2
$\begingroup$

So for a Diffie-Hellman problem, I am given the prime $p$ for the Diffie-Hellman exchange. I am also given $g$, secret number for machine as $A$, secret number for station as $B$, a Diffie-Hellman shielded Login Name $V$, and Diffie-Hellman shielded password $W$.

For this problem, I am given three users and see which one accessed files to which they had no clearance to. So I computed $x=g^A\pmod p$ and $y=g^B\pmod p$, then $x^A\pmod p$ and $y^B\pmod p$ which gave me my secret common key. Where I am confused is as to how to now unshield the DHS key and how I can use $V$ and $W$ to do so.

What do I do next?

According to a textbook I am reading, it says that the equation $DHS*u = 1\pmod p$ has a solution in $\mathbb N_p$ and this is the solution for $UDHS$, or the unshielding of DHS. But I am confused by what this means. Help is appreciated.

$\endgroup$

closed as unclear what you're asking by Squeamish Ossifrage, Maeher, kelalaka, Maarten Bodewes Mar 12 at 9:56

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Actually, you should be computing $x^B \bmod p$ and $y^A \bmod p$ to derive the shared secret. $\endgroup$ – poncho Oct 22 '13 at 19:26
  • $\begingroup$ Within the DH protocol, there's no standard way to do "Diffie-Hellman shielded Logic name" and "password". It is certainly possible to design a protocol that uses DH which does it, however the details of that protocol are outside the Diffie-Hellman protocol. $\endgroup$ – poncho Nov 4 '13 at 17:14