Re: Boundary conditions

"I am looking for those items in Check that are not member from Control"

unique_member(Item, Check, Control) :-

% translating your words directly into Prolog:
member(Item, Check), \+ member(Item, Control).


To get all of them:
setof(Item, (unique_member(Item, Check, Control)), Items).

 Of course if you don't want  to use setof,bagof or findall then your approach is okay except for two things:
1) the last clause accidentally is called uniques_List.  (That's why it doesn't work.)
2) Clause 3 overlaps with clause 2, leaving a choice point.  This means your answer would be all of these:
U = [6, 5] ;
U = [5] ;
U = [6] ;
U = [] ;
false.

You probably only want the first.  Inserting a ! after the \+ member(H, Control) will solve this.


Alan Baljeu


________________________________
 From: Bengbers <[email protected]>
To: [email protected] 
Sent: Wednesday, November 13, 2013 12:46:24 PM
Subject: [SWIPL] Boundary conditions
 

I am a novice to Prolog so please excuse this maybe stupid question.

I have two lists:
Control = [1,2,3,4]
Check = [4,5,6]
and I am looking for those items in Check that are not member from Control
([5,6])

I have written the following code:

1 uniques_list( Control, Check, Unique) :- uniques_list( Control, Check, [],
Unique).

2 uniques_list( Control, [H|Tail], Accu, Unique) :-
    \+member( H, Control), Accu2=[H|Accu], uniques_list( Control, Tail,
Accu2, Unique).
3 uniques_list( Control, [_|Tail], Accu, Unique) :-
    uniques_list( Control, Tail, Accu, Unique).
4 uniques_List( Control, [], Unique, Unique). 

When calling uniques_list( [1,2,3,4], [ 4,5,6], X) and tracing the code, I
see at some moment that Accu contains [5,6] and that the tail is [].
I would expect that the flow continues to line 4 and that the call returns
but instead it continues with line 2.

My questions are:
- How can I fix this (and let X unify with [5,6])?
- What are the boundary conditions (why doesn't the flow jump to line 4)?

Ben 



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