Constructions of PRF (Pseudo Random Function)The Goldreich-Goldwasser-Micali Construction with bad PRGSSecurity of KDF1 and KDF2 (hash based KDF's)Implementing a pseudo random function in practiceCryptanalysis of Marvin32 compared to SipHashPseudo Random FunctionWhy is this function pseudo random (PRF)?Decentralized consent over a random numberExistence of PRF $implies$ existence of PRGDoes a distinguisher for an PRF based on a hash make the hash function insecure?Can we convert a pseudorandom function (PRF) to an Oblivious PRF (OPRF) through an Oblivious Transfer (OT) protocol?
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Constructions of PRF (Pseudo Random Function)
The Goldreich-Goldwasser-Micali Construction with bad PRGSSecurity of KDF1 and KDF2 (hash based KDF's)Implementing a pseudo random function in practiceCryptanalysis of Marvin32 compared to SipHashPseudo Random FunctionWhy is this function pseudo random (PRF)?Decentralized consent over a random numberExistence of PRF $implies$ existence of PRGDoes a distinguisher for an PRF based on a hash make the hash function insecure?Can we convert a pseudorandom function (PRF) to an Oblivious PRF (OPRF) through an Oblivious Transfer (OT) protocol?
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I was taught only GGM based PRF construction in class. It's very inefficient. I am just curious about various PRF constructions from standard assumptions. Please provide a few PRF constructions from various assumptions.
pseudo-random-generator pseudo-random-function
$endgroup$
add a comment |
$begingroup$
I was taught only GGM based PRF construction in class. It's very inefficient. I am just curious about various PRF constructions from standard assumptions. Please provide a few PRF constructions from various assumptions.
pseudo-random-generator pseudo-random-function
$endgroup$
$begingroup$
Well, one standard assumption is that the SHA-256 compression function is a PRF, from which we can conclude that HMAC-SHA256 is a PRF (and a reasonably efficient one at that), but maybe you meant to restrict the domain of ‘standard assumptions’?
$endgroup$
– Squeamish Ossifrage
3 hours ago
add a comment |
$begingroup$
I was taught only GGM based PRF construction in class. It's very inefficient. I am just curious about various PRF constructions from standard assumptions. Please provide a few PRF constructions from various assumptions.
pseudo-random-generator pseudo-random-function
$endgroup$
I was taught only GGM based PRF construction in class. It's very inefficient. I am just curious about various PRF constructions from standard assumptions. Please provide a few PRF constructions from various assumptions.
pseudo-random-generator pseudo-random-function
pseudo-random-generator pseudo-random-function
asked 4 hours ago
satyasatya
441317
441317
$begingroup$
Well, one standard assumption is that the SHA-256 compression function is a PRF, from which we can conclude that HMAC-SHA256 is a PRF (and a reasonably efficient one at that), but maybe you meant to restrict the domain of ‘standard assumptions’?
$endgroup$
– Squeamish Ossifrage
3 hours ago
add a comment |
$begingroup$
Well, one standard assumption is that the SHA-256 compression function is a PRF, from which we can conclude that HMAC-SHA256 is a PRF (and a reasonably efficient one at that), but maybe you meant to restrict the domain of ‘standard assumptions’?
$endgroup$
– Squeamish Ossifrage
3 hours ago
$begingroup$
Well, one standard assumption is that the SHA-256 compression function is a PRF, from which we can conclude that HMAC-SHA256 is a PRF (and a reasonably efficient one at that), but maybe you meant to restrict the domain of ‘standard assumptions’?
$endgroup$
– Squeamish Ossifrage
3 hours ago
$begingroup$
Well, one standard assumption is that the SHA-256 compression function is a PRF, from which we can conclude that HMAC-SHA256 is a PRF (and a reasonably efficient one at that), but maybe you meant to restrict the domain of ‘standard assumptions’?
$endgroup$
– Squeamish Ossifrage
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The most common efficient PRFs from specific assumptions are:
The Naor-Reingold PRF, which is based on the decision Diffie-Hellman assumption (DDH), and
The BPR PRF, which is based on the learning with error assumption (LWE).
Perhaps slightly less well-known is the NRR PRF, which is based on the hardness of factoring.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The most common efficient PRFs from specific assumptions are:
The Naor-Reingold PRF, which is based on the decision Diffie-Hellman assumption (DDH), and
The BPR PRF, which is based on the learning with error assumption (LWE).
Perhaps slightly less well-known is the NRR PRF, which is based on the hardness of factoring.
$endgroup$
add a comment |
$begingroup$
The most common efficient PRFs from specific assumptions are:
The Naor-Reingold PRF, which is based on the decision Diffie-Hellman assumption (DDH), and
The BPR PRF, which is based on the learning with error assumption (LWE).
Perhaps slightly less well-known is the NRR PRF, which is based on the hardness of factoring.
$endgroup$
add a comment |
$begingroup$
The most common efficient PRFs from specific assumptions are:
The Naor-Reingold PRF, which is based on the decision Diffie-Hellman assumption (DDH), and
The BPR PRF, which is based on the learning with error assumption (LWE).
Perhaps slightly less well-known is the NRR PRF, which is based on the hardness of factoring.
$endgroup$
The most common efficient PRFs from specific assumptions are:
The Naor-Reingold PRF, which is based on the decision Diffie-Hellman assumption (DDH), and
The BPR PRF, which is based on the learning with error assumption (LWE).
Perhaps slightly less well-known is the NRR PRF, which is based on the hardness of factoring.
answered 3 hours ago
Geoffroy CouteauGeoffroy Couteau
9,29011834
9,29011834
add a comment |
add a comment |
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$begingroup$
Well, one standard assumption is that the SHA-256 compression function is a PRF, from which we can conclude that HMAC-SHA256 is a PRF (and a reasonably efficient one at that), but maybe you meant to restrict the domain of ‘standard assumptions’?
$endgroup$
– Squeamish Ossifrage
3 hours ago