I wrote a computer program simply trying to:
1- Create all possible matrices which represents our $a$, $b$, $c$ values in a summation by $1$ or $0$ if they exist in the sum, such as below:
$\begin{bmatrix}
0 &0 &0 &1 &1 &0 &1 \\
0 &0 &0 &0 &0 &0 &1 \\
1 &0 &0 &1 &0 &0 &1 \\
0 &0 &0 &0 &0 &1 &1 \\
0 &1 &1 &0 &1 &1 &0 \\
1 &0 &0 &1 &1 &0 &0 \\
1 &0 &0 &1 &1 &0 &0
\end{bmatrix}\begin{bmatrix}
a\\
b\\
c\\
d\\
e\\
f\\
g
\end{bmatrix}= \begin{bmatrix}
S_1\\
S_2\\
S_3\\
S_4\\
S_5\\
S_6\\
S_7
\end{bmatrix}$
So I represent everything as a matrix. In my program, It tries to create all possible combinations by using $1$ and $0$ and the program makes sure that every column has at least one $1$ and none of the rows are the same.
2- Secondly, I choose two values for our values ($a$,$b$,$c$,... etc.) and find our $S$ values for each possible two values. so there are $2^7$ possible values you can choose for your chosen matrix above. and every values I find $S$ sums. Let's say I choose $2$,$2$,$4$,$2$... etc
3- By using these $S$ values and the original matrix, I choose another two values which are different than our $a$, $b$, $c$,... (let say $3$,$1$,$1$,$3$...), and find other $S_t$ values.
4- Lastly, In part 3 if I encounter $S_t$ values which are the same $S$ values for each sum in part 2, I conclude that that matrix is not the matrix I need to use to find $a$,$b$,$c$, ... values since there could be other possible values satisfies the same conditions/equations.
If you are looking for 7 value and 5 equations, one of the possible answer becomes:
$\begin{bmatrix}
1 &1 &0 &1 &0 &0 &0\\
1 &1 &1 &0 &0 &0 &1\\
0 &1 &1 &1 &0 &1 &0\\
1 &0 &0 &1 &1 &1 &1\\
1 &1 &1 &0 &1 &1 &0\\
\end{bmatrix}$
Here is the code, you can change ks and d values to find for different results, such as if you input ks as 5, and d as 4. You will get instant answers, but it takes about an hour to find a 7 and 5. The program can be optimized better, but it works without any problem at the moment.
Here is another solution:
$\begin{bmatrix}
1 &1 &1 &1 &1 &1 &0\\
1 &1 &1 &1 &0 &0 &1\\
0 &0 &0 &1 &0 &1 &1\\
0 &0 &1 &0 &1 &0 &1\\
0 &1 &0 &1 &1 &0 &0\\
\end{bmatrix}$
Previous Solution
First weigh the first three:
$ w_1=a+b+c$
Then the other three:
$ w_2=d+e+f$
Then the some mix:
$w_3=a+e+f$
then the last mix:
$w_4=b+d+e$
we also have differences as an extra information among these $w$s. but what can we do with these differences, what kind of information can be gained?
Let's do the math:
There are a few possibilities for the $w$s: $3x$, $2x+y$, $x+2y$ and $3y$ and the differences could be $0$, $x-y$, $2x-2y$ and $3x-3y$. If you notice, the differences have some relation, the ratio between them is constant. So by using those $w$'s, we have the lots of relation we can find among $a,b,c,d,e,f$ except $g$ since we did not weigh it at all. Lastly we weigh our $g$ and find one of the weights! Then the rest becomes easily and quickly, for example let say:
$ w_1=7$
Then the other three:
$ w_2=15$
Then the some mix:
$w_3=11$
then the last mix:
$w_4=11$
and last weight:
$g=5$
from the difference between $w_2$ and $w_3$, it can be concluded that $a=1$ and $d=5$ since $g=5$. Moreover, from $g=5$, so you can easily say that $e=5$ and $f=5$. The only problem becomes $b$ and $c$. But since we have seperate results, $w_1$ and $w_4$, $b$ becomes $1$ and as a result $c=5$.
This is applicable to most kind of combinations.
