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PORecurrences using Substitution Method
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  1. PORecurrences using Substitution Method
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    copied!<p>Determine the positive number c &amp; n0 for the following recurrences (Using Substitution Method):</p> <ol> <li><p>T(n) = T(ceiling(n/2)) + 1 ... Guess is Big-Oh(log base 2 of n)</p></li> <li><p>T(n) = 3T(floor(n/3)) + n ... Guess is Big-Omega (n * log base 3 of n)</p></li> <li><p>T(n) = 2T(floor(n/2) + 17) + n ... Guess is Big-Oh(n * log base 2 of n).</p></li> </ol> <p>I am giving my Solution for Problem 1:</p> <p>Our Guess is: T(n) = O (log_2(n)). By Induction Hypothesis assume T(k) &lt;= c * log_2(k) for all k &lt; n,here c is a const &amp; c > 0</p> <pre><code> T(n) = T(ceiling(n/2)) + 1 &lt;=&gt; T(n) &lt;= c*log_2(ceiling(n/2)) + 1 &lt;=&gt; " &lt;= c*{log_2(n/2) + 1} + 1 &lt;=&gt; " = c*log_2(n/2) + c + 1 &lt;=&gt; " = c*{log_2(n) - log_2(2)} + c + 1 &lt;=&gt; " = c*log_2(n) - c + c + 1 &lt;=&gt; " = c*log_2(n) + 1 &lt;=&gt; T(n) not_&lt;= c*log_2(n) because c*log_2(n) + 1 not_&lt;= c*log_2(n). </code></pre> <p>To solve this remedy used a trick a follows:</p> <pre><code> T(n) = T(ceiling(n/2)) + 1 &lt;=&gt; " &lt;= c*log(ceiling(n/2)) + 1 &lt;=&gt; " &lt;= c*{log_2 (n/2) + b} + 1 where 0 &lt;= b &lt; 1 &lt;=&gt; " &lt;= c*{log_2 (n) - log_2(2) + b) + 1 &lt;=&gt; " = c*{log_2(n) - 1 + b} + 1 &lt;=&gt; " = c*log_2(n) - c + bc + 1 &lt;=&gt; " = c*log_2(n) - (c - bc - 1) if c - bc -1 &gt;= 0 c &gt;= 1 / (1 - b) &lt;=&gt; T(n) &lt;= c*log_2(n) for c &gt;= {1 / (1 - b)} so T(n) = O(log_2(n)). This solution is seems to be correct to me ... My Ques is: Is it the proper approach to do? Thanks to all of U. </code></pre>
 

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