Segregating Lists in Prolog

Question

I am having a really hard time understanding how to get my code to show my segregated lists consisting of even and odd numbers. I am not even sure what my understanding is lacking. I am new to this language obviously and must use it for school. My imperative and functional mind won't let me know what the hell is going on with this lol.

Now, no I am not asking you to do my homework! I am simply asking you to help me see what my lack of understanding is. I have also looked up similar answers but I cannot convert them to the way I am supposed to write this function.

Please, once more, do not bash me for this like I have previously usually been bashed. Please just help me see what my understanding is lacking. Do not just give me answers and code snippets without explaining it please.

Here it is:

is_even(H) :-
   0 is mod(H, 2).

segregate(List, Even, Odd) :- segregator(List, Even, Odd).

segregator([], [], []).
segregator([H|T], E, O) :-
    is_even(H),
    % I feel here is where I am supposed to build the list, 
    % but I have no clue how since Even or Odd has not been unified.
    segregator(T, E, O),
    write('Even is '), write(E), nl.
segregator([H|T], E, O) :-
    % Same here as above.
    segregator(T, E, O),
    write('Odd is '), write(O), nl.

Answer 1

A logically pure implementation is very straight-forward, thanks to clpfd:

:- use_module(library(clpfd)).

list_evens_odds([],[],[]).
list_evens_odds([X|Xs],[X|Es],Os) :-
   X mod 2 #= 0,
   list_evens_odds(Xs,Es,Os).
list_evens_odds([X|Xs],Es,[X|Os]) :-
   X mod 2 #= 1,
   list_evens_odds(Xs,Es,Os).

Some sample queries we expect to succeed (with a finite sequence of answers):

?- Xs = [1,2,3,4,5,6,7], list_evens_odds(Xs,Es,Os).
Xs = [1,2,3,4,5,6,7],
Es = [  2,  4,  6  ],
Os = [1,  3,  5,  7] ;
false.

?- list_evens_odds(Ls,[2,4],[1,3]).
Ls = [2,4,1,3] ? ;
Ls = [2,1,4,3] ? ;
Ls = [2,1,3,4] ? ;
Ls = [1,2,4,3] ? ;
Ls = [1,2,3,4] ? ;
Ls = [1,3,2,4] ? ;
no

What about queries we expect to fail?

?- list_evens_odds(Ls,[2,4,5],[1,3]).
no
?- list_evens_odds(Ls,[2,4],[1,3,6]).
no
?- list_evens_odds([_,_,_],[2,4],[1,3]).
no

At last, the most general query:

?- assert(clpfd:full_answer).
yes

?- list_evens_odds(Ls,Es,Os).
Ls = [],   Es = [],   Os = []                              ? ;
Ls = [_A], Es = [_A], Os = [], _A mod 2#=0, _A in inf..sup ? ...

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Edit 2015-05-06

Here's another way to do it with logical-purity!

Use the meta-predicate tpartition/4 together with zeven_t/2 or zodd_t/2.

bool01_t(1,true).
bool01_t(0,false).

zeven_t(Z,Truth) :- Z mod 2 #= 0 #<==> B, bool01_t(B,Truth).

%zodd_t(Z,Truth) :- Z mod 2 #= 1 #<==> B, bool01_t(B,Truth).
zodd_t(Z,Truth)  :- Z mod 2 #=         B, bool01_t(B,Truth). % tweaked

zeven_t/2 reifies the evenness of an integer, zodd_t/2 the oddness.

With everything in place, let's run some queries!

?- tpartition(zeven_t,[1,2,3,4,5,6,7],Es,Os).
Es = [2,4,6], Os = [1,3,5,7].
?- tpartition(zodd_t ,[1,2,3,4,5,6,7],Os,Es). % argument order differs
Es = [2,4,6], Os = [1,3,5,7].

Both succeed deterministically. The equivalent query using list_evens_odds/3 does not.