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    <p>Here's a simple tail-recursive solution:</p> <pre><code>def fibonacci(n: Long): Long = { def fib(i: Long, x: Long, y: Long): Long = { if (i &gt; 0) fib(i-1, x+y, x) else x } fib(n, 0, 1) } </code></pre> <p>The solution you posted takes <em>exponential</em> time since it creates two recursive invocation trees (<code>fibonacci(n - 1)</code> and <code>fibonacci(n - 2)</code>) at each step. By simply tracking the last <em>two</em> numbers, you can recursively compute the answer without any repeated computation.</p> <hr> <blockquote> <p>Can you explain the middle part, why <code>(i-1, x+y, x)</code> etc. Sorry if I am asking too much but I hate to copy and paste code without knowing how it works.</p> </blockquote> <p>It's pretty simple—but my poor choice of variable names might have made it confusing.</p> <ul> <li><code>i</code> is simply a counter saying how many steps we have left. If we're calculating the <em>Mth</em> (I'm using <code>M</code> since I already used <code>n</code> in my code) Fibonacci number, then <code>i</code> tells us how many more terms we have left to calculate before we reach the <em>Mth</em> term.</li> <li><code>x</code> is the <em>mth</em> term in the Fibonacci sequence, or <em>F<sub>m</sub></em> (where <code>m = M - i</code>).</li> <li><code>y</code> is the <code>m-1th</code> term in the Fibonacci sequence, or <em>F<sub>m-1</sub></em> .</li> </ul> <p>So, on the first call <code>fib(n, 0, 1)</code>, we have <code>i=M</code>, <code>x=0</code>, <code>y=1</code>. If you look up the <a href="http://en.wikipedia.org/wiki/Fibonacci_number#List_of_Fibonacci_numbers" rel="nofollow">bidirectional Fibonacci sequence</a>, you'll see that F<sub>0</sub> = 0 and F<sub>-1</sub> = 1, which is why <code>x=0</code> and <code>y=1</code> here.</p> <p>On the next recursive call, <code>fib(i-1, x+y, x)</code>, we pass <code>x+y</code> as our next <code>x</code> value. This come straight from the definiton:</p> <p><em>F<sub>n</sub> = F<sub>n-1</sub> + F<sub>n-2</sub></em></p> <p>We pass <code>x</code> as the next <code>y</code> term, since our current F<sub>n-1</sub> is the same as F<sub>n-2</sub> for the next term.</p> <p>On each step we decrement <code>i</code> since we're one step closer to the final answer.</p>
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    1. COThanks it works! But how are x and y passed as parameters?
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    2. CO@user2947615 - `fib` is a nested function definition. It's just declared as a helper. It's invoked on the last line: `fib(n, 0, 1)`. So, you call `fibonacci(someNumber)`, and that call will delegate the computation to `fib` by calling `fib(n, 0, 1)`, where `n` is `someNumber`.
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    3. COThanks! Can you explain the middle part, why (i-1, x+y, x) etc. Sorry if I am asking too much but I hate to copy and paste code without knowing how it works
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