Add my (quite C-like and densely-packed) solution to the lock problem.

solutions/jre21/lock.py

@@ -0,0 +1,48 @@
+import re
+
+lock = ['BDMJPRSTLN', 'AEIOUYRTLH', 'ACDEORSTLN', 'DHKYRSTLNE']
+
+words = set()
+
+# Create a hash table of all four-letter words which use only standard
+# (unaccented) characters.
+with open('/usr/share/dict/words', 'r') as f:
+    word = f.readline()
+    while (word):
+        word = word.strip().upper()
+        if re.match('^[A-Z]{4}$', word):
+            words.add(word)
+        word = f.readline()
+
+solutions = []
+
+# Iterate through all possible positions of the bottom three reels
+# relative to the top reel.  Then, for each position, enumerate the
+# columns which form real words.  Save all sets of 3+ words.
+for second in range(len(lock[1])):
+    for third in range(len(lock[2])):
+        for fourth in range(len(lock[3])):
+            found = []
+            for index in range(len(lock[0])):
+                word = ''.join([
+                                lock[0][index],
+                                lock[1][(second+index)%len(lock[1])],
+                                lock[2][(third+index)%len(lock[2])],
+                                lock[3][(fourth+index)%len(lock[3])]
+                                ])
+                if word in words:
+                    found.append(word)
+            if len(found) >= 3:
+                found.sort()
+                solutions.append(tuple(found))
+
+# Print the solutions, alphabetized by first word.
+solutions = list(set(solutions))
+solutions.sort(key=lambda x: x[0])
+for solution in solutions:
+    print(solution)
+
+# Print in alphabetical order all words which appear in a triplet.
+answers = list(set([x for y in solutions for x in y]))
+answers.sort()
+print(answers)