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I have a scheme and I don't know that this scheme is semantic security or not.

-Secret keys $\{s_1,s_2, x_1,x_2\}$ and public parameter $p$ (large prime number)

-Encryption E(m): $C_1 = s_1m + x_1k \bmod p$ and $C_2 = s_2m + x_2k \bmod p$ where $k$ is pseudo-random number.

Thanks.

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And how do you decrypt it? Or, is how $k$ is generated also part of the key? –  poncho Jul 3 at 4:07
    
$k$ is a pseudo-random number as the output of one-way function e.g., $k \leftarrow F_K(id_m)$ where $K$ is a secret key and $id_m$ is an identify of message $m$. Thanks. –  Jenifer Tran Jul 3 at 4:27
    
If $k$ is pseudorandom and not truly random, this is not a one-time pad. OTP is a very specific cryptosystem; if you don't have a truly random key which is never used for more than one message, you don't have it. –  cpast Jul 3 at 4:29
    
Ok, thanks @cpast, I edited the title of question. It is not OTP, I mean this scheme is very similar to the OTP encryption. –  Jenifer Tran Jul 3 at 5:05
    
You can't say "pseudorandom as output of one-way function". You can say "pseudorandom as output of pseudorandom function". –  Yehuda Lindell Jul 3 at 6:43

1 Answer 1

[Note: This answer is based on k being generated by applying a pseudorandom function to a unique message-ID (counter) each time.]

It depends how many times you want to encrypt with it.

If you want it as a complicated OTP, then it's secure. In order to see this, just ignore the x parts and note that $s_1m \bmod p$ and $s_2m \bmod p$ are to independent encryptions under random keys $s_1,s_2$ (note that $s_1$ and $s_2$ have to be uniform in $\{0,\ldots,p-1\}$. For this, you don't need $p$ to be large (just big enough to contain $m$).

If you want to encrypt many times, then it is certainly no longer information theoretically secure. However, just ignore $s_1,x_1,s_2,x_2$ and think of them as public constants. Then, you are encrypting by $m+k \bmod p$ where $k$ is pseudorandomly generated. This is like a type of "counter mode". As long as the RANGE of the pseudorandom function is $\{0,\ldots,p-1\}$ you should be able to prove it.

In both cases, it is overly complicated...

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"Also, the question should" ... $\;$ –  Ricky Demer Jul 3 at 7:01

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