Mixing
Distribution of the math operations results with variables from different sets.
Clash Royale CLAN TAG#URR8PPP
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There is a simple equation $ax = bc$, where $a$ and $b$ are sampled uniformly at random from the space of all possible $k$-bits values $mathcalK$. Also, we know, that $c$ is sampled uniformly at random from the space of all possible $j$-bits values $mathcalJ$.
What can be said about $x$ distribution?
I'm stuck. Intuitively it seems that $x$ should be a random value from $mathcalK$ space. However, I don't know how to prove this or how to modify the equation to make it true. I would appreciate any ideas and hints.
Here is my thought process:
The original equation I'm trying to solve is from modular arithmetic $G = g^axbmod p = g^bcbmod p$, where $g$ is a generator of a cyclic group $mathbbG$ of a prime order $q$ and $p=2q+1$ is also prime.
So, I have fixed $G=g^bcbmod p$ and want to find $x$ such that $G$ can be also computed as $g^axbmod p$. Log problem is hard, but we know $a,b,c$, so we can find $x$ such that $ax = bc$.
Observation 1: $a,b$ are $k$-bits outputs of a secure $PRF$, which implies that there is no efficient algorithm can distinguish this output and a truly random one. Since $c$ is also chosen uniformly at random, $x = fracbca$ is random.
Observation 2: By definition of the bit size of an integer (at least, a prime in a cryptographic context) $2^k-1 leq a < 2^k$, also $2^k-1 leq b < 2^k$ and $2^j-1 leq c < 2^j$. Thus, $ 2^k-1 + j-1 - (k-1) leq fracbca < 2^k +j - k$, which means $2^j-1 leq x < 2^j$.
This implies that a) $x$ is $j$-bits number b) $x$ is random since $fracbca$ is random. Though, I'm not sure it couns as a proof of $x$ being uniform and random in space $mathcalJ$.
uniform-distribution
edited Jul 25 at 12:00
TZakrevskiy
19.8k12253
asked Jul 24 at 12:58
pintor
11
add a comment |Â
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There is a simple equation $ax = bc$, where $a$ and $b$ are sampled uniformly at random from the space of all possible $k$-bits values $mathcalK$. Also, we know, that $c$ is sampled uniformly at random from the space of all possible $j$-bits values $mathcalJ$.
What can be said about $x$ distribution?
I'm stuck. Intuitively it seems that $x$ should be a random value from $mathcalK$ space. However, I don't know how to prove this or how to modify the equation to make it true. I would appreciate any ideas and hints.
Here is my thought process:
The original equation I'm trying to solve is from modular arithmetic $G = g^axbmod p = g^bcbmod p$, where $g$ is a generator of a cyclic group $mathbbG$ of a prime order $q$ and $p=2q+1$ is also prime.
So, I have fixed $G=g^bcbmod p$ and want to find $x$ such that $G$ can be also computed as $g^axbmod p$. Log problem is hard, but we know $a,b,c$, so we can find $x$ such that $ax = bc$.
Observation 1: $a,b$ are $k$-bits outputs of a secure $PRF$, which implies that there is no efficient algorithm can distinguish this output and a truly random one. Since $c$ is also chosen uniformly at random, $x = fracbca$ is random.
Observation 2: By definition of the bit size of an integer (at least, a prime in a cryptographic context) $2^k-1 leq a < 2^k$, also $2^k-1 leq b < 2^k$ and $2^j-1 leq c < 2^j$. Thus, $ 2^k-1 + j-1 - (k-1) leq fracbca < 2^k +j - k$, which means $2^j-1 leq x < 2^j$.
This implies that a) $x$ is $j$-bits number b) $x$ is random since $fracbca$ is random. Though, I'm not sure it couns as a proof of $x$ being uniform and random in space $mathcalJ$.
uniform-distribution
edited Jul 25 at 12:00
TZakrevskiy
19.8k12253
asked Jul 24 at 12:58
pintor
11
@JoséCarlosSantos done.
– pintor
Jul 24 at 13:36
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
There is a simple equation $ax = bc$, where $a$ and $b$ are sampled uniformly at random from the space of all possible $k$-bits values $mathcalK$. Also, we know, that $c$ is sampled uniformly at random from the space of all possible $j$-bits values $mathcalJ$.
What can be said about $x$ distribution?
I'm stuck. Intuitively it seems that $x$ should be a random value from $mathcalK$ space. However, I don't know how to prove this or how to modify the equation to make it true. I would appreciate any ideas and hints.
Here is my thought process:
The original equation I'm trying to solve is from modular arithmetic $G = g^axbmod p = g^bcbmod p$, where $g$ is a generator of a cyclic group $mathbbG$ of a prime order $q$ and $p=2q+1$ is also prime.
So, I have fixed $G=g^bcbmod p$ and want to find $x$ such that $G$ can be also computed as $g^axbmod p$. Log problem is hard, but we know $a,b,c$, so we can find $x$ such that $ax = bc$.
Observation 1: $a,b$ are $k$-bits outputs of a secure $PRF$, which implies that there is no efficient algorithm can distinguish this output and a truly random one. Since $c$ is also chosen uniformly at random, $x = fracbca$ is random.
Observation 2: By definition of the bit size of an integer (at least, a prime in a cryptographic context) $2^k-1 leq a < 2^k$, also $2^k-1 leq b < 2^k$ and $2^j-1 leq c < 2^j$. Thus, $ 2^k-1 + j-1 - (k-1) leq fracbca < 2^k +j - k$, which means $2^j-1 leq x < 2^j$.
This implies that a) $x$ is $j$-bits number b) $x$ is random since $fracbca$ is random. Though, I'm not sure it couns as a proof of $x$ being uniform and random in space $mathcalJ$.
uniform-distribution
edited Jul 25 at 12:00
TZakrevskiy
19.8k12253
asked Jul 24 at 12:58
pintor
11
There is a simple equation $ax = bc$, where $a$ and $b$ are sampled uniformly at random from the space of all possible $k$-bits values $mathcalK$. Also, we know, that $c$ is sampled uniformly at random from the space of all possible $j$-bits values $mathcalJ$.
What can be said about $x$ distribution?
I'm stuck. Intuitively it seems that $x$ should be a random value from $mathcalK$ space. However, I don't know how to prove this or how to modify the equation to make it true. I would appreciate any ideas and hints.
Here is my thought process:
The original equation I'm trying to solve is from modular arithmetic $G = g^axbmod p = g^bcbmod p$, where $g$ is a generator of a cyclic group $mathbbG$ of a prime order $q$ and $p=2q+1$ is also prime.
So, I have fixed $G=g^bcbmod p$ and want to find $x$ such that $G$ can be also computed as $g^axbmod p$. Log problem is hard, but we know $a,b,c$, so we can find $x$ such that $ax = bc$.
Observation 1: $a,b$ are $k$-bits outputs of a secure $PRF$, which implies that there is no efficient algorithm can distinguish this output and a truly random one. Since $c$ is also chosen uniformly at random, $x = fracbca$ is random.
Observation 2: By definition of the bit size of an integer (at least, a prime in a cryptographic context) $2^k-1 leq a < 2^k$, also $2^k-1 leq b < 2^k$ and $2^j-1 leq c < 2^j$. Thus, $ 2^k-1 + j-1 - (k-1) leq fracbca < 2^k +j - k$, which means $2^j-1 leq x < 2^j$.
This implies that a) $x$ is $j$-bits number b) $x$ is random since $fracbca$ is random. Though, I'm not sure it couns as a proof of $x$ being uniform and random in space $mathcalJ$.
uniform-distribution
edited Jul 25 at 12:00
TZakrevskiy
19.8k12253
asked Jul 24 at 12:58
pintor
11
edited Jul 25 at 12:00
TZakrevskiy
19.8k12253
edited Jul 25 at 12:00
TZakrevskiy
19.8k12253
edited Jul 25 at 12:00
TZakrevskiy
19.8k12253
19.8k12253
asked Jul 24 at 12:58
pintor
11
asked Jul 24 at 12:58
pintor
11
asked Jul 24 at 12:58
pintor
11
11
@JoséCarlosSantos done.
– pintor
Jul 24 at 13:36
add a comment |Â
@JoséCarlosSantos done.
– pintor
Jul 24 at 13:36
@JoséCarlosSantos done.
– pintor
Jul 24 at 13:36
@JoséCarlosSantos done.
– pintor
Jul 24 at 13:36
add a comment |Â
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@JoséCarlosSantos done.
– pintor
Jul 24 at 13:36