Key Exchange Protocol attack
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I am working on the exercise below which ask about whether it is possible to attack the following key exchange protocol on sharing session key $K_s$ between user $X$ and $Y$:
$X rightarrow Y : X | r$
$Y rightarrow X : E (r | K_s, K_{xy})$
$X rightarrow Y : E (r, K_s)$
where $K_{xy}$ is a pre-shared secret key between user $X$ and $Y$, $K_s$ is a session key, $E(m, k)$ is symmetric key encryption on message $m$, with key $k$
It appears to me that it is secure. Could any one can give me a hand or some hints on possible attack ?
cryptography
edited Dec 7 '18 at 17:10
kelalaka
329212
asked Dec 7 '18 at 12:57
MluiMlui
112
$endgroup$
add a comment |
$begingroup$
I am working on the exercise below which ask about whether it is possible to attack the following key exchange protocol on sharing session key $K_s$ between user $X$ and $Y$:
$X rightarrow Y : X | r$
$Y rightarrow X : E (r | K_s, K_{xy})$
$X rightarrow Y : E (r, K_s)$
where $K_{xy}$ is a pre-shared secret key between user $X$ and $Y$, $K_s$ is a session key, $E(m, k)$ is symmetric key encryption on message $m$, with key $k$
It appears to me that it is secure. Could any one can give me a hand or some hints on possible attack ?
cryptography
edited Dec 7 '18 at 17:10
kelalaka
329212
asked Dec 7 '18 at 12:57
MluiMlui
112
$endgroup$
$begingroup$
How does $Y$ know the $r$ in order to carry out step 2?
$endgroup$
– user10354138
Dec 7 '18 at 17:19
$begingroup$
@user10354138 It's sent in step 1.
$endgroup$
– Henno Brandsma
Dec 8 '18 at 6:42
add a comment |
$begingroup$
I am working on the exercise below which ask about whether it is possible to attack the following key exchange protocol on sharing session key $K_s$ between user $X$ and $Y$:
$X rightarrow Y : X | r$
$Y rightarrow X : E (r | K_s, K_{xy})$
$X rightarrow Y : E (r, K_s)$
where $K_{xy}$ is a pre-shared secret key between user $X$ and $Y$, $K_s$ is a session key, $E(m, k)$ is symmetric key encryption on message $m$, with key $k$
It appears to me that it is secure. Could any one can give me a hand or some hints on possible attack ?
cryptography
edited Dec 7 '18 at 17:10
kelalaka
329212
asked Dec 7 '18 at 12:57
MluiMlui
112
$endgroup$
I am working on the exercise below which ask about whether it is possible to attack the following key exchange protocol on sharing session key $K_s$ between user $X$ and $Y$:
$X rightarrow Y : X | r$
$Y rightarrow X : E (r | K_s, K_{xy})$
$X rightarrow Y : E (r, K_s)$
where $K_{xy}$ is a pre-shared secret key between user $X$ and $Y$, $K_s$ is a session key, $E(m, k)$ is symmetric key encryption on message $m$, with key $k$
It appears to me that it is secure. Could any one can give me a hand or some hints on possible attack ?
cryptography
cryptography
edited Dec 7 '18 at 17:10
kelalaka
329212
asked Dec 7 '18 at 12:57
MluiMlui
112
edited Dec 7 '18 at 17:10
kelalaka
329212
asked Dec 7 '18 at 12:57
MluiMlui
112
edited Dec 7 '18 at 17:10
kelalaka
329212
edited Dec 7 '18 at 17:10
kelalaka
329212
edited Dec 7 '18 at 17:10
kelalaka
329212
329212
asked Dec 7 '18 at 12:57
MluiMlui
112
asked Dec 7 '18 at 12:57
MluiMlui
112
asked Dec 7 '18 at 12:57
MluiMlui
112
112
$begingroup$
How does $Y$ know the $r$ in order to carry out step 2?
$endgroup$
– user10354138
Dec 7 '18 at 17:19
$begingroup$
@user10354138 It's sent in step 1.
$endgroup$
– Henno Brandsma
Dec 8 '18 at 6:42
add a comment |
$begingroup$
How does $Y$ know the $r$ in order to carry out step 2?
$endgroup$
– user10354138
Dec 7 '18 at 17:19
$begingroup$
@user10354138 It's sent in step 1.
$endgroup$
– Henno Brandsma
Dec 8 '18 at 6:42
$begingroup$
How does $Y$ know the $r$ in order to carry out step 2?
$endgroup$
– user10354138
Dec 7 '18 at 17:19
$begingroup$
How does $Y$ know the $r$ in order to carry out step 2?
$endgroup$
– user10354138
Dec 7 '18 at 17:19
$begingroup$
@user10354138 It's sent in step 1.
$endgroup$
– Henno Brandsma
Dec 8 '18 at 6:42
$begingroup$
@user10354138 It's sent in step 1.
$endgroup$
– Henno Brandsma
Dec 8 '18 at 6:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are two problems in this protocol.
An attacker can start the protocol instead of $X$ since the id $X$ sent unencrypted and the attacker can generate a random $r$. $Y$ generates the key sends back to $X$. In the first hand, there is no rejection for $Y$
The attacker can store all transmitted $E (r | K_s, K_{xy})$ and transmitted encrypted messages with $K_{s_i}$for a future usage. When he successfully attacked one of the $X$ and $Y$, he can decrypt all the messages first by decrypting the stored $E (r | K_s, K_{xy})$. In short, this protocol has no forward secrecy.
answered Dec 13 '18 at 21:46
kelalakakelalaka
329212
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are two problems in this protocol.
An attacker can start the protocol instead of $X$ since the id $X$ sent unencrypted and the attacker can generate a random $r$. $Y$ generates the key sends back to $X$. In the first hand, there is no rejection for $Y$
The attacker can store all transmitted $E (r | K_s, K_{xy})$ and transmitted encrypted messages with $K_{s_i}$for a future usage. When he successfully attacked one of the $X$ and $Y$, he can decrypt all the messages first by decrypting the stored $E (r | K_s, K_{xy})$. In short, this protocol has no forward secrecy.
answered Dec 13 '18 at 21:46
kelalakakelalaka
329212
$endgroup$
add a comment |
$begingroup$
There are two problems in this protocol.
An attacker can start the protocol instead of $X$ since the id $X$ sent unencrypted and the attacker can generate a random $r$. $Y$ generates the key sends back to $X$. In the first hand, there is no rejection for $Y$
The attacker can store all transmitted $E (r | K_s, K_{xy})$ and transmitted encrypted messages with $K_{s_i}$for a future usage. When he successfully attacked one of the $X$ and $Y$, he can decrypt all the messages first by decrypting the stored $E (r | K_s, K_{xy})$. In short, this protocol has no forward secrecy.
answered Dec 13 '18 at 21:46
kelalakakelalaka
329212
$endgroup$
add a comment |
$begingroup$
There are two problems in this protocol.
An attacker can start the protocol instead of $X$ since the id $X$ sent unencrypted and the attacker can generate a random $r$. $Y$ generates the key sends back to $X$. In the first hand, there is no rejection for $Y$
The attacker can store all transmitted $E (r | K_s, K_{xy})$ and transmitted encrypted messages with $K_{s_i}$for a future usage. When he successfully attacked one of the $X$ and $Y$, he can decrypt all the messages first by decrypting the stored $E (r | K_s, K_{xy})$. In short, this protocol has no forward secrecy.
answered Dec 13 '18 at 21:46
kelalakakelalaka
329212
$endgroup$
There are two problems in this protocol.
An attacker can start the protocol instead of $X$ since the id $X$ sent unencrypted and the attacker can generate a random $r$. $Y$ generates the key sends back to $X$. In the first hand, there is no rejection for $Y$
The attacker can store all transmitted $E (r | K_s, K_{xy})$ and transmitted encrypted messages with $K_{s_i}$for a future usage. When he successfully attacked one of the $X$ and $Y$, he can decrypt all the messages first by decrypting the stored $E (r | K_s, K_{xy})$. In short, this protocol has no forward secrecy.
answered Dec 13 '18 at 21:46
kelalakakelalaka
329212
answered Dec 13 '18 at 21:46
kelalakakelalaka
329212
answered Dec 13 '18 at 21:46
kelalakakelalaka
329212
answered Dec 13 '18 at 21:46
kelalakakelalaka
329212
329212
add a comment |
add a comment |
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$begingroup$
How does $Y$ know the $r$ in order to carry out step 2?
$endgroup$
– user10354138
Dec 7 '18 at 17:19
$begingroup$
@user10354138 It's sent in step 1.
$endgroup$
– Henno Brandsma
Dec 8 '18 at 6:42