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2. We assume AES with 192-bit key length. Furthermore, let us assume an ASIC which can check $3 * 10^7$ keys per second. If we use 100,000 such ICs in parallel, how long does the entire key search take for all possible keys? 3. Galois Counter Mode (GCM) provides authentication using GHASH unit. Suppose we get the polynomial for the 128-bit hash subkey (H) as: $H(x) = x^{27}+x^{25}+x^{20}+x^4+x+1$. Also, as you know, the irreducible polynomial for GCM is: $P(x) = x^{128}+x^7+x^2+x+1$. Find the polynomial representing $\beta(x)=(M_1.H+M_2).H \mod P(x)$ assuming the two 128-bit input blocks to GHASH are as follows: $M_1 = x^{89}+x^{23}+x^{10}$, $M_2 = x^{93}+x^{24}+x^{10}+x$. $\beta(x)=(M_1.H+M_2).H \mod p(x)$ is what you have to derive in $GF(2^{128})$ after reduction. This is a very simple calculation which is done in practice thousands of times for eventually deriving a tag.

Name: 2 we assume aes with 192 bit key length furthermore let us assume an asic which can check 3 107 keys per second if we use 100000 such ics in parallel how long does the entire key search take 59018
Uploaded: 2023-07-03T21:11:48-08:00
Duration: 3 min 43 s
Channel: Akash M
Description: 2 we assume aes with 192 bit key length furthermore let us assume an asic which can check 3 107 keys per second if we use 100000 such ics in parallel how long does the entire key search take 59018

          2. We assume AES with 192-bit key length. Furthermore, let us assume an ASIC which can check $3 * 10^7$ keys per second. If we use 100,000 such ICs in parallel, how long does the entire key search take for all possible keys?

3. Galois Counter Mode (GCM) provides authentication using GHASH unit. Suppose we get the polynomial for the 128-bit hash subkey (H) as: $H(x) = x^{27}+x^{25}+x^{20}+x^4+x+1$. Also, as you know, the irreducible polynomial for GCM is: $P(x) = x^{128}+x^7+x^2+x+1$. Find the polynomial representing $\beta(x)=(M_1.H+M_2).H \mod P(x)$ assuming the two 128-bit input blocks to GHASH are as follows: $M_1 = x^{89}+x^{23}+x^{10}$, $M_2 = x^{93}+x^{24}+x^{10}+x$.

$\beta(x)=(M_1.H+M_2).H \mod p(x)$ is what you have to derive in $GF(2^{128})$ after reduction. This is a very simple calculation which is done in practice thousands of times for eventually deriving a tag.

2 we assume aes with 192 bit key length furthermore let us assume an asic which can check 3 107 keys per second if we use 100000 such ics in parallel how long does the entire key search take 59018

Added by Shelby E.

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