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    <p>Your first recurrence relation is normally used to describe running time of <strong>divide-and-conquer</strong> algorithms. <code>a</code> here shows how many parts you are dividing your data to, <code>1/b</code> shows what piece of original data is used in each part, and <code>f(n)</code> shows how much time you need on each "level". For example, in QuickSort algorithm you divide your data (array or list) into 2 parts, each of which is exactly half (1/2) of original data, and on each "level" of dividing you need to go through all <code>n</code> elements 1 time. So recurrence relation is </p> <pre><code>T(n) = 2T(n/2) + n </code></pre> <p>(which evaluates to O(n * lg(n))) And in binary search you divide data into 2 parts, but take only 1 of them, and time on each "level" is constant, so relation is:</p> <pre><code>T(n) = T(n/2) + 1 </code></pre> <p>However, your function in C doesn't follow divide-and-conquer strategy. Instead it demonstrates <strong>mathematical induction</strong>. You define start conditions: if <code>l</code> equals <code>NULL</code>, then length is 0, and if it <code>l-&gt;next</code> equals <code>NULL</code> (you also define condition for 1 element in list). Then you define a kind of inductive step - you define how to compute function for nth element, if you know function value for (n - 1)th element. So, knowing value of a function for 1st element you can apply this rule and get value for 2nd element, then for 3rd and so on. </p> <p>So you can see that induction method is recurrence algorithm by the nature. Here we have 2 times to consider. First is a time to compute value at the start point (or end point - it depends on the order you are looking at the list). In your function you just return this value, so it is constant (<code>C1</code>). Second is a time to make 1 step. In your function it is also constant (<code>C2</code>). Intuitively you should see that execution time of this algorithm will be: </p> <pre><code>T(n) = C1, when n = 1 T(n) = T(n - 1) + C2, when n &gt; 1 </code></pre> <p>Thus, unless <code>n = 1</code>, you define time to execute algorithm on <code>n</code> element through time to execute it on <code>n - 1</code> elements. In BigO notation any constant is defined as <code>1</code> and case of 1 element is not considered, so your final recurrence relation is: </p> <pre><code>T(n) = T(n - 1) + 1 </code></pre> <p>(which evaluates to <code>T(n) = T(n - 1) + 1 = T(n - 2) + 2 = ... = 1 + n</code>, or <code>O(n)</code>)</p> <p><strong>Further reading</strong>: </p> <ul> <li><a href="http://rads.stackoverflow.com/amzn/click/0262033844" rel="noreferrer">Introduction to Algorithms</a> by Thomas H. Cormen. </li> <li><a href="http://rads.stackoverflow.com/amzn/click/0201485419" rel="noreferrer">Art of Computer Programming</a> by Donald Knuth.</li> </ul>
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