XOR-free sets: Maximum density?
$begingroup$
It is known that sum-free
subsets of $mathbb{N}$ can have
natural density at most
$frac{1}{2}$. This density is achieved by the odd numbers: the sum of two
odd numbers is even.
I ask now a similar question for XOR rather than addition.
For $a,b in mathbb{N}$, define $a oplus b$ as follows:
Represent $a ge b$ as binary numbers, pad the smaller $b$ with zeros,
and take the bit-wise XOR of the binary representation.
For example, $35 oplus 15 = 44$:
begin{eqnarray}
35 = & ;100011\
15 = & ;001111\
oplus = & ;101100
end{eqnarray}
The condition for a set $S subset mathbb{N}$ to be XOR-free is that, for any $a,b in S$, $a oplus b notin S$.
Q. What is the largest density of an XOR-free subset $S$ of $mathbb{N}$?
Again the odd numbers with density $frac{1}{2}$ are XOR-free.
I am not seeing an argument that $frac{1}{2}$ is the maximum possible density.
nt.number-theory integer-sequences
asked 2 hours ago
Joseph O'RourkeJoseph O'Rourke
85.3k16233700
$endgroup$
add a comment |
$begingroup$
It is known that sum-free
subsets of $mathbb{N}$ can have
natural density at most
$frac{1}{2}$. This density is achieved by the odd numbers: the sum of two
odd numbers is even.
I ask now a similar question for XOR rather than addition.
For $a,b in mathbb{N}$, define $a oplus b$ as follows:
Represent $a ge b$ as binary numbers, pad the smaller $b$ with zeros,
and take the bit-wise XOR of the binary representation.
For example, $35 oplus 15 = 44$:
begin{eqnarray}
35 = & ;100011\
15 = & ;001111\
oplus = & ;101100
end{eqnarray}
The condition for a set $S subset mathbb{N}$ to be XOR-free is that, for any $a,b in S$, $a oplus b notin S$.
Q. What is the largest density of an XOR-free subset $S$ of $mathbb{N}$?
Again the odd numbers with density $frac{1}{2}$ are XOR-free.
I am not seeing an argument that $frac{1}{2}$ is the maximum possible density.
nt.number-theory integer-sequences
asked 2 hours ago
Joseph O'RourkeJoseph O'Rourke
85.3k16233700
$endgroup$
$begingroup$
Maybe a parity argument helps? As a guide, consider all numbers with an odd number of 1-bits. Gerhard "Are Parity Arguments Also Even?" Paseman, 2019.02.24.
$endgroup$
– Gerhard Paseman
2 hours ago
1
$begingroup$
You can get close. For discussion sake, focus on the last bit. You can't have two pairs of numbers which differ only in the last bit as both pairs sum to 1. This means out of 2^n bit patterns, at most n + 2^(n-1) will avoid having two pairs sum to a single bit. Gerhard "At Least We're Asymptotically Right" Paseman, 2019.02.24.
$endgroup$
– Gerhard Paseman
1 hour ago
$begingroup$
Related question (bounded numbers in {0,1,..,2^n-1}): mathoverflow.net/questions/293198/…
$endgroup$
– kodlu
1 hour ago
add a comment |
$begingroup$
It is known that sum-free
subsets of $mathbb{N}$ can have
natural density at most
$frac{1}{2}$. This density is achieved by the odd numbers: the sum of two
odd numbers is even.
I ask now a similar question for XOR rather than addition.
For $a,b in mathbb{N}$, define $a oplus b$ as follows:
Represent $a ge b$ as binary numbers, pad the smaller $b$ with zeros,
and take the bit-wise XOR of the binary representation.
For example, $35 oplus 15 = 44$:
begin{eqnarray}
35 = & ;100011\
15 = & ;001111\
oplus = & ;101100
end{eqnarray}
The condition for a set $S subset mathbb{N}$ to be XOR-free is that, for any $a,b in S$, $a oplus b notin S$.
Q. What is the largest density of an XOR-free subset $S$ of $mathbb{N}$?
Again the odd numbers with density $frac{1}{2}$ are XOR-free.
I am not seeing an argument that $frac{1}{2}$ is the maximum possible density.
nt.number-theory integer-sequences
asked 2 hours ago
Joseph O'RourkeJoseph O'Rourke
85.3k16233700
$endgroup$
It is known that sum-free
subsets of $mathbb{N}$ can have
natural density at most
$frac{1}{2}$. This density is achieved by the odd numbers: the sum of two
odd numbers is even.
I ask now a similar question for XOR rather than addition.
For $a,b in mathbb{N}$, define $a oplus b$ as follows:
Represent $a ge b$ as binary numbers, pad the smaller $b$ with zeros,
and take the bit-wise XOR of the binary representation.
For example, $35 oplus 15 = 44$:
begin{eqnarray}
35 = & ;100011\
15 = & ;001111\
oplus = & ;101100
end{eqnarray}
The condition for a set $S subset mathbb{N}$ to be XOR-free is that, for any $a,b in S$, $a oplus b notin S$.
Q. What is the largest density of an XOR-free subset $S$ of $mathbb{N}$?
Again the odd numbers with density $frac{1}{2}$ are XOR-free.
I am not seeing an argument that $frac{1}{2}$ is the maximum possible density.
nt.number-theory integer-sequences
nt.number-theory integer-sequences
asked 2 hours ago
Joseph O'RourkeJoseph O'Rourke
85.3k16233700
asked 2 hours ago
Joseph O'RourkeJoseph O'Rourke
85.3k16233700
asked 2 hours ago
Joseph O'RourkeJoseph O'Rourke
85.3k16233700
asked 2 hours ago
Joseph O'RourkeJoseph O'Rourke
85.3k16233700
asked 2 hours ago
Joseph O'RourkeJoseph O'Rourke
85.3k16233700
85.3k16233700
$begingroup$
Maybe a parity argument helps? As a guide, consider all numbers with an odd number of 1-bits. Gerhard "Are Parity Arguments Also Even?" Paseman, 2019.02.24.
$endgroup$
– Gerhard Paseman
2 hours ago
1
$begingroup$
You can get close. For discussion sake, focus on the last bit. You can't have two pairs of numbers which differ only in the last bit as both pairs sum to 1. This means out of 2^n bit patterns, at most n + 2^(n-1) will avoid having two pairs sum to a single bit. Gerhard "At Least We're Asymptotically Right" Paseman, 2019.02.24.
$endgroup$
– Gerhard Paseman
1 hour ago
$begingroup$
Related question (bounded numbers in {0,1,..,2^n-1}): mathoverflow.net/questions/293198/…
$endgroup$
– kodlu
1 hour ago
add a comment |
$begingroup$
Maybe a parity argument helps? As a guide, consider all numbers with an odd number of 1-bits. Gerhard "Are Parity Arguments Also Even?" Paseman, 2019.02.24.
$endgroup$
– Gerhard Paseman
2 hours ago
1
$begingroup$
You can get close. For discussion sake, focus on the last bit. You can't have two pairs of numbers which differ only in the last bit as both pairs sum to 1. This means out of 2^n bit patterns, at most n + 2^(n-1) will avoid having two pairs sum to a single bit. Gerhard "At Least We're Asymptotically Right" Paseman, 2019.02.24.
$endgroup$
– Gerhard Paseman
1 hour ago
$begingroup$
Related question (bounded numbers in {0,1,..,2^n-1}): mathoverflow.net/questions/293198/…
$endgroup$
– kodlu
1 hour ago
$begingroup$
Maybe a parity argument helps? As a guide, consider all numbers with an odd number of 1-bits. Gerhard "Are Parity Arguments Also Even?" Paseman, 2019.02.24.
$endgroup$
– Gerhard Paseman
2 hours ago
$begingroup$
Maybe a parity argument helps? As a guide, consider all numbers with an odd number of 1-bits. Gerhard "Are Parity Arguments Also Even?" Paseman, 2019.02.24.
$endgroup$
– Gerhard Paseman
2 hours ago
1
1
$begingroup$
You can get close. For discussion sake, focus on the last bit. You can't have two pairs of numbers which differ only in the last bit as both pairs sum to 1. This means out of 2^n bit patterns, at most n + 2^(n-1) will avoid having two pairs sum to a single bit. Gerhard "At Least We're Asymptotically Right" Paseman, 2019.02.24.
$endgroup$
– Gerhard Paseman
1 hour ago
$begingroup$
You can get close. For discussion sake, focus on the last bit. You can't have two pairs of numbers which differ only in the last bit as both pairs sum to 1. This means out of 2^n bit patterns, at most n + 2^(n-1) will avoid having two pairs sum to a single bit. Gerhard "At Least We're Asymptotically Right" Paseman, 2019.02.24.
$endgroup$
– Gerhard Paseman
1 hour ago
$begingroup$
Related question (bounded numbers in {0,1,..,2^n-1}): mathoverflow.net/questions/293198/…
$endgroup$
– kodlu
1 hour ago
$begingroup$
Related question (bounded numbers in {0,1,..,2^n-1}): mathoverflow.net/questions/293198/…
$endgroup$
– kodlu
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $ain S$ be some fixed element. Note that $aoplus b le a + b$. Let $N$ be some big number. Put $M = [1, ldots , N]cap S$. We have $aoplus M cap M = varnothing$. We also have $aoplus M subset [1, ldots , N + a]$. Therefore $2|M| le N + a$ or $|M| le frac{N}{2} + frac{a}{2}$. Since $a$ is fixed taking limit $Nto infty$ yields the desired result.
UPD Here is an answer for a perhaps more interesting question: what is limsup of the biggest density of the subset of $[0, ldots , N]$ which is XOR-free. Note that if $N = 2^k - 1$ for some $kin mathbb{N}$ then $|S| le 2^{k-1}$. Indeed, for any $a, bin [0, ldots, N]$ we have $aoplus b in [0, ldots , N]$ . Since $|Soplus S| ge |S|$ and $Scap (Soplus S) = varnothing$ we get $2|S| le (N+1)$ or $|S| le 2^{k-1}$.
For the general case assume that $2^k le N < 2^{k+1} - 1$. Then on the one hand $|S| le 2^k$ since $Ssubset [0, ldots , 2^{k+1}-1]$ and on the other hand $|S| le 2^{k-1} + (N - 2^k) = N - 2^{k-1}$ since $|Scap [0, ldots , 2^k - 1]| le 2^{k-1}$. Therefore $3|S| le 2N$ or $|S| le frac{2N}{3}$.
Here is an example (found partly via computer search) which shows that density $frac{2}{3}$ is possible: put $N = 6*2^k$ and let $S$ be a set of all numbers consisting of $k + 3$ digits such that first three of them is from the following set $M = {(0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1)}$. Note that $Soplus S cap S = varnothing$, all elements of $S$ are not greater than $N$ and $|S| = 4*2^k$. Therefore $frac{|S|}{N} = frac{2}{3}$.
answered 1 hour ago
Aleksei KulikovAleksei Kulikov
1,2061310
$endgroup$
2
$begingroup$
The example can be described more simply as the integers whose fist two digits are $(0,1)$ or $(1,0)$, that is, $S = [2^{k+1}, 3 cdot 2^{k+1})$. Those digits sum to $1$, so in any element of $S oplus S$ they sum to $0$, whence $S$ is disjoint from $S oplus S$.
$endgroup$
– Noam D. Elkies
47 mins ago
add a comment |
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1 Answer
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$begingroup$
Let $ain S$ be some fixed element. Note that $aoplus b le a + b$. Let $N$ be some big number. Put $M = [1, ldots , N]cap S$. We have $aoplus M cap M = varnothing$. We also have $aoplus M subset [1, ldots , N + a]$. Therefore $2|M| le N + a$ or $|M| le frac{N}{2} + frac{a}{2}$. Since $a$ is fixed taking limit $Nto infty$ yields the desired result.
UPD Here is an answer for a perhaps more interesting question: what is limsup of the biggest density of the subset of $[0, ldots , N]$ which is XOR-free. Note that if $N = 2^k - 1$ for some $kin mathbb{N}$ then $|S| le 2^{k-1}$. Indeed, for any $a, bin [0, ldots, N]$ we have $aoplus b in [0, ldots , N]$ . Since $|Soplus S| ge |S|$ and $Scap (Soplus S) = varnothing$ we get $2|S| le (N+1)$ or $|S| le 2^{k-1}$.
For the general case assume that $2^k le N < 2^{k+1} - 1$. Then on the one hand $|S| le 2^k$ since $Ssubset [0, ldots , 2^{k+1}-1]$ and on the other hand $|S| le 2^{k-1} + (N - 2^k) = N - 2^{k-1}$ since $|Scap [0, ldots , 2^k - 1]| le 2^{k-1}$. Therefore $3|S| le 2N$ or $|S| le frac{2N}{3}$.
Here is an example (found partly via computer search) which shows that density $frac{2}{3}$ is possible: put $N = 6*2^k$ and let $S$ be a set of all numbers consisting of $k + 3$ digits such that first three of them is from the following set $M = {(0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1)}$. Note that $Soplus S cap S = varnothing$, all elements of $S$ are not greater than $N$ and $|S| = 4*2^k$. Therefore $frac{|S|}{N} = frac{2}{3}$.
answered 1 hour ago
Aleksei KulikovAleksei Kulikov
1,2061310
$endgroup$
2
$begingroup$
The example can be described more simply as the integers whose fist two digits are $(0,1)$ or $(1,0)$, that is, $S = [2^{k+1}, 3 cdot 2^{k+1})$. Those digits sum to $1$, so in any element of $S oplus S$ they sum to $0$, whence $S$ is disjoint from $S oplus S$.
$endgroup$
– Noam D. Elkies
47 mins ago
add a comment |
$begingroup$
Let $ain S$ be some fixed element. Note that $aoplus b le a + b$. Let $N$ be some big number. Put $M = [1, ldots , N]cap S$. We have $aoplus M cap M = varnothing$. We also have $aoplus M subset [1, ldots , N + a]$. Therefore $2|M| le N + a$ or $|M| le frac{N}{2} + frac{a}{2}$. Since $a$ is fixed taking limit $Nto infty$ yields the desired result.
UPD Here is an answer for a perhaps more interesting question: what is limsup of the biggest density of the subset of $[0, ldots , N]$ which is XOR-free. Note that if $N = 2^k - 1$ for some $kin mathbb{N}$ then $|S| le 2^{k-1}$. Indeed, for any $a, bin [0, ldots, N]$ we have $aoplus b in [0, ldots , N]$ . Since $|Soplus S| ge |S|$ and $Scap (Soplus S) = varnothing$ we get $2|S| le (N+1)$ or $|S| le 2^{k-1}$.
For the general case assume that $2^k le N < 2^{k+1} - 1$. Then on the one hand $|S| le 2^k$ since $Ssubset [0, ldots , 2^{k+1}-1]$ and on the other hand $|S| le 2^{k-1} + (N - 2^k) = N - 2^{k-1}$ since $|Scap [0, ldots , 2^k - 1]| le 2^{k-1}$. Therefore $3|S| le 2N$ or $|S| le frac{2N}{3}$.
Here is an example (found partly via computer search) which shows that density $frac{2}{3}$ is possible: put $N = 6*2^k$ and let $S$ be a set of all numbers consisting of $k + 3$ digits such that first three of them is from the following set $M = {(0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1)}$. Note that $Soplus S cap S = varnothing$, all elements of $S$ are not greater than $N$ and $|S| = 4*2^k$. Therefore $frac{|S|}{N} = frac{2}{3}$.
answered 1 hour ago
Aleksei KulikovAleksei Kulikov
1,2061310
$endgroup$
2
$begingroup$
The example can be described more simply as the integers whose fist two digits are $(0,1)$ or $(1,0)$, that is, $S = [2^{k+1}, 3 cdot 2^{k+1})$. Those digits sum to $1$, so in any element of $S oplus S$ they sum to $0$, whence $S$ is disjoint from $S oplus S$.
$endgroup$
– Noam D. Elkies
47 mins ago
add a comment |
$begingroup$
Let $ain S$ be some fixed element. Note that $aoplus b le a + b$. Let $N$ be some big number. Put $M = [1, ldots , N]cap S$. We have $aoplus M cap M = varnothing$. We also have $aoplus M subset [1, ldots , N + a]$. Therefore $2|M| le N + a$ or $|M| le frac{N}{2} + frac{a}{2}$. Since $a$ is fixed taking limit $Nto infty$ yields the desired result.
UPD Here is an answer for a perhaps more interesting question: what is limsup of the biggest density of the subset of $[0, ldots , N]$ which is XOR-free. Note that if $N = 2^k - 1$ for some $kin mathbb{N}$ then $|S| le 2^{k-1}$. Indeed, for any $a, bin [0, ldots, N]$ we have $aoplus b in [0, ldots , N]$ . Since $|Soplus S| ge |S|$ and $Scap (Soplus S) = varnothing$ we get $2|S| le (N+1)$ or $|S| le 2^{k-1}$.
For the general case assume that $2^k le N < 2^{k+1} - 1$. Then on the one hand $|S| le 2^k$ since $Ssubset [0, ldots , 2^{k+1}-1]$ and on the other hand $|S| le 2^{k-1} + (N - 2^k) = N - 2^{k-1}$ since $|Scap [0, ldots , 2^k - 1]| le 2^{k-1}$. Therefore $3|S| le 2N$ or $|S| le frac{2N}{3}$.
Here is an example (found partly via computer search) which shows that density $frac{2}{3}$ is possible: put $N = 6*2^k$ and let $S$ be a set of all numbers consisting of $k + 3$ digits such that first three of them is from the following set $M = {(0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1)}$. Note that $Soplus S cap S = varnothing$, all elements of $S$ are not greater than $N$ and $|S| = 4*2^k$. Therefore $frac{|S|}{N} = frac{2}{3}$.
answered 1 hour ago
Aleksei KulikovAleksei Kulikov
1,2061310
$endgroup$
Let $ain S$ be some fixed element. Note that $aoplus b le a + b$. Let $N$ be some big number. Put $M = [1, ldots , N]cap S$. We have $aoplus M cap M = varnothing$. We also have $aoplus M subset [1, ldots , N + a]$. Therefore $2|M| le N + a$ or $|M| le frac{N}{2} + frac{a}{2}$. Since $a$ is fixed taking limit $Nto infty$ yields the desired result.
UPD Here is an answer for a perhaps more interesting question: what is limsup of the biggest density of the subset of $[0, ldots , N]$ which is XOR-free. Note that if $N = 2^k - 1$ for some $kin mathbb{N}$ then $|S| le 2^{k-1}$. Indeed, for any $a, bin [0, ldots, N]$ we have $aoplus b in [0, ldots , N]$ . Since $|Soplus S| ge |S|$ and $Scap (Soplus S) = varnothing$ we get $2|S| le (N+1)$ or $|S| le 2^{k-1}$.
For the general case assume that $2^k le N < 2^{k+1} - 1$. Then on the one hand $|S| le 2^k$ since $Ssubset [0, ldots , 2^{k+1}-1]$ and on the other hand $|S| le 2^{k-1} + (N - 2^k) = N - 2^{k-1}$ since $|Scap [0, ldots , 2^k - 1]| le 2^{k-1}$. Therefore $3|S| le 2N$ or $|S| le frac{2N}{3}$.
Here is an example (found partly via computer search) which shows that density $frac{2}{3}$ is possible: put $N = 6*2^k$ and let $S$ be a set of all numbers consisting of $k + 3$ digits such that first three of them is from the following set $M = {(0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1)}$. Note that $Soplus S cap S = varnothing$, all elements of $S$ are not greater than $N$ and $|S| = 4*2^k$. Therefore $frac{|S|}{N} = frac{2}{3}$.
answered 1 hour ago
Aleksei KulikovAleksei Kulikov
1,2061310
edited 1 hour ago
answered 1 hour ago
Aleksei KulikovAleksei Kulikov
1,2061310
answered 1 hour ago
Aleksei KulikovAleksei Kulikov
1,2061310
answered 1 hour ago
Aleksei KulikovAleksei Kulikov
1,2061310
1,2061310
2
$begingroup$
The example can be described more simply as the integers whose fist two digits are $(0,1)$ or $(1,0)$, that is, $S = [2^{k+1}, 3 cdot 2^{k+1})$. Those digits sum to $1$, so in any element of $S oplus S$ they sum to $0$, whence $S$ is disjoint from $S oplus S$.
$endgroup$
– Noam D. Elkies
47 mins ago
add a comment |
2
$begingroup$
The example can be described more simply as the integers whose fist two digits are $(0,1)$ or $(1,0)$, that is, $S = [2^{k+1}, 3 cdot 2^{k+1})$. Those digits sum to $1$, so in any element of $S oplus S$ they sum to $0$, whence $S$ is disjoint from $S oplus S$.
$endgroup$
– Noam D. Elkies
47 mins ago
2
2
$begingroup$
The example can be described more simply as the integers whose fist two digits are $(0,1)$ or $(1,0)$, that is, $S = [2^{k+1}, 3 cdot 2^{k+1})$. Those digits sum to $1$, so in any element of $S oplus S$ they sum to $0$, whence $S$ is disjoint from $S oplus S$.
$endgroup$
– Noam D. Elkies
47 mins ago
$begingroup$
The example can be described more simply as the integers whose fist two digits are $(0,1)$ or $(1,0)$, that is, $S = [2^{k+1}, 3 cdot 2^{k+1})$. Those digits sum to $1$, so in any element of $S oplus S$ they sum to $0$, whence $S$ is disjoint from $S oplus S$.
$endgroup$
– Noam D. Elkies
47 mins ago
add a comment |
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$begingroup$
Maybe a parity argument helps? As a guide, consider all numbers with an odd number of 1-bits. Gerhard "Are Parity Arguments Also Even?" Paseman, 2019.02.24.
$endgroup$
– Gerhard Paseman
2 hours ago
1
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You can get close. For discussion sake, focus on the last bit. You can't have two pairs of numbers which differ only in the last bit as both pairs sum to 1. This means out of 2^n bit patterns, at most n + 2^(n-1) will avoid having two pairs sum to a single bit. Gerhard "At Least We're Asymptotically Right" Paseman, 2019.02.24.
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– Gerhard Paseman
1 hour ago
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Related question (bounded numbers in {0,1,..,2^n-1}): mathoverflow.net/questions/293198/…
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– kodlu
1 hour ago