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
    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
    2 hours ago












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












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






  • 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
    2 hours 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
2 hours 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
2 hours 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












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











$endgroup$









  • 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
    55 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











$endgroup$









  • 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
    55 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











$endgroup$









  • 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
    55 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 1 hour ago

























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
    55 mins ago














  • 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
    55 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
55 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
55 mins ago


















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Knooppunt Holsloot

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Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
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Altaar (religie)

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Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
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Gregoriusmis

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Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
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