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    <p>Let's start with the first four lines of the function's code:</p> <pre><code>def isprime(n): if n == 2: return True if n == 3: return True if n % 2 == 0: return False if n % 3 == 0: return False </code></pre> <p>The function tests to see if <code>n</code> is equal to 2 or 3 first. Since they are both prime numbers, the function will return <code>True</code> if <code>n</code> is equal to either.</p> <p>Next, the function tests to see if <code>n</code> is divisible by 2 or 3 and returning <code>False</code> if either is true. This eliminates an extremely large amount of cases because half of all numbers above two are not primes - they are divisible by 2. The same reason applies to testing for divisibility by 3 - it also eliminates a large number of cases.</p> <p>The trickier part of the function is in the next few lines:</p> <pre><code>i = 5 w = 2 while i * i &lt;= n: if n % i == 0: return False i += w w = 6 - w return True </code></pre> <p>First, <code>i</code> (or index) is set to 5. 2 and 3 have already been tested, and 4 was tested with <code>n % 2</code>. So, it makes sense to start at 5.</p> <p>Next, <code>w</code> is set to 2. <code>w</code> seems to be an "incrementer". By now, the function has tested for all even numbers (<code>n % 2</code>), so it would be faster to increment by 2.</p> <p>The function enters a <code>while</code> loop with the condition <code>i * i &lt;= n</code>. This test is used because <a href="http://www.millersville.edu/~bikenaga/number-theory/primes/primes.html" rel="nofollow">every composite number has a proper factor less than or equal to its square root</a>. It wouldn't make sense to test numbers after the square root because it would be redundant.</p> <p>In the <code>while</code> loop, if <code>n</code> is divisible by <code>i</code>, then it is not prime and the function returns <code>False</code>. If it is not, <code>i</code> is incremented by the "incrementer" <code>w</code>, which, again, is faster.</p> <p>Perhaps the trickiest part of the function lies in the second-to-last line: <code>w = 6 - w</code>. This causes the "incrementer" <code>w</code> to toggle between the values 2 and 4 with each pass through loop. In cases where <code>w</code> is 4, we are bypassing a number divisible by 3. This is faster than remaining at 2 because the function already tested for divisibility by both 2 and 3.</p> <p>Finally, the function returns <code>True</code>. If the function hasn't detected any cases where <code>n</code> is divisible by something, then it must be a prime number.</p>
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