dcg

We have a relatively simple assignment that I understand in theory but I think I just don't quite understand Prolog's syntax enough to get that into code. Basically, we have a list of English notations that represent operations in C. They're stored as a list when they're passed to our Prolog program. For example: add 4 to 3 is [add, 4, to, 3] We need to write a function that takes that list an returns the equivalent. So if I called english2C([add,4,to,3], C). C = 4+3 It would bind C to the result. So the data structure itself would be something like +(4(3)). We have a list of such English

Possible Duplicate: Read a file line by line in Prolog I found the following prolog code which reads one character at a time and prints out. process(File) :- open('C:/Users/BHARAT/Desktop/a.txt', read, In), get_char(In, Char1), process_stream(Char1, In), close(In). process_stream(end_of_file, _) :- !. process_stream(Char, In) :- print(Char), get_char(In, Char2), process_stream(Char2, In). But if the file has multiple lines is there a way to read 1 whole line at a time so that it will be easy for tokenizing. You say you want to tokenize the input - the best way to do this are definite clause

As a follow up to this question which poses the problem Return count of items in a list but if two identical items are next to each other then don't increment the count. This code is the closest I came to solving this with DCG and semicontext. lookahead(C),[C] --> [C]. % empty list % No lookahead needed because last item in list. count_dcg(N,N) --> []. % single item in list % No lookahead needed because only one in list. count_dcg(N0,N) --> [_], \+ [_], { N is N0 + 1 }. % Lookahead needed because two items in list and % only want to remove first item. count_dcg(N0,N) --> [C1], lookahead(C2), {

This is a follow-on question from an earlier question from Mat's answer Starting with this e([number(0)] , t1 , Uc0 , Uc0, Bc0 , Bc0) --> []. e([number(1)] , t2 , Uc0 , Uc0, Bc0 , Bc0) --> []. e([number(2)] , t3 , Uc0 , Uc0, Bc0 , Bc0) --> []. e([op(neg),[Arg]] , u1(E) , [_|Uc0], Uc1, Bc0 , Bc1) --> [_], e(Arg , E , Uc0, Uc1, Bc0, Bc1). e([op(ln),[Arg]] , u2(E) , [_|Uc0], Uc1, Bc0 , Bc1) --> [_], e(Arg , E , Uc0, Uc1, Bc0, Bc1). e([op(add),[Left,Right]], b1(E0,E1) , Uc0 , Uc2, [_|Bc0], Bc2) --> [_,_], e(Left, E0, Uc0, Uc1, Bc0, Bc1), e(Right, E1, Uc1, Uc2, Bc1, Bc2). e([op(sub),[Left,Right]],

We well know inorder implementation for BST tree. inorder(nil, []). inorder(t(Root, L, R), List) :- inorder(L, ListLeft), inorder(R, ListRight), append(ListLeft, [Root|ListRight], List). However, is it possible to do it for list ? I mean reconstruct all possible BST trees, for example: inorder(X, [1,2,3]). X = t(1, nil, t(2, nil, t(3, nil, nil)))); X = t(3, t(2, t(1, nil, nil), nil), nil), nil); X = t(2, t(1, nil, nil), t(3, nil, nil)); false. It seems to be impossible for me. First, let us use a Definite Clause Grammar ( dcg ) for relating trees to lists: inorder(nil) --> []. inorder(t(Root,