I want to implement a very simple automaton that restrict the number of consecutive 1s in a list of ones and zeros (e.g. [0,1,1,0,1,1,1]).
My automaton looks like this:
% 'Day' is a list of clpfd variables
% 'Allowed' is an integer
%
% consecutiveOnes(+Day, +Allowed)
consecutiveOnes(Day, Allowed) :-
automaton(Day, _, Day,
[source(n)],
[
arc(n, 0, n, [0] ),
arc(n, 1, n, [C+1])
],
[C],
[0],
[_N]
).
% example 1:
% consecutiveOnes([0,0,0,1,1,1], 2) -> there are three consecutive 1s and we allow only 2 -> Fail.
% example 2:
% consecutiveOnes([0,1,1,1,0,0], 2) -> there are three consecutive 1s and we allow only 2 -> Fail.
% example 3:
% consecutiveOnes([0,1,1,0,0,0], 2) -> there are only two consecutive 1s and we allow 2 -> OK
How can I add the constraint for counter C specifying C <= Allowed to the Prolog code above?
It may be best to formulate this with additional states. For example, for at most two consecutive 1s:
:- use_module(library(clpfd)).
at_most_two_consecutive_ones(Day) :-
automaton(Day,
[source(n),sink(n),sink(n1),sink(n2)],
[arc(n, 0, n),
arc(n, 1, n1),
arc(n1, 1, n2),
arc(n1, 0, n),
arc(n2, 1, false),
arc(n2, 0, n)
]).
Example queries:
?- at_most_two_consecutive_ones([0,0,0,1,1,1]).
false.
?- at_most_two_consecutive_ones([0,1,1,0,1,1]).
true.
?- at_most_two_consecutive_ones([0,1,1,0,1,0]).
true.
For a more general solution, you have to build an automaton on demand when given the maximal length of a run.