http://code.activestate.com/recipes/tags/square/new/2014-09-16T22:27:04-07:00ActiveState Code Recipes2014-09-16T22:27:04-07:00Barry Walkerhttp://code.activestate.com/recipes/users/4177147/http://code.activestate.com/recipes/578934-a-fun-perfect-square-checker-using-integer-arithme/

<p style="color: grey"> Bash recipe 578934 by <a href="/recipes/users/4177147/">Barry Walker</a> (<a href="/recipes/tags/arithmetic/">arithmetic</a>, <a href="/recipes/tags/bash/">bash</a>, <a href="/recipes/tags/check/">check</a>, <a href="/recipes/tags/checker/">checker</a>, <a href="/recipes/tags/cygwin/">cygwin</a>, <a href="/recipes/tags/macbook_pro/">macbook_pro</a>, <a href="/recipes/tags/math/">math</a>, <a href="/recipes/tags/mathematics/">mathematics</a>, <a href="/recipes/tags/perfect/">perfect</a>, <a href="/recipes/tags/root/">root</a>, <a href="/recipes/tags/square/">square</a>). </p> <p>A recent Python upload here gave me the inspiration to do a bash version... This is a little tongue-in-cheek but an enjoyable bit of fun.</p> <p>It took around 11 seconds to prove 90000000000 had a perfect square of 300000...</p> <p>It is a stand alone program and has a degree of INPUT error correction...</p> <p>It was done on a MacBook Pro, OSX 10.7.5, default bash terminal and should work on Linux flavours but it is untested...</p> <p>Enjoy finding simple solutions to often very difficult problems...</p> <p>Bazza...</p>

2014-09-10T02:21:27-07:00Ethan Hannhttp://code.activestate.com/recipes/users/4190746/http://code.activestate.com/recipes/578931-perfect-square-checker/

<p style="color: grey"> Python recipe 578931 by <a href="/recipes/users/4190746/">Ethan Hann</a> (<a href="/recipes/tags/check/">check</a>, <a href="/recipes/tags/checker/">checker</a>, <a href="/recipes/tags/math/">math</a>, <a href="/recipes/tags/mathematics/">mathematics</a>, <a href="/recipes/tags/root/">root</a>, <a href="/recipes/tags/square/">square</a>). </p> <p>This program will take input from the user and either check if a number is a perfect square or square a number, depending on user's choice.</p>

2011-08-04T05:29:16-07:00Steven D'Apranohttp://code.activestate.com/recipes/users/4172944/http://code.activestate.com/recipes/577821-integer-square-root-function/

<p style="color: grey"> Python recipe 577821 by <a href="/recipes/users/4172944/">Steven D'Aprano</a> (<a href="/recipes/tags/integer/">integer</a>, <a href="/recipes/tags/isqrt/">isqrt</a>, <a href="/recipes/tags/math/">math</a>, <a href="/recipes/tags/mathematics/">mathematics</a>, <a href="/recipes/tags/maths/">maths</a>, <a href="/recipes/tags/root/">root</a>, <a href="/recipes/tags/square/">square</a>). </p> <p>The <em>integer square root</em> function, or isqrt, is equivalent to floor(sqrt(x)) for non-negative x. For small x, the most convenient way to calculate isqrt is by calling int(x**0.5) or int(math.sqrt(x)), but if x is a large enough integer, the sqrt calculation overflows.</p> <p>You can calculate the isqrt without any floating point maths, using just pure integer maths, allowing the function to operate with numbers far larger than possible with floats:</p> <pre class="prettyprint"><code>&gt;&gt;&gt; n = 1234567*(10**1000) &gt;&gt;&gt; n2 = n*n &gt;&gt;&gt; math.sqrt(n2) Traceback (most recent call last): File "&lt;stdin&gt;", line 1, in &lt;module&gt; OverflowError: long int too large to convert to float &gt;&gt;&gt; isqrt(n2) == n True </code></pre>