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<div align="center"><img align="middle" src="doctitle.png"/></div>
<hr/>
<center>
<h2>RCM3720 Cryptography, Network and Computer Security</h2>
<h3>Laboratory Class 8: Modes of Encryption</h3>
</center>
<hr/>
We will investigate the different modes of encryption using the Hill
(matrix) cryptosystem. Start off by entering some matrices:
<ul>
<li>
<span class="cmd">
M:=matrix([[15,9,21],[2,10,7],[16,11,12]])::Matrix ZMOD 26
</span>
</li>
<li>
<span class="cmd">
MI:=matrix([[7,17,19],[24,0,23],[12,25,10]])::Matrix ZMOD 26
</span>
</li>
</ul>
Check that you've entered everything correctly with
<ul>
<li> <span class="cmd">M*MI</span></li>
</ul>
Note that because the matrices were defined in terms of numbers mod 26,
their product is automatically reduced mod 26.
Now enter the following column vector:
<ul>
<li>
<span class="cmd">
zero31:=matrix([[0],[0],[0]])::Matrix ZMOD 26
</span>
</li>
</ul>
For this lab, rather than fiddling about with translations between
letters and numbers, all our work will be done with numbers alone
(in the range 0..25).
<br/><br/>
<b>ECB</b>
<br/><br/>
For electronic codebook mode, encryption is performed by multiplying each
plaintext block by the matrix, and decryption by multiplying each ciphertext
block by the inverse matrix:
<pre>
-1
C =M.P , P =M C
i i i i
</pre>
where all arithmetic is performed mod 26.
<ul>
<li> Start by entering a plaintext, which will be a list of column vectors:
<ul>
<li>
<span class="cmd">
P:=[matrix([[3*i],[3*i+1],[3*i+2]]) for i in 0..7]
</span>
</li>
</ul>
</li>
<li> and a list which will receive the ciphertext:
<ul>
<li> <span class="cmd">C:=[zero31 for i in 1..8]</span></li>
</ul>
</li>
<li> and encrypt it:
<ul>
<li> <span class="cmd">for i in 1..8 repeat C.i:=M*P.i</span></li>
</ul>
</li>
<li> Now decrypt (first make an empty list <tt>D</tt>):
<ul>
<li> <span class="cmd">D:=[zero31 for i in 1..8]</span></li>
<li> <span class="cmd">for i in 1..8 repeat D.i:=MI*C.i</span></li>
</ul>
</li>
<li> If all has worked out, the list <tt>D</tt> should be the same
plaintext you obtained earlier.
</li>
<li> Now change one value in the plaintext:
<ul>
<li> <span class="cmd">Q:=P</span></li>
<li> <span class="cmd">Q.3:=matrix([[6],[19],[8]])</span></li>
</ul>
</li>
<li> Now encrypt the new plaintext <tt>Q</tt> to a ciphertext <tt>E</tt>. How
does this ciphertext differ from the ciphertext <tt>C</tt> obtained from
<tt>P</tt>?
</li>
<li> Check that you can decrypt <tt>E</tt> to obtain <tt>Q</tt>.</li>
</ul>
<br/><br/>
<b>CBC</b>
<br/><br/>
For cipherblock chaining mode, the encryption formula for the Hill
cryptosystem is
<pre>
C =M(P +C )
i i i-1
</pre>
and decryption is
<pre>
-1
P =M C -C
i i i-1
</pre>
<ul>
<li> To enable us to use these formulas, we shall first add an extra column
to the front of <tt>P</tt> and <tt>C</tt>:
<ul>
<li> <span class="cmd">P:=append([zero31],P)</span></li>
<li> <span class="cmd">C:=append([zero31],C)</span></li>
</ul>
</li>
<li> And we need to create a initialization vector:
<ul>
<li> <span class="cmd">IV:=matrix([[random(26)] for i in 1..3])</span></li>
</ul>
</li>
<li> Now for encryption:
<ul>
<li> <span class="cmd">C.1:=IV</span></li>
<li>
<span class="cmd">
for i in 2..9 repeat C.i:=M*(P.i+C.(i-1))
</span>
</li>
</ul>
</li>
<li> Let's try to decrypt the ciphertext, using the CBC formula:
<ul>
<li> <span class="cmd">D:=[zero31 for i in 1..9]</span></li>
<li>
<span class="cmd">
for i in 2..9 repeat D.i:=MI*(C.i)-C.(i-1)
</span>
</li>
</ul>
</li>
<li> Did it work out?</li>
<li> As before, change one value in the plaintext:
<ul>
<li> <span class="cmd">Q:=P</span></li>
<li> <span class="cmd">Q.4:=matrix([[6],[19],[8]])</span></li>
</ul>
</li>
<li> Now encrypt <tt>Q</tt> to <tt>E</tt> following the procedure outlined
above. Compare <tt>E</tt> with <tt>C</tt>---
how much difference is there?
How does this difference compare with the differences of ciphertexts
obtained with ECB?
</li>
<li> Just to make sure you can do it, decrypt <tt>E</tt> and make sure you
end up with a list equal to <tt>Q</tt>.
</li>
</ul>
<br/><br/>
<b>OFB</b>
<br/><br/>
Output feedback mode works by creating a <i>key stream</i>, and then adding
it to the plaintext to obtain the ciphertext. With the Hill system, and an
initialization vector <tt>IV</tt>:
<pre>
k =IV, k =Mk
1 i i-1
</pre>
and then
<pre>
c =p +k
i i i
</pre>
<ul>
<li> First, the key stream:
<ul>
<li> <span class="cmd">K:=[zero31 for i in 1..9]</span></li>
<li> <span class="cmd">K.1:=IV</span></li>
<li> <span class="cmd">for i in 2..9 repeat K.i:=M*K.(i-1)</span></li>
</ul>
</li>
<li> and next the encryption:
<ul>
<li> <span class="cmd">for i in 2..9 repeat C.i:=K.i+P.i</span></li>
</ul>
</li>
<li> What is the formula for decryption?
Apply it to your ciphertext <tt>C</tt>.
</li>
</ul>
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