Has My Pie Been Bisected?
J, 18 bytes

5 bytes thanks to Dennis.

+/e.[:,/2*+/\-/+/\


@HelkaHomba: Nope.

Usage

&gt;&gt; f =: +/e.[:,/2*+/\-/+/\
&gt;&gt; f 6 6 12 12 12 11 1 12
&lt;&lt; 4
&gt;&gt; f 10 20 10 1
&lt;&lt; 0


Ungolfed

black_magic  =: +/\-/+/\
doubled_bm   =: 2 * black_magic
flatten      =: ,/
sum          =: +/
is_member_of =: e.
f =: sum is_member_of monadic flatten doubled_bm




Previous 23-byte version:

[:+/[:+/+/=+:@-/~@(+/\)


Usage

&gt;&gt; f =: [:+/[:+/+/=+:@-/~@(+/\)
&gt;&gt; f 6 6 12 12 12 11 1 12
&lt;&lt; 4
&gt;&gt; f 10 20 10 1
&lt;&lt; 0


Ungolfed

black_magic =: -/~@(+/\)
double      =: +:
equals      =: =
sum         =: +/
monadic     =: [:
of          =: @
f =: monadic sum monadic sum (sum equals double of black_magic)




Explanation

The sum of all substrings is calculated by the black_magic. The +/\ calculate the partial sums.

For example, a b c d becomes a a+b a+b+c a+b+c+d.

The -/~ then constructs a subtraction table based on the input, so x y z becomes:

x-x x-y x-z
y-x y-y y-z
z-x z-y z-z


When applied to a a+b a+b+c a+b+c+d, the result would be:

    0  -b -b-c -b-c-d
    b   0   -c   -c-d
  b+c   c    0     -d
b+c+d c+d    d      0


This calculated the sums of all the substrings which does not contain a.

This guarantees to be enough, since if one bisection contains a, the other bisection will not contain a and also will not wrap around.