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copied!<p>Based on the error you posted, I think this should help.</p> <p>In Python, you can have local and global variables of the same name. The <code>freq</code> in the function is local, and thus the initialization of the global <code>freq</code> doesn't touch the local one. To use the global <code>freq</code>, you have to tell the function to do so, via the <a href="http://docs.python.org/release/2.4/ref/global.html" rel="nofollow">global statement</a>. This is explained a little more in the <a href="http://docs.python.org/2/faq/programming.html#what-are-the-rules-for-local-and-global-variables-in-python" rel="nofollow">Python FAQs</a>.</p> <p>That should be enough to get you back on track.</p> <p><strong>EDIT:</strong> Below is an edit of your <code>decrypt</code> function, with some explanations of the changes</p> <pre><code>def decrypt(): # we don't need the zero start value, that's the default # test all possible shifts for i in range(27): # initialize the array freqlist = [] # shift all the letters in the input for a in inpt: b = ord(a) b -= i if b > ord("Z"): b -= 26 elif b < ord("A"): b+=26 freqlist.append(b) # now put the shifted letters back together shifted = "" for a in freqlist: d = chr(a) # append the shifted letter onto our output shifted += d # after we put the decrypted string back together, print it # note this is outside the letter loops, # but still inside the possible shifts loop # thus printing all possible shifts for the given message print(d) </code></pre>

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