broken our problem into 3 pieces:
a special case (the empty list)
the normal case (a non-empty list)
what to do with each element in the list (not yet defined).
Let's move on to the implementation of how to flatten a single list element. That's easy, too. We've got two cases, here: the list item might be a list, or it might be something else.
First, the list element might be an unbound variable. We don't want untowards behaviour, like unbounded recursion happening, so let's take care of that straightaway, by disallowing unbound terms (for now). If the element is bound, we try to flatten it by invoking our public interface, flatten\2 again (oooooooh...more recursion!)
This accomplishes two things
First, it tells us whether we've got a list or not: flatten/2 fails if handed something other than a list.
Second, when it succeeds, the job of flatten_head/2 is done.
Here's the code:
flatten-head( X , Y ) :-
nonvar(X) ,
flatten( X , Y )
.
Finally, the last case we have to consider is the case of list elements that aren't lists (unbound vars, atoms or some other prolog term). These are already "flat"...all we need to do is wrap them as a single element list so that the caller (flatten\3) gets consistent semantics for its "return value":
flatten-head( X , [X] ).
Here's the complete code:
flatten ( X , R ) :-
flatten ( X , [] , R )
.
flatten( [] , X , X ) .
flatten( [X|Xs] , T , Y ) :-
flatten_head(X,X1) ,
append( T,X1,T1) ,
flatten( Xs , T1 , Y )
.
flatten-head( X , Y ) :-
nonvar(X) ,
flatten( X , Y )
.
flatten-head( X , [X] ) .
Each individual step is simple. It's identifying the pieces and weaving them together that's difficult (though sometimes, figuring out how to stop the recursion can be less than obvious).
Some Learning Resources
To understand recursion, you must first understand recursion—anonymous
Eric Roberts' Thinking Recursively (1986) is probably the best (only?) book specifically on developing a recursive point-of-view WRT developing software. There is an updated version Thinking Recursively With Java, 20th Anniversary Edition (2006), though I've not seen it.
Both books, of course, are available from the Usual Places: Powell's, Amazon, etc.
http://www.amazon.com/Thinking-Recursively-Eric-S-Roberts/dp/0471816523
http://www.amazon.com/Thinking-Recursively-Java-Eric-Roberts/dp/0471701467
http://www.powells.com/biblio/61-9780471816522-2
http://www.powells.com/biblio/72-9780471701460-0
You might also want to read Douglas Hofstadtler's classic Gödel, Escher, Bach: An Eternal Golden Braid Some consider it to be the best book ever written. YMMV.
Also available from the Usual Suspects:
http://www.powells.com/biblio/62-9780140289206-1
http://www.amazon.com/Godel-Escher-Bach-Eternal-Golden/dp/0465026567
A new book, though not directly about recursive theory, that might be useful, though I've not seen it (it's gotten good reviews) is Michael Corballis' The Recursive Mind:
The Origins of Human Language, Thought, and Civilization