XOR-free sets: Maximum density?












4












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
    1 hour 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
    58 mins ago


















4












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
    1 hour 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
    58 mins ago
















4












4








4





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour 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
    1 hour 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
    58 mins 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
    1 hour 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
    58 mins 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
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
1 hour 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
58 mins ago






$begingroup$
Related question (bounded numbers in {0,1,..,2^n-1}): mathoverflow.net/questions/293198/…
$endgroup$
– kodlu
58 mins ago












1 Answer
1






active

oldest

votes


















4












$begingroup$

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 $frac{|S|}{N} = frac{2}{3}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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
    9 mins ago











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









4












$begingroup$

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 $frac{|S|}{N} = frac{2}{3}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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
    9 mins ago
















4












$begingroup$

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 $frac{|S|}{N} = frac{2}{3}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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
    9 mins ago














4












4








4





$begingroup$

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 $frac{|S|}{N} = frac{2}{3}$.






share|cite|improve this answer











$endgroup$



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 $frac{|S|}{N} = frac{2}{3}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 31 mins ago

























answered 1 hour ago









Aleksei KulikovAleksei Kulikov

1,2061310




1,2061310








  • 1




    $begingroup$
    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
    9 mins ago














  • 1




    $begingroup$
    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
    9 mins ago








1




1




$begingroup$
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
9 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
9 mins ago


















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Tantalo (mitologio)

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Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,...
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Franz Schubert

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Franz Schubert Franz Schubert, pentrita en 1875 fare de Wilhelm August Rieder Bazaj informoj Naskiĝo 31-an de januaro 1797 ( 1797-01-31 ) en Himmelpfortgrund Morto 19-an de novembro 1828 ( 1828-11-19 ) (31-jara) en Vieno Ĝenroj Klasika muziko Profesio komponisto Aktivaj jaroj 1808-1828 v   •   d   •   r Franz SCHUBERT ([ˈfʁants ˈʃuːbɛɐ̯t], esperante Franz' Ŝuberto ; naskiĝis la 31-an de januaro 1797 je Himmelpfortgrund en Vieno, kiu nun estas parto de la 9-a viena distrikto Alsergrund, kaj mortis la 19-an de novembro 1828 en Vieno) estis aŭstra komponisto de klasika kaj fruromantisma muziko. Li komponis ĉirkaŭ 600 lidojn (germanaj kantoj) kaj estas nomata la "reĝo de lido". Enhavo 1 Vivo 1.1 Infaneco 1.2 Junaĝo 1.3 Zelis 1.4 Jaroj de maturiĝo kaj matureco 1.5 La lastaj jaroj 2 Postefiko kaj akceptado 3 Honoradoj 4 Verkoj 4.1 Eklezia muziko 4.2 Dram...
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