Symetric key encryption is not Authentication?
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Is there any way to show CCA-secure symmetric-key encryption does not have to be an authenticated encryption?
I know you can create a setup that isn't authentication, but how will you go about it?
authenticated-encryption chosen-ciphertext-attack
edited 4 hours ago
kelalaka
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asked 5 hours ago
Uhntiss
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up vote
2
down vote
favorite
Is there any way to show CCA-secure symmetric-key encryption does not have to be an authenticated encryption?
I know you can create a setup that isn't authentication, but how will you go about it?
authenticated-encryption chosen-ciphertext-attack
edited 4 hours ago
kelalaka
4,84121837
asked 5 hours ago
Uhntiss
112
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Uhntiss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is there any way to show CCA-secure symmetric-key encryption does not have to be an authenticated encryption?
I know you can create a setup that isn't authentication, but how will you go about it?
authenticated-encryption chosen-ciphertext-attack
edited 4 hours ago
kelalaka
4,84121837
asked 5 hours ago
Uhntiss
112
New contributor
Uhntiss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is there any way to show CCA-secure symmetric-key encryption does not have to be an authenticated encryption?
I know you can create a setup that isn't authentication, but how will you go about it?
authenticated-encryption chosen-ciphertext-attack
authenticated-encryption chosen-ciphertext-attack
edited 4 hours ago
kelalaka
4,84121837
asked 5 hours ago
Uhntiss
112
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Uhntiss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 4 hours ago
kelalaka
4,84121837
asked 5 hours ago
Uhntiss
112
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edited 4 hours ago
kelalaka
4,84121837
edited 4 hours ago
kelalaka
4,84121837
edited 4 hours ago
kelalaka
4,84121837
4,84121837
asked 5 hours ago
Uhntiss
112
New contributor
Uhntiss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Uhntiss
112
asked 5 hours ago
Uhntiss
112
112
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Uhntiss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Uhntiss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Uhntiss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
1 Answer
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Consider a CCA2 secure secret key encryption scheme $mathcal{E}$ with an exponentially large randomness space. Turn this into an encryption scheme $mathcal{E'}$ where we extend the randomness space of $mathcal{E}$ with one more element. Let's think of it as $0$ (we can get there by rearranging the randomness space). For randomness $0$ the encryption in $mathcal{E'}$ becomes the identity for every element.
Now, I claim $mathcal{E'}$ is still CCA secure as the probability that for the challenge query randomness 0 is chosen is negligible. However, $mathcal{E'}$ is not a secure MAC as everybody knows a valid tag for any message.
answered 2 hours ago
mephisto
2,2471226
Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
– Maeher
37 mins ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Consider a CCA2 secure secret key encryption scheme $mathcal{E}$ with an exponentially large randomness space. Turn this into an encryption scheme $mathcal{E'}$ where we extend the randomness space of $mathcal{E}$ with one more element. Let's think of it as $0$ (we can get there by rearranging the randomness space). For randomness $0$ the encryption in $mathcal{E'}$ becomes the identity for every element.
Now, I claim $mathcal{E'}$ is still CCA secure as the probability that for the challenge query randomness 0 is chosen is negligible. However, $mathcal{E'}$ is not a secure MAC as everybody knows a valid tag for any message.
answered 2 hours ago
mephisto
2,2471226
Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
– Maeher
37 mins ago
add a comment |
up vote
2
down vote
Consider a CCA2 secure secret key encryption scheme $mathcal{E}$ with an exponentially large randomness space. Turn this into an encryption scheme $mathcal{E'}$ where we extend the randomness space of $mathcal{E}$ with one more element. Let's think of it as $0$ (we can get there by rearranging the randomness space). For randomness $0$ the encryption in $mathcal{E'}$ becomes the identity for every element.
Now, I claim $mathcal{E'}$ is still CCA secure as the probability that for the challenge query randomness 0 is chosen is negligible. However, $mathcal{E'}$ is not a secure MAC as everybody knows a valid tag for any message.
answered 2 hours ago
mephisto
2,2471226
Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
– Maeher
37 mins ago
add a comment |
up vote
2
down vote
up vote
2
down vote
Consider a CCA2 secure secret key encryption scheme $mathcal{E}$ with an exponentially large randomness space. Turn this into an encryption scheme $mathcal{E'}$ where we extend the randomness space of $mathcal{E}$ with one more element. Let's think of it as $0$ (we can get there by rearranging the randomness space). For randomness $0$ the encryption in $mathcal{E'}$ becomes the identity for every element.
Now, I claim $mathcal{E'}$ is still CCA secure as the probability that for the challenge query randomness 0 is chosen is negligible. However, $mathcal{E'}$ is not a secure MAC as everybody knows a valid tag for any message.
answered 2 hours ago
mephisto
2,2471226
Consider a CCA2 secure secret key encryption scheme $mathcal{E}$ with an exponentially large randomness space. Turn this into an encryption scheme $mathcal{E'}$ where we extend the randomness space of $mathcal{E}$ with one more element. Let's think of it as $0$ (we can get there by rearranging the randomness space). For randomness $0$ the encryption in $mathcal{E'}$ becomes the identity for every element.
Now, I claim $mathcal{E'}$ is still CCA secure as the probability that for the challenge query randomness 0 is chosen is negligible. However, $mathcal{E'}$ is not a secure MAC as everybody knows a valid tag for any message.
answered 2 hours ago
mephisto
2,2471226
answered 2 hours ago
mephisto
2,2471226
answered 2 hours ago
mephisto
2,2471226
answered 2 hours ago
mephisto
2,2471226
2,2471226
Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
– Maeher
37 mins ago
add a comment |
Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
– Maeher
37 mins ago
Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
– Maeher
37 mins ago
Two comments/nitpicks: 1. You don't need to add an additional randomness value, you can just repurpose an existing one (e.g. the all 0 string). 2. For that scheme to be be complete you need to mark from which mode of the scheme the ciphertext resulted. You can do that e.g. by prefixing actual encryptions with 0 and the identity with 1.
– Maeher
37 mins ago
add a comment |
Uhntiss is a new contributor. Be nice, and check out our Code of Conduct.
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