reversible modulo equation
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I came across an equation regarding some cryptography article that said for each $i$
$z_i= mod( lfloor(s_i*f)rfloor+p_i, f)$ is reversible, i.e given $f, z,s$ we can get back $p$. My question is I doubt that this equation is invertible, but if at all this equation is invertible what makes it invertible, because there is a modulo operation involved and a flooring function which are not one-one
number-theory modular-arithmetic cryptography
asked Dec 23 '18 at 14:01
UpstartUpstart
1,706517
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add a comment |
$begingroup$
I came across an equation regarding some cryptography article that said for each $i$
$z_i= mod( lfloor(s_i*f)rfloor+p_i, f)$ is reversible, i.e given $f, z,s$ we can get back $p$. My question is I doubt that this equation is invertible, but if at all this equation is invertible what makes it invertible, because there is a modulo operation involved and a flooring function which are not one-one
number-theory modular-arithmetic cryptography
asked Dec 23 '18 at 14:01
UpstartUpstart
1,706517
$endgroup$
$begingroup$
Maybe the article means this: although the $p$ you get back may not be unique, the uniqueness does not matter, i.e., the $p$ you obtain is unique up to this modulo operation; and this is sufficient for the cryptography to work. This is my hunch.
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– user122049
Dec 23 '18 at 14:06
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But how do you get back `a' $p$
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– Upstart
Dec 23 '18 at 14:10
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What are the ranges/domains of of your functions/variables? What is the excact statement of the claim in the article?
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– gammatester
Dec 23 '18 at 14:42
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$f=8$, $0leq p_ileq 255,~forall~ i$, $0leq s_ileq 1$
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– Upstart
Dec 23 '18 at 14:59
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Consider operations in $Z/fZ$ simply
$endgroup$
– Damien
Dec 23 '18 at 17:58
add a comment |
$begingroup$
I came across an equation regarding some cryptography article that said for each $i$
$z_i= mod( lfloor(s_i*f)rfloor+p_i, f)$ is reversible, i.e given $f, z,s$ we can get back $p$. My question is I doubt that this equation is invertible, but if at all this equation is invertible what makes it invertible, because there is a modulo operation involved and a flooring function which are not one-one
number-theory modular-arithmetic cryptography
asked Dec 23 '18 at 14:01
UpstartUpstart
1,706517
$endgroup$
I came across an equation regarding some cryptography article that said for each $i$
$z_i= mod( lfloor(s_i*f)rfloor+p_i, f)$ is reversible, i.e given $f, z,s$ we can get back $p$. My question is I doubt that this equation is invertible, but if at all this equation is invertible what makes it invertible, because there is a modulo operation involved and a flooring function which are not one-one
number-theory modular-arithmetic cryptography
number-theory modular-arithmetic cryptography
asked Dec 23 '18 at 14:01
UpstartUpstart
1,706517
asked Dec 23 '18 at 14:01
UpstartUpstart
1,706517
asked Dec 23 '18 at 14:01
UpstartUpstart
1,706517
asked Dec 23 '18 at 14:01
UpstartUpstart
1,706517
asked Dec 23 '18 at 14:01
UpstartUpstart
1,706517
1,706517
$begingroup$
Maybe the article means this: although the $p$ you get back may not be unique, the uniqueness does not matter, i.e., the $p$ you obtain is unique up to this modulo operation; and this is sufficient for the cryptography to work. This is my hunch.
$endgroup$
– user122049
Dec 23 '18 at 14:06
$begingroup$
But how do you get back `a' $p$
$endgroup$
– Upstart
Dec 23 '18 at 14:10
$begingroup$
What are the ranges/domains of of your functions/variables? What is the excact statement of the claim in the article?
$endgroup$
– gammatester
Dec 23 '18 at 14:42
$begingroup$
$f=8$, $0leq p_ileq 255,~forall~ i$, $0leq s_ileq 1$
$endgroup$
– Upstart
Dec 23 '18 at 14:59
$begingroup$
Consider operations in $Z/fZ$ simply
$endgroup$
– Damien
Dec 23 '18 at 17:58
add a comment |
$begingroup$
Maybe the article means this: although the $p$ you get back may not be unique, the uniqueness does not matter, i.e., the $p$ you obtain is unique up to this modulo operation; and this is sufficient for the cryptography to work. This is my hunch.
$endgroup$
– user122049
Dec 23 '18 at 14:06
$begingroup$
But how do you get back `a' $p$
$endgroup$
– Upstart
Dec 23 '18 at 14:10
$begingroup$
What are the ranges/domains of of your functions/variables? What is the excact statement of the claim in the article?
$endgroup$
– gammatester
Dec 23 '18 at 14:42
$begingroup$
$f=8$, $0leq p_ileq 255,~forall~ i$, $0leq s_ileq 1$
$endgroup$
– Upstart
Dec 23 '18 at 14:59
$begingroup$
Consider operations in $Z/fZ$ simply
$endgroup$
– Damien
Dec 23 '18 at 17:58
$begingroup$
Maybe the article means this: although the $p$ you get back may not be unique, the uniqueness does not matter, i.e., the $p$ you obtain is unique up to this modulo operation; and this is sufficient for the cryptography to work. This is my hunch.
$endgroup$
– user122049
Dec 23 '18 at 14:06
$begingroup$
Maybe the article means this: although the $p$ you get back may not be unique, the uniqueness does not matter, i.e., the $p$ you obtain is unique up to this modulo operation; and this is sufficient for the cryptography to work. This is my hunch.
$endgroup$
– user122049
Dec 23 '18 at 14:06
$begingroup$
But how do you get back `a' $p$
$endgroup$
– Upstart
Dec 23 '18 at 14:10
$begingroup$
But how do you get back `a' $p$
$endgroup$
– Upstart
Dec 23 '18 at 14:10
$begingroup$
What are the ranges/domains of of your functions/variables? What is the excact statement of the claim in the article?
$endgroup$
– gammatester
Dec 23 '18 at 14:42
$begingroup$
What are the ranges/domains of of your functions/variables? What is the excact statement of the claim in the article?
$endgroup$
– gammatester
Dec 23 '18 at 14:42
$begingroup$
$f=8$, $0leq p_ileq 255,~forall~ i$, $0leq s_ileq 1$
$endgroup$
– Upstart
Dec 23 '18 at 14:59
$begingroup$
$f=8$, $0leq p_ileq 255,~forall~ i$, $0leq s_ileq 1$
$endgroup$
– Upstart
Dec 23 '18 at 14:59
$begingroup$
Consider operations in $Z/fZ$ simply
$endgroup$
– Damien
Dec 23 '18 at 17:58
$begingroup$
Consider operations in $Z/fZ$ simply
$endgroup$
– Damien
Dec 23 '18 at 17:58
add a comment |
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$begingroup$
Maybe the article means this: although the $p$ you get back may not be unique, the uniqueness does not matter, i.e., the $p$ you obtain is unique up to this modulo operation; and this is sufficient for the cryptography to work. This is my hunch.
$endgroup$
– user122049
Dec 23 '18 at 14:06
$begingroup$
But how do you get back `a' $p$
$endgroup$
– Upstart
Dec 23 '18 at 14:10
$begingroup$
What are the ranges/domains of of your functions/variables? What is the excact statement of the claim in the article?
$endgroup$
– gammatester
Dec 23 '18 at 14:42
$begingroup$
$f=8$, $0leq p_ileq 255,~forall~ i$, $0leq s_ileq 1$
$endgroup$
– Upstart
Dec 23 '18 at 14:59
$begingroup$
Consider operations in $Z/fZ$ simply
$endgroup$
– Damien
Dec 23 '18 at 17:58