Prolog finding target number in a list by applying operations (+,-,/,*) - prolog

Question

The task is to find if there exists at least one combination of numbers from given list and operators to obtain target number. No usemodule allowed
numbers([3,4,1,2], 7) → true. (cause 4 + 3 = 7)
numbers([1,7,7,3], 24) → true. (cause (7 - 3) * (7 - 1) = 24)
Tried member if target is already in the list. But lost further

Answer 1

Building upon https://professor-fish.blogspot.com/2009/11/countdown-with-prolog.html
solve_countdown(Ns, SumWanted, TsUniq) :-
findall(T, (
any_comb(Ns, Sub),
num_combs(Sub, T),
SumWanted is T
), Ts),
sort(Ts, TsUniq).
any_comb(_, []).
any_comb([H|T], [E|Comb]) :-
select(E, [H|T], Lst0),
any_comb(Lst0, Comb).
num_combs([N], N).
num_combs(As, T) :-
split_list_in_2(As, As1, As2),
num_combs(As1, T1),
num_combs(As2, T2),
% Break symmetry, since 5+2 is same as 2+5
( T1 #=< T2,
( T = T1 + T2
; T1 > 1, T = T1 * T2
)
; T = T1 - T2
; T = T1 / T2
),
R is T,
integer(R),
R #> 0.
split_list_in_2([H1, H2|T], [H1|Start], Remainder) :-
split_list_in_2_(T, H2, Start, Remainder).
split_list_in_2_(L, H2, [], [H2|L]).
split_list_in_2_([H|T], H2, [H2|Start], Remainder) :-
split_list_in_2_(T, H, Start, Remainder).
Result in swi-prolog:
?- time(solve_countdown([1,7,7,3], 24, Ts)).
% 10,996 inferences, 0.002 CPU in 0.002 seconds (99% CPU, 5236642 Lips)
Ts = [3*(1+7), (7-1)*(7-3)].
select/3 code is here.

Answer 2

Squeezing the lemon. I get a slightly faster version.
The logic replicates this paper:
https://www.cs.nott.ac.uk/~pszgmh/countdown.pdf
But adds forward checking:
% solve(+Integer, -Term, +Integer, +List, -List)
solve(1, N, N, P, Q) :- !, select(N, P, Q).
solve(K, G, N, P, Q) :-
J is K-1,
between(1, J, I),
L is K-I,
solve2(I, E, A, P, H),
forward(E, A, F, B, G, N),
solve(L, F, B, H, Q).
forward(E, A, F, B, E+F, N) :- N > A, B is N-A, A =< B.
forward(E, A, F, B, E-F, N) :- A > N, B is A-N.
forward(E, A, F, B, E*F, N) :- N mod A =:= 0, B is N div A, A =< B, A =\= 1.
forward(E, A, F, B, E/F, N) :- A mod N =:= 0, B is A div N, B =\= 1.
% solve2(+Integer, -Term, -Integer, +List, -List)
solve2(1, N, N, P, Q) :- !, select(N, P, Q).
solve2(K, G, N, P, Q) :-
J is K-1,
between(1, J, I),
L is K-I,
solve2(I, E, A, P, H),
solve2(L, F, B, H, Q),
combine(E, A, F, B, G, N).
combine(E, A, F, B, E+F, N) :- A =< B, N is A+B.
combine(E, A, F, B, E-F, N) :- A > B, N is A-B.
combine(E, A, F, B, E*F, N) :- A =< B, A =\= 1, N is A*B.
combine(E, A, F, B, E/F, N) :- B =\= 1, A mod B =:= 0, N is A div B.
Example run with SWI-Prolog 9.1.4:
% time((between(1,6,N), solve(N, E, 999, [1,3,5,10,25,50], _), fail; true)).
% % 2,618,953 inferences, 0.234 CPU in 0.242 seconds (97% CPU, 11174199 Lips)
% true.