Reverse Littlewood-Offord problem: lower bound for the number of choices of signs such that $|pm a_1dotspm...
$begingroup$
Let $n$ be a positive integer. For $mathbf{a}inmathbb R^n$ let $N(mathbf{a})$ be the number of choices of $epsilonin{-1,1}^n$ such that
$$Biggl|sum_{i=1}^nepsilon_ia_iBiggr|leq max_{1leq ileq n}|a_i|.$$
Example. for $n$ even and $a=(1,dots,1),$ we are counting $epsilon$ with $|sum_{i=1}^nepsilon_i|leq 1,$ which forces $epsilon$ to have exactly $n/2$ entries of $-1,$ so $N(1,dots,1)$ is the central binomial coefficient $binom{n}{n/2}.$
Is it true that $N(mathbf a)geq binom{n}{n/2}$ for even $n$?
I also have a conjectured lower bound for odd $n$ if you're interested. Define $$B_n=N(1,dots,1,tfrac12)=2binom{n-1}{lfloor (n-1)/2rfloor}=
begin{cases}
binom{n}{n/2}&text{ for $n$ even}\
2binom{n-1}{(n-1)/2}&text{ for $n$ odd.}
end{cases}
$$
I've confirmed $N(mathbf a)geq B_n$ by picking random vectors $mathbf{a}$ uniformly at random, but these tests aren't very convincing since it doesn't even find $B_n$ for $ngeq 9.$
For $n=3$ we always have $N(mathbf{a})geq B_3=4$: assume $1=a_1geq a_2geq a_3geq 0,$ then $a_1-a_2-a_3$ and $a_1-a_2+a_3$ and their negations all lie in $[-1,1].$
This would be a reversed form of the Littlewood-Offord problem, where Erdős showed there are at most $binom{n}{lfloor n/2rfloor}$ vectors $epsilonin{-1,1}^n$ such that $Bigl|sum_{i=1}^nepsilon_ia_iBigr|leq color{red}{min}_{1leq ileq n}|a_i|.$
It's a version of Tomaszewski's Problem with the $ell^2$ norm replaced by $ell^infty.$
I believe a positive answer would also provide a lower bound for the Minimum number of balanced partitions.
combinatorics inequality
asked Jan 14 at 15:25
DapDap
16.4k738
$endgroup$
add a comment |
$begingroup$
Let $n$ be a positive integer. For $mathbf{a}inmathbb R^n$ let $N(mathbf{a})$ be the number of choices of $epsilonin{-1,1}^n$ such that
$$Biggl|sum_{i=1}^nepsilon_ia_iBiggr|leq max_{1leq ileq n}|a_i|.$$
Example. for $n$ even and $a=(1,dots,1),$ we are counting $epsilon$ with $|sum_{i=1}^nepsilon_i|leq 1,$ which forces $epsilon$ to have exactly $n/2$ entries of $-1,$ so $N(1,dots,1)$ is the central binomial coefficient $binom{n}{n/2}.$
Is it true that $N(mathbf a)geq binom{n}{n/2}$ for even $n$?
I also have a conjectured lower bound for odd $n$ if you're interested. Define $$B_n=N(1,dots,1,tfrac12)=2binom{n-1}{lfloor (n-1)/2rfloor}=
begin{cases}
binom{n}{n/2}&text{ for $n$ even}\
2binom{n-1}{(n-1)/2}&text{ for $n$ odd.}
end{cases}
$$
I've confirmed $N(mathbf a)geq B_n$ by picking random vectors $mathbf{a}$ uniformly at random, but these tests aren't very convincing since it doesn't even find $B_n$ for $ngeq 9.$
For $n=3$ we always have $N(mathbf{a})geq B_3=4$: assume $1=a_1geq a_2geq a_3geq 0,$ then $a_1-a_2-a_3$ and $a_1-a_2+a_3$ and their negations all lie in $[-1,1].$
This would be a reversed form of the Littlewood-Offord problem, where Erdős showed there are at most $binom{n}{lfloor n/2rfloor}$ vectors $epsilonin{-1,1}^n$ such that $Bigl|sum_{i=1}^nepsilon_ia_iBigr|leq color{red}{min}_{1leq ileq n}|a_i|.$
It's a version of Tomaszewski's Problem with the $ell^2$ norm replaced by $ell^infty.$
I believe a positive answer would also provide a lower bound for the Minimum number of balanced partitions.
combinatorics inequality
asked Jan 14 at 15:25
DapDap
16.4k738
$endgroup$
add a comment |
$begingroup$
Let $n$ be a positive integer. For $mathbf{a}inmathbb R^n$ let $N(mathbf{a})$ be the number of choices of $epsilonin{-1,1}^n$ such that
$$Biggl|sum_{i=1}^nepsilon_ia_iBiggr|leq max_{1leq ileq n}|a_i|.$$
Example. for $n$ even and $a=(1,dots,1),$ we are counting $epsilon$ with $|sum_{i=1}^nepsilon_i|leq 1,$ which forces $epsilon$ to have exactly $n/2$ entries of $-1,$ so $N(1,dots,1)$ is the central binomial coefficient $binom{n}{n/2}.$
Is it true that $N(mathbf a)geq binom{n}{n/2}$ for even $n$?
I also have a conjectured lower bound for odd $n$ if you're interested. Define $$B_n=N(1,dots,1,tfrac12)=2binom{n-1}{lfloor (n-1)/2rfloor}=
begin{cases}
binom{n}{n/2}&text{ for $n$ even}\
2binom{n-1}{(n-1)/2}&text{ for $n$ odd.}
end{cases}
$$
I've confirmed $N(mathbf a)geq B_n$ by picking random vectors $mathbf{a}$ uniformly at random, but these tests aren't very convincing since it doesn't even find $B_n$ for $ngeq 9.$
For $n=3$ we always have $N(mathbf{a})geq B_3=4$: assume $1=a_1geq a_2geq a_3geq 0,$ then $a_1-a_2-a_3$ and $a_1-a_2+a_3$ and their negations all lie in $[-1,1].$
This would be a reversed form of the Littlewood-Offord problem, where Erdős showed there are at most $binom{n}{lfloor n/2rfloor}$ vectors $epsilonin{-1,1}^n$ such that $Bigl|sum_{i=1}^nepsilon_ia_iBigr|leq color{red}{min}_{1leq ileq n}|a_i|.$
It's a version of Tomaszewski's Problem with the $ell^2$ norm replaced by $ell^infty.$
I believe a positive answer would also provide a lower bound for the Minimum number of balanced partitions.
combinatorics inequality
asked Jan 14 at 15:25
DapDap
16.4k738
$endgroup$
Let $n$ be a positive integer. For $mathbf{a}inmathbb R^n$ let $N(mathbf{a})$ be the number of choices of $epsilonin{-1,1}^n$ such that
$$Biggl|sum_{i=1}^nepsilon_ia_iBiggr|leq max_{1leq ileq n}|a_i|.$$
Example. for $n$ even and $a=(1,dots,1),$ we are counting $epsilon$ with $|sum_{i=1}^nepsilon_i|leq 1,$ which forces $epsilon$ to have exactly $n/2$ entries of $-1,$ so $N(1,dots,1)$ is the central binomial coefficient $binom{n}{n/2}.$
Is it true that $N(mathbf a)geq binom{n}{n/2}$ for even $n$?
I also have a conjectured lower bound for odd $n$ if you're interested. Define $$B_n=N(1,dots,1,tfrac12)=2binom{n-1}{lfloor (n-1)/2rfloor}=
begin{cases}
binom{n}{n/2}&text{ for $n$ even}\
2binom{n-1}{(n-1)/2}&text{ for $n$ odd.}
end{cases}
$$
I've confirmed $N(mathbf a)geq B_n$ by picking random vectors $mathbf{a}$ uniformly at random, but these tests aren't very convincing since it doesn't even find $B_n$ for $ngeq 9.$
For $n=3$ we always have $N(mathbf{a})geq B_3=4$: assume $1=a_1geq a_2geq a_3geq 0,$ then $a_1-a_2-a_3$ and $a_1-a_2+a_3$ and their negations all lie in $[-1,1].$
This would be a reversed form of the Littlewood-Offord problem, where Erdős showed there are at most $binom{n}{lfloor n/2rfloor}$ vectors $epsilonin{-1,1}^n$ such that $Bigl|sum_{i=1}^nepsilon_ia_iBigr|leq color{red}{min}_{1leq ileq n}|a_i|.$
It's a version of Tomaszewski's Problem with the $ell^2$ norm replaced by $ell^infty.$
I believe a positive answer would also provide a lower bound for the Minimum number of balanced partitions.
combinatorics inequality
combinatorics inequality
asked Jan 14 at 15:25
DapDap
16.4k738
asked Jan 14 at 15:25
DapDap
16.4k738
edited 2 days ago
Dap
asked Jan 14 at 15:25
DapDap
16.4k738
asked Jan 14 at 15:25
DapDap
16.4k738
asked Jan 14 at 15:25
DapDap
16.4k738
16.4k738
add a comment |
add a comment |
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