Termination criteria
Terminate when a specified position is reached
The pattern for this is:
% 1 terminating condition
rule(1, Elem, [Elem|_]).
% 2 recursive
rule(Cnt, Elem, [_|Tail]) :-
Cnt1 is Cnt - 1,
rule(Cnt1, Elem, Tail).
Now, this is only a pattern: the procedures we'll see in this section will all have
this pattern as part of them, although there may be additions.
position/3
position/3 is really just the basic pattern with no additions. The idea is that we
know we have got to a particular position when the counter has been reduced to '1':
% 1 terminating condition
position(1, Elem, [Elem|_]).
% 2 recursive
position(Cnt, Elem, [_|Tail]) :-
Cnt1 is Cnt - 1,
position(Cnt1, Elem, Tail).
Try the following goals:
| ?- position(0, Elem, [a,b,c]).
| ?- position(1, Elem, [a,b,c]).
| ?- position(2, Elem, [a,b,c]).
| ?- position(3, Elem, [a,b,c]).
| ?- position(4, Elem, [a,b,c]).
delete_nth/3
This procedure is true if one list is the same as a second list, except that
the nth element is deleted. As with count_elem/2, values to be
passed back (such as the second list with the element deleted)
have to be included in the argument list.
% 1 terminating condition
delete_nth(1, [Elem|Tail], Tail).
% 2 recursive
delete_nth(Cnt, [Head|Tail1], [Head|Tail2]) :-
Cnt1 is Cnt - 1,
delete_nth(Cnt1, Tail1, Tail2).
Notice in the recursive clause how the Head of the first list is
unified with the Head of the second list. You could use the
Prolog tracer to follow how the variables are instantiated or add your
own write/1 clauses to display variables during recursion.
Try the following goals:
| ?- delete_nth(0,[a,b,c], List).
| ?- delete_nth(1, [a,b,c], List).
| ?- delete_nth(2, [a,b,c], List).
| ?- delete_nth(3, [a,b,c], List).
| ?- delete_nth(4, [a,b,c], List).
Take time to work through Self-Test 6.
These Pages are maintained by
Dr Peter Hancox
Last updated October 1998