Can a zero nonce be safely used with AES-GCM if the key is random and never used again? ...
Models of set theory where not every set can be linearly ordered
Check which numbers satisfy the condition [A*B*C = A! + B! + C!]
3 doors, three guards, one stone
How can I fade player when goes inside or outside of the area?
Does surprise arrest existing movement?
Right-skewed distribution with mean equals to mode?
Bonus calculation: Am I making a mountain out of a molehill?
Is it true that "carbohydrates are of no use for the basal metabolic need"?
How to recreate this effect in Photoshop?
Is it true to say that an hosting provider's DNS server is what links the entire hosting environment to ICANN?
How does cp -a work
How do I keep my slimes from escaping their pens?
What happens to sewage if there is no river near by?
How do I mention the quality of my school without bragging
Do you forfeit tax refunds/credits if you aren't required to and don't file by April 15?
What makes black pepper strong or mild?
Sorting numerically
How discoverable are IPv6 addresses and AAAA names by potential attackers?
How widely used is the term Treppenwitz? Is it something that most Germans know?
How can I make names more distinctive without making them longer?
How does a Death Domain cleric's Touch of Death feature work with Touch-range spells delivered by familiars?
Do I really need recursive chmod to restrict access to a folder?
How do I stop a creek from eroding my steep embankment?
What are the motives behind Cersei's orders given to Bronn?
Can a zero nonce be safely used with AES-GCM if the key is random and never used again?
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Multi-target attacks on AES-CTR with a random nonceAES-GCM and its IV/nonce valuenonce of AES-GCM in SSLCan we use the authentication tag as Nonce / IV for the next message?Is it acceptable to write the nonce to the encrypted file during AES-256 GCM?Using AES-CTR to generate AES subkeys from a master key and nonceNonce for AES GCM to prevent replay attacksSafety of random nonce with AES-GCM?Can I use a deterministic NONCE for AES-GCM file encryption if I generate “fresh” keys for each encrypted fileIs AES-GCM with static key and dynamic salt safe to reuse IV/nonceWhat Are the Risks of AES-GCM [Key, Nonce, Message] where Nonce = Message
$begingroup$
I could generate a random nonce and prepend it to the ciphertext, but storage space is at a premium and the only constraint AES-GCM has on the nonce (if I'm reading correctly) is that the same nonce must never be paired with the same key for a second encryption.
The encryption key is randomly generated, used for a single encryption, split using Shamir's Secret Sharing Scheme, and discarded. When the key is reconstructed for decryption, there is no chance that it can be fed back through to encrypt again; a new random key is always generated for each encryption.
If that's the only constraint, then twelve zero bytes are as safe as twelve random bytes prepended to the ciphertext. I'm reading that the AES-GCM nonce is used as the IV for AES in CTR mode. It's okay to use a zero IV for AES-CTR as long as the key is never reused, but I don't want to assume without confirmation that AES-GCM does nothing relevant with the nonce besides passing it to AES CTR. Am I missing anything?
aes initialization-vector gcm nonce aes-gcm
$endgroup$
add a comment |
$begingroup$
I could generate a random nonce and prepend it to the ciphertext, but storage space is at a premium and the only constraint AES-GCM has on the nonce (if I'm reading correctly) is that the same nonce must never be paired with the same key for a second encryption.
The encryption key is randomly generated, used for a single encryption, split using Shamir's Secret Sharing Scheme, and discarded. When the key is reconstructed for decryption, there is no chance that it can be fed back through to encrypt again; a new random key is always generated for each encryption.
If that's the only constraint, then twelve zero bytes are as safe as twelve random bytes prepended to the ciphertext. I'm reading that the AES-GCM nonce is used as the IV for AES in CTR mode. It's okay to use a zero IV for AES-CTR as long as the key is never reused, but I don't want to assume without confirmation that AES-GCM does nothing relevant with the nonce besides passing it to AES CTR. Am I missing anything?
aes initialization-vector gcm nonce aes-gcm
$endgroup$
add a comment |
$begingroup$
I could generate a random nonce and prepend it to the ciphertext, but storage space is at a premium and the only constraint AES-GCM has on the nonce (if I'm reading correctly) is that the same nonce must never be paired with the same key for a second encryption.
The encryption key is randomly generated, used for a single encryption, split using Shamir's Secret Sharing Scheme, and discarded. When the key is reconstructed for decryption, there is no chance that it can be fed back through to encrypt again; a new random key is always generated for each encryption.
If that's the only constraint, then twelve zero bytes are as safe as twelve random bytes prepended to the ciphertext. I'm reading that the AES-GCM nonce is used as the IV for AES in CTR mode. It's okay to use a zero IV for AES-CTR as long as the key is never reused, but I don't want to assume without confirmation that AES-GCM does nothing relevant with the nonce besides passing it to AES CTR. Am I missing anything?
aes initialization-vector gcm nonce aes-gcm
$endgroup$
I could generate a random nonce and prepend it to the ciphertext, but storage space is at a premium and the only constraint AES-GCM has on the nonce (if I'm reading correctly) is that the same nonce must never be paired with the same key for a second encryption.
The encryption key is randomly generated, used for a single encryption, split using Shamir's Secret Sharing Scheme, and discarded. When the key is reconstructed for decryption, there is no chance that it can be fed back through to encrypt again; a new random key is always generated for each encryption.
If that's the only constraint, then twelve zero bytes are as safe as twelve random bytes prepended to the ciphertext. I'm reading that the AES-GCM nonce is used as the IV for AES in CTR mode. It's okay to use a zero IV for AES-CTR as long as the key is never reused, but I don't want to assume without confirmation that AES-GCM does nothing relevant with the nonce besides passing it to AES CTR. Am I missing anything?
aes initialization-vector gcm nonce aes-gcm
aes initialization-vector gcm nonce aes-gcm
asked yesterday
jnm2jnm2
31738
31738
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Usually. However, if you are using 128-bit AES in CTR mode (remember that GCM is essentially just CTR with authentication), then a kind of attack called a multi-target attack can become possible. This attack is realistic when an attacker has a huge amount of stored ciphertext, each with a random key. While breaking a specific key requires performing up to 2128 operations, breaking any key is significantly easier. This attack can be mitigated either by using a larger key size, or by using a random nonce.
From the above-linked blog post by DJB:
What the attacker hopes to find inside the AES attack is a key collision. This means that a key guessed by the attack matches a key chosen by a user. Any particular guessed key has chance only 1/2128 of matching any particular user key, but the attack ends up merging costs across a batch of 240 user keys, amplifying the effectiveness of each guess by a factor 240.
$endgroup$
$begingroup$
Good to know! I'm using a 256-bit key and there is not much ciphertext per key (a few kilobytes). It sounds like my particular situation is safe, but can you quantify "huge" to shed light on the decision-making process if I find myself in a similar but different scenario?
$endgroup$
– jnm2
12 hours ago
$begingroup$
And the sole benefit of a random IV is that it avoids collisions using 256+96 random bits rather than just 256 random bits (or 128+96 instead of 128)?
$endgroup$
– jnm2
12 hours ago
$begingroup$
@jnm2 How huge depends on how much advantage you are OK with the attacker getting. The attack starts to become significantly easier than brute force after around $2^{40}$ keys. And if you are using a 256-bit key, then there is no reason to use a random nonce, as long as the key is unique and always random.
$endgroup$
– forest
52 mins ago
add a comment |
$begingroup$
Am I missing anything?
No, you are not; if you use a key only once, that is, to encrypt a single message, and never use it to encrypt anything else, then it doesn't matter what nonce you use. An implicit 'all-00' nonce is as good as any.
BTW: AES-GCM also uses the nonce as a part of the transform that generates the integrity tag; however, that addition does not complicate the fact that an all-00 nonce is fine, as long as you use the key once.
$endgroup$
$begingroup$
I'm not sure that this is entirely correct, due to the risk of multi-target attacks on AES128.
$endgroup$
– forest
18 hours ago
add a comment |
$begingroup$
Does your random generator guarantee (with sufficient confidence) that it won't generate the same random key a second time?
As you correctly stated, as long as the same nonce and key are never re-used, everything is fine. But a randomly generated key does not by itself have such an assurance.
There are two simple ways you can take:
a) accept the risk. Make a quick calculation based on your RNG what the probability is that a key will be repeated and then decide that this chance is acceptable (or not).
b) instead of using a zero nonce, use a simple counter. That's what many implementations actually do. The nonce can be predictable, that's ok.
The decision in a) largely depends on the number of messages you are going to send. If the number is low, the risk is most likely acceptable. If we're talking millions-plus messages, you might find the probability of an identical key (remember the birthday paradox!) too high for comfort.
$endgroup$
$begingroup$
The advantage of using a fixed nonce, is that you don't need to transmit it or store it. I presume this is enough of an advantage for the OP.
$endgroup$
– Martin Bonner
16 hours ago
$begingroup$
@MartinBonner You can usually derive the nonce from the same master secret that the key is derived from.
$endgroup$
– forest
16 hours ago
$begingroup$
@MartinBonner - as written: weigh the advantage against the risk and make a decision. The OP doesn't specify his use case, which makes it difficult to be specific on the threat level.
$endgroup$
– Tom
16 hours ago
$begingroup$
Good reminder. I can't use a counter since there is no context saved from one encryption to the next. This is a standalone tool with no central server to house a counter. The only options I know of are fixed nonce (e.g. zeros) and random nonce. The key is 256 bits and encryption will be occasional. The generator isBCryptGenRandom
withBCRYPT_USE_SYSTEM_PREFERRED_RNG
on Windows and OpenSSL on Unix.
$endgroup$
– jnm2
12 hours ago
$begingroup$
@jnm2 you could use a trivial counter, such as the timestamp (rounded to full seconds or even minutes, if both systems are time-synchronized) or even just the day-of-year (if not and the edge case of one message being not decryptable because it was sent at just the right second doesn't matter). This would already dramatically reduce the chances of a chance repetition.
$endgroup$
– Tom
7 hours ago
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "281"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcrypto.stackexchange.com%2fquestions%2f68774%2fcan-a-zero-nonce-be-safely-used-with-aes-gcm-if-the-key-is-random-and-never-used%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Usually. However, if you are using 128-bit AES in CTR mode (remember that GCM is essentially just CTR with authentication), then a kind of attack called a multi-target attack can become possible. This attack is realistic when an attacker has a huge amount of stored ciphertext, each with a random key. While breaking a specific key requires performing up to 2128 operations, breaking any key is significantly easier. This attack can be mitigated either by using a larger key size, or by using a random nonce.
From the above-linked blog post by DJB:
What the attacker hopes to find inside the AES attack is a key collision. This means that a key guessed by the attack matches a key chosen by a user. Any particular guessed key has chance only 1/2128 of matching any particular user key, but the attack ends up merging costs across a batch of 240 user keys, amplifying the effectiveness of each guess by a factor 240.
$endgroup$
$begingroup$
Good to know! I'm using a 256-bit key and there is not much ciphertext per key (a few kilobytes). It sounds like my particular situation is safe, but can you quantify "huge" to shed light on the decision-making process if I find myself in a similar but different scenario?
$endgroup$
– jnm2
12 hours ago
$begingroup$
And the sole benefit of a random IV is that it avoids collisions using 256+96 random bits rather than just 256 random bits (or 128+96 instead of 128)?
$endgroup$
– jnm2
12 hours ago
$begingroup$
@jnm2 How huge depends on how much advantage you are OK with the attacker getting. The attack starts to become significantly easier than brute force after around $2^{40}$ keys. And if you are using a 256-bit key, then there is no reason to use a random nonce, as long as the key is unique and always random.
$endgroup$
– forest
52 mins ago
add a comment |
$begingroup$
Usually. However, if you are using 128-bit AES in CTR mode (remember that GCM is essentially just CTR with authentication), then a kind of attack called a multi-target attack can become possible. This attack is realistic when an attacker has a huge amount of stored ciphertext, each with a random key. While breaking a specific key requires performing up to 2128 operations, breaking any key is significantly easier. This attack can be mitigated either by using a larger key size, or by using a random nonce.
From the above-linked blog post by DJB:
What the attacker hopes to find inside the AES attack is a key collision. This means that a key guessed by the attack matches a key chosen by a user. Any particular guessed key has chance only 1/2128 of matching any particular user key, but the attack ends up merging costs across a batch of 240 user keys, amplifying the effectiveness of each guess by a factor 240.
$endgroup$
$begingroup$
Good to know! I'm using a 256-bit key and there is not much ciphertext per key (a few kilobytes). It sounds like my particular situation is safe, but can you quantify "huge" to shed light on the decision-making process if I find myself in a similar but different scenario?
$endgroup$
– jnm2
12 hours ago
$begingroup$
And the sole benefit of a random IV is that it avoids collisions using 256+96 random bits rather than just 256 random bits (or 128+96 instead of 128)?
$endgroup$
– jnm2
12 hours ago
$begingroup$
@jnm2 How huge depends on how much advantage you are OK with the attacker getting. The attack starts to become significantly easier than brute force after around $2^{40}$ keys. And if you are using a 256-bit key, then there is no reason to use a random nonce, as long as the key is unique and always random.
$endgroup$
– forest
52 mins ago
add a comment |
$begingroup$
Usually. However, if you are using 128-bit AES in CTR mode (remember that GCM is essentially just CTR with authentication), then a kind of attack called a multi-target attack can become possible. This attack is realistic when an attacker has a huge amount of stored ciphertext, each with a random key. While breaking a specific key requires performing up to 2128 operations, breaking any key is significantly easier. This attack can be mitigated either by using a larger key size, or by using a random nonce.
From the above-linked blog post by DJB:
What the attacker hopes to find inside the AES attack is a key collision. This means that a key guessed by the attack matches a key chosen by a user. Any particular guessed key has chance only 1/2128 of matching any particular user key, but the attack ends up merging costs across a batch of 240 user keys, amplifying the effectiveness of each guess by a factor 240.
$endgroup$
Usually. However, if you are using 128-bit AES in CTR mode (remember that GCM is essentially just CTR with authentication), then a kind of attack called a multi-target attack can become possible. This attack is realistic when an attacker has a huge amount of stored ciphertext, each with a random key. While breaking a specific key requires performing up to 2128 operations, breaking any key is significantly easier. This attack can be mitigated either by using a larger key size, or by using a random nonce.
From the above-linked blog post by DJB:
What the attacker hopes to find inside the AES attack is a key collision. This means that a key guessed by the attack matches a key chosen by a user. Any particular guessed key has chance only 1/2128 of matching any particular user key, but the attack ends up merging costs across a batch of 240 user keys, amplifying the effectiveness of each guess by a factor 240.
answered 18 hours ago
forestforest
4,96011744
4,96011744
$begingroup$
Good to know! I'm using a 256-bit key and there is not much ciphertext per key (a few kilobytes). It sounds like my particular situation is safe, but can you quantify "huge" to shed light on the decision-making process if I find myself in a similar but different scenario?
$endgroup$
– jnm2
12 hours ago
$begingroup$
And the sole benefit of a random IV is that it avoids collisions using 256+96 random bits rather than just 256 random bits (or 128+96 instead of 128)?
$endgroup$
– jnm2
12 hours ago
$begingroup$
@jnm2 How huge depends on how much advantage you are OK with the attacker getting. The attack starts to become significantly easier than brute force after around $2^{40}$ keys. And if you are using a 256-bit key, then there is no reason to use a random nonce, as long as the key is unique and always random.
$endgroup$
– forest
52 mins ago
add a comment |
$begingroup$
Good to know! I'm using a 256-bit key and there is not much ciphertext per key (a few kilobytes). It sounds like my particular situation is safe, but can you quantify "huge" to shed light on the decision-making process if I find myself in a similar but different scenario?
$endgroup$
– jnm2
12 hours ago
$begingroup$
And the sole benefit of a random IV is that it avoids collisions using 256+96 random bits rather than just 256 random bits (or 128+96 instead of 128)?
$endgroup$
– jnm2
12 hours ago
$begingroup$
@jnm2 How huge depends on how much advantage you are OK with the attacker getting. The attack starts to become significantly easier than brute force after around $2^{40}$ keys. And if you are using a 256-bit key, then there is no reason to use a random nonce, as long as the key is unique and always random.
$endgroup$
– forest
52 mins ago
$begingroup$
Good to know! I'm using a 256-bit key and there is not much ciphertext per key (a few kilobytes). It sounds like my particular situation is safe, but can you quantify "huge" to shed light on the decision-making process if I find myself in a similar but different scenario?
$endgroup$
– jnm2
12 hours ago
$begingroup$
Good to know! I'm using a 256-bit key and there is not much ciphertext per key (a few kilobytes). It sounds like my particular situation is safe, but can you quantify "huge" to shed light on the decision-making process if I find myself in a similar but different scenario?
$endgroup$
– jnm2
12 hours ago
$begingroup$
And the sole benefit of a random IV is that it avoids collisions using 256+96 random bits rather than just 256 random bits (or 128+96 instead of 128)?
$endgroup$
– jnm2
12 hours ago
$begingroup$
And the sole benefit of a random IV is that it avoids collisions using 256+96 random bits rather than just 256 random bits (or 128+96 instead of 128)?
$endgroup$
– jnm2
12 hours ago
$begingroup$
@jnm2 How huge depends on how much advantage you are OK with the attacker getting. The attack starts to become significantly easier than brute force after around $2^{40}$ keys. And if you are using a 256-bit key, then there is no reason to use a random nonce, as long as the key is unique and always random.
$endgroup$
– forest
52 mins ago
$begingroup$
@jnm2 How huge depends on how much advantage you are OK with the attacker getting. The attack starts to become significantly easier than brute force after around $2^{40}$ keys. And if you are using a 256-bit key, then there is no reason to use a random nonce, as long as the key is unique and always random.
$endgroup$
– forest
52 mins ago
add a comment |
$begingroup$
Am I missing anything?
No, you are not; if you use a key only once, that is, to encrypt a single message, and never use it to encrypt anything else, then it doesn't matter what nonce you use. An implicit 'all-00' nonce is as good as any.
BTW: AES-GCM also uses the nonce as a part of the transform that generates the integrity tag; however, that addition does not complicate the fact that an all-00 nonce is fine, as long as you use the key once.
$endgroup$
$begingroup$
I'm not sure that this is entirely correct, due to the risk of multi-target attacks on AES128.
$endgroup$
– forest
18 hours ago
add a comment |
$begingroup$
Am I missing anything?
No, you are not; if you use a key only once, that is, to encrypt a single message, and never use it to encrypt anything else, then it doesn't matter what nonce you use. An implicit 'all-00' nonce is as good as any.
BTW: AES-GCM also uses the nonce as a part of the transform that generates the integrity tag; however, that addition does not complicate the fact that an all-00 nonce is fine, as long as you use the key once.
$endgroup$
$begingroup$
I'm not sure that this is entirely correct, due to the risk of multi-target attacks on AES128.
$endgroup$
– forest
18 hours ago
add a comment |
$begingroup$
Am I missing anything?
No, you are not; if you use a key only once, that is, to encrypt a single message, and never use it to encrypt anything else, then it doesn't matter what nonce you use. An implicit 'all-00' nonce is as good as any.
BTW: AES-GCM also uses the nonce as a part of the transform that generates the integrity tag; however, that addition does not complicate the fact that an all-00 nonce is fine, as long as you use the key once.
$endgroup$
Am I missing anything?
No, you are not; if you use a key only once, that is, to encrypt a single message, and never use it to encrypt anything else, then it doesn't matter what nonce you use. An implicit 'all-00' nonce is as good as any.
BTW: AES-GCM also uses the nonce as a part of the transform that generates the integrity tag; however, that addition does not complicate the fact that an all-00 nonce is fine, as long as you use the key once.
answered yesterday
ponchoponcho
94.2k2148247
94.2k2148247
$begingroup$
I'm not sure that this is entirely correct, due to the risk of multi-target attacks on AES128.
$endgroup$
– forest
18 hours ago
add a comment |
$begingroup$
I'm not sure that this is entirely correct, due to the risk of multi-target attacks on AES128.
$endgroup$
– forest
18 hours ago
$begingroup$
I'm not sure that this is entirely correct, due to the risk of multi-target attacks on AES128.
$endgroup$
– forest
18 hours ago
$begingroup$
I'm not sure that this is entirely correct, due to the risk of multi-target attacks on AES128.
$endgroup$
– forest
18 hours ago
add a comment |
$begingroup$
Does your random generator guarantee (with sufficient confidence) that it won't generate the same random key a second time?
As you correctly stated, as long as the same nonce and key are never re-used, everything is fine. But a randomly generated key does not by itself have such an assurance.
There are two simple ways you can take:
a) accept the risk. Make a quick calculation based on your RNG what the probability is that a key will be repeated and then decide that this chance is acceptable (or not).
b) instead of using a zero nonce, use a simple counter. That's what many implementations actually do. The nonce can be predictable, that's ok.
The decision in a) largely depends on the number of messages you are going to send. If the number is low, the risk is most likely acceptable. If we're talking millions-plus messages, you might find the probability of an identical key (remember the birthday paradox!) too high for comfort.
$endgroup$
$begingroup$
The advantage of using a fixed nonce, is that you don't need to transmit it or store it. I presume this is enough of an advantage for the OP.
$endgroup$
– Martin Bonner
16 hours ago
$begingroup$
@MartinBonner You can usually derive the nonce from the same master secret that the key is derived from.
$endgroup$
– forest
16 hours ago
$begingroup$
@MartinBonner - as written: weigh the advantage against the risk and make a decision. The OP doesn't specify his use case, which makes it difficult to be specific on the threat level.
$endgroup$
– Tom
16 hours ago
$begingroup$
Good reminder. I can't use a counter since there is no context saved from one encryption to the next. This is a standalone tool with no central server to house a counter. The only options I know of are fixed nonce (e.g. zeros) and random nonce. The key is 256 bits and encryption will be occasional. The generator isBCryptGenRandom
withBCRYPT_USE_SYSTEM_PREFERRED_RNG
on Windows and OpenSSL on Unix.
$endgroup$
– jnm2
12 hours ago
$begingroup$
@jnm2 you could use a trivial counter, such as the timestamp (rounded to full seconds or even minutes, if both systems are time-synchronized) or even just the day-of-year (if not and the edge case of one message being not decryptable because it was sent at just the right second doesn't matter). This would already dramatically reduce the chances of a chance repetition.
$endgroup$
– Tom
7 hours ago
add a comment |
$begingroup$
Does your random generator guarantee (with sufficient confidence) that it won't generate the same random key a second time?
As you correctly stated, as long as the same nonce and key are never re-used, everything is fine. But a randomly generated key does not by itself have such an assurance.
There are two simple ways you can take:
a) accept the risk. Make a quick calculation based on your RNG what the probability is that a key will be repeated and then decide that this chance is acceptable (or not).
b) instead of using a zero nonce, use a simple counter. That's what many implementations actually do. The nonce can be predictable, that's ok.
The decision in a) largely depends on the number of messages you are going to send. If the number is low, the risk is most likely acceptable. If we're talking millions-plus messages, you might find the probability of an identical key (remember the birthday paradox!) too high for comfort.
$endgroup$
$begingroup$
The advantage of using a fixed nonce, is that you don't need to transmit it or store it. I presume this is enough of an advantage for the OP.
$endgroup$
– Martin Bonner
16 hours ago
$begingroup$
@MartinBonner You can usually derive the nonce from the same master secret that the key is derived from.
$endgroup$
– forest
16 hours ago
$begingroup$
@MartinBonner - as written: weigh the advantage against the risk and make a decision. The OP doesn't specify his use case, which makes it difficult to be specific on the threat level.
$endgroup$
– Tom
16 hours ago
$begingroup$
Good reminder. I can't use a counter since there is no context saved from one encryption to the next. This is a standalone tool with no central server to house a counter. The only options I know of are fixed nonce (e.g. zeros) and random nonce. The key is 256 bits and encryption will be occasional. The generator isBCryptGenRandom
withBCRYPT_USE_SYSTEM_PREFERRED_RNG
on Windows and OpenSSL on Unix.
$endgroup$
– jnm2
12 hours ago
$begingroup$
@jnm2 you could use a trivial counter, such as the timestamp (rounded to full seconds or even minutes, if both systems are time-synchronized) or even just the day-of-year (if not and the edge case of one message being not decryptable because it was sent at just the right second doesn't matter). This would already dramatically reduce the chances of a chance repetition.
$endgroup$
– Tom
7 hours ago
add a comment |
$begingroup$
Does your random generator guarantee (with sufficient confidence) that it won't generate the same random key a second time?
As you correctly stated, as long as the same nonce and key are never re-used, everything is fine. But a randomly generated key does not by itself have such an assurance.
There are two simple ways you can take:
a) accept the risk. Make a quick calculation based on your RNG what the probability is that a key will be repeated and then decide that this chance is acceptable (or not).
b) instead of using a zero nonce, use a simple counter. That's what many implementations actually do. The nonce can be predictable, that's ok.
The decision in a) largely depends on the number of messages you are going to send. If the number is low, the risk is most likely acceptable. If we're talking millions-plus messages, you might find the probability of an identical key (remember the birthday paradox!) too high for comfort.
$endgroup$
Does your random generator guarantee (with sufficient confidence) that it won't generate the same random key a second time?
As you correctly stated, as long as the same nonce and key are never re-used, everything is fine. But a randomly generated key does not by itself have such an assurance.
There are two simple ways you can take:
a) accept the risk. Make a quick calculation based on your RNG what the probability is that a key will be repeated and then decide that this chance is acceptable (or not).
b) instead of using a zero nonce, use a simple counter. That's what many implementations actually do. The nonce can be predictable, that's ok.
The decision in a) largely depends on the number of messages you are going to send. If the number is low, the risk is most likely acceptable. If we're talking millions-plus messages, you might find the probability of an identical key (remember the birthday paradox!) too high for comfort.
answered 18 hours ago
TomTom
24716
24716
$begingroup$
The advantage of using a fixed nonce, is that you don't need to transmit it or store it. I presume this is enough of an advantage for the OP.
$endgroup$
– Martin Bonner
16 hours ago
$begingroup$
@MartinBonner You can usually derive the nonce from the same master secret that the key is derived from.
$endgroup$
– forest
16 hours ago
$begingroup$
@MartinBonner - as written: weigh the advantage against the risk and make a decision. The OP doesn't specify his use case, which makes it difficult to be specific on the threat level.
$endgroup$
– Tom
16 hours ago
$begingroup$
Good reminder. I can't use a counter since there is no context saved from one encryption to the next. This is a standalone tool with no central server to house a counter. The only options I know of are fixed nonce (e.g. zeros) and random nonce. The key is 256 bits and encryption will be occasional. The generator isBCryptGenRandom
withBCRYPT_USE_SYSTEM_PREFERRED_RNG
on Windows and OpenSSL on Unix.
$endgroup$
– jnm2
12 hours ago
$begingroup$
@jnm2 you could use a trivial counter, such as the timestamp (rounded to full seconds or even minutes, if both systems are time-synchronized) or even just the day-of-year (if not and the edge case of one message being not decryptable because it was sent at just the right second doesn't matter). This would already dramatically reduce the chances of a chance repetition.
$endgroup$
– Tom
7 hours ago
add a comment |
$begingroup$
The advantage of using a fixed nonce, is that you don't need to transmit it or store it. I presume this is enough of an advantage for the OP.
$endgroup$
– Martin Bonner
16 hours ago
$begingroup$
@MartinBonner You can usually derive the nonce from the same master secret that the key is derived from.
$endgroup$
– forest
16 hours ago
$begingroup$
@MartinBonner - as written: weigh the advantage against the risk and make a decision. The OP doesn't specify his use case, which makes it difficult to be specific on the threat level.
$endgroup$
– Tom
16 hours ago
$begingroup$
Good reminder. I can't use a counter since there is no context saved from one encryption to the next. This is a standalone tool with no central server to house a counter. The only options I know of are fixed nonce (e.g. zeros) and random nonce. The key is 256 bits and encryption will be occasional. The generator isBCryptGenRandom
withBCRYPT_USE_SYSTEM_PREFERRED_RNG
on Windows and OpenSSL on Unix.
$endgroup$
– jnm2
12 hours ago
$begingroup$
@jnm2 you could use a trivial counter, such as the timestamp (rounded to full seconds or even minutes, if both systems are time-synchronized) or even just the day-of-year (if not and the edge case of one message being not decryptable because it was sent at just the right second doesn't matter). This would already dramatically reduce the chances of a chance repetition.
$endgroup$
– Tom
7 hours ago
$begingroup$
The advantage of using a fixed nonce, is that you don't need to transmit it or store it. I presume this is enough of an advantage for the OP.
$endgroup$
– Martin Bonner
16 hours ago
$begingroup$
The advantage of using a fixed nonce, is that you don't need to transmit it or store it. I presume this is enough of an advantage for the OP.
$endgroup$
– Martin Bonner
16 hours ago
$begingroup$
@MartinBonner You can usually derive the nonce from the same master secret that the key is derived from.
$endgroup$
– forest
16 hours ago
$begingroup$
@MartinBonner You can usually derive the nonce from the same master secret that the key is derived from.
$endgroup$
– forest
16 hours ago
$begingroup$
@MartinBonner - as written: weigh the advantage against the risk and make a decision. The OP doesn't specify his use case, which makes it difficult to be specific on the threat level.
$endgroup$
– Tom
16 hours ago
$begingroup$
@MartinBonner - as written: weigh the advantage against the risk and make a decision. The OP doesn't specify his use case, which makes it difficult to be specific on the threat level.
$endgroup$
– Tom
16 hours ago
$begingroup$
Good reminder. I can't use a counter since there is no context saved from one encryption to the next. This is a standalone tool with no central server to house a counter. The only options I know of are fixed nonce (e.g. zeros) and random nonce. The key is 256 bits and encryption will be occasional. The generator is
BCryptGenRandom
with BCRYPT_USE_SYSTEM_PREFERRED_RNG
on Windows and OpenSSL on Unix.$endgroup$
– jnm2
12 hours ago
$begingroup$
Good reminder. I can't use a counter since there is no context saved from one encryption to the next. This is a standalone tool with no central server to house a counter. The only options I know of are fixed nonce (e.g. zeros) and random nonce. The key is 256 bits and encryption will be occasional. The generator is
BCryptGenRandom
with BCRYPT_USE_SYSTEM_PREFERRED_RNG
on Windows and OpenSSL on Unix.$endgroup$
– jnm2
12 hours ago
$begingroup$
@jnm2 you could use a trivial counter, such as the timestamp (rounded to full seconds or even minutes, if both systems are time-synchronized) or even just the day-of-year (if not and the edge case of one message being not decryptable because it was sent at just the right second doesn't matter). This would already dramatically reduce the chances of a chance repetition.
$endgroup$
– Tom
7 hours ago
$begingroup$
@jnm2 you could use a trivial counter, such as the timestamp (rounded to full seconds or even minutes, if both systems are time-synchronized) or even just the day-of-year (if not and the edge case of one message being not decryptable because it was sent at just the right second doesn't matter). This would already dramatically reduce the chances of a chance repetition.
$endgroup$
– Tom
7 hours ago
add a comment |
Thanks for contributing an answer to Cryptography Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcrypto.stackexchange.com%2fquestions%2f68774%2fcan-a-zero-nonce-be-safely-used-with-aes-gcm-if-the-key-is-random-and-never-used%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown