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<title>Axiom Documentation</title>
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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 4: Simple Cryptosystems</h3>
</center>
<hr/>
We have experimented with the Caesar cipher and the more general
translation cipher. We shall start looking at the Vigenère cipher.
The trick is to add the correct letter of the code to the letter of
the key:
<pre>
Index of plain text i: 1 2 3 4 5 6 7 8
Plaintext: W I T H D R A W
Key: C O D E C O D E
Index of key j: 1 2 3 4 1 2 3 4
</pre>
The indices of the key repeat 1, 2, 3, 4. We can get a repetition of
length four by using a modulus of 4:
<pre>
i: 1 2 3 4 5 6 7 8
i (mod 4): 1 2 3 0 1 2 3 0
</pre>
What we need to do is to subtract one before the modulus, and add one after:
<pre>
i: 1 2 3 4 5 6 7 8
i-1: 0 1 2 3 4 5 6 7
i-1 (mod 4): 0 1 2 3 0 1 2 3
i-1 (mod 4) + 1: 1 2 3 4 1 2 3 4
</pre>
This means that in the Vigenère cipher, we add the <i>i</i>-th
character of the plaintext, and the <i>j</i>-th character of the key, where
<pre>
j=i-1 (mod n) + 1
</pre>
with <i>n</i> being the length of the key.
<ul>
<li> First read in the <tt>rcm3720.input</tt> file you have created:
<ul>
<li> <span class="cmd">)read rcm3720</span></li>
</ul>
You may have to include the full path here.
</li>
<li> Enter a plaintext
<ul>
<li> <span class="cmd">plaintext:="WITHDRAWONEHUNDREDDOLLARS"</span></li>
</ul>
</li>
<li> and a keyword:
<ul>
<li> <span class="cmd">key := "CODE"</span></li>
</ul>
</li>
<li> Now we can obtain the lengths of the plaintext and key with the hash
symbol:
<ul>
<li> <span class="cmd">pn:=#plaintext</span></li>
<li> <span class="cmd">kn:=#key</span></li>
</ul>
</li>
<li> Turn both plaintext and key into lists of numbers:
<ul>
<li> <span class="cmd">pl:=str2lst(plaintext)</span></li>
<li> <span class="cmd">kl:=str2lst(key)</span></li>
</ul>
</li>
<li> Now we can add them using the formula for <tt>j</tt> above to obtain
the list corresponding to the ciphertext:
<ul>
<li>
<span class="cmd">
cl:=[(pl.i+kl.((i-1) rem kn+1))::ZMOD 26 for i in 1..pn]
</span>
</li>
</ul>
</li>
<li> And obtain the ciphertext (we need to convert our list to a list of
integers first):
<ul>
<li> <span class="cmd">ciphertext:=lst2str(cl::List INT)</span></li>
</ul>
</li>
<li> Try a few other Vigenère encryptions.</li>
<li> To decrypt, we just <i>subtract</i> the key value from the ciphertext
value:
<ul>
<li>
<span class="cmd">
pl:=[(cl.i+kl.((i-1) rem kn+1))::ZMOD 26 for i in 1..pn]
</span>
</li>
<li> <span class="cmd">lst2str(pl::List INT)</span></li>
</ul>
</li>
<li> Now for the Hill (matrix) cipher. We shall use <tt>3 x 3</tt>
matrices, so first create a plaintext whose length is a multiple of 3:
<ul>
<li> <span class="cmd">plaintext:="WITHDRAWONEHUNDREDDOLLARSXX"</span></li>
<li> <span class="cmd">pl:=str2lst(plaintext)</span></li>
<li> <span class="cmd">r:=3</span></li>
<li> <span class="cmd">c:INT:=#pl/r</span></li>
</ul>
</li>
<li> The values <tt>r</tt> and <tt>c</tt> are the row and column numbers
of the plaintext matrix.
</li>
<li> Now put all the plaintext values into a <tt>r x c</tt> matrix:
<ul>
<li>
<span class="cmd">
S:=matrix([[pl.(r*(i-1)+j) for i in 1..c] for j in 1..r])
</span>
</li>
</ul>
</li>
<li> Create the key matrix:
<ul>
<li>
<span class="cmd">
M:Matrix ZMOD 26:=matrix([[22,11,19],[15,20,24],[25,21,16]])
</span>
</li>
</ul>
</li>
<li> Multiply the two matrices:
<ul>
<li> <span class="cmd">C:=M*S</span></li>
</ul>
</li>
<li> Notice how the results are automatically reduced modulo 26,
because that is how the matrix <tt>M</tt> was defined.
</li>
<li> Now we have to read off the elements of <tt>C</tt> into a single list;
this can be done by transposing the matrix, and reading off the rows as
lists:
<ul>
<li>
<span class="cmd">
CL:=concat(transpose(C)::List List ZMOD 26)
</span>
</li>
</ul>
</li>
<li> And finally turn into ciphertext:
<ul>
<li> <span class="cmd">lst2str(CL::List INT)</span></li>
</ul>
</li>
<li> Finally, here's how we can invert our matrix <tt>M</tt> modulo 26:
<ul>
<li> <span class="cmd">adj:=adjoint(M).adjMat</span></li>
<li> <span class="cmd">invdet:=recip(determinant(M))</span></li>
<li> <span class="cmd">MI:=invdet*adj</span></li>
</ul>
</li>
<li> Or alternatively, as one command:
<ul>
<li>
<span class="cmd">
MI:=recip(determinant(M))*adjoint(M).adjMat
</span>
</li>
</ul>
</li>
<li> Check the result:
<ul>
<li> <span class="cmd">M*MI</span></li>
</ul>
</li>
</ul>
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