Im studying prolog in a course. For an exercise, I need in part to generate all subtrees of a given tree. The problem is it generates false subtrees
Here's the code
build_tree3(T):-
T=t(t(t(nil, -5, nil), 4, t(t(nil, 15, nil), -20, t(nil, 10, nil))), 2, t(nil, -8, t(t(nil, 9, nil),12,t(nil, 10, nil)))).
get_sol(Tree, List, N):-
without_root(Tree, List1, N1),
with_root(Tree, List2, N2),!,
max_set(List1, N1, List2, N2, List, N).
max_set(List1, N1, List2, N2, List, N) :-
(N1>N2,List=List1,N=N1;
N2>N1,List=List2,N=N2;
N2=:=N1,N=N1,(List=List1;List=List2)).
without_root(nil, _, 0).
without_root(t(L, _, R), List, N):-
get_sol(L, LeftList, LeftN),
get_sol(R, RightList, RightN),
append(LeftList, RightList, List),
N is LeftN + RightN.
with_root(nil, [], 0).
with_root(t(L, X, R), [X|List], N):-
with_root(L, LeftList, LeftN),
with_root(R, RightList, RightN),
append(LeftList, RightList, List),
N is LeftN + RightN + X.
Here's the console
build_tree3(T), get_sol(T, L, N).
The outcome of it is L = [15,10,12,9,10] , N = 56 ; when it is supposed to be L = [12,9,10] , N = 31;
Your solution returns L=[15,10,12,9,10] because it finds maximum independent set , it is not forced to return a subtree. You could change some parts as the code below:
sol_tree(Tree,List,N):-
sol_tree_noroot(Tree,L1,N1),
sol_tree_withroot(Tree,L2,N2),!,
max_set(L1,N1,L2,N2,List,N).
max_set(List1, N1, List2, N2, List, N) :-
(N1>N2,List=List1,N=N1;
N2>N1,List=List2,N=N2;
N2=:=N1,N=N1,(List=List1;List=List2)).
sol_tree_noroot(nil,[],0).
sol_tree_noroot(t(L,_,R),List,N):-
sol_tree(L,L1,N1),sol_tree(R,R1,N2),!,
max_set(L1, N1, R1, N2, List, N).
sol_tree_withroot(nil,[],0).
sol_tree_withroot(t(L,X,R),[X|List],N3):-
sol_tree_withroot(L,L1,N1),sol_tree_withroot(R,R1,N2),
max_set2(L1,N1,R1,N2,List,N),
N3 is N+X.
max_set2(L1,N1,L2,N2,List,N):-
(N1>0,N2>0,N is N1+N2,append(L1,L2,List);
N1>=0,N2<0,N is N1 ,List=L1;
N1<0,N2>=0,N is N2 ,List=L2;
N1<0,N2<0,N1<N2,N is N2 ,List=L2;
N1<0,N2<0,N1>N2,N is N1 ,List=L1;
N1>0,N2=0,N is N1,(List=L1;append(L1,L2,List));
N1=0,N2>0,N is N2,(List=L2;append(L1,L2,List));
N1=0,N2=0,N is N1,(List=L1;List=L2;append(L1,L2,List))).
The idea is that you can skip the root to find a subtree or you can keep the root where in this case you can skip a root because then it will not be a subtree.