Symetric key encryption is not Authentication?
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 Dec 17 '18 at 7:29
kelalaka
6,10022142
asked Dec 17 '18 at 6:20
UhntissUhntiss
162
add a comment |
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 Dec 17 '18 at 7:29
kelalaka
6,10022142
asked Dec 17 '18 at 6:20
UhntissUhntiss
162
add a comment |
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 Dec 17 '18 at 7:29
kelalaka
6,10022142
asked Dec 17 '18 at 6:20
UhntissUhntiss
162
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 Dec 17 '18 at 7:29
kelalaka
6,10022142
asked Dec 17 '18 at 6:20
UhntissUhntiss
162
edited Dec 17 '18 at 7:29
kelalaka
6,10022142
asked Dec 17 '18 at 6:20
UhntissUhntiss
162
edited Dec 17 '18 at 7:29
kelalaka
6,10022142
edited Dec 17 '18 at 7:29
kelalaka
6,10022142
edited Dec 17 '18 at 7:29
kelalaka
6,10022142
6,10022142
asked Dec 17 '18 at 6:20
UhntissUhntiss
162
asked Dec 17 '18 at 6:20
UhntissUhntiss
162
asked Dec 17 '18 at 6:20
UhntissUhntiss
162
162
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 Dec 17 '18 at 8:55
mephistomephisto
2,2571226
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
Dec 17 '18 at 11:08
True. While 1. i just an efficiency improvement, 2. is indeed necessary to allow for decryption of the added ciphertexts.
– mephisto
Dec 18 '18 at 8:43
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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active
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active
oldest
votes
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 Dec 17 '18 at 8:55
mephistomephisto
2,2571226
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
Dec 17 '18 at 11:08
True. While 1. i just an efficiency improvement, 2. is indeed necessary to allow for decryption of the added ciphertexts.
– mephisto
Dec 18 '18 at 8:43
add a comment |
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 Dec 17 '18 at 8:55
mephistomephisto
2,2571226
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
Dec 17 '18 at 11:08
True. While 1. i just an efficiency improvement, 2. is indeed necessary to allow for decryption of the added ciphertexts.
– mephisto
Dec 18 '18 at 8:43
add a comment |
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 Dec 17 '18 at 8:55
mephistomephisto
2,2571226
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 Dec 17 '18 at 8:55
mephistomephisto
2,2571226
answered Dec 17 '18 at 8:55
mephistomephisto
2,2571226
answered Dec 17 '18 at 8:55
mephistomephisto
2,2571226
answered Dec 17 '18 at 8:55
mephistomephisto
2,2571226
2,2571226
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
Dec 17 '18 at 11:08
True. While 1. i just an efficiency improvement, 2. is indeed necessary to allow for decryption of the added ciphertexts.
– mephisto
Dec 18 '18 at 8:43
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
Dec 17 '18 at 11:08
True. While 1. i just an efficiency improvement, 2. is indeed necessary to allow for decryption of the added ciphertexts.
– mephisto
Dec 18 '18 at 8:43
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
Dec 17 '18 at 11:08
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
Dec 17 '18 at 11:08
True. While 1. i just an efficiency improvement, 2. is indeed necessary to allow for decryption of the added ciphertexts.
– mephisto
Dec 18 '18 at 8:43
True. While 1. i just an efficiency improvement, 2. is indeed necessary to allow for decryption of the added ciphertexts.
– mephisto
Dec 18 '18 at 8:43
add a comment |
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