Blom's scheme

The key exchange protocol involves a trusted party (Trent) and a group of ${\displaystyle \scriptstyle n}$ users. Let Alice and Bob be two users of the group.

Trent chooses a random and secret symmetric matrix ${\displaystyle \scriptstyle D_{k,k}}$ over the finite field ${\displaystyle \scriptstyle GF(p)}$, where p is a prime number. ${\displaystyle \scriptstyle D}$ is required when a new user is to be added to the key sharing group.

For example:

${\displaystyle {\begin{aligned}k&=3\\p&=17\\D&={\begin{pmatrix}1&6&2\\6&3&8\\2&8&2\end{pmatrix}}\ \mathrm {mod} \ 17\end{aligned}}}$

New users Alice and Bob want to join the key exchanging group. Trent chooses public identifiers for each of them; i.e., k-element vectors:

${\displaystyle I_{\mathrm {Alice} },I_{\mathrm {Bob} }\in GF^{k}(p)}$.

For example:

${\displaystyle I_{\mathrm {Alice} }={\begin{pmatrix}1\\2\\3\end{pmatrix}},I_{\mathrm {Bob} }={\begin{pmatrix}5\\3\\1\end{pmatrix}}}$

Trent then computes their private keys:

${\displaystyle {\begin{aligned}g_{\mathrm {Alice} }&=DI_{\mathrm {Alice} }\\g_{\mathrm {Bob} }&=DI_{\mathrm {Bob} }\end{aligned}}}$

Using ${\displaystyle D}$ as described above:

${\displaystyle {\begin{aligned}g_{\mathrm {Alice} }&={\begin{pmatrix}1&6&2\\6&3&8\\2&8&2\end{pmatrix}}{\begin{pmatrix}1\\2\\3\end{pmatrix}}={\begin{pmatrix}19\\36\\24\end{pmatrix}}\ \mathrm {mod} \ 17={\begin{pmatrix}2\\2\\7\end{pmatrix}}\ \\g_{\mathrm {Bob} }&={\begin{pmatrix}1&6&2\\6&3&8\\2&8&2\end{pmatrix}}{\begin{pmatrix}5\\3\\1\end{pmatrix}}={\begin{pmatrix}25\\47\\36\end{pmatrix}}\ \mathrm {mod} \ 17={\begin{pmatrix}8\\13\\2\end{pmatrix}}\ \end{aligned}}}$

Each will use their private key to compute shared keys with other participants of the group.

Now Alice and Bob wish to communicate with one another. Alice has Bob's identifier ${\displaystyle \scriptstyle I_{\mathrm {Bob} }}$ and her private key ${\displaystyle \scriptstyle g_{\mathrm {Alice} }}$.

She computes the shared key ${\displaystyle \scriptstyle k_{\mathrm {Alice/Bob} }=g_{\mathrm {Alice} }^{T}I_{\mathrm {Bob} }}$, where ${\displaystyle \scriptstyle T}$ denotes matrix transpose. Bob does the same, using his private key and her identifier, giving the same result:

${\displaystyle k_{\mathrm {Alice/Bob} }=k_{\mathrm {Alice/Bob} }^{T}=(g_{\mathrm {Alice} }^{T}I_{\mathrm {Bob} })^{T}=(I_{\mathrm {Alice} }^{T}D^{T}I_{\mathrm {Bob} })^{T}=I_{\mathrm {Bob} }^{T}DI_{\mathrm {Alice} }=k_{\mathrm {Bob/Alice} }}$

They will each generate their shared key as follows:

${\displaystyle {\begin{aligned}k_{\mathrm {Alice/Bob} }&={\begin{pmatrix}2\\2\\7\end{pmatrix}}^{T}{\begin{pmatrix}5\\3\\1\end{pmatrix}}=2\times 5+2\times 3+7\times 1=23\ \mathrm {mod} \ 17=6\\k_{\mathrm {Bob/Alice} }&={\begin{pmatrix}8\\13\\2\end{pmatrix}}^{T}{\begin{pmatrix}1\\2\\3\end{pmatrix}}=8\times 1+13\times 2+2\times 3=40\ \mathrm {mod} \ 17=6\end{aligned}}}$

In order to ensure at least k keys must be compromised before every shared key can be computed by an attacker, identifiers must be k-linearly independent: all sets of k randomly selected user identifiers must be linearly independent. Otherwise, a group of malicious users can compute the key of any other member whose identifier is linearly dependent to theirs. To ensure this property, the identifiers shall be preferably chosen from a MDS-Code matrix (maximum distance separable error correction code matrix). The rows of the MDS-Matrix would be the identifiers of the users. A MDS-Code matrix can be chosen in practice using the code-matrix of the Reed–Solomon error correction code (this error correction code requires only easily understandable mathematics and can be computed extremely quickly).^[4]