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Answer updated to match the changes in the original problem
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Seva
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Consider the addition Cayley graph $\Gamma$ induced by $F$ on $\mathbb Z_2^n$ (which is the graph with the vertex set $\mathbb Z_2^n$, with the vertices $u,v\in\mathbb Z_2^n$ adjacent whenever $u+v\in F$). A set $A$ with $(A+A)\cap F=\emptyset$ is an independent set in $\Gamma$. Since $\Gamma$ is regular of degree $|F|=n$$|F|$, it has an independent set of size at least $2^n/(n+1)$$2^n/(|F|+1)$.

In general, this bound is best possible: it is attained when $n=2^k-1$, the set $F$ is a subgroup of $\mathbb Z_2^n$ of rank $k$ with the zero element removed (so that $|F|=2^k-1$, where $k$ is the rank of $F$), and $A$ contains a unique element from each $F$-coset.

It is, however, possible to give betterBetter bounds can be given if some information about the set $F$ is available. Say, if $F$$F=\{f_1,\dotsc, f_m\}$ is an independent set, then weone can takefind a subgroup $A$ to be the set of all even-weight linear combinations of the elements of$H<\mathbb Z_2^n$ such that $F$$\mathbb Z_2^n=\langle F\rangle\oplus H$, toand take $$ A := \{ c_1f_1+\dotsb+c_mf_m+h\colon c_1+\dotsb+c_m=0,\ h\in H \} $$ to have $|A|=2^{n-1}$.

Consider the addition Cayley graph $\Gamma$ induced by $F$ on $\mathbb Z_2^n$ (which is the graph with the vertex set $\mathbb Z_2^n$, with the vertices $u,v\in\mathbb Z_2^n$ adjacent whenever $u+v\in F$). A set $A$ with $(A+A)\cap F=\emptyset$ is an independent set in $\Gamma$. Since $\Gamma$ is regular of degree $|F|=n$, it has an independent set of size at least $2^n/(n+1)$.

In general, this bound is best possible: it is attained when $n=2^k-1$, the set $F$ is a subgroup of $\mathbb Z_2^n$ of rank $k$ with the zero element removed, and $A$ contains a unique element from each $F$-coset.

It is, however, possible to give better bounds given some information about the set $F$. Say, if $F$ is an independent set, then we can take $A$ to be the set of all even-weight linear combinations of the elements of $F$, to have $|A|=2^{n-1}$.

Consider the addition Cayley graph $\Gamma$ induced by $F$ on $\mathbb Z_2^n$ (which is the graph with the vertex set $\mathbb Z_2^n$, with the vertices $u,v\in\mathbb Z_2^n$ adjacent whenever $u+v\in F$). A set $A$ with $(A+A)\cap F=\emptyset$ is an independent set in $\Gamma$. Since $\Gamma$ is regular of degree $|F|$, it has an independent set of size at least $2^n/(|F|+1)$.

In general, this bound is best possible: it is attained when $F$ is a subgroup of $\mathbb Z_2^n$ with the zero element removed (so that $|F|=2^k-1$, where $k$ is the rank of $F$), and $A$ contains a unique element from each $F$-coset.

Better bounds can be given if some information about the set $F$ is available. Say, if $F=\{f_1,\dotsc, f_m\}$ is an independent set, then one can find a subgroup $H<\mathbb Z_2^n$ such that $\mathbb Z_2^n=\langle F\rangle\oplus H$, and take $$ A := \{ c_1f_1+\dotsb+c_mf_m+h\colon c_1+\dotsb+c_m=0,\ h\in H \} $$ to have $|A|=2^{n-1}$.

added 370 characters in body
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Seva
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  • 59
  • 141

Consider the addition Cayley graph $\Gamma$ induced by $F$ on $\mathbb Z_2^n$ (which is the graph with the vertex set $\mathbb Z_2^n$, with the vertices $u,v\in\mathbb Z_2^n$ adjacent whenever $u+v\in F$). A set $A$ with $(A+A)\cap F=\emptyset$ is an independent set in $\Gamma$. Since $\Gamma$ is regular of degree $|F|=n$, it has an independent set of size at least $2^n/(n+1)$.

In general, this bound is best possible: it is attained when $n=2^k-1$, the set $F$ is a subgroup of $\mathbb Z_2^n$ of rank $k$ with the zero element removed, and $A$ contains a unique element from each $F$-coset.

It might beis, however, possible to give a better boundbounds given some information about the structureset $F$. Say, if $F$ is an independent set, then we can take $A$ to be the set of all even-weight linear combinations of the elements of $F$, to have $|A|=2^{n-1}$.

Consider the addition Cayley graph $\Gamma$ induced by $F$ on $\mathbb Z_2^n$ (which is the graph with the vertex set $\mathbb Z_2^n$, with the vertices $u,v\in\mathbb Z_2^n$ adjacent whenever $u+v\in F$). A set $A$ with $(A+A)\cap F=\emptyset$ is an independent set in $\Gamma$. Since $\Gamma$ is regular of degree $|F|=n$, it has an independent set of size at least $2^n/(n+1)$.

It might be possible to give a better bound given some information about the structure of $F$.

Consider the addition Cayley graph $\Gamma$ induced by $F$ on $\mathbb Z_2^n$ (which is the graph with the vertex set $\mathbb Z_2^n$, with the vertices $u,v\in\mathbb Z_2^n$ adjacent whenever $u+v\in F$). A set $A$ with $(A+A)\cap F=\emptyset$ is an independent set in $\Gamma$. Since $\Gamma$ is regular of degree $|F|=n$, it has an independent set of size at least $2^n/(n+1)$.

In general, this bound is best possible: it is attained when $n=2^k-1$, the set $F$ is a subgroup of $\mathbb Z_2^n$ of rank $k$ with the zero element removed, and $A$ contains a unique element from each $F$-coset.

It is, however, possible to give better bounds given some information about the set $F$. Say, if $F$ is an independent set, then we can take $A$ to be the set of all even-weight linear combinations of the elements of $F$, to have $|A|=2^{n-1}$.

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Seva
  • 23k
  • 2
  • 59
  • 141

Consider the addition Cayley graph $\Gamma$ induced by $F$ on $\mathbb Z_2^n$ (which is the graph with the vertex set $\mathbb Z_2^n$, with the vertices $u,v\in\mathbb Z_2^n$ adjacent whenever $u+v\in F$). A set $A$ with $(A+A)\cap F=\emptyset$ is an independent set in $\Gamma$. Since $\Gamma$ is regular of degree $|F|=n$, it has an independent set of size at least $2^n/(n+1)$.

It might be possible to give a better bound given some information about the structure of $F$.