问题
Let's say you have a list in Prolog such as: [3,4,2,2,1,4]. How would one go about generating a list of lists of all possible patterns that start at the first element of the list, then either go to the i + 2th element, or the i + 3rd element, and so on from there.
Example: Say I have [3,4,2,2,1,4,8]. I want to be able to generate a list of lists such as: [[3,2,1,8], [3,2,4], [3,2,8]]
I.e. all possibilities of either every other element or every i+3 element, or any other combination, such as i+2,i+3,i+2,i+2, etc.
I've implemented my own version of a powerset, but I can't seem to figure out where to start.
回答1:
gen([], []).
gen([A], [A]).
gen([A, _ | T], [A | Xs]) :- gen(T, Xs).
gen([A, _, _ | T], [A | Xs]) :- gen(T, Xs).
results in
?- gen([3,4,2,2,1,4,8], X).
X = [3, 2, 1, 8] ;
X = [3, 2, 1] ;
X = [3, 2, 4] ;
X = [3, 2, 4] ;
X = [3, 2, 8] ;
false.
You can use findall/3 to get all results
?- findall(X, gen([3,4,2,2,1,4,8], X), Z).
Z = [[3, 2, 1, 8], [3, 2, 1], [3, 2, 4], [3, 2, 4], [3, 2, 8]].
来源：https://stackoverflow.com/questions/36579234/prolog-generating-every-possibility-of-a-list-given-a-pattern