What is a mathematical solution to this problem? (Project Euler #106) I've already asked this question before, but then I realized that wording was, unfortunately, quite confusing.
The statement of the problem is following:

Let $S(A)$ represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, $B$ and $C$, the following properties are true:
$1.$ $S(B) ≠ S(C$); that is, sums of subsets cannot be equal.
$2.$ If B contains more elements than $C$ then $S(B) > S(C)$.
For this problem we shall assume that a given set contains n strictly increasing elements and it already satisfies the second rule.
Surprisingly, out of the 25 possible subset pairs that can be obtained from a set for which $n = 4$, only 1 of these pairs need to be tested for equality (first rule). Similarly, when $n = 7$, only 70 out of the $966$ subset pairs need to be tested.
For $n = 12$, how many of the $261625$ subset pairs that can be obtained need to be tested for equality?

Problem statement specifies that, if size of subset $B$ doesn't equal to the size of the subset $C$, then theirs sums will not be equal by default. So when testing set $A$ for the equality, we only consider subsets with the same size.
The main question is, for arbitrary set $A$ with size $n$ which satisfies conditions specified in the problem, how many pairs of subsets with the same size are needed to be tested?
I couldn't have come up with purely mathematical solution myself, so I checked answers provided by the users in the discussion thread.
A lot of them mentioned so called "grid method", for example, one of the posts:

First, some observations. If elements of the set are assigned in ascending order to subset $B$, subset $C$, or discarded, and every element of $B$ can be paired with an element of $C$ that was selected later, then B's sum will be smaller than $C$'s, and the comparison will not be necessary.
If you imagine a walk on a grid from upper left to lower right, where selecting an element for subset $B$ is like walking East, and selecting an element for subset $C$ is like walking South, and selecting an element for neither subset is effectively the same as selecting it for both (East, then South), then a walk that crosses the diagonal from north to south (with this direction being the first diagonal crossing) corresponds exactly to a subset pair that must be compared.

And one more

Clearly we only need to test equal size groups $(k)$. If we select $2k$ elements, we only need to test some partition of that into $2k$-size groups if there is an $m$-smallest number in the group with the smallest element that is bigger than the $m$-smallest number in the other group. This can be modeled as a path across a $k×k$ grid that crosses the diagonal, so we can use Catalan numbers $C_k$
(which count paths that don't cross the diagonal) and half the total number of paths across the grid to get the number of diagonal-crossing paths (half = given start direction).Then the selection of the initial $2k$
set is a binomial coefficient, and sum across values of $k$.

Can some explain what is this "grid" they are referring to? And how do you solve the problem using this method?
 A: A grid is just this regular pattern of horizontal and vertical lines:

In red, this image also shows one of those paths: 
It consists of $9$ steps, hence is about $n=9$. It has $3$ east and $3$ south steps, so $|B|=|C|=3$ (where $|B|=|C|$ is equivalent to the path ending on the dotted diagonal; and as the path is not purely on the diagonal, $B,C$ are non-empty). The first step away form the diagonal is east, meaning that $B$ contains the smallest used element - we can assume this by symmetry. But at some point, the path goes below the diagonal - if this were not the case, the path would correspond to a case where trivially $S(B)<S(C)$.
(For reference, the path depicts $B=\{a_2,a_7,a_9\}\, C=\{a_3,a_5,a_6\}$).

Maybe put it in a different form:
Consider all strings of length $n$ that can be formed from "(", ")", and "-" such that at least one "(" and at least one ")" occur. This can encode the disjoint non-empty subsets $B$ and $C$ of the ordered set $\{a_1,a_2,\ldots, a_n\}$, namely we let $B$ be the set of all $a_i$ where our string has "(" in position $i$, and similarly $C$ for ")".
There are 
$$3^n-2^{n+1} +1$$
such strings (so for $n=4$: $50$ strings; to arrive at the $25$ from the problem statement, we can make use of the symmetry $B$ vs. $C$, which I'll do further down).
To begin with, we need only perform our test for cases with equal number of "(" and ")", for in all other cases condition 2 applies. By symmetry (i.e., because otherwise we can simply swap $B$ and $C$), we may assume without loss of generality that the first bracket of our string is an opening bracket. 
By these conditions, we would have to test
$$\tag1 \frac12\sum_{k=1}^{\lfloor n/2\rfloor}{n\choose 2k}{2k\choose k}=\frac12\sum_{k=1}^{\lfloor n/2\rfloor}{n\choose k}{n-k\choose k}=\frac12\sum_{k=1}^{\lfloor n/2\rfloor}\frac{n!}{(n-2k)!\,k!^2}$$
cases (so for $n=4$: $9$ strings).
Also, whenever the brackets are "properly nested" (for $n=4$ this means "(())", "(--)", "(-)-", "()()", "()--", "-(-)", "-()-", "--()"), we need not perform a test - because we can pair of each "(" with the corresponding ")" and thereby pair off all elements of $B$ with elements of $C$ such that the former are smaller than the latter each time and hence trivially $S(B)<S(C)$. How many tests do we get rid of this way? If it were note for the "-", this would be counted by the Catalan numbers $C_k=\frac1{k+1}{2k\choose k}$. Due to the interspersed "-"'s we remove
$$\tag2\sum_{k=1}^{\lfloor n/2\rfloor}{n\choose 2k}C_k =\sum_{k=1}^{\lfloor n/2\rfloor}\frac1{k+1}{n\choose 2k}{2k\choose k}=\sum_{k=1}^{\lfloor n/2\rfloor}\frac{n!}{(n-2k)!\,(k+1)!\,k!}$$
tests instead. By subtracting $(2)$ from $(1)$, we are left with
$$\tag3\sum_{k=1}^{\lfloor n/2\rfloor}\frac{(k-1)\,n!}{2\,(n-2k)!\,k!\,(k+1)!}.$$
However - can be be sure that no further reduction of tests is possible?
Yes, we can. 
Given an string of "(", ")", "-" as above, let $k_1>1$ be the position of the first ")" not matching a previous "(", and $k_2$ the position of a later "(". 
Set $a_1=1$, and then recursively $a_k=a_{k-1}+\alpha_k$ where $\alpha_k$ is an irrational number $\Bbb Q$-linearly independent of all previous $\alpha_i$ and such that $0<\alpha_k<\frac1{n^2}$  -except that for $x=k_1$ and for $k=k_2$ we set $a_k=a_{k-1}+X_1$ and $a_k=a_{k-1}+X_2$, respectively.
The condition that $S(B)=S(C)$ then becomes an equation of the form $$\tag4X_1+c_1=X_2+c_2$$
where $0\le c_1,c_2<\frac1n$. It is possible to find solutions to $(4)$ with $0<X_1,X_2<\frac1n$. We still have enough leeway to make the smaller of the two variables irrational and $\Bbb Q$-linearly independent from all $\alpha_i$ previously chosen. Then $(4)$ (i.e., the application of our test to $a_1,\ldots, a_n$) is up to rational multiples the only valid equation with rational coefficients among the $a_i$. Therefore no other test will  result in equality. Hence condition 1 holds for all choices of $B,C$ except the one choice corresponding to our test (or its negative, i.e., swapping $B$ and $C$). Moreover, as all $a_i$ are between $1$ and $1+\frac 1n$, it follows that $\lfloor S(X)\rfloor = |X|$ for all subsets, hence condition 2 also holds.
