Differential and Linear trail propagation in Noekeon
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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: $A = operatornameLambda(a)$. 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 Mar 17 at 16:42
YuonYuon
837
$endgroup$
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: $A = operatornameLambda(a)$. 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 Mar 17 at 16:42
YuonYuon
837
$endgroup$
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: $A = operatornameLambda(a)$. 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 Mar 17 at 16:42
YuonYuon
837
$endgroup$
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: $A = operatornameLambda(a)$. 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 Mar 17 at 16:42
YuonYuon
837
asked Mar 17 at 16:42
YuonYuon
837
edited Mar 17 at 17:11
Yuon
asked Mar 17 at 16:42
YuonYuon
837
asked Mar 17 at 16:42
YuonYuon
837
asked Mar 17 at 16:42
YuonYuon
837
837
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 $mathbbF_2^n$, 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 $x in mathbbF_2^n$ 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 Mar 17 at 20:04
AlephAleph
1,3161220
$endgroup$
$begingroup$
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
Mar 17 at 21:43
$begingroup$
@Yuon You can think of the output mask $u_2$ as a selection of some of the output bits, so you want $u_2^top (Lx)$ because $Lx$ is the output.
$endgroup$
– Aleph
Mar 18 at 10:03
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 $mathbbF_2^n$, 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 $x in mathbbF_2^n$ 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 Mar 17 at 20:04
AlephAleph
1,3161220
$endgroup$
$begingroup$
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
Mar 17 at 21:43
$begingroup$
@Yuon You can think of the output mask $u_2$ as a selection of some of the output bits, so you want $u_2^top (Lx)$ because $Lx$ is the output.
$endgroup$
– Aleph
Mar 18 at 10:03
add a comment |
$begingroup$
This is due to the duality between linear and differential trails.
Let $L$ be an invertible linear map on $mathbbF_2^n$, 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 $x in mathbbF_2^n$ 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 Mar 17 at 20:04
AlephAleph
1,3161220
$endgroup$
$begingroup$
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
Mar 17 at 21:43
$begingroup$
@Yuon You can think of the output mask $u_2$ as a selection of some of the output bits, so you want $u_2^top (Lx)$ because $Lx$ is the output.
$endgroup$
– Aleph
Mar 18 at 10:03
add a comment |
$begingroup$
This is due to the duality between linear and differential trails.
Let $L$ be an invertible linear map on $mathbbF_2^n$, 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 $x in mathbbF_2^n$ 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 Mar 17 at 20:04
AlephAleph
1,3161220
$endgroup$
This is due to the duality between linear and differential trails.
Let $L$ be an invertible linear map on $mathbbF_2^n$, 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 $x in mathbbF_2^n$ 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 Mar 17 at 20:04
AlephAleph
1,3161220
answered Mar 17 at 20:04
AlephAleph
1,3161220
answered Mar 17 at 20:04
AlephAleph
1,3161220
answered Mar 17 at 20:04
AlephAleph
1,3161220
1,3161220
$begingroup$
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
Mar 17 at 21:43
$begingroup$
@Yuon You can think of the output mask $u_2$ as a selection of some of the output bits, so you want $u_2^top (Lx)$ because $Lx$ is the output.
$endgroup$
– Aleph
Mar 18 at 10:03
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
Mar 17 at 21:43
$begingroup$
@Yuon You can think of the output mask $u_2$ as a selection of some of the output bits, so you want $u_2^top (Lx)$ because $Lx$ is the output.
$endgroup$
– Aleph
Mar 18 at 10:03
$begingroup$
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
Mar 17 at 21:43
$begingroup$
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
Mar 17 at 21:43
$begingroup$
@Yuon You can think of the output mask $u_2$ as a selection of some of the output bits, so you want $u_2^top (Lx)$ because $Lx$ is the output.
$endgroup$
– Aleph
Mar 18 at 10:03
$begingroup$
@Yuon You can think of the output mask $u_2$ as a selection of some of the output bits, so you want $u_2^top (Lx)$ because $Lx$ is the output.
$endgroup$
– Aleph
Mar 18 at 10:03
add a comment |
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