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Let $Z_2=\{0,1\}$, $Z_2^r=Z_2\times Z_2 \times...\times Z_2,$ $S_1\subset Z_2^{n_1}$ and $S_2\subset Z_2^{n_2}$, $S=(S_2,S_1)\subset Z_2^{n_1+n_2}$, I'd like to construct $S_1$ and $S_2$ s.t the following equations hold for $S$:

For any $x=(x_{n_1+n_2},...,x_1)\in Z_2^{n_1+n_2}$

1) $\Sigma_{1 \leq i \leq n_1+n_2}(-1)^{x_i}=0,$

2) $\Sigma_{1 \leq i,j \leq n_1+n_2,i\neq j}(-1)^{x_i+x_j}=0,$

$‎\vdots$‎

l) $\Sigma_{1 \leq i_1,i_2,...,i_l \leq n_1+n_2}(-1)^{x_{i_1}+x_{i_2}...+x_{i_l}}=0,$ for some $1 \leq l \leq n_1+n_2$,

The sum is "+"over real feild,like "0+1+0+0+1+1+1=4" How should I form $S_1$ and $S_2$ s.t above equations hold? Does anyone have any idea or seen any references? I should say it's about Boolean functions.

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  • $\begingroup$ The sets $S_k$ don't appear in your equations. $\endgroup$
    – kodlu
    Commented Oct 25, 2016 at 6:22
  • $\begingroup$ What is $Z_2$, and what is $(S_2,S_1)$? $\endgroup$
    – Emil Jeřábek
    Commented Oct 25, 2016 at 7:41
  • $\begingroup$ If you take the square of the first equation, you get $0 = n_1+n_2 + \sum_{i\neq j} (-1)^{x_i+x_j}$, thus the first two equations contradict each other. Unless you consider equations in $Z_2$. If this is the case these equations either contradict each other, or the first trivially implies the second, depending on whether $n_1+n_2$ is odd or even. $\endgroup$
    – Jan-Christoph Schlage-Puchta
    Commented Oct 25, 2016 at 11:05
  • $\begingroup$ I still don't see the subsets $S_1,S_2$ used anywhere in the equations, how are we to find them using the equations? $\endgroup$
    – kodlu
    Commented Oct 25, 2016 at 19:25

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