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<!DOCTYPE html>
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<title>CDHA1-256 Statistics & Proof</title>
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<ul style="list-style-type: none;margin: 0;padding: 0;">
<li style="display: inline;margin-right:10px;"><a href="https://andrew-musa.github.io/cpsc329/index.html"
style="color: white;text-decoration: none;">Home</a></li>
<li style="display: inline;margin-right:10px;"><a href="https://andrew-musa.github.io/cpsc329/overview.html"
style="color: white;text-decoration: none;">Overview</a></li>
<li style="display: inline;margin-right:10px;"><a href="https://andrew-musa.github.io/cpsc329/statistics.html"
style="color: white;text-decoration: none;">Statistics & Proof</a></li>
<!-- <li><a href="https://andrew-musa.github.io/cpsc329/python-implementation.html"
style="color: white;text-decoration: none;">Python Implementation</a></li> -->
<li style="display: inline;margin-right:10px;"><a
href="https://andrew-musa.github.io/cpsc329/supplementary-materials.html"
style="color: white;text-decoration: none;">Supplementary Materials</a></li>
</ul>
</div>
<center>
<!--<div class="thesquare" style="text-align: center;">-->
<div class="solution" style="text-align: left;">
<h1 id="single-thread-brute-force-attack-time-analysis">Single Thread
Brute Force Attack Time Analysis</h1>
<p><span class="math display">
\begin{align}
&\text{Number of Hashes Computed: } 526044 \text{~~hashes} \\
&\text{Time Elapsed: } 21660.5585421999 \text{~~sec} \\
&\text{Average Time Per Hash } = \frac{21660.5585421999}{526044} =
0.0411763246842468 \text{~~sec/hash} \\
&\text{Time to Hash All 8 character Passwords } =
\frac{21660.5585421999}{526044} \times \frac{95^8}{60 \cdot 60 \cdot 24
\cdot 365} \\
&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=
8662232.1 \text{~~ years} \\
\end{align}
</span></p>
<h1 id="hashing-similar-messages">Hashing Similar Messages</h1>
<pre
class="py"><code># Base message
CDFA1_256_Hash("The quick brown fox jumps over the lazy dog")
PRGSIAVAAGRRDVRVRPAISCS_QFVLLKLAILRGSMLSERRLAHCVWGVSLMLIYPRSGSVVFLKSSLT_DVRAWSPIGSIGYVCPSYPPTRPRRRCCDAEKCSSLGTFFKTE_IVTRGCPQTRLYY_AAHMHVVLSPAVSMRRVIRPDYVSPGTMKRQYSSNTISHSL_RHEIVS_AYFQLLSSWIDGLLLGRREQLLAGENMRLYQIYSMKGL_YAPIRVERVSEIIKGRNWTTTGQD_GTPRKPRYRLCFH</code></pre>
<pre
class="py"><code># Adding a period to the message
CDFA1_256_Hash("The quick brown fox jumps over the lazy dog.")
QSQRIREASCSSSCSLIR_WQDN_L_RCPRPLSHGCSCVPK_IR_RTCTQREQGLLAKVGKLLSKRSSSGMAAQMYFMLASFMANVIRSLASISDGSNSTFSFVPTKVAEPRHG_CIIMPVKNTITRVITTLPGRHQSISRGVSMKDNSCNTRVQTDAVEGLPYAFDS_YLWIPYPPDNKPTWIAHIFP_HWDWCFTMNVERIHRVETATTEALASRKISELPRSQRVRTATSSYRHINSLFPLTPGERTHTHGNY</code></pre>
<h1 id="probability-of-collision-assuming-uniform-distribution">Probability
of Collision Assuming Uniform Distribution</h1>
<ul>
<li>Suppose we have <span class="math inline">n</span> possible <span class="math inline">8</span> character
passwords and <span class="math inline">m</span> possible <span class="math inline">256</span> character
hashes. What is the probability
that at least two passwords will collide and have the same hash?
<ul>
<li><span class="math inline">\text{n passwords}</span>
<ul>
<li>ASCII has <span class="math inline">95</span> printable
characters.</li>
<li>Since there are <span class="math inline">8</span> character is a
password <span class="math inline">n = 95^8</span>.</li>
</ul>
</li>
<li><span class="math inline">\text{m hashes}</span>
<ul>
<li>There are <span class="math inline">20</span> different amino acids
and <span class="math inline">1</span> nonsense codon thus there are
<span class="math inline">21</span> possible characters that can be used
in the hash.
<ul>
<li><span class="math inline">\text{char in hash} \in
\{A,R,N,D,C,E,Q,G,H,I,L,K,M,F,P,S,T,W,Y,V,\_\}</span></li>
</ul>
</li>
<li>Since there are 256 characters in a hash <span class="math inline">m
= 21^{256}</span>.</li>
</ul>
</li>
<li>Suppose the passwords are labeled from <span class="math inline">1...n</span>.</li>
<li>Let <span class="math inline">E_{i,j}:\text{``password i and j
collide", } (i \ne j)</span>
<ul>
<li><span class="math inline">E_{i,j} = E_{j,i}</span></li>
</ul>
</li>
<li>Assuming uniform distribution:
<ul>
<li><span class="math inline">p(E_{i,j}) = \frac{1}{m}:</span> as
regardless of what the hash of the first password is the second password
must have the same hash.</li>
</ul>
</li>
<li><span class="math inline">E: \text{"at least two passwords
collide"}</span>
<ul>
<li>In other words a collision.</li>
<li>The event <span class="math inline">E</span> that at least two
passwords collide also includes the cases where more than two passwords
collide. <span class="math display">E = \bigcup_{1 \le i \lt j \le
n}E_{i,j}</span></li>
</ul>
</li>
</ul>
</li>
</ul>
<p><span class="math display">
\begin{align}
p(E) = p(\bigcup_{1 \le i \lt j \le n}E_{i,j}) &\le \sum_{1 \le i
\lt j \le n}p(E_{i,j}) \quad \text{(by the Union Bound)}\\
&\le \sum_{1 \le i \lt j \le n}\frac{1}{m} \\
&\le {n \choose 2} \cdot \frac{1}{m} \\
&\le \frac{n!}{2!(n-2)!} \cdot \frac{1}{m} \\
&\le \frac{n(n-1)(n-2)!}{2(n-2)!} \cdot \frac{1}{m} \\
&\le \frac{n(n-1)}{2} \cdot \frac{1}{m} \\
&\le \frac{n(n-1)}{2m} \\
&\lt \frac{n^2}{2m} \\
&\lt \frac{(95^8)^2}{2 \cdot 21^{256}} \approx 7.1517 \times
10^{-308}
\end{align}
</span></p>
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