Differential and Linear trail propagation in Noekeon
$begingroup$
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 7 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 7 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 7 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 7 hours ago
YuonYuon
737
asked 7 hours ago
YuonYuon
737
edited 7 hours ago
Yuon
asked 7 hours ago
YuonYuon
737
asked 7 hours ago
YuonYuon
737
asked 7 hours ago
YuonYuon
737
737
add a comment |
add a comment |
1 Answer
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$begingroup$
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 4 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
2 hours ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 4 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
2 hours 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 4 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
2 hours 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 4 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 4 hours ago
AlephAleph
1,2961220
answered 4 hours ago
AlephAleph
1,2961220
answered 4 hours ago
AlephAleph
1,2961220
answered 4 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
2 hours 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
2 hours 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
2 hours 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
2 hours ago
add a comment |
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