Minimum distance between polynomials in ring-LWE
Let $R_q=mathbb{Z}_q[x]/langle f(x)rangle$ where $f(x)=x^n+1$, as in the ring-LWE problem.
Let $a(x)$ be chosen uniformly at random from $R_q$.
Question: Is there any theorem that lower bounds the distance between any two polynomials of the form $a(x)s_1(s)$ and $a(x)s_2(x)$?
In other words, what is the value of $d$ such that $$||a(x)s_1(x)-a(x)s_2(x)||geq d$$ except with negligible probability, for any two polynomials $s_1(x),s_2(x)in R_q$ and where $||cdot||$ is the usual $L_2$ norm?
lattice-crypto lwe ring-lwe
asked Nov 24 at 16:33
P.B.
1476
add a comment |
Let $R_q=mathbb{Z}_q[x]/langle f(x)rangle$ where $f(x)=x^n+1$, as in the ring-LWE problem.
Let $a(x)$ be chosen uniformly at random from $R_q$.
Question: Is there any theorem that lower bounds the distance between any two polynomials of the form $a(x)s_1(s)$ and $a(x)s_2(x)$?
In other words, what is the value of $d$ such that $$||a(x)s_1(x)-a(x)s_2(x)||geq d$$ except with negligible probability, for any two polynomials $s_1(x),s_2(x)in R_q$ and where $||cdot||$ is the usual $L_2$ norm?
lattice-crypto lwe ring-lwe
asked Nov 24 at 16:33
P.B.
1476
Hello. It is a good question, but the $L_2$ norm is defined over vectors and it is not clear how you are embedding the polynomials in a vector space. Are you just representing the polynomials as vectors with their coefficients? (So, for instance, $2x^3 -1$ becomes the vector $(2, 0, 0, -1)$).
– Hilder Vítor Lima Pereira
Nov 24 at 17:19
Yes, I am thinking of the canonical embedding
– P.B.
Nov 24 at 17:26
Well, the canonical embedding is the one that uses isomorphisms to embed the polynomials. The one I've described is the coefficient embedding...
– Hilder Vítor Lima Pereira
Nov 24 at 17:31
Sorry. I mean the coefficient embedding then
– P.B.
Nov 24 at 17:46
How do you define "negligible probability" in this case?
– kodlu
Nov 24 at 21:19
add a comment |
Let $R_q=mathbb{Z}_q[x]/langle f(x)rangle$ where $f(x)=x^n+1$, as in the ring-LWE problem.
Let $a(x)$ be chosen uniformly at random from $R_q$.
Question: Is there any theorem that lower bounds the distance between any two polynomials of the form $a(x)s_1(s)$ and $a(x)s_2(x)$?
In other words, what is the value of $d$ such that $$||a(x)s_1(x)-a(x)s_2(x)||geq d$$ except with negligible probability, for any two polynomials $s_1(x),s_2(x)in R_q$ and where $||cdot||$ is the usual $L_2$ norm?
lattice-crypto lwe ring-lwe
asked Nov 24 at 16:33
P.B.
1476
Let $R_q=mathbb{Z}_q[x]/langle f(x)rangle$ where $f(x)=x^n+1$, as in the ring-LWE problem.
Let $a(x)$ be chosen uniformly at random from $R_q$.
Question: Is there any theorem that lower bounds the distance between any two polynomials of the form $a(x)s_1(s)$ and $a(x)s_2(x)$?
In other words, what is the value of $d$ such that $$||a(x)s_1(x)-a(x)s_2(x)||geq d$$ except with negligible probability, for any two polynomials $s_1(x),s_2(x)in R_q$ and where $||cdot||$ is the usual $L_2$ norm?
lattice-crypto lwe ring-lwe
lattice-crypto lwe ring-lwe
asked Nov 24 at 16:33
P.B.
1476
asked Nov 24 at 16:33
P.B.
1476
asked Nov 24 at 16:33
P.B.
1476
asked Nov 24 at 16:33
P.B.
1476
asked Nov 24 at 16:33
P.B.
1476
1476
Hello. It is a good question, but the $L_2$ norm is defined over vectors and it is not clear how you are embedding the polynomials in a vector space. Are you just representing the polynomials as vectors with their coefficients? (So, for instance, $2x^3 -1$ becomes the vector $(2, 0, 0, -1)$).
– Hilder Vítor Lima Pereira
Nov 24 at 17:19
Yes, I am thinking of the canonical embedding
– P.B.
Nov 24 at 17:26
Well, the canonical embedding is the one that uses isomorphisms to embed the polynomials. The one I've described is the coefficient embedding...
– Hilder Vítor Lima Pereira
Nov 24 at 17:31
Sorry. I mean the coefficient embedding then
– P.B.
Nov 24 at 17:46
How do you define "negligible probability" in this case?
– kodlu
Nov 24 at 21:19
add a comment |
Hello. It is a good question, but the $L_2$ norm is defined over vectors and it is not clear how you are embedding the polynomials in a vector space. Are you just representing the polynomials as vectors with their coefficients? (So, for instance, $2x^3 -1$ becomes the vector $(2, 0, 0, -1)$).
– Hilder Vítor Lima Pereira
Nov 24 at 17:19
Yes, I am thinking of the canonical embedding
– P.B.
Nov 24 at 17:26
Well, the canonical embedding is the one that uses isomorphisms to embed the polynomials. The one I've described is the coefficient embedding...
– Hilder Vítor Lima Pereira
Nov 24 at 17:31
Sorry. I mean the coefficient embedding then
– P.B.
Nov 24 at 17:46
How do you define "negligible probability" in this case?
– kodlu
Nov 24 at 21:19
Hello. It is a good question, but the $L_2$ norm is defined over vectors and it is not clear how you are embedding the polynomials in a vector space. Are you just representing the polynomials as vectors with their coefficients? (So, for instance, $2x^3 -1$ becomes the vector $(2, 0, 0, -1)$).
– Hilder Vítor Lima Pereira
Nov 24 at 17:19
Hello. It is a good question, but the $L_2$ norm is defined over vectors and it is not clear how you are embedding the polynomials in a vector space. Are you just representing the polynomials as vectors with their coefficients? (So, for instance, $2x^3 -1$ becomes the vector $(2, 0, 0, -1)$).
– Hilder Vítor Lima Pereira
Nov 24 at 17:19
Yes, I am thinking of the canonical embedding
– P.B.
Nov 24 at 17:26
Yes, I am thinking of the canonical embedding
– P.B.
Nov 24 at 17:26
Well, the canonical embedding is the one that uses isomorphisms to embed the polynomials. The one I've described is the coefficient embedding...
– Hilder Vítor Lima Pereira
Nov 24 at 17:31
Well, the canonical embedding is the one that uses isomorphisms to embed the polynomials. The one I've described is the coefficient embedding...
– Hilder Vítor Lima Pereira
Nov 24 at 17:31
Sorry. I mean the coefficient embedding then
– P.B.
Nov 24 at 17:46
Sorry. I mean the coefficient embedding then
– P.B.
Nov 24 at 17:46
How do you define "negligible probability" in this case?
– kodlu
Nov 24 at 21:19
How do you define "negligible probability" in this case?
– kodlu
Nov 24 at 21:19
add a comment |
1 Answer
1
active
oldest
votes
I'm assuming $n$ is a power of $2$ and that $q$ is an odd prime larger than $n$. I'm discarding the trivial case $s_1 = s_2$.
If you consider everything $mod q$, then it is most likely over the choice of $a$ that there exists $s_1 neq s_2$ such that $|a s_1 - a s_2| = sqrt{n}$. Indeed, $a$ is invertible in $R_q$ with probability about $1 - n/q$. Take $s_2 = s_1 - a^{-1}$, then you have $a s_1 - a s_2 = 1 mod q$ and the embedding norm of $1$ is $sqrt{n}$.
If you do not consider this $mod q$, i.e. you work in $R=mathbb Z[x]/⟨f(x)⟩$, then you are precisely asking for the minimal distance $lambda_1(mathfrak I)$ of the ideal lattice $mathfrak I$ generated by $a$. For such an ideal lattice, we can estimate rather precisely this minimal distance. A simple lower bound is
$lambda_1(mathfrak I) geq Delta_K^{1/2n} cdot N(a)^{1/n}$, where $N$ denotes the algebraic norm of $a$ (that is, the product of all its embeddings), and $Delta_K$ is the discriminant of field $K = mathbb Q(x)/(x^n+1)$. The reason is that the minimal vector $x$ must generate a subideal of $mathfrak I$, so $N(x) geq N(a)$, and $|x|^n geq Delta_K^{1/2} N(x)$. An upper bound is also given by Minkowski's theorem.
edited Nov 24 at 22:21
Ella Rose♦
15.1k44179
answered Nov 24 at 21:30
LeoDucas
43028
Shouldn't the embedding norm of 1 be 1?
– P.B.
Nov 25 at 0:26
Well, each embedding of 1 is one, so we are looking at the norm of (1, 1, ... 1), right ?
– LeoDucas
Nov 25 at 7:54
So this is the canonical embedding we are talking about right? In this case, we have that $||as_1-as_2||geq sqrt{n}$? Or are you just saying that there are $s_1,s_2$ such that $||as_1-as_2||= sqrt{n}$ but there could be others $r_1,r_2$ for which the differente between $ar_1$ and $ar_2$ is even lower?
– P.B.
Nov 25 at 13:36
There could be $0$ as well ;). BBut nothing in between indeed, at least for that particular ring.
– LeoDucas
Nov 25 at 16:26
Thank you for your help! Can you provide me some references about these facts? It would be very useful.
– P.B.
Nov 25 at 16:52
|
show 1 more comment
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I'm assuming $n$ is a power of $2$ and that $q$ is an odd prime larger than $n$. I'm discarding the trivial case $s_1 = s_2$.
If you consider everything $mod q$, then it is most likely over the choice of $a$ that there exists $s_1 neq s_2$ such that $|a s_1 - a s_2| = sqrt{n}$. Indeed, $a$ is invertible in $R_q$ with probability about $1 - n/q$. Take $s_2 = s_1 - a^{-1}$, then you have $a s_1 - a s_2 = 1 mod q$ and the embedding norm of $1$ is $sqrt{n}$.
If you do not consider this $mod q$, i.e. you work in $R=mathbb Z[x]/⟨f(x)⟩$, then you are precisely asking for the minimal distance $lambda_1(mathfrak I)$ of the ideal lattice $mathfrak I$ generated by $a$. For such an ideal lattice, we can estimate rather precisely this minimal distance. A simple lower bound is
$lambda_1(mathfrak I) geq Delta_K^{1/2n} cdot N(a)^{1/n}$, where $N$ denotes the algebraic norm of $a$ (that is, the product of all its embeddings), and $Delta_K$ is the discriminant of field $K = mathbb Q(x)/(x^n+1)$. The reason is that the minimal vector $x$ must generate a subideal of $mathfrak I$, so $N(x) geq N(a)$, and $|x|^n geq Delta_K^{1/2} N(x)$. An upper bound is also given by Minkowski's theorem.
edited Nov 24 at 22:21
Ella Rose♦
15.1k44179
answered Nov 24 at 21:30
LeoDucas
43028
Shouldn't the embedding norm of 1 be 1?
– P.B.
Nov 25 at 0:26
Well, each embedding of 1 is one, so we are looking at the norm of (1, 1, ... 1), right ?
– LeoDucas
Nov 25 at 7:54
So this is the canonical embedding we are talking about right? In this case, we have that $||as_1-as_2||geq sqrt{n}$? Or are you just saying that there are $s_1,s_2$ such that $||as_1-as_2||= sqrt{n}$ but there could be others $r_1,r_2$ for which the differente between $ar_1$ and $ar_2$ is even lower?
– P.B.
Nov 25 at 13:36
There could be $0$ as well ;). BBut nothing in between indeed, at least for that particular ring.
– LeoDucas
Nov 25 at 16:26
Thank you for your help! Can you provide me some references about these facts? It would be very useful.
– P.B.
Nov 25 at 16:52
|
show 1 more comment
I'm assuming $n$ is a power of $2$ and that $q$ is an odd prime larger than $n$. I'm discarding the trivial case $s_1 = s_2$.
If you consider everything $mod q$, then it is most likely over the choice of $a$ that there exists $s_1 neq s_2$ such that $|a s_1 - a s_2| = sqrt{n}$. Indeed, $a$ is invertible in $R_q$ with probability about $1 - n/q$. Take $s_2 = s_1 - a^{-1}$, then you have $a s_1 - a s_2 = 1 mod q$ and the embedding norm of $1$ is $sqrt{n}$.
If you do not consider this $mod q$, i.e. you work in $R=mathbb Z[x]/⟨f(x)⟩$, then you are precisely asking for the minimal distance $lambda_1(mathfrak I)$ of the ideal lattice $mathfrak I$ generated by $a$. For such an ideal lattice, we can estimate rather precisely this minimal distance. A simple lower bound is
$lambda_1(mathfrak I) geq Delta_K^{1/2n} cdot N(a)^{1/n}$, where $N$ denotes the algebraic norm of $a$ (that is, the product of all its embeddings), and $Delta_K$ is the discriminant of field $K = mathbb Q(x)/(x^n+1)$. The reason is that the minimal vector $x$ must generate a subideal of $mathfrak I$, so $N(x) geq N(a)$, and $|x|^n geq Delta_K^{1/2} N(x)$. An upper bound is also given by Minkowski's theorem.
edited Nov 24 at 22:21
Ella Rose♦
15.1k44179
answered Nov 24 at 21:30
LeoDucas
43028
Shouldn't the embedding norm of 1 be 1?
– P.B.
Nov 25 at 0:26
Well, each embedding of 1 is one, so we are looking at the norm of (1, 1, ... 1), right ?
– LeoDucas
Nov 25 at 7:54
So this is the canonical embedding we are talking about right? In this case, we have that $||as_1-as_2||geq sqrt{n}$? Or are you just saying that there are $s_1,s_2$ such that $||as_1-as_2||= sqrt{n}$ but there could be others $r_1,r_2$ for which the differente between $ar_1$ and $ar_2$ is even lower?
– P.B.
Nov 25 at 13:36
There could be $0$ as well ;). BBut nothing in between indeed, at least for that particular ring.
– LeoDucas
Nov 25 at 16:26
Thank you for your help! Can you provide me some references about these facts? It would be very useful.
– P.B.
Nov 25 at 16:52
|
show 1 more comment
I'm assuming $n$ is a power of $2$ and that $q$ is an odd prime larger than $n$. I'm discarding the trivial case $s_1 = s_2$.
If you consider everything $mod q$, then it is most likely over the choice of $a$ that there exists $s_1 neq s_2$ such that $|a s_1 - a s_2| = sqrt{n}$. Indeed, $a$ is invertible in $R_q$ with probability about $1 - n/q$. Take $s_2 = s_1 - a^{-1}$, then you have $a s_1 - a s_2 = 1 mod q$ and the embedding norm of $1$ is $sqrt{n}$.
If you do not consider this $mod q$, i.e. you work in $R=mathbb Z[x]/⟨f(x)⟩$, then you are precisely asking for the minimal distance $lambda_1(mathfrak I)$ of the ideal lattice $mathfrak I$ generated by $a$. For such an ideal lattice, we can estimate rather precisely this minimal distance. A simple lower bound is
$lambda_1(mathfrak I) geq Delta_K^{1/2n} cdot N(a)^{1/n}$, where $N$ denotes the algebraic norm of $a$ (that is, the product of all its embeddings), and $Delta_K$ is the discriminant of field $K = mathbb Q(x)/(x^n+1)$. The reason is that the minimal vector $x$ must generate a subideal of $mathfrak I$, so $N(x) geq N(a)$, and $|x|^n geq Delta_K^{1/2} N(x)$. An upper bound is also given by Minkowski's theorem.
edited Nov 24 at 22:21
Ella Rose♦
15.1k44179
answered Nov 24 at 21:30
LeoDucas
43028
I'm assuming $n$ is a power of $2$ and that $q$ is an odd prime larger than $n$. I'm discarding the trivial case $s_1 = s_2$.
If you consider everything $mod q$, then it is most likely over the choice of $a$ that there exists $s_1 neq s_2$ such that $|a s_1 - a s_2| = sqrt{n}$. Indeed, $a$ is invertible in $R_q$ with probability about $1 - n/q$. Take $s_2 = s_1 - a^{-1}$, then you have $a s_1 - a s_2 = 1 mod q$ and the embedding norm of $1$ is $sqrt{n}$.
If you do not consider this $mod q$, i.e. you work in $R=mathbb Z[x]/⟨f(x)⟩$, then you are precisely asking for the minimal distance $lambda_1(mathfrak I)$ of the ideal lattice $mathfrak I$ generated by $a$. For such an ideal lattice, we can estimate rather precisely this minimal distance. A simple lower bound is
$lambda_1(mathfrak I) geq Delta_K^{1/2n} cdot N(a)^{1/n}$, where $N$ denotes the algebraic norm of $a$ (that is, the product of all its embeddings), and $Delta_K$ is the discriminant of field $K = mathbb Q(x)/(x^n+1)$. The reason is that the minimal vector $x$ must generate a subideal of $mathfrak I$, so $N(x) geq N(a)$, and $|x|^n geq Delta_K^{1/2} N(x)$. An upper bound is also given by Minkowski's theorem.
edited Nov 24 at 22:21
Ella Rose♦
15.1k44179
answered Nov 24 at 21:30
LeoDucas
43028
edited Nov 24 at 22:21
Ella Rose♦
15.1k44179
edited Nov 24 at 22:21
Ella Rose♦
15.1k44179
edited Nov 24 at 22:21
Ella Rose♦
15.1k44179
15.1k44179
answered Nov 24 at 21:30
LeoDucas
43028
answered Nov 24 at 21:30
LeoDucas
43028
answered Nov 24 at 21:30
LeoDucas
43028
43028
Shouldn't the embedding norm of 1 be 1?
– P.B.
Nov 25 at 0:26
Well, each embedding of 1 is one, so we are looking at the norm of (1, 1, ... 1), right ?
– LeoDucas
Nov 25 at 7:54
So this is the canonical embedding we are talking about right? In this case, we have that $||as_1-as_2||geq sqrt{n}$? Or are you just saying that there are $s_1,s_2$ such that $||as_1-as_2||= sqrt{n}$ but there could be others $r_1,r_2$ for which the differente between $ar_1$ and $ar_2$ is even lower?
– P.B.
Nov 25 at 13:36
There could be $0$ as well ;). BBut nothing in between indeed, at least for that particular ring.
– LeoDucas
Nov 25 at 16:26
Thank you for your help! Can you provide me some references about these facts? It would be very useful.
– P.B.
Nov 25 at 16:52
|
show 1 more comment
Shouldn't the embedding norm of 1 be 1?
– P.B.
Nov 25 at 0:26
Well, each embedding of 1 is one, so we are looking at the norm of (1, 1, ... 1), right ?
– LeoDucas
Nov 25 at 7:54
So this is the canonical embedding we are talking about right? In this case, we have that $||as_1-as_2||geq sqrt{n}$? Or are you just saying that there are $s_1,s_2$ such that $||as_1-as_2||= sqrt{n}$ but there could be others $r_1,r_2$ for which the differente between $ar_1$ and $ar_2$ is even lower?
– P.B.
Nov 25 at 13:36
There could be $0$ as well ;). BBut nothing in between indeed, at least for that particular ring.
– LeoDucas
Nov 25 at 16:26
Thank you for your help! Can you provide me some references about these facts? It would be very useful.
– P.B.
Nov 25 at 16:52
Shouldn't the embedding norm of 1 be 1?
– P.B.
Nov 25 at 0:26
Shouldn't the embedding norm of 1 be 1?
– P.B.
Nov 25 at 0:26
Well, each embedding of 1 is one, so we are looking at the norm of (1, 1, ... 1), right ?
– LeoDucas
Nov 25 at 7:54
Well, each embedding of 1 is one, so we are looking at the norm of (1, 1, ... 1), right ?
– LeoDucas
Nov 25 at 7:54
So this is the canonical embedding we are talking about right? In this case, we have that $||as_1-as_2||geq sqrt{n}$? Or are you just saying that there are $s_1,s_2$ such that $||as_1-as_2||= sqrt{n}$ but there could be others $r_1,r_2$ for which the differente between $ar_1$ and $ar_2$ is even lower?
– P.B.
Nov 25 at 13:36
So this is the canonical embedding we are talking about right? In this case, we have that $||as_1-as_2||geq sqrt{n}$? Or are you just saying that there are $s_1,s_2$ such that $||as_1-as_2||= sqrt{n}$ but there could be others $r_1,r_2$ for which the differente between $ar_1$ and $ar_2$ is even lower?
– P.B.
Nov 25 at 13:36
There could be $0$ as well ;). BBut nothing in between indeed, at least for that particular ring.
– LeoDucas
Nov 25 at 16:26
There could be $0$ as well ;). BBut nothing in between indeed, at least for that particular ring.
– LeoDucas
Nov 25 at 16:26
Thank you for your help! Can you provide me some references about these facts? It would be very useful.
– P.B.
Nov 25 at 16:52
Thank you for your help! Can you provide me some references about these facts? It would be very useful.
– P.B.
Nov 25 at 16:52
|
show 1 more comment
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Hello. It is a good question, but the $L_2$ norm is defined over vectors and it is not clear how you are embedding the polynomials in a vector space. Are you just representing the polynomials as vectors with their coefficients? (So, for instance, $2x^3 -1$ becomes the vector $(2, 0, 0, -1)$).
– Hilder Vítor Lima Pereira
Nov 24 at 17:19
Yes, I am thinking of the canonical embedding
– P.B.
Nov 24 at 17:26
Well, the canonical embedding is the one that uses isomorphisms to embed the polynomials. The one I've described is the coefficient embedding...
– Hilder Vítor Lima Pereira
Nov 24 at 17:31
Sorry. I mean the coefficient embedding then
– P.B.
Nov 24 at 17:46
How do you define "negligible probability" in this case?
– kodlu
Nov 24 at 21:19