Does adding complexity mean a more secure cipher?
$begingroup$
I have a cryptography workshop question I'm having trouble with as follows;
Person A creates a cipher $E_k(m)$ which produces a ciphertext from message "m" using key "k". The function inside E is kept secret but the length of $E_K(m)$ is known.
Person B recommends "increasing" security of the cipher by instead doing :
$(E_k(m) oplus m) || (E_k(m) oplus 1111...11)$
Does this in fact increase security of the cipher or increase new problems.
My thinking is, depending on the function within E, xoring the output of the cipher with the plaintext message could expose the key, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.
Unfortunately the above context is all I have been provided for this question.
encryption
$endgroup$
migrated from stackoverflow.com 2 days ago
This question came from our site for professional and enthusiast programmers.
add a comment |
$begingroup$
I have a cryptography workshop question I'm having trouble with as follows;
Person A creates a cipher $E_k(m)$ which produces a ciphertext from message "m" using key "k". The function inside E is kept secret but the length of $E_K(m)$ is known.
Person B recommends "increasing" security of the cipher by instead doing :
$(E_k(m) oplus m) || (E_k(m) oplus 1111...11)$
Does this in fact increase security of the cipher or increase new problems.
My thinking is, depending on the function within E, xoring the output of the cipher with the plaintext message could expose the key, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.
Unfortunately the above context is all I have been provided for this question.
encryption
$endgroup$
migrated from stackoverflow.com 2 days ago
This question came from our site for professional and enthusiast programmers.
add a comment |
$begingroup$
I have a cryptography workshop question I'm having trouble with as follows;
Person A creates a cipher $E_k(m)$ which produces a ciphertext from message "m" using key "k". The function inside E is kept secret but the length of $E_K(m)$ is known.
Person B recommends "increasing" security of the cipher by instead doing :
$(E_k(m) oplus m) || (E_k(m) oplus 1111...11)$
Does this in fact increase security of the cipher or increase new problems.
My thinking is, depending on the function within E, xoring the output of the cipher with the plaintext message could expose the key, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.
Unfortunately the above context is all I have been provided for this question.
encryption
$endgroup$
I have a cryptography workshop question I'm having trouble with as follows;
Person A creates a cipher $E_k(m)$ which produces a ciphertext from message "m" using key "k". The function inside E is kept secret but the length of $E_K(m)$ is known.
Person B recommends "increasing" security of the cipher by instead doing :
$(E_k(m) oplus m) || (E_k(m) oplus 1111...11)$
Does this in fact increase security of the cipher or increase new problems.
My thinking is, depending on the function within E, xoring the output of the cipher with the plaintext message could expose the key, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.
Unfortunately the above context is all I have been provided for this question.
encryption
encryption
edited 2 days ago
Ella Rose♦
17.1k44584
17.1k44584
asked 2 days ago
melloncolliemelloncollie
417
417
migrated from stackoverflow.com 2 days ago
This question came from our site for professional and enthusiast programmers.
migrated from stackoverflow.com 2 days ago
This question came from our site for professional and enthusiast programmers.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
xoring the output of the cipher with the plaintext message
Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.
If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".
$(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$
The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.
In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.
$$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_{text{a}} = E_k(m) oplus m\c_{text{b}} = E_k(m) oplus 1111 dots 11\c' = c_{text{b}} oplus 1111dots11\m = c_{text{a}} oplus c'$$
The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$
So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$
This scheme is completely broken.
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.
It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)
Does this in fact increase security of the cipher or increase new problems.
It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.
It helps to list:
- What you have
- What your adversary can do
- What you want to accomplish (in very specific terms).
If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)
$endgroup$
2
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
2 days ago
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
2 days ago
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
2 days ago
add a comment |
$begingroup$
This is indeed a example of complexity not adding security and actually weakening it.
The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.
Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
And we can easily get $m$ as $m' oplus 111ldots 11$
$endgroup$
add a comment |
$begingroup$
Even ignoring the complete and total brokenness of the full cipher, the $(E_k(m) oplus 1111...11)$ component keeps any possible first half from adding security. Since XORing the ciphertext with all 1
s is an operation independent of either the key or the message, the cipher
$(f(x)) || (E_k(m) oplus 1111...11)$
permits the recovery of $E_k(m)$ for any possible $f(x)$.
$endgroup$
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%2f68705%2fdoes-adding-complexity-mean-a-more-secure-cipher%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
$begingroup$
xoring the output of the cipher with the plaintext message
Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.
If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".
$(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$
The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.
In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.
$$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_{text{a}} = E_k(m) oplus m\c_{text{b}} = E_k(m) oplus 1111 dots 11\c' = c_{text{b}} oplus 1111dots11\m = c_{text{a}} oplus c'$$
The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$
So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$
This scheme is completely broken.
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.
It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)
Does this in fact increase security of the cipher or increase new problems.
It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.
It helps to list:
- What you have
- What your adversary can do
- What you want to accomplish (in very specific terms).
If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)
$endgroup$
2
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
2 days ago
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
2 days ago
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
2 days ago
add a comment |
$begingroup$
xoring the output of the cipher with the plaintext message
Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.
If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".
$(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$
The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.
In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.
$$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_{text{a}} = E_k(m) oplus m\c_{text{b}} = E_k(m) oplus 1111 dots 11\c' = c_{text{b}} oplus 1111dots11\m = c_{text{a}} oplus c'$$
The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$
So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$
This scheme is completely broken.
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.
It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)
Does this in fact increase security of the cipher or increase new problems.
It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.
It helps to list:
- What you have
- What your adversary can do
- What you want to accomplish (in very specific terms).
If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)
$endgroup$
2
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
2 days ago
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
2 days ago
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
2 days ago
add a comment |
$begingroup$
xoring the output of the cipher with the plaintext message
Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.
If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".
$(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$
The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.
In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.
$$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_{text{a}} = E_k(m) oplus m\c_{text{b}} = E_k(m) oplus 1111 dots 11\c' = c_{text{b}} oplus 1111dots11\m = c_{text{a}} oplus c'$$
The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$
So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$
This scheme is completely broken.
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.
It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)
Does this in fact increase security of the cipher or increase new problems.
It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.
It helps to list:
- What you have
- What your adversary can do
- What you want to accomplish (in very specific terms).
If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)
$endgroup$
xoring the output of the cipher with the plaintext message
Xoring the message into the ciphertext removes the ability to decrypt the ciphertext.
If all you have is $k, c = E_k(m) oplus m$, then you need to know $m$ in order to strip the external $m$ off of $E_k(m)$ before you can apply $m = D_k(E_k(m))$; Basically, you would need to know the message in order to "decrypt" the message, but since you already know the message, then there would be no knowledge gained from "decrypting".
$(E_k(m)oplus m)||(E_k(m) oplus 1111...11)$
The previous section was striked out, because there was more to the suggestion than simply xoring the message into the ciphertext.
In fact, the complete suggestion is far, far worse then simply implying the inability to decrypt a ciphertext: Anyone can decrypt a ciphertext from this scheme without requiring the key.
$$c = (E_k(m) oplus m) || E_k(m) oplus 1111dots 11)\c_{text{a}} = E_k(m) oplus m\c_{text{b}} = E_k(m) oplus 1111 dots 11\c' = c_{text{b}} oplus 1111dots11\m = c_{text{a}} oplus c'$$
The value $1111dots11$ is known to all, so anyone can compute $$E_k(m) = E_k(m) oplus 1111dots11 oplus 1111dots11$$
So $E_k(m)$ is effectively public knowledge, so again anyone can compute $$m = E_k(m) oplus m oplus E_k(m)$$
This scheme is completely broken.
I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.
See the previous section - "encryption" is an invertible process: You have to be able to undo the transformation so that you can retrieve the plaintext from any given ciphertext.
It is completely broken, so you won't find anything similar to it (other than situations where people asked this same question and found it to be broken)
Does this in fact increase security of the cipher or increase new problems.
It's a lot easier to determine whether or not doing x or y will increase/decrease security once you have a clear goal of what it means to be secure. This is a context dependent notion.
It helps to list:
- What you have
- What your adversary can do
- What you want to accomplish (in very specific terms).
If you don't know where to start, then look for the standard notions of security that cryptographers have already worked out for the context of interest (e.g. indistinguishability for ciphers)
edited 2 days ago
answered 2 days ago
Ella Rose♦Ella Rose
17.1k44584
17.1k44584
2
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
2 days ago
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
2 days ago
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
2 days ago
add a comment |
2
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
2 days ago
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
2 days ago
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
2 days ago
2
2
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
2 days ago
$begingroup$
In this case we actually could decrypt the message. The problem is everybody can :-(. Because in addition to $E_k(m) oplus m$ we also get $E_k(m) oplus 111ldots 11$
$endgroup$
– Marc Ilunga
2 days ago
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
2 days ago
$begingroup$
@MarcIlunga I actually failed to pay attention to the mathjax part of the question, which is different than what the text part asks! Thanks for bringing that to my attention
$endgroup$
– Ella Rose♦
2 days ago
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
2 days ago
$begingroup$
Thank you to both of you. Greatly appreciate the help. Definitely need to work on my number theory.
$endgroup$
– melloncollie
2 days ago
add a comment |
$begingroup$
This is indeed a example of complexity not adding security and actually weakening it.
The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.
Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
And we can easily get $m$ as $m' oplus 111ldots 11$
$endgroup$
add a comment |
$begingroup$
This is indeed a example of complexity not adding security and actually weakening it.
The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.
Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
And we can easily get $m$ as $m' oplus 111ldots 11$
$endgroup$
add a comment |
$begingroup$
This is indeed a example of complexity not adding security and actually weakening it.
The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.
Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
And we can easily get $m$ as $m' oplus 111ldots 11$
$endgroup$
This is indeed a example of complexity not adding security and actually weakening it.
The second encryption can be written as $c = c_1|| c_2$, where $c_1 = E_k(m) oplus m$ and $c_2 = E_k(m) oplus 111ldots11$.
Now observe that $m' = c_1 oplus c_2 = m oplus 111ldots11$.
And we can easily get $m$ as $m' oplus 111ldots 11$
answered 2 days ago
Marc IlungaMarc Ilunga
42817
42817
add a comment |
add a comment |
$begingroup$
Even ignoring the complete and total brokenness of the full cipher, the $(E_k(m) oplus 1111...11)$ component keeps any possible first half from adding security. Since XORing the ciphertext with all 1
s is an operation independent of either the key or the message, the cipher
$(f(x)) || (E_k(m) oplus 1111...11)$
permits the recovery of $E_k(m)$ for any possible $f(x)$.
$endgroup$
add a comment |
$begingroup$
Even ignoring the complete and total brokenness of the full cipher, the $(E_k(m) oplus 1111...11)$ component keeps any possible first half from adding security. Since XORing the ciphertext with all 1
s is an operation independent of either the key or the message, the cipher
$(f(x)) || (E_k(m) oplus 1111...11)$
permits the recovery of $E_k(m)$ for any possible $f(x)$.
$endgroup$
add a comment |
$begingroup$
Even ignoring the complete and total brokenness of the full cipher, the $(E_k(m) oplus 1111...11)$ component keeps any possible first half from adding security. Since XORing the ciphertext with all 1
s is an operation independent of either the key or the message, the cipher
$(f(x)) || (E_k(m) oplus 1111...11)$
permits the recovery of $E_k(m)$ for any possible $f(x)$.
$endgroup$
Even ignoring the complete and total brokenness of the full cipher, the $(E_k(m) oplus 1111...11)$ component keeps any possible first half from adding security. Since XORing the ciphertext with all 1
s is an operation independent of either the key or the message, the cipher
$(f(x)) || (E_k(m) oplus 1111...11)$
permits the recovery of $E_k(m)$ for any possible $f(x)$.
answered 2 days ago
MarkMark
1885
1885
add a comment |
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%2f68705%2fdoes-adding-complexity-mean-a-more-secure-cipher%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