reworked these examples a bit based on feedback in class

[wordplay-cdsw-solutions] / solution_6_advanced.py

diff --git a/solution_6_advanced.py b/solution_6_advanced.py
index 36aa0d2d78390f5032ba646676d70433eaca8d0c..43653c0f6188f0eb1ca274f8fc76fc3a73e227aa 100644 (file)
--- a/solution_6_advanced.py
+++ b/solution_6_advanced.py
@@ -1,17 +1,15 @@
 import scrabble
 
-# Print the longest word where every character is unique.  I used a
-# Set for this which we'll talk about today.
+# Print the longest word where every character is unique.  I used a Set for
+# this which we'll talk about today.
 
-# Don't worry: you didn't miss anything if you don't know what a set
-# is. We didn't teach it, but if you are reading this, you get a
-# bonus!
+# Don't worry: you didn't miss anything if you don't know what a set is. We
+# didn't teach it, but if you are reading this, you get a bonus!
 
-# A set is a container like a list or a dict, except that *each
-# element can be stored only once*.  Think of it like the keys of a
-# dict, except there isn't any value associated with each key.  I use
-# Sets to count digits below. Feel free to look up the Set online and
-# try it in the interpreter.
+# A set is a container like a list or a dict, except that *each element can be
+# stored only once*.  Think of it like the keys of a dict, except there isn't
+# any value associated with each key.  I use Sets to count digits below. Feel
+# free to look up the Set online and try it in the interpreter.
 
 new_words = []
 for word in scrabble.wordlist:
@@ -19,8 +17,8 @@ for word in scrabble.wordlist:
     if len(word) == len(set(word)): # Wait what!? See if you can figure out why this works.
         new_words.append(word)
 
-# Reuse my code for longest word, (in this case, the code to track all occurences) from the
-# advanced solution.
+# Reuse my code for longest word, (in this case, the code to track all
+# occurences) from the advanced solution.
 
 longest_so_far = []
 length_of_longest_word = 0