Prolog Permutations

Question

I'm trying to get more into learning prolog as I'll be taking an AI class at school next semester. I've been able to get down the basics down and can do relation based stuff, however, I've been trying to learn permutations and combinatronics and they seem pretty straightforward, but it led me to a question that I can't figure out how to solve. Say I wanted to know the permutations of 1's and 0's with a certain condition that there must be atleast 4 1's in a row. I have no idea where I would start to try and find a solution for this, but in the end I want the code to do something like this:

?- placeOnesAndZeros(9,X).
% where 9 is the length of the list/array and X is the permutations

[0,0,0,0,0,0,0,0,0]
[1,1,1,1,0,0,0,0,0]
[0,1,1,1,1,0,0,0,0]
[0,0,1,1,1,1,0,0,0]
[0,0,0,1,1,1,1,0,0]
[0,0,0,0,1,1,1,1,0]
[0,0,0,0,0,1,1,1,1]
[1,1,1,1,0,1,1,1,1]
[1,1,1,1,1,0,0,0,0]
[0,1,1,1,1,1,0,0,0]
[0,0,1,1,1,1,1,0,0]
[0,0,0,1,1,1,1,1,0]
[0,0,0,0,1,1,1,1,1]
[1,1,1,1,1,1,0,0,0]
[0,1,1,1,1,1,1,0,0]
[0,0,1,1,1,1,1,1,0]
[0,0,0,1,1,1,1,1,1]
[1,1,1,1,1,1,1,0,0]
[0,1,1,1,1,1,1,1,0]
[0,0,1,1,1,1,1,1,1]
[1,1,1,1,1,1,1,1,0]
[0,1,1,1,1,1,1,1,1]
[1,1,1,1,1,1,1,1,1]

Thank you in advance!

EDIT CODE:

printList([ ]).
printList([H|T]) :- print(H), nl, printList(T).

eval([],_).
eval([H|T],[1,0]):-member(H,[1,0]),eval(T,[1,0]).

placeOnesAndZeros(N, L):-length(L,N), eval(L,[1,0]).

Answer 1

Generate and test it's the name of the basic technique used to search a solution space. In Prolog, it's practically built in. Just provide a filter discarding what is not required:

?- placeOnesAndZeros(9,L),once(append(_,[1,1,1,1|_],L)).

once/1 is required, otherwise append/3 could succeed multiple times. To check the correctness of the approach, here is how to count how many solutions we get:

?- aggregate(count,L^H^T^(placeOnesAndZeros(9,L),once(append(H,[1,1,1,1|T],L))),C).
C = 111.

The quantification on variables L,H,T (these last 2 being introduced only to aggregate) can be avoided using aggregate_all :

?- aggregate_all(count,(placeOnesAndZeros(9,L),once(append(_,[1,1,1,1|_],L))),C).
C = 111.

edit

As @lurker noted, my filter isn't correct. Try instead

atLeastFourOnes(L) :- memberchk(1,L), atLeastFourOnes_(L).

atLeastFourOnes_([]).
atLeastFourOnes_([0|L]) :- atLeastFourOnes_(L).
atLeastFourOnes_([1,1,1,1|L]) :- stripOnes(L,R), atLeastFourOnes_(R).

that yields

?- placeOnesAndZeros(9,L),atLeastFourOnes(L).
L = [1, 1, 1, 1, 1, 1, 1, 1, 1] ;
L = [1, 1, 1, 1, 1, 1, 1, 1, 0] ;
L = [1, 1, 1, 1, 1, 1, 1, 0, 0] ;
...

?- aggregate(count,L^(placeOnesAndZeros(9,L),atLeastFourOnes(L)),C).
C = 22.