GNU Prolog - Build up a list in a loop

Question

I need to build a new list with a "loop". Basically i can't use recursion explicitly, so i am using append to go through lists of list.

I can get the element. Problem is i need to check this element and if something is true it returns another element i need to put back into the list. It does check correctly and it changes correctly.

Problem i am having is how do i create a completely new list.

So, if i had

[[1,1,1],[2,6,2],[3,3,3]]

I go through each element. say i get to the 6 and it changes. So i need to create a new list like so,

[[1,1,1],[2,10,2],[3,3,3]].

Right now my main problem is just creating each row. If i can create each row, i will be able to create a list of lists.

So to break this down a little more, lets just worry about [1,1,1].

I go through each element while appending the new element to a newlist. the new list is now [1,1,1]

I have this:

set(Row,Col,Bin,TheEntry,Bout) :- 
 append(ListLeft, [R|_], Bin),
  append(ListLeft2, [C|_], R),
   length(ListLeft, LenR),
   length(ListLeft2,LenC),
   CurrRow is LenR + 1,
   CurrCol is LenC + 1,
   getChar(C, Row, Col, CurrRow, CurrCol,TheEntry, NewC),
            appendhere?.

I need to create a new list there with the character returned from NewC. Not sure how to do this.

Any clues?

Thanks.

Answer 1

If you cannot use recursion and have to do it with backtracking you should do something like this: Assume Bin is a list of lists (each item is a full row) ~ Split input Bin in three parts (a list of 'left' rows, a Row, and a list of remaining rows). This can be done using append/3 with something like append(Left, [Item|Rest], Rows) ~ Now obtain the length of the 'left' rows ~ Test the length using 'is' operator to check wether the left list has Row - 1 items ~ Do the same but now with the Item, i.e. split it in three parts (LeftColums, ColumItem and Rest) ~ Test now the length against the required Column ~ Now you have the Item to change so all you need to do is rebuild a list using two appends (one to rebuild the chosen row and another to rebuild the output list).

So from your code you wouldn't use unnamed variables (_). Instead of that you have to use a named variable to be able to rebuild the new list with the item changed.

Answer 2

To give you an idea about how to use append/3 to extract an item from a list of lists, consider the following predicate called replace/2:

replace(In, Out) :-
    append(LL, [L|RL], In), 
    append(LE, [E|RE], L),
    replaceElement(E, NewE), !,
    append(LE, [NewE|RE], NewL),
    append(LL, [NewL|RL], Out).
replace(In, In).

This non-recursive predicate takes, as Input, a list of lists, and backtracks to find an element E within an inner list L that can be replaced via replaceElement/2; if so, it is replaced by constructing the inner list first (NewL), then uses this new list in the construction of the new outer list (Out), as the result.

Note that this simply serves to demonstrate how to use append/3 to break apart a list of lists to retrieve individual elements as you need via backtracking, and not recursion, as requested. Once an element E is found to be replaceable by NewE via replaceElement/3, it is used in the construction of the list again using append/3 as shown.

Also note that this suggestion (which is intended to help you, not be your final answer) also happens to replace only a single element within an inner list, if any at all. If you want to do multiple replacements of the input list in a single call to replace/2 or similar using this technique, then you will almost certainly need a recursive definition, or the ability to use the global database via assert. I'm happy to be corrected if someone else can provide a definition as a counterexample.

With this example predicate replace/2, together with, say, the following fact:

replaceElement(6, 10).

Executing the following gives us your required behaviour:

1 ?- replace([[1,1,1],[2,6,2],[3,3,3]], Out).
Out = [[1, 1, 1], [2, 10, 2], [3, 3, 3]] ;
false.

If you cannot use cut (!), it is fine to omit it, but note that the second clause replace(In, In) will cause all calls to replace/2 to backtrack at least once to give you the input list back. If this behaviour is undesirable, omitting this second clause will cause replace/2 to fail outright if there is no replacement to be made.