Differential and Linear trail propagation in Noekeondifferential and linear cryptanalysisDifference between...
In Aliens, how many people were on LV-426 before the Marines arrived?
Practical application of matrices and determinants
Probably overheated black color SMD pads
Can you move over difficult terrain with only 5' of movement?
Turning a hard to access nut?
Dual Irish/Britsh citizens
Comment Box for Substitution Method of Integrals
How can my new character not be a role-playing handicap to the party?
What does Deadpool mean by "left the house in that shirt"?
What is the term when voters “dishonestly” choose something that they do not want to choose?
How to generate binary array whose elements with values 1 are randomly drawn
Wrapping homogeneous python objects
Is there a term for accumulated dirt on the outside of your hands and feet?
How to define limit operations in general topological spaces? Are nets able to do this?
Do I need to be arrogant to get ahead?
A Ri-diddley-iley Riddle
Geography in 3D perspective
Replace four times with sed
Using Past-Perfect interchangeably with the Past Continuous
How are passwords stolen from companies if they only store hashes?
Can other pieces capture a threatening piece and prevent a checkmate?
Hausdorff dimension of the boundary of fibres of Lipschitz maps
Maths symbols and unicode-math input inside siunitx commands
Deletion of copy-ctor & copy-assignment - public, private or protected?
Differential and Linear trail propagation in Noekeon
differential and linear cryptanalysisDifference between linear cryptanalysis and differential cryptanalysisWhat is the complexity for attacking 3DES in linear or differential cryptanalysis?Differential CryptanalysisDifferential & linear characteristics for integer multiplicationWhat is the meaning of Maximum Expected Differential/Linear Probability (MEDP/MELP)?Understanding the wide trail design strategyit is possible to use quantum algorithm search (Grover's algorithm) for new searching strategies for differential and linear attacksLinear cryptanalysis and number of linear approximationsHow does linear vs. non-linear operations relate to cryptographic security and differential cryptanalysis?
$begingroup$
In the Noekeon Cipher Specification they write the following :
The propagation through Lambda is denoted by $(a rightarrow A)$, also called a
step. Because of the linearity of Lambda it is fully deterministic:
both for LC and DC patterns, we have: $A = operatorname{Lambda}(a)$. The fact that the
relation is the same for LC and DC is thanks to the fact that the
Lambda is an orthogonal function. If represented in a matrix, its
inverse is its transpose.
I'm having a hard time understanding why the orthogonality of Lambda affects the relation with regards to selection patterns (LC).
Why does the orthogonality of Lambda make it so that the relationship is the same as for DC ? How would the selection pattern propagate through the linear layer if Lambda was not orthogonal ?
cryptanalysis block-cipher linear-cryptanalysis differential-analysis
asked 6 hours ago
YuonYuon
737
$endgroup$
add a comment |
$begingroup$
In the Noekeon Cipher Specification they write the following :
The propagation through Lambda is denoted by $(a rightarrow A)$, also called a
step. Because of the linearity of Lambda it is fully deterministic:
both for LC and DC patterns, we have: $A = operatorname{Lambda}(a)$. The fact that the
relation is the same for LC and DC is thanks to the fact that the
Lambda is an orthogonal function. If represented in a matrix, its
inverse is its transpose.
I'm having a hard time understanding why the orthogonality of Lambda affects the relation with regards to selection patterns (LC).
Why does the orthogonality of Lambda make it so that the relationship is the same as for DC ? How would the selection pattern propagate through the linear layer if Lambda was not orthogonal ?
cryptanalysis block-cipher linear-cryptanalysis differential-analysis
asked 6 hours ago
YuonYuon
737
$endgroup$
add a comment |
$begingroup$
In the Noekeon Cipher Specification they write the following :
The propagation through Lambda is denoted by $(a rightarrow A)$, also called a
step. Because of the linearity of Lambda it is fully deterministic:
both for LC and DC patterns, we have: $A = operatorname{Lambda}(a)$. The fact that the
relation is the same for LC and DC is thanks to the fact that the
Lambda is an orthogonal function. If represented in a matrix, its
inverse is its transpose.
I'm having a hard time understanding why the orthogonality of Lambda affects the relation with regards to selection patterns (LC).
Why does the orthogonality of Lambda make it so that the relationship is the same as for DC ? How would the selection pattern propagate through the linear layer if Lambda was not orthogonal ?
cryptanalysis block-cipher linear-cryptanalysis differential-analysis
asked 6 hours ago
YuonYuon
737
$endgroup$
In the Noekeon Cipher Specification they write the following :
The propagation through Lambda is denoted by $(a rightarrow A)$, also called a
step. Because of the linearity of Lambda it is fully deterministic:
both for LC and DC patterns, we have: $A = operatorname{Lambda}(a)$. The fact that the
relation is the same for LC and DC is thanks to the fact that the
Lambda is an orthogonal function. If represented in a matrix, its
inverse is its transpose.
I'm having a hard time understanding why the orthogonality of Lambda affects the relation with regards to selection patterns (LC).
Why does the orthogonality of Lambda make it so that the relationship is the same as for DC ? How would the selection pattern propagate through the linear layer if Lambda was not orthogonal ?
cryptanalysis block-cipher linear-cryptanalysis differential-analysis
cryptanalysis block-cipher linear-cryptanalysis differential-analysis
asked 6 hours ago
YuonYuon
737
asked 6 hours ago
YuonYuon
737
edited 6 hours ago
Yuon
asked 6 hours ago
YuonYuon
737
asked 6 hours ago
YuonYuon
737
asked 6 hours ago
YuonYuon
737
737
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is due to the duality between linear and differential trails.
Let $L$ be an invertible linear map on $mathbb{F}_2^n$, think of it as a matrix for convenience.
In general, a nonzero differential $Delta_1 to Delta_2$ over $L$ must satisfy
$$Delta_2 = L,Delta_1.$$
A nonzero linear approximation $u_1 to u_2$, however, must satisfy
$$u_2 = L^{-top},u_1$$
An elementary way to see this is to observe that $u_1^top x = u_2^top (Lx)$ is equivalent to $u_1^top x = (L^top,u_2)^top x$. This holds for all $x in mathbb{F}_2^n$ whenever $u_2 = L^{-top},u_1$, and otherwise for half (some hyperplane) the $x$.
If $L$ is orthogonal, then $L^{-T} = L$. So then we have both $Delta_2 = LDelta_1$ and $u_2 = L u_1$.
answered 3 hours ago
AlephAleph
1,2961220
$endgroup$
$begingroup$
I suspected it was because of something like that. Could you just give some intuition as to why we want $u^T_1 x = u^T_2(Lx)$ in first place ? If I had to come up with that, I'd think it's the other way around $u^T_2 x = u^T_1 (Lx)$ just like the differential case.
$endgroup$
– Yuon
1 hour 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
});
}
});
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%2f68085%2fdifferential-and-linear-trail-propagation-in-noekeon%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is due to the duality between linear and differential trails.
Let $L$ be an invertible linear map on $mathbb{F}_2^n$, think of it as a matrix for convenience.
In general, a nonzero differential $Delta_1 to Delta_2$ over $L$ must satisfy
$$Delta_2 = L,Delta_1.$$
A nonzero linear approximation $u_1 to u_2$, however, must satisfy
$$u_2 = L^{-top},u_1$$
An elementary way to see this is to observe that $u_1^top x = u_2^top (Lx)$ is equivalent to $u_1^top x = (L^top,u_2)^top x$. This holds for all $x in mathbb{F}_2^n$ whenever $u_2 = L^{-top},u_1$, and otherwise for half (some hyperplane) the $x$.
If $L$ is orthogonal, then $L^{-T} = L$. So then we have both $Delta_2 = LDelta_1$ and $u_2 = L u_1$.
answered 3 hours ago
AlephAleph
1,2961220
$endgroup$
$begingroup$
I suspected it was because of something like that. Could you just give some intuition as to why we want $u^T_1 x = u^T_2(Lx)$ in first place ? If I had to come up with that, I'd think it's the other way around $u^T_2 x = u^T_1 (Lx)$ just like the differential case.
$endgroup$
– Yuon
1 hour ago
add a comment |
$begingroup$
This is due to the duality between linear and differential trails.
Let $L$ be an invertible linear map on $mathbb{F}_2^n$, think of it as a matrix for convenience.
In general, a nonzero differential $Delta_1 to Delta_2$ over $L$ must satisfy
$$Delta_2 = L,Delta_1.$$
A nonzero linear approximation $u_1 to u_2$, however, must satisfy
$$u_2 = L^{-top},u_1$$
An elementary way to see this is to observe that $u_1^top x = u_2^top (Lx)$ is equivalent to $u_1^top x = (L^top,u_2)^top x$. This holds for all $x in mathbb{F}_2^n$ whenever $u_2 = L^{-top},u_1$, and otherwise for half (some hyperplane) the $x$.
If $L$ is orthogonal, then $L^{-T} = L$. So then we have both $Delta_2 = LDelta_1$ and $u_2 = L u_1$.
answered 3 hours ago
AlephAleph
1,2961220
$endgroup$
$begingroup$
I suspected it was because of something like that. Could you just give some intuition as to why we want $u^T_1 x = u^T_2(Lx)$ in first place ? If I had to come up with that, I'd think it's the other way around $u^T_2 x = u^T_1 (Lx)$ just like the differential case.
$endgroup$
– Yuon
1 hour ago
add a comment |
$begingroup$
This is due to the duality between linear and differential trails.
Let $L$ be an invertible linear map on $mathbb{F}_2^n$, think of it as a matrix for convenience.
In general, a nonzero differential $Delta_1 to Delta_2$ over $L$ must satisfy
$$Delta_2 = L,Delta_1.$$
A nonzero linear approximation $u_1 to u_2$, however, must satisfy
$$u_2 = L^{-top},u_1$$
An elementary way to see this is to observe that $u_1^top x = u_2^top (Lx)$ is equivalent to $u_1^top x = (L^top,u_2)^top x$. This holds for all $x in mathbb{F}_2^n$ whenever $u_2 = L^{-top},u_1$, and otherwise for half (some hyperplane) the $x$.
If $L$ is orthogonal, then $L^{-T} = L$. So then we have both $Delta_2 = LDelta_1$ and $u_2 = L u_1$.
answered 3 hours ago
AlephAleph
1,2961220
$endgroup$
This is due to the duality between linear and differential trails.
Let $L$ be an invertible linear map on $mathbb{F}_2^n$, think of it as a matrix for convenience.
In general, a nonzero differential $Delta_1 to Delta_2$ over $L$ must satisfy
$$Delta_2 = L,Delta_1.$$
A nonzero linear approximation $u_1 to u_2$, however, must satisfy
$$u_2 = L^{-top},u_1$$
An elementary way to see this is to observe that $u_1^top x = u_2^top (Lx)$ is equivalent to $u_1^top x = (L^top,u_2)^top x$. This holds for all $x in mathbb{F}_2^n$ whenever $u_2 = L^{-top},u_1$, and otherwise for half (some hyperplane) the $x$.
If $L$ is orthogonal, then $L^{-T} = L$. So then we have both $Delta_2 = LDelta_1$ and $u_2 = L u_1$.
answered 3 hours ago
AlephAleph
1,2961220
answered 3 hours ago
AlephAleph
1,2961220
answered 3 hours ago
AlephAleph
1,2961220
answered 3 hours ago
AlephAleph
1,2961220
1,2961220
$begingroup$
I suspected it was because of something like that. Could you just give some intuition as to why we want $u^T_1 x = u^T_2(Lx)$ in first place ? If I had to come up with that, I'd think it's the other way around $u^T_2 x = u^T_1 (Lx)$ just like the differential case.
$endgroup$
– Yuon
1 hour ago
add a comment |
$begingroup$
I suspected it was because of something like that. Could you just give some intuition as to why we want $u^T_1 x = u^T_2(Lx)$ in first place ? If I had to come up with that, I'd think it's the other way around $u^T_2 x = u^T_1 (Lx)$ just like the differential case.
$endgroup$
– Yuon
1 hour ago
$begingroup$
I suspected it was because of something like that. Could you just give some intuition as to why we want $u^T_1 x = u^T_2(Lx)$ in first place ? If I had to come up with that, I'd think it's the other way around $u^T_2 x = u^T_1 (Lx)$ just like the differential case.
$endgroup$
– Yuon
1 hour ago
$begingroup$
I suspected it was because of something like that. Could you just give some intuition as to why we want $u^T_1 x = u^T_2(Lx)$ in first place ? If I had to come up with that, I'd think it's the other way around $u^T_2 x = u^T_1 (Lx)$ just like the differential case.
$endgroup$
– Yuon
1 hour 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%2f68085%2fdifferential-and-linear-trail-propagation-in-noekeon%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
