double encryption - One Time Pad The 2019 Stack Overflow Developer Survey Results Are...
The difference between dialogue marks
What does "fetching by region is not available for SAM files" means?
Worn-tile Scrabble
Are children permitted to help build the Beis Hamikdash?
Where to refill my bottle in India?
Is there a symbol for a right arrow with a square in the middle?
Are there any other methods to apply to solving simultaneous equations?
Why do UK politicians seemingly ignore opinion polls on Brexit?
What does ひと匙 mean in this manga and has it been used colloquially?
Did Section 31 appear in Star Trek: The Next Generation?
FPGA - DIY Programming
Protecting Dualbooting Windows from dangerous code (like rm -rf)
How to support a colleague who finds meetings extremely tiring?
Landlord wants to switch my lease to a "Land contract" to "get back at the city"
Why isn't the circumferential light around the M87 black hole's event horizon symmetric?
"as much details as you can remember"
Building a conditional check constraint
What is the meaning of the verb "bear" in this context?
double encryption - One Time Pad
Write faster on AT24C32
Does the shape of a die affect the probability of a number being rolled?
What do hard-Brexiteers want with respect to the Irish border?
Why isn't airport relocation done gradually?
How to manage monthly salary
double encryption - One Time Pad
The 2019 Stack Overflow Developer Survey Results Are In“Padless” One-time-Pad encryptionHave these compositions of block ciphers the same security?One-time pad mistake in the transmission?OTP same message encrypted twice (with different keys)Breaking One Time Pad with CCAOne-time pad using RSA and Diffie-Hellman functionsDoes sending the same message with 2 different keys in OTP leak information?One Time Pad Alphabet SizeSafely using ciphers that take small key sizesIs one time pad cipher reuse and random key secure?
$begingroup$
Can the security of the encryption system be improved by using double encryption with two randomly chosen keys?
By double encryption I mean using two keys K1 and K2, each 20-bit long, to obtain ciphertext $C = ( M ⊕ K_1 ) ⊕ K_2$.
one-time-pad multiple-encryption
New contributor
$endgroup$
add a comment |
$begingroup$
Can the security of the encryption system be improved by using double encryption with two randomly chosen keys?
By double encryption I mean using two keys K1 and K2, each 20-bit long, to obtain ciphertext $C = ( M ⊕ K_1 ) ⊕ K_2$.
one-time-pad multiple-encryption
New contributor
$endgroup$
1
$begingroup$
An ideal one time pad already has perfect security, you can't just get twice as perfect. The second layer doesn't add much.
$endgroup$
– Natanael
1 hour ago
1
$begingroup$
Hi. Is the size of the 20 part within your control? 20 bits = 1 million values. That can easily be enumerated. A hundred keys won't prevent that. And what Natanael says.
$endgroup$
– Paul Uszak
1 hour ago
1
$begingroup$
I don't see how enumerating the key values makes any difference for the one-time pad. Even a single bit key for a single bit message would be secure.
$endgroup$
– Maarten Bodewes♦
19 mins ago
add a comment |
$begingroup$
Can the security of the encryption system be improved by using double encryption with two randomly chosen keys?
By double encryption I mean using two keys K1 and K2, each 20-bit long, to obtain ciphertext $C = ( M ⊕ K_1 ) ⊕ K_2$.
one-time-pad multiple-encryption
New contributor
$endgroup$
Can the security of the encryption system be improved by using double encryption with two randomly chosen keys?
By double encryption I mean using two keys K1 and K2, each 20-bit long, to obtain ciphertext $C = ( M ⊕ K_1 ) ⊕ K_2$.
one-time-pad multiple-encryption
one-time-pad multiple-encryption
New contributor
New contributor
edited 58 mins ago
Ella Rose♦
17k44483
17k44483
New contributor
asked 3 hours ago
MinaMina
61
61
New contributor
New contributor
1
$begingroup$
An ideal one time pad already has perfect security, you can't just get twice as perfect. The second layer doesn't add much.
$endgroup$
– Natanael
1 hour ago
1
$begingroup$
Hi. Is the size of the 20 part within your control? 20 bits = 1 million values. That can easily be enumerated. A hundred keys won't prevent that. And what Natanael says.
$endgroup$
– Paul Uszak
1 hour ago
1
$begingroup$
I don't see how enumerating the key values makes any difference for the one-time pad. Even a single bit key for a single bit message would be secure.
$endgroup$
– Maarten Bodewes♦
19 mins ago
add a comment |
1
$begingroup$
An ideal one time pad already has perfect security, you can't just get twice as perfect. The second layer doesn't add much.
$endgroup$
– Natanael
1 hour ago
1
$begingroup$
Hi. Is the size of the 20 part within your control? 20 bits = 1 million values. That can easily be enumerated. A hundred keys won't prevent that. And what Natanael says.
$endgroup$
– Paul Uszak
1 hour ago
1
$begingroup$
I don't see how enumerating the key values makes any difference for the one-time pad. Even a single bit key for a single bit message would be secure.
$endgroup$
– Maarten Bodewes♦
19 mins ago
1
1
$begingroup$
An ideal one time pad already has perfect security, you can't just get twice as perfect. The second layer doesn't add much.
$endgroup$
– Natanael
1 hour ago
$begingroup$
An ideal one time pad already has perfect security, you can't just get twice as perfect. The second layer doesn't add much.
$endgroup$
– Natanael
1 hour ago
1
1
$begingroup$
Hi. Is the size of the 20 part within your control? 20 bits = 1 million values. That can easily be enumerated. A hundred keys won't prevent that. And what Natanael says.
$endgroup$
– Paul Uszak
1 hour ago
$begingroup$
Hi. Is the size of the 20 part within your control? 20 bits = 1 million values. That can easily be enumerated. A hundred keys won't prevent that. And what Natanael says.
$endgroup$
– Paul Uszak
1 hour ago
1
1
$begingroup$
I don't see how enumerating the key values makes any difference for the one-time pad. Even a single bit key for a single bit message would be secure.
$endgroup$
– Maarten Bodewes♦
19 mins ago
$begingroup$
I don't see how enumerating the key values makes any difference for the one-time pad. Even a single bit key for a single bit message would be secure.
$endgroup$
– Maarten Bodewes♦
19 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The answer is we cannot improve the security of the one-time pad in this manner. Intuitively the reason is that the double one-time pad is just a less efficient one time pad.
The security of the traditional xor-based one-time pad is requires that the key $K$ is chosen uniformly at random for each message and that the key is at least as large as the message.
Because the xor operation is associative, we could re-write the double one-time pad as $C = M oplus(K_1 oplus K_2)$ or $ C = M oplus K$, where $K = K_1 oplus K_2$. Now if $K_1$ and $K_2$ are uniform, it is easy to show that $K$ is also uniform. Assuming the key and the message have length $n$
$P_K[k] = sum_{x in {0,1}^n}P_{K_1,K_2}[k_1 = x, k_2 = k oplus k_1] = frac{1}{2^n}$
In other words $K_1 oplus K_2$ could be simply replaced by a single uniform key, therefore $K_1$ is enough. Moreover the double one-time pad is inefficient because it requires two xor operations.
Regarding cascade encryption, Maurer and Massey showed that cascade encryption is as strong as the first cipher. Matthew Green's blog post is a nice (easy to read) summary of multi-encryption security.
$endgroup$
add a comment |
$begingroup$
For the one-time-pad the answer is no, since it already achieves information-theoretic security (meaning that the ciphertext is statistically independent from the plaintext).
So, applying it twice doesn’t add any extra security (and may actually worsen security if the second key is not independent from the first).
New contributor
$endgroup$
add a comment |
$begingroup$
The intention of a good cipher is to remove all orders and bit arrangements in a plaintext and produce output ciphertext in which there is no distinguishable orders and arrangements obtainable.I mean that the ciphertext should be a uniform distribution.
By considering this remark, if we have a good encrypted output(means plaintext which is encrypted with a good encryption algorithm ex AES),we can not find bit orders on it and re-encrypting the same output is ineffective and useless. In other words, imposing more security affections on the second encrypted ciphertext based on using the same cipher with another key is not remarkable, However we can not give a global rule for this.
An obvious example of this cipher types is 2DES (or even 3DES). This cipher, encrypts a plaintext with 2 different keys. Again the security of ciphertext rely on the DES structure. but we should consider that the whole system(2DES or 3DES) is vulnerable to meet-in-the-middle attack and in this situation and also the key space is not the sum of keys( k1+k2).
Therefore using ciphers in this manner is not more common. but if we want to encrypt a plaintext with two different ciphers with different keys, the whole structure gives more stamina to ciphertext against cryptanalysis.
And about One-time pad encryption, this cipher has a perfect security, so imposing another encryption with different key is waste of time and resources.
$endgroup$
$begingroup$
I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
$endgroup$
– Maarten Bodewes♦
21 mins ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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
});
}
});
Mina is a new contributor. Be nice, and check out our Code of Conduct.
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%2f68685%2fdouble-encryption-one-time-pad%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
$begingroup$
The answer is we cannot improve the security of the one-time pad in this manner. Intuitively the reason is that the double one-time pad is just a less efficient one time pad.
The security of the traditional xor-based one-time pad is requires that the key $K$ is chosen uniformly at random for each message and that the key is at least as large as the message.
Because the xor operation is associative, we could re-write the double one-time pad as $C = M oplus(K_1 oplus K_2)$ or $ C = M oplus K$, where $K = K_1 oplus K_2$. Now if $K_1$ and $K_2$ are uniform, it is easy to show that $K$ is also uniform. Assuming the key and the message have length $n$
$P_K[k] = sum_{x in {0,1}^n}P_{K_1,K_2}[k_1 = x, k_2 = k oplus k_1] = frac{1}{2^n}$
In other words $K_1 oplus K_2$ could be simply replaced by a single uniform key, therefore $K_1$ is enough. Moreover the double one-time pad is inefficient because it requires two xor operations.
Regarding cascade encryption, Maurer and Massey showed that cascade encryption is as strong as the first cipher. Matthew Green's blog post is a nice (easy to read) summary of multi-encryption security.
$endgroup$
add a comment |
$begingroup$
The answer is we cannot improve the security of the one-time pad in this manner. Intuitively the reason is that the double one-time pad is just a less efficient one time pad.
The security of the traditional xor-based one-time pad is requires that the key $K$ is chosen uniformly at random for each message and that the key is at least as large as the message.
Because the xor operation is associative, we could re-write the double one-time pad as $C = M oplus(K_1 oplus K_2)$ or $ C = M oplus K$, where $K = K_1 oplus K_2$. Now if $K_1$ and $K_2$ are uniform, it is easy to show that $K$ is also uniform. Assuming the key and the message have length $n$
$P_K[k] = sum_{x in {0,1}^n}P_{K_1,K_2}[k_1 = x, k_2 = k oplus k_1] = frac{1}{2^n}$
In other words $K_1 oplus K_2$ could be simply replaced by a single uniform key, therefore $K_1$ is enough. Moreover the double one-time pad is inefficient because it requires two xor operations.
Regarding cascade encryption, Maurer and Massey showed that cascade encryption is as strong as the first cipher. Matthew Green's blog post is a nice (easy to read) summary of multi-encryption security.
$endgroup$
add a comment |
$begingroup$
The answer is we cannot improve the security of the one-time pad in this manner. Intuitively the reason is that the double one-time pad is just a less efficient one time pad.
The security of the traditional xor-based one-time pad is requires that the key $K$ is chosen uniformly at random for each message and that the key is at least as large as the message.
Because the xor operation is associative, we could re-write the double one-time pad as $C = M oplus(K_1 oplus K_2)$ or $ C = M oplus K$, where $K = K_1 oplus K_2$. Now if $K_1$ and $K_2$ are uniform, it is easy to show that $K$ is also uniform. Assuming the key and the message have length $n$
$P_K[k] = sum_{x in {0,1}^n}P_{K_1,K_2}[k_1 = x, k_2 = k oplus k_1] = frac{1}{2^n}$
In other words $K_1 oplus K_2$ could be simply replaced by a single uniform key, therefore $K_1$ is enough. Moreover the double one-time pad is inefficient because it requires two xor operations.
Regarding cascade encryption, Maurer and Massey showed that cascade encryption is as strong as the first cipher. Matthew Green's blog post is a nice (easy to read) summary of multi-encryption security.
$endgroup$
The answer is we cannot improve the security of the one-time pad in this manner. Intuitively the reason is that the double one-time pad is just a less efficient one time pad.
The security of the traditional xor-based one-time pad is requires that the key $K$ is chosen uniformly at random for each message and that the key is at least as large as the message.
Because the xor operation is associative, we could re-write the double one-time pad as $C = M oplus(K_1 oplus K_2)$ or $ C = M oplus K$, where $K = K_1 oplus K_2$. Now if $K_1$ and $K_2$ are uniform, it is easy to show that $K$ is also uniform. Assuming the key and the message have length $n$
$P_K[k] = sum_{x in {0,1}^n}P_{K_1,K_2}[k_1 = x, k_2 = k oplus k_1] = frac{1}{2^n}$
In other words $K_1 oplus K_2$ could be simply replaced by a single uniform key, therefore $K_1$ is enough. Moreover the double one-time pad is inefficient because it requires two xor operations.
Regarding cascade encryption, Maurer and Massey showed that cascade encryption is as strong as the first cipher. Matthew Green's blog post is a nice (easy to read) summary of multi-encryption security.
edited 28 mins ago
answered 40 mins ago
Marc IlungaMarc Ilunga
36617
36617
add a comment |
add a comment |
$begingroup$
For the one-time-pad the answer is no, since it already achieves information-theoretic security (meaning that the ciphertext is statistically independent from the plaintext).
So, applying it twice doesn’t add any extra security (and may actually worsen security if the second key is not independent from the first).
New contributor
$endgroup$
add a comment |
$begingroup$
For the one-time-pad the answer is no, since it already achieves information-theoretic security (meaning that the ciphertext is statistically independent from the plaintext).
So, applying it twice doesn’t add any extra security (and may actually worsen security if the second key is not independent from the first).
New contributor
$endgroup$
add a comment |
$begingroup$
For the one-time-pad the answer is no, since it already achieves information-theoretic security (meaning that the ciphertext is statistically independent from the plaintext).
So, applying it twice doesn’t add any extra security (and may actually worsen security if the second key is not independent from the first).
New contributor
$endgroup$
For the one-time-pad the answer is no, since it already achieves information-theoretic security (meaning that the ciphertext is statistically independent from the plaintext).
So, applying it twice doesn’t add any extra security (and may actually worsen security if the second key is not independent from the first).
New contributor
New contributor
answered 1 hour ago
guilhermemtrguilhermemtr
1114
1114
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
The intention of a good cipher is to remove all orders and bit arrangements in a plaintext and produce output ciphertext in which there is no distinguishable orders and arrangements obtainable.I mean that the ciphertext should be a uniform distribution.
By considering this remark, if we have a good encrypted output(means plaintext which is encrypted with a good encryption algorithm ex AES),we can not find bit orders on it and re-encrypting the same output is ineffective and useless. In other words, imposing more security affections on the second encrypted ciphertext based on using the same cipher with another key is not remarkable, However we can not give a global rule for this.
An obvious example of this cipher types is 2DES (or even 3DES). This cipher, encrypts a plaintext with 2 different keys. Again the security of ciphertext rely on the DES structure. but we should consider that the whole system(2DES or 3DES) is vulnerable to meet-in-the-middle attack and in this situation and also the key space is not the sum of keys( k1+k2).
Therefore using ciphers in this manner is not more common. but if we want to encrypt a plaintext with two different ciphers with different keys, the whole structure gives more stamina to ciphertext against cryptanalysis.
And about One-time pad encryption, this cipher has a perfect security, so imposing another encryption with different key is waste of time and resources.
$endgroup$
$begingroup$
I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
$endgroup$
– Maarten Bodewes♦
21 mins ago
add a comment |
$begingroup$
The intention of a good cipher is to remove all orders and bit arrangements in a plaintext and produce output ciphertext in which there is no distinguishable orders and arrangements obtainable.I mean that the ciphertext should be a uniform distribution.
By considering this remark, if we have a good encrypted output(means plaintext which is encrypted with a good encryption algorithm ex AES),we can not find bit orders on it and re-encrypting the same output is ineffective and useless. In other words, imposing more security affections on the second encrypted ciphertext based on using the same cipher with another key is not remarkable, However we can not give a global rule for this.
An obvious example of this cipher types is 2DES (or even 3DES). This cipher, encrypts a plaintext with 2 different keys. Again the security of ciphertext rely on the DES structure. but we should consider that the whole system(2DES or 3DES) is vulnerable to meet-in-the-middle attack and in this situation and also the key space is not the sum of keys( k1+k2).
Therefore using ciphers in this manner is not more common. but if we want to encrypt a plaintext with two different ciphers with different keys, the whole structure gives more stamina to ciphertext against cryptanalysis.
And about One-time pad encryption, this cipher has a perfect security, so imposing another encryption with different key is waste of time and resources.
$endgroup$
$begingroup$
I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
$endgroup$
– Maarten Bodewes♦
21 mins ago
add a comment |
$begingroup$
The intention of a good cipher is to remove all orders and bit arrangements in a plaintext and produce output ciphertext in which there is no distinguishable orders and arrangements obtainable.I mean that the ciphertext should be a uniform distribution.
By considering this remark, if we have a good encrypted output(means plaintext which is encrypted with a good encryption algorithm ex AES),we can not find bit orders on it and re-encrypting the same output is ineffective and useless. In other words, imposing more security affections on the second encrypted ciphertext based on using the same cipher with another key is not remarkable, However we can not give a global rule for this.
An obvious example of this cipher types is 2DES (or even 3DES). This cipher, encrypts a plaintext with 2 different keys. Again the security of ciphertext rely on the DES structure. but we should consider that the whole system(2DES or 3DES) is vulnerable to meet-in-the-middle attack and in this situation and also the key space is not the sum of keys( k1+k2).
Therefore using ciphers in this manner is not more common. but if we want to encrypt a plaintext with two different ciphers with different keys, the whole structure gives more stamina to ciphertext against cryptanalysis.
And about One-time pad encryption, this cipher has a perfect security, so imposing another encryption with different key is waste of time and resources.
$endgroup$
The intention of a good cipher is to remove all orders and bit arrangements in a plaintext and produce output ciphertext in which there is no distinguishable orders and arrangements obtainable.I mean that the ciphertext should be a uniform distribution.
By considering this remark, if we have a good encrypted output(means plaintext which is encrypted with a good encryption algorithm ex AES),we can not find bit orders on it and re-encrypting the same output is ineffective and useless. In other words, imposing more security affections on the second encrypted ciphertext based on using the same cipher with another key is not remarkable, However we can not give a global rule for this.
An obvious example of this cipher types is 2DES (or even 3DES). This cipher, encrypts a plaintext with 2 different keys. Again the security of ciphertext rely on the DES structure. but we should consider that the whole system(2DES or 3DES) is vulnerable to meet-in-the-middle attack and in this situation and also the key space is not the sum of keys( k1+k2).
Therefore using ciphers in this manner is not more common. but if we want to encrypt a plaintext with two different ciphers with different keys, the whole structure gives more stamina to ciphertext against cryptanalysis.
And about One-time pad encryption, this cipher has a perfect security, so imposing another encryption with different key is waste of time and resources.
edited 1 hour ago
answered 1 hour ago
Arsalan VahiArsalan Vahi
1169
1169
$begingroup$
I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
$endgroup$
– Maarten Bodewes♦
21 mins ago
add a comment |
$begingroup$
I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
$endgroup$
– Maarten Bodewes♦
21 mins ago
$begingroup$
I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
$endgroup$
– Maarten Bodewes♦
21 mins ago
$begingroup$
I'm not convinced that reencrypting the output of AES cannot strengthen the ciphertext in case weaknesses are found. For 2DES: even 2DES is more secure than single DES. It is not as strong as it should be given the doubling of the key size (which is actually also true for 3DES, only less so). So I'm finding myself disagreeing with a large portion of the answer (and agreeing with the last two sections).
$endgroup$
– Maarten Bodewes♦
21 mins ago
add a comment |
Mina is a new contributor. Be nice, and check out our Code of Conduct.
Mina is a new contributor. Be nice, and check out our Code of Conduct.
Mina is a new contributor. Be nice, and check out our Code of Conduct.
Mina is a new contributor. Be nice, and check out our Code of Conduct.
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%2f68685%2fdouble-encryption-one-time-pad%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
1
$begingroup$
An ideal one time pad already has perfect security, you can't just get twice as perfect. The second layer doesn't add much.
$endgroup$
– Natanael
1 hour ago
1
$begingroup$
Hi. Is the size of the 20 part within your control? 20 bits = 1 million values. That can easily be enumerated. A hundred keys won't prevent that. And what Natanael says.
$endgroup$
– Paul Uszak
1 hour ago
1
$begingroup$
I don't see how enumerating the key values makes any difference for the one-time pad. Even a single bit key for a single bit message would be secure.
$endgroup$
– Maarten Bodewes♦
19 mins ago