Split list into 2 lists of equal sum erlang - erlang

Question

I want to know how to split a given list into two lists such that both lists have the same sum. I want to do that by using concurrency. I am doing this in erlang.
So, I'm doing something like this:
Read the list, if its sum is even, then proceed else fail. Take the first element of the list and check if it is greater than half of the sum, if not, then I add this element to a new list. Next, I take the second element of the list, check the sum of this element and that of the new list and do the same operation. And so on.. Such that when the sum in the new list is equal to half of the sum of the first list, it calls another function to send the remaining elements.
-module(piles_hw).
-compile(export_all).
start([]) -> 0;
start(List) ->
Total = lists:foldl(fun(X, Sum)-> X+Sum end,0,List),
if (Total rem 2) == 0 ->
Total/2,
copy_to_list_one([],List,start(List));
true ->
func_fail()
end.
copy_to_list_one(L1,[H|T],X)->
Y =lists:sum(L1)+H,
if Y<X ->
copy_to_list_one(lists:append(L1,[H]),lists:delete(H,[H|T]),X);
Y==X ->
take(lists:append(L1,[H]));
Y>X ->
copy_to_list_one(L1,lists:delete(H,[H|T]),X)
end;
copy_to_list_one(L1,[],X)->
copy_func_two([1,2,3,4,19,20,28,14,11],X).
copy_func_two([H|T],X)->
copy_to_list_one([],lists:append(T,[H]),X).
take(L3)->
io:format("~w",[L3]).
func_fail() ->
io:format("~n fail ~n").
But, in this way I go into an infinite loop sometimes. Could somebody help?

Answer 1

Edit:
Pascal was entirely correct: there is no algorithm (at least not that I could come up with) that can solve certain sets by running down the list one item at a time. (In particular when half the sum of the list equals X * N where X is present in the list N times.) I initially put a flawed algorithm here.
That got me excited in the nerdiest of ways, so here is an exhaustive algorithm involving the pairs of [{P, (List - P)} || P <- powerset(List)].
There are some lists:usort/1 shenanigans in there that I didn't clean up to uniquify the list prior to the final comparison (otherwise you get duplicate similar pairs, which is ugly). Anyway, ugly, but now correct:
comblit(List) ->
Power = powerset(List),
Lists = lists:usort([lists:sort([Z, lists:subtract(List, Z)]) || Z <- Power]),
Pairs = lists:map(fun([H|[B|[]]]) -> {H, B} end, Lists),
[{Z, X} || {Z, X} <- Pairs, lists:sum(Z) == lists:sum(X)].
powerset([H|T]) ->
Part = powerset(T),
powerset(Part, H, Part);
powerset([]) -> [[]].
powerset(A, Part, [H|T]) ->
powerset([[Part|H]|A], Part, T);
powerset(A, _, []) -> A.
This is still not a concurrent solution, but the path to making it concurrent is a lot more obvious now.
Thanks for pointing that out, Pascal. That was sort of fun.

Answer 2

I have this solution that is not concurrent:
-module(split).
-export([split/1,t_ok/0,t_too_long/0,t_fail/0,t_crash/0]).
%% [EDIT]
%% Don't use this code, it fails with negative integers!
% Exported
%% take a list and split it in 2 list which sum are equals
split(L=[_|_]) ->
T2 = lists:sum(L),
{ok, TRef} = timer:send_after(20000,too_long),
R = case T2 rem 2 of
1 -> {error,fail};
0 -> split(tl(L),[hd(L)],[],T2 div 2,hd(L),0)
end,
timer:cancel(TRef),
R.
% test
t_ok() -> split([1,2,3,4,5,6,7]).
t_too_long() -> split(lists:seq(1,3+4*100000)).
t_fail() -> split([2,4,6,10000,8,6]).
t_crash() -> split([]).
% private
split([H|Q],A,B,T,Asf,_Bsf) when H + Asf == T -> {ok,{[H|A],B ++ Q}};
split([H|Q],A,B,T,_Asf,Bsf) when H + Bsf == T -> {ok,{A ++ Q,[H|B]}};
split([H|Q],A,B,T,Asf,Bsf) when H + Asf > T, H + Bsf < T -> c_split(Q,A,[H|B],T,Asf,Bsf+H);
split([H|Q],A,B,T,Asf,Bsf) when H + Asf < T, H + Bsf > T -> c_split(Q,[H|A],B,T,Asf+H,Bsf);
split([H|Q],A,B,T,Asf,Bsf) when H + Asf < T, H + Bsf < T ->
case c_split(Q,A,[H|B],T,Asf,Bsf+H) of
{error,fail} -> c_split(Q,[H|A],B,T,Asf+H,Bsf);
R -> R
end;
split([],A,B,_T,_T,_T)-> {ok,{A,B}};
split(_,_,_,_,_,_) -> {error,fail}.
c_split(L,A,B,T,Asf,Bsf) ->
receive
too_long -> {error,too_long}
after 0 ->
split(L,A,B,T,Asf,Bsf)
end.
To turn it concurrent, you could replace the line 0 -> split(tl(L),[hd(L)],[],T2 div 2,hd(L),0) by a call to a function which spawn_link several processes (as much as there are core available) which start the split/6 function with different initial conditions. The split/6 must have a 7th parameter: the Pid of the main process where it will send back its answer. The main process wait for answers and stop
if a solution is found
if all processes fail to find one
if the time out occurs
I have edited the code following #Odobenus remark (but it still fail on [] -> {ok,[],[]} :o), and I also made a concurrent version. The funny thing is that for this kind of problem, and with the input list I use (a lists:seq) there are so many solution that any start sequence I choose can give a solution, so the concurrent version is slower.