What does “constant rate” mean in universal composable commitment scheme?
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I'm wondering what does the "constant rate" mean in universal composable commitment scheme? I have known the rate of a commitment scheme is message length divided by the communication complexity of the scheme. What's the "constant" mean here? Must the constant be a number less than 1?
commitments
asked Dec 18 '18 at 17:14
CryptoLoverCryptoLover
402212
$endgroup$
add a comment |
$begingroup$
I'm wondering what does the "constant rate" mean in universal composable commitment scheme? I have known the rate of a commitment scheme is message length divided by the communication complexity of the scheme. What's the "constant" mean here? Must the constant be a number less than 1?
commitments
asked Dec 18 '18 at 17:14
CryptoLoverCryptoLover
402212
$endgroup$
$begingroup$
Can you please post a reference to a paper or something where you encountered this term?
$endgroup$
– SEJPM♦
Dec 18 '18 at 17:21
add a comment |
$begingroup$
I'm wondering what does the "constant rate" mean in universal composable commitment scheme? I have known the rate of a commitment scheme is message length divided by the communication complexity of the scheme. What's the "constant" mean here? Must the constant be a number less than 1?
commitments
asked Dec 18 '18 at 17:14
CryptoLoverCryptoLover
402212
$endgroup$
I'm wondering what does the "constant rate" mean in universal composable commitment scheme? I have known the rate of a commitment scheme is message length divided by the communication complexity of the scheme. What's the "constant" mean here? Must the constant be a number less than 1?
commitments
commitments
asked Dec 18 '18 at 17:14
CryptoLoverCryptoLover
402212
asked Dec 18 '18 at 17:14
CryptoLoverCryptoLover
402212
asked Dec 18 '18 at 17:14
CryptoLoverCryptoLover
402212
asked Dec 18 '18 at 17:14
CryptoLoverCryptoLover
402212
asked Dec 18 '18 at 17:14
CryptoLoverCryptoLover
402212
402212
$begingroup$
Can you please post a reference to a paper or something where you encountered this term?
$endgroup$
– SEJPM♦
Dec 18 '18 at 17:21
add a comment |
$begingroup$
Can you please post a reference to a paper or something where you encountered this term?
$endgroup$
– SEJPM♦
Dec 18 '18 at 17:21
$begingroup$
Can you please post a reference to a paper or something where you encountered this term?
$endgroup$
– SEJPM♦
Dec 18 '18 at 17:21
$begingroup$
Can you please post a reference to a paper or something where you encountered this term?
$endgroup$
– SEJPM♦
Dec 18 '18 at 17:21
add a comment |
1 Answer
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Constant rate in general means that the overhead from a non-secure method is constant. So, in a simple way, if I am committing to an $ell$-bit message, then the size of the commitment is $O(ell)$. In some cases, however, one also allows an additive factor that is independent of the message size. Thus, for example, it could be that to commit to a message of size $ell$ the amount is $O(ell)+{rm poly}(n)$, where $n$ is the security parameter.
Note that typically these are measured in an amortized manner. So, you have to send many commitments (or a long message) for it to be true. But, again, this depends on the exact scheme, so you'll have to read the details.
answered Dec 18 '18 at 17:34
Yehuda LindellYehuda Lindell
18.6k3661
$endgroup$
$begingroup$
Thanks for your answer. I have one more question: how large the size of the commitment is would be regarded as bad? For example, if the size of the commitment is O(nl) for committing l-bit message, where n is security parameter, is it bad?
$endgroup$
– CryptoLover
Jan 10 at 16:50
$begingroup$
Good or bad depends on your setting and what you want. In general, $O(ncdotell)$ is of course much worse than $O(ell)+poly(n)$, but if you want to commit to a little bit only, sometimes a simpler scheme is better.
$endgroup$
– Yehuda Lindell
Jan 11 at 1:23
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Constant rate in general means that the overhead from a non-secure method is constant. So, in a simple way, if I am committing to an $ell$-bit message, then the size of the commitment is $O(ell)$. In some cases, however, one also allows an additive factor that is independent of the message size. Thus, for example, it could be that to commit to a message of size $ell$ the amount is $O(ell)+{rm poly}(n)$, where $n$ is the security parameter.
Note that typically these are measured in an amortized manner. So, you have to send many commitments (or a long message) for it to be true. But, again, this depends on the exact scheme, so you'll have to read the details.
answered Dec 18 '18 at 17:34
Yehuda LindellYehuda Lindell
18.6k3661
$endgroup$
$begingroup$
Thanks for your answer. I have one more question: how large the size of the commitment is would be regarded as bad? For example, if the size of the commitment is O(nl) for committing l-bit message, where n is security parameter, is it bad?
$endgroup$
– CryptoLover
Jan 10 at 16:50
$begingroup$
Good or bad depends on your setting and what you want. In general, $O(ncdotell)$ is of course much worse than $O(ell)+poly(n)$, but if you want to commit to a little bit only, sometimes a simpler scheme is better.
$endgroup$
– Yehuda Lindell
Jan 11 at 1:23
add a comment |
$begingroup$
Constant rate in general means that the overhead from a non-secure method is constant. So, in a simple way, if I am committing to an $ell$-bit message, then the size of the commitment is $O(ell)$. In some cases, however, one also allows an additive factor that is independent of the message size. Thus, for example, it could be that to commit to a message of size $ell$ the amount is $O(ell)+{rm poly}(n)$, where $n$ is the security parameter.
Note that typically these are measured in an amortized manner. So, you have to send many commitments (or a long message) for it to be true. But, again, this depends on the exact scheme, so you'll have to read the details.
answered Dec 18 '18 at 17:34
Yehuda LindellYehuda Lindell
18.6k3661
$endgroup$
$begingroup$
Thanks for your answer. I have one more question: how large the size of the commitment is would be regarded as bad? For example, if the size of the commitment is O(nl) for committing l-bit message, where n is security parameter, is it bad?
$endgroup$
– CryptoLover
Jan 10 at 16:50
$begingroup$
Good or bad depends on your setting and what you want. In general, $O(ncdotell)$ is of course much worse than $O(ell)+poly(n)$, but if you want to commit to a little bit only, sometimes a simpler scheme is better.
$endgroup$
– Yehuda Lindell
Jan 11 at 1:23
add a comment |
$begingroup$
Constant rate in general means that the overhead from a non-secure method is constant. So, in a simple way, if I am committing to an $ell$-bit message, then the size of the commitment is $O(ell)$. In some cases, however, one also allows an additive factor that is independent of the message size. Thus, for example, it could be that to commit to a message of size $ell$ the amount is $O(ell)+{rm poly}(n)$, where $n$ is the security parameter.
Note that typically these are measured in an amortized manner. So, you have to send many commitments (or a long message) for it to be true. But, again, this depends on the exact scheme, so you'll have to read the details.
answered Dec 18 '18 at 17:34
Yehuda LindellYehuda Lindell
18.6k3661
$endgroup$
Constant rate in general means that the overhead from a non-secure method is constant. So, in a simple way, if I am committing to an $ell$-bit message, then the size of the commitment is $O(ell)$. In some cases, however, one also allows an additive factor that is independent of the message size. Thus, for example, it could be that to commit to a message of size $ell$ the amount is $O(ell)+{rm poly}(n)$, where $n$ is the security parameter.
Note that typically these are measured in an amortized manner. So, you have to send many commitments (or a long message) for it to be true. But, again, this depends on the exact scheme, so you'll have to read the details.
answered Dec 18 '18 at 17:34
Yehuda LindellYehuda Lindell
18.6k3661
answered Dec 18 '18 at 17:34
Yehuda LindellYehuda Lindell
18.6k3661
answered Dec 18 '18 at 17:34
Yehuda LindellYehuda Lindell
18.6k3661
answered Dec 18 '18 at 17:34
Yehuda LindellYehuda Lindell
18.6k3661
18.6k3661
$begingroup$
Thanks for your answer. I have one more question: how large the size of the commitment is would be regarded as bad? For example, if the size of the commitment is O(nl) for committing l-bit message, where n is security parameter, is it bad?
$endgroup$
– CryptoLover
Jan 10 at 16:50
$begingroup$
Good or bad depends on your setting and what you want. In general, $O(ncdotell)$ is of course much worse than $O(ell)+poly(n)$, but if you want to commit to a little bit only, sometimes a simpler scheme is better.
$endgroup$
– Yehuda Lindell
Jan 11 at 1:23
add a comment |
$begingroup$
Thanks for your answer. I have one more question: how large the size of the commitment is would be regarded as bad? For example, if the size of the commitment is O(nl) for committing l-bit message, where n is security parameter, is it bad?
$endgroup$
– CryptoLover
Jan 10 at 16:50
$begingroup$
Good or bad depends on your setting and what you want. In general, $O(ncdotell)$ is of course much worse than $O(ell)+poly(n)$, but if you want to commit to a little bit only, sometimes a simpler scheme is better.
$endgroup$
– Yehuda Lindell
Jan 11 at 1:23
$begingroup$
Thanks for your answer. I have one more question: how large the size of the commitment is would be regarded as bad? For example, if the size of the commitment is O(nl) for committing l-bit message, where n is security parameter, is it bad?
$endgroup$
– CryptoLover
Jan 10 at 16:50
$begingroup$
Thanks for your answer. I have one more question: how large the size of the commitment is would be regarded as bad? For example, if the size of the commitment is O(nl) for committing l-bit message, where n is security parameter, is it bad?
$endgroup$
– CryptoLover
Jan 10 at 16:50
$begingroup$
Good or bad depends on your setting and what you want. In general, $O(ncdotell)$ is of course much worse than $O(ell)+poly(n)$, but if you want to commit to a little bit only, sometimes a simpler scheme is better.
$endgroup$
– Yehuda Lindell
Jan 11 at 1:23
$begingroup$
Good or bad depends on your setting and what you want. In general, $O(ncdotell)$ is of course much worse than $O(ell)+poly(n)$, but if you want to commit to a little bit only, sometimes a simpler scheme is better.
$endgroup$
– Yehuda Lindell
Jan 11 at 1:23
add a comment |
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$begingroup$
Can you please post a reference to a paper or something where you encountered this term?
$endgroup$
– SEJPM♦
Dec 18 '18 at 17:21